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is the oriented edge from the vertex P, to P2 , and is the oriented triangle bounded by the oriented edges , , . We identify the triangle with — and the edge with — (P 2 ,P i ).. An n-chain (n = 0,1,2) is a finite linear combination of n-simplices with integer coefficients. We define an operator (5 from n-chains to n — 1 chains as follows: For n = 0, we define (5
=0. For n = 1, we define b = — . For n = 2, we define (5 = — + . The preceding defines (5 on an n-simplex and we extend the definition to n-chains by linearity. It is clear that the set of n-chains forms a group under addition and that (5 is a group homomorphism of the group of n-chains to the group of (n — 1)chains. We denote the group of n-chains by C„ = 101 for n> 2). Let Z„ denote the kernel of 6:C„ C„_1 . Furthermore, let B„ denote the image of C,,.,. 1 in C. under 6. Since 6 2 = 0, it is clear that B„ is a subgroup of Z„, and in fact since all groups in sight are abelian, a normal subgroup. It therefore follows that C,,/Z„ is isomorphic to B„_,. The group we are interested
1.2. Topology of Riemann Surfaces
15
in is, however, H(M) = H. = Z./B. and we call this group the nth simplicial homology group (with integer coefficients). (By definition H„ = MI for n > 2.) Let us now denote by [z] the equivalence class in H„ of z e Z. We shall say that [:], j = 1, , /3„, are a basis for H„ provided each element of H. can be written as an integral linear combination of the [z ] and provided the integral equation > 1 ai[zi] = 0 implies a; = O. In this case we shall call the number /3„ the nth-Betti number of the triangulated manifold. It is very easy to describe the groups H o and H2, and thus the numbers
/30 and /32 . In fact it is apparent that /30 = 1, and that H o is isomorphic to the integers. As far as H2 is concerned, a little thought shows that there are exactly two possibiliti;§. If M is compact, then H2 is isomorphic to the integeis and ,62 = 1. If M is not Compact, then H2 is trivial and /32 = O. The only non-trivial case to consider is H 1 and ,6 1 . We have seen in the previous paragraph that H o and H2 are independent of the triangulation. The same is true for H, although this is not at all apparent. One way to see this is to recall the fact that H, is isomorphic to the abelianized fundamental group. We shall not prove this result here. Granting the result, however, and using the normal forms for compact surfaces to be described in 1.2.5, it will be easy to compute H 1 (M) and hence fi, for compact surfaces M.
1.2.4. Covering Manifolds. The manifold M* is said to be a (ramified) covering manifold of the manifold M provided there is a continuous surjective map (called a (ramified) covering map) f:M* M with the following property: for each P* e M* there exist a local coordinate z* on M* vanishing at P*, a local coordinate z on M vanishing at f(P), and an integer n > 0 such that f is given by z = z" in terms of these local coordinates. Here the integer
n depends only on the point P* e M*. If n> 1, P* is called a branch point of order n — 1 or a ramification point of order n. (Compare these definitions with those in 1.1.6.) If n = 1, for all points P* e M* the cover is called smooth or unramified. EXAMPLE. Proposition 1.1.6 shows that every non-constant holomorphic mapping between compact Riemann surfaces is a finite-sheeted (ramified) covering map. Continuing the general discussion, we call M* an unlimited covering manifold of M provided that for every curve c on M and every point P* with f(P*) = c(0), there exists a curve c* on M* with initial point P* and f(c*) = c. The curve c* will be called a lift of the curve c. There is a close connection between zr 1 (M) and the smooth unlimited covering manifolds of M. If M* is a smooth unlimited covering manifold of M, then zt,(M*) is isomorphic to a subgroup of n,(M). Coversely, every subgroup of x 1 (M) determines a smooth unlimited covering manifold 14* with 7r 1(M*) isomorphic to the given subgroup. (Conjugate subgroups determine homeomorphic covers.)
16
I Ricmann
Surfaces
This is also a good place to recall the monodromy theorem which states: Let M* be a smooth unlimited covering manifold of M and cl , e 2 two curves on M which are homotopic. Let cf, cl he lifts of c 1 , c 2 with the saine initial point. Then el' is homotopic to el'. In particular, the curves cr and el' must have the same end points. If M* is a covering manifold of M with covering map f, then a homeomorphism h of M* onto itself with the property that foh=f is called a covering transformation of M*. The set of covering transformations forms a group, which is called transitive provided that whenever f(Pf) = f(PT) there is a covering transformation h which maps Pt onto P. For the smooth unlimited case, the group of covering transformations is transitive if and only if n i(M*) is isomorphic to a normal subgroup of t 1(M) and in this case the group of covering transformations is isomorphic to n i (M)ht i (M*). Let us now assume that M is a Riemann surface. Then the definition of covering manifold shows that M 5 has a unique Riemann surface structure on it which makes f a holomorphic map. Furthermore, the group of covering transformations consists now of conformal self maps of M. In the converse direction things are not quite so simple. If M is a Riemann surface and G is a group of conformal self maps of M, it is not necessarily the case that the orbit. space M/G is even a manifold. However, if the group G operates discontinuously on M, then M1G is a manifold and can be made into a Riemann surface such that the natural map f :M M 1G is an analytic map of Riemann surfaces. More details about these ideas will be found in IV.5 and IV.9. 1.2.5. Normal Forms of Compact Orientable Surfaces. Any triangulation of a compact manifold is necessarily finite. Using such a triangulation we can proceed to simplify the topological model of the manifold. We can map successively each triangle in the triangulation onto a Euclidean triangle and by auxilliary topological mappings obtain at each stage k, a regular (k + 2)gon, k 2. This (k + 2)-gon has a certain orientation on its boundary which is induced by the orientation on the triangles of the triangulation. When we are finished with this process we have an (n + 2)-gon (n being the number of triangles in the triangulation). Since each side of this polygon is identified with precisely one other side, the polygon has an even number of sides. This polygon with the appropriate identifications gives us a topological model of the manifold M. In order to obtain the normal form we proceed as follows: We start with an edge of the triangulation which corresponds to two sides of the polygon. The edge can be denoted by , where both P and Q correspond to two vertices of the polygon. In traversing the boundary of the polygon we cross the edge once and the edge (Q,P> once. We will label one of these In this way we can associate a letter with edges by c and the other by each side of the polygon, and call the word obtained by writing the letters in the order of traversing the boundary the symbol of the polygon. The remainder of the game is devoted to simplifying the symbol of the polygon.
17
1.2. Topology of Riemann Surfaces
If the sides a and a -1 follow one another in the polygon, and there is at least one other side then you can remove both sides from the symbol and the new symbol still is the symbol of a polygon which is a topological model for M. The polygon's sides have been labeled and so have the vertices of the polygon. We now wish to transform the polygon into a polygon with all vertices identified. This is done by cutting up the polygon and pasting in a fairly straight forward fashion. To illustrate, suppose we have a vertex Q not identified with a vertex P, as in Figure I.1. Make a cut joining R to P and paste back along b to obtain Figure 1.2. We note that the number of Q vertices has been decvased by one. Continuing, we end up after a finite numbér of steps with a triangulation with-all the vertices identified.
Figure 1.2
The final simplification we need involves the notion of linked edges. We say that a pair of edges a and b are linked if they appear in the symbol of the polygon in the order a • • • • • • a' • • b" • • • . It is easy to see that each edge of the polygon is necessarily linked with some other edge (unless we are in the situation that there are only two sides in the polygon). We can then transform the polygon, by a cutting and pasting argument similar to the one used above so that the linked pair is brought together as aba-1 1) -1 . We finally obtain the normal form of the surface. The normal form of a compact orientable surface is a polygon whose symbol is aa' or b i- • • 4,41; 1 6 1 . In the former case we say that the genus of M is zero and in the latter case we say that the genus is g. It is clear that g is a complete topological invariant for compact orientable surfaces. From the normal form we can, of course, reconstruct the original surface by a "pasting" process. Figures 1.3 and 1.4 explain the procedure. In particular, a surface of genus g is topologically a sphere with g handles. A surface of genus 0 is topologically (also analytically—but this will not be
b,
.b, 1511 (
)
■
)
a,
-
Figure 1.3.
Surface of genus 1.
I Riemann Surfaces
18
Figure 1.4. Surface of genus 2. seen until 111.4 or IV.4) a sphere. A surface of genus 1 is topologically a torus. There are many complex tori. (In fact, as will be seen in IV.6, a one-complexparameter family of tori.) Using the common vertex of the normal form as a base point for the fundamental group, one shows that it i (M) is generated by the 2g closed loops al, , ar b i, . subject to the single relation IP1,. aibia; 1 1); = 1. Hence I-1 1(M) is the free group on the generators [ai], [b1], j = 1, . . . , g. In particular for a compact surface of genus g, H 1 (M) 24_ Z29 and /3 1 = 2g. Remark. The "pasting" process is, of course, not uniquely determined by the symbol. For example, in the case of genus 1, after joining side a' a l , we may twist the resulting cylinder by 2ir radians before identifying b -1 with 6 1 . The "twisted" surface is, of course, homeomorphic to the "untwisted" one. The homeomorphism is known as a Dehn twist. 1.2.6. Euler-Poincaré. The Euler-Poincaré characteristic z of compact surfaces of genus g is given by x = ao - a 1 + a 2 , where ak is the number of k simplices in the triangulation. A triangulation of the normal form gives = 2 - 2g. The Euler-Poincaré characteristic is also given by fit, - fi1 + /3 2, where fik is the kth Betti number. Thus the computations of the Betti numbers in 1.2.4 and 1.2.6 yield an alternate verification of the value of x. 1.2.7. As an application of the topological invariance of the Euler- Poincaré characteristic, we establish a beautiful formula relating various topological indices connected with a holomorphic mapping between compact surfaces. Consider a non-constant holomorphic mapping between compact Riemann surfaces given by (1.5.1). Assume that M is a compact Riemann surface of genus g, N is a compact surface of genus y. Assume that f is of degree n (that is, f -1 (Q) has cardinality n for almost all Q e N). We define the total branching number (recall definition preceding Proposition 1.1.6) of f by B=
E
b
PEM
Theorem (Riemann-Hurwitz Relation). We have g = n(y - 1) + 1 + B/2.
19
1.3. Differential Forms
PROOF. Let S = {f(x); x e M and b f (x)> 01. Since S is a finite set, we can triangulate N so that every point of S is a vertex of the triangulation. Assume that this triangulation has F faces, E edges, and V vertices. Lift this triangulation to M via the mapping f. The induced triangulation of M has nF faces, flE edges, and n V — B vertices. We now compute the Euler—Poincaré characteristic of each surface in two ways: F— E + V = 2 — 2y nF — nE + nV — B = 2 — 2g.
From the above we obtain
1 — g n(1 —
13/2.
1.2.8. We record now (using the same notation as above) several immediate consequences. Corollary 1. The total branching number B is always even. Corollary 2. Assume that f is unramified. Then
a. g.--0n=landy=O. y = 1 (n arbitrary). c. g > 1 g = y for n = 1. g > y> 1 for n> 1 (and n divides g — 1).
b. g = 1
Corollary 3 a. If g = 0, then y = O. b. If 1 g = y, then either n = 1 and (thus) B = 0 or g = 1 and (thus) B = O.
1.3. Differential Forms We assume that the reader is familiar with the theory of integration on a differentiable manifold. We briefly review the basic facts (to fix notation), and make use of the complex structure on the manifolds under consideration to simplify and augment many differential-geometric concepts. The necessity of introducing differential forms stems from the desire to have an object which we can integrate on the surface. The introduction of 1-forms allows us to consider line integrals on the surface, while the introduction of 2-forms allows us to consider surface integrals. Various operators on differential forms are introduced, and in terms of these operators we define and characterize different classes of differentials. Remark. We shall use interchangeably the terms "form", "differential", and
"differential form".
20
I Riemann Surfaces
1.3.1. Let M be a Riemann surface. A 0-form on M is a function on M. A 1-form co on M is an (ordered) assignment of two continuous functions f and g to each local coordinate z (=x + iy) on M such that (3.1.1)
f dx + g dy
is invariant under coordinate changes; that is, if 2' is another local coordinate on M and the domain of Y intersects non-trivially the domain of z, and if co assigns the functions j, "d. to Z, then (using matrix notation) ax
.R2)]
ay (3.1.2)
[
[
g(z(1)
[
)
on the intersection of the domains of z and Y. The 2 x 2 matrix appearing in (3.1.2) is, of course, the Jacobian matrix of the mapping 1—■ z. A 2-form Q on M is an assignment of a continuous function f to each local coordinate z such that f dx dy (3.1.3) is invariant under coordinate changes; that is, in terms of the local coordinate Z we have a(x,y)
= .gz(2)) em57) ,
(3.1.4)
where 0(x,y)/a(50 is the determinant of the Jacobian. Since we consider only holomorphic coordinate changes (3.1.4) has the simpler form
= f(z( 2))
dz 2 •
(3.1.4a)
1.3.2. Many times it is more convenient to use complex notation for (differential) forms. Using the complex analytic coordinate z, a 1-form may be written as u(z) dz + v(z) dY, (3.2.1) where dz = dx + i dy,
di = dx i dy, and hence comparing with (3.1.1) we see that f=u+ g = i(u — y).
Similarly, a 2-form can be written as g(z) dz
df.
(3.2.2)
21
1.3. Differential Forms
It follows from (3.2.2) that d:A
d
=
—
2i dx dy.
(3.2.3)
1.3.3. Remark. To derive (3.2.3), we have made use of the "exterior" multiplication of forms. This multiplication satisfies the conditions: dx A dx = 0 = dy A dy, dx A dy = —dy A dx. The product of a k-form and an /-form is a k + 1 form provided k + I < 2 and is the zero form (still k + 1) for k + 1 > 2. In view of the last remark, we let Ak denote the vector space of k-forms, we see that Ak is a miklule over A° and that Ak = tol for k > 3. Further
A = Me/VeA2 is a graded anti-commutative algebra under the obvious multiplication of forms. 1.3.4. A 0-form can be "integrated" over 0-chains; that is, over a finite set of points. Thus, the "integral" of the function f over the 0-cycle
ENP„
P„e M, ni Z
is En,f(P2). A 1-form co can be integrated over 1-chains (finite unions of paths). Thus, if the piece-wise differentiable path c is contained in a single coordinate disc z = x + iy, c: I M (I = unit interval [0,1]), and do is given by (3.1.1), then ri dx dy w = Jo ff(x(t),y(t)) Tt + g(x(t),y(t))--d7} dt. By the transition formula for o), (3.1.2), the above integral is independent of choice of z, and by compactness the definition can be extended to arbitrary piece-wise differentiable paths. Similarly, a 2-form Q can be integrated over 2-chains, D. Again, restricting to a single coordinate disc (and Q given by (3.1.3)),
L = ff„, f(x,y) dx
A
dy.
The integral is well defined and extends in an obvious way to arbitrary 2-chains. It is also sometimes necessary to integrate a 2-form Q over a more general domain D. If D has compact closure, there is no difficulty involved in extending the definition of the integral. For still more general domains D, one must use partitions of unity. Remark. We will see in IV.5 that evaluation of integrals over domains on an
arbitrary surface M can always be reduced to considering integrals over plane domains.
22
I Riemann Surfaces
1.3.5. For C'-forms (that is, forms whose coefficients are C' functions), we introduce the differential operator d. Define df = fx dx + fy dy
for C' functions f. For the C' 1-form co given by (3.1.1) we have (by definition) dco = d(f dx) + d(g dy) = df A dx + dg A dy = (fx dx + f, dy) A dx + (g x dx + g, dy)A dy = (gx — fy) dx A dy.
For a 2-form Q we, of course, have (again by definition) (IQ = 0.
The most important fact concerning this operator is contained in Stokes' theorem. lf u) is a k-form (k = 0,1,2) and D is a (1 + k)-chain, then
= fD do). (Of course, the only non-trivial case is k = 1.) Note also that
d2 = 0, whenever c1 2 is defined.
1.3.6. Up to now we have made use only of the underlying C structure of the Riemann surface M—except, of course, for notational simplifications provided by the complex structure. Using complex analytic coordinates we introduce two differential operators a and by setting for a C' function f, af = dz and
= df;
and setting for a C 2 1-form co = u dz + y di,
ea, = Cu A dz + ey A clf= y, dz A cif, = -"eu A dz + v A cif = uf d2- A dz = — a! dz A where
fi = i(fx + For 2-forms, the operators a and .0 are defined as the zero operator. Recall. The equation fi = 0 is equivalent to the Cauchy-Riemann equations for Re f, Im f ; that is, fi = 0 if and only if f is holomorphic. It is easy to check that the operators on forms we have defined satisfy d =a
23
1.3. Differential Forms
It is also easy to see that
a2 = a + o = = 0, whenever these operators are defined.
1.3.7. In the previous paragraph the complex structure on M was still not used in any essential way. We shall now make essential use of it to define the operation of conjugation on smooth (C I or C2—as is necessary) differential forms. We introduce the conjugation operator * as follows: For a 1-form co given by (3.1.1), we define *co =
—
g dx
dy.
(3.7.1)
This is the most important case and the only one we shall need in the sequel. To define the operator * on functions and 2-forms, we choose a non-vanishing 2-form 2 (z) dx A dy on the surface. (Existence of such a canonical 2-form will follow trivially from IV.8.) Iff is a function, we set
*f f(z)(4z) dz
A dy).
For a 2-form ,Q, we set
*S2 = 012(z) dx A dy. It is clear that for each k = 0, 1, 2, *:Ak
and ** = ( — 1)k. Further, if co is given in complex notation by (3.2.4 then
*a) = — iu(z) dz + iv(z) dT.
(3.7.2)
The operation * defined on 1-forms co has the following geometric interpertation. Iff is a C' function and z(s) = x(s) + iy(s)is the equation of a curve parametrized by arc-length, then the differential df has the geometric interpretation of being (0110-c) ds, where Offe-t is the directional derivative of f in the direction of the tangent to the curve z. In this context, *df has the geometric interpertation of being ( Of/an) ds, where Of left is the directional derivative of f in the direction of the normal to the curve z. (Note that ds = Idzi in this discussion.) Remark. The reader should check that the above definitions (for example (3.7.1)) are all well defined in the sense that *co is indeed a 1-form (that is, it transforms properly under change of local coordinates).
1.3.8. Our principal interest will be in 1-forms. Henceforth all differential forms are assumed to be 1-forms unless otherwise specified. A form co is called exact if a) = df for some C2 function f on M; a) is called co-exact if *a) is exact (if and only if co = *df for some C2 function f). We say that co is closed provided it is C' and dco = 0; we say co is co-closed provided *ai is closed.
24
I Riemann Surfaces
Note that every exact (co-exact) differential is closed (co-closed). Whereas on a simply connected domain, closed (co-closed) differentials are exact (co-exact). Hence, closed I co-closed) differentials are locally exact (co-exact). If f is a C2 function on M, we define the Laplacian of f, Af in local coordinates by Af = (fx„ + o f ) dx A dy. The function f is called harmonic provided tlf = 0. A 1-form co is harmonic provided it is locally given by df with f a harmonic function. Remarks
1. It is easy to compute that for every C2 function f,
— 2i f = 4f= d*df
(3.8.1)
2. It must, of course, be verified that the Laplacian operator A is well defined. (Here, again, the fact that we are dealing with Riemann surfaces, and not just a differentiable surface, is crucial.) 3. The concept of harmonic function is, of course, a local one. Thus we know that locally every real-valued harmonic function is the real part of a holomorphic function. Further, real valued harmonic functions satisfy the maximum and minimum principle (that is, a non-constant real-valued harmonic function does not achieve a maximum nor a minimum at any interior point). Proposition. A differential a) is harmonic (land only (fit is closed and co-closed. PROOF. A harmonic differential is closed (since d 2 = 0). It is co-closed by (3.8.1). Conversely, if co is closed, then locally co -- df with f a 0 function. Since co is co-closed, d(*df) = 0. Thus, f is harmonic. 0
At this point observe that we have a pairing between the homology group H„ and the group of closed n-forms of class C'. This pairing is interesting for n = 1, and we describe it only in this case. If c is a 1-cycle and co is a closed 1-form of class C2, define = $, co.The homology group H, is defined by Z,/B 1 , where Z 1 is the kernel of 6 and 13 1 is the image of the 2-chains in the
one chains. The operator d on functions of class C2 and differentials of class C2 gives rise also to subgroups of the group of closed 1-forms. In particular the exact differentials are precisely the image of C2 functions (0-forms) in the group of 1-forms, and the closed forms themselves are the kernel of d operating on C2 1-forms. Hence the quotient of closed 1-forms by exact 1-forms is a group and we have for compact surfaces a nonsingular pairing between H I and this quotient group. We shall soon see that this quotient group is isomorphic to the space of harmonic differentials. (See 11.3.6.) L3.9. A 1-form to is called holomorphic provided that locally co = df with
f holomorphic.
25
J.3. Differential Forms
Proposition. a. If u is a harmonic function on M, then (3u is a holomorphic differential. b. A differential co = u dz + y di is holontorphic if and only if y = 0 and u is a holomorphic function (of the local coordinate).
PROOF. If u is harmonic, then (31.4 = O. Since au = uz dz, we see that uz is holomorphic (u„ = 0) and thus it suffices to prove only part (b), which is, of course, trivial (since we can integrate power series term by term). CI
1.3.10. Assume a) = udz + v cif is a C' differential. Then (using d = a +0) we see-that da) = Oh — vz )df
A
dz,
and d*co = —
+ vz)clf
A
dz.
Thus, we see that a) is harmonic if and only if u and T7 are holomorphic. Hence, we see that if co is harmonic, there are unique holomorphic differentials co l and a)2 such that w = (Di + (D2-
1.3.11. Theorem. A differential form co is holomorphic if and only if co = a + i*a for some harmonic differential a. PROOF. Assume that a is harmonic. Then
a = ca l + (7)2 with co -(j = 1,2) holomorphic. Thus
* or =
+ 02.
Thus,
a + i*a = 2a) i is holomorphic. Conversely, if a) is holomorphic, then a) and cr) are harmonic and so is
a
— = 2 •
Further —10) — itr3
2 Thus CO
= + i *CZ.
Corollary. A differential o) is holomorphic ff and only if *co = —ico.
it is closed and
26
I Riemann Surfaces
The forward implication has already been verified (every holomorphic differential is harmonic). For the reverse, note that if co is given by (3.2.1) and *o.) = ico, then (3.7.2) implies that co = u dz. Since do) = 0, u is PROOF.
holomorphic.
1.4. Integration Formulae In this section we gather several useful consequences of Stokes' theorem. 1.4.1. Theorem (Integration by Parts). Let D be a relatively compact region on a Riemann surface M with piecewise differentiable boundary. Let f be a C' function and co a C' 1-form on a neighborhood of the closure of D. Then
(4.1.1) PROOF. Apply Stokes' theorem to the 1-form fco and observe that d(jio)= f do) + df A co.
Corollary 1. If co is a closed (in particular, holomorphic) 1-form, then (under the hypothesis of the theorem) IÔD CI)
PROOF.
Take f to be the constant function with value 1.
Corollary 2. Let f be C' function and w a C' 1-form on the Riemann surface M. If either f or co has compact support, then
ffm f do)— Sim o) df = 0.
(4.1.2)
If M is not compact, then take D to be compact and have nice boundary so that either f or co vanishes on bD and use (4.1.1). If M is compact cover M by a finite number of disjoint triangles Ll, j = 1, . . . , n. Over each triangle (4.1.1) is valid. We obtain (4.1.2) by noting that PROOF.
•E
fro = 0,
J=1
since each edge appears in exactly 2 triangles with opposite orientation.
D
1.4.2. We fix a region D on M. By a measurable 1-form w on D, we mean
a 1-form (given in local coordinates by) co = u dz + y dT, where u and y are measurable functions of the local coordinates. As usual, we agree to identify two forms if they coincide almost everywhere (sets of Lebesgue measure zero are well defined on M!). We denote by L 2(D) the
27
1.4. Integration Formulae
complex Hilbert space of 1-forms co with
IfreV = If, co A *53
E)• Let ti be a C function with support in D i _ 2,. Consider
d
0(z) =.'ff, ( 0(1C -- zi)ii(C)
2i ,
ZE
C.
(2.1.5)
It is clear that ik is a continuous function on C with support in D I Observe that we may integrate over C instead of over D, and that by a change of variable (after extending i to be zero outside its support)
oz)
ssca)(1012(c
z) E1C A2idr.
Thus
- - SS, 0(1(1);i p(c
i z) dc—^24
= fjc co(Koîz p(c z)422": —
=
a
A
Similarly,
aly
a
A
dr
dC —2i OC
Thus Ili is C. We claim that 4
a2q,
araz =
Note that
4/(z) =
+
4A
v(zatco - 2..
1dC A 4. zi 0 be chosen so that the closed ball of radius e about z is contained in B. Let B, be the complement in B of this ball. Start with Stokes' theorem PROOF.
= 5.1.B. :c (cu(C)z )4 cg,
35
11.2. Weyl's Lemma
and obtain
ff.., _ ,
=
z d-c•A
2w
f B r, — Z —
0
4
L47.
ze ie) dO.
Letting e ■ 0, we obtain (2.2.1). -
11.2.3. Proof of Weyl's Lemma (Conclusion). To verify formula (2.1.9), it suffices to assume that p has support in — zl < e/2, and thus that the integral in (2.1.8) extends over C. To check this last claim, define _ v(0 = P(2 IC - z1),40, E C.
lc
Note that
ff
p(C) dC A dr rr v -2i - ..1.31;-:1<e/2 —
zl a/2, f is harmonic and h — 0 = O. Thus u is harmonic on M \ {Po } . We summarize our conclusions in the following Theorem. Let M be a Riemann surface with z a local coordinate vanishing at
Po E
M.
There exists on M a fiinction u with the following properties:
u is harmonic on M\ (,P01,
(4.1.4)
u — z - " is harmonic on every sufficiently small neighborhood N of Po , (4.1.5)
ffm,N du A
(4.1.6)
< oo, and
(du,c/f) = 0 = (du,*df) for all smooth functions f on M that have compact support and vanish on a neighborhood of Po.
(4.1.7)
PROOF. Only (4.1.7) needs verification, and this follows because du is in H with respect to the surface M',Cl N. El
11.4.2. Note that condition (4.1.5) is invariant under a limited class of coordinate changes. (Determine this class!) It can, of course, also be replaced by (4.2.1) u — Re z - " (resp., u — Im z") is harmonic in a neighborhood N of Po. In this case we may require u to be real valued. Observe also that if M is compact, then u is unique up to an additive constant. 11.4.3. We note that our arguments to prove Theorem 11.4.1 did not depend on the choice of the particular form of the function h with singularity as long as h is harmonic in {a/2 < z < a}, and *dh = 0 on {izi = a } . Thus, another candidate for h is the function h(z) = log
z — z 1 z a2 1.7, z — z 2 z — a 2/ 2 —
1z1
a, 0 < lz 11,1z21 < a/2.
47
11.4. Harmonic Differentials
Theorem. Let M be a Riemann surface and P 1 and P2 two points on M. Let (j = 1.2) be a local coordinate vanishing at Pi . There exists on M a realvalued function u with the following properties: u is harmonic on MqP 1 ,P2 },
(4.3.1)
u — loglz i l is harmonic in a neighborhood of P 1 and u + loglz2 1 is harmonic in a neighborhood of P2,
(4.3.2)
ffm\N du A *du 0 discs have been removed. Consider now two copies D and D' of D. We shall construct a compact Riemann surface M = Cl D u D' known as the double of D. We use the usual local coordinates on D. A function z is a local coordinate at P' E D' if and only if is a local coordinate at the corresponding point P E D. We now identify each point P E 3D with the corresponding point P'E SD'. To construct local coordinates at points of 3D, we map a neighborhood U of P E 3E) by a conformal mapping z into the
48
11 Existence Theorems
closed upper half plane such that U SD goes into a segment of the real axis. By the reflection principle z is a local coordinate at P E M. Note that M is a compact Riemann surface of genus 2g ± n — 1, and that there exists on M an anti-conformal involution j such that j(D) = D' and j(P) = P for all P E (5D. From now on we may forget about Mo and think of D as a domain on its double M. (The above discussion is not strictly necessary for what follows. It was introduced for its own sake.) We consider now the following problem: Fix a C2 function cp o defined on a neighborhood of Cl D. Among all functions cp, C2 on a neighborhood of Cl D, find (if it exists) one u that has the same boundary values (on SD) as cpo and minimizes 11411 0 over this class. Assume that u is a harmonic function with the same boundary values as (p o . We c. , ;npute for arbitrary (p with the same boundary values;
11d(o11 2 = (d(cp — u) + du, d(cp — u) + du) = — 01 2 + 11(142 + 2 Re(d(q) — u),du). Now use Proposition 1.4.4 to conclude that (d(cp — u),du) = 0 (since — u = 0 on (5D and du = 0 on D). Thus
11411 2 = 11d((p — u)11 2
+ I1d u 112 Ildu112.
Thus our problem has a unique solution—if we can find a harmonic function with the same boundary values as (po . We shall in 1V.3 solve this problem by Perron's method. Here we outline how the decomposition of L2(D) given by (3.2.1) can be used to solve our problem. Finding a function for which a given non-negative function on L 2(D) achieves a minimum is known as the Dirichlet Principle. Consider the function 9 0 . Since dcpo is exact (and thus closed), dcp o e E (1) H. Now let co be the orthogonal projection of dcp onto H. We have already seen that co is exact and harmonic. Thus co = du, for some harmonic function u on D. Now d(u — cp 0) e E. Thus
(d(u — (p0),c() = 0, all a E H. Let up,p2 be the function produced by Theorem 11.4.3. By letting a run over the set of differentials
{dup,p2; P 1 E D', P2 E D'1, one can show that u — q,,, is C 2 on a neighborhood of Cl D and that u —q, 0 = 0 on SD.
11.5. Meromorphic Functions and Differentials Using the results of the previous section, we construct first meromorphic differentials on an arbitrary Riemann surface M and then (non-constant) meromorphic functions.
11.5. Meromorphic Functions and Differentials
49
11.5.1. By a meromorphic differential on a Riemann surface we mean an assignment of a meromorphic function f to each local coordinate z such that f(z) dz
is invariantly defined. Theorem a. Let P E M and let z be a local coordinate on M vanishing at P. For every integer n > 1, there exists a meromorphic differential on M which is holomorphic on MVP} and with singularity 1/z"+ at P. b. Give* two distinct pobtts P, and,P 2 on M and local coordinates z; vanishing at pi, j = 1, 2, there exists a meromorphic differential a), holomorphic on MVP,,P,}, with singularity 1/: at P 1 and singularity —1/z 2 at P2. PROOF. Let 1 = du where u is the function whose existence is asserted by Theorem 11.4.1 for part (a) and by Theorem 11.4.3 for part (b). In the former case set
—1 2n
w = —( + i*a), and in the latter
11.5.2. Let q be an integer. By a (meromorphic) g-differential a) on M we mean an assignment of-a meromorphic function f to each local coordinate z on M so that f(z)
(5.2.1)
is invariantly defined. For q = 1, these are just the meromorphic differentials previously considered, and they are called abelian differentials. If w is a q-differential on M, and z is a local coordinate vanishing at P E M, and w is given by (5.2.1) in terms of z, then we define the order of ta at P by ordp w = ord o f. f(z)= zng(:) with g holomorphic and non-zero at z = 0, then (If we write ord o f = n.) It is an immediate consequence of the fact that local parameters are homeomorphisms that the order of a q-differential at a point is well defined. Note that {P e M; ordp w 0 } is a discrete set on M; thus a finite set, if M is compact. If w is an abelian differential, then we define the residue of w at P by resp co = a_ 1 ,
where w is given by (5.2.1) in terms of the local coordinate z that vanishes at P, and the Laurent series of f is f(z) =
E n= N
11
50
Existence Theorems
The residue is well-defined since resp co =
1 2711
where c is a simple closed curve in M that bounds a disk D containing P such that c has winding number 1 about P and co is holomorphic in CID \ 11.5.3. Theorem. Let P1,. , Pk be k> 1 distinct points on a Riemann surface M. Let c l, , ck be complex numbers with El= ci = 0. Then there exists a ,P} with meromorphic abelian differential co on M, holomorphic on MVP 1 , ordpj to = — 1,
resp2 to =- ci.
PROOF. Let Po E M, Po Pi, j = 1, . • , k. Choose a local coordinate , k. For j = 1, . . . , n, let co; be an abelian differvanishing at Pi, j = 0, ential with singularities 1/z i at Pi and — 1/z0 at Po and no other singularities. Set =
E
ci
1=1
Corollary. Every Riemann surface M carries non-constant meromorphic functions.
PROOF. Let P1 , P2, P, be three distinct points on M. Let co, be a differential holomorphic on MVP 1 ,P2 } with ordp, Co 1 = —1 = ordp2 coi resp2 co l = — 1. resp, co l = +1, Let oh be a differential holomorphic on MVP2,P3 } with ordp2 w2 = —1 = ordp, w2, reSp3 CO2 = —1.
reSp2 CO2 = 1,
Set f = co1 /co2 . Note that f has a pole at P 1 and a zero at P3. 11.5.4. Proposition. Let M be a compact Riemann surface and co an abelian differential on M. Then
E
resp co = O.
(5.4.1)
PeM
PROOF. Triangulate M so that each singularity of co is in the interior of one triangle. Let A 1 , 4 2, , Ak be an enumeration of the 2-simplices in the triangulation. Then
E resp co = Pet,'
1 2rti
E f64i CO, I
.1=
(5.4.2)
11.5. Meromorphic Functions and Differentials
51
where 64 ; is the (positively oriented) boundary of 4, . Since each 1-simplex appears twice, with opposite signs, in the sum (5.4.2), we conclude that (5.4.1) holds. EXERCISE Using only formal manipulations of power series show that the residue of a meromorphic abelian differential is well defined.
Remark. The above proposition shows that the sufficient condition in
Theorem 11.5.3 is also necessary if the surface M is compact. EXERCISE Give an alternate proof of Proposition 1.1.6 in the special case that N=Cv too). First reduce to showing that it suffices to establish that f has as many poles as zeros. Then relate ordp f to resp(dflf).
11.5.5. Remark. The existence of meromorphic functions shows at once that every compact Riemann surface M is triangulable. Let f
C u (c}
be a non-constant meromorphic function. Triangulate C u {oo} with a triangulation d i , 4 2, , zi k such that the image under f of every ramified point (points P e M with bf (P) > 0) is a vertex of the triangulation and such that f restricted to the interior of each component off -1 (A ) is injective. It is clear that this triangulation of C u {co} lifts to a triangulation of M. The existence of meromorphic functions has many other important consequences. We will discuss these in IV.3 and IV.5. Also, in IV.3 we will establish the existence of meromorphic functions without relying on integration (for which triangulations are needed). Thus we will be able to derive the above topological facts from the complex structure on the Riemann surface.
CHAPTER
m
Compact Riemann Surfaces
This is one of the two most important chapters of this book. In it, we prove. on the existence theorems of the previous chapter) the three most (based important theorems concerning compact Riemann surfaces: the RiemannRoch theorem, Abel's theorem, and the Jacobi inversion theorem. Many applications of these theorems are obtained; and the simplest compact Riemann surfaces, the hyperelliptic ones, are discussed in great detail.
III.!. Intersection Theory on Compact Surfaces We have shown (in 1.2) that a single non-negative integer (called the genus) yields a complete topological classification of compact Riemann surfaces. Every surface of genus Ois topologically the sphere, while isurface of positive genus, g, can be obtained topologically by identifying in pairs appropriate sides of a 4g-sided polygon. The reader should at this point review Figures 1.3 and 1.4 in 1.2.5. Thus, with each surface of genus g > Owe associate a 4g-sided polygon with symbol a,biai- b i-1 agbga.; 1 1); 1 (as in 1.2.5). The sides of the polygon correspond to curves (homology classes) on the Riemann surface. These curves intersect as shown in the figures referred to above. Our aim is to make this vague statement precise. This involves the introduction of a cononical homology basis (basis for H 1 ) on M. • • •
111.1.1. Let c be a simple closed curve on an arbitrary Riemann surface M. We have seen (11.3.3) that we may associate with c a (real) smooth closed differential ric with compact support such that .1: = (oc,* tic)
= — ffm cx A ti c ,
(
1.1.1)
53
111.1. Intersection Theory on Compact Surfaces
for all closed differentials 1. Since every cycle c on M is a finite sum of cycles corresponding to simple closed curves, we conclude that to each such c, we can associate a real closed differential tic with compact support such that (1.1.1) holds. Let a and b be two cycles on the Riemann surface M. We define the intersection number of a and b by a' b=
S$M
A
(1.1.2)
tlb= ('1., —* %).
Proposition. The intersection number is well defined and satisfies the following properties (here a, b, c are cycles on M): . a • b depends only on the homology classes of a and b, (1.1.3) a • = —b a, (1.1.4)
(1.1.5)
(a+b).c=a•c+b-c,
and a •b
(1.1.6)
Furthermore, a • b "counts" the number of times a intersects b. PROOF. To show that (1.1.2) is well defined, choose (1.1.1). We must verify
and ti;, also satisfying
A
Note that while ti„ and are only closed, their difference pi', — = df, where f is a C function with compact support. Thus it suffices to show
ff. df
= 0,
all f as above.
= —(df,*ri b)= 0, because df E E and *rib e E1 (by Proposition 11.3.2). The verification of (1.1.3), (1.1.4), and (1.1.5) is straightforward. To check (1.1.6), it suffices to assume that a and b are simple closed curves. Therefore we have (as in 11.3.3), But ffm df A
a•b=
Jim
A
lib
=—
ffM 'lb '"
lu
=(nb,*n.)= fa llb•
But Sa th contributes + 1 for each "intersection" of a with b. (This can be verified as in 11.3.7 using Figure III.1.) 111.1.2. We now consider a compact Riemann surface of genus g
represent this surface by its symbol aa' (genus 0),
fi
i=t
(t • bia;
(genus g
1).
0, and
Ill Compact Riemann Surfaces
54
The sides of the polygon corresponding to the symbol give a basis for the homology, H, = 11 1( M ), of M. Assume now that g 1. It is easy to check that this basis has the following intersection properties • bk =
CI
ik =
0 j
=
k
a; • ak = 0 = bi bk . We can represent all this information in an intersection matrix J. This J is a 2g x 2g matrix of integers. If we label = ai, j = 1,
, g and
= bJ_ g, j = g + 1,
then the ( j,k)-entry of J is the intersection number X; • form 0 [-
Nk.
, 2g,
(1.2.1)
Thus J is of the
11
ï OT
where 0 is the g x g zero matrix and I is the g x g identity matrix. Any basis {N I, ... ,N 29) of H 1 with intersection matrix J will be called a canonical homology basis for M. Given a canonical homology basis we can use (1.2.1) to define the "a" and "b" curves. Note that we do not claim that these curves come from a polygon in normal form.
111.2 Harmonic and Analytic Differentials
on Compact Surfaces We compute the dimensions of the spaces of holomorphic and harmonic differentials on a compact Riemann surface. Certain period matrices are introduced and we determine some of their basic properties. The key tool is Theorem 11.3.5. 111.2.1. Theorem. On a compact Riemann surface M of genus g, the vector space H of harmonic differentials has dimension 2g.
55
111.2. Harmonic and Analytic Differentials on Compact Surfaces
The theorem is easy to prove if g = O. For in this case, let a be a harmonic differential on M. Fix Po E M and define PROOF.
u(P) =
Pe
a,
M.
The function u is well defined (since M is simply connected). By the maximum principle for harmonic functions, u is constant. Thus a = O. (The maximum principle implies, of course, that there are no non-constant harmonic functions on any compact Riemann Surface.) ,N29) be a canonical homology basis on M. Assume g> O. Let IN,, Construct a map &H—C -2 or
depending on whether we are interested in complex-valued harmonic differentials or real-valued harmonic differentials, by sending a E H into the 2g-tuple
--
U
N, œ ' • »f.„, ' If dim H > 2g, then cf) has a non-trivial kernel; that is, there exists an a E H, all of whose periods are zero. Such an a must be the differential of a harmonic function. Since there are no non-constant harmonic functions on a compact surface, dim H 2g. The above argument also establishes the injectivity of 0. It remains to verify surjectivity. As we saw in the previous chapter, (Theorem 11.3.7) it suffices to find a closed differential with period 1 over a cycle j and period 0 over cycles Nk, k j. Let j=1,...,g,
= ?hp
Œj =
j = g + 1, . . . , 2g.
Then we see (by(1.1.1)) that for k = 1, . . . , g,
L
a; = _j'Ç nj Aqak
=
= ff„
tak • bi = bkj , —(ak • ni- g) = 0,
j = 1, , g, , 2g, = g + 1,
and
ffm
Sbk Œj _
f
œf A qbk
=
f$M
bk
A Œi
bk • b; = 0,
j = 1, .
, g,
j =g+1,...,2g.
We summarize the above information in sak = 101, k = j, , otherwise,
k = j — g, fbk cc i = {01 ise. , otherw
111 Compact Riemann Surfaces
56 In other words,
ftti, cci = bg„,
j,
(2.1.1) 0
k = 1, . .. , 2g.
Thus we have proven the following. Corollary. Given a canonical homology basis {X i , ... X 4} for HAM) there ,c(29 } of H; that is, a basis satisfying (2.1.1). is a unique dual basis {a t , Furthermore, each aj is real. Thus given any set of 2g complex numbers Co . ,C29,
2g
CC
=
E
cpj
J-1
is the unique harmonic differential whose N,, period (that is fN,, a) is ck , for k = 1, , 2g.
111.2.2. We have seen that the 2g x 2g matrix with (k, j)-entry identity. We note that
j = 1,
ak A
Li
= ffm Mk A nKi =
iL CCk
A /j- g,
a, is the
, g,
= g + 1, . . . , 2g.
From this it is evident that the matrix whose (k, j)-entry is JIM; cc.; is of the form [I fl = J. We conclude that
— *c()) = Xj• We will investigate a companion matrix F whose (k,j)-entry is (a,,a;) = ilk( ak A *at We note immediately that F is a real matrix. Further from 1.4, 0940 = (*ccp *Œk) = ffm * CCj A **Mk = ffm Mk A * Mj = (CCOC(j). Thus, I' is symmetric. Before continuing our investigation of the matrix F, we establish the following
111.23. Proposition. If 0 and are two closed differentials on M (compact of genus g), then
ofai 6]. ffm 0A 6- =,i[faf of 6—$ 61 61
(2.3.1)
PROOF. The right-hand side of (2.3.1) is obviously unchanged if we replace 0 by 0 + df with f a C2-function. The left-hand side is also unchanged since
1(0 + df) A = (0 + d f , — = (0, — = SSm 0 A a,
57
111.2. Harmonic and Analytic Differentials on Compact Surfaces
by Proposition 11.3.2. Replacing thus 0 and g by harmonic differentials with the same periods, we may assume
2g =E
2g -= E j= 1
1=1
Tipp
pi, j e C,
(2.3.2)
where { ai} is the basis dual to the canonical homology basis {a 1 , 61, = t.t , ,N29). Thus,
SI( 0
2g A
E /LA Jim a; A
0
2g
k,1 j=
••
k. j=
g`•
■
pi/74Ni
=E 1
2g
=E g pii-40-i(Ni •E j=1 .1=g+
• Ni-d-
(2.3.3)
Now it follows immediately from (2.3.2) that pi = ft4j 0 and
(2.3.4)
=
We now substitute (2.3.4) into (2.3.3) and obtain, using the fact that for , g, (N; • Ng+ = 1, and for j = g + 1, , 2g, (N; N i _ g) = — 1,
j = 1,
ff„
()AU=
j=1
fx
J
0
fx +
g J
=t f of ai
bi
2g
—
J=0+1
a—fbi ofa i
•
Corollary. If 0 is a harmonic 1-form on M, then
11011 2= PROOF. Since
i=
[Jrdi Ofbi * 0 —
bi
of *
0]
•
(2.3.5)
0 is harmonic, *3 is also closed. Thus, we compute 11002 =
ff„
0 A *0
by (2.3.1) and obtain (2.3.5). Remark. We may view the Riemann surface M as a polygon .1( whose 1 6; 1 . Since .11 is simply connected 0 = dfon N. Thus, symbol is 1T 1
1 OA
df AO = ffa d(frh
=
Let Zo E
[fa
+ fb j g +
= , be arbitrary, then for z E
f(z)= f :00.
= + fb;i fti
58
III Compact Riemann Surfaces
Letting z and z' denote two equivalent points on the sides ai and al of &It we see that
fa, fa+ s fa= false: o — da = fa, — [f„fd — fb, fa,
•
The remaining terms can be treated similarly to obtain an alternate proof of Formula (2.3.1). III.2.4. We return to the matrix
Yki =
ff„ A
r and note that its (k, j)-entry yki is given by ai
1=1
bt
k af
1,
k
= fb,, *cci {
f I f *2.
-
—5 *ce — f *2 i=
-
. , g,
k = g + 1,
AZ1,- g
]
, 2g.
We show next that the real and symmetric matrix r is positive definite (r > 0). Let 2g
2g
k=1
lo= 1
E ift etik with k E C, E Rok i2 >0.
=
Recall that the differentials ak are real and that 11011 0 O. Thus, by (2.3.5), g
E
0
[
J-1
J
,
2g
E
k= 1
2g
=
S
20
2g
b, '1= 1
E k=1
Zi*cti
g
E
k.1=1
kCX/c
E .1= 1
fbj * at —
.11
Ski f.j * Celt
scock
f
2g
E
ai 1=1
Zi*at]
C41]
20
= k,1= E1
7"
kl•
It is now convenient to write
=
A Bi C Di
[
with A, B, C, D, g x g matrices. Note that we have established (since T = and r>0) these are real with
B = 'C,
A
= 'A,
D
= 'D,
(2.4.1)
and
A> 0,
D > 0.
(2.4.2)
111.2. Harmonic and Analytic Differentials on Compact Surfaces
59
111.2.5. Let us consider * as an operator on the space of real valued harmonic differentials. It is clearly C-linear and *2 = —I. We represent * by a 2g x 2g real (since * preserves the space of real harmonic forms) matrix with respect to the basis x i, , 22g:
k,i =1,• • • , 2g,
= (AO,
Thus
2g * Crk
=
k = 1,
E j .i
, 2g.
If we represent by d the column vector of the basis elements 2 19 • • • 9 ct2g9 then we can consider V ols defined by the equation = Ws/. Since *2 = —I, g2 = —I. We wish to compute the matrix /. We note that ( 2g 4/1k
E
= ( 100 = ( *219 *2k) =
1= 1 2g
2g
= E )-0(2.,,*ceo=j=E
Ali
1
fist Mk A Œ.
1 In 111.2.2 we saw that, the matrix with (k, j)-entry ff 2k A x; is given by J = [_°, 10 ]. It therefore follows that the above equation can be written as j=
=
(2.5.1)
If we now write A.2 1
=
13 24S [11 we find as a consequence of (2.5.1) that
-1 i r = r —23 We therefore conclude, because of (2.4.1) and (2.4.2), that
=
12
= t1 2,
2 3 = '1 3,
(2.5.2)
and
12 >
0,
—23 > 0.
(2.5.3)
Since / 2 1- /2 9 = 0, we see that / satisfies the additional equations: + 1213 + 1 g = 0,
(2.5.4) 131 1 = `À.,1 3 . 42 2 = 12`1,, the space of real-valued essentially used only 111 2.6. Up to now we have harmonic forms. (A basis over l for the space of real-valued harmonic forms is also a basis over C for the space of complex-valued harmonic forms.) We construct the holomorphic differentials co; = 2; + i*ai,
j = 1, . , 2g,
m Compact Riemann Surfaces
60
and a matrix whose (k,j)-entry is Raik,c0;) = 4(20)) + (2k.i *tx;) + 1(i *Œost) + = (4,2) — i( 70*2) = (11,1 + 02j) = (Yk,q•
We see from the above that this is the matrix F + if,
and since (recall 1.3.11)
(ark /a/j) = i Jim 2k A
Fli cck
1=1
faj e5i]
, g,
k = 1,
— fak . 9
—
L
k = g + 1, , 2g,
(.15
this matrix can be viewed as a period matrix. Observe also that Rcok,(0.1) = i(co") = (ce") i fbj Wk,
f
j
Wk,
= = g + 1 , . . . , 2 9-
Before continuing our investigation of the matrix
(2.6.1)
r + if, we establish
111.2.7. Proposition. On a compact Riemann surface of genus g, the rector space ye = dr1(M) of holomorphic differentials has dimension g. Furthermore, {co l , ,a),} forms a basis for dr. We show that we have a direct sum decomposition of the space H of complex-valued harmonic forms
PROOF.
(2.7.1)
where
represents the anti-holomorphic (= complex conjugates of holomorphic) differentials. It is obvious that Yf n 17 = {01. It remains to verify that the decomposition (2.7.1) is possible. If a E H, then a + i*a e Ye, — i * 2 E (since + i e de ), and a -1(a + i*a) + f(a — i*I). onto Y.?, it follows from Since wi--+ C5 is an 51-linear isomorphism of (2.7.1) that dime ye = dimR ye = 1 4
2
To show that {ah, 09 } form a basis for .rf it suffices to show that this is a linearly independent set. Let = (c i ,
,c,), with Ci = (oh,
'911 = (œ1, • • • ad,
C,
A = Re C, B = Im C,
'91 2 = 019+1, • • • 912g).
111.2. Harmonic and Analytic Differentials on Compact Surfaces
61
Assume (recall that A and ;12 have been defined in 111.2.5) 0 = 'CO = ('A + i'13)(1 1 + i*91 1 ) = A + irB)[91 +
+ ;.2 91 2)]
— `B(A 1 i11 3 + A2 91 2 ) + i[ 'B211 + `A(A i % J. + ).2/1 2)]. Thus, we obtain two equations: CA. —`13A 1)9.1 3 = '&2912 ,
('B + tAA1) 9.1 1=
Since the differentials in 21, are linearly independent from the differentials in 9{2, we conclude that 'BA; = 0 = Since %2 is non-singular, '13 =0 ='A. 111.2.8. We return now to the matrix r + if. In particular, we restrict our attention to the first g rows of this matrix; that is, the g by 2g matrix ().2 ,
—
A 1 + i/).
We have shown by (2.6.1) that the (j,k)-entry of ).2 is — fbk COi; and the (j,k)-entry of —A, + ills i fak coj. We conclude that if we consider ico i , , icog as a basis for the vector space 0, of holomorphic differentials on M, then the period matrix (whose (j,k)-entry is f„,, icc),) with respect to this basis is (—A, + il, —A i). Similarly, the holomorphic differentials cog+1, (02, are linearly independent (over C), and the last g rows of the matrix r + Li is the g x 2g matrix — il, —A 3).
(
Thus, the period matrix of hog+ . . . , io129 (whose (j,k)-entry is fxj, is of the form (
—A3, 'A 1
We now make another change of basis. Let
3' = VI,
• • • ,
C9) = - ;.3) - (0(.09 + 1, • • • jr- ) 29).
With respect to the basis S of .r, we obtain the period matrix (1,( — A3) - 1 `,1 t + 4 — A3) -1 ) =
(2.8.1)
for the space of holomorphic abelian differentials (= space .ir) with the property J ., Cit = Furthermore, for this basis, the matrix II = (nik) with nik = Ck is symmetric with positive definite imaginary part. Proposition. There exists a unique basis 1C 1 ,
62
III Compact
Riemann Surfaces
We must only verify that // is symmetric and 1m 17 > 0. To show that H is symmetric it suffices (because ';.3 = %.3) to show that ( ).1 ) -1 `;., is symmetric. But (recall (2.5.4)) PROOF.
2 3) -1 = ( -11 3) -1 definiteness is not an issue. Note that since Im 17 = ( ).3)', positive
'[( -2 3) -
= Ai( —9, 3) - 1 =
Cl
Remark. Note that ( 1 3) = I if and only if for k, j = g + 1, . . . , 2g, 1 (oh " co.) = -2
w=
1,
k=j k
j'
if and only if (coo. 1 /12, • • • ,(02 is an orthonormal basis for A' viewed as a Hilbert subspace of OM). Can this happen? Yes, if and only if Im TI = I (see (2.8.1)).
111.3. Bilinear Relations We start with a compact Riemann surface M of positive genus g > 0, and a canonical homology basis tch, ,ag,b i , ,bal on M. Let IC I , be the dual basis for holomorphic differentials (that is, f„, ;.; = e•fi ). Represent the Riemann surface M by a 4g-sided polygon di with identification. Our starting point is Formula (2.3.1) for the "inner product" of closed differentials. Let 0 and g be closed differentials. Since 4 is simply connected, 0 = df on 4 with f a smooth function on di (note that at equivalent points on 54, f need not take on the same value). As we have seen in the remark in 111.2.3, formula (2.3.1), may be viewed as a consequence of two identities:
ffh,
and
A U=
e f, — fb, e L rd.
g
fa=
(3.0.1)
E
(3.0.2)
We shall see in this section that normalizing a set of meromorphic differentials (with or without singularities) forces certain identities between their periods. These are the "bilinear relations" of Riemann. They will turn out to be useful in the study of meromorphic functions on M. 1113.1. Let us assume that 0 and g are holomorphic differentials, then 0=
[Se,
e
e
(3.1.1)
Equation (3.1.1) follows from (3.0.2) by Cauchy's theorem, or from (2.3.1) since 0 A = 0 for holomorphic 0 and O. We now let 0 = Ci and g = Ck, and
63
111.3. Bilinear Relations
obtain (3.1.2) Of course, (3.1.2) is just another way of saying that the matrix fl introduced in 111.2.8 is symmetric.
a =L. We note that (df = Ci on Ji) by
1113.2. Next we let 0 = C; and Stokes' theorem (Formula (3.0.1))
55„, = IL /zit
Cj A Zk •
We conclude that k
bk
= —2i(Im nik) (as usual it is the ( j,k)-entry of the matrix II). In particular, 1m 7r1; > 0. Applying the same argument to O=
E Ckck,
Ck
E C,
k= 1
we conclude that Im 11 > 0. These facts have already been established in 111.2. 111.33. Proposition. Let 0 be a holomorphic differential. Assume either a. all the "a" periods of 0 are zero (that is, fa, 0 = 0,j = 1, . . . , g), or b. all the periods of 0 are real. Then U = 0. PROOF. We compute 110112 = o A '10 = i o A 6
p.m
=i
L, 0 _
01, 01.
In either case (a) or (b) we conclude that 11011 2 = 0, and hence 0 = 0.
El
Remark. The above observation (plus the fact that dim Ye = g, where Ye = dri(M)) can be used to give an alternate proof of the theorem that there is a basis for Ye dual to a specific canonical homology basis. (The reader is invited to do so!) 1113.4. We shall now adopt the following terminology: Recall that meromorphic one-forms are called abelian differentials. The abelian differentials which are holomorphic will be called of the first kind; while the
111 Compact Riemann Surfaces
meromorphic abelian differentials with zero residues will be called of the second kind. Finally, a general abelian differential (which may have residues) will be called of the third kind. Before proceeding let us record the following consequence of the previous proposition.
Corollary. We can prescribe uniquely either a. the "a" periods, or b. the real parts of all the periods for an abelian differential of the first kind. PROOF. Consider the maps dr and .re 51 20 defined by 91—• (1,,, 9, • • • ja, (p) and 9 f .(iin fa, 9, • • . Jin fa, 9,Im lb, 9, • • ,Im lb. 9), re—
spectively. The map Ye Cg is a linear transformation of .e viewed as a g-dimensional vector space over C, and the map if •-■ R 20 is a linear transformation of Yt' viewed as a 2g-dimensional vector space over R. The proposition tells us that these maps have trivial kernels and thus are isomorphisms (since the domains and targets have the same dimensions). 0 Remark. Parts (a) of the proposition and its corollary have previously been established (Proposition 111.2.8).
111.3.5.
Let us consider abelian differentials of the third kind on M. Choose two points P and Q on M. It involves no loss of generality to assume that the canonical homology basis has representatives that do not contain the points P and Q. Let us consider a differential r regular, (= holomorphic) on M\{P,Q}, with ordp t = — 1 = orda r, (3.5.1) r esp t = 1, res T = — 1.
Let c be a closed curve on M. It is, of course, no longer true that fc r depends only on the homology class of the curve c. However (assuming P, Q are not on the curves in question) if c and c' are homologous, then there is an integer n such that
f
t —f i= 27rin.
(3.5.2)
e
The easiest way to see (3.5.2) is as follows: Let 0 be an arbitrary abelian differential of the third kind on M. Let P,, , P, (k > 1) be the singularities of O. Assume that the canonical homology basis for M does not contain any of the points Pi. Let ci, j = 1, . k, be a small circle about Pi. We may assume that the curves a1,
, ag ,
61 ,
, bg,
el,
, Ck
are mutually disjoint. It is easy to see that on M' = M\{P i , . ,P, } , c, is homotopic to
11 j=I
k—1
1 a.b.a7 1 b7 J
nc
1= 1
65
111.3. Bilinear Relations
and that the curves
a1,
, ag,
6 1 ,b 9 r
C1, . .
Ck_
form a basis for H 1(M'). The differential O is closed on AL Hence, if c is a curve on M which is homologous to zero, then on M' it is homologous to a linear combination of c 1,...,ck _ i . Thus there are integers ni with c homologous to Elizi nick Hence
fO=
E
ni
0 = 2ni
E
n; resp, 0.
(3.5.3)
J=
Clearly (3.5.2) is a special case of (3.5.3). In 'Particular, for r ar before, fit., T is defiffed only modulo 27ri7L (hereafter, mod 2ni). 111.3.6. To get around the above ambiguity, we consider M as represented by the polygon with identifications. We choose two points P and Q in the interior of ,,if. Let T be a differential of the third kind satisfying (3.5.1). By subtracting an abelian differential of the first kind, we normalize t so that j= 1,...,g,
T
Or
fx,
T
is
j = 1, . . . , 2g.
purely imaginary,
(3.6.1) (3.6.2)
We denote the (unique) differential r with the first normalization by and the one with the second normalization by con .
TN
Warning. In (3.6.1) we think of ai as a definite curve—not its homology class. If we want to think of it as a homology class, (3.6.1) must be replaced
by j = 1, . . . , g,
r = 0 (mod ai),
(3.6.1a)
and "Cm is no longer unique.
Applying (3.0.2) with 0 = Ç and
L ai
fTPQ
=
TpQ,
=
fb
,
we obtain T
PQ'
The line integral around the boundary of .ift can be evaluated by the residue
theorem, since f(z) = with the path of integration staying inside It. Thus we obtain or
f,„„ frpQ = 2ni(f(P) — AO), 2ni LP C.; =
pQ,
as long as we integrate C.; from Q to P along a path lying in .11.
(3.6.3)
HI Compact Riemann Surfaces
Similarly,
2ni where
nil
sQ = fbj — ,E_ n.0 copQ,
is the UM-entry of the period matrix H.
111.3.7. We treat now the case where
= Tits
0=
(P, Q, R, S are all interior points of di). Here we cannot assert that 0 = df on ilft. To get around this little obstruction, we cut it by joining a point 0 on bdi to P by one curve and to Q by another curve. We obtain this way a simply connected region 4 ' (see Figure 111.2). In 4', 0 = df where
f(z) = 5:0 0.
)
Now a simple calculation yields
= 2ni[resR ftj + ress fê] = 27ri(f(R) — f(S)) = 2rri Et 0. The formula analogous to (30.2) is
=
LE,
0
L,
fb, 0 fa, 01 + .1:
where c is a curve from 0 to Q back to 0 ("on the other side") to P and back to O. Now the value of f on the + side of c differs from the value of f on the — side by 2rti (by the residue theorem). Thus
fei = 2ni[f: —
fP a
We summarize our result in .
f: Tpa =
TRs
b, Figure 111.2. Illustration for genus 2.
67
111.4. Divisors and the Riemann—Roch Theorem
(where as before each path of integration is restricted to lie in .It" = {lines joining R and S to O}). For the differential copQ, a similar formula can be derived; namely Re
s
= Re f coRs .
cop
1 11 .3.8. Let P E M ancrae a local coordinate z vanishing at P. We have seen that there exists on M {P} a holomorphic differential whose singularity at P is of the form x dz
=
n > 2.
— -A '
,
a
Assume that is normalized so that it has zero periods over the cycles a ,, a. Let
==E
1=0
40i)dz
at P.
(3.8.1)
Then, as before,
fb, 6 = n - 1 cz("j1 We will denote the differential a considered above by the symbol
(3.8.2)
(n)
Tp
and note it depends on the choice of local coordinate vanishing at P. (For our applications, this ambiguity will not be significant.)
111.3.9. A few other possibilities remain. We will not have any use for other bilinear relations. The reader who is looking for further amusement may derive more such relations.
111.4. Divisors and the Riemann—Roch Theorem We come now to one of the most important theorems on compact Riemann surfaces—the Riemann-Roch theorem, which allows us to compute the dimensions of certain vector spaces of meromorphic functions on a compact Riemann surface. The beauty and importance of this theorem will become apparent when we start deriving its many consequences in subsequent sections. As immediate corollaries, we obtain the fact that every surface of genus zero is conformally equivalent to the Riemann sphere and give a new (analytic) proof of the Riemann-Hurwitz relation. Although many definitions will make sense on arbitrary Riemann surfaces, most of the results will apply only to the compact case. We fix for the duration of this section a compact Riemann surface M of genus g > O. We let ye (M ) denote the field of meromorphic functions on M.
68
10 Compact Riemann Surfaces
111.4.1. A divisor on M is a formal symbol 21 = PP
(4.1.1)
- • IT,
with P E M, x i e Z. We can write the divisor 21 as
21 =
Pz(P) ,
(
4.1 .1a)
PEA!
with a(P) E Z, a(P) 0 for only finitely many P E M. We let Div (M) denote the group of divisors on M; it is the free commutative group (written multiplicatively) on the points in M. Thus, if 91 is given by (4.1.1a) and
=
pP(P), PEM
then and
9.1$ =
n pa,p)+1„p), p.m
21 -1 = n p—a(P) . PeM
The unit element of the group Div(M) will be denoted by 1. For 91 E DiN .1,1) given by (4.1.1a), we define
E
deg 91
x(P).
PE U
It is quite clear that deg establishes a homomorphism deg: Div(M) Z from the multiplicative group of divisors onto the additive group of integers. If f E .ir(M)\ {0}, then f determines a divisor (f )E Div(M) by
(4.1.2)
(f) = n PmdPf' PE M
It is clear that we have established a homomorphism
( ):X*(M)*
Div(M)
from the multiplicative group of the field X(M) into the subgroup of divisors of degree zero (see Proposition 1.1.6). A divisor in the image of ( ) is called principal. The group of divisors modulo principal divisors is known as the divisor class group. It is quite clear that the homomorphism, deg, factors through to the divisor class group. The (normal) subgroup of principal divisors introduces an equivalence relation on Div(M). Two divisors 91, 23 are called equivalent (21 $) provided that 21/23 is principal. If f E ir(110C, then
fi(cc)=
fl
PM
pmax{—ordp
f,0)
69
111.4. Divisors and the Riemann-.Roch Theorem
defines the divisor of poles of (or polar divisor of) f. Similarly,
f _, (0)=
pmaxtordp f .01 PcM
defines the divisor of zeros of (or zero divisor of) f. Both divisors have the same degree as the function f, and since
j:1(0) , f 1 (x) they are equivalent. More generally, for c E C, (f)=
f 1 (c),= ( f — c): 1 (0).
Thus the divisor class of f '(c) is independent of ce Cu ool. (It will be clear from the context when f `(c) stands for a divisor or just for the underlying point set.) One more remark about equivalent divisors: Let fi and f2 be meromorphic non-constant functions on M. Assume that fl = A o f2 for some Miibius transformation A. Then f t (x)= f ' (A '(c)). Thus f '(oc) If 0 0 co is a meromorphic q-differential, then we define the divisor of co in anology to (4.1.2) by
= fl
pordp
PEM
A divisor of a meromorphic q-differential is called a g-canonical divisor, or simply a canonical divisor if g = 1. We note that if co l and co2 are two non-(identically) zero meromorphic q-differentials, then co i/w2 e Y(M)* and hence the divisor class of (oh) is the same as the divisor class of (w2). We will call it the g-canonical class (canonical class, if g = 1). Since abelian differentials exist, the q-canonical class is just the qth power of the canonical class.
111.4.2. The divisor 91 of (4.1.1a) is integral (in symbols, 91 >_ 1) provided a(P) > 0 for all P. If, in addition, 91 1, then 91 is said to be strictly integral (in symbols, 91> 1). This notion introduces a partial ordering on divisors; thus 91 >_ 93 (or 21> 93) if and only if 9193' 1 (or 9193' > 1). A function oofeir(m) (resp., a meromorphic q-differential O co) is said to be a multiple of a divisor 91 provided (f)91'' > 1 (resp., (co)91 - > 1). In order not to make exceptions of the zero function and differential, we will introduce the convention that (0)91' > 1, for all divisors 21 E Div(M). Thus, f is a multiple of the divisor 91 of (4.1.1a) provided f = 0 or ordpf a(P), all P E M.
Hence such an f must be holomorphic at all points P e M with a(P) = 0; f must have a zero of order a(P) at all points P with (P)> 0; and f may have poles of order —a(P) at all points P with a(P) < O.
III Compact Riemann
Surfaces
111.4.3. For a divisor 11 on M, we set L(21) = {f E X(M); (f) It is obvious that LI91) is a vector space. Its dimension will be denoted by 491), and we will call it the dimension of the divisor 91. Proposition. Let 91, B E Div(M). Then
93 21
L(S) c L(21).
PROOF. Write
= 913 with 3 integral. If f E L(23), then (f)93 -1
1. But
(f )91 - 1 = ( f)93 -
3 1.
Thus, f E L(91).
111.4.4. Proposition. We have L(1) = C, and thus r(1) = 1. PROOF.
li f e L(1), then (f)
1; that is,
ordi, f > 0 all P E M. Since f has no poles it is constant by Proposition 1.1.6.
III.4.5. Proposition. If 91 e Div(M) with deg 91 > 0, then r(91) = 0. PROOF. If 0
f e L(91), then deg(f)
deg 91 > 0, contradicting Proposition
1.1.6. 111.4.6. For 91 E Div(M), we set Q(91) = {co; co is an abelian differential with (co)
91 } ,
and
491) = dim Q(91). We call i(91), the index of specialty of the divisor 91. Theorem. For 91E DiV(M), r(91) and 1(21)depend only on the divisor class of 91. Furthermore, if 0 to is any abelian differential, then i(21) = r(21(to) 1 ) •
PROOF. Let 91 1 be equivalent to 912 (that is, 211 912-1 is a principal divisor or 91,912-1 = (f) for some 0 f e AIM)). Then the mapping
L(912) a
hf e L(91 1 )
establishes a C-linear isomorphism, proving that
r(91 1 ) = / ( 912). Next, the mapping
12(91)
— co e L(9.1(o.))-1)
(4.6.1)
71
111.4. Divisors and the Riemann-Roch Theorem
also establishes a C-linear isomorphism proving that
491) = r(91(w) - 1 ).
(4.6.2)
Finally, if 91 1 is equivalent to 91 2 , then (4.6.1) and (4.6.2) yield (upon choosing a non-zero abelian differential co)
= r( 21 1(co) 1 ) = r(212(co) 1 ) = 012). 111.4.7. It is quite easy to verify the following Proposition. We have g(1) = ye (M ), the space of holomorphic abelian differentials (see Proposition 111.2.7) and !him 7(1) = g.
111.4.8. We will first prove our main result in a special case. Theorem (Riemann-Roch). Let M be a compact Riemann surface of genus g and 21 an integral divisor on M. Then
(4.8.1)
r(91 - = deg 91 — g + 1 + i(91).
PROOF. Formula (4.8.1) holds for 91 = 1 by Proposition 111.4.4 and Proposition 111.4.7. Thus, it suffices to assume that 91 is strictly integral.
91 = P",' • • P:r,
Pi e M, nj G
deg 91 =
E
> 0,
n1 > O.
.1=
Note that C L(91 -1 ). Furthermore, if f E L(91 - 1 ), then f is regular on MVP2 , ,P„,} and f has at worst a pole of order ni at P. Choosing a local coordinate zi vanishing at Pi, we see that for such an f, the Laurent series expansion at P is of the form
Consider the divisor 21 r = pne +1
pn.,,,+ 1 .
Let {a 1, 1, ,b,} be a canonical homology basis on M. We think of the elements of this basis as fixed curves in their homology classes and assume they avoid the divisor 91. Let 00( 21' - 1 ) be the space of abelian differentials of the second kind that are multiples of the divisor 9.1' -1 and have zero "a" periods. We can easily compute the dimension of 00(91' Recall the abelian differentials 411 introduced in 111.3.8. For P = Pi and 2 < n < ni + 1, 1(; ) E 00(91' 1 ) • Thus
-
dim 00(91— 1 )
E n; = deg 9.1. j= 1
111 Compact Riemann Surfaces
Conversely, if w e 00(21-1 ), then
w =E
dikz.,)dz i with d _ 1 =
in terms of the local coordinates 71. Thus we define a mapping s: r20(91 ,- 1 )
cdeg 41
by setting (CO
(d1,-2, • • •
1422— 2, • • • 42,
41, — n,
—
n2 — 11 • • •
,d„,,
-
2> • • •
,d,„, — n„,— 1)•
If co E Kernel S, then w is an abelian differential of the first kind with zero "a" periods, and hence co = 0 by Proposition 111.3.3. Hence ,§ is injective, dim 120(91— 1 ) = deg 21, and every c0 E Q0(91") can be written uniquely as
w=E
j= 1
—2
E
dikr(p.
(4.8.2)
k=nj1
We consider the differential operator d: L(21 -1 ) -+ 12 0(21"). Since Kernel d = C, it is necessary in order to compute r(21 - I ) to characterize the image of d. Now we dL(21 -1 ) if and only if a) has zero "b" periods. We conclude that dim Image d deg 91 — g (4.8.3) (since each "b" curve imposes precisely one linear condition). Using the "classical" linear algebra equation r(91 - 1 ) = dim Image d + 1,
(4.8.4)
we obtain from (4.8.3), the Riemann inequality, r(9.1 -1 )
deg 21 — g + 1.
To obtain the Riemann-Roch equality we must evaluate dim Image d. We use bilinear relation (3.8.2), to observe that the differential w of (4.8,2) has zero bi period if and only if —2
m
276.
E
E
j= 1 k= —ni—i k + 1 dika
2(P. = 0, )
(4.8.5)
where {C I, ... ,C,} is a basis for the holomorphic differentials of the first kind dual to our canonical homology basis and the power series expansion of CI in terms of z; is given (in analogy to (3.8.1)) by =
E s=0
ce1)(p.01)dz,
at Pr
73
111.4. Divisors and the Riemann—Roch Theorem
We recognize from (4.8.5) that the image of d is the kernel of a certain operator from Cdeg to C. Consider the operator deg 9I T:(2(1) defined by T(co) = (to a 1.09 • • • 9ei.n i - 1 ,6'2,02 • • • ,em,O, • • • ,em.n - 1), where co E Q(1) has Taylor series expansion -
E
dz.; at
k= 0
Represent the linear operator T with respect to the basis {Ç 1 , and the standard basis Cde" as the matrix , 42)(P1) ceon(P i) • %(A 13 1)
ag 1 (P2)
ar(P2)
Œg )(P.)
ceg)(P.)
,C9} of Q(1)
1(P.) cc,1-1(P.) ' • • 49,!, i(Pm )
Thus we recognize that Image d = Kernel 'T. We thus conclude that dim Image d = dim Kernel 'T = deg 91 — dim Image 'T = deg 91 — dim Image T = deg 91 — (dim Q(1) — dim Kernel T). Since Kernel T = Q(9I), and dim Q(1) = g, we have shown that dim Image d = deg 91 — g
+ i(91),
and (4.8.4) yields (4.8.1). 111.4.9. We collect some immediate consequences of our theorem.
Corollary 1. If the genus of M is zero, then M is conformally equivalent to the complex sphere C u PROOF. Consider the point divisor, P e M. Then r(P- 1 ) = 2.
Thus, there is a non-constant meromorphic function z in L(P -1 ). Such a function provides an isomorphism between M and C u tool by Proposition 1.1.6.
III Compact Riemann Surfaces
/.4
Corollary 2. The degree of the canonical class Z is 2g - 2. PROOF. If g = 0, then compute the degree of the divisor dz (which is regular except for a double pole at x)). Thus we may assume g > O. Since the space of holomorphic abelian differentials has positive dimension, we may choose one such; say C. Since (C) is integral
r((C)') = deg() - g + 1 + i((C)). By Theorem 111.4.6 we have r((C)') = i(1) and i((C)) = r(1). Thus we have g = deg() - g + 1 + 1, and hence the corollary follows. Corollary 3. The Rientann-Roch theorem holds for the divisor 91 provided that
a. 91 is equivalent to an integral divisor, or b. Z/21 is equivalent to an integral divisor for some canonical divisor Z. PROOF. Statement (a) follows from the trivial observation that all integers appearing in the Riemann-Roch theorem depend only on the divisor class of 21 (by Theorem 111.4.6). Thus, to verify (b), we need verify Riemann-Roch for 91 provided we know it for Z/91. Now
491) .--- r(91/Z) = deg Z/21 - g + 1 + i(Z/9.1) = deg Z - deg 91 g + 1 + r(2l = 2g - 2 - deg 91 - g + 1 + / ( 21 -1 ). Hence, we have proven the Riemann-Roch theorem for 91. 111.4.10. To conclude the proof of the Riemann-Roch theorem (for arbitrary divisors), it suffices to study divisors 21 such that neither 91 nor Z/21 is equivalent to an integral divisor, for all canonical divisors Z. Proposition. If 1 ( 21') > 0 for 21e Div(M), then 91 is equivalent to an integral divisor. PROOF. Let f E L(21 -1 ). Then (f)% is integral and equivalent to Corollary. If 421) > 0 for 91 E Div(M), then Z/21 is equivalent to an integral divisor. PROOF. Use 421) = r(21/Z). If 21 E Div(M), and neither 21 nor Z/21 is equivalent to an integral divisor, then i(21) =-- 0 = r(91 - 1 ). Thus, the Riemann-Roch theorem asserts in this case (to be proven, of course) deg If
g - 1.
(4.1 1.1)
Thus verification of (4.1 1.1) for divisors 91 as above will establish the following theorem.
75
1114. Divisors and the Riemann—Roch Theorem
Theorem. The Riemann-Roch theorem holds for every divisor on a compact surface M. PROOF.
We write the above divisor 21 as
21 =21 1/91 2 , with 91, (j = 1,2) integral and the pair relatively prime (no points in common). We now have deg 91 = deg 21, - deg 212, and by the Riemann inequality 1
) _)L- deg 91, - g
1 = deg 212 + deg 21 -g + 1.
Assume that deg 21 We then have deg 212 + 1.
r(91,- 1 )
Thus, we can find a function f E L(211— 1 ) that vanishes at each point in 912 . (Vanishing at points of 212 imposes deg 91 2 linear conditions on the vector space L( 21 1 ) • If r(21T ') is big enough, linear algebra provides the desired function.) Thus f E L(212/111) = L(21 - '), which contradicts the assumption that r(91 -1 ) = O. Hence deg 21
g - 1.
Since 0 = 421) = r(21/2), it follows that
deg(Z/21)
g - 1,
deg 21
g - 1,
or concluding the proof of the Riemann-Roch theorem.
El
111.4.12. As an immediate application of Corollary 2 to Theorem 111.4.8 we give another proof of the Riemann-Hurwitz formula (Theorem 1.2.7). Our first proof was topological. This one will be complex analytic. Let f be an analytic map of a compact Riemann surface M of genus g onto a surface N of genus y. Let n = deg f. Let co be a meromorphic g-differential on N. We lift co to a meromorphic g-differential Q on M as follows: Let z be a local coordinate on M and a local coordinate on N. Assume that in terms of these local coordinates we have
= f(z).
If w is
h(0 dCq in terms of C, then we set Q to be h(f(z))f
dzq
(4.12.1)
76
III Compact Riemann Surfaces
in terms of z. Note that if z is replaced by z 1 , with z = w(z,), then in terms of z 1 the map f is given by (f
0:1),
and thus we assign to z i lw(z Orwlzir dz
htf(w(z
which shows that Q is indeed a meromorphic g-differential. Without loss of generality, we may assume that for P E M, z vanishes at P, C vanishes at _ zh,-(p)+1 .
From this and (4.12.1) we see that ordp Q = (b1(P) + 1) ordftp) CO
qb f (P).
(4.12.2)
Formula (4.12.2) will be very useful in the sequel. For the present we merely use it for q = 1. We rewrite it as (for q = 1) ordp Q = (bf (P) + 1) ord
w + bf (P).
(4.12.3)
Let us choose an abelian differential co that is holomorphic and non-zero at the images of the branch points of f. (With the aid of Riemann-Roch (and allowing lots of poles) the reader should have no trouble producing such an co. If 7> 0, 0,) can be chosen to be holomorphic.) We wish to add (4.12.3) over all P e M. Observe that Corollary 2 of Theorem 111.4.8 gives
E
ordp (2 = 2g — 2,
PeU
and recall that by definition
E
b f (P) = B.
Pe M
We need only analyze
E PeM
(b 1(P) + 1) ordf(p)
ord f(p) CO PeM
E
n ordQ co = n(27 — 2).
QeN
This last equality is a consequence of the fact that each Q
E
M with
ord(2 w o 0 is the image of precisely n points on M.
111.5. Applications of the Riemann—Roch Theorem What can we say about the meromorphic functions on the compact Riemann surface M with poles only at one point? What is the lowest degree of such a function? In this section we shall show that on a compact Riemann surface of genus g 2, there are finitely many points P e M (called Weierstrass
111.5. Applications of the Riemann-Roch Theorem
77
points) such that there exists on M a meromorphic function f regular on M\IPI with deg f g. We shall see when we study hyperelliptic surfaces (in 111.7) and when we study automorphisms of compact surfaces (in Chapter V) that these Weierstrass points carry a lot of information about the Riemann surface. Throughout this section, M is a compact Riemann surface of genus g (usually positive), and Z E Div(M) will denote a canonical divisor.
111.5.1. We recall (to begin) that the Riemann-Roch theorem can be written for D E Div(M) as r(D - 1 ) = deg D - g + 1 + r(D/Z).
(5.1.1)
-
Furthentore,
r(D) = 0 provided deg D > 0,
(5.1.2)
i(D) = 0 provided deg D > 2g - 2.
(5.1.2a)
and If deg D = 0, then
r(D)
1,
(5.1.3)
and r(D) = 1 0
g=1 g> 1
all q q1 PROOF. Let D = Zq
(2q - 1)(g - 1)
in (5.1.1), then r(Z) = (2g - 1)(g - 1) + r(Z4- 1 ).
From (5.1.5), the dimension to be computed is r(Z - q).
(5.2.1)
78
III Compact Riemann Surfaces
Assume that g > 1. If q < O, then deg Z = — q(2g — 2) > 0, and thus r(Z) = 0 by (5.1.2). We already know that r(1) = 1, and that r(Z -1 )= g. For q > 1, r(Z")= 0 (by what was said before), and (5.2.1) gives a formula for r(Z Next for g = 1, (5.2.1) reads — r(Z - q)= r(Z4-- i
(5.2.2)
).
If() w is a holomorphic q-differential, it must also be free of zeros (deg(w) = 0). Thus w ' isa ( — g)-differential. Hence, (5.2.3) r(Z) = r(Z) We conclude from ( 5.2.2) and ( 5.2.3), that r(Z) = r(Zq we conclude that r(Z) = 1, by induction on q. Finally, for g = 0, deg Z = —2. Thus r(Z")= 0 for n
From r(1) = 1, —1, and (5.2.1)
yields the required results. EXERCISE Establish the above proposition for g = 0 and 1 without the use of the Riemann-Roch Thus f is a theorem. (For g = O. write any co e i(M) as w = f de with f e rational function. Describe the singularities of f.)
111.5.3. Theorem (The Weierstrass "gap" Theorem). Let M have positive genus g, and let P e M be arbitrary. There are precisely g integers 1 = ni < n 2 < • - •
O. Thus n i = 1, as asserted. The answer to Question "j" is yes if and only if r(D7 1 )— r(191_1,)= 1 (the answer is no if and only if r(Di 1 )— r(D»,)= 0). From the Riemann-Roch theorem (5.4.1) r(D1 1 )— r(D;1.1 1 )= 1 + i(DJ)— Thus for every k > 1: r(14-1 )— r(Di)-1 ).=
E
i=t
(r(D1 1)— r(D7- 1 1))
(i(D;)— i(D ; _ 1 ))= k + i(D,)— = r(D; — 1 = k + i(Dk)— g,
=k+ Or
and this number is the number of "non-gaps" 2g — 2 (thus deg Dk > 2g — 2 and i(Dk) = 0) k — (number of "gaps" s k)= k — g. Thus there are precisely g "gaps" and all of them are _s 2g — 1. 111.5.5. We now begin a more careful study of the Weierstrass "gaps". Let P E M be arbitrary, and let 1 < < a2 < • • • < = 2g be the first g "non-gaps".
80
III Compact Riemann Surfaces
Proposition. For each integer j, 0 <j < g, we have a; +
;
2g.
PROOF. Suppose that xi + < 2g. Thus for each k j we would also have ak + < 2g. Since the sum of "non-gaps" is a "non-gap", we would have at least j "non-gaps" strictly between ag _ i and ag. Thus at least (g — j) + j + 1 = g + 1 "non-gaps" 52g, contradicting the fact that there are only g such "non-gaps".
111.5.6. Proposition. If a, = 2, then ; = 2j and ; + a9 _ 1 = 2g for 0 <j < g. PROOF. If a l = 2, then 2, 4, . , 2g are g "non-gaps" 5 2g, and hence these are all the "non-gaps" 5 2g.
111.5.7. Proposition. If a i > 2, then for some j with 0 <j < g, we have a; + Œg _; > 2g. PROOF. We assume that ai + = 2g for all j with 0 < j < g. For q E let [q] be the greatest integer 5. q. Then x i , 2;, . . . , [2g/;]; are "non-gaps" 2, then the above accounts for at most ig < g "non-gaps", and there must be another one 5.2g. Let a be the first "non-gap" not appearing in our previous enumeration. For some integer r, 1 r [2g/a i ] < g, we must have ra, < a < (r + 1)a i . Thus we have "non-gaps"
11, 2 2 = 2a1, and by our assumption
1 = cc, = 2g — ra i ,
_ 0. 4 = 2g — a
(note that if r + 1 = g, then a = 2g and the last equality reads ao = 0—which can be added consistently to our data). These are all the "non-gaps" which _ " and 52g. It follows that are a l + ag _ (, + = a, + 2g — a = 2g — (a — a i) > 2g — ra i = It therefore follows that there is a "non-gap" < 2g, greater than ag _ , and not in the list a9-19 • • • 9 a9-(9.+1)• This is an obvious contradiction. Corollary. We have
g—1
E ce; I
g (g — 1 ),
with equality if and only if a, = 2. PROOF. From Proposition 111.5.5, 2 ai _>.: 2g(g — 1). Furthermore if a l = 2, then we have equality in the above by Proposition 111.5.6. If x i > 2, we must have strict inequality by Proposition 111.5.7.
81
111.5. Applications of the Riemann-Roch Theorem
111.5.8. We have seen in 111.5.4, that j
1 is a "gap" at P e M if and only if r(P- — r(P - -1+1) = 0,
if and only if
i(P11 ) —
= 1;
that is, if and only if there exists on M an abelian differential of the first kind with a zero of order j — 1 at P. Thus the possible orders of abelian differential of the first kind at P are precisely
0 = n i — 1 < n, — 1 < • • • < rt, — 1 2g — 2, where tEe nis are the "gâps" at P-(appearing in the list (5.3.1)). The above situation is a special case of a general phenomenon, which we shall proceed to study. Before proceeding to study the general situation, let us observe that we have established the following basic Fact. Given a point P on a compact Riemann surface M of genus g> 0, then there exists an abelian differential Co of the first kind (ro E l (M)) that does not vanish at P (that is, ordp co = 0). Let A be a finite-dimensional space of holomorphic functions on a domain D c C. Assume that dim A = n > 1. Let z E D. By a basis of A adapted to z, we mean a basis {(p i, ,cp.} with
ordz tP i < ordz 2 p i } is an (n — 1)-dimensional subspace of A, and we can set
P2 = min l ord (pl. we Ai
By induction, we can now construct the basis satisfying (5.8.1). The basis adapted to z is (of course) not unique. We can make it unique as follows: Let p; = ordz cp; . Consider the Taylor series expansion of (pi at z (in terms of C)
= k=0 E akX
z)k.
We may and hereafter do require that
1, k = j,
a- 10 k
where j, k = 1, . , n. Remark. On a Riemann surface "the unique" basis adapted to a point will, of course, depend on the choice of local coordinate.
III Compact Riemann Surfaces
82
It is obvious that yi j — 1. We define the weight of z with respect to A by
TG7) =
(5.8.2)
Oki — j +
Proposition. Let {( p 1 , ,cp„} be any basis for A. Consider the holomorphic function (the Wronskian) [
(1)(z)= det
(Pi(z)
Vi(z)
•• ' •• •
q(z) (P,(z)
(5.8.3)
Then
ordz P = r(:). PROOF. It is easy to see that a change of basis will lead to a non-zero constant multiple of O. Hence we may assume, whenever necessary, that the basis used is adapted to the point z. Let us abbreviate equation (5.8.3) by 0(z) = det[(p i(z), In order to prove the proposition we derive some easy properties of the function 0. First: for every holomorphic function f,
det[fip,,
, f(p„] = f" det[T„
(5.8.4)
,cp„].
This follows from well-known properties of determinants. Explicitly,
det[fip,,
,f(p„] 101 +f'41 = det fqi; + 2f + f"cp1 JOT — 11 +
...
f In 1)(p
[(Pi f(p't- + f'(p 1 = f det f(Pr ." + • ' • + f (" -1)(Pi
.fiP. f(P;. + f*, + f"('. Jig + f9 (nn— 1)
[.fi7. 1 fil; T - " + • • • + f ("- "(pi
+ fog
—
1),„
(p. f(p,,+ f 4.
Mr-1) + • • • + f(n-1) (Pn_ (19 n
= f det
±
—
.10'n
f(p.
ri ) + - - • + f (" - i )(p„__
where the last equality arises from multiplying the first row of the determinant by —f' and adding the result to the second row. In a similar fashion we can
111.5. Applications of the Rieinann—Roch Theorem
83
remove from each column the appropriate multiple of (pi leaving us with the previous expression equal to •" f(p'.
det ftp7 +2f'tp'i
••'
.fiPT -11 + ' • +
— ilf (" -2),P'i
fig + 21"0.
• ' • f(P(- 1) -4- • ' • +
= ff det f(PV + 2f'(p'1 1.-
• ••
(o.
- • •
(p'.
'•
fip (r -i) + • • + ( n — 1) P-2)(p', ••
—
t " --2) (1):.
+ 2f(p'.
f9„ -i) + • • + (n
—
1)f" 21
We now repeat the same procedure to remove from each column the appropriate multiple of cp'j . This clearly terminates with the preceding equal to det[cp„
We now turn our attention to the proof of the proposition which shall be by induction on n. Clearly the result is true for n = 1. Let us now assume that the proposition is true for n = k. Explicity we are thus assuming that ordz det[T i ,
,(pk]
=
E (pi -i+
where pj = ord., pj. Consider now det[(Pi, preceding remarks that
,(Pk+1].
1),
It is clear from the
det[ 1 ,02/T1, • • •
det[Ti, • • • ,49k4-1] =
Now the right-hand side is simply cp1+ 1 det[(T 2 /(p 1 )', induction hypothesis now gives that
,frpk+ i ftp i r]. The
ordz {cpki + 1 det[(T2/TIY .•.,(4'k+ l(pi )'i} ,
k+1
= (k + 1)/./ 1 +
E
— 1) —
2)/
j=2 k+1
= 11 1 +
E
1=2
j
+
provided that for each j, p. — (j — 1) — p, O. Since the {(pi} are a basis adapted to z, this inequality is always satisfied, and we have k+1
ordz det[q i , • • • ,g9x+ 1] =
E
i=1
This concludes the proof of the proposition.
84
111 Compact Riemann
Surfaces
Remark. Let {cp„ ,T,,} be any set of n holomorphic functions of D. We ,(pJ. Our argument here shows that 410 is idencan define 0 = detkp „ ,cp„ are linearly dependent. tically zero if and only if the functions (p i , Corollary 1. Let A be a finite-dimensional space of holomorphic functions on a domain D c C. The set of z e D with positive weight with respect to A is discrete. Corollary 2. Under the hypothesis of Corollary 1, for an open dense set in D, the basis {(p i, ,q)„} of A adapted to z has the property ordz (p; = j — 1.
PROOF. By the hypothesis ord. cp) = pi. It follows from Corollary 1 that (jm; — j + 1) = o for an open dense set. Since (as we have preT(z)= viously remarked) pi j — 1 we have p = j — 1 for each j, on this open dense set. 111.5.9. The considerations of the last paragraph apply (of course) to the space YE(AI) of holomorphic q-differentials (q 1) on a compact Riemann surface of genus g 1. A point P E M will be called a q-Weierstrass point provided its weight with respect to yeq(m) is positive. A 1-Weierstrass point is called simply a Weierstrass point or a classical Weierstrass point. It is clear from the ideas in the previous paragraph that we have the following
Proposition. A point P on a Riemann surface M of genus g 2 is a q-Weierstrass point if and only if there exists a (not identically zero) holomorphic qdifferential on M with a zero of order dim .V(M) at P. For q = 1, this condition is equivalent to either (and both) a. i(P9)> 0, or b. r(P 9) 2 (that is, at least one of the integers 2, ... , g is not a "gap"). 111.5.10. There are clearly no q-Weierstrass points (for any q 1) on a surface of genus 1. We assume thus that g > 2.
Proposition. For g 2, q 1, let -OP) be the weight of P E M with respect to ifq(M). Let IV, be the Wronskian of a basis for ifg(M). Set d = d, = dim Yeq(M). Then W„ is a (non-trival) holomorphic m = medifferential where m = (d/2)(2q — 1 + d). Hence
E
t(p)= (g — 1)d(2q — 1 + d).
PeA1
PROOF. We must merely verify that the determinant 0 defined by (5.8.3) transforms as an m-differential under changes of coordinates. Explicitly, let {C„ ,C,,} be a basis for .10(M). Let z and 2' be local coordinates with
85
111.5. Applications of the Riemann—Roch Theorem 7 = f(z) on the overlap of their respective domains. Assume that
= cpi(z)dzq = (that is, =
in terms of the local coordinates z and F. We must show that (det
2, • • • , (Pdi)de' = (det Pi, • • • :0,11)dr.
(5.10.1)
But it is easy to verify that det [4p„
,T,4 ] = det [41 °
• ACP-a ° f)(f ')g]
= (fr(det[01, • • • ,e7,d]
f),
which is equivalent to (5.10.1). Corollary. For g PROOF.
2 there
are q- Weierstrass points for every q
1.
Any mg-differential always has zeros.
111.5.11. We now finish the study of classical Weierstrass points (the case q = 1).
Theorem.
For g > 2, the weight of a point with respect to the holomorphic abelian differentials is g(g — 1)/2. This bound is attained only for a point P
where the "non-gap" sequence begins with 2.
PROOF. We have seen that (Proposition 111.5.10) (5.1 1.1)
r(P) = (g — 1)g(g + 1). PeU
The above, of course, gives a trivial estimate on -OP). We need a better one. Let 2 < cc i < Œ2 < • • • 2, we will have to rely on different methods (involving more analysis and topology). These methods, which will be applicable to all surfaces, will be treated in IV.4 and IV.5. Throughout this section M represents a compact Riemann surface of genus g > O.
111.6.1. We start with {a,
,b9) = {a,b},
a canonical homology basis on M, and ir
= {41,
the dual basis for Ie l (M); that is,
S
Qk
Ci
jk ,
j, k = 1, 2, .
, g.
We have seen in 111.2, that the matrix H with entries
= fbk Ci,
j, k = 1, .
, g,
87
111.6. Abel's Theorem and the Jacobi inversion Problem
is symmetric with positive definite imaginary part. Let us denote by L = L(M) the lattice (over Z) generated by the 2g-columns of the g x 2g matrix (1,11). Denote these columns (they are clearly linearly independent over Ft) by e4 ", , eg), , 7t(g). A point of L can be written uniquely as
E new+ E
nirtu), with m, n e Z,
i=
Or
Im + [In with m = '(m 1 ,
,m9) E Zg and n =
,ndE lg.
We shall call J(M) = «Y/L(M) the Jacobian variety of M. It is a compact, commutative, g-dimensional complex Lie group. We define a map (f):M --• J(M) by choosing a point Po E M and setting (P(P)= fpPoic =t(ipPoci,
fpPoc.)
Proposition. The map y) is a well defined holomorphic mapping of M into J(M). It has maximal rank.
PROOF. Let c and c2 be two paths joining P o to P, then c,c2- is homologous to (a,b)[] for some m, n e z9. Thus
sc, •c — L = 1m + lin e L(M). If z is a local coordinate vanishing at P and q', of cp (in Cg), then writing Ci = ?viz, we have
, (I), are the components
i (z) dz,
(pi(z)= ipPo C.; and we see that &pi =
az
i(z).
Thus q) would not have maximal rank if there were a point at which all the abelian differentials of the first kind vanished. Since this does not occur (recall 111.5.8), the rank of cp is constant and equals one.
111.6.2. Let, for every integer n > 1, Mn denote the set of integral divisors of degree n. We extend the map cp: cp:M„ --■ J(M) by setting for D=131 •••P „,
(p(D)=
E
T(P).
88
Ill Compact Riemann Surfaces
Note that (since
q(D) = cp(P 0D)) 90ln+ t) 49 (Mn) D • • D
i)
We can also obtain a map that does not depend on the base point Po. Let Div(°)(M) denote the divisors of degree zero of M; define 9:Div(M) J(M)
by setting
E
49 (D) =
E
(p(P)—
1=1
q(Q)
i=1
for Pi • • Pr/Qi • • Q..
D
It is clear that if r = s,
cp(D)is
independent of the base point; that is, the map
cp:Div1°) (M)
J(M)
is independent of the base point.
111.63. Theorem (Abel). Let D e Div(M). A necessary and sufficient condition for D to be the divisor of a meromorphic function is that p(D) = 0 mod (L(M))
and deg D = 0.
(6.3.1)
PROOF. Assume that f is a meromorphic function. Let D = (f ). We have seen (Proposition 1.1.6) that deg D = 0. Since for D = 1, Abel's theorem trivially holds, we assume that f C. Write D = Pi' • • • FT/VI' • • • Or,
k > 1, r >
1,
(6.3.2)
with Pi # all j, I; PPi and Qi 0 Q1 ,
j 0
•
J=1
œi
=
E E 13P. 1 ).
1;
(6.3.3)
J=1
Without loss of generality, we may assume that none of the points Pi, Q. lie on the curves representing the canonical homology basis. Recall the normalized abelian differentials tiv of the third kind introduced in 111.3. Observe that df/f is an abelian differential of the third kind with simple poles and df resp — =
ordp f, for all P E m.
Thus df
f
CE
E
.1=
Aitzpo)
(where P o is not any of the P. or Q. nor on the curves ai, bi) is an abelian differential of the first kind. Hence, we can choose constants ci, j = 1, . . . , g,
89
111.6. Abel's Theorem and the Jacobi Inversion Problem
such that df
E
L
J
j= 1
E q.
QiP0
j=
1= 1
It follows that df
.=
7 Jaj
and by the bilinear relations for the normalized differentials rpQ ,
df
Pi
= 2nii( E
E fli fQ") p.
J=1
E
j=1
Since df/f = d logf, we see that df
df 27rin 1 , f =
. = brim! ,
where mi, ni are integers. It follows that the /th component of co(D) is o c o, .•
CP.; j= 1
2i JPo
JP0
j= 1
=
1
df
1
‘I
Jbl
7
j= 1
Cinil
= ni J= 1
and thus q(D) = 0 mod(L(M)). We have used Po as the base point for cp. Recall that since D is of degree 0, (MD) is independent of Po . To prove the converse, we let D be given by (6.3.2) subject to (6.3.3). Choose a point Q0 not equal to Po , nor any of the Pi nor any of the Q.; nor lying on any of the curves aj, bp Set k
f(P) = exp( E ai f
r
P tp p o
Qo '
.i= 1
—
g
P
rP r \
E fl•f TQ ± E c;k 0 6j) 1 Qo 1 p°
j= 1
= exp fP T, Qo
are to be determined. It is clear that f is a where the constants meromorphic function with (f) = D, provided f is single-valued. We compute
f p,po E fif j= 1
fa; = i=E1
=
TQjpo
and
5,1 r = E ai L lc
= 2ni
E 1=1
-rp,p. Pi
— E si L, •r
9 tQfp.
+ E C,ICi7
Q .;
ai fp. i", : — 2ni E 13; fpc, CI + 1=1
g
E cinfi. 1=1
III
90
Compact Riemann Surfaces
For f to be single-valued, we must have that J utf ,- ,b, - are of the form 2nin, n E Z. Now (6.3.1) yields,
sp
20
p:
t=
p(
,
2 = nt +
E
1=1
with n1, m; E Z, I = 1, . . . , g, j = 1, . . . , g. This means we can choose paths yi and 7; joining Po to P; and Q; respectively, such that integrating over these paths, leads to the above equation. It is clear that if we choose c; = —2nim i, f will be single-valued. 0
EXERCISE We outline an alternate proof of the necessity part of Abel's theorem. Let f be a nonconstant meromorphic function on M. For each aeCuf on), let f '(2) be viewed as the integral divisor of degree deg f, consisting of the preimage of a. Then a cp(f ' (a)) is a holomorphic mapping of C In) into J(M). Since C uIx) is simply connected, it lifts to a holomorphic mapping of C u { co} into C°. Since C u {co } is compact this mapping is constant and thus (p( f '(0))=
111.6.4. For g = 1, the following
J(114)
is of course a compact Riemann surface. We have
Corollary 1. If M is of genus 1, then J(M)
9:M
is an isomorphism (conformal homeomorphism). PROOF. Clearly cp is surjective (since it is not constant). Let P, Q E M, P 0 Q. If (p(P) = 9(Q), then P/Q is principal by Abel's theorem. Thus there is a meromorphic function on M with a single simple pole. This contradiction shows that cp is injective. 0 Corollary 2. If M has genus >1, then tp is an injective holomorphic mapping of M onto a proper sub-manifold (p(M) of J(M). 111.6.5. Let D be an integral divisor of degree g on M; that is, D = P1 • Pg
(P E M, j = 1, . . . ,g).
Then by the Riemann-Roch theorem r(D - 1 ) = 1 + i(D) 1. We call the divisor D special provided r(D -1 )> 1.
(6.5.1)
91
111.6. Abel's Theorem and the Jacobi Inversion Problem
Theorem. Let D E Div(M) with D > 1 and deg D = g. There is an integral divisor D' of degree g close to D such that D' is not special. Further D' may be chosen to consist of g distinct points. Remark. Write D as in (6.5.1), and choose a neighborhood L Ji of P. The condition that D' be close to D is that D' = • • Pg with P, E U. PROOF OF THEOREM. Note that for divisors of degree g, i(D)= O.
r(D -1 )= 1
Assume that D is given by (6.5.1) and define V.-
(j= 1,
D. = P,• • •-Pi
,g).
Thus Dg = D. Set D o = 1. We prove by induction that for j 1, we can find a divisor D ;. of the form 1 13; (D', = 1) with P'.; arbitrarily close to P and i(D'i)= g — j, j = 1, . , g. The Riemann-Roch theorem (or linear algebra) implies that i(D) g — j (for any integral divisor D'; of degree j). Note that for arbitrary P, (as we have seen before),
ff/3 1 ) = r(P-1- 1 ) — 1 + g — 1 = r(Pi-1 ) + g — 2 = g — 1 (since C = L(PT `), because a surface of positive genus does not admit a meromorphic function with a single pole). Thus, it suffices to take = Assume that for some 1 j < g, we have constructed a D of the required ,(pg _ i} be a basis for the abelian differentials of the first type. Let Ig) i , kind vanishing at Pi, , P. Look at cp,. If cp,(Pi+ ,) 0 0, then i(D'iPi .„ ,) g — (j + 1), and we are done. If tp 1 (13». 1 ) = 0, then arbitrarily close to PJ.,. 1 there is a point P +1 with cp,(Pi+ ,) 0 and once again i(D'irp-t) g — (j + 1). El
111.6.6. We consider now the map cp:Mg J(M). We will show that this map is always surjective (generalizing the first corollary in 111.6.4). To this end let D, = P, • • • Pg with Pi E M, Pi Pk for j k be such that i(D o) = O. Let U.; be a coordinate disk around Pi with local coordinate ti vanishing at P1. Let K = cp(P, • • Pi). In terms of the local coordinates ti we write Ck = riki(t)dti and thus in terms of the local coordinates ti at Pi the mapping cp in a neighborhood of the origin (0, ,0) is simply K+ ,z g), , cpg(z 1 ,... ,z g)), (z 1 , where ,z g) =
E
1 zj
li.(t.)dt..
J=
Thus Ozk
(0, • • • ,0)= nik(0) =
92
111 Compact Riemann Surfaces
(the last equality is merely a convenient abbreviation). Thus the Jacobian of the map 9 at P, • • • Pg is [ . '
WI)
— Ci( 1 g) . • • • • • C 9 (P)
The condition ((P i , ,Pg) = 0, however, gives us immediately that the rank of the Jacobian is g. Therefore, the map 9: U, x • • • x Ug
Cg
,0) by the inverse function is a homeomorphism in a neighborhood of (0, theorem; and thus covers a neighborhood U of K (in Cg or in J(M)). Let ,cy) a Cg. Then for a sufficiently large integer N, K + e/N e U, and thus there exists a Q 1 , such that 9(Q 1 • • • Q.) = K + c/N Or
N(9(Q1 ' • • Qd — K) = c.
Thus to show that c E Image cp, it suffices to show that there exists an integral divisor D of degree g such that (P(D)= N(4)(Q1,
Consider the divisor (P i • • PON/W I rem gives ( (P1
• •
,Q9) — K).
Mil. The Riemann-Roch theo-
)= 1 + i(( Q' • •• • 1211)N1:1 )> 1. (P1 • • P VQ1' • • Q1Pgol Hence, there is an f E L((P, • • P1/(Q,• • W P) ); that is, there is an r
• • Pg)
integral divisor D of degree g such that (f) =
D(P, • • • I'd' (Q1 . • 42,)."11 .
By Abel's theorem 9(D) + N91P1 ' • • Pd= N(49 (Qi • • Qd.
We have solved the Jacobi inversion problem: Theorem (Jacobi Inversion). Every point in J(M) is the image of an integral divisor of degree g. Corollary. As a group J(M) is isomorphic to the group of divisors of degree zero modulo its subgroup of principal divisors. Remark. We have shown (Corollary 1 of 111.6.4) that every surface of genus one can be realized as C modulo a lattice. This is the uniformization theorem
93
111.7. Flyperelliptic Riemann Surfaces
for tori. We shall rcturn to this topic (including uniformization theorems for surfaces of genus > 1) in the next chapter. 111.6.7. Exercise We have seen that every torus M is conformally equivalent to C/G where G is the group generated by two elements zi— ■ z + 1 and z + T, Tm T > O. Further, the origin of C/G may be made to correspond to any point in M. (a) Every meromorphic function f on M can be viewed as a doubly periodic function f (this identification should cause no confusion) on C; that is, a meromorphic function J on C with le`
f(z + 1) = f(z) f(: + s), all e C. The IVeierstrass p-function is defined by p(z) =
1
1 1 2 ), - e C. — a — ms) 2 (n + ms)
E
()*(O O) (n.no)e Z.2
Note that p is an even function. Let P, Q e M. Let f e L(1/PaC. Show that there exists x e Aut M (the group of conformal automorphisms 01M), 13 e Aut(C u {col), such that f=fiodoom (For a discussion of Aut M, see V.4.) (b) Show that every meromorphic differential on M is of the form f(z)dz where f is a doubly periodic function. If we choose the loop corresponding to z z + 1 as the "a-curve" and z 1--. z + r as the "b-curve", then we have a canonical homology basis on M. The basis for .e(M) dual to this canonical homology basis is then fdzI. (c) From what we said above, p(z)dz is an abelian differential of the third kind with zero residue and singularity 1/z 2 at the origin. Hence there exists a meromorphic function on C such that C' = — p. This function cannot be doubly periodic (why?). However, satisfies for all z E C
;(z + 1) = ;(z) + 1 , ;(z + r) ;(z) + r12 , where
and g 2 satisfy Legendre's equation th r — 77 2 = 2ra.
(6.7.1)
Derive (6.7.1) from (3.8.2).
111.7. Hyperelliptic Riemann Surfaces In this section we study hyperelliptic Riemann surfaces—the simplest surfaces. These are the two-sheeted (branched) coverings of the sphere. We shall see that there exist hyperelliptic surfaces of each genus g, and that these
94
III Compact Riemann Surfaces
surfaces are the ones for which the number of Weierstrass points is precisely 2g + 2. These surfaces thus show that the lower bound obtained in the
Corollary to Theorem 111.5.11 is sharp.
111.7.1. A compact Riemann surface M is called hyperelliptic provided there exists an integral divisor D on M with deg D = 2,
r(D 1 )
2.
Equivalently, M is hyperelliptic if and only if M admits a non-constant meromorphic function with precisely 2 poles. If M has such a function, then each ramification point has branch number 1, and hence the genus g of M and the number B of branch points (= ramification points) off are related by (using Riemann -Hurwitz) B = 2g + 2. Remarks 1. We can hence describe a hyperelliptic surface of genus g as a two-sheeted covering of the sphere branched at 2g + 2 points. 2. Some authors restrict the term hyperelliptic to surfaces of genus 2 that satisfy the above condition.
111.7.2. Proposition. Every surface of genus PROOF. Let D
2 is hyperelliptic.
be an integral divisor of degree 2. Riemann-Roch yields r(D - 1 ) = 2 — g + 1 + i(D).
Thus r(D - 1 )
(7.2.1)
2 for g 1, and the only issue is g = 2.
First proof for g = 2: Let P be a Weierstrass point on a surface of genus 2. Then there is a non-constant f e L(P - 2)• (This function cannot have degree 1.) Second proof for g = 2: Choose co 0 0, an abelian differential of the first kind on M. Then, since deg(w) = 2, (0 ) = PQ.
Since i(PQ) = 1, (7.2.1) yields r(P-1 Q -1 )= 2. Remark. Surfaces of genus 1 are also called elliptic (tori). Surfaces of genus zero admit, of course, functions of degree 1. Thus, hyperelliptic surfaces are those which admit functions of lowest possible degree.
111.7.3. Let M be a hyperelliptic Riemann surface of genus > 2. Choose a function : of degree 2 on M. Let f be another such function. We claim that f is a Miibius transformation of z. Let the polar divisor of z be P ,Q, and the polar divisor of f be P2 Q2. It suffices to show that P1 Q 1 P2Q 2 . For then there is an h e YOM), such that multiplication by h establishes an isomorphism between L(PT 1 Q t- ') and L(P 2-1 Q2-1 ). Since {1,4 and {1,f} are bases for
95
Hyperelliptic Riemann Surfaces
these spaces, there are constants
13, 7, such that 1 Œh + 13/iz f = 711 + Shz 2,
or
y + Sz
f = a + 13z .
To establish the above equivalence, we observe first that the branch points of f are precisely the Weierstrass points of M. To see this let P a M be a branch point of f. Then f is locally two-to-one at P. Thus if f(P) = co, f has a pole of order 2 at P (avl no other poles) .and - then P is a Weierstrass point. If f(P) co, then f — f(P) has a pole of order 2 at P, proving that P is a Weierstrass point on M. It now follows by Proposition 111,5.6 that the "gap" sequence at any of the 2g + 2 branch points of f is
1, 3, ... , and thus the weight of any of these points is
(2k — 1) —
k = g2 2 2 k= Thus these 2g + 2 points contribute g(g2 — 11 to the sum of the weights of the Weierstrass points. Since the sum of the weights of all the Weierstrass points is precisely g(g2 — 1) there are no other Weierstrass points. Let us choose any Weierstrass point P on M. We claim that the polar divisor of f is equivalent to P2. If f(P) = co, there is nothing to prove. Otherwise, look at the function 1/(f — f(P) = F. Its polar divisor is P2 which is equivalent to the polar divisor of f since F is a Mdbius transformation of f ( f l(co) f (N. We have therefore established most of the following Theorem. Let M be a hyperelliptic Riemann surface of genus g 2. Then the function z of degree 2 on M is unique up to fractional linear transformations. Furthermore, the branch points of z are precisely the Weierstrass points of M. The hyperelliptic surfaces of genus g 2 are the only ones with precisely 2g + 2 Weierstrass points. Only the last statement needs verification. It is clear that if a surface has precisely 2g + 2 Weierstrass points, then the weight of each such point must be --g(g — 1) by 111.5.10 and 111.5.11, and thus, as we saw there, the "non-gap" sequence must begin with 2. PROOF.
Remark. We show next that on a hyperelliptic surface each of the Weierstrass points is also a g-Weierstrass point for every q > 1. It follows from the fact
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III Compact Riemann Surfaces
that the "gap" sequence is 1, 3, ... , 2g — 1 for each Weierstrass point P, that there is an abelian differential cp of the first kind with divisor P29- 2. Thus cpq is a holomorphic g differential with divisor Pq(2g - 2) . Since g(2g — 2) > (2g — 1)(g — 1) —1, P must also be a g-Weierstrass point.
111.7.4. We now wish to construct another function on a hyperelliptic surface of genus g 1. Let z be a function of degree 2. Let P 1 , , P2 9 +2 be the branch points of z. Without loss of generality we assume that z(P;)
j = 1,
co,
, 2g + 2.
Consider "the function"
n (z - z(p)).
20-1=
(7.4.1)
Remark. We have introduced above a multivalued function which we will show to be single-valued. Multivalued functions are treated in 111.9. In IVA 1, we will show how the function w can be obtained without the use of multivalued functions. Proposition. The above defines w as a meromorphic function on M whose divisor is Pt • ' P2,4-2 (7.4.2) Q1 +10+ 1 where Q 1 Q 2 is the polar divisor of z. PROOF. Since all the branch points of z have ramification number 2, w locally defines a meromorphic function on M which is two-valued. We must show we can choose a single valued branch. It is convenient at this point to introduce a "concrete" representation of the surface M as a two-sheeted covering of the sphere Cu{co}. If P e M, z(P) se and P is not a branch point of z, then z — z(P) is a local coordinate vanishing at P. If z(P) = oo (recall we have assumed that P is not a branch point), then 1/z is a local coordinate vanishing at P. If P is a branch point, then either branch of — z(P) is a local coordinate vanishing at P. With slight and obvious modification, the above procedure could have been carried out with any meromorphic function on any (compact or not) surface (recall Remark 3 in 1.1.6). We define now e • = z(Pi). Then these ei are distinct; and z 1 (a) consists of precisely two points on M for all a E Cu fooNe l , • • • ,e2g+21, whereas z' '(es) consists only of the point P. We picture now two copies of the sphere. We label these two copies sheet I and sheet II. On each sheet for each k = 1, . . . , g + 1, we draw a smooth curve called a "cut" joining e2&_1 to e2k• We may assume that the els have been ordered so that these cuts do not intersect. Each "cut" is considered to have two banks; an N-bank and an S-bank. We construct a Riemann surface kr by joining every S-Bank on sheet I to an N-bank of the corresponding "cut" on sheet II, and then
97
111.7. Hyperelliptic Riemann Surfaces
_ Figure 111.3. Cross "cuts" for a hyperelliptic surface of genus 2.
joining the corresponding S-bank on sheet II to the N-bank of the corresponding "cut" on sheet I. It is quite clear that It'd'. is a compact Riemann surfacéand z is a meronorphic function briM. . Thus .f).4 is indeed a concrete model for M. We remark that a simple closed curve around a point ei, that does not go around any ek with k j, may be pictured as beginning in one sheet say at R„ continuing around until the point, R2, on the second sheet with z(R i ) = :(R2) and returning back to R t . We now construct a canonical homology basis for M using this two sheeted representation. Draw simple smooth closed curves ak , k = 1, . . . , g, winding once around the "cut" from e2k_ l to e 2k in one sheet of M oriented as indicated in Figure 111.3. This curve ak exists because we cross between sheets only through the "cuts". Next choose curves bk , k = 1, . , g, starting from a point on the "cut" from e2k _ 1 to e2k going on the first sheet to a point on "cut" from e2g , , to e20. 2 and returning on the second sheet (indicated in Figure 111.3 by dotted lines) to the original point. The orientation of the b-curves is again illustrated in Figure 111.3. Let us stop to analyze what is happening on the surface itself. The reader should convince himself (or herself) that the picture on the surface (Figure 111.4) is actually the lift of the picture in the extended plane via the two-sheeted covering z. (Note that all we are using is that a curve in the plane passing through ei can be lifted in two ways as a curve in M passing through Pi.) We have actually constructed a canonical homology basis, since by inspection the intersection matrix is of the form
ab a b
0 —1
0
Figure 111.4. The hyperelliptic model in genus 2.
tit
98
Compact Riemann Surfaces
We return—after this lengthy digression—to investigate the behavior of our function w. We need only look at what happens in the plane if we continue analytically a branch of J The analytic continuation of this function element around a closed path changes sign if and only if the path has odd winding number about ei (if it has even winding number, we return to the original function element). Thus w changes sign if we continue it along a simple closed path in Cu{co} that encloses an odd number of the ep If a curve in C {ao}\{e 1 ....,e29 .,. 2 } begins at z and returns to this point, enclosing an odd number of ei, then this curve must cross one of the "cuts" and thus its lift to M is a curve joining the points P and Q, P Q, on M with z(P) = z(Q). Thus w can be continued analytically along all paths in M. Furthermore, continuation along any closed path in M (which must encircle an even number of els when viewed in the plane) leads back to the original value of w. Thus w is single-valued on M, and for any P, Q on M with P Q, we have z(P) w(P) = — w(Q). The fact that (w) is given by (7.4.2) is an immediate consequence of (7.4.1). .
111.7.5. The above proposition has some immediate consequences. Corollary 1. The g differentials zi dz
j=0,...,g--1,
(7.5.1)
form a basis for the abelian differentials of the first kind on M. PROOF. Without loss of generality z(Pi) 0 0, and Q3 124
(z) = QiQz . It is clear of course that the differentials in (7.5.1) are linearly independent, and all we must show is that they are holomorphic. Since (dz) = Pi
we see that
• • P2g+2
Q?■21
dz\ w )
rid ni V
V g2
V3V4.
This divisor is integral as long as ) < g — 1 and therefore these differentials are holomorphic. Corollary 2. On a hyperelliptic surface of genus g 2 the products of the holomorphic abelian differentials (taken 2 at a time) form a (2g — 1)-dimensional
99
111.7. Hyperelliptic Riemann Surfaces
subspace of the (3g — 3)-dimensional space of all holomorphic quadratic differentials.
PROOF. Using the basis constructed in Corollary 1, the span of the products has a basis consisting of zi(dz) 2 w2
j = 0,
, 2g — 2.
(7.5.2)
Remarks
1. Note that 2g — 1 = 3g —3 if and only if g = 2. 2. Torobtain a basis fet the holomorphic quadratic differentials for a hyperelliptic surface of genus g> 2 we must add to the list in (7.5.2), the
differentials
:Ad: 12 w
j = 0,
, g — 3.
EXERCISE
Obtain a basis for the holomorphic g-differentials on a hyperelliptic surface. What is the dimension of the span of the homogeneous polynomials (of degree q in g variables) of the abelian differentials of the first kind?
111.7.6. Recall the injective holomorphic mapping cp:M J(M)
of a Riemann surface M into its Jacobian variety introduced in 111.6.1. Let n be a positive integer. Since J(M) is a (commutative) group, we say that a point e e J(14) is of order n, provided ne = O.
2. Choose a Weierstrass point as a base point for the map q4). Then p(P) is of order 2 whenever P is a Weierstrass point. Proposition. Let M be a hyperelliptic Riemann surface of genus
PROOF. Let Po be the base point (a Weierstrass point) and P another Weierstrass point. Since 9(1'0= 0, we may assume P Po . We have seen that there is a meromorphic function f on M whose polar divisor is P2 . Since Po is a branch point of f, (f — f(P 0)) = POI'. Thus P1 — P2 and by Abel's theorem, 2 40(P)
= 49 (P2) = 49 (N) = a
EXERCISE M be an arbitrary compact Riemann surface of genus g I. Let cp:M --+ J(M) be the embedding of M into its Jacobian variety with base point Po . Assume that OP) has order n, for some P E M, P Po . Show that there exists an f e or(M) with (1' = Pyrio . In particular if g> 1 and 2 n < g, then both P and Po are Weierstrass points on M. Let
)
100
HI Compact Riemann Surfaces
For g = 1, return to the Weierstrass 0-function considered in Exercise 111.6.7. Show that the 4 branch points of 0 are precisely the 4 half periods of J(M).
111.7.7. If M is an arbitrary Riemann surface, we denote by Aut M the group of conformal automorphisms of M. Let H a Aut M be a finite subgroup. For P E M, set H p the H;h(P) = 13}. Then Hp is clearly a subgroup of H. We claim that Hp is cyclic. This is a consequence of the following Proposition. Let h 1 ,. , h„ be n holotnorphic functions defined in a neighborhood of the origin. Assume that hi(0)= 0,j = 1, . . . , n, and that these fl-functions form a group H under composition. Then II is a rotation group (that is, there is a simply connected neighborhood D of the origin and a conformal mapping f of the unit disk A onto D such that f(0) = 0, hi(D)= D and f o hi o f is a rotation for j = 1, , n). Remark. The existence of a simply connected D invariant under H implies the rest of the proposition. In this case, choose f to be a Riemann map of zi onto D with f(0) = 0. Then hi = f1 o h of is a conformal self-mapping of the unit disk that fixes 0, and hence of the form
hi(z) =
0 < O. < 2n.
Choosing the smallest positive O. and calling it (3, we see that h(z) generates the group f ' Hf. Thus we also have
e= 2xte
Corollary. The group H, is cyclic. PROOF OF THE PROPOSITION. Let h be a typical element of H. Then h'(0)* 0, since h is invertible. We claim that there is an E > 0 such that h maps every disk { Izi r < a} onto a convex region. Such a region is convex if and only if c = (h{izi = r}) is a convex curve if and only if the direction of the tangent vector to c is a monotonically increasing function of arg z; that is, if and only if the angle (in + arg z + arg h'(z)) is an increasing function of arg z on Izi = r). A simple calculation shows that the hypothesis (h'(0) 0 0) guarantees the monotonicity of the direction of the tangent line. The derivative of the above angle with respect to arg z is (recall arg h'(z) = Re( — i log h')) 1 + Re
zh"
and this derivative is positive as long as izi is small.
Now choose a so small, so that letting A = Iti(A E) is convex for j = 1,
{Z E C; IZI
0 for only finitely many P e M), (8.3.1)
then
gcd(D 1 ,D2) = (DI,D2) =JJ Pe M
Pnin(21(P)A2(P)).
111.8. Special Divisors on Compact Surfaces
105
Similarly, the least common multiple (lcm) of D, and integral divisor D satisfying: a. D > D1 and D D2, and b. whenever fi is an integral divisor with D > D I and If Di is given by (8.3.1), then
D2
is the unique
D2,
then D > D.
lcm(D1,D2) = fl pmaxt3t(P).22(P)). Pe M
Furthermore, lcm(D 1,D2) gcd(D 1 ,D2) = D 1 D2 . Proposition.
Let D I , r (i)
PROOF.
D2
be integral divisors."Set D = (DI ,D2). Then
1) + r(D- 1)_ r(D
D
r(151-152) .
Observe that L(D5
c L(DDT 1 DI 1 ).
This inclusion follows from the fact that D I D2 /D is integral and is for j = 1, 2. Thus L(D,- ) y L(D 2--1) L(DDT 1 D2-1 ),
(8.3.2)
where y denotes linear span. Next we show L(D 1 ) n L(D2-1 )= L(D-1 ).
(8.3.3)
To verify (8.3.3) assume Di is given by (8.3.1). If f e L(Di 1) n L(Di 1 ) has a pole at P of order a 1, then a 5 ai(P),
j= 1, 2.
Thus also a 5 ininfai(P),OE2(P)}, and f e L(D'). We have established L(13 -," 1 ) n L(DV) c L(D'). The reverse inclusion follows from the fact that Di > D, j = 1, 2. This verifies (8.3.3). Finally, using (8.3.2) and (8.3.3) and a little linear algebra, r(DT 1 )+r(Dji )—r(D - 1 )= dim L(DT 1 )+ dim L(D 2--1 )—dim(L(DT I )nL(DV)) = dim(L(D ') y L(132- 1 )),.. dim L(DD I-1 D2-1 ) = r(DDT 1 D 2-- 1 ). Corollary 1. Under the hypothesis of the proposition, c(D i ) + c(D2) c(D) + c(D 1D2D -1 ). PROOF. The proof is by direct computation.
(8.3.4) El
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III Compact Riemann Surfaces
Corollary 2. If D is a special divisor with complementary divisor D*, then c(D) -_;.! c((D,D*)). PROOF. By Corollary I and Proposition 111.8.2, 2c(D)= c(D)+ c(D*) c((D,D*)) + c ("* = 2c((D,D*)). 111.8.4. Theorem (Clifford). Let D he a special diiisor on M. We have: a. c(D)
0.
b. If deg D = 0 or deg D = 2g — 2, then c(D) = O. c. If c(D) = 0, then deg D = 0 or deg D = 2g — 2 unless the Riemann surface M is hyperelliptic.
PROOF. Since c(D) = c(D*) and deg D + deg D* = 2g — 2, where D* is a complementary divisor of D, the theorem need be verified only for special divisors of degree < g — 1. If deg D = 0, then r(D")= 1 and c(D) = 0. Thus part (b) is verified. Next, if deg D = 1, then also c(D) = 1. We proceed by induction. Suppose now that 1 < deg D g — 1, and c(D) < 0. Thus r(D') >1 deg D + 1
2.
Thus there is a non-constant function in L(D" 1 ). Let D* be a complementary divisor. Replacing D by an equivalent divisor (it has the same Clifford index as D), we may assume that (D,D*) 0 D. (This last assertion is of critical importance. It will be used in ever more sophisticated disguises. We shall hence verify it in detail. We must show that there is P e M which appears in D* with lower multiplicity than in an integral divisor equivalent to D. Let co be an abelian differential of the first kind such that (0 ) = DD*. Let f e L(D - ')\C. Then for all c E C, ( f — c)D is integral and equivalent to D. By properly choosing c (for example, f '(c) should contain a point not in D*), (f — c)D will contain a point not in D*. Now (f — c)D and D* are still complementary divisors since (f — c)DD* = ((f — c)w).) By Corollary 2 to our previous proposition, c((D,D*)) < 0. But deg(D,D*) < deg D and (D,D*) is a special divisor. By repeating the procedure we ultimately arrive at a divisor of degree zero or 1, which is a contradiction. This establishes (a). It remains to verify (c). We have seen that if 0 < deg D 3. All we want is a function in L(131 -1 ) that vanishes at some point of D* and at one point not in D*. Let f be such a function. Then (f)D is integral, equivalent to D, and satisfies (8.4.2). We have produced a special divisor (13,D*) with 0 < deg(D,D*) < deg D and c((D,D*)) = O. Thus 2 < deg(D,D*) 15. deg D — 2, and we can arrive at a special divisor of degree 2, with Clifford index zero. 111.8.5 We now derive some consequences of Clifford's theorem. If D e Div (.%I) is arbitrary with 0 < deg D :5_ 2g — 2, then c(D)> deg D unless r(D - > 2. In the latter case (since there is a non-constant function f in L(D')) we have D' =.(f)D is integral and equivalent to D. Since linearly equivalent divisors have the same Clifford index, we have almost obtained Corollary 1. Let D be a divisor on M with 0 < deg D < 2g — 2. Then c(D) > 0. Equality occurs if and only if D is principal or canonical, unless M is a hyperelliptic Riemann surface.
PROOF. The remarks preceding the statement of the corollary show that unless r(D - 1 ) > 2, c(D) deg D. If r(D -1 ) 2, we have D equivalent to an integral divisor D' of the sanie degree. Proposition 111.8.2 allows us to assume with no loss of generality that deg D g — 1. Now D' is a special divisor. The fact that c(D') = c(D) > 0 follows from Clifford's theorem. If D is neither principal nor canonical (and since as already stated we may restrict our attention to divisors of degree 4. Let D I and D2 be two inequivalent integral divisors of degree
3 such that r(D; I ) =
2 --- r(D 2-1 ). Then M is hyperelliptic. PROOF. Choose non-constant functions fi e L(DI '),j = 1, 2. We may assume that each fi is of degree 3 as otherwise there is nothing to prove. Since D I and D2 are inequivalent, f, t cf2 for any C E C. As a matter of fact, we have that f1 A o f2 for any Möbius transformation A (of course, A o f2 = (af2 + b)/(cf2 + d) where a, b, c, de C, ad — bc 0 0). For if fi = A of2, then the divisor of poles of f, would be equivalent to the divisor of poles of 12
108
III Compact Riemann Surfaces
(since the polar divisor of f2 is equivalent to the polar divisor of A f 2 ). This would contradict the fact that fi is of degree 3, and that the two divisors Di are inequivalent. Thus, we have 4 linearly independent functions in L(D D 2"), namely: 1, f f 2 , fi f2 . For if co + c 1 f1 + c2 f2 + c3 f1 f2 = 0 with c C, then f1 is a Maius transformation of f2 . Thus r(1/D 1 D 2 ) 4 = I deg D 1 D 2 + 1, and c(D 1 D 2) = 0. Since deg D ID2 = 6 I. For almost all (to be defined in the proof of the assertion) integral divisors D of positive degree s t, it is possible to find integral divisors A, B satisfying B'
(8.10.1)
(A,B)/D is integral,
(8.10.2)
r(A -1 B') — s = r(DA'B').
(8.10.3)
A — A',
B
and To verify the above claim, we begin by showing that for every s t + 1 there is an integral divisor D = P, • • P, such that r(1/A'B') — s = r(D/A'B'). This is clearly equivalent to showning that the matrix (fic(Pi)), k = 1, . , d, j = 1, . . . , s, with f1 , . . , fd a basis for L(1/A'B'), has rank s. Note first that d > t + 1> s. We choose P, such that P, does not appear in A'B' and such that fi(P i) 0. Consider now the meromorphic function of P: det
f2(P1)1 fi(P)
f2(P)
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111 Compact Riemann Surfaces
Since {f 1 ,f2 } are linearly independent, the determinant is not identically zero, and thus we can find P2 such that P2 does not appear in A'B' and such that the determinant does not vanish at P2. Hence fa(Pi) fa(Pk)i
rank [ fi(P1) f1(P2) •
2
Having now chosen P,,.. ,P,,_ such that no PI appears in A'B' and such that det(f„,* (P;)), m = 1, , k — 1,j = 1, . . . , k — 1 (k s) does not vanish, we consider • • • fk(P1) .1.01) det i)
.îi(P)
' • • fk(Pi---11 • _
,f,) assures us that the determinant The linear independence of ffi, does not vanish identically so that we can choose P, not to appear in A'B' and such that det(f„,(P i)), m = 1, . . . , k, j =1, . . . , k does not vanish. Hence 1) • • • fd( 1)
= k.
rank [ f,(Pk)
•••
fd(Pk)
This verifies the existence of a divisor D of degree s with the required properties and in fact, shows that almost all divisors D of degree s would work, Finally, we need show the existence of divisors A — A' and B B' such that (A,B)/D is integral. To this end consider L(D/A'). Clearly (for s r(DIA') t + 1 — 1, and thus there is a non-constant function f E L(D/ A') such that (f)= DI,/A' and we may take A = D11. Similarly there is an integral divisor 12 such that B may be chosen as D/ 2 . 111.8.11. Theorem. Let M be a compact Riemann surface of genus g > 4. Assume that M is not hyperelliptic. Let 91 be an integral divisor of degree 5g — 1 with c(91) = 1. Then r(9.1 - I) 5 2 (and thus deg 91 3) except possibly if g = 6. In this case it is possible that r(21-1 ) = 3 and 91 2 is canonical. PROOF. Suppose that r(91 - ') > 2. Let s = r(21 -1 ) — 2. Choose integral divisors A, B, D with deg D = s such that they satisfy (8.10.1), (8.10.2), and (8.10.3) with A' = 21 = B'. We use now (8.3.4) to obtain
AB c((A ,B)
+ c((A,B)) c(A) + c(B)= 2.
From (8.10.2) we see that 1 s = deg D < deg(A,B) deg A
g— 1,
(8.11.1)
115
111.8. Special Divisors on Compact Surfaces
and hence,
AB deg( --.) 2g < — 3. (A ,B)
1 Thus, by Clifford's theorem,
AB c((A ,B)
= 1 = c((A,B)).
We now consider cases:
Case I: r(11(A,B))= 1. From-the definition (*Clifford index,
1 ---- c((A,B))= deg(.4,B). But Proposition 111.8.9 implies that
c( 4112) = c(AB)
deg(A,B) — c((A,B)) + 2(1 — s) = 2(1 — s).
Since s > 1, we see by Clifford's theorem that s = 1,20 Z, and r(91 - 1 ) = 3. Furthermore, 1 = c(21) = deg 91 — 4, shows that deg Z = 2 deg 91 = 10 or g = 6. Case II: r(11(A,B))._ 2 (thus deg(A,B) 3).
In this case we may assume that (A,B) is the polar divisor of a function. (Note first that if P E AI appears in the divisor (A,B), then
r
((A,B)) = r ((A
1\ ,B)
1.
Otherwise, =
P
deg(A,B) 1
2r(1 2 = c((A,B)) — 1 = 0, (A,B))+
contradicting that M is not hyperelliptic. A function belonging to
GAPA)
GA1,B)) \ PeY4,B) L (this is non-empty since every term in the finite union is of codimension 1 in L(1/(A,B)) will necessarily have polar divisor (A,B).) We thus apply Corollary 1 to Proposition 111.8.8, 2= 2c(21)
c(91(A,B)) + c (()). ,
(8.1 1.2)
Since 2 and (A,B) have odd Clifford index, they also have odd degree. Hence 21(A,B) and 2I/(A,B) have even degree and even Clifford index. Thus one of the terms on the right hand side of (8.1 1.2) must be zero. Since neither 91(A,B) nor 211(A,B) is principal or canonical we are done. (If, for example,
116
III Compact Riemann Surfaces
21/(A,B) were principal, then A — B 91— (A,B). Thus A =(J./3) B, which clearly can be avoided from the beginning because r(DI9.1) r(1/%) — s = 2.) 111.8.12. To change the pace slightly, we prove a result in linear algebra. For the proposition of this section, we will need some elementary results from algebraic geometry. To begin with, the set of r x r symmetric matrices can be viewed in a natural way as a vector space of dimension 112-r(r + 1) which we identify with Ç(2)(+ 1) Proposition. The set Vp of r x r symmetric matrices of rank
2. PROOF.
From the Riemann-Roch theorem, r(A - = 2n — g + 1.
Let r = r(A - 1 ). Let {f1 , . . ,f,} be a basis for L(A -1 ). Then fifk E L(A for 1 j k r, and these ir(r + 1) elements are dependent (because ir(r + 1) :2_ , 6(2n + 8 — g)(2n + 12 — g) > 2n — g + 1), and satisfy at least k = ir(r + 1) — (2n — g + 1) linearly independent symmetric relations of the form .
aikfifk = O. j.k=1
By the corollary to the previous proposition, a necessary condition for the existence of a non-trivial rank 4 relation among these is that ir(r + 1) < k + 10 + 4(r — 4),
which is precisely the condition imposed on r = r(A -1 ). By a change of basis in L(A'), we may assume the relation is of the form (it cannot be of rank 4. Then M carries a non-constant tneromorphic function of degree 1(3g + 1). PROOF. Let D be a special divisor of degree n with ig < n 2g - 2. The Riemann-Roch theorem implies that / ( D') = n - g + 1 + i(D) n - g + 2.
The assumption that n > ig implies n - g + 2> 1(2n - g + 7) and allows us to apply the theorem. We conclude that there exists an integral divisor B of degree 5 4,-n, with L(B ') > 2. Choosing the integer n as low as possible, the Corollary follows. 0 Remark. The above result generalizes Proposition 111.8.6. 111.8.14. We consider a special case of the preceding theorem. Let Z be an integral canonical divisor (then r(Z - f) = g). Note r(Z - 1 )> ;1(2 deg Z + 7 - g) if and only if g> 3. Hence we obtain, in this case, from Theorem 111.8.13, an integral divisor B of degree - 1 with r(11B) > 2. Thus, also i(B) 2. (The above also follows from the corollary to Theorem 111.8.13.) We have produced the divisor B as a consequence of a quadric relation of rank 54 among products of meromorphic functions. It could have equivalently been produced as a consequence of a quadric relation of rank 54 among products of abelian differentials of the first kind. Now, there are ig(g + 1) such products and 3g - 3 linearly independent holomorphic quadratic differentials. Thus there are at least ig(g + 1) - (3g - 3) = 1(g - 2)(g - 3) linearly independent symmetric relations
E
afimg , = o
j.k=
among products of a basis tcp i , ,cp,1 of abelian differentials of the first kind. The space of symmetric g x g matrices of rank s if and only if D has s — 1 "free points", and gives a precise meaning to this statement. Remark. The lemma could have been established at the end of 111.4. We have delayed its appearance because its proof uses techniques of this section. In fact, we have already used and proved part of the lemma in 111.8.10. Leming. Let D he an integral divisor on M. A necessary and sufficient condition for r(1/D)> s is: given any integral divisor D' of degree — 1, there is an
integral divisor D" such that D'D" D. Further, for sufficiency it suffices to assume that D' is restricted to any open subset U of Mcs _ 1) , the (s — 1)symmetric product of M. PROOF. To prove the sufficiency, we assume r(1/D) = d and that (f1 , ,fd ) is a basis for .1(1/D). As in 111.8.10, we construct a divisor D' = Q, • • • Q,_ such that rank (fk (Qi)), k = 1, . , d, j = 1, . , s — 1, is precisely min {d, s — 1) and such that none of the points Oi appear in D. If d s — 1 we would have that the only function which vanished at Q1,..., Q 1 in L(1/D) would be the zero function contradicting the fact that we can choose an integral divisor D" such that D'D" D. Hence d > s. To show necessity assume that r(D -1 )= s> 1. Note that for arbitrary divisor 21 and arbitrary point P E M, r(P91 1 ) > r( 21 - 1) — 1. Now let D' be integral of degree — 1. By the above remark, r(D'/D) s — (s — 1) = 1. Now choose f e L(D'/D). Thus there exists a divisor D" such that D'D" D.
111.9. Multivalued Functions We have on several occasions referred to certain functions as being (perhaps)
multivalued, and then proceeded to show that they were single-valued. Examples of this occurred in the proof of the Riemann-Roch theorem, in the proof of the sufficiency part of Abel's theorem (Theorem 111.6.3), and in the section on hyperelliptic surfaces. In this section we give a precise meaning to the term multivalued function and generalize Abel's theorem and the Riemann-Roch theorem to include multivalued functions. Multivalued functions will also be treated in IV.4 and IV.11. Analytic continuation (= multivalued functions) is one of the motivating elements in the development of Riemann surface theory. Riemann surfaces are the objects on which multivalued functions become single-valued. This aspect of multivalued functions will be explored in IV.1 1.
120
III Compact Riemann Surfaces
111.9.1. Let M be a Riemann surface. By a functilm elenOtt (f,U) on M, we mean an open set U M and a meromorphic function f:U -+C
Two function elements (f,U)and (g,V) are said to be equivalent at PEUn V, provided there is an open set W with P E WC U n V and flIV = gii.V.
The equivalence class of (f,U) at P E U is called the germ of ( f,U) at P and will be denoted by (f,P). It is obvious that the set of germs of all function elements at a point P E M is in bijective correspondence with the set of convergent Laurent series at P (in terms of some local parameter) with finite singular parts. We topologize the set of germs as follows. Let (f,P) be a germ. Assume it is the equivalence class of the function element (f,U) with P E U. By a neighborhood of (f,P), we shall mean the set of germs (f,Q) with Q E U. It is obvious that each connected component Sr-, of the set of germs, is a Riemann surface equipped with two holomorphic maps C
where
L.)
{ c;c}
eval( f,P) = f(P) proj(f,P) = P.
The verification of the above claims is routine, and hence left to the reader. In this connection the reader should see also IVA.
111.9.2. We shall be interested exclusively in a restricted class of components g7 as above. Namely, we require that i. proj be surjective, and that ii. for every path c:I —> M, and every f e ...42-• with proj(f) = c(0), there exists a (necessarily) unique path a:I 3-- with E(0) = f and c = proj 0 E. We shall call e the analytic continuation of E(0) along c. Note that by the Monodromy theorem E(1) depends only on the homotopy class of the path c and the point E(0). (In the language of 1.2.4, we are considering only those components Y which are smooth unlimited covering manifolds of M.)
111.9.3. Let M be a compact Riemann surface and rc,(M), its fundamental group. By a character x on ic,(M) we mean a homomorphism of ic,(M) into the multiplicative subgroup of C, C* = C\10). Since the range is commutative, a character x is actually a homomorphism x : Ii i(M) C*.
111.9. Multivalued Functions
121
The character x is normalized if it takes values in the unit circle {z c C; ,N 2g} = {a 1 , . izi = 1. If M has positive genus g, and {N,, b1, is a canonical homology basis on M, then x is determined uniquely by its values on a canonical homology basis, and these values may be arbitrarily assigned. The set of characters on rz 1 (M) forms an abelian group (under the obvious multiplication) that will be denoted by Char M. 111.9.4. We are finally ready to define a multiplicative multivalued function belonging to a character .2t on M. By this we mean a component 5 of the germs ofrf meromorphic rinctions on M satisfying the two properties listed in 111.9.2 and the following additional property:
iii. the continuation of any f e 5 along a closed curve c leads to a point fi with eval f1 = x(c) eval f. 111.9.5. We examine more closely a multiplicative function 5 belonging to a character x. Let c be a closed path in M and t a path in 5 lying above it (proj o = c). For each point t E [0,1], the point (t) is represented by a function element (f„U(t)) with U(t) open in M and c(t) E U(t). By the continuity of the map "c":/ 5 and the compactness of I, we can find a subdivision 0= to < t i < t2 < • • • < t„ < =1
and function elements ( 4 .U0), (f1 ,U 1 ), , (f,„U„) such that(t) is the equivalence class of (fi,U;) for t E [tpti+ 1 1, j = 0, , n. Furthermore, if c is a closed path, c(0) = c(1), we may assume without loss of generality that U0 = Un . We say that the function element (f„,U„) has been obtained from the function element (f0,U0) by analytic continuation along the curve c. Since 5 belongs to the character x, we see that
= x(e)fo. Thus, we may view a multiplicative function 5 belonging to a character x as a collection of function elements (f,U) on M with the properties (i) given two elements (f1 ,U 1) and (f2 ,U2) in 9, then (f2,U2) has been obtained by analytic continuation of (f1 ,U 1) along some curve c on M, and (ii) continuation of a function element (f,U) in F along the closed curve c leads to the function element (x(c)f,U). 111.9.6. To define multiplicative differentials belonging to a character, we proceed as follows: Consider the triples (w,U,z) where U is an open set in M, z is a local coordinate on U, and w is a meromorphic function of z. If (0)1, 17,C) is another such triple and PE Un V, then the two triples are said to be equivalent at P provided there is an open set W U r) V, with P e W, and for all Q E W
c1C — = w(z), dz
z = z(Q).
III
122
Compact Riemann Surfaces
Repeating the previous arguments, with this equivalence relation, one arrives at multiplicative (or Prim) differentials belongingto a character x. To fix ideas, this involves a collection of triples (o.U,..:). IT (w0 ,U 0 ,z0) and (a)„,U„,z„) are two such triples, then we can find a chain of triples (a);,Upzi),
j = 0,
, n,
such that n
# 0,
j
0, . , n — 1,
and coi(zi(zi .„ „))
dzi
— o f ,(zif ,), zit :J .,. 1 (Q), QEUi U j+ -J+ Furthermore, if we continue an element (a),U,z) along a closed curve c to the element (w 1 , U,;), then co i(z)= x(c)a)(z),
z=
Q E U.
111.9.7. It is quite clear that if ff(j = 1,2) is a multiplicative function belonging to the character xi, then 1112 is a multiplicative function belonging to the character X X2 (L/f2 is a multiplicative function belonging to x,x2provided 12 0 0). Similarly, if coi (j = 3,4) is a multiplicative differential belonging to the character xi, then f1 (o3 is a multiplicative differential belonging to the character x i x 3 and 0)3/0)4 is a multiplicative function belonging to the character x 3x," j, provided a),, # O. Proposition. If f r 0 is a multiplicative function belonging to the character
X, then df is a multiplicative differential belonging to the same character and dflf is an abelian differential. PROOF. We need only assure the reader that df is exactly what one expects it to be. If f is represented by (f,U) on the domain of the local parameter z, then df is represented by (f '(z),U,z). 0
111.9.8. It is quite obvious that a multiplicative function (and differential) has a well-defined order at every point, and thus we can assign to it a divisor. Corollary 1. If f
0 is a multiplicative function, then deg( f) = O.
The order of f at P, ord,, f, just as in the ordinary case is given by the residue at P of df/f. Since df/f is an abelian differential, the sum of its residues is zero. PROOF.
Corollary 2. If co 0 0 is a multiplicative differential, then deg(w) = 2g — 2.
PROOF. Choose an abefian differential a), on M, a), 0 O. Then a)/a), is a multiplicative function belonging to the same character as a). Since deg(a) 1 ) = 2g — 2, Corollary 1 yields Corollary 2.
123
111.9. Multivalued Functions
111.9.9. If f is a multiplicative function without zeros and poles, then dfif = d(log f ) is a holomorphic abelian differential. Letting jC i , ,} be a basis for the abelian differential of the first kind on M dual to the canonical homology basis, we see that df
E
- = d log f f
and thus f(P)
= f(P0) eXP 5P
CI:»
Po
with
ci E C.
(9.9.1)
x of the function f is then given by x(ak )= exp ck , k = 1, . . . , g,
(9.9.2)
z(bk) = exp(E cirrik),
(9.9.3)
The character
k=
.= I
We shall call a character
x as above inessential, and
f as above a unit.
Proposition. If y is an arbitrary character, then there existsa unique inessential
character y i such that y/y, is normalized. PROOF.
Assume X(ak) = esk+ irk,
sk , tk E
x(bk) =
uk ,
01,
E
for k = 1, . . , g. To construct an inessential character x, with lx,(c)i = Ix(c)I for all c E ,(M), we choose constants ck = Œ + i/3,, so that (compare with 9.9.1, 9.9.2, and 9.9.3),
= esk or ak = Sk ,
and
t
cgr,
= v -,
Re c,x ,
= euk
or
E
Re cirri, =
uk .
(9.9.4)
1=1
To see that this cho'ce is indeed possible and in fact unique, recall that we can write the matrix 17 = (ink) k ) as 11= X + iY with Y positive definite, and thus non-singular. To solve (9.9.4) we write c = r(c,, ,eg), = t(a l , ,;), etc.,. , and note that we want to solve
Re[(X + iY)(a + ill)] = u. Since we have already chosen a = s the equation we wish to solve is
Re[(X + iY)(s + il3)] = u,
124
Ill Compact Riemann Surfaces
or what amounts to the same thing
Yfl = — u + Xs. Since Y is non-singular there is a unique fI which solves this system of equations.
Corollary. A normalized inessential character is trivial.
111.9.10. Theorem. Every divisor D of degree zero is the divisor of a unique (up to a multiplicative constant) multiplicative function belonging to a unique normalized character.
, r > O. If PROOF. Let D = P, • • • PA, • • • Q, with Pi Qk all j, k = 1, fi is a multiplicative function belonging to the normalized character zi (j 1,2), and (fi) = D, then f 1 /f2 is a multiplicative function without zeros and poles. Thus, 7.17.2-1 is inessential and normalized; hence trivial. Thus it suffices to prove existence for the divisor D = Pi/Q,, with P,.Q, e M, P, Q 1 . Recall the normalized abelian differential Tp,Q, introduced in 111.3.6. Define f(P)= exp Tp i Q i . The character to which f belongs is not necessarily normalized. But the arguments in 111.9.9 showed how to get around this obstacle.
Corollary. Every divisor D of degree 2g — 2 is the divisor of a unique (up to a multiplicative constant) multiplicative differential belonging to a unique normalized character. PROOF. Let Z be a canonical divisor, and apply the theorem to the divisor of degree zero D/Z. 0
111.9.11. Theorem. Every character z is the character of a multiplicative function (that does not vanish identically). PROOF. Let {a 1, b1, ,b,} be a canonical homology basis on M. It is obvious that x may be replaced (without loss of generality) by xx, with Xi inessential (a unit always exists with character x i ). Since the value of an inessential character can be prescribed arbitrarily on the "a" periods (see 111.9.9), we may assume
x (a) = 1,
= 1,
(9.1 1.1)
Furthermore, writing
z(b) = exp(fli),
, g,
j = 1,
and fixing e> 0, we may assume in addition
< e,
j = I, •
g•
(9.1 1.2)
125
111.9. Multivalued Functions
(For if z is arbitrary and satisfies (9.11.1), then choose an integer N > 0 so that
111j, NI
2 2rci f 1') r,k
(9.11.3)
by the bilinear relation (3.6.3), where {C,, ..; g } is the normalized basis for the abelian differentials of the first kind dual to the given holomogy basis. Now (9.11.3) obviously defines a holomorphic mapping 40: Ug Cg
= (0 1, • • • , ( Pg) (Pk (2) = exp
E1 27ri
j=
11(:1,
:g)
=
(1,
ri
,1).
The theorem will be established if we show that 0 covers a neighborhood in Cg of (1, . ,1). By the inverse function theorem it suffices to show that the Jacobian of 0 at (z 1 , ,zg) is non-singular. Write 27riCi(z) = cpi(z)dz in terms of the local coordinate z, and note that
00 k
xi
= (pk (zi),
j, k = 1, . . , g.
Thus, we have to show that the matrix 01(7, 2)
•••
91(z9 ) 92(z g)
9g(zg)_
126
III
Compact Riemann Surfaces
is non-singular. If this matrix were singular, then a non-trivial linear combination of the rows would be the zero vector in Cq. That is (since , Cg are linearly independent), there would be a non-zero abelian differential of the first kind vanishing at P 1, , P,„ contradicting the assumption that the divisor D is not special. 0
Remark. The reader should notice the similarity between the proof of the above theorem and the proof of the Jacobi inversion theorem.
111.9.12. Let 21 be an arbitrary divisor on M. In analogy to our work in 111.4, we define for an arbitrary character x, 4(91 ) = (multiplicative functions f belonging to the character
x
such that
rx(91) = dim L,(21), Q1(91) = {multiplicative differentials w on M belonging to the character such that (w) 21), and
x
ir(21) = dim f2x(21). Theorem (Riemann-Roch). Let M be a compact surface of genus g, and x a character on M. Then for every divisor II on M, we have r(91') deg 21 — g + 1 +
(9.12.1)
PROOF. Choose a multiplicative function 0 f belonging to the character (Theorem 111.9.11 gives us the existence of such an f.) Then h e L,(21 -1 ).=. h/f
E
L(It - t ( f)
x.
(9.12.2)
and
e
cof e Q(21(f)).
(9.12.3)
Thus
r,(21 1 ) = r((21(f ))') and j7 - i(21) = 491( f)).
(9.12.4)
We apply now the standard Riemann-Roch theorem (4.11) to the divisor 21(f ) and obtain
r((21(f )) -1) = deg((91(f ))) — g + 1 +1 (21(f)), which is equivalent to (9.12.1) in view of (9.12.4) and the fact that
deg(21(f )) = deg 21 + deg( f) = deg 21. Remark. The isomorphisms established in (9.12.2) and (9.12.3) are also useful in their own right.
111.9.13 An important class of characters are the so-called nth-integer characteristics. Let {a i, b 1 , . . ,b,} be a canonical homology basis on
127
111.9. Multivalued Functions
M. Consider an integer n > 2 and a 2 x g matrix [el = I- & Lin, ... ,e,/n1 Le' in Le',/n, . . . ,e;inj
with ere; integers between 0 and n — 1. The symbol [], will be called an nth-integer characteristic. It determines a normalized character x on M via
X(a) = exp(27tiej/n), x(b3) = exp(27ris;/n), le' _
j = I, . . . , g. • -.' It is clear that for every multiplicative function f belonging to such a character x, Ifl is a (single-valued) function on M, and f itself lifts to a function P.
on an n-sheeted covering surface of M. Furthermore, we can construct such an f by taking nth roots of a meromorphic function h on M provided (h) is an nth power of a divisor on M (that is, provided the order of zeros and poles of h are integral multiples of n).
111.9.14. Proposition. For x E Char M,
ix (1 ) = { gg _
if x is inessential, I if x is essential.
PROOF. The Riemann-Roch theorem says: r
- i(l) = —g + 1 + i,(1).
Thus, i1(1) ..- g — 1, and ir(1) __ g. Furthermore. i,(1) = g if and only if there is an f E L 1 _ 1 (1), f # O. Since such an f must be a unit if it is not identically zero, we are done.
111.9.15. We end this section with the statement of Abel's theorem for multiplicative functions belonging to a character x. The proof is omitted since it is exactly the same as the proof already studied (in 111.6.3). Theorem (Abel). Let D E Div(M), x e Char M. A necessary and sufficient condition for D to be the divisor of a multiplicative function belonging to the character x is 1g
(PP)
27ti .41 log X (L)e
co
1g log
and
deg D = O.
111.9.16. Exercise Define a mapping
9:Char M -■ J(M)
x(c
(mod L(M)),
III Compact Riemann Surfaces
128
as follows. For X E Char M. select a non-constant meromorphic function f belonging to the character x. Let rp(x) = g)( ( f)), where, as before, ça(( f)) is the image of the divisor (f) in the Jacobian variety. Show that: (1) The mapping (p is a well defined group hornomorphism. (2) The mapping q) is surjective.
(3) x e Kernel q) if and only if x is an inessential character. Cone-ride that as groups J(M)-1- Char Al/Unessential characters on M).
(4) Obtain an alternate proof of the Jacobi inversion theorem as follows. First show that every e E J(M) is the image of a divisor of degree zero. (We established this fact by the use of the implicit function theorem.) Thus every e e J(M) is the image of some y e Char M. Now use Theorem 111.9.12 (Riemann-Roch) to show that rx il : P4)= 1. Thus there is a multipheative function belonging to the character x with < g poles. Jacobi inversion follows from this observation. The same method (see HD I) can also determine the dimension of the space of divisors of degree g that have image e. Remark. The reader should at this point review the remark at the end of 111.9.11.
111.9.17. Exercise (1) Let P E M and let z be a local coordinate vanishing at P. Let n e Z, n > 1. By a principal part (of a meromorphic function) at P we mean a rational function of z of the form akz4.
f( ,)= -n
,P„,} be in distinct points and zi a local coordinate vanishing at P. (2) Let {P 1, Let fi be a principal part of a meromorphic function at Pi (in terms of the local coordinate zi). The collection F = (f1..... f,,,} will be called a system of principal parts at (I' 1,
,P.}.
(3) Let F be a system of principal parts at {P 1 , ... ,P„,}. Let gl be an abelian differential on M that is regular at Pi lor i = 1, . , rn. Then F(v)=
E
Resp, fjgr
.1= I
is a well-defined linear functional on the space of all such cp. In particular, F induces a linear function on ifl(M), the vector space of holomorphic differentials on M. (4) Let ye Hom(H,(M),C); that is, x is a homomorphism from the first homology group of M into the complex numbers. By an additive multivalued function belonging to y we mean a component 5Fof the germs of meromorphic functions on M satisfying the two properties listed in 111.9.2, and (iii)' the continuation of any f E ..5F along a closed curve c leads to f, e 3'7 with eval f1= eval f y(c).
Show that for every additive function ..5r, dgr is a (well-defined) holomorphic differential. Further .F defines a system of principal parts and this induces a linear functional on
111.10. Projective Imbeddings
129
(5) Using the normalized abelian differentials of the second kind introduced in 111.3, show that (a) Every system of principal parts is the system of principal parts of an additive function. (b) For every x e Hom(11 1 (M),C) there is an additive function belonging to z. (c) Let { al, b1, ,b9} be a canonical homology basis on M. A homomorphism x e Hom(H,(M).C) is called normalized if x(c(i) = 0, j = 1, , g. Show that every system of principal parts belongs to a unique normalized homomor-
phism. (6) Prove that a system F of principal parts is the system of principal parts of a (singleval ued) mcromorphic (OM ion on M if and on ly if F induces the zero linear functional on (7) Use the notation of 111.3, and show that an _ 2
J.P C too
R
n—1 where TN=
E
(ajzi)dz
at P,
jO
and R. P, Q, are three distinct points on M.
(8) Let F be a system of principal parts at {P I , ... ,P.}. Let Q be arbitrary but distinct from Pi, j = 1, , m. Define for P # P, P # 0, E(P)= — F(T pi!), Show that E agrees on M additive function with F normalized homomorphism.
,P„„Q} with the unique (up to additive constant)
as
its system of principal parts and belonging to a
111.10. Projective Imbeddings Throughout this section, M is a compact Riemann surface of genus g 2, q is an integer 1, and d = dim O (M ) (=g for q = 1, =(2q — 1)(g — 1) for q > 2). We show that every such surface M can be realized as a submanifold of three-dimensional complex projective space. We have seen that for each P E M, there is a a.) e Ye l (M) with co(P) 0 0 (thus also ail E Ya(M) with w(P) 0 0). Let {C I, ... ,Cd} be a basis for Yeq(M). The preceding remark shows that we have a well-defined holomorphic mapping, called the g canonical mapping, of M into complex projective space -
0:M
pd
130
III Compact Riemann Surfaces
where 0(P) is given by (C 1 (P), ,:"J(P)) in homogeneous coordinates. We are using here our usual conventions: Let z be any local coordinate on M. Express the differential = cpi(z)dzq in terms of this local coordinate. Then 0(z) = (9 1(z),
,cpa(z))
in terms of the local coordinate z. The image of P E M in Pd-1 is clearly independent of the choice of local coordinate used, and a change of basis of feq(M) leads to projective transformation of Pd-1 • Hence the map 0 is canonical, up to the natural self-maps of projective space.
111.10.2. Theorem. The g-canonical holomorphic mapping 0:M -, 11' is injective (and of maximal rank) unless g = 1 and M is hyperelliptic or g = 2 = q. In these exceptional cases 0 is two-to-one. PROOF. Assume that for P e M, 2 is a "q-gap" (that is, there exists a holomorphic q-differential with a simple zero at P). Then by choosing a basis for 09(M) adapted to P, we see that (z = local coordinate vanishing at P)
z — ■ 0, z
91(z) = (1 + 0(IzI))dzg, 92( z ) = (z 0(1.71 2 ))dzq, Thus 0,
z 0, 01 and we conclude that 0 has a non-vanishing differential at P (z = 0) and is, hence, of maximal rank at P. Now assume there exist P and Q E M, P Q, with 0(P). 0(Q). By choosing a basis adapted to P as above, we see that in homogeneous coordinates 0(P) = (1 0 0). (d — 1)-times
Thus, if 0(P) = 0(Q), we must have, 0(Q) =
,0),
A
0.
(d — 1)-times In particular, we see that every holomorphic q-differential that vanishes at P also vanishes at Q, or L(Z-q P) = L(Z -4 PQ), (10.2.1) since L(Z - qP) and L(Z - qPQ) are isomorphic to the vector spaces of holomorphic q-differentials which vanish at P and P and Q, respectively. Hence r(Z - qP) = r(Z- VQ) = d — 1.
(10.2.2)
131
111.10. Projective lmbeddings
Let us assume q = 1, and use (10.2.2) and Riemann-Roch to compute = 2 — g + 1 + r(PQ/Z)= 2. 1) Thus M must be hyperelliptic. Next we assume q> 1, and compute r(Z - qPQ)= q(2g — 2) — 2— g + 1 + r(Z9-1 /PQ) = (2q — 1)(g — 1) — 2 + r(Zq - 2 /PQ). Now for (10.2.2) to hold we must have that r(Zi - 1 /PQ) =I; which implies that deg(Z"/PQ) 0. This last statement is equivalent to (q — 1)(g — 1)
1.
Since g 2 and q> 2, this is only possible for g = 2 = q. Note that the above argument also establishes that for each P E M, 2 is a "q-gap" except if g = 2 = q or q = 1 and M is hyperelliptic. 111.10.3. To study the excluded cases, we represent a hyperelliptic surface M by w2 = (z — el ) • • (z — e29 + 2) with distinct ei. We have seen that a basis for dri(m) is then
j dz
zdz
zg dz}
tw w
Thus in affine coordinates 0(P) = (1,z(P),
,z(P)6-1 ).
Thus 0 is clearly two-to-one in this case (since z is two-to-one), and not of maximal rank at the Weierstrass points z -1 (ei). For the other excluded case (q = 2 and g = 2), M is hyperelliptic again. A basis for .0 2(M) is IdZ 2 (12 2 2 C/Z 2 W2 W 2 w2
1'
Again 0(P) = (1,z(P),z(P)2) is independent of w, and the 2-canonical map (for genus 2) is two-to-one, and not of maximal rank at the Weierstrass points. 0 111.10.4. Since the image of a compact manifold under an analytic mapping of maximal rank is a sub-manifold, we have obtained
132
111 Compact Riemann Surfaces
Corollary 1. Every Riemann surface of genus >_ 2 is a submanifold of a complex
projective space. Corollary 2 a. Every non-hyperelliptic surface of genus 3 is a sub-manifold of b. Every other surface of genus g 2 is a sub-manifold of P 3.
P2.
Part (a) has already been proven. We have also seen that every surface is a submanifold of for large d. Now assume that a Riemann surface M can be realized as a submanifold of Pk for k> 3. Call a point x E Pk\M good, provided for all P e M, the (projective) line joining x to P is neither tangent to M at P, nor does it intersect M in another point. Say that we can find a good point x. We may assume that PROOF.
x = (1,0 ,
0)
k-times A line L through x may be represented thus by any other point y on it. Since y 0 x, y = (4, ). 1 , ).,), where (;. 1 , ,;.k ) = y(L), determines a well defined (depending only on L) point in P j. Now the mapping
M 9 P y(L)
E
Pk-1 ,
where L is the line joining x to P, is a one-to-one holomorphic mapping of maximal rank. Thus we are reduced to finding good points. The tangent lines to M form a two-dimensional subspace of Pk, and the lines through two points of M form a three dimensional subspace of Pk. Thus a good point (a point not in the union of these two subspaces) can certainly be found provided Pk has dimension >4. We also saw in the above proof that the embedding of M as a submanifold of P3 can be achieved by using 3 or 4 linearly independent holomorphic q-differentials (g = 1 except in the few exceptional cases). We will show in 1V.11 that every two meromorphic functions on M are algebraically dependent. Hence we will also have Corollary 3. Every compact Riemann surface of genus g > 2 can be realized as an algebraic submanifold of P 3.
111.1 1. More on the Jacobian Variety This section is devoted to a closer study of various spaces of divisors on a compact Riemann surface M, and the images of these spaces in the Jacobian variety J(M). We show that J(M) satisfies a universal mapping property (Proposition 111.11.7). The technical side involves the calculation of dimensions of subvarieties of J(M). The most important consequence is Noether's theorem (111.11.20). Throughout the section, we assume that the
133
111.11. More on the Jacobian Variety
reader is familiar with elementary properties of complex manifolds of dimension> 1.
I11.1 1.1. By a complex torus T, we mean the quotient space, T= C"/G, where G is a group of translations generated by 2n R-linearly independent vectors in C". A torus T is thus a group (under addition modulo G) and a complex analytic manifold with the natural projection T, a holomorphic local homeomorphism. We introduce now some notation that we will follow throughout this section..If u E T, then iitE Cn will denote a point with p(fi) = u. The gener, /1' 2") E C. The ators of G will be denoted by the column vectors //"), j-th component of the vector Irk) will be denoted by trik . The n x 2n matrix OW= II will be called the period matrix of T. Our first observation is that the 2n x 2n matrix [fi] is non-singular. To see this, assume that there exists a vector c E C 2" such that [ ]c = [ifcc] = O. Then I7(c +?) 0 = — Z). Thus Re c = 0 = Im c, by the R-linear independence of the columns of H. Thus c = 0. Hence we have for `x, 'y E Cm, E C 2" the equation (x,y)[0] = c has a unique solution. In particular if c E C 2", then xI7 +yfi = c =Z = 17 +Vii, Or
(x — y)/7
(y — x)n = o.
Thus x = y, and
(11.1.1)
c = 2 Re(x/7).
111.1 1.2. Since G acts fixed point freely on C", n i (T) -1-• G, and since G is abelian 1/ 1 (T,7L):,,_-: G. As a matter of fact, the paths corresponding to the columns of H (that is, t e [0,1 ], k --= 1,
t/r),
. , 2n)
project to a basis of H,(T,Z). We observe next that the holomorphic differentials, dû are invariant under G, and hence project to holomorphic 1-differentials du; on T.
Remark Let V be a complex manifold of dimension m. By definition, a holomorphic 1-form to on V is one that can be written locally as dF for some locally defined holomorphic function F on V, or in terms of local coordinates, z = (z 1, . ,z,„), as
w = E fi(z)dzi, with
is, holomorphic. In particular, to is closed. Since I71
dco
111
=EE
af
J=1 k=1 a2k
dzk A dz
= o,
134
Ill Compact Riemann Surfaces
we also have
= (14- , all k, j = 1, . . . , m. ez, cz j Lemma. The projections to T of the differentials dû 1 ,. , di.1„ form a basis for the holomorphic 1-differentials on T, that will be denoted by tdu,, , du}. PROOF. Let {a 1 , ators of G. Then
,a2„ } be a basis for H 1 ( T,Z) corresponding to the gener-
jai, dui = rcik ,
j = 1,
n, k = 1, . . . , 2n.
Thus, letting e E R2", it follows by (11.1.1) that there is a unique vector such that
Re
ak
E xi du, = ck ,
k = 1,.. . , 2n.
E
C"
(11.2.1)
Now (11.2.1) shows that the differentials du,, du,, , du„ are linearly independent over C. Furthermore, given any holomorphic 1-differential on T, x dui such that there exists a differential co = Re
f
= Re
L1/4
k= 1, . . . , 2n.
to,
(11.2.2)
Define a function F on T by
F(P) = Re j: ((5 — to),
P E T.
By (11.2.2), F is well-defined. Since (5— cu is holomorphic, F is locally the real part of a holomorphic function. Thus F satisfies the maximum (and minimum) principle. Since T is compact, F is constant and thus = O. By the Cauchy—Riemann equations
1m J" (ô —
CO)
=
and thus 3 = w. Definition. By du we will denote the column vector of 1-differentials
`{du„
,du}.
111.1 1.3. Let V be any connected compact m-dimensional complex analytic manifold, and let (1): V --. T be a holomorphic mapping. Let z = (z 1 ,. ,z) be a local coordinate at a point (P(P) E T. If co is a holomorphic 1-form on T, then locally
w=E 1=1
f3(z)dz.
III] I.
135
More on the lacobian Variety
Let C = ( ( 1 .... ,C„,) be a local coordinate at P on V. The pullback of co via (P, tb*co, is the holomorphic 1-form defined in terms of the local coordinate C by ns
SP * 0)
E
=
where we write z = h(( ) = (ha), . ,h„(C)), and oh k g = E ji(h(C))—
=
k= 1
Let 1 , . e-. , 3„ denote the,pullbacks of du 1 , . du,, via 0. Let .5 be the column vector formed by the 3.1 , and p be, as before, the projection from Ca -+ T. Proposition. Let P0 e V. Then
(11.3.1)
OW) = (P(Po) + (9 (fpPo (5). PROOF. Since (5 = 0* du, and P
S I:4P)
SP0
o(ro)
(11.3.2)
du = 0(P)— 0(P0)
modulo periods, the result is clear.
111.11.4. Corollary. Let 01 :V T he holomorphic mappings for j = 0,1. Assume 00 is homotopic to 0 1 . Then there exists a co e T such that 0 0(P)= PROOF.
01(P) + co, aIlP e V.
Since 00 is homotopic to eb„ there exists a continuous function 0:V xi
T
(1= [0,1]) such that 0 o,
0(', 1) = 0 1.
Thus for every closed curve a in V, 00(a) is homotopic to 0.1 (a). Let (5‘1) = Ofdu. From (11.3.2) we see that (51,1) (since they have the same periods over every closed curve a on V). Thus 0 0 and 0 1 satisfy (11.3.1) with the 0 same (5.
c) =
111.1 1.5. Let T 1 = Cm 1G 1 and T2 = Cn/G2 be two complex tori. Let 0: T i --+ T2 be a holomorphic mapping. As above, let (5 = eb* du. Then there exists an n x m matrix A such = A dv where dv =`{dv i , ,dv,„} is a basis for the holomorphic 1-forms on Ti . Hence, as a consequence of the previous proposition, the map 0 can be written in the form 0 (P) = P2 ° A ° P1-1 (P) + co,
PE
Tip
136
111 Compact Riemann Surfaces
for some co E T2. (Since A is an n x m matrix it also represents a linear transformation from Cm into C".) We have thus established the following
Proposition. The only holomorphic maps of a complex torus into a complex torus are the group homomorphisms coin posed with translations. 1II.11.6. By an underlying real structure for the complex torus T = C"/G we mean the real torus R 2"/Z 2" together with the map R2 "g 2n T induced by the linear map R 2" 17X E Cm. We have seen (as a consequence of Proposition 111.1 1.5) that any endomorphism of T is induced by a linear transformation A: C" C" that preserves periods. Thus if A represents the matrix of this linear transformation with respect to the canonical basis for C", then there exists a 2n x 2n integral matrix M such that All = TIM. The matrix M now induces an endomorphism of the underlying real structure. The endomorphism A is completely determined by M (since [ff] is nonsingular), and the following diagrams commute:
R2"
C"
which is simply another way of saying All = 17M. It follows immediately from the previous remarks that a holomorphic injective map of one complex torus into another torus of the same dimension is necessarily biholomorphic. If H i and n, are the period matrices of the two tori, the map can be represented by a matrix A :C" C" such that A/7 2 = 11 2 M for some M as above. It is necessarily the case that both A and M are non-singular. Thus we have established the following
Proposition. Two complex tori T 1 = C"/G 1 and T2 = Cm/G2 are holomorphically equivalent if and only if n = m and there are matrices A E SL(n,C), M E SL(2n,7L) such that All = 112M where II is a period matrix of T j = 1, 2. Let A represent an endomorphism of a complex torus A: C"/G -+ C"/G. Let /7 2, /72 be two period matrices for this torus. Then for some M. e GL(2n,Z), All, = 17 1 M i and
An, = n2m2.
137
111.11. More on the Jacobian Variety
Further,
M E SL(2n,Z).
11 2
Thus 1 1 1M = Afl ,M = A17 2 = 112 M 2 =
and hence (since H
"
C" is an isomorphism)
MiM = MM2. It thus follows, since M is non-singular, that trace M i = trace M2. Corollarj: The trace of tlfé endomorphism A is-well defined by setting it equal to trace M 1 (or M 2). 111.1 1.7. We return now to the situation of 111.6. Let M be a compact Riemann surface of genus g > 0, `{a,b} a canonical homology basis on M, and {C } the dual basis for ye1 (M ). As before II denotes the period matrix of the surface. (Here II is ag x g matrix.) Let J(M) be the Jacobian variety of M; it is, of course, a complex torus (with period matrix (LIM. Let 0 be the mapping of M into J(M) with base point Po previously defined. Note that 9 *dui = Ci, j = 1, . . . , g. Now let 4): M T be any holomorphic mapping of M into a complex torus T = C"/G. Let dv 1 , ,dv„} be a basis for the holomorphic differentials on T. Let (5.1 = tb*dvi E Je (M ). Since {C} is a basis for Ye l (M), there are unique complex numbers aft such that 3,- Ejkk, k=1
(or 3 = AC, 6 =` {6 1 ,
J = 1, . . . n,
,6„}, A = (aft )). Then p(Ail)+ (P(P0) E T
defines a unique mapping ili:J(M)-4 T. It now follows from (11.3.1) that
= i,1, 0 (p.
(11.7.1)
We have established the following Proposition. Let ck:M T be a holomorphic mapping of a Riemann surface
M into a complex torus T, then there exists a unique holomorphic mapping 0:J(M)-4 T such that (11.7.1) holds. The above proposition shows that J(M) is determined by M up to a canonical isomorphism. The mapping 9 is, of course, determined up to an additive constant by the canonical homology basis on M.
138
III Compact Riemann Surfaces
111.1 1.8. We have seen in 111.6, that the mapping 9 extends to divisors on M. Let W. be the image in J(M) of the integral divisors of degree n. Set by definition Wc, = {0}. It is thcn clear that WN WN+1,
(because tp(D) = cp(DP0) for every divisor D) and 147, = J(M)(Jacobi inversion theorem). Let W„ be the set of points in J(M) which are images of integral divisors D that satisfy the two conditions deg D = n, and r(D - ')> r + 1. Denote by K the image under of the canonical (integral) divisors. By Abel's theorem, K consists of one point. Proposition. 1K) = PROOF. Let D be an integral with deg D = 2g — 2. Then r(D') > g if and only if i(D)> 1 if and only if D is canonical.
111.1 1.9. Let M be a Riemann surface and n > 0 an integer. The n-fold Cartesian product, M", is naturally a complex manifold of dimension n. By M. we denote the set of integral divisors of degree n on M. We topologize and give a complex structure to M. as follows: Let Y„ be the symmetric group on n letters. An element o- E .99„ acts on M" by: C(P1 ,
-
••
,P.) = (Paw, • • • 'Poo).
As point sets M. and M"/Y„ can clearly be identified. We shall show that much more is true. A function f on M. is said to be continuous (holomorphic) provided f o p is continuous (holomorphic) on M", where p:M" M„ is the canonical projection. This gives a natural topology and complex structure to M„. To see that with this structure M„ is a complex manifold, ,P„) E M" and let zj be a local coordinate at P. Define the elelet (P 1 , mentary symmetric functions of zi :
= ( — 1) E
= ( _ 1)2 E
ZiZk Pck („ = ( - 1rZi • :in .
Notice that C.; is (locally) a holomorphic function on M. and that the ordered set {C I , ,C„} determines the unordered set {z i, ,z„} uniquely as the roots of the polynomial z" + C i f + • • • + C„ = O. Hence = is a local homeomorphism of a neighborhood of p(P 1 , 'P.) E M„ and ,P„) E M„. thus serves as a local coordinate at p(P 1 ,
111.1 1.
More on
139
the Jacobian Variety
Another useful set of local coordinates on M„ is obtained by considering the elementary symmetric functions of the second kind : t„
k = 1,
, n.
j=1
An easy induction argument shows that for each k, 1 k n, {t 1 ,. ,tk } uniquely determines and is uniquely determined by {CI, :k 1 • We remark that with the choice of local coordinates made, p is obviously a holomorphic map (between complex manifolds) and that M„ is compact if and ably if M is. We Mlle established the following • • >1
Proposition.
If M is a compact Riemann surface, then M,, can he given a unique n-dimensional complex structure so that the natural projection p: M" M is holomorphic. Further, if Visa complex manifold, and the diagram
M"
V
■
M„ commutes, then f is holomorphic if and only if fi is. Remarks
1. Let f be a holomorphic function in a neighborhood of 0 in C. Thus .0
f(z) =
E
akzk.
k=0
,z„) = For (z 1 , ,z„) e CI with 14 sufficiently small F(.7 1 , f(z i) defines a symmetric holomorphic function in a neighborhood of the origin in C. On the quotient space (C"1,99 we can write n
F(z,,
,z„) = k=0
j=1
and conclude that in terms of the local coordinate (t 1 , OF at
1 dif dZi
,t„) on C'115' „,
= a.. z=0
2. If D = P 1 • • • P„ e M„ and the Pi are distinct, we can identify a neighborhood of D in M,, with a neighborhood of (P1 , ,P„) E Mn. Thus the symmetric group and symmetric functions enter only at those points D E M„ which contain non-distinct entries. It is also clear from this remark that in general we can describe a neighborhood of D by considering blocks consisting of the distinct points of D listed according to their multiplicities.
140
in Compact Riemann Surfaces
111.11.10. Let b be a holoinorphic 1-differential on J(M). Let a E AM) and set cps(P) = cp(P) + a, P e M. Then S' = 9:6 is a holomorphic 1differential on M, and is independent of a. Thus there is a canonical identification via cp of the space of holomorphic 1-differentials on M and J(M). Let Q1,. , Q,, be distinct points on M and let 4 be a local coordinate vanishing at Qk. Assume s 15_ n, and let m i , , ms be positive integers with
E
m
= n.
I
Consider the map
: M,, in a neighborhood of the n-tuple A made up of elch Qs appearing ink times. Let 6" be 3 pulled back to M,,. A neighborhood of the point A consists of points Qi, • • • , Q,1 2 , •••,Qi, s • • • Qfn,,
Q1, Q1, • • .
with Q, near Qk . Consider the point = Q1, • • • , Q1, Q2, ••• Q2' Q3' • • ,
Qs,
•••,
Qs 6 Mn•
ms-times mi - times m2-times A local coordinate (on Mil vanishing at this point is therefore given by (z11, • • • >Zino., Zi2, • • • ,7m 2 2, • . •
where zik(Q) = zk (Q),
j = 1, . . . , mk,
k = 1, . . . , s.
Let h,,(4) be a local primitive of 3' near Qk on M. Without loss of generality h,,(0) = 0. The sum
E
mkhk(zk)
k= 1
defines a function on M„ whose differential is 3". (To see this recall the definition of the pullback of a differential. The differential can be written as df with f a linear function on Cg. Now hk is simply f o cp. The differential of f ocps is 3", where 9„ is (I) viewed as a map of M„ into J(M). Since f is linear, f (9 JP • • P.)) = f(9( 13 1) + • • • + (p(P.)) = .1((P(P 1)) + " • + f(T(P s)).) Assume now that .5" vanishes at the point A. Since the pullback of 3" to M" vanishes at we must have dh k/dz k = 0 at zk = 0 for k = 1, . . . , s. We need more accurate information about the differential 6' in the case that at least one mk > 1. In this case, the function hk(zu) + • • • + hk(z.,k)
(11.10.1)
can be expressed as a power series in the elementary symmetric functions of the second kind of the Ink variables z1 . If we write hk(zk) =
E
v=1
141
111.11. More on the Jacobian Variety
we see that the coefficient of the /-th symmetric function of the second kind in (11.10.1) is precisely a,. We see therefore that
6" =
d(
E
Mkhk),
k= 1
vanishes at A = • Qs?". if and only if dhk vanishes at zero to order m„. Thus the differential 3' vanishes at Q,, to order > mk . We have thus proved the following. Lemma. If 6 is a holomorphic 1-differential on J(M), lose pullback to M„ via themap 9: M n JeVf) vanishes at a preimage of a point U E 147„, corresponding to an integral divisor D of degree n on M, then the holomorphic differential 6' on M which corresponds to 6 under the canonical map 9:M J(M) has the property that WO is integral.
111.11.11. Before proceeding we define a filtration of M,, that is analogous to the filtration of W„ by W;, described in 111.11.8. Let
M;, =
{D e Mn ; r(D -1 )_ r +
1).
(We abbreviate M,? by M„). It is obvious that
M; = (P -1 (Win. Proposition a. Let De M„. The Jacobian of the mapping 9:M„ W„c J(M) at D has rank equal to n + 1 — r(D -1 ). b. The fiber 9 -1 (9(D)) of the mapping is an analytic subvariety of M„ of dimension y = r(D -1 ) — 1, which can be represented as a one-to-one analytic image of P. c. The mapping 9:M„ W„ establishes an analytic isomorphism between M„',114,1, and W„\W l .
PROOF. Let us examine the differential d(pD of the analytic mapping 9:M „ —■ = QV • • Q'sn' (where m 1 + • • • + m, = n). It is, of course, a linear mapping between tangent spaces J(M) at a point D
d90 :TD (M„)-- ■
Let us denote by Z a local coordinate vanishing at Qk and let Ci, the j-th normalized differential of first kind, have the power series expansion
E (
arzt dzk
I= 0
in a neighborhood of 12,,. The matrix which represents d90 is then 11
11 12 am,_ tao
gl
91 a mi
• •
a° [a0
-
1
12 mz -
1 • • • ak
142
III Compact Riemann Surfaces
or interpreting the number ce in the obvious manner the matrix can be written (up to certain obvious constants) as ; i(Q 1) • • •
,;(1"" — n (Q I)
• • • C i( Qs) • • •
,',11"' "(QJ]
[i9 02 1) — • C (',"' —1) (Q1) • • • C 9 02 s) • • • C!',"' — 1) (Q s) •
(1" 1.1)
The rank of the above matrix is g — i(D), which by Riemann-Roch equals
n + 1 — r(D - I ). Thus (a) holds. Let Di E Ai n A1,./ = 1, 2. If cp(D,)= 9(102), then by Abel's theorem D i/D, is principal. Hence D, = D2 unless r(D2-1 ) > 2. The latter would imply that D2 E M. We have thus established (c).
Since 9: M. --* AM) is analytic, the fiber tp '((p(D)) is a subvariety of M.. Thus it suffices, in order to establish (b), to produce a one-to-one surjective holomorphic mapping tli: P' —•. (p'((p(D)). By Abel's theorem, 9 -1 (9(D)) consists of integral divisors of degree n equivalent to the divisor D. Let D' be an arbitrary integral divisor of degree n equivalent to D, then D'/D is principal and we conclude that every such D' is of the form (f )D for some f e L(D'). Let {f0 = 1, fi , ... ,f, I be a basis for L(D'). Send the point 0 # (co ,c i , . . . ,c„) = c onto the divisor P(c)= D(f) with f =Y:i=.,, cif/ . Note that if P(c) = .1, to, then (f)/(j) is the unit divisor and' is a constant multiple of f. Thus we have produced a well defined one-to-one surjective mapping W:P v --■ 9 -1 (9(D)) c M..
It remains to show that this mapping is holomorphic. Let us write D=
E
fl PatP),
Pe Al
cc(P) = n.
Pe M
Note that W(c) = 11 pord, + * 11) . PEM
Fix a point 0 # (cg, 4) = C° E Cv +1 . Let (co, ... ,c„) = c be sufficiently close to c°. Let f ° = Let { P1, ,P,,} be a list consisting of the (distinct) zeros off ° and the (distinct) points in D. Let 7. be a local coordinate vanishing at pi defined in the closure of U. The image of c ° in M. is, of course, (t i , ,t.) = (0, ,0). (We are using the symmetric functions of the second kind as local coordinates on M. at P(c °).) If c is sufficiently close to c° the function
E;., 0 cm.
f=
E j= 0
will have zeros only close to the zeros off ° or close to the distinct points of D. To see that this is the case, we first observe that if P is a zero of f° of multiplicity m then f has m zeros in a neighborhood of P. It is, however, possible that new zeros are introduced which are not in these neighborhoods. We claim that the new zeros must be in neighborhoods of the distinct points of D. (Delete from M the open neighborhoods of the distinct zeros of f ° and the
143
111.1 1. More on the Jacobian Variety
distinct points of D. On m\ur,,, (4, If°1 is bounded and bounded away from zero. For c sufficiently close to c° the sanie is true for f. Hence the only place the zeros of f can be are in the sets Uk . k = 1, , p.) We compute P(c) in tcrms of the local coordinates =
E
j= 1,
k=0
Clearly
1
,A r
zk (c)= — L , 2ni k=1 da" f(=k) t= and we have ti as a comptex analytic function of c. ti(c)=
E
n.
zikdzk,
Remarks
1. We saw in the proof that the rank of the Jacobian of ço at D E M. can be written as g — i(D). 2. A "generic" (see 111.6.5) divisor D E M„, n < g, has index of specialty i(D)= g — n. Thus W. for n g has dimension precisely n (as expected), by Remark (1). 111.11.12. By the inverse function theorem it follows that at the image of an integral divisor D of degree n < g with r(D - = 1, there are local coordinates • , .79 for J(M) so that the points of W„ are given by the equations • = • • • = = 0. We say in this case that W„ is regularly embedded at 9(D). Conversely, if W„ is regularly embedded at a point u e J(M), then there are g — n linearly independent holomorphic differentials on J(M) whose pullbacks to M. vanish at the pre-image of u on M.. We have seen (Lemma 111.1 1.10) that such a holomorphic differential corresponds to a differential 6 on M such that (3) is a multiple of D for all D E 9 -1 (u). Now if u E W,1„ then for any Q E M there is a divisor D of degree n containing Q such that (p(D)= u. Hence the differential 6 must vanish identically on M. But then it vanishes identically on J(M). Now if g — n 1, and W. is regularly imbedded at u, there is at least one differential on J(M) that vanishes on W. and is not identically zero. We have established the Proposition. If n s g — 1, then u E W„ is a singularity of W„ if and only if
u
E 141!.
Remarks
1. The above considerations allow us to describe more intrinsically the tangent space to J(M). Let u 1 . .... u9 be the canonical coordinates on Cg. Then these also serve as local coordinates at an arbitrary point x e J(M)= Co/G. In terms of these coordinates, a natural basis for the (complex) tangent space Tx (J(M)) of the manifold J(M) at the point x is given by the vectors a/aui, , a/aug. On the cotangent space 'T(.1(M)), the
144 covectors du i ,
III Compact Riemann Surfaces
, du, provide a basis. The dual pairing Tx (J(M)) x T,I(J(M))
C,
(11.12.1)
is then given by g y a ai .---)x bi du)= (E b E ai i, ( i=i u uj „,=„1 i= 1
g
E
here ai, bi E C. Of course, every cotangent vector may also be viewed as a
holomorphic 1-form on J(M) and conversely. As we have already seen the holomorphic 1-forms on J(M) restrict to holomorphic 1-forms on W, c J(AI) and pull back via an isomorphism to elements of .fe'(M). Thus the cotangent space to the complex manifold J(M) at any point x is naturally identified with the abelian differentials of the first kind on M. We return briefly to the differential &pp discussed in 111.11.11. The image of &pp is spanned by the column vectors of the matrix (11.11.1). The linear subspace of T:(D)(J(M)) dual to the image of dcp, consists (by definition) of those cotangent vectors (E5= 1 bi dui) that annihilate the image of &pp under the pairing (11.12.1); in other words, of those covectors ( du i + • • + bg clug) which are annihilated by the transpose of the above matrix. Under the natural identification of T:(D)(J(M)) with yew), these covectors correspond to those co E yet(t4) with (w) > D. 2. If V c J(M) is any analytic subvariety, then R(V) denotes the regular points of V; that is, those points I; E V at which V is an analytic submanifold of J(M). The remaining points are called singular, and the set of singular points is denoted by S( V). To any point x E V, there is associated a linear subspace T:(V)c T(,I(M)) spanned by those covectors of the form dfx , where f is any analytic function in an open neighborhood of the point x in J(M) which vanishes (identically) on V. The natural dual to the subspace Tt(V) is a linear subspace Tx(V)c T x(J(M)) called the tangent space to the variety V c J(M) at the point x, and dim Tx(V) is called the embedding dimension of the variety V at the point x. The embedding dimension is the dimension of the smallest submanifold of J(M) which contains the intersection of V with a small neighborhood of x in J(M). It is thus clear that the embedding dimension of x e R(V) is precisely the local dimension of x at V. The proposition preceding these remarks has shown that W„ has embedding dimension g precisely at the points of W,1, for n < g — 1. (Hence for n g as well.)
III.11.3. We introduce now some useful operations on subsets of J(M). If S c J(M) and a E J(M), then we set S + a = ts + a; ses), —S = {—S; S E
111.1 1.
More on the
Jacobian
145
Variety
If S, Tc J(M), then we set SQ T= u{S + a; a E T } = {s+ t; s E S, t E T}, e T= nst S — a; a c T.
A subvaricty S of J(M) is irreducible if every meromorphic function on J(M) that vanishes on an open subset of S vanishes on S. Proposition 111.11.11 has as an immediate consequence the following Corollary. For each n n.
g, I4
an irreducible subvariety of J(M) of dimension
111.1 1.14-. Proposition. For any a EJ(M), • —W9 1 a
-
a — K.
PROOF. For any integral divisor D of degree g — 1, we can find an integral divisor D' of the same degree such that DO' is canonical. Thus
cp(D) + (p(D') = K, showing that
W9 _ 1 = K — W9 _ 1 . 111.1 1.15. Proposition. Let 0 r < t < g — 1. Let a e J(M), b e J(M). Then (W, + h).=> a e (W,_, + b)•= e (— W,_,. + a).
(W, +
PROOF. First note that a E (PK, + b)-4.> a = X + b, x e
xe aE
E (—
b = a — x,
+ a). Hence the last equivalence is trivial. Also if
+ b), it is easy to see that (W, + a) c ( + h).
Assume now that (W, + a) c (141, + h). Thus for every integral divisor D of degree < r there is a D' E M, such that p(D) + a = (p(D') + b.
(11.15.1)
We need show that there is an A E M,_, such that a = ç(A) + b. Now the Jacobi inversion theorem implies that there is an integral divisor B such that a = (p(B) + b, and (11.15.1) implies that deg B :5_ t. Let A be a divisor of minimal degree such that a = q)(A) + b. It follows that A is unique, since r(A') is necessarily equal to one. If r(A")> 1, then A would be equivalent to a divisor which contains P o . Thus (p(A)= 024') = (p(A'); contradicting minimality of the degree of A. We now wish to show that deg A < t — r. We now have for every D e M,, there is a D' e M, such that 9(D) + cp(A) = (p(D'),
where deg A = s < t.
(11.15.2)
To show that s < t — r, we assume that s > t — r. We can now choose a divisor D of degree t + 1— s< r such that P o D and such that r(A 'D -1 ). 1. = 1.) Now, (We are here using the facts that deg A + deg D < g and !IA (11.15.2), r(A - 1 D - 1 ) = 1, and Abel's theorem give us AD = D'Po ; which is a contradiction since Po tt A, P 0 0 D. Hence we conclude s t — r and thus El + b. aE
III Compact Riemann Surfaces
146
Remark. The proposition shows that the sobvarieties W J(M) for t < g are as far from being translation invariant as possible. Specifically an inclusion 141, + u c Wu W, = {0 } . 111.11.16. Proposition
a. For any r, t 0 and any a, b e J(M), we have (W, + a) + (W, + h)
+ a + b).
b. For 0 r t g — 1 and any a, b E J(M), (W, + a)e (W,. + b) =
PROOF. Part (a) is a triviality. To prove (b) we use Proposition 111.11.15 to obtain u + (a — b)) (b — a) E (W,_, — U) 4=> (W,. ± (b — a)) c (W, — u)-=. (14; +h) c (W, + a— u). Assume now that u + a — b) [and thus (W, + b) c (W, + a — u)] and V e (W,. b). Thus v E (W, + a — u), or u E W, + a — v for all v (W, + b). Hence u (W, F a) C)(W, + b). Conversely, if u (W, + a) e (W, + h) then u W, + a — v for all ye W, + b or v (W, + a — u). Hence (W, + b) c (W, + a — u) or u (VV,, + a — 17). -
Remarks 1. The condition that t < g — 1 in (b) is necessary because W, = J(M) for t g. Hence for any S c J(M), J(M) for t g. 2. A special case of the proposition is also worthy of mention here:
es=
=
Wg-1
e
K/9_ 1 - u.
Wg-2 = 1,6 W9-2
Thus W1 M can be recovered from 14/9 _ 2 and Wg_ determined by W8 _ 1 and Wg _ 2.
Hence M is
111.11.17. Proposition. For any n > 0 and any r > 0, Wr„ = W„_,
e(-- W,),
whenever r < n, and
W rn = Ø whenever r > n.
(11.17.1) (11.17.2)
PROOF. Let x E W. Then x = (p(D), D E M., and r(D') r + 1. The Riemann—Roch theorem gives r(D") = n — g + 1 + i(D). This is clearly impossible when r > n. We have previously seen (Lemma 111.8.15) that u E Wr„ if and only if for every v E W, there is a if E W, such that u = u + e. Hence =
n
ye W,.
(w„_,
y).
(11.17.3)
Remarks
—(W, — u) c W,,„. 1. Formula (11.17.3) can be reformulated as u E Clifford's theorem gives a necessary condition that Wr„ not be empty.
147
111.11. More on the Jacobian Variety
The condition is that 2r < n or that r < n — r. Hence the above gives a geometric interpretation of this theorem. 2. It is an immediate consequence of the proposition (in the form of Equation (11.17.3)) that the subsets W: are complex analytic subvarieties of J(M). 3. If we define W„ to be the empty set for n < O, then (11.17.2) becomes a special case of (11.17.1), and (11.17.3) is always valid. We shall adopt this convention.
111.1 I.18. Proposition. Let 1 n g — 1 and let x = q(Q) with Q (H'
x)= (14/„_ + x) u WL. 1 .
Po . Then
(11.18.1)
PROOF. Assume u E W„ n (IV„ + x). Then
u
ço(Q i
• Q.) + 4)(Q) = (P(Pi ' • ' Thus QQ1 ' • ' Q.— PoPi • • P„. If — can be replaced by =, then Q appears among P1 ,. , P„ and we may assume Q = P„. Thus u = cp(P • • • P .1) + (P(Q) 6 (Wm— +
X).
If the two divisors are not identical, then r(P,77 • • P; > 2 and u E W , 1 . The reverse inclusion is trivial. Remark. A non-empty component of WL. cannot be a subset of W„_ + x
for all x e W. For if V is a component of WL. 1 and V c 14/„_ + x for all x E W1 , then
v.
n ( w„_,+x) = KI,.
XE W1
Now Vc W certainly implies that V + W1 c WL. I . Since V + W1 is an irreducible subvariety of J(M) contained in W +1 (it is the image of the irreducible subvariety V x W1 under the mapping (x,y)t---+x + y) that contains V, it must agree with V. Thus in particular we have W1 invariant under translations by ve V. This contradicts the remark following
Proposition 111.11.15.
III.I 1.19. Theorem a. Let r be the dimension of a non-empty component of W, 1, +1 with 1 n :5 g — 1. Then
2n — g < r < n — 1. b. A component of W2,-1, 1 < n < g — 1, has dimension n — 1 if and only if M is hyperelliptic. PROOF. Let V be a component of WL. 1 . By the above remark we can choose an X E W1 such that V± (W„_ + x). By (11.18.1), V must be a component
148
III Compact Riemann Surfaces
of the intersection W, n (W. + x). Hence dim V > 2n — g. Since W„ is irreducible, and W„ + x o W„, for each component V of the intersection, we must have dim V < n — 1. Assume now that M is hyperelliptic. Then WI is non-empty and consists of a single point x. Hence x + W„_, is a component of WL., of dimension n — 1. Conversely, assume W,1, 4. 1 has a component V of dimension n — 1. If n = 1, then WI is not empty and M is hyperelliptic. Thus assume n > 2. Let us choose a point X E W1, and a point u E R(V) c W„ n (W„ + x) that is a manifold point of the two varieties of the intersection. (This is possible since V has dimension n — 1 and each of the two singular sets + have dimensions at most n — 2.) The tangent space to at u has dimension n — I. Thus the dual cotangent space has dimension g — n + 1. Now the point u E k n (I,V„ + x ) is given by u = 49 (Qi ' • • Q + T(Q) = 9(13 • • P.) + (P0), )
where x 9(Q). Assume now that P 1 • • • P„Po is the polar divisor of a function. We can then choose Q so that neither Q nor any of the Q. appear among the Pk. The space of differentials vanishing at u on WL., is the span of those vanishing at u on PK and on W. + x, which are, respectively those vanishing at P1,..., P„ and Q1,..., 0„. However, r(P1-1 • • • P; 1 P0-1 )= r(13 1-1 • • P; + 1 implies that i(P, • - • P„Po) = i(P, • • P„). Thus every differential vanishing at P I , . , P„ also vanishes at P o . Similarly, every differential vanishing at Q ,, . . . , Q„ also vanishes at Q. Each of these spaces are of dimension g — n. Their span must have dimension g — n + 1. Thus their intersection has dimension 2(g — n) — (g — n + 1) = g — n
1.
The intersection is S2(Q, • • Q.QPI • • P„Po) = Cl(D). Now r(D - = 2n + 2 — g + 1 + (g — n — 1) = n + 2,
and hence c(D) = 2n + 2 — 2(n + 2) + 2 = O.
By Clifford's theorem, M is hyperelliptic unless D — Z. If D Z, then 2n + 2 = 2g — 2 or n = g — 2. In this case 2u = K, which has only finitely many solutions in J(M). But n 2 implies (g 4 and) dim V = n — 1 1. Thus V is a continuum and u could be chosen so that 2u K (from the beginning). There remains the possibility that D = P1 • • • P„Po is not the polar divisor of a function for every D E T - 1 (V) with D containing P o . Note that every D' E g + , is equivalent to a D E ML. 1 that contains P o . Thus for every D e cp -1 (V) there is a P c D, such that r(PD- = r(D -1 ); showing that every u E V can be written in the form u = y + y with y e 147„1 and y e W,. Thus WI must have a component of dimension > n — 2. By induction this implies the hyperellipticity of M.
111.11. More
on the Jacobian Variety
149
111.11.20. An application of the above results on the dimension of WL., yields the following. Theorem (Noether). On a non-hyperelliptic surface of genus g 3, for every g 2, the g-fold products of the abelian differentials of the first kind span the space of holomorphic g-differentials. Before we prove Noether's theorem we make the following useful observation: we can always find two elements of .Ye9, (m) that have no common zeros. Choose any oh E ye '(1t4) and let P I , , P, be its distinct zeros. If the result were not true, eery other- element of if i (M) would have to vanish at one or more of the points P ....., P,. Consider, however, S2(P,) k.) • • Q(P,.). Each set is g — 1 dimensional, so that the union is not all of Ye l(M). Thus there is an co, e lf 1 (M) which does not vanish at any of the points P 1, • • • ,
Pr•
The preceding implies that the function f = co l/a)2 gives rise to a (2g — 2)sheeted cover of the sphere with the property that for each zeCu {cc}, f "(z) is a canonical integral divisor. Since the function f is branched over only finitely many points we can even assume that the differentials oh and co 2 have only simple zeros. A similar argument applied to S2(P0) shows that we can always find two elements co l , co 2 e '(M) with precisely one common zero at P,. We shall use these results in the proof of Noether's theorem. PROOF OF THEOREM. Assume that M is a non-hyperelliptic surface. Thus there exists on M an integral divisor D of degree g — 2, such that i(D)= 2 and for each Q E M we have i(DQ)--= 1. Assume, on the contrary, that for every D e M9 _ 2, there is aQEM such that i(DQ)._ 2. Thus for all v there is an x e W, such that v + x e W, , orvE — x c ® (— W„). OE) (— W,), and it must be the case that dim 147 , But then W. 2 C g — 3. Theorem 111.11.19 implies that dim 14/1,_, < g — 3, and that M is hyperelliptic. Let cop co2 span Q(D) and let 0) 3 be any holomorphic differential which does not vanish at any of the zeros of co l co,. Our assumptions on D give that co l /w 2 is a meromorphic function on M with precisely g poles and i((co 2)/D) = 1. The function 0) 1/(0 2 has precisely g poles because co, and co2 have no common zeros other than in D. The assertion on the index of (0) 2)/D follows from the equality i((co 2 )/D) = r(D')= —1 + i(D). Let {(9 1 , be a basis for .)F 1 (M) and consider the 2g elements of 3t02(M): (0 1 0 1, • • •
WA;
w2 0 1, • • •
(0 2 09.
Note that the first g products are linearly independent and so are the last g products. Let A, and A2 denote the subspaces of d 02 (M ) spanned by these products. Now dim(A, n A 2) = dim A, + dim A2 — dim(A, v A 2), and dim(A, n A 2) = I. To verify the last assertion write (coi) = Dpi,j = 1, 2, and note that D, and D2 are relatively prime integral divisors of degree g
150
In
Compact Riemann Soi faces
(with (co 1 /co 2 ) = D 1 /D 2 ). Now if ti eA l n A 2 , then rr= (icop j = 1, 2, with /co In pare Yel(m) (these ‘`,"; need not be normalized). Thusco 2t,l 1,-2. ticular, (C,) = /5D 2 and () = [W I for some 15 E 2, Since i(D 1 ) = 1 = we must have C,/co, and L:2/co 1 E C and A, n A2 is spanned by (.0 1 0) 2 . Thus dim(A, v A 2) = 2g — 1, and the 2g products span a subspace of 02(M) of dimension 2g — 1. We now adjoin w 3 0 1 , (9 309 to our list, and denote the subspace of 2 (M) spanned by these g linearly independent products by A3. Once again
dim((A, v.4 2 ) n A 3 ) = dim(A, v A 2 ) + dim A 3 — dim(A, = 2g — 1 + g — dim(A v A2 V A 3).
V A2 V A3)
We show that dirnt (A, v A 2) n A 3) = 2. and in fact the space in question is spanned by w 1 w 3 , co,w 3 . Note that if i) e(.4 1 v A,) n A3, then q = (:30) 3 = co, ± ," 2a) 2 . Since the right-hand side vanishes at D, it follows that (C 3) = DD. Thus 4:3 = X1011 ± X2(02 (with Xi E C) and the dimension is 2, as claimed. Thus dim(A, v A2 V A3) = 3g — 3 = dim y" 2(M). This concludes the proof for g = 2. For the case g = 3, we start by choosing two elements co l , (0 2 E Ye i (M) such that co, and co, have precisely one common zero at Po , and a) 3 E Ye l (M) such that (03 does not vanish at the zeros of w 1 co 2 . Let {f„ ,f39 _ 3 } be a basis of if 2 (M) with each fi a 2-fold product of elements of Ye i (M). Let A ; he the (3g — 3)-dimensional space spanned by 3. As before
dim(A n A 2) = dim A, + dim
A2 —
dim(A, v A 2).
Ifwq 1 = (0 2/12 6 A 1 n A 2 , then 1) 2 /co 1 is an abelian differential with at most a single pole at Po and therefore holomorphic at Po . Hence q 2 can be written as )1 2 = co,co with CO E air j (M). This shows that dim(A, n A 2 ). g, and thus that dim(A, v A 2) = 5g — 6. Clearly co3 does not vanish at 13 0 , while every element in A, V A2 does. Hence co3 is not in A 1 y A2. Adjoining it to our list, we have a space of dimension 5g — 5 = dim 03(M), spanned by three-fold products. This concludes the proof for g = 3. For g 4 we now can proceed by induction. We let co„ co, E OW be such that co, and co, have no common zeros. Let If1 , , x9 _ „I be a basis for Yem(M), with m > 3, composed of m-fold products of elements of ..e l (M). Let A ; denote the space spanned by f,•, " aV(2m - 1)(6— 1)• Then dim(A, n A 2) = 2(2m — 1)(g — 1) — dim(A, v A 2 ). Clearly however dim(A, n A2)= (2m — 3)(g — 1), and thus dim(A, v A 2) = (2m + 1)(g — 1) = dim °' 4 '(M). EXERCISE Recall the exercises of 111.7.5 and complete the above discussion for hyperelliptic surfaces (including the case g = 2).
CHAPTER IV
Uniformizati on
This chapter has two purposes. The first and by far the most important is to prove the uniformization theorem for Riemann surfaces. This theorem describes all simply connected Riemann surfaces and hence with the help of topology, all Riemann surfaces. The second purpose is to give different proofs for the existence of meromorphic functions on Riemann surfaces. These proofs will not need the topological facts we assumed in Chapter II (triangulability of surfaces). As a matter of fact, all the topology can be quickly recovered from the complex structure. This chapter also contains a discussion of the exceptional surfaces (those surfaces with abelian fundamental groups), an alternate proof of the Riemann—Roch theorem, and a treatment of analytic continuation (algebraic functions on compact surfaces).
IV. 1. More on Harmonic Functions (A Quick Review) In this paragraph we establish some of the basic properties of harmonic functions. The material presented here is probably familiar to most readers.
IV.!.!. We begin by posing a problem that will motivate the presentation of this section. Details will only be sketched. For more see Ahlfors' book Complex Analysis. Let D be a region on a Riemann surface M with boundary SD. Let f be a continuous function on SD. Does there exist a
DIRICHLET PROBLEM.
152
IV Uniformization
continuous function F defined on D u 6D such that j. F I D is harmonic, and
FISD = f? 1V.1.2. Let g be a real-valued harmonic function in {1z1 < p}. There exists, of course, a holomorphic function f defined on {1z1 < 19 ) with g = Re f. We may define f by Im f(z)= jz *dg,
and observe that the integral is independent of the path (because {1z1 < pl is simply connected). The Taylor series expansion of f about the origin is pre') =
E a„rnel",
r < p.
0
n=0
Without loss of generality a, E R. Hence
1 g(ren = (f(re'e) + f(ren) = a, +
1
E n1L
ianeine+ Le'
ine).
Multiplying the above by e -i" and integrating, we get
=1 —Ir 0 g (a, re) dB, 2
1 fit g(ren an= dO it 0 (reier Thus for Izi < r, we have f(z)= 17, f 2x g(ren[l + 2 =
n > 1.
E re()
tz
1
re' + z j ' g(rele) . dB 2ir 0 2 re' - z '
(1.2.1)
and g(z)= Re f(z)= — s g( re") Re w dB re - z 2n 0 r2 - 1:1 2 1 $27r 2 dO. = — g(ren .0 Ire' - z1 27r 0
(1.2.2)
The expression (r2 - 1z1 2)/Irew - z1 2 is known as the Poisson kernel (for the disc of radius r about the origin). It has the following important properties:
1z12
2. r2 d19 = 1, 2n 10 J - 21 2
Izi
0 for fzi < r.
r2 -
9-
fr-?-1 8°I>—
12 dB = 0
for 0 < I
u(K). Thus, if u is both, u = u. Since harmonization does not affect such a function u, it must be harmonic. The converse is, of course, as previously remarked, trivial.
IV.2.5. Proposition. Let u E C2(M). Then u is subharmonic if and only if du> O. PROOF. Since subharmonicity is a local property, it involves no loss of
generality to assume that M is the unit disc {z = x + iy; Izi < I}. Furthermore, du > 0 is a well-defined concept on any Riemann surface (because we are interested only in complex analytic coordinate changes). We view d as an operator from functions to functions:
az u azu du = — + —
ax 2
(3y 2.
Say du > O. Thus u has no maximum on M. (If u had a relative maximum at P,, then a2u ax2 (PO) < 0
52 u
v-oy 2- (Po)
0 or au(120).o.)
If h is harmonic on M, then tl(u + h) > 0, and, as seen above, this implies that u + h has no maximum on M. Thus u is subharmonic. Suppose now du _>. O. Let e> 0 be arbitrary. Set
v(x,y) = u(x,y) + &(x 2 + y2).
159
IV.2. Subharmonic Functions and Perrons Method
Then r E C 2(M) and „1r = 4.1u + 4E> 0. Thus, by the first part, y is subharmonic. Hence, for every K, a conformal disc on M, y< or u (K) e > noc) e(x 2 y 2 K) > u t (x 2 ± y2), ,
)
or (by letting e approach zero) u < u(K) ; that is, u is subharmonic. Conversely, say u is subharmonic and du < 0 at some point. Then also in a neighborhood D of this point. By the above, u is superharmonic in D. Hence, we conclude that such a u must be harmonic in D, and we arrive at the contradiction du = 0 in D. IV.2.6. A family 97 of tubharmonic functions on a Riemann suface M is called a Perron family (on M) provided: • is non-empty, (2.6.1) for every conformal disc K c M and every u e 3F, there is a y E Y such that v I K is harmonic and y > u,
(2.6.2)
for every u1 e ,F and every u2 E 37, there is a E ..9-7 such that y > maxtu 1 ,u 2 1.
(2.6.3)
and
Remarks 1. In most applications the functions y satisfying (2.6.2) and (2.6.3) will be u( K) and max{u 1 ,142 }, respectively. 2. If F is a Perron family on M, if K is a conformal disc in M, and if u; E 37-", j = 1, , n, then there is a V e Y such that viK is harmonic and y = 1, . , n. Theorem (Perron's Principle). Let 3-47 be a Perron family and define
u(P) = sup v(P),
P M.
ve..*
Then either u
+ op or u is harmonic.
PROOF. Cover M by a family of discs {DJ. If we have the theorem for discs, we have it for all of M. We claim that if u is harmonic on one disc in the cover, say D i, then u is harmonic on all discs, Let D, be another such disc. Since M is connected we can find a chain of discs
D 2 ,D 3 ,
,D
1 =D
with Di n D + 1 non-empty for j = 1,
(2.6.4)
, n.
By the theorem for discs u I Di is either harmonic or E + oo. Since we have (2.6.4), it is impossible for uD 1 to be harmonic and ul D,1 to be + oo. Thus we may assume K = {z
C; izi
Pm,
and vrni,
v, +1
all m < n + 1, all I < n + 1.
We now observe that vn(zi)
yik (z,),
for n
k j,
and thus
lim v(z) = sup v(z) = u(.7;). ra-.33
Assume u + cc. Without loss of generality we assume that u(z i ) < + cc. By IIarnack's principle W = lim yk (2.6.5) is harmonic in K. We must verify W = u. Since Vk E 3(7, vk < u, and hence W < u by (2.6.5). Further W = u on a dense set (we do not know yet however that u is even continuous). Thus W > y on a dense set for all y E .F. Since all u are continuous, W > u for all V E F. Thus, W > u on K.
Remark. The above proof also showed how to obtain the function u. Let K be an arbitrary compact subset of M. Cover K by finitely many discs {K; = 1, . . ,n}. For each j, there is a sequence of increasing harmonic functions {VA } such that
u = lirn Vi,, on Choose
uk E
K
such that yk max v1,,; j = 1, ... ,n}. Then u = lirn vk
uniformly on K. In general, the functions yk are only subharmonic on K (that is, not always harmonic). IV.2.7. We now return to the Dirichlet problem introduced in IV. 1.1. We take a region D with boundary SD on a Riemann surface M.
161
IV.2. Subharinonic Functions and Perron's Method
Let P e D. We shall say that # is a barrier at P provided there exists a neighborhood N of P such that
/3 E C(CI(D n N)),
(2.7.1)
fl is superharmonic on Int(D n N),
(2.7.2)
I3(P) = 0,
(2.7.3)
/3(Q)> 0 for Q P, Q E CHD n N).
(2.7.4)
and Rentark.rif P e SD can tte. reached by an analytic arc (= image of straight line under holomorphic mapping). with no Points in common with Cl D, then there exists a barrier at P. PROOF. This is a local problem and so we may assume that P can be reached by a straight line with no points in common with D and D c C. Furthermore, we may assume P = 0, and the line segment is y = 0, x < O. Choose a singlevalued branch of „G in the complement of this segment and set I3(z) = Re Writing z = re', we see that 13(z)= r112 cos 0/2 with — n < 0 < it, and is thus a barrier at O. We need a slight (free) improvement. Let fl be a barrier at P E S'D with N (as in the definition) a relatively compact neighborhood of P. Let us choose any smaller neighborhood N o of P with CI No c Int N. Set
m = min (/3(Q); Q
E (CI(N\No) n Cl D)} >
O.
Then set
imin{m4Q)},
QeN n D, Q E Cl(g,N).
0, #(Q) = 0 if and only if Q = P, and 13- is Then /7 is continuous on D, superharmonic on D. Further ifini is again a barrier at P with Film = 1 outside N. (It is defined on all of Cl D.) We shall call 137m a normalized barrier at P. IV.2.8. Let D c M. A point P e SD is called a regular point (for the Dirichlet
problem) if there exists a barrier at P. A solution u to the Dirichlet problem for a bounded f e C (Ô D) is called proper provided
inf {f(P); P E 6D)
u(Q)_ sup{ f(P); P e SD}, all Q eCI D.
Theorem. The following are equivalent for D c M: a. There exists a proper solution for every bounded f b. Every point of SD is regular.
E
C (5 D).
PROOF. (a) (b): Let P e SD. It is easy to construct an f E C (5 D) with f 1 and f(Q)= 0 if and only if Q = P. Let u be a proper solution to the Dirichlet problem with boundary value f. Then 0 < u < 1. We claim
0
162
Iv Uniformization
u > 0 in D. Otherwise, u 0 in D (by the minimum principle) and f Since u is harmonic, it is superbarmonic, and thus a barrier at P. (a): Let (b) = v e C(C1 D); u is subharmonic in D, sup f = in 1 and v(Q) m o = inf f for all Q e ,
O.
f(Q)
Clearly the function which is identically m o is in F. If u,, u 2 E F, then max{u 1 ,u2 } E F. Also for every u e 5 and every conformal disc K in D, u(K) e F. Thus 5 is a Perron family on D. Let
u(Q) = sup v(Q),
Q E D.
ve.F
Then u is harmonic in D and m o u < mt. We verify two statements for P e (5D: 1. lim inf u(Q) f(P), and 2. lim sup u(Q) f(P). Q
PROOF OF (1). If f(P) = mo , there is nothing to prove. So assume f(P) > trio . Choose e> 0 such that f(P) — e > m o . There is then a neighborhood N(P) such that
f(Q) f(P) — &
for all Q E N(P) n SD.
Let # be a normalized barrier at P which is 1 outside N(P). Set w(Q) = — (f(P) — mo — 6)P(Q) +f(P) — e,
Q E CI D.
Clearly, 11? E C(CI D) and w is subharmonic. For Q E Cl D,
w(Q) f(P) — a < mi, and
w(Q) = mofl(Q) + (f(P) — 0(1 — I3(Q)) moll(Q) + mo( 1 — fl(Q)) = mo. Finally, for Q E SD,
w(Q) = mo f(Q),
Q N(P),
and
w(Q) f(P) — & f(Q), We have shown that w E F. Hence, w(Q)
Q E N(P). u(Q), all Q E D, and thus
lim inf u(Q) w(P) = f(P) — e. Since & is arbitrary, (1) follows.
163
IV.3. A Classification of Rieinann Surfaces
PROOF OF (2). There is nothing to verify if f(P) = in,. Assume f(P) < m,. Choose e > 0 so that f(P) + e < in s . Choose N =- N(P), a relatively compact neighborhood of P, such that
f(0). f(P) + e
for all Q e N(P) n D.
Let y cF be arbitrary. We claim
r(Q) — (m, — f(P) — e)fl(Q) f(P) + E,
Q
E
N(P) n D.
(2.8.1)
(Here # is again a normalized barrier of P, that is F_-- 1 outside N(P).) Since the function on the left of the inequality is subharmonic, it suffices to check the inequality on ii(N D). If Q e iS(N D), then either (i) Q e 6N n Cl D or (ii) Q E Cl N n 6D. In case (i), the left-hand side of (2.8.1) satisfies
=r(Q)
nil +f(P) + e
f(P) + e.
In case (ii) we have the estimates (for the same quantity)
v(Q)
f(Q) f(P) -F e.
We have thus verified (2.8.1). Thus for Q E N(P) n D
v(Q) fiP) +
(M1 — f(P)
6)13
and hence also 1 4 0)
.f(P) + & +
f(P)
From this last inequality we conclude him sup u(Q) 5_ f(P) + e. Since e can be chosen arbitrarily small, we have (2).
IV.3. A Classification of Riemann Surfaces In this section we partition the family of all Riemann surfaces into three mutually exclusive classes: compact (=elliptic), parabolic, and hyperbolic. The partition depends on the existence or non-existence of certain subharmonic functions. It will turn out (next section) that each of these classes contains precisely one simply connected Riemann surface. Perhaps of equal importance is the fact that this classification also enables us to construct non-constant meromorphic functions on each Riemann surface. The constructions in this section differ in a few important respects from the constructions in 11.4. We do not need here the topological facts that were previously used (triangulability of surfaces and existence of partitions of unity), and we get sharper information about the meromorphic functions that we construct (see, for example, Theorem IV.3.11).
164
IV 1;niformization
IV.3.1. Lemma. Let K be a compact connected region on a Riemann surface M, with K M. There exists a domain D on M such that K D,
(311)
Cl D is compact,
(3.1.2)
6D consists of finitely many closed analytic curves.
(3.1.3)
and PROOF. If M is compact the lemma is trivial. Choose P K and a small disc U about P in NI\ K. In the general case, every P E K is included in a conformal disc. Finitely many such discs will cover K. Let D, be the union of these discs. By changing the radii of the discs, if necessary, we may assume that the discs intersect only in pairs and non-tangentially. Delete from D, a small disc U in D i\K and call the resulting domain D2. Soke the Dirichlet problem with boundary %aloes I on 6L'. 0 on 6D 1 (note !ii.zt i), , D 1 Cl U has a regular boundary). Let D be the component containing K of
1P E D2; u(P) > e > 01, where e < min tu(P); P e K1. The critical points of the harmonic function u (the points with du = 0) form a discrete set. By changing e we eliminate them from 6D. Thus, the domain D satisfies (3.1.1)-(3.1.3).
IV.3.2. Let M be a Riemann surface. We will call M elliptic if and only if M is compact ( = closed). We will call M parabolic if and only if .11 is not compact and M does not carry a negative non-constant subharmonic function. We will call M hyperbolic if and only if .11 does carry a negative non-constant subhartnonic function. Remark. It is obvious that a hyperbolic surface cannot be compact (by the maximum principle for sulpharmonic functions), and thus we have divided Riemann surfaces into three mutually exclusive families.
IV.3.3. Subharmonic functions on a parabolic surface satisfy a strong maximum principle.
Theorem. Let D be an open set on a parabolic surface M. Let u E C(C1 D). Assume u is subhartnonic in D. Furthermore, assume there exist m l , m2 E R such that u < m 1 on SD and u < m2 on Cl D. Then u m l on Cl D. PROOF. Assume m2 > m, (otherwise there is nothing to prove). Choose e, 0 < e <m2 — m i . Define V
=
{max f,u, m, + el — m2
+e—m2
in D, m, in M\D.
It is clear that y is subharmonic on M, y < O. Thus y is constant; that is,
maxtu,:n i + el = m, + e in D.
165
1V.3. A Classification of Riemann Surfaces
Hence
u < m,
e,
and since e is arbitrary, we are donc. Corollary 1. Let u be a bounded real-valued harmonic function on an open set D on a parabolic surface M. Assume that u E C(Ci D),with m o = inf{u(P);P e SD} and m 1 = sup {u(P); P E SD}. Then m, u m (on Cl D). Corollary 2. On a parabolic Riemann surface, the Dirichlet problem has at
most one bounded solution. IV.3.4. On hyperbolic surfaces we do not, in general, have uniqueness of solutions to the Dirichlet problem. To see this we first establish the following Theorem. Let M be a hyperbolic Riemann surface and K a compact subset
with b(M\K) regular and M\K connected. Then there exist a function E C(CI(W,K)) such that i. to is harmonic on M\K, = 1 on (5(M\K), and iii. O <w < 1 on M\K. Remark. We will call the smallest co as above, the harmonic measure of K. PROOF OF THEOREM. Let 4/ 0 be a non-constant superharmonic function on AI with oo > O. Let mo = min(lpo l K) and tp i = tP o/mo . Then is superharmonic on M, non-constant, tP i > 0 and 0 1 1 K 1. We claim there exists P o E SK such that tP i(Po) = I. Clearly, since K is compact we can find such a Po E K. If Po E Int K, then tP i is constant on the component of K containing Po , and thus also on the boundary of that component. Thus we can find a Po e 6K at which tp, has the value 1. There exists a Q 0 E M\K with tp 1 (Q 0) < 1. Otherwise IP I 1 and since iP i(Po) = 1, this would mean that tP i is constant. Finally, we set ip = min {1,tp i }, and note that 1P is superharmonic on M,
0 < 11/
il/(Qo) < 1, 11/1K = 1.
We define
= tveC (C1(M\K)); v is subharmonic on M\K and u
l(M\K)}.
The family is clearly closed under formation of maxima and under harmonization. Thus 37 is a Perron family on M\K provided it is not empty. Clearly, —t/' e However, we have to show that .F has more interesting functions. Choose a domain D K with Cl D compact, and SD consisting of (finitely many) Jordan curves (this is possible by Lemma IV.3.1). Let vo be the solution to the Dirichlet problem on D\K with boundary values 1 on
166
IV Uniformization
(5K and 0 on SD. Clearly, 0 < v o < 1 on Cl(DK). Extend vo to be zero on M\D and observe that ro is subharmonic on Al (by the Corollary to Proposition IV.2.2). We show that vo E .f. We must verify that vo < ti/ on M\K. Note that — ill is subharrnonic on M, and ro — if( < 0 on (5D, and also on M\D. By the maximum principle vo — 5 0 on D because Cl D is compact. Define P E Cl(M\K),
co(P) = sup v(P), then V0
and in particular co is harmonic on .44\ K, and to = 1 on 6K (and also co e C(C1(M \K))). The function to is non-constant since (( ( O()) < tproo) < I. Furthermore, 0 < < 1 on C1(.11 K) and thus 0 < w < 1 on M K.
IV.3.5. Important Addition. We can obtain the harmonic measure of K with slight modification of the above argument. Set {v
E :1-7 ; V
has compact support)
and define P E Cl(M\K).
co i(P) = sup v(P), E .Y71
Note that 4r., contains ro and is hence non-empty. Furthermore, co, has all the properties of co. We claim that co l < (.75 for all C15 enjoying the properties of the theorem. Consider
ve Assume that v I M\K' = 0 for some compact set K'. Thus on 5(K",K) (since Co 0 on SK', u = 0 on 61(' ; & > v on SK). Hence Co' > v on (K'\K) k.) (M\K'), and since v is arbitrary (4) 1 on M\K. Corollary. Let M be a hyperbolic Riemann surface and D a domain in M with regular boundary such that M\D is compact. Then we have non-uniqueness of
solutions for the Dirichlet problem for D. PROOF. If u is any solution to a Dirichlet problem, then so is (1 — (o) + u, where co is the harmonic measure for M\D.
Remark. Any D as above cannot have compact closure in M. If it did, we would have a unique solution to the Dirichlet problem for D. IV.3.6. Let M be a Riemann surface and P E M. A function g is called a Green's function for M with singularity at P provided: g is harmonic in M\ {P}.
(3.6.1)
167
IV.3. A Classification of Riemann Surfaces
g > 0 in MVP}.
(3.6.2)
If : is a local parameter vanishing at P, then
g(z) + logizi is harmonic in a neighborhood of P. If is another function satisfying (3.6.1)-(3.6.3), then
(3.6.3) g.
(3.6.4)
Remark. Condition (3.6.3) is independent of the choice of local parameter vanishing at P (that is, the condition makes sense on M). For if ; is another such parameter, then (with at 0 0) C(z)=„a iz + ci2 z 2 + • • = a l z(1 +
at
z + • • -)= a I zf(z),
where f is holomorphic near z = 0 and f(0) # O. We write f(z) = ez ), with h holomorphic near z = 0, and conclude that
= logla i l + logizi + Re h(z). Since Re h(z) is harmonic, we see that g(z) + logizi is harmonic if and only if
+ logj is. A CLASSICAL EXAMPLE. Let D be a domain in C with Cl D compact and SD regular (for example (SD consisting of finitely many analytic arcs). Let z o E D. By solving the Dirichlet problem we can find a function y E C(C1 D) such that y is harmonic in D and y(z) = loglz — zol,
z e SD.
Set g(z) = —loglz — zol + y(z),
z
E
D.
Show that g is the Green's function for D with singularity at z o . (The proof of this assertion follows the proof of Lemma IV.3.8) Formulate and prove a general theorem so that this example and Lemma IV.3.8 become special cases of the theorem. EXERCISE , Cn. Let D c C be a domain bounded by finitely many disjoint Jordan curves C 1 , The harmonic measure of CJ(j = 1, . ,n) with respect to D is classically defined as the unique harmonic function co) in D that is continuous on Cl D and has boundary values k = 1, • • . , n. Show that this definition makes sense and relate this concept to the one introduced in IV.3.4. Show that co l + • • • + con = 1 and discuss the properties of the (n — 1) by (n — 1) "period" matrix Li *dcok.
P1 .3.7. Theorem. Let M be a Riemann surface. There exists a Green's function on M (with singularity at some point P E M) if and only if M is hyperbolic.
168
IV Uniformization
PROOF. Let g be a Green's function on M with singularity at P. Let m> 0. Set f = min {m,g}. Then f is clearly superharmonic in M\{P}. Since g(Q) + oo as Q P, f ni in a neighborhood of P. Thus f is super harmonic on M. Also f> O. If f were constant, then g > in, and thus g — m would be another candidate for the Green's function with g — in < g. This contradiction shows that f cannot be constant and hence M is hyperbolic. Conversely, assume that M is hyperbolic. Let P e M. Let K be a conformal disc about P with local coordinate z. Set
= {y; y is subharmonic in MVP), 0, y has compact support, and v(z) + log1:1 is subharmonic in Izi < 11. To note that F is non-empty, we define
vo(z) = { —
log1:1, 0,
0 < 1:1 < 1 1z1 1,
and observe that vo e F. It is easy to see that F is a Perron family (on
M \ {P}). We establish now the following Lemma. Outside every neighborhood of P, .F is uniformly bounded. PROOF OF LEMMA. Choose r, 0 < r < 1. Let a), be the harmonic measure of {1z1 r}. (It is only here that we use the fact that M is hyperbolic.) Thus a), is harmonic on Iz1> r, a), = 1 on 1z1 = r, and 0 < a), < 1 for 1z1 > r. Let A, = max{a),(z);1z1 = 1 } . Thus 0 < ;., < 1. For u e F,let it, = max{u(z); z = r). We claim r. (3.7.1) urco„ — u 0 for Clearly urai„ — u is superharmonic on 1z1 > r, and u,a), u 0 on 1z1 = r. Let .Y/' be the support of u. Then u = 0 on , and thus (3.7.1) holds on Hence (3.7.1) holds on AP\ {1z1 r} by the minimum principle for superharmonic functions. Inequality (3.7.1) obviously holds on MVC. Finally, u(z) + logizi is subharmonic in 1z1 < 1 and continuous on 1z1 _5 1. Thus
syr.
up + log r = max u + log r 5 max u + log 1 = max u
1=1= r
=1
ur).,
Iz1 = 1
(the last inequality is a consequence of (3.7.1)). Thus
U,
0. For each Q e iSD, there exists a neighborhood UQ of Q in Cl D\{ Q0} such that gr, < e on UQ . Set U=
U U.
QeSD
Then on D\U we have — gp> —e since we have this estimate on S(D\U)= SU. In particular, j(Q0) — gD(P,Q0) —E. (Intuitively ge, is necessarily the right choice since it has the smallest possible value on SD, namely 0.)
IV.3.9. Lemma. In addition to the hypothesis of the previous lemma, assume that SD consists of closed analytic arcs, and P. Q E D. We have g v(P,Q) = gv(Q,P), all P, Q e D. PROOF. Let U1, U2 be two small disjoint discs about P and Q respectively. Let 13 = D\(U1 u U2). For R e 1.:0,P.Q1 set
u(R) = gD(P.R) r(R)= go(Q,R). Then by 1 (4.4.2),
0 = ffb. (u de — v 4u) =
Li„
(it *dv — v*du)
= —ib u 1 +6u 2
(u *dv — v *du).
(3.9.1)
Note that by the reflection principle for harmonic functions, we may assume that u anduare C 2 in a neighborhood of the closure of D. We introduce now a conformal disc (U 1 ) at P with local coordinate z = re'e. In terms of this local coordinate, we write u(z) = fi(z)— log r with Et harmonic in z < I. Thus
Lut u*dv — v
*du =— log r) *dv — y*d(ii — log r)
Lui
= f6u rt*dv — v*dfi — =
ui 2
= f v(rel°) o
Similarly,
fatf2
—v
(log r)*dv — v*d(logr)
1 — Lui (log r)*dv — v(re i9)- rd0 r
dB = 27tv(0)
27rg0(Q,P).
u*dv — v*du = —2ngD(P,Q).
171
IV.3. A Classification of Ritrnatm Surfaces
IV.3.10. Theorem. If M is hyperbolic, then g(P,Q)= g(Q,P), all P, Q e M.
PROOF. Fix a point P E M. Consider the collection 3 of all domains D c M such that 1. P E D,
2. Cl D is compact, and 3. 6D is analytic. Note trat if Di e 9 (j 4.'1,2), then there is also a D E 9 with D i U D2 c D. Extend each of the Green's functions go (P,•) to be identically zero outside D. Let = {ga(P,•); De f2}. The last corollary in IV.2.2 shows that the functions in F are subharmonic on MVP. Let K be any conformal disc on M\ {P}. Let D E 9. Choose D* e 3 such that D* c D t.) K. Thus ga.(P,•) ge,(P,•) and ga.(P,•)lInt K is harmonic. Similarly, if D i , D2 e 9, we can choose D E with D D 1 U D2 and observe that gb(P,•) goi(P,•), j = 1, 2. Thus the family F is a Perron family. By Perron's Principle y(Q)= sup fgp(P,Q)1,
Q E M \ (,11
DE 9
is harmonic on MVP}, since g(P,•)
shows that 7(Q) < + cc, all Q e M \ {P } . The last inequality also shows that y
g(P,•)
on MVP}.
Since y is a competitor for the Green's function (it clearly satisfies (3.6.2) and (3.6.3)), y g(P,•) on M \{P}. We have thus shown that 7(Q) = g(P,Q) = sup {g D(P,Q)},
Q e M.
Deg,
Hence for Q, P e D go (P,Q) = gp(Q,P) g(Q,P).
Now fixing P and taking the supremum over D E 9, we see that g(Q,P).
Reversing the roles of P and Q gives the opposite inequality. Thus the proof of the theorem is complete.
172
IV Uniforrnization
EXERCISE
Show that the Green's function is a conformal invariant; that is, if f :AI --■ N
is a conformal mapping and g is the Green's function on N with singularity at f(P) for some P e M, then g 0 f is the Green's function on M with singularity at P. DIGRESSION. Let D be a domain on a Riemann surface M with SD consisting of finitely many analytic arcs. Let us generalize the Dirichlet problem and (recall 11.4.5) our earlier discussion of the Dirichlet principle. Let f? be a 2-form which is C2 on a neighborhood of the closure of D. (We shall show in IV.8 that there is a Coe 2-form flo on M that never vanishes. Thus Q = FS20 for some function Fe C2 (C1 D).) Let f e COD). We want to solve the boundary-value problem for u E C 2 (C1 D):
= 14
1 6D
=
f.
Let Po e D be arbitrary and assume our problem has a solution u. Let us take a conformal disc z centered about Po and let D, = DV1z1 r},
0 S 1 .
(4.2.2)
In case (b), (4.2.1) is replaced by (Pi = p + i(c) with x(c)e R, and (4.2.2) is replaced by a homomorphism CI Corollary. Let M be a Riemann surface with H i (M). r,01. Let a- be a discrete subset of M, and u a harmonic function on M cr. If u is locally the real part (respectively, the log modulus) of a meromorphic function on M, then u is
globally so. IV.4.3. As an application of the above corollary we establish the following Proposition. Let D be a domain in the extended plane C u {co} = 5 2 with H 1 (D) = {0}. Then D is hyperbolic unless S 2\D contains at most one point. Remarks 1. The two sphere 5 2 is of course elliptic. If ,S 2 \D is a point, then (via a Möbius transformation) D C. The plane C is parabolic. (Prove this directly. This is also a consequence of Theorem IV.4.5.)
182
IV Li itiform izat i on
2. The proposition can he viewed as a consequence of the Riemann mapping theorem. It can also be obtained by examining any of the classical proofs of the Riemann mapping theorem. The proposition together with Theorem IV.4.4, imply the Riemann mapping theorem. PROOF OF PROPOSITION. Say S2\D contains 2 points. By replacing D by a conformally equivalent domain, we may assume that S 2\ D contains the points 0, x. We now construct a single valued branch of \!: on D. Observe that loglzi is a harmonic function on D. Thus there exists a holomorphic function f on D with logIf(s)1 = 4 logizl,
z a D.
logif(z)1 2 =
z e D,
Thus and there is a 0 e
J'(..)2 = eiez,
Set
g(z)= e -1912f(z),
z e D.
Then
g(7)2 = 7, ,
Z
e D.
Let a e D and g(a)= b. We claim that g does not take on the value —b on D. For if 0.7.) —b, then z = g(z) 2 =- 112 = g(a)2 = a. This contiadiction shows that g misses a ball about —b (using the same argument—since g takes on all values in a ball about b). We define a bounded holomorphic
function
h(z)=
1 g(z)+
z e D.
Since D carries a non-constant bounded analytic function, it is hyperbolic. IV.4.4. Theorem. If M is a hyperbolic Riemann surface with H 1(M)= { 0 } , then M is conformally equivalent to the unit disc 4.
P E M. Let g(P,•) be the Green's function on M with singularity at P. The function g(P,.) satisfies the hypothesis (a) of Theorem IV.4.2. (The only issue is at P. Let z be a local coordinate vanishing at P. Then g(P,z) + logizi = v(z) is harmonic in a neighborhood of z = O. Choose a harmonic conjugate v* of v, write f = e"'. Then PROOF. Choose a point
g(P,z) + logizi = Re log f(z)= logif(z)I, near 0, or
g(P,z)= log
f(z)
, near O.
183
IV.4. The Uniformization Theorem for Simply Connected Surfaces
By Corollary 1V.4.2, (applied to - g(P,-)) there exists a meromorphic
function, f(P,•) on M such that
1ogif(P,Q)1 = - g(P,Q), all Q e M. The function f(P,.) is holomorphic: For Q P,g(P,Q) > 0, thus I f(P,Q)I 0. 2
1,
and A is elliptic.. AI = 1,
;1. 0 1.
IV.6.3. Theorem. The only Riemann surface M which has as universal covering the sphere, is the sphere itself. PROOF. The covering group of M would necessarily have fixed points if
i(A1)
{ 1 }.
111 .6.4. The fixed point free elements in Aut C are of the form z Z + a, a E C. Since a covering group of a Riemann surface must be discrete, we see that (consult Ahlfors' book Complex Analysis) we have the following
Theorem. If the (holomorphic) universal covering space of M is C, then M is conformally equivalent to C, C*, or a torus. These correspond to r i (M) being trivial, and 2 e Z. Note that if 7r1 (M) Z, then we can take as generator for the covering group of M, the translation z z + 1. The covering map it: C -- ■ C* is given by n(z) = exp(2xiz).
IV.6.5. All surfaces except those listed in Theorems 1V.6.3 and IV.6.4 have the unit disc or equivalently the upper half plane U as their holomorphic universal covering space. We have shown in 111.6.4, that every torus has C as its holomorphic universal covering space (we will reprove this below). Since no surface can have both U and C as its holomorphic universal covering space, a torus cannot be written as U modulo a fixed point free subgroup of Aut U. More, however, is true (see Theorem IV.6.7).
194
IV Uniformization
Since Aut U PL(2,R), we see that each elliptic element A e A ut U fixes a point z e U (and as well). Conversely, every A E Aut U that fixes an element of U is elliptic. Thus a covering group of a Riemann surface cannot contain elliptic elements. (As an exercise, prove that a discrete subgroup of Aut U cannot contain elliptic elements of infinite order.) Since every element of Aut U of finite order is elliptic, we have established the following topological result. Proposition. The fundamental group of a Riemann surface is torsion free.
Remark. Using Fuchsian groups, one can determine generators and relations for fundamental groups of surfaces. IV.6.6. If A e Aut U is loxodromic, we can choose an element B E Aut U such that C(0) = B . A . B -1 (0) = 0,
and C(x)) = B o A o B 1 (ci) = x). Thus C(z) = Az with A e R, A > 0, A o 1. Thus Aut U does not contain any non-hyperbolic loxodromic elements, and hence the covering group of a Riemann surface (whose universal covering space is U) consists only of parabolic and hyperbolic elements.
Lemma. Let A and B be Mafia transformations which commute, with neither A nor B the identity. Then we have: a. If A is parabolic, so is B and both have the same fixed point. b. If one of A and B is not parabolic (then neither is the other by (a)), then either A and B have the same fixed points, or both of them are elliptic of order 2 and one permutes the fixed points of the other. Say A0B=B.A. If A is parabolic, it involves no loss of generality to assume A(z) = z + 1 (by conjugating A and B by the same element C). E S L(2,C), A = [(1) 1]. Thus the statement A commutes Write B = with B gives
PROOF.
[
a + c b + d] d
[a a + c c+d
in PL(2,C).
From which we conclude that c = 0, thus ad = 1 and d -= a; showing that B(z) = z + If neither A nor B have precisely one fixed point, then by conjugation we may assume A(z)= Az with A 0 0, 1. The commutativity relation now reads [Aa Ab ] = [Aa b ] . in c d Ac d
195
IV.6. The Exceptional Riemann Surfaces
There are now two possibilities: e 0 or e = O. If c the equality [ i za b] = [.2a
L
;al
0, then we must have
;Ix di .
Thus a = 0 and if we assume that 2 —1, also b = 0 (hence ad — bc o 1). Thus this case is impossible and 2 = —1. The matrix identity now reads
Fo L—c
bi _ o d
from which we concludra = 0 = d and be = —1; from which we conclude B(z) = k/z. The remaining case is c = 0. Thus d O 0 and the matrix equality becomes a ,.b1 Fa bl d i — LO cli •
r0
Thus b = 0 and B(z) = kz.
Corollary. Two commuting loxodromic transformations have the same fixed point set. IV.6.7. Theorem. Let M be a Riemann surface with ir,(M) Z G31 Z, then the holomorphic universal covering space of M is C. PROOF. Assume the covering space is U. The covering group of M is an abelian group on two generators. Let A be one of the generators. If A is parabolic, then we may assume A(z) = z + 1. Let B be another free generator. Since B commutes with A, B must also be parabolic. Thus B(z)= z + fi E RAI The group generated by A and B is not discrete. So assume A is hyperbolic (it cannot be anything else if it is not parabolic). We may assume A(z) = Iz it> L The other generator B must be of the form B(z) = pz with p > 1 and 2" # pm for all (n,m) E Z Z \ {O}. Again, such a group cannot be discrete (take logarithms to transform to a problem D in discrete modules). ,
IV.6.8. To finish the proof of Theorem IV.6.1. we must establish the following
Theorem. Let M be a Riemann surface with holomorphic universal covering space U. Then M zi zl*, or LI„ provided n,(M) is commutative. ,
PROOF. We have seen in the proof of the previous theorem that if M is covered by U and n 1 (M) is commutative, then the covering group G of M must be cyclic. The two possibilities (other than the trivial one) are the z + 1) or generator A of G is parabolic (without loss of generality hyperbolic (zi— ■ :41z, A> 1). In the first case the map U
196
IV liniformization
is given by
• expi2ni.:). For z = re', 0 < 0 < 27, we define log z = log r + i0 (the principal branch of the logarithm). The covering map (for the case of hyperbolic generator) U —3 1.1, is then given by
zl--4 exp ( 2tri
log
:
log,.
.
From this we also see that r
exp(-21r 2/log
(6.8.1)
IV.6.9. As an application we prove the following Theorem. Let D be a domain in C u (t x} such that C u f re, `•, D consists of two components ; )3. Then D _+-L- C*, 4*, or LI,. C u -.c ) is simply connected ( A C u i and thus conformally equivalent to C or 4. Thus it suffices to assume a is a point or the unit circle. Now (C u {o)\) is also equivalent to C or 4. Thus we are reduced to the case where 6D consists of points or analytic arcs. In either case, it is now easy to see that ic i (D) Z, and the result follows by our classification of Riemann surfaces with commutative tr,(M). PROOF. The complement of a (in
IV.7. Two Problems on Moduli The general problem of moduli of Riemann surfaces may be stated as follows: Given two topologically equivalent Riemann surfaces, find necessary and sufficient conditions for them to be conformally equivalent. What does the "space" of conformally inequivalent surfaces of the same topological type look like? The solution to this general problem is beyond the scope of this book; however, two simple cases (the case of the annulus and the case of the torus) will be solved completely in this section. IV.7.1. Let Mi be a Riemann surface with tr.) :
M., the holomorphic universal covering map (j = 1,2). First, recall that the covering group Gi of Mi is determined up to conjugation in Aut M1. Furthermore, if f:mi-+ M2
IV.7. Two Problems on Moduli
197
is a topological, holomorphic, conformal, etc., map, then there exists a map
1: SI, -4 .171 2 of the same type so that n 2 el= fo
1.
(Of course G., is the set of lifts of the identity map M; M;.) The map/ is not uniquely determined. It may be replaced by A2 .1. A 1 with A i E Gi. 11/.7.2. How many conformally distinct annuli are there? W have seen that every annulus can be written as U/G where G is the group generated by :i—z, A e > 1jf A l and A2 are two such annuli with the corresponding ,1 1 and A2 , when do they determirre the same conformal annulus? They determine the same annulus if and only if there is an element T a Aut U such that T(2 i z) = A2T(z), z e U. From this it follows rather easily that A i =A. (by direct examination or using the fact that the trace of a Möbius transformation is invariant under conjugation). We conclude Theorem. The set of conformal equivalence classes of annuli is in one-to-one canonical correspondence with the open real interval (1,x). Of course, we could substitute for the word "annuli" the words "domains in C u apl with two non-degenerate boundary components." IV.7.3. Let T, and T2 be two conformally equivalent tori. As we have seen, the covering group G. of Ti may be chosen to be generated by the translations z z + 1, z' •—■ z + r; with Tm r; > O. Thus we are required to find when two distinct points in U determine the same torus. Let f be the conformal equivalence between T 1 and T2 and letf be its lift to the universal covering space: •C °i
.
gll '
T2
The mappingf is conformal (thus affine: that is, z' --4 az + b), and it induces an isomorphism g e G2. G1 g We abbreviate the Mtibius transformation
z + c by c. Thus
a = 9(1) = al + [3'1. 2 with a, /3 e
at, = 0(r 1 ) = 71 + 6r2 with y, Se Z.
(7.3.1)
198
iv uraformization
n is invertible. Thus Further, since O is an isomorphism the ma trix — 137 = + 1. Considering the symbols in (7.3.1) once again as numbers, we see that y + br 2 T= n Œ + pre 2 and since both r i and T2 E U we see that fly — ctb = —1. Conversely, two es related as above correspond to conformally equivalent tori. Let F denote the unimodular group of Möbius transformations; that is, PL(2,2). It is easy to see that F is a discrete group (hence discontinuous), and with some work that U/F;1'.- C. Thus we have Theorem. The set of equiralence classes of tori is in one-to-one canonical
correspondence with the points in C.
IV.8. Riemannian Metrics In this section we show that on every Riernann surface we can introduce a complete Riemannian metric of constant (usually negative) curvature. We also develop some of the basic facts of non-Euclidean geometry that will be needed in IV.9.
IV.8.1. We introduce Riemannian metrics on the three simply connected Riemann surfaces. The metric on M will be of the form 2(z)idzi,
z e M.
(8.1.1)
We set
2(z) =
2 1+
, for M = C L.) {
).(z) = 1, for M = C, 2 2(z) = 1 _ 1z12 , for M =
= {z e C; izi < 1}.
o. The definition of A for M = C L.) {co} is, of course, only valid for z At infinity, invariance leads to the form of 2 in terms of the local coordinate = 1/z. We will now explain each of the above metrics.
IV.8.2. The compact surface of genus 0, C u {co}, has been referred to (many times) as the Riemann sphere; but up to now no "sphere" has appeared. Consider hence the unit sphere .52, ± r 2 + C2 1
199
IV.8. Riemannian Metrics
in R 3. We map .52\ {(0,0,1)} onto C by (stereographic projection) 4 ,0 1---■
+
=
1—
It is a trivial exercise to find an inverse and to show that the above map establishes a diffeornorphism between S2\{(0,0,1)} and C that can be extended to a diffeomorphism between S2 and C u {co}. The inverse is z Fie( 2 Re z
izi2 (Write x = Re z, y
2 Im z tz1 2 — 1 friz + 1 , z 12 + 1). 1
(8.2.1)
Im z.) The Euclidean metric on R3 ds z =
de 4. de + dc2
induces a metric on S2, which in turn induces a metric on C u { Do } . Equation (8.2.1) shows that
cg =
2(1 — x 2 + y 2) dx — 4xy dy (1 + 1z1 2)2
x2 — y2) dy 2(1+ 4.xy dx
du — dC —
(1 + 4x dx + 4y dy (1 + lz1 2)2 •
A lengthy, but routine, calculation now shows that
4(dx2 + dy 2) ds2 — — (1 + )zI 2)2 The curvature K of a metric (8.1.1) is given by K—
log ). 42
(here 4 is the Laplacian, not the unit disc). A calculation shows that K= +1,
and that the area of C
u {00
}
in the metric is
Area(C u loop = ffc ).(z)2 dx dy = fo2n Sox' Proposition. An element T e Aut(C u {,,0 (2/(1 + 1z1 2))1dzi if and only if as a matrix
T=
[a +? c 5
}
)
lar
4
+ r2)2 rdrd9 = 4n.
is an isometry in the metric
ic1 2
=
1'
(8.2.2)
200
IV Uniformization
PROOF. Write
ra bl Lc
di'
ad — bc = 1.
Of course, T is an isometry if and only if
il(Tz)IT'(z)1 =
ZEC u {X},
or equivalently if and only if
1 1 Cz + dj 2 + laz + 612 = 1 +
(8.2.3)
If (8.2.2) holds, then so does (8.2.3), as can be shown by the calculation
1 + 41 2 + laz — 12 = (CZ +
1
1
÷ a) + (az - --e)(ZE — c) = 1 +
Conversely, if (8.2.3) holds, we have
1+
lc: + d 2 +
+ hr.
Expanding the right-hand side gives 1
+ 1z1 2 = Oar + Ici 2)1.71 2 + ca + ab)z + (d-C +
161 2 + id1 2,
and the only way this identity can hold is if lal 2 + 1c1 2 = 1, 161 2 +1c112 = 1, + as = 0. We now add to this the determinant condition — cb + ad = 1. The last four equations yield c = —5, a =
a
Remark Note that for T not the identity Trace T = a + =2 Re a,
and (since Re a = +1 implies that T is the identity)
—2 < Trace T < 2. Hence T is elliptic. As a matter of fact, T is an arbitrary elliptic element fixing an arbitrary point z and —r
EXERCISE Determine the geodesics of the metric of constant positive curvature on the sphere.
IV.8.3. The metric idzi on C needs little explanation. It is, of course, the Euclidean metric. It has constant curvature zero. Geodesics are the straight lines; and automorphisms of C of the form
z are isometries in the metric.
+ b,
6 e R, b e C,
201
IV.8. Riemannian Metrics
IV.8.4. We turn now to the most interesting case: the unit disk 4. Let us try to define a metric invariant under the full automorphism group Aut 4. The metric should be conformally equivalent to the Euclidean metric; hence of the form ,1.(z)Id4 Set 40) = 2 and note that ;1(A0)124'(0)1 = ).(0) for all A e Aut 4 that fix 0 (since A(z)= e lez). Let z o e 4 be arbitrary and choose A e Aut J such that A(0) = z o. Define (to have an invariant metric) = 2144'(0)1 - 1 . Note that A(z) = (z + z0)/(1 + 70z) (it suffices to consider only motions of this form). From which we conclude that
._ 2
2
1
A simple calculation shows that this metric is indeed invariant under Aut 4. (In particular, the elements of Aut A are isometries in this metric.) Before proceeding let us compute the formula for the metric on the upper half plane U. Since we want the metric X(z)id:( to be invariant under Aut U, we must require
X(z) = 2IA'(0)1 -1 , where A is a conformal map of 4 onto U taking 0 onto zo . A calculation similar to our previous one shows that
1(z) = iml Computations now show that the metric we have defined has constant curvature — 1. Let c: I —* A be a smooth path. Then the length of this path 1(c) is defined by
1(c). sol .1.(c(t))1e(t)idt.
(8.4.1)
We can now introduce a distance function on J by defining
d(a,b)= inf{/(c); c is a piecewise smooth path joining a to b in J } .
(8.4.2)
Take a = 0 and b = x, real, 0 x < 1, and compute the length 1(c) of an arbitrary piecewise smooth path c joining 0 to x,
2
,
1(c) = fo 1 _ ic(t)12 ic (oldt
2 Ça > $o1 1 — = c(t)2 t = = log
1+x 1—x
log
1+x . 1—x
2 _ u2
202
IV Uniformization
We conclude tint
d(0,.x) = log
+ 1—
x
and that c(t) = tx, 0 t < 1, is the unique geodesic joining 0 to x. We now let 0 # b = zed be arbitrary (and keep a = 0). Since e
y
(0 = arg z) is an isometry in the non-Euclidean metric i(z)Idzi, we see that d(0„-..) = log
1 + IzI
(8.4.3)
1 — lzr and 00= I:, 0
t
1, is the unique geodesic joiniru0 to
Proposition a. The distance induced by the metric 2: (1 !:i 2 );ii.:1 is ef,mplete. b. The topology induced by this metric is equivalent to the usual topology on J. c. The geodesics in the metric are the arcs of circles orthogonal to the unit circle {z C; lzi = 1). d. Given any two points a, b E .1, a 0 b, there is a unique geodesic between them. PROOF. Note first that (8.4.31 shows that every Euclidean circle with center at the origin is also a non-Euclidean circle and vice-versa. Also note that
. d(0,z) lim
Izi
showing that at the origin the topologies agree. Since there is an isometry taking an arbitrary point z e d to the origin, the topologies agree everywhere. This establishes (b). We have already shown that the unique geodesic between 0 and b is the segement of the straight line joining these two points. Now let z i and z 2 be arbitrary. Choose A E Aut d such that A(: 1 ) = 0. Let b = A(z 2). The geodesic between z, and z2 is A 1 (c), where c is the portion of the diameter joining 0 to b. Since A' preserves circles and angles, (c) and (d) are verified. Now let {z„} be a Cauchy sequence in the d-metric. Then {d(0,z)} is bounded. Hence {z„) is also Cauchy in the Euclidean metric and tiza ll is bounded away from 1. This completes the proof of the proposition. IV.8.5. If D is any Lebesgue-measurable subset of d, then Area D =
1
defines the non-Euclidean area of D.
).(z) 2 1dz A dui
203
IV.8. Riemannian Metrics
The following technical proposition will be needed in the next section. Proposition. The area of a non-Euclidean triangle (that is, a triangle whose sides are geodesics) with angles a, fl, 7 is
— (a + fi + y). It is convenient to work with the upper half plane U instead of the unit disk. We will allow "infinite triangles"—that is, triangles one or more of whose sides are geodesics of infinite length (with a vertex on Ô U = 51 u If a triangle has a vertexepn 5U, it must make a zero degree angle (because the geodesics in U are semi-circles (including straight lines = semi-circle through x,) perpendicular to FI). We call such a vertex a cusp. PROOF.
(see Figure IV.1), and one right angle. The angles of the triangle are hence: 0, it/2, a. (Note we are not excluding the possibility that x = 0; that is, a cusp at B.) The area of the triangle is then
Case I. The triangle has one cusp at
dx
dy
a Jo
dx .or —
-- arc sin -
P
= 7r —
Case 11. The triangle has a cusp at co. (See Figure IV.2.) It has angles 0, a, )3.
We are not excluding the possibility that either a = 0 or /3 = 0 or both = fi = 0. Choose any point D on the arc AB and draw a geodesic "through
-P a
Figure 1V.2
204
IV Uniform ization
D and co." The area of the triangle is the sum of the areas of two triangles: 2 The case in which A and B lie on the same side of D is treated similarly. Case III. If the triangle has a cusp, then it is equivalent under Aut U to the one treated in Case II. Case IV. The triangle has no cusps. (See Figure IV.3.) Extend the geodesic AB until it intersects at D. Join C to D by a non-Euclidean straight line. The area in question is the difference between the areas of two triangles and thus equals
Figure IV.3 IV.8.6. Theorem. Let M be an arbitrary Riemann surface. We can introduce on M a complete Riemannian metric of constant curvature: the curvature is positive for M=Cu }, zero if C is the holomorphic universal covering space of M, and negative otherwise. PROOF. Let IR be the holomorphic universal covering space of M and G the corresponding group of cover transformations. We have seen that G is a group of isometrics in the metric of constant curvature we have introduced. Since the metric is invariant under G, it projects to a metric on M. Since curvature is locally defined, it is again constant on M. (The reader should verify that the curvature is well-defined on M; that is, that it transforms correctly under change of local coordinates.) The (projected) Riemannian metric on M (denoted in terms of the local coordinate z by )(z)Idzi) allows us to define length of curves on M by formula (8.4.1) and hence a distance function (again denoted by d) by (8.4.2). Let
n:
M
be the canonical holomorphic projection. Let x, y
E
M. We claim that
d(x4) = inf {d(,n); 7r( ) = x, 11(n) =
(8.6.1)
Note that we have shown that in g/-1 the infimum in (8.4.2) is always assumed for some smooth curve c. Let d be the infimum in (8.6.1); that is, the right-hand side. Let c be any curve joining to 71 in g/I- with it() = x, it(i) = y. Then
IV.9. Discontinuous Groups and Branched Coverings
205
n c is a curve in M joining x to y. Clearly, l(rr c) 1(c). Thus d(x, y) :5_ 1(n c c) = 1(c), and since c is arbitrary, Finally, since
d(x,Y) q are arbitrary in the preimage of x, y, d(x,y) 5_ d.
Next, if c is any curve in M joining x to y, it can be lifted to a curve "C" joining to q (5,g) = x, n(q) = 4),of the same length (because rr is a local isometry). Thus, 1(c) = 1(e) d(,ry) d. Since c is arbitrary, we conclude d(x,y)
d.
It is now easy to show that the metric d on M is complete. We leave it as an exercise to the reader. Remarks
I. If a Riemann surface M carries a Riemannian metric, then the metric can be used to define an element of area dA and a curvature K. The Gauss— Bonnet formula gives
ff. K dA = 2742 — 2g), if M is compact of genus g. Thus, in particular, the sign of the metric of constant curvature is uniquely determined by the topology of the surface. We will not prove the Gauss—Bonnet theorem in this book. The next section will, however, contain similar results. 2. Let M be an arbitrary Riemann surface with holomorphic universal covering space M. On gl A(z) 2 dz (1-2 is a nowhere vanishing 2-form that is invariant under the covering group of it-4 M. Thus the form projects to a nowhere vanishing smooth 2-form on M. (Recall that in 1.3.7, we promised to produce such a form.)
IV.9. Discontinuous Groups and Branched Coverings In this section we discuss the elementary Kleinian groups and establish a general uniformization theorem. We show that a Riemann surface with ramification points can be uniformized by a unique (up to conjugation)
206
IV Uniformization
Fuchsian or elementary group. This establishes an equivalence between the theory of Fuchsian (and elementary) groups and the thcory of Riemann surfaces with ramification points. We have not always included the complete arguments in this section. The reader should fill in the details where only the outline has been supplied.
IV.9.1. Let M be a Riemann surface. We shall say that M is of finite type if there exists a compact Riemann surface M such that M \III consists
of finitely many points. The genus of M is defined as the genus of M (it is well defined). By a puncture on a Riemann surface M, we mean a domain Do M with D o conformally equivalent to {Z E C; O < < 1} and such that every sequence { z„} with z,, —■ 0 is discrete on M. We shall identify z = 0 with the puncture D o . Let D be a connected, open, G-invariant subset of Q(G), where G is a Kleinian group. Let { x be the set of points that are either punctures on DIG or points x E DIG with the canonical projection n:D DIG ramified at it - '(x). Let vj = ramification index of ir -1 (xj) for such an xj, and let vi = Do for punctures. The set {xi} is at most countable, since the punctures and ramified points each form discrete sets. Let us now assume that DIG is of finite type. If in addition, it is ramified over only finitely many points, then we shall say that G is of finite type urer D. In this latter case, we let p be the genus of DIG. We may arrange the sequence xj so that
< v i < v2 < • < v„ < oo (here n = cardinality of Ix 1 ,x2 ,
< oo), and we call
(P;v1,1'2, • • • ,v,) the signature of G (with respect to D). We also introduce the rational number, called the characteristic of G (with respect to D),
1 j=1
v
,
)
where, as usual, 1/oo = 0, whenever G is of finite type over D and x = oo, otherwise.
IV.9.2. Definition. Let G be a Kleinian group. Let D c Q(G) be a G-invariant open set. By a fundamental domain for G with respect to D we mean an open subset o) of D such that i. no two points of to are equivalent under G, ii. every point of D is G-equivalent to at least one point of Cl to, iii. the relative boundary of to in D, 6w, consists of piecewise analytic arcs, and iv. for every arc c St.o, there is an arc c' c Sw and an element g e G such that gc = c'.
207
IV.9. Discontinuous Groups and Branched Coverings
By a fundamental domain for G we mean a fundamental domain for G with respect to SAG). The next proposition will only be needed to discuss examples. We will hence only outline its proof. Proposition. Every Kleinian group G has a fundamental domain.
Let it: C2 --> Q/G = U ; S be the canonical projection. Let Q; be a component of 7r - '(Si), and let G; be the stabilizer of Q./. If co; is a fundamental domain for G. with respect to f2;, then U ; co; is a fundamental domain for G. Thus it suffices to consider only the case where Q is connected. We ci-n, therefore, asstme that-we are given a discontinuous group G of automorphisms of a domain D c Cu {Do . We now introduce a construction that is very useful in studying function groups. (These are groups that have an invariant component.) Let PROOF.
}
p:b *1) be the holornorphic universal covering map of D and define F
{yeAutb;poy=gopforsomegeGI.
It is now easy to check that 5fr DIG (thus G is discontinuous if and only if F is). Furthermore, define a homomorphism p*:r --+ G
by p 0 y = p *( 7) 0 p,
y E
F,
and let H = Kernel p*. Then {1} —J1
F G -+ {1 }
is an exact sequence of groups and group homomorphisms. Now let iZ be a fundamental domain for F with respect to f It is quite easy to show that = p(c7. ) is a fundamental domain for G in D. Thus we need only establish fundamental domains for groups acting discontinuously on C u {Do}, C, or U. The first case (Cu cop is trivial (because the group is finite). The second case (C) will be exhausted by listing all the examples. If b = u, then we choose a point ze E U with trivial stabilizer and set = {z e U; d(z,z 0) < d(gz,z 0) all g e G, g 0 1). It must now be shown that co is indeed a fundamental domain. We omit (the non-trivial) details. 0 Definition. The group F constructed above will be called the Fuchsian equivalent of G (with respect to D). We now take a slight detour and on the way encounter some interesting special cases.
208
IV Uniformization
IV.9.3. Let us now classify the discontinuous groups G with 12(G) = C Since C x} is compact, G is finite (let N = la!) and M = (C u {co})/G is compact. Let n:C
{co} M
be the canonical projection. Let (7; y 1 ,. ,v„) be the signature of G. Clearly 2 < v; < N and each vi divides N. The Riemann-Hurwitz relation (Theorem 1.2.7; see also V.1.3) reads 1 (1 - -). v 1•
— 2 = N(27 — 2) + N
(9.3.1)
In particular, the characteristic of G,
2
x= — < is negative. Since (1 — 11v1) > 0 all j, we conclude that y = 0. Thus (9.3.1) becomes
2— 1
— Vj
.1=1
Assume N> 1 (discarding the trivial case). If N = 2, then vi = 2, all j and n = 2. Assume now N> 2. Then
1 and (n > 0) since
1 1 - 5 1 — — < 1, v
2
we conclude that
n < 4.
n > 1, Thus n = 2 or n = 3. If n = 2, then
1
21 = y1
;72:
Now since 1/v; 1/N it must be that v i = v 2 = N. If n = 3, then 21 1 1 1+=—+—+ 1/1
112
V3
Observe (again) that 25
1 4),
1
12 1. 2 2 N If v 2 = 2, then v 3 is (apparently) arbitrary 2. If v 2 > 2, then v 2 = 3 and hence 121 Thus 11 6 1, 3" —
v3.
Thus v 3 = 3,4 or 5. We summarize below the signatures and cardinalities of the groups G that could possibly act on Cu { x }. Signature of G
IG1
(O;—) (0; v,v) t0;2.2,v) (0;2,3,3) (0;2,3,4) (0:2,3,5)
1 y (25v 0 and Im K 0, we conclude that m = 1 and K = T (also p> 2). Since zi—■ Kz generates the stabilizer of 0, p = v3.
213
1V.9. Discontinuous Groups and Branched Coverings
Fundamental domain for {: -.+ K"z + n + mK}, K = e 2 "" (Figure IV.8)
Im z = 0 Re z=0 Figure IV.8. Signature (02,3.6) where the two Corresponds to the point z = the three corresponds to the point z = (K + 1)/3 and the six corresponds to the point = 0. Fundamental domain _An- {z 1—* K": + n + niK} K = e 2 ,a14 = i (Figure IV.9)
Im z =0 Re Z=0 Figure IV.9. Signature (0;2,4,4) where the two corresponds to z = I, one four corresponds to z = (1 + 0/2 and the last four corresponds to z = 0. + n + mK}, K = e2 ' 1" 3 (Figure IV.10) Fundamental domain for {zi—+
Im z =0 Re Z =0 Figure IV.I0. Signature (0;3,3,3) where the threes correspond to the points z = 0, z = (K + 2)/3, z = (2K + 1) 73. (Each of these three points is equivalent to a corresponding point in the figure.)
Remark. We have established above the existence and uniqueness of all the elementary groups with one limit point. We have also determined all possible signatures that yield zero characteristic.
214
IV Uniformization
IV.9.6. To classify the remaining elementary groups, we must study the groups with two limit points; without loss of generality the limit points are
0 and oc. The elements of such a group G are of the form zi--+kz
and
z
k/z.-
By passing to the Fuchsian equivalent of the group G, it is easy to classify these groups. The details arc left to the reader.
IV.9.7. Let G be a Kleinian group. We have seen that y O. In addition, we want to interpret the characteristic in geometric terms. We need some preliminaries. Lemma. Let G he a Fitehsian group operating on the upper half plane U. Assume that P is a puncture on U/G, and that the natural projection it is unratnified over a deleted neighborhood of P. Then there exists a deleted neighborhood D of P in U/G, a disc b in U, and a parabolic element T e G such that D .5/G 0 , where G o is the cyclic group generated by T, and two points in b are equivalent under G if and only if they are equivalent under G o . PROOF. Let D u PI he a simply connected neiphbovhood of the puncture P such that it is unramified over D. Let b be a connected component of it '(D) and let G o be the stabilizer of b in G; G0 =
G;
= 151.
b/Go = D. Let B be the holomorphic universal covering space of 15 and let & be the Fuchsian model of Go with respect to D. Then do is infinite cyclic and hence Go (being a homomorphic image of G- 0) is also cyclic. It cannot be generated by an elliptic element. If Go were generated by an elliptic element, the fixed point of its generator would have to be on the real axis. But an elliptic element with a fixed point on R u {co} cannot fix U. Thus Go is generated by a parabolic or hyperbolic element T. We clearly have an inclusion We must, of course, have that
B/G 0 c+ U/Go . If T were hyperbolic, then U/G o would be an annulus that contains the punctured disc b/G o. Thus the puncture P on ZS/G o would be a point in U/Go and hence would be the image of a point z e U under the map no : U -+ U/Go . We conclude that D must contain many punctured discs. But & contains as a subgroup. This is an obvious contradiction. Thus T is parabolic. By conjugating G in the group of real Möbius transformations, we may assume that T(z)= z + 1. By shrinking D, we may assume that b = tz E C; Z > 14 for some b > O. Definition. We shall call
, a disc (half plane) corresponding to
the puncture P.
215
1V.9. Discontinuous Groups and Branched Coverings
IV.9.8. To establish the converse of the above lemma, and to show that D can always be chosen to be the half plane ft z e C; 1m:> I} we prove the following
Lemma. Let A(z)=(az + b)(cz + d)' with a, b, c, d e C and ad — bc =1. Let T(z)= z +1. Then the group generated by A and T is not discrete if 0 < lc! < 1. PROOF. Let A o = A and for n e Z, n > 0, set
We show that if 0 < ici < 1, then
lim A„ = T. n-•
3.3
We write as matrices in SL(2,C), [an b„
c„ d„1, and compute An+1
[a„ b„][1 11[ d„ —17,1 c„ d„ 0 —c„ a, [1 — anc„ a! ] = [a„, 2 —c, 1 + an Cni LCrt +i
b„,. 1 1 d.+
We conclude that
c„ = — c2" and
(9.8.1)
lim c„= 0
(in particular, the sequence A„) is distinct). Now we choose 7 so that 0 < (1 — < y and lai < y. By choice, la c)! =la! < y. Suppose Ian; 2. Thus Z = z' is a local coordinate at n(0). Since
A,(z) —
2 (1 —
we see that
.T.(Z) =
21Z1(1/")-1 v(1 —
Since (1/v) — 1 > —1, ;7(Z) 2 IdZ A d.ZI is integrable in a sufficiently small neighborhood of Z = O. It remains to investigate a neighborhood of a puncture. Here it is more convenient to assume that G acts on U, and that the puncture corresponds
218
IV Uniformization
to the cyclic subgroup of G generated by zi-4z + 1. Thus Z local coordinate on U/G vanishing at the puncture, and
= e 2rdz
is a
IZI log which is readily seen to be square integrable. We are thus able to define
Area(U/G) = IfuiG
2
(Z)
dZ d.2
Theorem. Let G be a Fuchsian group operating on U. Assume that G is of finite type orer U with characteristic x, then
Area(U/G) = 2xx. PROOF. Consider the natural projection surface II, G so that
7C:
U U/G. Triangulate the compact
i. each ramified point is a vertex of the triangulation (however, there may (and as a reFult of (ii) there must) be other vertices, ii. every triangle of the triangulation contains at most one ramified point, iii. if a triangle 4 contains no ramified points, then there exists a connected neighborhood 4 in U so that n(d) 4 and nij is one-to-one, iv. if a triangle 4 contains a finite ramified point with ramification number v, then there exists a connected neighborhood 3- in U so that 7r(4-) and nij is v-to-one, and v. if a triangle J contains an infinite ramification point, then each component of n'(A) is contained in a half plane corresponding to the puncture. Having constructed a triangulation as above, it is easy to replace it by another triangulation in which each triangle is the image under it of a non-Euclidean triangle in U. Let
(P; vi, • • • ,v) be the signature of G with respect to U. Let us assume that in the above triangulation we have ci j-simplices (j = 0,1,2). Let Mk, flu, yk be the three angles of the kth triangle, k = 1, , c 2 . Then C2
Area(UfG) =
— yk) k= 1 co
= nc 2 — E
bk,
k= 1
where bk = sum of the angles at the kth vertex.
219
1V.9. Discontinuous Groups and Branched Coverings
Now bk = 27r, if the kth vertex is an unramified point,
= 27r/v, if the kth vertex is a point of finite ramifi cation number v, and = O. if the kth vertex is an infinite ramification point. Hence
E k=
= 27r(c 0 - n) +
E 20k .
k=1
1
Thus we see that
Area(U/G),F 27r( c 2 - co) + 127r
1 E (1 - -).
Since each edge in the triangulation appears in exactly two triangles, we have
3c2 = 2c 1 , and hence (using the fact that for a compact surface of genus p, c, - c + c, =
2 - 2p) Area(U/G) = 27r(2p - 2 +(1 - —1 )). k=1
0
Vk
Corollary. Let G be a group of conformal automorphisms of a Riemann surface D. If D is simply connected and G operates discontinuously on D, it is possible to define the characteristic x of G with respect to D. Furthermore:
x 0 if and only if D A. IV.9.12. Theorem. Let M be a Riemann surface and Ix 1,x 2 , .} a discrete sequence on M. To each point xk we assign the symbol v k which is an integer 2 or co. If M = C u tool we exclude two cases: j.
{x 5 ,x 2 , .} consists of one point and v o oo, and {x 1 ,x2 , ..} consists of two points and v o v 2 .
fxkl, M" = M\U„ tx kl. Then there exists a simply Let M' = M connected Riemann surface It-1, a Kleinian group G of self mappings of fa such that a. It-1/G M', 2i71,/G M", where ft-4G is lt-4 with the fixed points of the elliptic elements of G deleted, and b. the natural projection n:Ii71 M' is unramified except over the points x k with v k < oo where bk(2)= v, - 1 for all 7t -1 ({x k }). Further, G is uniquely determined up to conjugation in the full group of automorphisms of 117. 1. The conformality type of ;VI is uniquely determined by the characteristic of the data: that is, by the genus p of M and the sequence
220 of integers v 1 .v,
IV UniCormi7ation
. 1. (If we set x = 2p — 2 +
(1 — 1 '1,1). then
x < 0 if and only if M- C u fx}. = 0 if and only if At E, and x > 0 if and only if U.) PROOF. Let fr i(M") = n i (M",b) be the fundamental group of M" with base point b e M". Let [or ] denote the homotopy equivalence class of the closed loop a on M" beginning at b. Let 4 denote a closed curve beginning at b extending to a point x near xk, (by x being near to x, we mean that a closed disc Dk around xk that contains x and does not contain any xl with 1 k) winding around xk exactly y times in Dk , and then returning from x to 1, along the original path.
Let ro be the smallest subgroup of n i(M") containing ", [41; vk < Lc} and let F be the smallest normal subgroup of rr i(M") containing F. Observe that for any k, [4] r for le Z with 0 < 1 < vk • The normal subgroup F of tr j (M") defines, of course, an unbounded, unrarnified. regular (Galois) covering (Si",fr) of M". We recall the (familiar) topological construction of (.1.4".7r). Consider the set (6 of curves on M" beginning, at b. Two such curves a, /3 are called equivalent ( ) provided both have the same end point and [4 -1 ] e F. The surface ,1-4" is defined as (6i —, and the mapping it sends a curve a into its end point. Choose g e SY" such that n(6) = b. 1 hen r 1 (1t71",E). A closed curve a on M" beginning at b lifts to a closed curve "i on M" beginning at i; if and only if x e F. The topology and complex structure of "6/ — are defined so that the natural projection it:/-.' is holomorphic Choose a point xk with yk < x. Let Dk be a small deleted disc around X. Consider a connected component bk of IC '(D„). Choose a point g e bk . We may assume that :)7.- corresponds to a curve c from b to x e A,. Let d be a closed loop around xk beginning and ending at x. Then tr(edv)= x, and [cdl, = 0, , yk — 1, represent vk distinct points in the same component of IT '(D,‘). Thus it ID,, is at least v,,-to-one. Since any closed path in Dk is homotopic to d' for some y, it is not too hard to see that by choosing Dk sufficiently small is precisely v,,-to-one. In particular, the lift of as; consists of a path from i? to a point 5c' with it() near x,, and then a homotopically nontrivial loop beginning and ending at g, and finally a path from g back to g (the original path in reverse direction). We want to show next that we can fill in a puncture in /3k over xk . Let G be the covering group of it; that is, G is the group of conformal automorphisms g of 1-l;f" such that it o g = 7r. Since V is a normal subgroup of It i(M"), G .1- tr,(M")/r, and 14 "/G M" (and G is transitive and acts fixed point freely on /I-4"). Let G be the Fuchsian equivalent of G with respect to gr. We assume that 6 acts on the upper half plane U (the cases where 0 acts on the plane or the sphere are left to the reader). Let
p:U
i./1"
and
p*:'6
G
IV.9. Discontinuous Groups and Branched Coverings
221
be the holomorphic covering map and the associated homornorphism of IV.9.3. Then U/(Kernel 1)*) Si" and U/G. M". Every "puncture" xk on M" with vk < cc is determined by a parabolic element of d by Lemma 1 V.9.6. Recall the exact sequence
1
(Kernel p*)464G— ■ { 1}.
We may assume that the half plane U 1 = Iz e C; Im z > 11 gets mapped by it o p onto the deleted neighborhood Dk of the puncture xk , and that the corresponding parabolic subgroup of G is generated by z + 1. The intersection of the subgtOup with the kernel .of p* is a cyclic subgroup. We z + vk . It is clear claim that it must be the cyclic subgroup generated by that p(U 1 ) is a component Bk of i( '(D1 ). Since p(U 1 ) is conformally equivalent taU 1 modulo a parabolic subgroup, Dk is indeed a punctured disc. Let Al be Si" together with all the punctures on the punctured discs 5„ corresponding to the punctured discs D,, corresponding to the punctures xk - is simply connected. Let be a closed with v„ 0,
> 0.
The dimension (over R) of the space H of harmonic functions on M with "poles" of order cc.; at P1 is precisely
E + i) 2( J= by Theorem IV.3.11. Not all the functions u e H have single valued harmonic conjugates. Let c; be a small circle around P. Notice that
ICI *du = 0,
j= 1, . , r, all u e H.
(To see this last equality, recall that the singularity of u at P.; is of the form Re z' or Im z - m, with 1 < m < ai. Thus the meromorphic differential du + i*du has no residues.) A necessary and sufficient condition for u to have a harmonic conjugate is that
f. *du = 0
Iv
224
Uniformization
for all closed curves a on M. Since Ii i(M) is generated by 2g elements, the dimension over of the space of harmonic functions on M with "poles" of orders < ai at Pi which have harmonic conjugates is
Ecxj + 1 — g]. 2[ i = Thus the dimension over C of the space of meromorphic functions on M with poles of order 1) with a simple pole at P o (2 &-t(P0) = 0). This contradiction (of Proposition 11.5.4) establishes (4). Definition. 9(D) = r(D') — deg D — r(DZ'). (5) For all Po e M, (p(DP0) _5. 9(D). We expand
OD!' 0) = r(D'P(7, 1 ) — deg(DP0) — r(DP0Z - 1 ) = r(D- 1 ) + E — deg D — 1— r(DZ') + = 9(D) + e + — 1 OD). (6) (p(D)
1 — g.
Choose k e Z large so that
deg(ZD'PV`) < where P o e M is arbitrary. By (5) and (1)
9(D) q(DP) = r(D- P,r; k) — deg(DPt). Hence, from (2) we get (6).
(7) 'p(D)
1 — g.
Choose k large so that deg(D/Pt) < O. Then by the argument which is established in (5) and (1),
p(D)
OD/Pt) = — deg(D/n) — r(D/PtZ).
226
IV
Uniformization
We shall now use (2) and Proposition 1V.10.1 to obtain
r(1)1 Pko Z)
deg(PtZ. 'D) + 1 — g deg(It/D) + (2g — 2) + 1 — g = dog(PD) — 1 + g,
establishing (7). The Riemann-Roch theorem follows from (6) and (7).
D
Remark. The above is certainly a shorter and more elegant proof of the
Riemann-Roch theorem than the one in 111.4. The earlier proof was, in our opinion, more transparent than this one. The reader should now re-establish (using the Riemann-Roch theorem) the following: (1) The dimension of the spec: of holomorphic abelian differentials is g. (2) The dimension of the space of harmonic differentials is 2g. (3) Let x i, , x„ be n distinct points of M and (c,, . . . ,c„) e C". A necessary and sufficient condition that there exist an abelian differential a), with a) regular on M \ {x i , ... ,x„1,
Res„ = ci
ordxi co = —1, is that
c. = o i= I
(4) Let x t , , n 1 distinct points on M. Let zi be a local coordinate vanishing at xi, and let
,a;,_ be complex numbers with ki —1, ki E Z, j = 1, . . . , n. If Erj= ai. _ = 0, then there exists a meromorphic differential a) regular on M \ tx„ ,x„ } such that
.,
w(z3) = ( E a.,,,.z;)dz i, v= k,
j = 1, . . . , n.
Furthermore, co is unique modulo differentials of the first kind.
IV. 11. Algebraic Function Fields in One Variable In this section, we explore the algebraic nature of compact Riemann surfaces. We show that there is a one-to-one correspondence between the set of conformal equivalence classes of compact Riemann surfaces and the set of birational equivalence classes of algebraic function fields in one variable. We determine the structure of the field of meromorphic functions on compact surfaces and describe all valuations on these fields.
IV.I I. Algebraic Function Fields in One Variable
227
1V.11.1. In 111.9 and 1V.4, we briefly considered the space of germs of holomorphic functions. We gave a topology and complex structure to the set . //Olt of germs of meromorphic functions over a Riemann surface M. Let us change our point of view slightly. The elements of . 11(M) will now merely represent convergent Laurent series with finitely many "negative terms". Let us assume M=Cu{ oo }. Thus we are dealing only with convergent series f of two forms
E
a(z — .7 0),
Zo E C, aN 0 0,
(11.1.1)
n=N
Or
(e•
N 4-1 )n ,
aN O.
(11.1.2)
n=
Recall the projection and evaluation functions introduced earlier. For f given by (11.1.1), proj f = :,); while for f defined by (11.1.2), proj f = Also, eval f = .:;o if N < O, eval f = ac, if N = 0, and eval f = 0 if N > O.
IV.11.2. Let us take a component F of . //IC which is "spread" over a simply connected domain Dc Cu ; that is, proj: is surjective and every curve c in D can be lifted to F. Thus every germ in can be continued analytically over D and the Monodromy theorem shows that ..-9; is determined by a merornorphic function on D; that is, proj is conformal and there exists an F e .f(D) such that for all : e D, proj' z is the germ of the function F at :. Let us now assume that D = d* is the punctured disc { 0 < Izi < 1 } . We keep all the other assumptions on -.9-7 (surjectivity of proj and path lifting property). Let us take an element f e F. It is represented by a convergent series of the form (11.1.1) with 0 < izol < 1. (Without loss of generality we assume izo i = z 0—this can be achieved by a rotation.) Since analytic continuation is possible over all paths, we may continue f along a generator c of xi (d*) which we take to be the curve :0e, 0 < 0 < 2n. There are now two possibilities: continuation of f along el' (k e Z) never returns to the original f (in which case the singularity of Jr. at 0 is algebraically essential) or there is a smallest positive k such that continuation of f around ck leads back to the original function element f. In this latter case, we consider a k-sheeted unramified covering of 4* by 4* given by the map ck. We define a function F on 4* as follows: If Co e d*, we join 4,1k to 'c:c, by a smooth curve y in 4* and set F ) to be the evaluation of the germ obtained by continuing f along p(y). It is clear that F is a locally well defined meromorphic function. We must show it is globally well defined. Let y I be another path joining ze to C o . Then p(yyr 1 ) is homotopic to a power of c" and hence continuation off along this path leads back (
o
228
IV Uniformization
to f. The function F is now represented by a convergent Laurent series
E
F( ) =
o O. Before proceeding, we must introduce an equivalence relation on de. We shall say that o 0 . //* is equivalent to (33 e de (given by (11.3.1) or (11.3.2) with coefficients rin, Y o, provided
proj = proj Co,
ram co = ram Co = k,
and there exists an E E C with ek = 1 such that =
n = N, N + 1, . .
From now on ../t* will stand for the previously defined object with the same symbol modulo this equivalence relation. Note that ram, proj, and even eval are well defined on de (the new di* !). We topologize de as follows. Let co c, be given by (11.3.1). Assume that the series coverges for If — Zol < a. Let 0 < r < a. We define a neighborhood U(coo ,r) of coo . Let zi e C with 0 < Iz i — zol < r. The "multivalued" function (Z — z 0) 1 3' determines k single valued analytic functions converging in {lz — z i I < a — r}—thus k distinct Taylor series in (z — z 1 ). Substituting these Taylor series into (11.3.1) we obtain k convergent Laurent series. The collection of all such Laurent series (for all z 1 with 0 1/e. Choose, again, O < r < E. For Iz i l > 1/r, the "multivalued" function z -13` determines k convergent Taylor series in (z — z 1 ) with radius of convergence tir — Va. Substitute these functions into (11.3.2) and proceed as before. We see that with this "topology" (we will show next that the sets we have introduced do form a basis for a topology), a deleted neighborhood of any point coo E de can be chosen to consist only of (.0 E .,/i * with proj w 0 00, eval ai co, and ram w = 1. Let CO E U(ai 1 ,r 1 ) n U(w 2 ,r2). Assume that proj w = zo, proj w 1 = z 1 , and proj co 2 = z2 all belong to C, and that co is given by (11.3.1). We must find an r> 0 such that U(ca,r) c U(ah,r,) n U(w 2 ,r2). Let us first assume that
230
IV Uniformization
w Thus ram w ----- I. Let s, he the radii of convergence of w. o.) ; (j = 1,2). Then 0 lz; — :01 < ri and Ci — rj (% = 1,2). Choose r > 0 such that — zol < r; flz — z,10 about z o . Since fi f2 (otherwise (0 1 = 0) 2), U(o) 1 r) n LT Ico 2,0= 0. (Otherwise, choose (0 in the intersection anti obtain that in some open subset of lz — z o l < r both f, and f2 agree with the function defined by (.0.) We are left to consider the case ram (0 1 > 1. Let 0, be given by (11.3.1), and (0 2 similarly with coefficients = zo. 17. We have seen that w i determines a single valued function F. on a punctured disc which covers the given punctured disk k(k) times. If F 1 * F,, then clearly we can choose non-intersecting neighborhoods of (0 1 and (0 2 by the above method. So let us assume that F 1 = F2. Since w 1 w2 we must have ram w, o ram w,. It is now easy once again to choose non-intersecting neighborhoods of w, and w 2 . (The case with proj w 1 E C u {CO} is treated similarly.) We have shown that 4* is a Hausdorff space. Now we introduce local coordinates in 4*. Let w be given by (11.3.1) we introduce a local coordinate t vanishing at w by (z — z o) = tram w .
If w is given by (11.3.2), t is defined by zt ram a) . It is clear that we have introduced a conformal structure on ,,f1*• Proj and eval are meromorphic functions in 4* (note that proj is no longer locally univalent.) Each component of 4* is a Riemann surface known as an analytic configuration.
An analytic configuration S is determined by a single w o E S. Look at all (0 E S with ram w -- 1. Two points in 4* belong to the same component S if and only if they can be joined by a curve in 4*. If the initial and terminal points of the curve are unramified, we may assume that the entire curve consists of unramified points. We thus see that an analytic configuration S can be described as follows: 1. Take a fixed germ (0 0 of a meromorphic function (= Laurent series). 2. Continue this germ ()Jo in all possible ways to obtain a Riemann surface So 3. Take the closure S of So in 4*.
IV. !. Algebraic Function Fields in One Variable
231
EXERCISE
Let f be a "multivalued merornorphic function". Then f -1 (the inverse function) is again a "multivalued rneromorphic function". Show that the map f f defines a conformal involution on JP'. lise this fact to prove that the analytic configuration of a function and its inverse are conformally equivalent. " 1
IV.11.4. Theorem. Let S be an analytic configuration. The Riemann surface
S is compact if and only if there exists an irreducible polynomial P in two variables such that P(proj a), a)) = 0, all co E S. PROOF. 'Ass urne S is a cbmpact analytic configuration. Let . , Co,. be the collection of points with either proj 5i = oo or ram (.5; > 1. (Since f = proj is a meromorphic function on the compact Riemann surface S, f is a finite sheeted covering. The points CO E S with ram a) > 1 correspond to points co with bf (co)> O.) Let n be the degree of the function proj. Let zi = proj (Jrj and call z 1 , , z,. (co is one of these points) the excluded points. If 7.0 e C is a non-excluded point, then there exist n distinct function elements oh, • • • , COn E S with proj a); zo . Form the elementary symmetric functions a0 , , a„ of w 1 ,. , con. Note that a); is a Laurent series in (z — z 0). Thus ,
a0(:) = 1 a t (z.) = — co i (z) — • • — co.(z) a2 (z) = (01(492(z) + co1(403(z)
+••+
0),,-1(40.(z)
= E como.) ;(z) ‘,/:; E
a(z) = (— re,
(0„,(440,,2(z)
•••
co„„(z)
nv = 1
n i < n2 1, z(t) = r"
or z(t) = z(P) + t", near P.
If
(11.5.1)
w(o= E
then we set tp(P) to be either
xiz - J " or
E 1,1z —
depending on which case applied in (11.5.1).
0
IV.11.6. Proposition. If M is compact and w, z e .f(M)\0, then there exists a polynomial P such that P(z,w)
O.
PROOF. Let (p be the map of the previous theorem. Since the image of tp is a compact component of .*(C u { xp}), z, w satisfy an algebraic equation by Theorem IV.11.4. 0 Remark. The mapping 4, of Theorem IV.11.5 is one-to-one if, for example, the pair z, w separate points on M. The converse is not true. We begin with a Definition. We say that z, w E X(M)\C form a primitive pair provided the map cp of Theorem IV. 11.5 is one-to-one.
IV.11.7. Proposition. If M is a compact Riemann surface and z then there exists w
E
E
JC(MK,
,r(M)\C such that z, w form a primitive pair.
z. Choose a e C such that z -1 (a) consists of n distinct points x l , , x„. By the Riemann inequality, there exists a non-constant meromorphic function wi on M such that
PROOF. Let n = deg
w; has a pole of order v; 1 at xi, and w; is holomorphic on Mqx;}.
IV
234 Choo;:e positive inte.,7.ers k t ,
Uniformization
k„ such that < k2v2 < • •
max{g,2g — 1 } . Then 1 — g > 1,
D
and r (P - = deg D — g > O. We choose ho E L(D 1 )\L(P0D - 1 ); that is, (ho)
P; • • • P!
• • • P,1Po
(ho)
Thus ordpi h o ?_; 2, j = 1, . . . , n, and ord o ho = O. If ord o dh o = 0, we are done (fo h0/h0(P 0)). Otherwise, construct h l such that (hi)
PoP
• • P!
pm+1
9
4
PP2 ...P2 pm+I
n (h1)
238
IV liniformintion
and set
fo =
ho
ho(Po)
+ h i.
IV.11.15. Theorem. Let M be a compact Riemann surface. Then v is a valuation on ir(M) if and only if there exists (a unique) x E M such that v(f) = ordx f ,
all f e .Yr(M).
(11.15.1)
PROOF. It is clear that (11.15.1) defines a valuation on iron, so we need only prove the converse. Let y be given. Choose f E ..K(M) such that v( f) = 1. Clearly f C. (IfA E C, then v(A) = nr(A"), all n e Z. Thus v().) = O.) Let r be a rational function, then
v(r( f)) = ord o r.
(11.15.2)
To establish (11.15.2), recall (that up to a non-zero constant multiple)
rf=
fl
of f -f - 13;
1, ,Y ; E C,
atui thus. v(r o f) = > v(f ;=o
E vcr —
j0
By Lemma IV.11.13, v(f — A) = 0 unless ). = 0, in which case v(f — = v(f ) = 1. This establishes (11.15.2). Let 13 1 , , P. be the zeros of f listed according to their multiplicities. Let F E It(M) be arbitrary. By Propositions 1V.11.6 and IV.11.9, F satisfies an equation n-
F" +
E ri( f)Fi = 0,
;=o
with r; rational functions.
Thus, by (11.15.2)
nv(F)
min tordo r; + jv(F), j = 0,
, n — 1).
If v(F) 0 0 .=-(Ff)(x) = A.
We claim that F* is a continuous mapping. Let { x„} be a sequence on M, with lim„ x„ = x. If F*x„ does not converge to F*x, then (since N is compact) there must be a subsequence that converges to y 0 F*x. We may assume that the entire sequence {F*x„} converges to y. Now there is an J e (N) with f(y) = 0 and f(F*x)= 1. Thus
f(F*x„)= (Ff)(x„)— ■ (Ff)(x)= f(F*x) = 1,
240
IV Uniformization
and
= 0. This contradiction establishes the continuity of F*. We show next that F* is holoinorphic. Let x e M be arbitrary. Choose an f E .f(N) such that f is univalent in a disc U about F*(x). Clearly Ff C. There is thus a disc V about x such that F*(V) c U
and
(Ff)(V) c f(U).
Thus F* =f1 oFf in V. Uniqueness of F* follows from the fact that Y(N) separates points. IV.11.17. The map F F-■ F* is a
0
functor, since we have
Corolla, y 1 a. Let F:.-K(M)-- ■ 11M) be the identity, then F*:M 114 is also the identity. b. If dr(M i ) 11: .Y((:t 2)L: A/(M 3 ) are C-algebra homomorphisms, then
(F2 o F 1 )* = F o F.
Corollary 2. If F is an isomorphism (surjective), then F* is a homeomorphism. Since the holomorphic mapping (11.16.2) between compact Riemann surfaces induces the homomorphism (11.16.1) between their function fields that is defined by (11.16.3), the functor under discussion establishes an equivalence between two categories.
CHAPTER V
Automorphisms of Compact Surfaces— Elementary Theory
In this chapter we develop the basic results on the automorphism group of a compact Riemann surface. continuing the study began in 111.7. Some of the deeper results will have to await the creation of more powerful machinery. Using quite elementary methods, we study the action of the group of automorphisms on various spaces of differentials.
VA. Hurwitz's Theorem Throughout this section, M is a compact Riemann surface of genus g usually (but not always) 2, and Aut M denotes the group (under composition) of conformal automorphisms of M. The most important result of this section is the theorem referred to in the title (Theorem V.1.3). We also obtain various bounds on the order of an automorphism of a compact Riemann surface of genus 2 in terms of the number of fixed points of the automorphism. We also generalize the concept of hyperellipticity. The methods of proof in this section are mostly combinatorial and involve the examining of many special cases.
V.1.1. The next result will be strengthened considerably in V.1.5. Proposition. If 1
T e Aut M, then T has at most 2g + 2 fixed points.
PROOF. Since M is compact and the fixed point set of T is discrete, the fixed point set is finite. Choose a PEM that is not a fixed point of T. There is a meromorphic function f on M whose polar divisor is I" with 1 r 5 g + 1 (r = g + 1 if g 2 and P is not a Weierstrass point or g 5 1, r < g, otherwise).
242
V Automorphisms of Compact Surfaces—Elementary Theory
Consider the function h= f — fo T. Its polar divisor is Pr(T -1 P)r. The ft:netion h thus has 2r 2g + 2 zeros. Each fixed point of T is a zero of h. Thus T has at most 2g + 2 fixed points.
V.1.2. We assume now that g >_ 2. Let W(M) be the (finite) set of Weierstrass points on M. Recall that W(M) consists of at least 2g + 2 points and precisely 2g + 2 points if and only if M is hyperelliptic. Proposition. If T e Aut M, then T(W(M))-= W(M). PROOF. As a matter of fact the gap sequences (with respect to q-differentials, q> 1) at P e M and at TP are the same. We define Perrn(14'(.14)) as the permutation group of the Weierstrass points on M. Corollary 1. There is a homomorphism
Aut M —■ Perm( W(M)). Furthermore, 2 is inject ire unless M is hyperelliptic, in which case Kernel A = <J>, where J is the hyperelliptic inuulutiun. PROOF. The existence of A follows from the proposition. If M is not hyperelliptic, then there are more than 2g + 2 Weierstrass points. By Proposition V.1.1, only the identity fixes all the Weierstrass points and thus A is injective. If M is hyperelliptic, then by Proposition 111.7.11, an element T <J> has at most 4( 2, then Aut M is a
finite group. PROOF. We have produced a homomorphism of Aut M into a finite group (the permutation group of a finite set), and the homomorphism has a finite kernel.
V.1.3. Having seen that Aut M is a finite group, we naturally want to get a bound on its order. Theorem (Hurwitz). Let N be the order of Aut M, where M is compact Riemann surface of genus g 2, then N
84(g — 1).
PROOF. Consider (recall the discussion in 111.7.8) the holomorphic projection (abbreviate Aut M by G) -+ M/G.
243
V. t. Hurwitz's Theorem
We know that it is of degree N, and M/G is a compact Riemann surface of genus y. The mapping rt is branched only at the fixed points of G and
b,c(P)= ord G p — 1,
all P
M.
Let /3 1 , , P. be a maximal set of inequivalent (that is, Pi 0 h(Pk ), all h e G, all ) k) fixed points of elements of G\ (1}. Let v; = ord G,»1. Then (using either group theory or covering space theory) there are div i distinct points on M equivalent under G to Pr—each with a stability subgroup of order v; (if h takes P to Q, then GQ = hGph -1 ). Thus the total branch nuber of it is given by N , L — (v i — 1) = N E ( i
,.1
1 B=
The Riemann-Hurwtiz relation now reads
1 2g — 2 = N(27 — 2) + N E (i v•
(1.3.1)
Note that v; > 2 and thus < 1 — 1/v; < 1. The rest of the proof consists of an analysis of (1.3.4 It is clear (since we may assume that N > 1) that g > y. We consider possibilities:
Case I: y 2. In this case we obtain from (1.3.1) that
2g — 2 > 2N or N
g — 1.
Case II: y = 1. In this case (1.3.1) becomes
1 2g — 2 = N E (i - - ).
(1.3.2)
If r = 0, then also g = 1 (we assumed g > 1). This is the basic fact (the left hand side of (1.3.1) is 2) that will be used repeatedly. Thus (1.3.2) implies
2g — 2
IN or
N
4(g — 1).
Case III: y = 0. We rewrite (1.3.1) as
2(g — 1) = N( E (i
1 _-_)_ 2 vi
and conclude that
r> 3 (since 2(g — 1) > 0, N> 1, and (1 — 1/v;) < 1 for each j).
(1.3.3)
V Automorphisms of Compact Surfaces—Elementary Theory
244
1f r> 5, then (1.3.1) gives or N 4(g — 1). If r = 4, then it cannot be that all the v; are equal to 2. Thus, at least one is 3, and (1.3.3) gives 2(g
2(g — 1)
—
1)
IN
N( + — 2) or N
12(g — 1).
It remains to consider the case r = 3. Without loss of generality we assume 2
v1
v2
v3.
(1.3.4)
Clearly v 3 > 3 (otherwise the right hand side of (1.3.3) is negative). Furthermore, v 2 3. 11v3 7, then (1.3.3) yields 2)
+ +
2(g — 1)
or N 84(g — 1).
If v 3 = 6 and v i — 2, then v 2 > 4 and N
24(g — 1).
If v 3 = 6 and v l ..=L- 3 (recall (1.3.4)), then
N 5 12(g — 1). If v 3 = 5 and v 1 — 2, then v 2 > 4 and N
40(g — 1).
N
15(g — 1).
N
24(g — 1).
If v 3 -- 5 and v 1 > 3, then If v 3 = 4, then v, > 3 and This exhaustion (of cases) completes the proof. EXERCISE We outline below an alternate proof of Hurwitz's theorem. (1) Represent M as U/F where U is the upper half plane and F is a fixed point free
Fuchsian group. (2) Show that N(F)=-- normalizer of F in A ut U is Fuchsian and that A ut M (3) Use the fact that it: finite type over U.
u/r
N(F)/F.
U/N(F) is holomorphic to conclude that N(F) is of
(4) Observe that deg it = [N(r):1],
where deg it = degree of 7C, and [N(F):
= the index of F in N(T).
(5) Show that Area(U/T) — [N(T):F]. Area(U/N(F))
245
V.I. Hurwitz's Theorem (6) Prove that for any Fuchsian group F of finite type over U, we have
Area(U/F)
H it .
(7) Since Area(G7T) = 4n(g — 1), conclude Hurwitz's theorem.
V.1.4. To simplify the statement of results, we introduce the following notation. For f a non-constant meromorphic function on a compact Riemann surface M(f e Y(M)\C), let deg f be the degree of f, then (of course)
deg f = deg D, ._ where D = f '(œ ). For T e Ant M, ord T will denote the order of T and T) the number of fixed points of T. Our next result is similar to Proposition V.1.1. Proposition. Let f E .if(M)\ C and 1 T e Aut M. Assume that v(T)> 2 deg f. Then f=foT and deg f is a multiple of ord T. If in addition, deg f is prime, then ord T is prime and M/:45 C u {X }.
Consider h=f—fo T. If h C, then deg h 5 2 deg f — r, where 0 < r deg f and r is the number of poles of f fixed by T. Each fixed point of T that is not a pole off is a zero of h. Thus, h has v(T) — r> 2 deg f — r zeros. This contradiction shows that h e C, and since T fixes points which are not poles of f, h = O. Thus, f projects to a well-defined function f on M /. In particular, deg f = (deg j)(ord T). PROOF.
If deg f is prime, then deg je = 1 and thus Al /115 Cu {co}.
CI
Remark. The fact that f is invariant under T shows that T has finite order.
V.1.5. In this section, M is a compact surface of genus g
2.
Proposition. For 1 Te Aut M, V(T) PROOF.
M
2+
2g ord T — l •
(1.5.1)
Apply the Riemann-Hurwitz relation to the natural projection
Mi. If the range has genus 7, then ord T — I
(2g
—
2) = (ord T)(27 — 2) +
E
v(Ti).
(1.5.2)
1= 1
We must explain the evaluation of the total branch number B of the projection appearing in the above formula. Clearly B is the weighted sum of the fixed points of ; each fixed point appearing one less time than the order of its stability subgroup. This is exactly the contribution for B in
V Automorphisms of Compact Surfaces— Elementary Theory
246
(1.5.2). We use now the obvious inequality v(T) and conclude
v(Ti),j = 1,
, ord T — 1,
( 241 — 2) (ord T)(27 — 2) + v(T)(ord T — 1), Or
v(T)
2+
2g (27)(ord T) ord T — 1 ord T — 1
from which (1.5.1) follows.
El
Corollary 1. If ord T is prime, then v(T) = 2 +
2g — 27(ord T) ord T — 1
where y is the geilits of M/(T). In this case, equality holds in (1.5.1) if and only if = O. PROOF. If ord T is prime, then v(T) = v(T-1 ), j = 1,
, ord T 1.
Remark. In general, (even if ord T is not prime) if equality holds in (1.5.1), then y = O. Corollary 2. In general, r(T)s- 2g — 1 if M is not hyperelliptie. Remark. For hyperelliptic surfaces we have, in general, better bounds (provided g -2: 3). See Proposition 111.7.11. PROOF OF COROLLARY. From Corollary 1, an automorphism of order 2 has precisely 2g + 2 — 47 fixed points. Since M is not hyperelliptic, y Hence for such an automorphism v(T) < 2g — 2. If ord T > 3, then v(T) 2 + g. Now 2g — 1 2 + g unless g = 2. But in genus 2, every surface is hyperelliptic. 0 V.1.6. We now give two examples to show that our results are sharp. EXAMPLE I. Consider the hyperelliptic Riemann surface of genus g 2, iv2 = (z — e 1) • • • (z — e29 +2), here e l , , e 2g+2 are 2g + 2 distinct complex numbers. The hyperelliptic involution is described by (z,w)i—■ (z,—w). It clearly has 2g + 2 fixed points: 7. -1 (0, j = 1, . . . , 2g + 2; and the points over infinity (that is, the two points in 7. - x7)) cannot be fixed. We introduce some symmetry now by setting eg÷iti = j=1,...,g +1.(Hence we are assuming also ei 0, for every j.) Thus the hyperelliptic surface is represented by w2 = (2 -
• • (z 2 -
247
V.1. Hurwitz's Theorem
Here we have the additional automorphism T of period 2 given by z,w). How many fixed points does this automorphism T have? Clearly the quotient surface is 2 • • (z — e 2 + 1) W 2 = (Z e) 9 h which is of genus (g — 1)/2 if g is odd, and of genus g11 2 if g is even. In the former case, v(T) = 4 and in the latter v(T) = 2. What are these fixed points? They clearly must be the two points over 0 in the later case, and the four points 'yips over 0 and ocein the former case. We now introduce hyperelliptic surface's i7vith another symmetry: Let s = exp(2rri/(2g + 2)) and let e; = j = 1, . . . , 2g + 2 and define an automorphisrn T of order 2g + 2 by (:,w)1— ■ The only possible fixed points are over 0 and oo. We will now examine the Taylor series of w at the origin and at co to show that T has 2 fixed points (lying over zero). Note that the Riemann surface is represented by the equation w 2 = z 2g+ 2 The two function elements lying over zero are then
ajz"g + 24) with ao = 1.
w(z) = +i( o
Clearly these two are fixed under T. At infinity (7, = 1/z is a good local coordinate, and the function elements over oo are
=
-g(
V bi(2g +2)i) o
with bc, = 1.
Thus T interchanges these two function elements. Similarly, Tk fixes the two points lying over 0, and Tk sends a function element lying over oo onto the element times s k(g +1). Thus Tk has 4 fixed points if and only if k 0 mod 2, and r has only 2 fixed points otherwise. EXAMPLE 2. Consider the Riemann surface defined by the equation W 3 = (Z -
e 1)2(z —
e2) • • • (z —
with r 2 and e l , , e, distinct complex numbers. We first compute its genus. The surface is a 3-sheeted covering of the sphere. It is branched (with each branch point of order 2) over the points e l, . , e,. It is branched over co if and only if r + 1 # 0 mod 3. Thus, the genus of the surface is r — 2,
if r
2 mod 3,
r — 1,
if r # 2 mod 3.
and
248
V Automorphisms of Compact Surfaces—Elementary Theory
An automorphism T of this surface :liven by
(z,w) 1—> (z,sw),
with s = exp(21 .
3
For r = 5 (thus g = 3) it has 5( = 2g — 1) fixed points. More generally, for r a.-- 2 mod 3, the surface has genus r — 2 and the automorphism has r fixed points (=g + 2). Similarly, if r # 2 mod 3 the surface has genus r — 1 and the automorphism has r + 1 fixed points (=g + 2).
V.1.7. Theorem. Let M be a compact surface of genus g 2. Let 1 Te Aut M. If v(T) > 4, then every fixed point of T is a Weierstrass point. PROOF. We may assume that n = ord T is prime (if not, there is a j such that Ti is of prime order, and clearly every fixed point of T is a fixed point of Ti). Let be the canonical projection, and let y be the genus of .11 The Riemann— Hurwitz formula yields
2g — 2 = n(27 — 2) + (n — 1 )v.(T).
(1.7.1)
Let PE M be a fixed point of T and is it(P) E /171. There is anon-constant function f Ar(k) such that fis holomorphic on :1-4- \ {P } and Jhas a pole at P of order Sy + 1. Let f = f o IC f is the lift of Ito Af). Then f e f is holomorphic on M\ `A, and ordp f n(y + 1). We now use (1.7.1) and the hypothesis v(T) > 4 to conclude (
2g — 2 -= n(2y — 2) + (n — 1)v(T) > n(2y — 2) + 4(n — 1), Or
g + 1 > n(y + 1).
It follows that P is a Weierstrass point on M.
V.1.8. Theorem. Let M be a compact surface of genus g 2. Let 1 T e Aut M with n = ord T, a prime. Let be the genus of fa = M/ (thus, v(T)= 2 + 2(g — nj)/(n — 1)). Suppose that g > n2g + (n — 1) 2, and that there is a 1 T E Aut M with v(T) > 2n('+ 1). Then: Each fixed point of
T is a fixed point of T
and ord T
If ord T = n, then T E
.
ord f = n. (1.8.1) (1.8.2)
is a normal subgroup of Aut M.
(1.8.3)
If n = 2, f is in the center of Aut M.
(1.8.4)
, P„(f) be the fixed points of T. As in the proof of the last PROOF. Let P 1 , theorem, for each j, there is a non-constant fi holomorphic on M\ {Pi} with a pole at Pi of order < n(g + 1). It therefore follows from Proposition
249
V.I. Hurwitz's Theorem
V.1.4 that f; = fi o T for each j and thus T(Pi) = P. We have thus verified the first part of (1.8.1). In particular, v(T) > v(T). We rewrite
g> n26. + (n — 1)2 in the form (add (n — 1)g to both sides and simplify)
g — ng g + n — 1 n— 1 > n
(1.8.5)
We continue with n—1 20 2g n— 1 = 2 +-- +-2-- > 2 +—. n (The first strict inequality being a consequence of (1.8.5).) By Proposition V.1.5, 2g v(T) 2+ ord T — and hence ord T n, finishing the proof of (1.8.1). Assume (ird T = ord T. 13kith T and 1' are in the stability subgroup (of Aut M) of P 1 . Since this stability subgroup is a cyclic rotation group (Corollary to Propoition 111.7.7), it is clear that = . To B. Then ord C = ord T; and, Next, let B e Aut M and set C = by the inequality on g, v(C) = v(D) = 2 + 2(g — n),/(n — 1) > 2n(j + 1). Thus by (1.8.2) C e or < t> is normal in Aut M. We have established (1.8.3) and hence also (1.8.4).
Remark. The above generalizes the fact that on a hyperelliptic surface M (of genus 2) any automorphism with more than 4( = 2 2(0 + 1)) fixed points must (be the hyperelliptic involution and hence must) commute with every element of Aut M. V.1.9. We wish to generalize slightly the concept of hyperellipticity. A compact Riemann surface M will be called y-hyperelliptic (y e Z, y 0) provided there is a compact Riemann surface fa of genus y and a holomorphic mapping of degree 2 7r:M Thus, a y-hyperelliptic surface is a two sheeted covering of a compact surface of genus y, with "0-hyperelliptic" corresponding to our previous notion of "hyperelliptic". It is immediately obvious that on every y-hyperelliptic surface M, there is a y-hyperelliptic involution; that is, a .1, e Aut M with ord
= 2,
v(.1,)= 2g + 2 — 4y.
250
V Automorphisms of Compact Surfaces—Elementary Theory
Theorem. Let M be a y-hyperelliptic surface of genus g> 4y + 1. We have: If! 0 Te Aut M with v(T) > 4(y + 1), then T =
(1.9.1)
.1 7 is in the center of Aut M.
(1.9.2)
If f eY(M)\C, and deg f < g + 1 — 2y, then deg f is even.
(1.9.3)
(1.9.1) and (1.9.2) are immediate consequences of the preceding theorem. Note that 2 deg f < 2g + 2 — 4y = v(J,), thus by Proposition V.1.4, f = f o J., and deg f is a multiple of ord J ., = 2. PROOF. Statements
Corollary 1. The y-hyperelliptic involution on a surface of genus g is unique (if it exists) provided g > 4y + 1. PROOF. The 7-hyperelliptic involution has 2g + 2 — 47 fixed points, and 2g + 2 — 4y > 4y + 4 if and only if g> 47 + 1. The uniqueness now follows from (1.9.1). Corollary 2. If 7, r are non-negative integers with r> 0 and g > 4 7 + 1 + 2r, then a surface of genus g cannot be both y-hyperelliptic and (y + r)-hyperelliptic. In particular for g>3, a surface of genus g cannot be both hyperelliptic and
Remark. In V.1.6 we have seen examples of surfaces of genus 2 and 3 that are both hyperelliptic and 1-hyperelliptic
PROOF OF COROLLARY 2. Suppose M is both (y + r)- and y-hyperelliptic, then J., + , has
2g + 2 — 4(7 + r) > 4(y + 1) fixed points. Thus Jy+, = .1 7 , which is only possible if r = 0.
0
V.1.10. Proposition (Accola). Let M be a compact surface of genus g 2. Let G, G he subgroups of Aut M such that G = U.1 =1 G. and Gi n G = {1} far i j. Assume ord G = n,, ord G = n, and let y = genus of M /G,y i = genus of MIG i. Then
(k — 1)g + fly = E
(1.10.1)
J=1 PROOF. Let us denote by B and Bi the total branch number of the canonical projections M M/G, M M/G i. Obviously B= E (ord Gp 1) (with Gp, the stability subgroup of G at P e M), with a similar formula for Bi. Now if Gp is nontrivial, it is cyclic and generated by some T e G. Clearly
251
V.I. Hurwitz's Theorem
Te (G)p for some j. It follows easily from these observations that B = X B.
We now use Rieniann-Hurwitz
2g - 2 = n(2y - 2) + B, 2g - 2 = ni(2yi - 2) + B. Summing the last expressions over ], subtracting the first, and using the fact that n = tti - (k - 1), we get (1.10.1). El Corollary. Let M be a surface of genus 3. Assume there is a subgroup G of
Aut M with G Z, Z,. and AVG of genus zero. If one element of G\{1} operates fixed point freely, then M is hyperelliptic. PROOF. The group G can be decomposed as in the theorem with k = 3 and
n, = n 2 = n 3 = 2. Hence 3
= yi + + 73*
Since one of the elements of G operates fixed point freely, one of the yi must be equal to 2. The other two cannot both be positive.
V.1.11. We record for future use the following simple Proposition. Let M be con ;pact Rieniannstuface of genus g 5 is of prime order n, then n 52g + 1.
2. If T e Aut M
PROOF. Let y = y(T) and use Riemann-Hurwitz with respect to the projection 7r:M = 114/ onto a surface of genus g
2g - 2 = n(24 -2) + v(n - 1). (1.11.1) Assume n > 2g. If g 2, then n(2g - 2) 2n 4g; a contradiction. If g= 1, then v 0 0 and v(n - 1) 2g -1: again a contradiction. If g = 0, then v > 3. Assume first that v > 4, then -2n + v(n - 1) > 2n - 4 > 4g - 4; a contradiction. If y = 3, then (1.11.1) reads 2g - 2 = n - 3 or n = 2g + 1. Remark. Proposition V.1.11 has shown that the maximal prime order of an automorphism is 2g + 1 and that in this case the automorphism T has 3 fixed points and Mg T> is the Riemann sphere. Examples of such Riemann surfaces are those defined by the equations w29+ 1
= (Z — e i )(z - e 2)22(z - e3)2',
where e l, e2 , e3 are distinct complex numbers and a,, a 2, 2 3 are positive integers such that a, + a2 + a, 0 mod(2g + 1), and 2g + 1 is a prime
V Autornorphisms of Compact Surfaces--Elementary Thcory
252
number. The automorphism T sends w into exp(2/ri/(20 + 1))w and fixes :. A natural question to ask is what are some other possible pri;,e orders. w e in the next section that the only prime order >g other than 2g + 1 shale is g + I. In this case 7 . will turn out to have 4 fixed points and once again .11/ will be the Riemann sphere.
V.2. Representations of the Automorphism Group on Spaces of Differentials Let M be a compact Riemann surface of genus g > 2 and Aut M the group of conformal autotnorphisms of M. In this section we study ihe representation of Aut M on the space of holomoi- phic ..1-dilTeretnials on M, J 4(M), g > O. With the aid of the representations, we will improve man results of the previous section. V.2.1. Let us begin by observing that a product of a holonlorpllic q 1 a holomorphic g 2-differential is a holomorphic (q t + TO--diferntialwth differential. Thus the direct sum (M) =
..rfq(M)
q=o is a commutative graded algebra. Let T e Aut M and 9 e )t‘g(M), then T acts on 9 to produce Tp = 9 0 T -1 .
(Say that 9 in terms of the local coordinate z vanishing at P e M is given by p(z)dz. Choose a local coordinate vanishing at TP. Assume that in terms of these local coordinates T . ' is given by z = f( . Then Ty) in terms of the local coordinate is given by p(f())f '(C)q cgq.) We note several properties of this action of T on differentials: )
a. For T 1 ,
T2
e Aut M, yo e Yeq(M), (T, 0 T 2)9 = T 2 (T 2 9).
b. For each T e Aut M, T:..Y0(M) Yfq(M)
is a C-linear isomorphism. c. For T e Aut M, (pi e Yfq ,(M), j = 1, 2, T(TIY)2)= (T(PI)(T 2, we compute using Riemann-Roch
= deg(
g + 1 + t --
Po)
Z (q'
= g(2ff — 2) +
1 [g(1 — — sf v(P)
E PE
— 1) +
Zq
Thus to complete the proof of the corollary, it suffices to show that i(Zq/ZN )) = 0 which follows from the inequality deg(11,,Z 41) > 2g — 2. We have already used the fact that 1
deg(-K g—) = q(24 — 2) ± E [41 — Z(q ) PE Thus the desired inequality clearly holds for g examine
2. For g
1, we must
z. -
(2.2.5)
If g = 1, then we saw in the proof of Hurwitz's theorem that for some P e v(P) 2, thus the sum in (2.2.5) is greater than or equal to [q/2] I. For g = 0, we must show that the sum in (2.2.5) exceeds 2(q — 1). Now
z Fq ( i PER
L
i1
v(P)A
z fq q p la 1 v(P)
v(P) 1
= (q l)
PL (1 — ;
j) .
But we saw in the proof of Hurwitz's theorem (equation (1.3.3)) that
Ep Esr, — (1/v(P))) > 2. Remark. Let g = 2 and let n = the number of points
1 e la with
then dim yew
= 3g — 3 +n
0).
v(P) > 1,
256
Y Automorphisms of Compact Surfaces—Elementary Theory
V.2.3. We start now a more detailed investigation of the action of T e Aut M on YO(114), g 1. Proposition. There exists a basis for ..)(4 (M ) such that the action of T on
.Yt'l(M) is represented by a diagonal matrix. PROOF. Since Aut M is a finite group, there is an integer n such that T" = 1. Thus (because Aut M is represented on Yeq(M)) T" = 1 as a matrix with respect to any basis for g'q(M). In particular, A." — 1 is an annihilating polynomial for T and the minimal polynomial (a factor of A" — 1) has distinct roots. Thus, by a well known result of linear algebra, T can be diagonalized. We can say something more. Each entry on the diagonal must be an n-th root of unity. We will continue with this analysis in the next section. Remark. We shall actu:.!ly be able (in most cases) to exhibit a basis for ,Yt'q( M) with respect to which T is diagonal.
V.2.4. We continue with the problem introduced in the last section. We wish to determine the eigenvalues (and their multiplicities) of T. Assume that T e Aut M, ord T =.n > 1, and (T> operates fixed point freely on M. (If n is prime, then operates fixed point freely if and only if T is fixed point free. If n is not prime, the backwards implication need not hold.) Let us recall that the eigenvalues must be of the form ci where e. = exp(2ni/n) with j = 0, , n — 1. The eigenspace of 1 is precisely „A"1 7.) (M) and thus has dimension equal to (2q — 1)(g — 1) for q> 1 and equal to g for g = 1, where as usual g = genus of Si = M/< T>. Next assume that A is any eigenvalue. Then for ça e .o(M) with Tcp = ;.(p, we have that (9) is invariant under T (and hence also under ). Let us denote by EA the
eigenspace of A. If we now choose any non-trivial 90 e EA we see that for ça e EA, (PAP° is a < T »invariant meromorphic function on M whose divisor is a multiple of (90) -1 . The divisor ((p o) projects (because it is < T »invariant) to an integral divisor D on M. It is thus easy to see that EA L(1/D). Now deg D = 2q(g — 1)/n and Riemann—Hurwitz (or elementary arithmetic) shows that (2g — 2) = n(2g. — 2), and thus deg D = 2g(g — 1). Hence by Riemann— Roch (recall that g 2) (1 r
=
fdeg D — (g - 1) + i (D) = (2q — 1) ( - 1), — 1 + i(D),
for q > 1, for q = 1.
For g = 1, i(D) = 1 if D is canonical and i(D)= 0 otherwise. Now the divisor D is canonical if and only if (po projects to a holomorphic differential on M; that is, if and only if (p o is G-invariant (if and only if A = 1). We conclude
(2q — 1)(g- 1), dim EA
={g, - 1,
q> 1, A" = 1,
g = 1, A = 1, g = 1, A 1, A" = 1.
257
V.2. Representations of the A utomorphism Group on Spaces of Differentials
Since EAa dim EA= dim AIM) and (g — 1) = n(g — 1), we see that all the n-th roots of unity must be eieenvalues and that their multiplicities are given by the above formulae. V.2.5. We extend next the considerations of the previous section to the general case (G = is assumed to have fixed points). The development given below is due to I. Guerrero. To fix notation (in addition to what was introduced in V.2.4), we let
j = 0,
ni = dim Eo ,
, n — 1.
Formulae (2.2.3) and (2.2.4) compute n,. To compute n j for 1 j we partition the branch vet X of 7r.; M M/G into a disjoint union
— 1,
n-1
X = U X i, 1=1
where X, = {P e M;
= P and Tk P P for 0 < k 51— 1}.
We claim that if Xi 0 0, then /In (1 divides n). Suppose P e X, and 2 5 1 n — 1. Write n = pl + k, 0 < k < 1 — 1. Then VT = P, and thus k = O. Hence we see 11n. Furthermore, the orbit of each point P e X, consists of the 1 points P, TP. . , T''P and are also in Xi. It is thus clear that if for non-empty X i , we set
n(X I ) = {Q.1; 1
x i},
1 < 1 < n — 1,
then n(X,) consists of x, distinct points and X, consists of lx, points. We define x, to be zero for empty X,. We can choose a local coordinate z for each point in it -1 (Q„„) such that is given by
where 1, is a primitive nfi-root of unity (note that n,,,, depends only on Q„,, and not on the choice of P E Suppose now that 0 o (po e E„, I <j n — 1. As before, in this case we also have
n = r(1 ), Di
where Di is a certain integral divisor on M/G. To find what Di is, let us consider the equation for 0 0 go e
T9 = gig). Thus we also have
749 = zfi(p.
(2.5.1)
Let us look at the Taylor series expansion of cp, at a point in it -1 (Qii):
9=
E k=0
safki)dzq.
258
y Automorphisrns of Compact Surfaces—Elementary Theory
Thus equation (2.5.1) reads
.511,111.i
X
=
k= 0
k
o
.144-7k .
In particular, for each k = 0, 1, 2, ... ,
= 0.
sik1nti q Choose the unique integer
such that
1 < 41j < n11 and q1 , = Note that /1„,1(n ,1) = n/I (this defines ).„,i„) and we set (for further use) ;.„,to = 0. We conclude that unless k + q ).„„.*mod -
=0
I n)
.
Let xm, be the order of rp o at ir -1 (Q„,,) (this integer only depends on Q„,, and not on the point P we choose in its preimage). By our previous remarks there is an integer /3„,,i such that ml
n q 0 = /3?NU• - +.n
(note that /3„,u 0). Let Q 1 , .. , Q, be the projections to M/G of the zeros of cp o not in X. Let = ord„, (00 ço o . We note that a function _re /G) is such that fcp c, e drq(M), where f = o rr, if and only if
/ is holomorphic except at the points ordok -xk, and iii. ord/ + (q - )..01014 j.
Qk,
Q„,,,
Now we know how to define the divisor Di (which depends on To): D
=
n 421,k Fl 11
k= I
QAmi . +Lams' — 9110/11]
ml
m= 1
•
Remark. With our convention for ] = 0, the formula for D
valid for all j,
0 < j < n - 1. Note that deg(rp o) = 2q(g - 1), and hence
n E ;+E
E /1„„ =
2q(g -
1).
(2.5.2)
) +[
•mi n/I
(2.5.3
lin m= I
k=1
Now from the definition of Di, deg 1); =
E k= I
+EE m= 1
13
V.2. Representations of the Autornorphism Group on Spaces of Differentials
259
Next we use Riemann-Hurwitz in the form
x,
E
2g — 2 = n(2g — 2) +
— 1) I
lin m= 1
= n(2g — 2) +
x,(n — I). l in
We return to (2.5.3) and continue our calculation (using (2.5.2) and RiemantiHurwitz)
deeD; =
E
+
k= 1
n
„
xi
E
2k
s =t
1 = ;i 2q(g
n11
n/1
lin m= 1
(n
q]) v.
)
+ lin m=1
L
"
lin m=1
1 ;77 1,
—1 nit
1 -r-1 (20 lin
1 + (2g — 2)(q — 1) = 2g— 2 + !(2 — 2)(q — 1)
>. 14. — 2. We have also shown that for q> 1 deg Di > 29
2.
-
We claim that even if q = land even if deg D = 2g — 2, i(DJ) = 0 for j > O. Otherwise Di is canonical. In this case, there exists an abelian differential on M/G such that (0) = D. Note that the only way for deg 1) = 2 - 2 is to have equality all along the way in our previous calculation. In particular, (0) — }l(A„, u — 1), and since 1 (n11), we conclude that /L I; = n11. Thus {
X
D=
fl 01/4 H II k=1 lin m= 1
The lift (p of gi will have a zero of order ak at each preimage of Qk, and a zero of order a n (n P .a.; + 1
/
= 1 2,21
at each point of 7t -1 (Q.). Thus 9 will have the same divisor as 0, [().„,u — 1)1(n11)] = O.
v
260
A utomorphistns of Compact Surfaces--Elementary Theory
Thus for j > 0, deg Di = 17,
2 ffi l
+y
k=1
lin m
+1—
n11
1
and from (2.5.2)
2g - 2 1 E E 1 (Anii - 1 ). n lin m=1 Xi
deg Di -
Hence we conclude that (using Riemann-Hurwitz)
1 1 r(-)= deg Di + 1 - g= 4 - 1 + - E (n Di n lin rn ,-1 Xi
XI
=g- i —
E Iii,
1 ;•ntlj•
(2.5.4)
m=1
We compute next
Recall that n, is a primitive (n//)-root of unity and that ii,trj efi. For j = 1, 2, ... , (n11), WI is the set of all (n//)-roots of unity. We thus conclude that 1-'4 1 mtj
is a permutation of 11, ... ,(n/01 with o-(nI1) = (n/1). Further for j = k(n11) + Therefore r, Amu =
E i=
A„,1; = /(1 + 2 + • • + -n\ -n ( 1 + LI\ 1) 2 1)
Thus n- 1
n-1
j=0
ii
(1)
x, (1 1 - ni + - n2 2 n lin m=1 2 1
= n(4 - 1) + 1 + (n - 1)
E x, - - y E tin
1 = n(g - 1) + 1 +(n-1):Ci L
= 9. We finally conclude that once again for each j, ei is an eigenvalue (and we have computed its multiplicity). Note that some of the multiplicities may be zero (for example, if g= 0). It involves no loss of generality to consider these as eigenvalues with zero multiplicity.
261
V.2. Representations of the Automorphism Group on Spaces of Differentials
V.2.6. Let us now consider the case q> 1. We calculate for 0
Di
j
n — 1,
= deg Di + 1 — 2 +
Xi
=E2k+EE(mi k=1 Iln m=1 x,„
2q(g — 1)
+E
E
mu+[na'
n/I
n/1
41)+1 —
;
( 1-
-
n/I
fin m=1
mu
+rmun/I-
1+1 -
x, )
n
il,,
[
nil
Iln m = 1
+1—g.
n/1
Next we compute n— 1
E
(
r
Ong — 1) + q E x i(n — I) —
= (2q
Ext (n- I). fin
tin
uj
J=
To explain the last term, we write
q = - + 011 1 '
v(1) e Z, p > 0, 1 < v(1)
e
(We will write y for y" ).) Thus we have
l
x'
E E lin
m=1
P+ x'
V') —
=E 1=i ti n m
[-A mu
).„,„
n/I
voi
± L 0 - P - n/ _I)
/vw— 2 A.l n/I P
+ .1„,,i — val nI 1 _
Recall now that {).„,u ; j = 1, . ,n/i} is a permutation of {1, ... ,n/1}. Thus the above summed from j = 1 to j = n/I yields
E (o) - 2 lin
+ I - (o) -
1)) x, = E (1 - (1+ Iln
2
Thus the sum from j = 0 to n — 1 yields
E (i - ( 1 +
ix, =
E iln
In
1 1)/x,--2
- 1
lin
In conclusion (using Riemann-Hurwitz once again)
i =o
D•
= (2q — 1)n(— 1) + (q — -1)E xi(n— I) 2 1 2 = (2q — 1)(g — 1).
/
Xi.
262
V Automorphisms of Compact Surfaces--Elementary Theory
We have just proved that in all cases
(1) r — = ni. Di
V.2.7. We now want to compute the trace of T(=tr T). Note that for q =1 n-1
E
tr T = g +
njsi.
J=1
First observe that
E d=
1,
-
1=1
and thus using (2.5.4)
F
trT=g+1
n lin m=1 j=1
fin 1
xi
n
!in
n-1
E
(2.7.1)
lin m=1 j=1
Calculate for ! > 1
As before, we fix m and f and compute Dr= Amii6i=j„,“&+;.,„,2& 2 + • • • + l= 1 ▪ 1. mi 1E
± )..
1+ 1 + • • • ± A rni(nI0e
,, e
-i)(P0)+1
20/1)
. • • + j1.1(0 )En
l-1
=EE&
k(n[1)-i-
k=0
E ckm/i)
nil
I— I
.11
k=o
=E (sinceEll', s4"in = 0). Hence for 1 > 1 n—
E 1
Amusi=
—2
m1(nlI) = i=
Finally, for 1=1, n- 1
n—
.i=1
i=1
E
(and because filmii ; j = 1,
E
Àrniinntti
,n — 11 is a permutation of 11, n— I
n
.J=1
—nmi
—
•
,n — 1))
263
V.2. Representations of the Autoniorphism Group on Spaces of Differentials
We now return to (2.7.1) and continue with our computation after changing notation: x1 = t = number of fixed points of T and nrni
tr T = 1 —E
n - -1 E'1
x, - -1 E E n
lin
I
in m=1
—n
nm 1 1—
= 1 - E x, + E x, + E 1 lin
tin
m=1 1 —
1 #1
=I—t+
m=1
1 _ ev„,
ev= E m = 1 1 - byV.2.8. As usual, we proceed to the case q> 1. Here we compute n-1
tr T = E J0
-
n1
= ((2q — 1)( — 1) +
+E
11 II.
(q
liz -. i
.= 1
J j= 0
2, mi . i
i
I li
E
xi(n —
n/L
j= 0
q'
j±
.
e.
n/I
The first sum, as before, is equal to zero. Thus we conclude (simplying further as in V.2.6) xr n- 1 1 tr T= -- E/ n lin m=1 j0
n-1
zi
v (I)
+
n/I
m=1 j= 0
= — 1 E 1(--f) x, n tin
1 x'
n
E
—
n
x' v v
n
v -1
— vo) .
0 m n/I
1—
1#1
As we remarked before, [
Y.1; =
2m"ti va) n/I
0 WO) < —1
if 0
1
E
ymud =
mIlE Ym12 62
ymi(n/I) 0
j= 1
+ Yrnzien")+1 + • • • + Y.10/1)E 2(n/i) Iwo+
= y(1 +
ell
. . • ± y rnione
+ 8 2(n/l) + • • • + 0 - 1)(n/1)) = 0,
f.
264
v A utomorphisms of Compact Surfaces—Elementary Theory
where y = y„,ii s +
y„,10,mend. Therefore for / > 1,
7,7,120 ± • ' n-1
E
ymuci= 7,m0— ym,„= —1•
j=0
For 1. 1 n-1
E j= 1
n-
Yrni.,e1
=E
j= 1
= E1
[11
I
""
1= 1 m-1
• J
=E
[
—
1
Vi ,..,1
n
v- I
q;',. 1
--
i= 1
—
1—
qm ,
We can finally complete the calculation: xi 1
tr T --
E x, + E
m. 1 1 — riml
II.
=E m=1
/in
m=1
1 — nm1
1* 1
1* 1
e
V,V
1 — em .
V.2.9. We summarize our results in the following Theorem (Eichler Trace Formula). Let T be an automorphism of order n > 1 of a compact Riemann surface M of genus g> 1. Represent T by a matrix via its action on .Yeq(M). Let t be the number offixed points of T. Let s = exp(2ni/n). Let P„ . . . , Pt be the fixed points of T. For each m --= 1, . . . , t choose a local coordinate z at P. and an integer v„, such that 1 < v. n — 1 and such that T -1 is given near P. by T -1 :z (note that v m must be relatively prime to n). Then
t Ev. tr T = 1 + m=11 E _ sv„, for q = 1, and
E
en v
for q> 1, m=1 1 — s v." where 0 < y n is chosen as the unique integer such that q = pn + v with p e Z. In each case the sum is taken to be zero whenever t = O.
tr T =
Note that only the fixed points of T contribute to tr T (not the points that are fixed by a power of T but not T itself).
V.2. Representations
of the Automorphism Group on Spaces of Differentials
265
Corollary (Lefsehetz Fixed Point Formula). For g = 1,
tr T + i T = 2 — t. PROOF. We merely note that for
E C,
0 0 1,101= 1, 2 Re(0,11 — 0)) = —1.
0 Remark. For g = 1, tr T+ tr T is the trace of the matrix representing T on the space of harmonic differentials—which is, of course, by duality related to the representation of T on H i(M). V.2.10. We now turn to the case where T has prime order n and at least one fixeSpoint. Say P E su is the fixed point of-T. Choose a basis {q, . ,q)d} of .Yel(,1 ,1) adapted to P(d = dim Yfq(M)); that is, in terms of some local coordinate z vanishing at P, we have (pi =(z' + 0(1z171)de, j=1, . . . , d, where 1 = y i 2. Then every fixed point of T is a q- Weierstrass point for every q> 1, q 1 mod n. Corollary. Let 1
T E Aut M. If v(T)> 2, then every fixed point of T is a q- Weierstrass point for some q. An alternate proof of the theorem is obtained by equating 2k(g - 1) + [(11n)(g - 1)] to dim 01 7)(M). This leads to the inequality v < 2 + 2/(n - 1). If n> 3, we find y = 2. If n = 2, we find v 2. A conformal involution on M has either no fixed points, only non-Weierstrass fixed points, or is the hyperelliptic involution J. Let T be an automorphism of prime order n with 2 < n have genus g. Then there are two possibilities: a. g = jn, v(T) = 2, and both fixed points are Weierstrass points, or b. n = (2g + 1)/(2j+ 1) (ff 1), v(T) = 3, and the other fixed points are P and J(P) with P not a Weierstrass point.
2. Assume T fixes a Weierstrass point. Represent T on .01 (M). The action is given by (see V.2.10)
PROOF. Let T be of order
diag( -1, . g-times
1).
V Automorphisms
268
of Compact Surfaces—Elementary Theory
This is, however, the representation off on ye' (AI). Since the representation of Aut M on Je l (M) is faithful T = J. Next we consider an automorphism T of prime order n, 2 1, then
Min IXal
xeSL(n,Z)
v.
Thus in particular for n = 2, the above minimum is 1 if and only if a has relatively prime components. We have also established necessity in general. Now assume that n> 2, and that a has relatively prime components. Suppose X is a solution to the minimum problem. Let b = Xa. Clearly b has relatively prime components. (Otherwise a = Xb would not have relatively prime components.) Assume that b has two non-zero components. We may assume that these are the first two components 6 1 , b,. Let y > 1 be the greatest common divisor of b 1 , b2 . Choose X E SL(2,Z) such that
46 1 ,1) 2) = 11(1,0). Let X' =
?„_J. Then X' Fx 0 t(b
E
SL(n,Z) and IX'Xal n
i,b2 ,
,b„) =
+ b and the natural projection it: M Let j = 1, 2, g + 1, or g + 2. Because of T(Œ) = cti, the differentials a; project to harmonic differentials itai on M. (This assertion is not quite obvious. Its verification is left to the reader.) If Zr is a closed curve on gr, then we lift it to a curve c on M beginning at some Q E M and ending at TIV. By (3.3.2) we see that
fe 7rŒi e Z,
j
1, 2, g + 1, g + 2.
(Note that since we are only interested in the homology classes of the curves, we may assume that c does not pass through any fixed points of T.) Let us on M and a construct a canonical homology basis {di ,
274
V Automorphisms of Compact Surfaces—Elementary Theory
basis (ff,, ,32 4. 1 of harmonic differentials on dual to the given homology basis. Since act, has integral periods over every cycle on M.- , there are integers nil, such that 2§
nai =
E
nikak,
j= 1,2,g+ 1,g+ 2.
k1
It follows from (3.3.1) (viewed on /171) that
Orap*nagi. e Z,
j = 1, 2.
It is also quite obvious that (for j =- 1,2) (ord T)(7r1;,*nct 9 , ;) = (cci,*29+ j) = — 1. Note that the second equality follows from (3.3.1). Thus we conclude that for the hypothesis of the theorem to hold, T must be fixed point free (if it is not the identity). To finish the proof, we must study fixed point free automorphisms. By our previous considerations no power of T(01) can have fixed points. Hence by Riemann-Hurwitz 2g — 2 = n(2g — 2), with g > 2 and g > 2. Thus we are studying smooth n-sheeted cyclic coverings of a surface :VI. of genus g > 2. A smooth n-sheeted covering is described by z- Z. the selection of a normal subgroup G of ,(k) such that n 1 (,171)1G ( = integers mod n). Thus G appears in a short exact sequence of groups and group homomorphisms (1} --■ G --o n i(k)-r*Z„—• {0}. Without loss of generality we assume that the canonical homology basis on gt comes from a set of generators for it 1 (A7/) denoted by the same symbols. The homomorphism 9 is described by its action on the generators. Setting 9(a) = c and p(b) = e"i, we see that 9 is described by the 2 x q matrix ,
X = [e
C]=[ b • '"]
of integers mod n. Since 9 is surjective the vector (s,e') has relatively prime components. We review the above situation in the language of multiplicative functions. Construct on Sa a multiplicative function f belonging to the n-characteristic x. This function lifts to a single valued function on the cover corresponding to the homomorphism By a change of homology basis (since Z„ is commutative we may work with 1-1 1 (1N) instead of It i (k)) we may assume that x is of the form [I; 8::: This follows immediately from our second lemma. More precisely, if x is then we can consider the change of homology basis effected by a matrix
V.3. Representation of Aut M
on Ii i (M)
`X E Sp(j,l) such that
275
te
X
r euri [ e'l = LO
Such an X exists by the lemma and the x for this new basis is [1:, Sj. The last assertion follows from the fact that X has integral entries. lie H 1 0) is written as nA + mjb , then X(c) = (EM [ n ]. (As usual n = t(n,,
,n4,2n = t(m,,
,m4-).) Thus
('Xe) = and in particular for k'i = Xi = (6,0( jth column of 'X) = jth entry of X[88,] [e 1)] = jth entry of
0
It is now a simple matter to complete the proof. We examine the cover corresponding to X = .:: g]. It is clear that if di, ... is the homology basis for :V/ such that X = B g ::: 8], then di , 2, , (n — WTI lift to open curves on M and nil, lifts to a closed curve on M. Further b 1 has n disjoint lifts all of which are homologous on M. Each of the other cycles also have n disjoint lifts on M; however, these are all homologously independent. The action of T on M fixes the lift of na, and permutes the lifts of ; however, since the lifts are all homologous we can say that (as far as the induced action on homology) T fixes the lift of b 1 . No other cycles are fixed. Hence we finally conclude that T must be the identity and this concludes the proof of the theorem. ,
V.3.4. In this paragraph we strengthen (in another direction) Theorem V.3.1. The group Aut M acts not only on H 1 (M) but also on H i(M,4), the first homology group on M with coefficients in the integers mod n. Choose T E Aut M. We have seen that T acts on H ,(M) by a square integral matrix A E Sp(g,Z) and A has finite order m. We first prove the following Lemma (Serre). Let A E SL(k,Z) have finite order in> 1. If A I mod n, then in = n = 2. PROOF. (Due to C. J. Earle). Let p be a prime factor of n. Then A I mod IA where ! 1. If we show that p = m = 2 and I = 1, then the lemma will follow easily. We break the argument down into a series of steps. a. If A I mod p', then ! = 1.
276
V Automorphisms of Compact Surfaces—Elementary Theory
PROOF. Let q be a prime factor of m. So m = qr. Then B = A' has order q, and B -a I mod pl. So B = (I + psX), where X is integral, with each element
of X prime to p, and s _>_ 1> 1. Now I = (I + p')09 = 1+ tip s x + p 2 3 y. Thus
qX = —psY, and so
q =p and X = — p' Y .
Again, since p does not divide X, s = 1 and since s 1 > 1, also 1 = 1. b. A has order p.
PROOF. Since p is prime, we need only show order, and by (a)
AP
= 1. Clearly AP has finite
I p2x + p 2 y = I AP = (I + pX)P =+ so
AP a. I
+ p2 y ,
mod p2. This contradicts (a) unless AP = I.
c. p = 2. PROOF. Otherwise p is an odd prime, and p ?._. 3, and I = (I + pX)P = I + ppX + P(P —
')2y2
+ p3y.
2
Thus —x = p (1)---
2
) X 2 + p Y.
But p is odd, and so p divides the right hand side. This is the final D contradiction. The above has as an immediate consequence the following Theorem. There is a natural homomorphism
h:Aut M — ■ Sp(g,Z.). For g > 2, n 3, h is injective. For g of order 2 are in the kernel of h.
2 and n = 2 only automorphisms
V.4. The Exceptional Riemann Surfaces We have seen in 1V.6, that there are a few Riemann surfaces with commutative fundamental groups—the so called "exceptional surfaces". We shall now encounter the same surfaces in a slightly different context. V.4.1. We want to describe all Riemann surfaces M for which Aut M is
not discrete. We shall not bother defining what we mean by Aut M not being
277
V.4. The Exceptional Riemann Surfaces
discrete. We will establish a result which will hold with any reasonable concept of discreteness for Aut M. We state it as a Theorem. A ut M is not a discrete group if and only if M is conformally equivalent to one of the following Riemann surfaces:
1. C fœl, 2. C, 3. 4 = {z E C, iZi < 1 } ,
4. C* =
(01,
5. A* ..4\{O}, 6. = {z E C, r < Izi < 1), 0 < r < 1, 7. a torus, C/G, where G is the free group generated by the two translations
7.1.—oz+ V.4.2. In any standard book on complex analysis, the reader will find a description of the automorphism groups of the three simply connected Riemann surfaces. We have used this information already in IV.5.2. We repeat it here: Aut(C {x}) PL(2,C) Aut C P4(2,C) Aut 4 Aut U PL(2,1/). The above are clearly not discrete subgroups of PL(2,C). As a matter of fact they are Lie groups. The first two are complex Lie groups (of complex dimensions 3 and 2, respectively). The last is a real Lie group (of real dimension 3).
V.4.3. It is clear that every automorphism A of C* extends to one of C {CO}. Furthermore, A must either fix 0 and oo or interchange these points. Thus the automorphism group of C* consists of Möbius transformations of the form z kz
or
Hence Aut C* is isomorphic to a 74 -extension of C*.
V.4.4. Every automorphism of 4* extends to an automorphism of 4 that fixes O. Thus Aut 4* is the rotation group z
eiez
which is isomorphic to the circle group .51 .
V.4.5. Let M be a Riemann surface with holomorphic universal covering map it: i17/ -+ M. Let G be the covering group of it. We claim that Aut M
N(G)/G,
278
V Automorphisms of Compact Surfaces—Elementary Theory
where N(G) is the normalizer of G in Aut k. The verification is easy and straightforward—and left to the reader. We shall show that in the nonexceptional cases N(G) is already a discrete subgroup of Aut M.
V.4.6. Let us consider the surface A,. Its holomorphic universal covering space is U and its covering group G is generated by C:z 1—* Az for some
A> 1 (the relation between A and r was given by IV(6.8.1)). Let A E N(G). Thus
A Co
= CI I ,
since conjugation by A is an automorphism of G. Note that A o Cod' =C implies (by Lemma IV.6.6) that A is hyperbolic with 0 and x as fixed points; that is, a mapping of the form kz,
k> 0.
(4.6.1)
We are left with the case A 0 C = C". Since A o C' = C, we conclude that A' commutes with C and is of the form (4.6.1). Thus N(G) is trivial or a 12-extension of the hyperbolic subgroup of Aut U fixing 0 and rx: (this group is isomorphic to the multiplicative positive numbers). To show we have a 1 2-extension, we must find just one Möbius transformation that conjugates C into C. Clearly, z 1-0 —(11z) will do this. We wish to see what N(G)/G looks like as a group of motions of 4„ First let us consider the mapping (4.6.1). Recall the holomorphic universal covering p:U —* A, is given by p(z) = exp( 2ni
log
log;,) •
Thus a (multivalued) inverse of p is given by
1 2ni
z )—* exp (- log A log z), From this it follows that z 1—* kz induces the automorphism (a rotation) z e acciz of d„ where ko = (2n log k)/log A. Note that these automorphisms are extendable to 5 4 , and fix each boundary component (as a set). The automorphism corresponding to z )—• — (11 z) is z rlz. This map also has a continuous extension to 54, and switches the boundary components. We see that Aut A, is isomorphic to a 12-extension of the circle group.
V.4.7. We examine next the torus. Let the covering group G of the torus T be generated by z 1-0 z + 1 and zi—• z + t 1m t > O. It is quite easy to see that A(z) = az + to (A e Aut C) is in N(G) if and only if a and at are generators for the lattice G. In particular, all the translations (a = 1) are in N(G). Thus, Aut T contains T(^.--t' C/G) as a commutative subgroup. (We have thus shown that Aut T is never discrete.) ,
279
V.4. The Exceptional Riemann Surfaces
When is Aut T a proper extension of T? First, since a and at are two linearly independent (over Z) periods that generate the group G we have
= 2 + f3T,
LIT
= 2T + fh 2 =
+ St,
(4.7.1)
with X =[; ,Pd E SL (2 2 ). In particular, we see that for (Aut T)IT to be non-trivial, T must satisfy the quadratic algebraic equation with integral coefficients pT 2 + (c( — syr — 7 = 0. (4.7.2) If /3 = 0, then y = 0 and a = ô = +1 and the equation (4.7.2) reduces to the trivial equation. Since a = 1 corresponds to a translation, the only new automo,rphism (a = — ljnin this case is the one corresponding to z 1— ■ —z. Thus Aut T always contains a 12-extension of T. If fi 0, the question becomes more complicated. We have of course shown that for most T (in particular for all transcendental t), Aut T is a 1 2-extension of T. Assume that # 0 0. Equation (4.7.2) asserts that the Meibius transformation corresponding to X E SL(2,Z)has a fixed point T in the upper half plane. Thus X must be an elliptic element. If A E Aut T, then we can write A uniquely as
A = A 1 0 A2, where .9, fixes 0 and A2 E T. To show that (Aut T)/T is finite we have to show that the stability subgroup of the origin is finite. We may assume that A E (Aut T) 0 is given by z at:. The number a E C * is, of course, subject to the restriction (4.7.1). Simple calculations show that
a = 1 X ± NI X 2 — 4 2
2
with x the trace of the elliptic element X e SL(2,7L)/+I. Since SL(2,1) has only finitely many conjugacy classes of elliptic elements, the number of a's is finite. (Since 0 < x2 G 4, we must have al = 1. As an exercise prove this another way.)
V.4.8. It remains to show that we have exhausted all M with Aut M not discrete. We may restrict our attention to those surfaces whose universal covering space is the unit disc (or the upper half plane U). Let G be the covering group of M. Suppose there exists a distinct sequence fn e N(G) such that lim„ f„-= 1. Choose 1 0 A E G and observe that lim„f„.Aof„-1 .21 -1 = 1. Since f, o A o f, o A'eG, there exists an N such that f„o A o f1 o A 1 = 1 for n > N. Thus f,, commutes with A for large n. If A is parabolic, then by Lemma IV.6.6 so is f„ for large n; and A and f„ have a common fixed point. Since 1 0 A E G was arbitrary, G is a commutative parabolic group and by Theorem IV.6.1, M is the punctured disc. Similarly, if A is hyperbolic, every element of G is hyperbolic and commutes with A. Appealing to the same earlier theorem, we see that M is an annulus.
CHAPTER vi
Theta Functions
We have seen in Chapters III and IV how to construct meromorphic functions on Riemann surfaces. In this chapter, we construct holomorphic functions on the Jacobian variety of a compact surface, and via the embedding of the Riemann surface into its Jacobian variety, multivalued holomorphic functions on the surface. The high point of our present development is the Riemann vanishing theorem (Theorem VI.3.5). Along the way, we will reprove the Jacobi inversion theorem.
VIA. The Riemann Theta Function In this section we develop the basic properties of Riemann's theta function. VL1.1. We fix an integer g > 1. Let eg denote the space of complex symmetric g x g matrices with positive definite imaginary part. Clearly, ag is a subset of the 1/2g(g + 1)-dimensional manifold X of symmetric g x g matrices. To show that Z, is a manifold, it suffies to show that it is an open subset of X. But a real symmetric matrix is positive if and only if all its eigenvalues are positive. Thus if r o E Ice r o has positive eigenvalues. Since this also holds for all r e X sufficiently close to r o, S9 is an open subset of X. The space CS g is known as the Siegel upper half space of genus g. (Note that E i is the upper half plane, U.) We define Riemann's theta function by
1 exp 2 14- iNTN +Wz), 2 Nag
o(z,i) = E
281
VI.I. The Riemann Theta Function
where z E Cg (viewed as a column vector) and r E Z 9, and the sum extends over all integer vectors in Cg. To show that 0 converges absolutely and uniformly on compact subsets of Cg x ag, we review some algebraic and analytic concepts. First of all, as we already observed, every (real) symmetric matrix has real eigenvalues. Next, every real symmetric positive definite matrix has positive eigenvalues. Furthermore, for a real symmetric matrix T, the smallest eigenvalue is given as the minimum of the quadratic form (here ( , ) is the usual inner product in Rg) X E. Rg, Ilxii = 1
(TX,X),
(we use to denote the Euclidean norm). we let A(T) = minimum eigenvalue of 1m T. Then A(r) > 0, For z e and if K is a compact subset of Eg we also have min {A(T)} A o > 0,
r ER
because
ag X R9 9 (T,X)1--).
r)x,x) e R
is a continuous function. To show convergence of 0 on compact subsets of Cg x Zg, it suffices to show convergence in a region K = {(z,T)E C
X
M and
eg ;
;(T)
> 0).
We note that
lexp 2ni(itNTN + 1Nz)1= exp Im(--R`NrN — 27eNz). Now by the characterization of minimum eigenvalues: / N N \ Im(-7eNTN)= —n((lm r)N,N) = u11N11 2 Vim 1. ) -) ,
IINII IINII
--/r1INPA(r) and by the Cauchy-Schwarz inequality
1m( — 270 s z)
2nMliNli.
24Nzi = 2n1(z,N)I
Thus an upper estimate for the absolute value of the general term of (1.1.1) is
exp(
+
= exp( — nlIN11 2.1.0 (1
Now only finitely many terms N e Zg satisfy the equation 2M
IINIP-0
1 2
IIN 2A110 )).
282
VI Theta Functions
(because N is an integer vector). Thus for all but finitely many terms, the general term of (1.1.1) is bounded in absolute value by
exp( — lir). 011N11 2), and our series is majorized by
E exp(--21 70.01!NI1 2) = E
N EZ0
exp(--,..1120
g
E
E
ni)
J=1
4
Neill
1 10. 04 2
exp(
J=1 nj=
which converges since each factor converges (by the Cauchy root test, for example). We have shown that a is a holomorphic function on C9 x We shall be mostly interested in this function with fixed "C E In this case, we will abbreviate 0(:,r) by 0(4
Remark. The one variable theta function 1 n't + nz], = E exp 2 n=—
C, E U,
Z
(1.1.2)
already has a non-trivial theory.
VI.1.2. We now proceed to study the periodicity of the theta function. Let I. = I denote the g x g identity matrix. Proposition. Let p, p' E Z. Then 0(z + IA' + rp,z) = exp
—`pz — 4jrri.d0(z,r),
(1.2.1)
for all z E Cg, all r E PROOF.
0(z +
We start with the definition (1.1.1)
+ -cp,t) 1 exp bar'NrN + tiV(z + I p' + Ty)] 2 Nezg
=E =E
1 1 exp 2.7ti[- '(N + p)-c(N + p)+ t(N + p)z — — `ptp + 2
Ne In
2
— 527.1
(to obtain the above identity use the fact that t t =
exp lni[—raz — pap] exp 27ri[i t(N + p)T(N + p)+`(N + p)z]
= Ne Zg
(because 'N
E Z)
= exp 2ni[—` /Az — 4tyrid0(z,r) (because (N + p) is just as good a summation index as N).
283
VI. I. The Riemann Theta Function
Corollary 1. Let eu ) be the j-th column of I. Then 0(z + eu),r)= 0(.7.T.),
: e C9, all T
PROOF. Take It' = eu),t = O.
the k-th column of s, and Tkk the (k,k)-entry of
Corollary 2. Let t
rkk 0(z + r(k),r)= exp 2ni[ - z - —]0(:,r), k
2
T.
Then
all : e tg, all r E S9, (1.2.3)
where z k is the k-th component of:. ..PROOF. Take y' = 0, y = e(".
0
Remark. It is also easy to verify that 0 is an even function of z. Thus in addition to our basic formula (1.2.1), we also have 0( - z,r) = 0(:,t),
all z e Cg, all
ag.
TE
(1.2.4)
VI.I.3. Formula (1.2.1) suggests that we should regard the 2g vectors eU ), j = 1, . . . , g, ru), j = 1, . . . , g, as periods of 0(:,r). Of course, only the first g vectors are actually periods; however, if we consider the g x 2g matrix, Q = (/,,T), we can rewrite (1.2.1) as 1 2
0(z + S2[11 ] r) = exp 2niRA[11 ])z - - 'pry] 0(:,t),
(1.3.1)
where A is the g x 2g matrix (0, If we now denote by e , k = 1, . , 2g, the 2g columns of the 2g x 2g identity matrix we can rewrite (1.2.2) and (1.2.3) as 0(z + Qe(k) ,r) = exp 2/ti[t(Ae (k) )z + yk]0(z,r), (1.3.2) where yk is the kth component of the vector y in C29 defined by
...
...
A holomorphic function, f, on C9 which satisfies the conditions fiz + Qe(k))= exp 27ri['(2ek) )z + 7k] f(z),
k =-- 1, . . . , 2g,
all z E C9, with Q and A g x 2g matrices and 7 E C 2g, is called a multiplicative function of type (52,A,y). What we have seen here is that 0(z,$) is a multiplicative function of type (Q,),,y) on C9 with Q = (LT), A = (0,- I), y = - Y(0, . ,O,r i 1 , ... ,T99). Given Q, A, and 7, we can ask whether there exist multiplicative functions of this type. We shall, however, not pursue this question any further. There is, of course, an intimate connection between these questions and function theory on complex tori (recall III.11).
VI.1.4. More generally, we shall consider in place of 0(z) the function obtained by taking a translate of z namely 0(z + e) for some e e C. It is clear that (1.4.1) 0(z + e + et")) = 0(z + e),
284
VI Theta Functions
and that Tkk
0(z + e + T( ) = exp 2n[ i - z k - ek - — 10(z + e). 2
(1.4.2)
The connection between theta functions and the theory of multiply periodic functions on Cg is seen rather clearly when one considers two distinct points, d and e E Cg, and sets
0(z + e)0(7 - e) = f(z). d)
(1.4.3)
It is clear from (1.4.1) and (1.4.2) that f(z) in (1.4.3) is a periodic meromorphic function on Cg with periods eti), rtn, j = 1, . . . , g. Recall that the columns of the g x 2g matrix (I,r) are linearly independent over R. Hence, every e e Cg can be uniquely expressed as e = le' + TE with e, e' points in R9. Hence 0(z + e) = 0(z + + TO. This suggests that for any e, e' E R9 we define a function on Cg xC.A. by
2s 1(z,r) = 0[2s'
1 exp 274- t(N + c)r(N + e) + `(N + e)(: + 2 NeZe
y
(1.4.4)
We observe at once that
= and in general
26 0[ 1(z r) = 2er
N
is
I exp 276[- :A/TN + ' N(z + 2
, 1, = exp .ari[- 'STE 2
t EZ
t E816(Z
1
— 'et& + 2 +
+ TE,T).
tez + (1.4.5)
From the last equality it follows that for integral p, p' e Zg, we have (because of (1.2.1)) 0[28 1(Z + lie
2s'
T/4,
1 = exp 27ri[-
2
ETE
t EZ
x 0(z + le' + te + 1
exp 21rik'sre +
t EE' t EGI ' + TA, i)
+ tee'] exp 21i[e(ii + Ty)
1 - '1(z + le' + Te) - - '114(1 0(z + le' + se, T) 2 1 2e = exp 24-'tap- f izz + tp's - tpte1[ 1 2 2e' (1.4.6)
285
V1.1. The Ricmann Theta Function
In particular, we have
26 0 [ 1(z + e(*),T) = exp 27ri[ek] Or]
2e
(1.4.7)
and
0[26 ](z + z (k),T) = exp 2e'
E 4 0 2-- e, (Z,T).
zk
(1.4.8)
VI.1.5. For us the most important case is when 24 and 24 E Z. Let e, e' e Z9. Then OM (z,t) is called the first order theta- function with integer characteristic D.1 Proposition. The first order theta function with integer characteristic [e] has the following properties: 6 ](z + e(*),T)= exp 2niNO[ E,](z,T), 2 e 0[c'
(1.5.1)
9[6 1(z + r( , = exp "'Tit — k —
(1.5.2)
0
— ]0[e ](z,-r), 2 s'
2
(_z,T)= exp 276[-10[8e ' (z,r), [:] ]
(1.5.3)
and
or+ 2v 1 (zr) = ex p 2 7t i — tcv' 0 (z 2 E' [e
(1.5.4)
(Here y and y' are integer vectors in Z9.) PROOF. Equations (1.5.1) and (1.5.2) are simply the restatements of (1.4.7) and (1.4.8). Equations (1.5.3) and (1.5.4) are derived in the same way as (1.4.5), (1.2.4), and (1.2.1). LI
The importance of the statements (1.5.3) and (1.5.4) is that we immediately have the following Corollary. Up to sign there are exactly 229 different first order theta functions with integer characteristics. Of these 29- '(29 + 1) are even functions, while 29- '(29 — 1) are odd. These 229 functions can be thought to correspond to the 229 points of order 2 in Cg/ means the group of translations of C9 generated by the columns of (./,T). PROOF. The only part needing some comment is the number of even and odd functions. This is established by an easy induction argument. LI
286
VI Theta Functions
EXAMPLE. When g -- 1 the three even functions arc: O[g](z,r), 0[?](:.-r), and OW(z,r). The odd function is 0[1](z,r). When g 2 the six odd functions are:
[i o ](:t)
0,,
o r0 1,i (z,r),
i o[i ].7A
o[
(
1 ](-7,0, o[
1
Remarks 1. For the first order theta functions with integer characteristic, 0[:.](::,r), the even ones are precisely those for which t e.E' 0 (mod 2) and the odd ones those for which iCe' -Z4 1 (mod 2). 2. Formula (1.4.5) applied to the case of integer characteristic yields [ ]( ,T)
1 re re = exp 2ni [— + — t+ — —Z
re
1,(
--
e
Z+ / — +T—
In particular, if 0[:,](z,r) is an odd function, we have 0 = 0[](0,r) = exp 2riRee/2)T(e/2) + (tel2)(672)]0(I(e72)+ r(e/2),r). The points 1(672) + r(e/2) are points of order 2 in Cyl,r> and will be called even or odd depending on whether Oraz,r) is an even or odd function. Hence we immediately see that the Riemann theta function 0(z,t) = 0[8](z,z) always vanishes at the odd points of order two or at what we shall call odd half-periods.
VI.2. The Theta Functions Associated with a Riemann Surface In the preceding section we have defined and derived some of the basic properties of first order theta functions with characteristics. In this section we show how to associate with a compact Riemann surface a collection of first order theta functions and study the behavior of these functions as "functions" on the Riemann surface. The basic result of this section is the following: A first order theta function either vanishes identically on the Riemann surface or else has precisely g zeros (g = genus of the surface). In the latter case, we can evaluate the zero divisor of the theta function (more precisely, we can determine the image of the divisor in the Jacobian variety of the surface).
VI.2.1. Let M be a compact Riemann surface of genus g > 1. Let t{a,b} ,691 be a canonical homology basis on M, and let = be a basis for dri(m) dual to the canonical homology basis (recall 111.6.1).
VI.2. The Theta Functions Associated with a Riemann Surface
287
We thus obtain a g x 2g matrix (1,17) with ell) = C, nu) = jb, C, where as usual eland 1t1» are the jth columns of the matrices 1 and 17 respectively. We have already seen that the matrix 17 is a complex symmetric matrix with positive definite imaginary part. It follows therefore, from VI.1.1, that we can define first order theta functions with characteristics using the matrix 17. We therefore obtain, as in VI.1.5, 229 theta functions 9 [e](z,17).
VI.2.2. There are, of course, many ways to choose a canonical homology basis on M. If {a'1 , ,k) = 1 {a',6'} is another canonical homology basis, and ttr,',, ... 4,1 = t: is the basis for if i(M) dual to this new basis, we obtain a new g x 2g matrix (I, 11'), and can define first order theta functions with characteristics using the matrix II'. We have seen in Chapter V that AB ird] Lid =Lc D. jLb with t] e Sp(g,Z). Thus using obvious vector notation
rdi F
a' = Aa + Bb, and hence
S . ; = S4a, Bb
A + Bll,
C' = (A + B17) -1 C, and 17' = C' =Sc. Db (A + BI7) -1 C = (A + B17) -1 (C + D17). Thus II and 17' are related by an element of the symplectic modular group of degree g. The relation among the corresponding theta functions involves then the study of modular forms for the symplectic modular group (which will not be pursued here).
VI.2.3. We shall for the remainder of this chapter assume that the Riemann surface M has on it a fixed canonical homology basis ' {a,b} = {a 1 , . ,b9}, and we will study the theta functions associated with M and qa,b1 as functions on M. We do this by recalling that we have introduced in 111.6 a map 9:M J(M) = Cg/L(M), where L(M) is the lattice (over Z) generated by the 2g columns of the matrix (1,17). Recall that q was defined by choosing a point Po e M and setting 9(P) = C, where C = t (C i , ,C9) is the basis of OW) dual to ' {a,b}. We now consider 0 0 9 and in this way view 0 or 0[ ] , where [ e ] is an integer characteristic, as a function on M. The reader is, of course, aware that 0 0 9 is not single valued on M, since 9 is not single valued (as a function into C9) on M, but depends on the path of integration. We have seen (Proposition 111.6.1) that the map 9 is well defined into J(M), and therefore the function 0 o q has a very simple multiplicative behavior. The behavior of this function is not quite as simple as the behavior of the functions considered in 111.9, since here the multiplier
VI Theta Functions
288
x acquired by continuation over a cycle depends on the variable z as well as the cycle. At any rate, it is immediate from Proposition V1.1.5, that continuation of O [ .] c (p along the closed curve a, (respectively, b,) beginning at a point P e M, multiplies it by x(a,) = exp 246,/2],
(2.3.1)
X(bi) = exp 2iti[ —4/2 — it,,/2 — (PAN,
(2.3.2)
or we may say more simply that analytic continuation of Oft] o (p over the curve a, (respectively, b,) beginning at P carries OM o y) into exp 2ni[el/2]O[,] o (r)
(respectively, exp 2iti[—s;/2 — 7r 11/2 — y),(P)]0[] o cp), where q), is the /th component of the map ça. It therefore follows that Oft] 0 ça is a holomorphic multiplicative function on M with multipliers given by (2.3.1) and (2.3.2). VI.2.4. We now wish to study the zeros of O[] o q) on M. We observe that even though Oftaz,/7) is not a single valued function on J(M) = Cg/L(M), the set of zeros of O [ .](:.11) is a well defined set on J(M) (Proposition VI.1.5). The set of zeros of Oftl(z,z), for any symmetric s with 1m r > 0, is an analytic set in Cg of codimension one and in particular for T = 17, the zeros of Ofti(z./7) form an analytic set of codimension one in J(M). The map (p, by Proposition 111.6.1, is a holomorphic mapping of maximal rank of M into J(M). The study of the zeros of O [ .] o q) on M is thus the study of the intersection of the image of M in J(M) under (f) and the analytic set consisting of the zeros of û[l(z,/7). Since M is compact there are only two possibilities. Either Oftl 0 q; vanishes identically on M or else Oft] o y) has only a finite number of zeros on M. Neglecting for the moment the former possibility, can we determine how many zeros does Oft] o y) have on M? It involves no loss of generality to assume that no zero of O [z .] o (p lies on any curve chosen as a representative for the canonical homology basis. In order to count the number of zeros, we need to evaluate 1
d log OM o (p,
where If is the polygon associated with M whose boundary consists of representatives for the canonical homology basis OM = [ljO I aibia7 '1); 1 ). In order to further simplify notation we shall denote O[] 0 q) by f. We now compute 27E1 h-a f
27ri 1
=
Jak+bo r
+
df f r (cif
df
+
ibk
df
289
VI.2. The Theta Functions Associated with a Riemann Surface
where f - denotes the value of f on the cycle ak-1 or bk-1 . It follows immediately from (2.3.1) and (2.3.2) that if P is a point on ak Ck f - (P) = exp 2n[ i ---
_.... -n kk
2
2
—.
while if P is a point on 6,, f(P) = exp 2701f (P). Thus for P a point of b - df /f )(P) = 0, while for P a point of ak, (df/f df /f )(P) = Ini9(P). It therefore -follows that (1/2ni) J (df/ f) = (1/2710 Ef.„ , fak 27Ci9;,(z)dz = g. We are here exploiting the fact that 91(z)dz is the kth element of the basis of ,r)(M) dual to the homology basis (a l , ,b9 }. At any rate, we have shown that if O[] o q, is not identically zero it has precisely g zeros on M (counting multiplicities). The next natural question which arises in regard to the zeros is: can we find the divisor of zeros? What we shall do is find the image of the divisor of zeros of OM o go under the induced map on integral divisors of degree g, 9: Mg J(M). To this end we consider (1/2ni) J,4 9(df/f), where 9 is the column vector `(9 ,q,) and f = Orj yo. We immediately find that
= f _ df = r 27ri k= 1 Jak+ bk+a + DA' 27ri f
=
41* 97 ."
[ r (4) df_ _
1
27ri kL : =1 Jak
f
df -)
9
r ibk
df 9
_ df f
-
where (p - plays the same role for 9 as f - served for f. We once again use the relation between f - and f and the fact that for P a point on a,,, 9 - = 9 n(k), while for P a point on bk, 9 = (p - e ). Thus 1
=
Fr
Liak 9
df
y—
fbk Op - e,‘L= ' a k
L
df
+ n (k))(i k dz) fd - 27d9' (z)
- 9 _ df -]-
n(k) dff + 2nin(k) 9(z)dz + 27ci99;,(z)dz
1 g f 00 df e +— f-• 27ri =
E
We now need to evaluate integrals of the form f,,k (df/f) and fbk (df/f). Since df/f = d log f, the integral in question is simply log f(P 1 ) - log where P 1 = P2 are the initial and terminal points of the cycle a„ (or kJ. (The difference is not zero, since we must use different "branches" of f.) We now once again use (2.3.1) and (2.3.2) to obtain for P i and P2 initial and
290
VI Theta Functions
terminal points of a„, f(P2)= exp[2ni(sk/2)]f(P,), and for P, and P2 initial and terminal points of b,./.(P2) = exp 2iti[ -(412 ) - Orkki 2 ) - (Pk(P fiP 1). This shows that
= 1
g -
t k, 1
9
E 4ni
k= 1
it
bri
2ni
+ 2nink + n(k)2ni +
f ahcp(p(z)dz
Ek nkk - pk) + 2itim k], e0127,i(--— 2 2
where mk and nk are integers. It therefore follows that
1
=— 27r1 Li g
9
df f
E
ek) s'
2
2
2kk + I (pcp'k dz - e( 5--
n(k) (1 — nk )
(pet)
or finally E
= - TI - - / - + lin + lm - K, 2 2 where 9
K=-
k
E [f
ak
7rkk cpcpkdz - em(-+ (p k(P,))].
2
(2.4.1)
Remark and Definition. The vector K depends, of course, on the choice of the canonical homology basis and the choice of the base point P o of the map (p. (There is an illusory dependence on the base point P, of n 1(M) which can be dispensed with by choosing P, = Po in the above.) The vector K is known as the vector of Riemann constants. We have proved the following Theorem. Let M be a compact Riemann surface of genus g 1 with canonical homology basis tfa,b1. Let O[e](z,17) be the first order theta function associated with (M,({a,b}) and let cp be the map M J(M). Then 0[:.] o p is either identically zero as a function on M or else has precisely g zeros on M. In this case let P, • • • Pg be the divisor of zeros. We then have (p(P, • • • Pd = - lit/2 - Is72 + lin +1m - K, where n and m are integer vectors and K is a vector which depends on the canonical homology basis `{a,b} and the base point of the map (p:M -* J(M). In particular, since fin + Im is the zero point of J(M), we have (P1 • • • Pg) + K = Ilc/2 - 1872. Remark Consider the "function" on M P 1-* 0((P) - e),
291
VI.3. The Theta Divisor
with e E Cg. If this "function" does not vanish identically on M, it has g zeros P,, , Pg. The theorem asserts (sec (1.4.5)), that y(P i • Pg) + K = e (where in the last equation we understand by e not the point in C 9, but its projection in Cg/L(M)). Our remark is valid because in the proof of the theorem € and e' were arbitrary points in tRg (we never used the fact that the characteristics were integral).
VI.2.5. We wish next to characterize the zero set of 0r,1(z,11) as a subset of J(M). In view of (1.4.540nd (1.4.6) it suffices to consider only the functions . of the form zi-■ 0(z - e), with fixed e e Cg. It is now convenient to review some concepts that we studied in 111.1 1. Recall that for every integer n > 1, M,, denotes the integral divisors of degree n on M (or equivalently the n-fold symmetric product of M). Using the mapping 9: 211„ J(M), we can view 0 as a locally defined holomorphic function on M„. Also, W„ denotes the image in J(M) of M„ under the mapping (p. Recall that for De M„ (Proposition III.11.11(a)), the Jacobian of the mapping cp has rank n + 1 - r(D - I), which by RiemannRoch equals g - i(D). Taking n = g and a non-special divisor D e M., we conclude that J(M) is a local homeomorphism at D. Thus since 0 # 0, 0 does not vanish identically on any open subset of Mg.
VI.3. The
Theta Divisor
In this section we begin a study of the divisor of zeros of the theta function, culminating in the Riemann vanishing theorem (Theorem VI.3.5), which prescribes in a rather detailed manner the zero set of the 0-function on V. We also obtain necessary and sufficient conditions for the 0-function to vanish identically on the Riemann surface (Theorem VI.3.3) and as a byproduct, an alternate proof of the Jacobi inversion theorem.
VI.3.1. Theorem. Let e e C°. Then 0(e) = 0 if and only if e e U'_ 1 + K, where K is the vector of constants of Theorem VI.2.4. PROOF. We first show that if e e Wg _ + K, then 0(e) = O. To this end, choose a point = P 1 • • • Pg in Mg with P, 0 pi for k éj such that i() = O. If' is any other point in Mg sufficiently close to then since 40 = i(P " ' Pp) = g - rank(CaP)), i(Ç) = 0 as well. Set e = to( ) + K and consider OP) = 0(9(P) - e) as a function of P e M.
292
VI Theta Functions
There are now two possibilities to consider. Either i/c is identically zero or not. In the former case, we have for each k = 1, . , g, 0((p(P,) — (OP, • • • P9) + K)) = O(—(P 1 • • • = 0((P(P • • • Pk
P„ • • • Pg)• • P g) +
K)
K)= 0
(where we have used symbol P 1 • • Pk • • • P9 to mean Pk does not appear in the divisor) because (by (1.2.4)) 0 is an even function. In the latter case, we have from Theorem VI.2.4 that ik has precisely g zeros on M, say, Q1, , Q9 , and that (p(Q, • • Q9) + K = e. Since ,`," = P, • • P9 was chosen so that i(P i P9) = 0, it follows from the Riemann-Roch theorem and Abel's theorem that Q, • • • Q9 P, • • • Pg. Therefore, even in this case we have 0((p(P2 ' • ' Pk • • P9) + K) = 0 for each k = 1, . , g. The remark at the beginning of this proof showed that the divisors of the form under consideration, = P, • • • P9, with Pi 0 P, for j o k and i(;)=-- 0, form a dense subset of M9 . Thus divisors of the form P, - • Pk • • • P9 form a dense subset of M9 _ Hence 0 vanishes identically on W9 _ 1 + K. Conversely, suppose 0(e) = O. Let s be the least integer such that — — e) 0 but 0( W, — W, — e) # O. Here W, — W, has the obvious meaning (e e W, — W, e = f — h with f,heiV,.). Our remarks at the end of VI.2.5 immediately give 1 s g. The hypothesis we have thus far made assures us of the existence of two points in M, say 4. = P1 • • • co = Q, • • • Qs such that 0((p(P 1 • • • Ps) — cp(Q, • • • Q 3)— e) O. We can assume without loss of generality that P1 0 Pi for k j, Q„ Q. for k é j, and P,, é Qi for all k, j. Consider now the function P 1—* 0((p(P)+ 4 (P2 • • • P,)— cp(Q, • - • Qs) — e). This function does not vanish identically as a function on M, since P = P, is not a zero. On the other hand, it is immediately clear that P = Qi, j = 1, . , s, is a zero of this function. It therefore follows from Theorem VI.2.4 that for some T, • • • 7"9 _, e M9 _, (AQI• • •
Q.) - (P(P2 . • •
PJ) + e = (P( 421 • •
42.Ti
9 „)
+ K (3.1.1)
or that e = cp(T, • • - T9 _,P2 • Ps) + K,
(3.1.2)
which is obviously a point of W9 _ 1 + K. (Note that the above arguments El work also for s = 1.) VI.3.2. We now observe that we can really prove a bit more than we have claimed in VI.3.1. The points = P, • • Ps and co Q, • • • Q, utilized in the above proof, were fairly arbitrary in the sense that any other points t;', co' e M, which are sufficiently close to and w would have worked as well. We also found that e = cp(T • • T 9 ,P2 • • • Ps) e M 9 _ 1 has at least s — 1 "arbitrary" points P2 , , Ps in it. Let D' = ' • • Ps be arbitrary but
293
V1.3. The Theta Divisor
close to P2 • • ' P,. The argument of VI.3.1 shows
e = (p(T, • • • T. s P2 • • • PJ+ K ça(T', • • • T'9 _,P, • • • P's) + K. Thus T, T._ sP, • • • Ps — T',- • • T._ 5 1Y2 - • • P', by Abel's theorem. Lemma 111.8.15 now implies that r(1/7- 1 • • T9 -5P2 ' P3) s and thus (by Riemann-Roch) that i(T, • - T 9 _ 5P2 • • • P5) s. On the other hand, suppose that e = (WO+ K with d e M9 _ 1 and i(d) = s. Then d has precisely s — 1 "arbitrary" points. Any point X of Ws _ — Ws _ — e is of the form cp(P 1 - • P„,) (p(Q, • • Q 3 _ 1 )— e, with e = 9(41+ K = (p(P, • .....P s _ 1 3) + K and e Mg ,. Therefore, we have P.--16) — K X = (P(Pi ' • ' — (1)(421 • Q5-1) -(P 1 • • •
= — 4) (Q1 • • Os--1 6) — K;
and X e —(14/9 _ + K): and 0(X) = O. We can therefore now strengthen slightly Theorem VI.3.1 to the following Theorem. For e e C9, 0(e) = 0 if and only if e e and e = (P(;) + K with i: e M 9 _ 1 and i() = s 1),
+ K. /f e
e 1479 _, + K,
then 0(Ws -- — W,_ — e) O. Conversely, if s is the least integer such that 0(W,, — Ws _ — 0 but 0(1K — W — e)# 0, then e = 9(;)+ K with e 9 _ 1 and i() = s.
PROOF. All but the last remark is contained in the discussion preceding the statement of the theorem. In the discussion we only showed i(C) s; however, the statement i(;) = s is an immediate consequence of the one preceding it. If i(C)> s, we would have O( W3 — W5 — e) 0 contrary to the hypothesis.
VI.3.3. In the above analysis we have several times encountered the multivalued holomorphic function on M :P
0(cp(P)
e),
with fixed e e J(M). (Recall that since cp depends on the choice of base point Po e M, so does the above function.) We have observed that ti/ either vanishes identically or else has precisely g zeros. When does each of these possibilities occur? The answer is contained in the following Theorem a. If e e J(M) and
-L=. 0, then e = cp(D) + K for some D s > 1, and s is the least integer such that
0 1 4 s+i —
—
e
M. with i(D)=
O.
b. If e = cp(D)+ K, D e M9, then tk # 0 if and only if i(D)= 0 and D is the divisor of zeros of O. In particular, III 7.. =. 0 if and only if i(D)> 0.
294
VI Theta Functions
Let s be the least integer such that 0( Ws , — W, — e) , Q, such that s 5 g — 1. Choose P 1, ... P,, i, Q 1 ,
PROOF.
0
019(P1 •
i) — (DWI • • Q.) —
O. Then
O.
Consider the function P
0(49(P) + (P(P 2 • ' ' Ps+ 1) — (1)(Q1
• Qs)
—
which does not vanish identically (it is non-zero at P1 ). It then has (by the minimality of s) g zeros Q i, • • •
QS ,
Ti, • • • , Tu-s•
Thus, again by Theorem VI.2.4, we conclude that 60(Qi • • 12,T or that
• • T g -,)+ K = (t)(421 • • Qs) — (P(P2' • Ps+0+ e;
e = 9(T i - • T g -sP2 • • " Ps4-1)+KeWg +K.
As before, the divisor D = T1 - • T g ...,P2 • • P„1 has s free points. Thus r(D -1 ) s + 1 and (by Riemann-Roch) i(D) s. We have established part (a), since ik a 0 implies s 1. To establish part (b), note that by (a) if i(D) = 0, then tp 0 O. Conversely, suppose i(D) > 0 and ik 0 O. Choose Q e M such that iii(Q) o O. Lemma 111.8.15, implies that there exists a D' e Mg _, such that 9(D) = 9(QD'). We now use Theorem VI.3.1 to conclude that 1,0(Q) = 0(9(Q) — e) = 0(9(D) — 9(D') — e) = 0( — 9(D') — K) = 0(9(D') + K)= O.
This contradiction shows that tfr O. To conclude the proof of part (b) we must verify that whenever e = 9(D) + K, D e Mg, and i(D) = 0, then D is indeed the divisor of zeros of Let D' be the divisor of zeros of then p(D) = 9(D'). Since i(D) = 0, this implies that D = D'. VI.3.4. As a consequence of Theorems VI.3.1 and V1.3.3, we obtain an alternate proof of the Jacobi inversion Theorem. Given a e J(M), there exists a D e Mg such that 9(D) = a. PROOF. View a as a point in C. Let e = a + K. Consider the function P 1-0 0(9(P) — e). If this function does not vanish identically, it has as its zero set a divisor I) e M g with 9(D)+K=e=a+K,by Theorem VI.2.4. If the function does vanish identically, then the last equation follows from Theorem V1.3.3. 0
VI.3.5. Lemma. The condition 0(W, — W, — e) derivatives of 0 of order O. We have 0(p(P 1 ' • • P.) — (P(Q, • " Q,)— e)= 0 for all D = Pi • • • Pr M„ all D' = Q i • • Q, e M,. Thus PROOF.
0(40(P) + 402 • • • Pd — (.0 (Qt • ' e) vanishes identically on M as a function of P. If we now expand this function in a Taylor series about the point Q, and set P = Q1, we find the coefficient of a local coordinate at Q i to be (130/az)((aP2 • • •
Pr) — (1)022* • • Qd —
e)C;( 121),
i=
which must vanish. Moreover, since Q 1 on M is arbitrary, we find that Carezi)((p(P2 • • • P,)— (p(Q 2 - • Q,)— e) = 0 (by the linear independence of the Ci's for j = 1, . , g), and this holds for all points P2 ' • P 0 Qr 6 M,_ 1 . We conclude that (00/0:•;)( W,_ i — — e) 0, for j = 1, . . . , g. We now simply repeat the argument for each of these functions and continue until we find all partial derivatives of 0 of order vanish at — e, and therefore also at e. D -
• • •
This lemma together with Theorem VI.3.2 gives us a substantial portion of Riemann's theorem. We have seen that any zero, e, of 0 is of the form = s, e = cp(C)+ K with C e M9 _ 1 . Furthermore, we have seen that if then O(Ws _ — Ws _ 1 — e) Es. 0, and therefore all partial derivatives of 0 of order less than s vanish at e. We can thus say that in this case û vanishes to order at least s at e. The Riemann vanishing theorem asserts that s is the precise order of vanishing at e; that is, at least one partial derivative of order s does not vanish at e. We now return to the hypothesis of the last part of Theorem VI.3.2 and observe that this hypothesis allows us to conclude the existence of three distinct points C, w, r e M,, with the additional property that the points in these divisors are all distinct, and 0((pg)— (p(co)— e)
0,
(3.5.1)
6VP()
0,
(3.5.2)
0.
(3.5.3)
(P(r) — e)
0(9(w)— 9(0 — e)
Equation (3.5.1) is a consequence of the hypothesis 0(W3 — W, — e) # 0, while (3.5.2) follows by continuity from (3.5.1) for all T near a). Finally, if for all r' near co, 0(cp(co)— (p(f)— e)= 0, then by varying co we would conclude once again that 9(W, — Ws — e) 0, contrary to hypothesis. Consider now
0 (9(P • • P.) — (P ( D) 0 (4)(PI' • P.) — (P(T) —
as a function on Ms in a neighborhood of. By (3.5.1) and (3.5.2) this function is not identically zero on M„ since neither numerator nor denominator
296
VI Theta Functions
vanish at ‘;'. We would rather view this for the moment as a function on M, and in order to do so we consider LP
0(4)(P) + (I)(P 2 ' • • PJ -
-
014)(P) + go(P2 ' • • P.) -
- e)
and observe that as a function of P on M it is also not identically zero. The same arguments which were used in V1.3.1 give that the divisor of zeros of the numerator is coy with y e M. 3, and the divisor of zeros of the denominator is r 6. Furthermore, Theorem V1.2.4 when applied to numerator and denominator separately gives e = cp(P2 • • • Ps) + (p(y) + K = cp(P2 • • • Ps) + (p(b) + K;
from which we conclude q(y) = We now claim y = & If y 0 ô then Abel's theorem gives us the existence of a non-constant function in L(y - 1 ). The Riemann-Roch theorem gives r(7 = g - s - g + 1 + i(y) = i(y) + 1 - s. Now r(y - I) 2 yields i(y) s + 1; which then yields i(wy) 1 (since i(wy) i(y) - deg co). We will obtain a contradiction by showing that i(coy) > 0 implies 0(4(P)+ cp(P2 • - P5) (p(co) - e) 0 as a function of P on M. We conclude from Theorem V1.2.4 that e + p(w) - cp(P2 • • - P5) = cp(coy) + K so that we may rewrite our function as P 1-4 000(P) - (49 (0)7) + K)).
By Theorem V1.3.3(b), this function vanishes identically, showing that y = 6. Therefore, the function f viewed as a (multiplicative) function on M vanishes at the points of w and has poles at the points of r (and is holomorphic and non-zero elsewhere). It is now necessary to analyze the multiplicative behavior of the function f on M. It follows immediately from (1.4.1) and (1.4.2) that continuation of this function over ak leads back to the original function while continuation over the cycle bk beginning at P leads to exp 27rikp 1(co) - cp k(t)] f(P), where cpk is the kth component of cp. Recall now the normalized differentials of third kind r,,Q with simple poles at P and Q with residue + 1 at P and - 1 at Q. We found (111(3.6.3)) that
1 2ni Jbk
t PQ = (Pk(P)
k(Q).
Let us now consider the function P
g(P) = exp(S P°
E
1
'Cm)
k
„ ex
(f"
TR
4
where co = R 1 • • • R„ T = T, • • • T 5. It is clear that g(P) and f(P) have the same zeros and poles and the same multiplicative behavior. Hence it follows immediately that f(P) = cg(P), with c a constant which may depend on
VI.3. The Theta Divisor
297
Ps , since both f(P) and g(P) depend on P2 • • • P. If we write P = P 1 , however, we observe that both f and g are symmetric in I) 1 , , Ps ; which implies that c is independent of the points P„ and thus is an absolute constant. We therefore have P2 ' ' •
0(01' 1 • • • Ps) — qi(co) — e) = c0 (q)(P " Ps) — 4)(0 — e)
11
exp[fpPk
k=1
°
r]. 1
(3.5.4)
j
We now differentiate (3.5.4) with respect to a local coordinate z at P, and set P =-R„ to obtain 9,i= ,
cz i
(q)(P2 • Ps) —4(R 2 • • Rs) — egi(Ri)
= c[E(R 1)
azi 0)(R P 2 ' • • P — Q(r) — eg;( 1? i)
(3.5.5)
+ 0 (tP(RiP • • • Ps) — (p(z) — e)dE(R
where E = fJ exPgt 7% R,T). Since R 1 is a simp e zero of E we have E(R 1 ) = 0 but dE(R,) 0 0, where we easily compute the derivative S
dE(R 1 ) =
ri
k= 2
ex p (IPo L
j= I
by properly choosing the local coordinate z at R, with respect to which we differentiate. We now continue the process by differentiating (3.5.5) with respect to a local coordinate at P2 and set P2 = R2 to obtain
E J. g
k=
32e (q)(P3 • • •
:k 1uZjC
Ps) — (1)(R 3 • • Rs) —
e)C(R 1 )Ck (R 2)
= c0(cp(R ,R 2 P 3 - • Ps) — T(T) — e)(d 2 E(R 1,R 2)).
(3.5.6)
We have used once again that R2 is a simple zero of E, and therefore O. We continue this process arriving finally at the sth stage and obtain
(dE(R,))(R 2)
ass?) . . .
= c0(9(a)) q(t) — e)(ds E(R 1 ,
••
Vgg(R s) (3.5.7)
It follows from (3.5.3) that the right side of (3.5.7) does not vanish. Hence the same is true for the left side and therefore it is surely not the case that all sth order partial derivatives of 0 vanish at — e (and therefore also at e).
298
VI Theta Functions
Theorem (Rieinann). Let s be the least integer such that û(H'_ — W, — e) 0 but 0(W3 — W — e) O. Then e = + K with Ce M g .., and i(C) = s. Furthermore, all partial derivatives of() of order less than s vanish at e while at least one partial of order s does not vanish at e. Conversely, i f e is a point such that all partials of order less than s vanish at e but one partial of order s does not vanish at e then we have e = cp(C) + K with C e Mg _ , and WO= s. The first claim in the theorem is precisely the last statement of Theorem VI.3.2. The Lemma then gives that all partial derivatives of order less than s vanish at e. The statement that at least one partial of order s does not vanish at e is precisely the remark following (3.5.7). We now turn to the second (converse) part of the theorem. We assume that all partials of Oof order less than s > 1 vanish at e. Hence by Theorem VI.3.1 we have e = 9() + K with E A/4_ 1 , and the only question is: What is ig) equal to? First, i(;) cannot be less than s. If it were, then by the first part of the theorem it could not be the case that all partial derivatives of order less than s vanish at e. Similarly i(;) cannot be greater than s. If it were, then by the first part of the theorem it would not be the case that there is a non-vanishing partial derivative of order s. Hence we conclude that i(C) = s. PROOF.
VI.3.6. We have been making reference to the vector K of Riemann constants since Theorem VI.2.4. In this section, we give a new characterization of K that partly explains its significance. Up to this point K has been defined simply by (2.4.1). We show here that the vector —2K is the image of the divisor of a meromorphic differential under (the extension to divisors of the map) cp. We have already seen that the divisor of a meromorphic differential on a compact surface of genus g has degree 2g — 2. Hence we now prove the following Theorem. Let d be a divisor of degree 2g — 2 on a compact Riemann surface, M, of genus g 1. Then d is the divisor of a meromorphic differential on M if and only if cp(11) = —2K. PROOF. We first show that — 2K is indeed the image of the divisor class of meromorphic differentials. It suffices to show, of course, that —2K is the image of the divisor of a holomorphic differential. To this end let be an arbitrary integral divisor of degree g — 1. It therefore follows from Theorem VI.3.2 that e = cp(C) + K is a zero of the theta function. Since the theta function is even, —e is also a zero and (again by Theorem VI.3.2) —e = q(co) + K, with co an integral divisor of degree g — 1 on M. It therefore follows that (,o(Cco) = — 2K. We now need show that Cai is the divisor of a holomorphic differential on M. We need only note that the divisor Ca) has g — 1 "arbitrary" points. It therefore follows (by Lemma 111.8.15 as in VI.3.2) that r(l/Cco) > g; which
299
VI.3. The Theta Divisor
by the Riemann- Roch theorem implies Kw) >: 1. and therefore igco) = 1. Hence ,:co is the divisor of a holomorphic differential. Conversely, suppose d is a divisor of degree 2g — 2 on M such that q(J) = —2K. It follows by Abel's theorem that there is a function f on M with divisor z1/(Œ), where is a holomorphic differential on M and (a) denotes the divisor of Thus fa is a differential on M with divisor (fa) = J. In the next chapter we will often make use of the following
Corollarz. The vector K of Riemann constants is a point of order 2 in J(M) if and only if P,14-2 is ..eanonical,- where PC is the base point of the snap (p: M J (M). PROOF. The point K is of order 2 if and only if — 2K = 0 ---- cp(Ng - 2 ). Now use the theorem. D
VI.3.7. We have encountered, in this chapter and in 111.1 1, three very important and beautiful subsets of J(M). The first, is the zero set of the 6-function eltrCI
= Wg- 1
K;
the second, is the singular set of the 0-function (the points where CI and its first partials vanish)
= W:-1 + K (see also VII.1.6); the third is the set of points e J(M) which lead to the
identically vanishing of the function ik of VI.3.3 ()
super Zero
=
+ K.
It is obvious that °zero D °super zero D esinz•
(3.7.1)
Note that _ super zero is always non-empty for g 2 (take a Weierstrass point, for example), while is non-empty for g 4 by the Corollary to Theorem 111.8.13 and Theorem 111.8.7 (also for hyperelliptic surfaces of genus 3). We conclude from Theorem 111.1 1.19 that () sin ,
dim O zer. = g — 1, dim esup„ zero = g — 2, and g— 4
dim esing g — 3,
and hence the inclusions of (3.7.1) are all proper. The points corresponding to vanishing of the 0-function, but not identically vanishing (on the surface), are in (417,\W) + K = (J(M)\W;) + K.
300
VI Theta Functions
only on the period matrix 11. The fact that The sets ezer,„ and E) such a set is non-empty can, of course, be translated by the Riemann vanishing theorem and the Riemann—Roch theorem into a statement about existence of meromorphic functions (of a certain type) on M. It should also be remarked that Ozer° and esing are independent of the base point of the map (p:M J(M). Thus describes those points e E J(M) corresponding to the identically vanishing of ti/ for all choices of the base point (see also VII.2.1).
°sing
CHAPTER VII
Examples
In this chapter we give applications and examples of the theory developed in the preceding chapters. The examples will consist mainly of taking concrete representations of compact Riemann surfaces given by algebraic functions, and computing on these surfaces various objects of interest. The applications will give rise to new characterizations of surfaces with some given property. The format of this chapter will differ considerably from the format of the preceding ones. Many details will be omitted. One of the aims of this chapter is to convince the reader that computations are quite often possible; and these computations yield beautiful results.
VII. 1. Hyperelliptic Surfaces (Once Again) Our first set of examples will continue to deal with the class of surfaces studied extensively in 111.7: the hyperelliptic surfaces. Recall that a hyperelliptic surface M of genus g > 2 has an essentially unique realization as a two sheeted cover of the sphere branched over 2g + 2 points. It is the Riemann surface of the algebraic curve
n
2g+ 2
=
(z —
e,,),
ek 0 ep for k j.
(1.0.1)
k= 1
VII.1.1. We adopt the following point of view in the above situation. We think of z as a variable point in C u { oo}, and then view M as the Riemann surface on which w is a well defined (single valued) meromorphic function. On this surface, which must be a two sheeted branched covering of C u the projection map (which we shall also denote by z) onto C u foe) is the
302
VII Examples
function of degree two, and we can think of w as a function of the point P e M in the sense that w sends Pc M into w(:(1') ). Note that as a function of w is two-valued: it is defined up to sign. Alternatively, we can think of M as a surface of genus g > 2 with a meromorphic function : of degree 2 and another function w satisfying (1.0.1). Theorem 111.7.3 established the result that the 2g + 2 points z'(e k), k = 1, , 2g + 2, are the Weierstrass points on M, and that each of these points has weight g(g 2 Proposition 111.7.6 showed that if we choose a Weierstrass point (say :: -1 (e 1 )) as the base point for the map (p of M into the Jacobian variety J(M), then for P e M. q(P) is of order 2 if and only if P is a Weierstrass point. We can actually say more. We can compute precisely which half-periods (points of order 2) are in the image of (9:.11 ---. AM). We need for this purpose a model for a canonical homology basis on M. Recall the model discussed in 111.7.4, and set .: -1 (ei) = Pp j = 1, 2, .. , 2g + 2. We wish to amplify (for the convenience of the calculations) that model slightly to the one given in Figure V11.1.
Figure VII.1. Hyperelliptic surface of genus 2.
The hyperelliptic involution J can be viewed as a rotation by it radians about an axis passing through the 2g + 2 Weierstrass points. For k = 1, . g + 1, let /3k be an oriented curve from P2k_i to P2k. Define bk as /3, followed by —J13k (that is, followed by Jp, is the opposite direction). The curve ak, k = 1, . . . , g, is all in "one sheet" and it joins a point on bk to a point on b9 _,. 1 and then returns to the point on bk . Whereas bk is invariant (as a point set) under J, a„ is not. The curves a,, , (79 , b 1 , , b, form a canonical homology basis on M. In the following calculations, we use standard notation: ICI = is the normalized basis of Yel(M) dual to the canonical homology basis, 7r(k) is the k-th column of the period matrix TI, etc... . , as well as the fact that J acts as multiplication by — 1 on Ye l (M). Now, for k = 1, . . . , g, bk
Pk
Pk
J0k
f
P2k
= 2 J. = 2 f., ç. r2k - 1 sk Next, the intersection numbers satisfy for = 1, . . . , g, )
ai . 1)0. 1 = —1,
bi - b9+1 = 0.
Pk
303
Hyperelliptic Surfaces (Once Again)
Thus up to homology from which we conclude that P2g +
2
= 1
or(1)
.
2
f P29+1
To compute some more integrals we introduce the curves joining P2j to P2+1 and set a; = ei; Yi p Again we compute intersection numbers = 1, , g and k= 1, . . . , g) ra; • b; = 0 f6r3 k,j k +1, Œa5 = 0, a; • b. = -1. 1.; • bi = +1, Thus we conclude that (in H 1 (.14)) for j =1..... g -1,x 9 = a.
Œ=a It now follows, as above, that P2k 4.
1
I
2
afP2k P29+
J P29
=
1
(e.u) - eu + ") = -(e(k) + d""),
k = 1, . . . ,g - 1,
1
We now combine all the above data in: (p(P 1 ) = (102) = q(P1 ) + f:: = 2) + 5:3 c = ' it(1) + e(i) + e(2)), (
qi(P =
1 2
(P ,51 ) = - (70) + • • • + it(k) + eu ) + e(k+1)), )
(PlP2k + = 1-2 (x( 1 + • • • + 7t(k+ "
(P(P2g+ i) = 1-2 (7 41P2, +2)
1
(1)
+•
+ 7/(g)
eu ) + e+1)),
eu )),
k = 1, 2, . . . , g - 1, k = 1, . , g - 1,
VII Examples
304
Notice that D 2 (p(p)= 0, because (P 2 • • • P294 2)10 +1 is the divisor of the function w (here we assume that z has a double pole at P I , and hence in (1.0.1) the product runs over the indices 2 to 2y + 2). We compute next
E J=1
= 2-1 (g7t(1) + (g —
1)2r(2)
22(g) +
gem + e(2) + • • • + e(0). (1.1.1)
Consider the map cp:Mg --+ J(M) at the point D = P3 P$ • • • P29 .". Its differential at this point is (the transpose of) .
2(133)
'•'
:i(P5)
Ca(P3)
••
(1.1.2)
CI(P29+ 1)
9( .13 29 1)..
A change of basis for abelian differentials of the first kind, multiplies the above matrix by a non-singular matrix. Hence to compute the rank of (1.1.2), we may choose a convenient basis for .e(M). We shall use the basis given by 111(7.5.1): zi dz
j = 0, . . . , g — 1.
As before we assume for convenience that z has a double pole at P1 , and vanishes at the point P2. Note then that (zidz)
w)=
p2g - 2 i -2 pi I 2•
In particular, dz/w does not vanish at compute the rank of rf1r
P21 41 ,
j = 1, . . . , g. Thus we want to z9-1 (P 3 )
1 z(P 3) 1 z(P $) • — (P 2J+ )
-
z' '(P5)
dz
J= I w 1 Z(P 29+ 1)
1 e3 T—r g dz
= 11 — (P2)+ i= 1 w
Z.- 1 (P 29+ 1)_
es1
1 e$ I)
• •
1 e294- 1
„..„ 9+ I c 294 1
The last matrix is, of course, the Vandermonde matrix which has already been encountered (in IV.11.10), and is non-singular. We have hence shown (III.11.11) that i(D) = 0, and that (p. establishes an isomorphism between a neighborhood of D in Mg and its image in J(M).
305
VII.! . Hyperelliptic Surfaces (Once Again)
VII.1.2. In VI.2.4, we have introduced the vector of Riemann constants It2kk (pcp'k dz — e (k) (-
K= — k-=1
where we assumed that the base point for 7t 1 (M) is chosen to agree with the base point of the map ça:M J(M)—which in our case was selected to be a Weierstrass, point P1 . It is rather difficult to evaluate K directly. We know (Theorem V1.3.6) that — 2K is the image in J(M) of the canonical divisor class. Since P?g' is the divisor of an abelian differential of the first kind, the canonical class gets nipped into 0 by (p. Thus K is a half-périod. We want to know which half-period. We shall shoir- that K=
2 i+ 1).
(1.2.1)
i=
Assume for the moment that we knew that 0(0) 0 0, for the 0-function introduced in Chapter VI. In this case 0 o cp does not vanish identically on M (this is the fact we really need), and hence 0 0 cp has g zeros Q1,. . . , Qg on M. These zeros satisfy f
J=1
+ K = 0.
(recall Theorem VI.2.4). We have seen that (P) is a half-period. Such a half-period can be written as PM + Hs), where rej is an integer characteristic. Thus the half-periods can be classified as odd or even, depending on the parity of the characteristic. Up to a constant non-zero factor, 19 06' +
—
We hence conclude that 0 vanishes at odd half-periods. We notice that (p(P2i , i ), = 1, . , g, is an odd half-period. (We also, for the future, notice that (p(P2), j = 1, . . . , g + 1, is an even half-period.) Thus cp(P21 ., i ) is a zero of the function 0 on M. Hence 9
E w (p2 ,0= —K
= K.
Note that conversely, the identity (1.2.1) shows that 0 does not vanish identically on M. For if 0 vanished identically on M, then (p(P3P3 • • • P20.1)-1K = 0 and i(P 3 • • P2 g+ j) > 0 by Theorem VI.3.3. This contradicts the previously established fact that i(P3 • • - P 2g+ ,) = 0. These remarks also show that 0 vanishes at 0 if and only if 0 vanishes identically on M. The fact that an even theta function vanishes at 0 if and only if it vanishes identically on M is also a consequence of the Riemann vanishing theorem.
306
VII Examples
Remark. There is an additional property that the 2g + 1 half-periods cp(p), j = 2, ... , 2g + 2 have. We have already mentioned the fact that cp(P21 ., j = 1, . . . , g, is odd and ça(P,i), j = 1, . . . , g + 1, is even. The origin (p(P,), is, of course, also even. If we use the notation (ç .) = + Ile) for halfperiods, and label the half period (p(Pk ), k = 2, ... , 2g + 2 by (j %), then the 2g + 1 half-periods (3!)) have the property that for all k I, s(k) • s'(1)± e(k) • s(I) = 1. Here c(k) = (s 1 (k), ,e,(k)), E'(k) = (8(k), ,s;(k)), and the operation ( ) is the usual inner product over Z 2 .
VII.1.3. We need to establish that 0 does not vanish at the origin. We postpone this proof to VII.1.9, since it is more convenient to have some further computations at hand. Note that we never use anything other than the fact that K is a half-period. We do not anywhere use the exact form of K (before VII.4).
VII.1.4. It is, of course, true that 0(K) = 0 by Theorem V1.3.1. The order of vanishing of 0 at K is determined by the Riemann vanishing theorem (VI.3.5). Write K = cp(C) + K, CeM Clearly we can choose C = PÇ . The order of vanishing of 0 at K is precisely i(:). Since P, is a Weierstrass point, the orders of zeros at P1 of abelian differentials of the first kind on M are (see 111.7.3):
0, 2, ... , 2g — 2. Thus
i(pv 1
)=
11(g + 1), if g is odd, if g is even. 19,
(Using the greatest integer notation, we can write that the order of vanishing is always [1(g + 1)].) For D e Mg _ 1 , r(D - 1 ) = i(D), and Clifford's theorem (111.8.4) shows that
—
2 i•
We have established that for a hyperelliptic surface there are points at which the 0-function vanishes to the maximum order possible. We have shown that such vanishing occurs at a point of order 2 in J(M). As a matter of fact, given any non-negative integer n [ -(g + 1)], there is a point e e J(M) such that e has order 2 and 0 vanishes at e to precisely order n. To see this, we have to construct for n> 0, divisors C E M9 _1 with C containing only Weierstrass points and i(C) = n. Then the point e = (p(() + K will have the desired property (for n = 0, we, of course, need a divisor C e Mg with i(C) = 0). The calculation in VII.1.1 established the following fact: If D e M, and D consists of distinct Weierstrass points, then i(D) = 0. From this observation it follows that if (for 0 r g)D E M9 _,. and
Hyperelliptic Surfaces (Once Again)
307
D consists of distinct Weierstrass points, then i(D)----- r. Let 1 < j i < j2 < • • •
0, and each base point. It then follows that 0(+ ---- 0 = (00/0t4;)( -+e), j = 1, . . . , g. This information is not new. It is a consequence of the Riemann vanishing theorem (V1.3,5). A new result is contained in Proposition. For e e e,ing , we have
.20
e
g
E
u JCUk
(±
(2.2.3)
= 0,
and
E=
A
a: ° Uk 041
OW/EC!
(2.2.4)
= 0.
PROOF. Equation (2.2.3) follows from (2.2.2) and the previous observation that the first partials of 0 vanish. The vanishing of the coefficient of z3 in (2.2.2) yields
a20
E a20 3
1,3o
E Gui C/Uk GUI (e), = , ( e)cick + j,k,i
E euiatik
030
+.. eU k j,k,1 aU aU
(2.2.5) 0.
(2.2.6)
If we subtract (2.2.6) from (2.2.5) and use the fact that 0 is an even function, then we obtain (2.2.4).
Remark. There are instances when the above proposition is of little value. For example, if e e &sing , and all the second partials of 0 at e vanish (e e W.2 _ 1 + K), then (2.2.3) yields no information. It could also happen that all
VII.2. Relations among Quadratic Differentials
313
the second order partial derivatives do not vanish at e, but all the third order partial derivatives do vanish. This takes place when e e °sing is an even point of order 2. Why? (EXERCISE.)
VII.2.3. We return now to case (c) of Theorem and consider a surface M of genus 4 where O m., consists of precisely two points e and —e (with e not a point of order 2 in J(M)). We have seen, in 111.10, that the abelian differentials of the first kind on M provide an embedding, M P 3, of the surface into projective space. Proposition VII.2.2 tells us that the image of M in P 3 is contained in the intersection of the quadric and cubic defined by
(2.2.3) she]. (2.2.4). Notethat (2.2.3) is not the trivial relation because the highest order of vanishing the 0-function for a surface of genus 4 is 2. It should be observed that the coefficients of the curve (given in this way) in P 3 depend only on e s,ng, which in turn depends only on the period matrix H of M. Hence the curve in projective space or equivalently the Riemann surface M can be recovered from the period matrix 17, whenever (2.2.3) and (2.2.4) are independent equations.
Remark. (For those familiar with elementary properties of curves.) The intersection of the quadric (2.2.3) and the cubic (2.2.4) certainly contains a curve of degree 2 • 3 = 6. The degree of the curve M in P 3 is 2 • 4 — 2 = 6. It is important to observe that while, as stated, the quadric in the above case is non-trivial, we have not really shown that the cubic is non-trivial or that the quadric is not contained in the cubic. In these situations the curve is not recoverable by the above procedure. EXERCISE How much of the above carries over to cases (a) and (b) of Theorem VII. 1.6?
VII.2.4. It is not our intention to develop here the theory of moduli of compact Riemann surfaces of genus g> 2. However, there is one interesting observation that can be made now. We know that the dimension of ye2(m), the space of holomorphic quadratic differentials on M, is 3g — 3. We also know that CiCh e .1e 2(M), j, k = 1, . . . , g. Hence, if the products of the holomorphic abelian differentials span ../e 2(M) (if and only if M is not hyperelliptic, by Noether's theorem (1 11 .11.20 ) ), there must be exactly 1(g — 3)(g — 2) relations among the products. When g = 4, this is precisely the relation (2.2.3). For hyperelliptic surfaces there are more relations. For hyperelliptic surfaces of genus 4, there are 10 — 7 = 3 linearly independent relations. We shall now exhibit two possible ways of obtaining these additional relations. One way is quite obvious. Since for hyperelliptic surfaces of genus 4, °sing is 1-dimensional, it seems reasonable to expect that one can find three
314
VII Examples
points ei e ()sin, such that 4 00
y ka-;'
1, 2, 3.
0,
CtikCU,
are linearly independent. We however, do not know how to effectively prove such independence. Another possible way is to use more essentially the fact that ° sin, is 1-dimensional. Let Po be the base point of cp:M AM), and let J:M M be the hyperelliptic involution (there should be no confusion with the same letter appearing with two meanings). For P e M, let P' = J(P). Choose P 1 e M and set ec, = (p(PI PI P0)+ K e (Note that that since every we yew) that vanishes at P c M also vanishes at P'; i(PP'Q) = i(PQ) 2.) The embedding of M into ()sing . J(M) is now given by M PI—+ cp(P,P'i P) (2.4.1) Let z be a local coordinate vanishing at P o . Equation (2.4.1) defines e(z) as a holomorphic function of z. By Proposition VII.2.2, 4
,12.0
Su , auk
(e(z))“:, = 0,
(2.4.2)
for all z in a neighborhood of 0. We now expand (52 0/eui euk)(e(z)) as a power series in z, and obtain 520 au,uuk
520
(e(z)) =
a30
4
(eo) + (E S UJ SIk SL (eo)(Po) atij aU k (v
v a3o 7, ufônk(e0)i(Pog„,(P0))fau,au, au„`"(P 0' 4- IL:. Otim auS41 -O 2 0 (2.4.3)
Inserting (2.4.3) into (2.4.2), we obtain 4
5
20
j,k= I aUjaUk(e°"k
= 0,
-3n
E E au, Su"SU,, (e ° g 1(Po)
C.4= 0,
and 030
(eAl(P0) E (E j,k 1 autaujauk
,940
E G um au,,,ui cuk (eogi(Po)m(P0)) CiCk = O. -
It seems reasonable to expect that one should be able to extract three linearly independent relations from the huge list obtained above (as one
315
VIII Examples of Non-hyperelliptic Surfaces
varies the base point Po). The problem of specifying three such relations is
apparently still open. EXERCISE Is the map M
esing, that we have constructed, surjective?
VII.2.5. We shall discuss briefly in this paragraph a question intimately related to the question of relations among the products of the normalized abelian differentials of the first kind. Let 110 e 4. be a period matrix of a compact Riemann surface of genus 4. (Recall that is the Siegel upper half space of genus 4 introduced in V1.1.1.) Theorem VII.1.6 guarantees that (51 ; is non-empty. Assume we are, again, in case (c). By the remark following 111.8.7, the rank of the associated relation (2.2.3) among the abelian differentials must be 4. Choose a (small) neighborhood N of Flo e Z4. What is a necessary condition for // E N to be a period matrix of a compact surface of genus 4? Clearly, the 0-function for II must have the property that esing 0 Ø. We proceed to write down an equation in 54 that expresses this condition. Consider the system of equations on C4 x
eo
ez.
,
u= 1. 2, 3, 4.
( 2.5.1)
The hypothesis that /70 is a period matrix tells us that (z°,I70) solves (2.5.1), whenever z° e for /70 . The condition that (2.2.3) have rank four, tells us that the Jacobian matrix of the system (2.5.1) with respect to the z-variables has rank 4, and thus the implicit function theorem asserts that we can solve for z = t(z 1,z 2 ,z 3 ,z4) as holomorphic functions of t in a neighborhood of (z °,17 0) and that
°sin ,
= 0 for n = 1, 2, 3, 4, t e N. A necessary analytic condition for t to be a period matrix of a compact surface is now easily written down:
F(r)= Ci(z(T),T) = 0,
t e N.
VII. 3. Examples of Non-hyperelliptic Surfaces We shall study in this section three sheeted covers of the sphere. The calculations will be considerably more involved than in the hyperelliptic case. Other special cases will also be described.
VII.3.1. In Chapter V, we began the study of automorphisms of (compact) Riemann surfaces. At the end of V.1.6, we gave some examples to show that the results of V.1.5 were sharp. We now return to a variation of the second of those examples.
316
VII Examples
Consider the Riemann surface M:
= z(z — 1)(z — ). 1 ) • • • (z — ). 3k _ 3),
k
2.
(3.1.1)
Here i.,, , i. 3,_ 3 are 3k — 3 distinct points in We view z as a meromorphic function on M of degree 3 (w is of degree 3k — 1). It is branched over 0, 1, co, A l , , ;.3k - 3 with branch number 2. For convenience we label:
Q 1 = z 1 (0),
Q2 = Z -1 (1),
I); =
Q3 =
j = 1, . . . ,
z -1 (oo),
3k — 3,
and note that the divisor of the function z is Q?
= 03 --• The Riemann-Hurwitz formula yields that the genus g of M is 3k — 2 ( >4). Our first task is to compute a basis for Ye"(M) —compare 111.7.5. Observe that ••'
(dz) =
3
and Q1Q2P1 ( w) =
3k- I Q3
•
From the above two observations it is easy to conclude that ,
j = 0,
, k — 2,
and zi
dz -v-v---2,
/ = 0, .. . ,
2k — 2,
form a basis for ie l (M); as a matter of fact
(zi dz) _ 0 ,,, , 3(,_ .0 _, n, v2v3 I- ' ' ' w and
e h
=
t
Q3lQ3(2k-I-2)
W2
We shall consider the special case k = 2 (hence, g = 4). Thus the basis for Je l (M) is: dz dz -, ww
dz
w
d-
(3.1.2)
The above basis is adapted (recall 111.5.8) to the point Q 1 . As a matter of fact, the Weierstrass "gap" sequence at Q i is:
1, 2, 4, 7.
(3.1.3)
VII.3. Examples of Non-hyperelliptic Surfaces
317
Note that the above is also the "gap" sequence at Q2, Q3 and Pi , j = 1, 2, 3. Thus each of these 6 points is a Weierstrass point of weight 4. We have thus accounted for 24 of the 60 Weierstrass Points (here we arc counting each Weierstrass point according to its multiplicity). The remaining 36 Weierstrass points will occur in groups of three. Each such group of three will project to the same point in C u { 3o} by the map z. This follows from the fact that M has an automorphism of period three that interchanges the sheets of the cover. We continue with the case k = 2 in (3.1.1). Since M is of genus 4, by Theorem VII.1.6, A7t is either hyperelliptic (impossible, because M carries a function of degree 3, by Proposition 111.7.10) or °sing consists of one point of order 2 or es,,,, consists of precisely two points (neither one of order 2). Since Qf is the divisor of the abelian differential z 2 (dz,42 ), the vector K of Riemann constants with respect to the base point Q 1 , is a point of order 2 of J(M) by Theorem VI.3.6. Further, since
K =9{Q?) ± K, K is a zero of the theta function. Also; looking at the "gap" sequence (3.1.3), we see that i(Q1) = 2,
°sins
which shows that K e 0,ing . We have shown that consists of the single point K. It must be the case that there is a relation of rank 3 among the products of the abelian differentials of the first kind. The relation is easily exhibited: ( 7. dZy = ( z2 dZ2 )(dZ2 ) . w2 W
(3.2.1)
Of course, alternatively, (3.2.1) can be used to conclude that of precisely one point. However, the conclusion (we obtained)
es ,„, consists
es ,„, = {K}, required a little more analysis. VII.3.3. We return to the general case (3.1.1) with k 2. Using the basis '(M), we see that the Weierstrass "gap" sequence at we constructed for (hence also at Q2, Q3, Pp j = 1, . . . , 3k — 3) is: 421
..°
{3 1 + 1; 1 = 0,
, 2k — 2} k..) {3j + 2; j = 0,
, k — 2}.
In particular, every function f, holomorphic on MqQ,}, with deg f 3k — 4 must satisfy deg f 0(mod 3). An easy calculation shows that the weight of each of these 3k Weierstrass points is 2k-2
E 1=0
3k-2
k-2
(31+ 1) +
(3j + 2) — j=0
E m = (3k — 2)(k — 1). m=0
318
vfi Examples
We have accounted for 3k(3k -- 2)(k — I) of the (3k — 3)(3k 2)(3k — I) Weierstrass points on M. The vector of Riemann constants K with respect to (2 1 is again a point of order 2 in 301). since ()`;' is a lOcal divisor. 3 ) = k = j(g + 2). The point K E e,in , because i(Q Finally, we leave it to the reader to explore the question of (linearly independent) relations among the abelian differentials of the first kind. For example, il k = 3, and if we label dz
91= ÇO4
dz =
W2
_ , 3 dz
-
95
93 = z
-7,
dz w-
dz (P6Th
dz
Ç07
Z
dz -,
then we can write down the following maximal set of linearly independent relations = (P1 (P 3 , =
q93(05,
(Pi =
(P293
(P295 = (,03(P7
9394,
9t 94,
(1)197
= 9496,
= (P2(P6,
92 ,P4 = ( f9
So2(P7
= iP3(p6,
94 ( 7 =
VII.3.4. The Riemann surface M of (3.1.1) is an example of a surface with an automorphism T (of period 3, in our case) such that M,i C u tool. We have seen that for these surfaces, contains a point of order 2. It may seem at first glance that the existence of points of order 2 in 0,ing is caused by the presence of the automorphism. This conclusion is, in general, false. Consider, for example, the surface M defined by () sin ,
1V3 = Z Z (
1)(z — ;. 1 )2(z — ;1 2)2(z — ). 3)2,
where A » j = 1, 2, 3, are three distinct points in C \ {0,1}. The surface M has again an automorphism T of order 3 such that A / is conformaly equivalent to C u {Do } . Note that M (again) has genus 4. Using notation as in VII.3.1, we conclude that
(z)= 42? Q3
(w) —
■
Q1Q2P PiP23
421
and
(dz)= QiQiPiPiPi
421
A basis for if l (M) is thus dz dz z w w
(z — ,1 1 )(z — ,1 2 )(z —
dz
z(z — 2 1 )(z — ). 2)(z —
dz
VII.3. Examples of Non-hyperelliptic Surfaces
319
From the above we see that the possible orders of zeros of abelian differentials of the first kind at Q 1 are: 0, 1, 3, 4, and thus the "gap" sequence at Q 1 is:
1, 2, 4, 5. In particular, there is no differential in 3r1 (M) all of whose zeros are at Q 1 . Now, the point K belongs to e sing (because f(Q1) = 2), and if °sing a point of order 2, K would have to be that point. Hence 2K = O. contaied By Theorem VI.3.6, weewould also have y9(0) = 0 = —2K or that 421 is canonical. We have seen that this does not occur in our example.
VII.3.5. There is, however, one case where the presence of an automorphism T produces points of order 2 in 0„,,g : the case when T has period 2. Suppose now that M is a surface of genus g 2, and T e Aut M has period 2 and v(T) fixed points. The genus 4-. of MgT> is computed (recall V.1.9) by = v(T) = 2g + 2 — (and M is called #-hyperelliptic).
To show that 0,, n, contains a point of order two, it suffices to find an w e Ye 1 (M) with even order zeros (0.)) =
Pi • • •
such that i(Pi • • P9 _ 1 )
2.
For then e cp(P, • • • P9 _ 1 ) + K e esing , and 2e = cp((co)) + 2K = O. If # = 0, then M is hyperelliptic. Then for g 3, there is a point of order 2 in °sin, and the order of vanishing of the 0-function at this point is 1(g + 1). Hence we now assume g> O. VII.3.6. Let us generalize by considering an automorphism T of prime order N with MI of positive genus. Consider the action of T on Y i(M). Since MgT> has positive genus #, 1 is an eigenvalue. Since M has bigger genus than M MgT>, there is another eigenvalue e = 1, a 0 1). In other words, there are holomorphic differentials co, co, e .e.(m) such that Tco = co,
Tco, = scoi ,
co
O.
The differential w projects to A7i; while co, projects to a multivalued differential on M. The function f = co i/co on M projects to an N-valued function on M. , P, To describe the structure of the divisors (w) and (co 1 ), we let P1 , be the fixed points of T (v = v(T) could be zero). Calculations similar to the ones performed in V.2 show that
ordp, (.0 = Ns; + N — 1,
r•eZ r•>- 0• 1
320
VII Examples
Hence we can write j( N-1,,
I...
( ( 9) =
(3.6.1)
where 4 is an integral divisor on M and 4 (4 = r(4). Note that J could contain some of the points 134 = 1, . . . , v. Similarly, ordp,cz = riN + ti - 1,
ti N - 1, rj > 0.
7.1 e Z, ti e Z, 1
Thus, as before, we can write = -1 . . pivv - t xx( 1) .
xev-1).
(3.6.2)
From (3.6.1) and (3.6.2) we can compute the divisor of the meromorphic function f = w 11 w on M, and the projected function Jon .1/:
X.
-..
(/) =
(3.6.3)
In (3.6.3) we have made an obvious identification of points on M with their images in Si-. Note that (/) contains fractional powers because locally / is given by a Puiseaux (not Taylor) series. The fractional divisors on .ii can be mapped (locally) into J(S-1). Since IN is single valued on S4, the image of the divisor of f in J(17/) is a point of order N. In order to study the converse, we begin by remarking that v(T)
R=
E (1 -
is an integer (EXERCISE) that satisfies
v(T)
N-1 1 v(T). R - -T-
(3.6.4)
Every multivalued function on A-4. whose divisor has the same fractional exponent at 11; as in (3.6.3) and that has the same image in J(M) as (1), lifts to a meromorphic function on M. Consider the divisor D on AI D=
••
131:1-i'4N .
We now pose the following problem: Does there exist an integral divisor
X on k such that v deg X =
-
t•
,
(where, as usual, [a] denotes the greatest integer in a), 9(x 2N) = op),
(3.6.5)
32 1
VII.3. Examples of Non-hyperelliptic Surfaces
and
4 (X 2) —
- • • .1)„1- '-"' LI)
(3.6.6)
equals the point of order N in J(.1) determined by the divisor of the function _T? The theory of the Jacobi inlersion problem asserts that we can always (1 - ti/N)] - 1 solve the above problem, and in fact do so with [-I free points. The condition that we shall need is
R=
(1 — -t--1)
2,
•■•• which occurs by (3.6.4) whenever
v(T) 2N. Furthermore, there are (2N)24 different solutions to (3.6.5) corresponding to the (2N) 24 points of order 2N in J(k). Only 22i of these solutions will satisfy (3.6.6). There are now two cases to consider:
R is an even integer ( 2),
(3.6.7)
R is an odd integer ( 3).
(3.6.8)
and In case (3.6.7), x2N
AN
prtl .
is the divisor of a meromorphic function on meromorphic function on M with divisor x2( x2)(1)
p - t,
(x2)(N
pNv -t„A zi (I)
Si
whose Nth-root lifts to a
— 1)
4 (N-1).
Thus we conclude that (multiply the above function by (o)
Z1 =
-1 . . . pr i x 2(x2)i) . por - 1.)
(3.6.9)
is a canonical divisor on M. In case (3.6.8) we choose P 1 e A4 as the base point of the map cp: We conclude that -
p+11x2N
prt,
prvr-tv A N
is the divisor of a meromorphic function on Al whose N-th root lifts to a a meromorphic function on M with divisor
prIx2(x2 )(1) . por -1)
322
VII Examples
In this case we conclude that
Z2 = Pr"- I
1 - • p t (X 2 )tXV 1) • • •
(
v s - i) )
(3.6.10)
is a canonical divisor on M. The sole purpose of the above series of exercises was to produce a holomorphic differential on M (with enough free points) all of whose zeros are of even order. We have succeeded in this whenever N = 2 (because ri = 1 for j = 1,..., v(T)). In this case R = E (i - = N
and thus X has precisely
2
r4_(7_111
free points. We conclude from (3.6.9) that v ( T) i(Z1)
with a similar result for (3.6.10). We have established the following Theorem. Let M he a g-hyperelliptic swface of genus g + 3. Then esin, contains 2 24 points of order 2 with the order of vanishing of the 0-function at these points greater than or equal to [ .-(2g + 2 — 4g)]. N = 2 and v = 0 does not give points of order two in esi„,. A slightly different and in some sense simpler analysis, however, does.
Remark. The above analysis for
VII.3.7. The Riemann surface M of w29 + 1 = z(z — 1)
affords another interesting example of a surface of genus g. It is an immediate consequence of the fact that w is a function of degree 2, that M is hyperelliptic. There is an obvious involution on M:
(z,w)i—* (1 — z w). ,
This is the hyperelliptic involution since it fixes the 2g + 1 points lying over z = 4 and the point z' ( e). The Riemann surface M of w4 =
z4
1
VII.3. Examples of Nun-hyperelliptic Surfaces
323
is of genus 3 which obviously has two cyclic groups of order 4 operating on it; the groups generated by
(i:.w).
(z.01--.(zjw),
Using obvious notational conventions, we record the following facts: o22Q3Q4
(-)— Q5Q6127128' (w) —
P iP 2P 3P 4
5Q6Q7Q8'
and
(dz)
P?P3P3P,3, - -- T.
12;MiQi
From the above we see that
d: d:dz z -7 h3 2, IV
form a basis for
yel(m). Note that (=
= Q1Q2023 ,24 =
i t%
More generally for any ceCk..) 13c1, '(e) is a canonical divisor. It follows from these observations that the "gap" sequence at any 1=1» = 1,2,3,4) is
1, 2, 5. (Note that the Pi are defined by z(Pi)4 = 1.) Thus each 13 is a Weierstrass point of weight 2. We have accounted for 8 of the 24 Weierstrass points. Similarly, the points Qi (j = 1,2,3,4) are Weierstrass points of weight 2, since
dw dw dw
z2 ' is also a basis of Ye l (M). Finally, the points Q. (j = 5,6,7,8) are also Weierstrass points of weight 2, since
z 1) (z,w) 1—*(—, — ww is an automorphism (of period 2) of the surface M that interchanges the zeros and poles of w. We have thus accounted for all the Weierstrass points on M. It is a surface of genus 3 with 12 distinct Weierstrass points each of weight 2.
VII.3.9. The next to last example of this section is the most complicated one. It will be used to show that the upper bound on the number of (distinct) Weierstrass points g3 — g on a surface of genus g (Theorem 111.5.11) is
324
VII Examples
attained. Our example will be for g = 3. Consider the Riemann surface M of the algebraic curve (19.1) = — 1) 2 . On the surface, the function z is of degree 7 (and w is of degree 3). The function z is ramified over 0, 1, co (with ramification number 7). A calculation using Riemann-Hurwitz now shows that M has genus 3. In our usual short hand: (0= (d-) =
Pi
P7131 Qt
and P i Pj I' (w)= — 42 31 From the above formulas we see that the differentials dz
dz
(z
dz — 1)-6
(3.9.2)
have (respectively) divisors PiQi, P1P 32, It is is immediate from this calculation that we have again produced a basis for Yr 1 (M) and that P„ P2, Q 1 are all simple ( = weight one) Weierstrass points. We claim that all the Weierstrass points are simple (hence we must have 24 distinct Weierstrass points). We will compute the Wronskian of our basis of Ye l(M). Since we are no longer interested in points lying over 0, 1, 30 (via z), the function z is a good local coordinate at such points. Using the notation of 111.5.8, we are computing 1 z — 1 z — 1] 6 37 5 7 w w w det[
1 =
det[w 3 , w(z
1), z — 1]
3! 1 (z3 — 8:2 + 5z + 1). = z3 :5(z — 1)5 (The computation is long and tedious, but routine.) Denoting the cubic polynomial by p(z) we see that p(-1)= —13,
p(0) = +1,
p(1) = —1,
p(8) = 41,
and hence p has three distinct real roots (in the open intervals ( —1,0), (0,1) and (1,8)). Each zero of the Wronskian corresponds to 7 distinct points on M. Thus we have produced a surface of genus 3 with 24 distinct (simple) Weierstrass points.
V11.3. Examples of Non-hyperelliptic Surfaces
325
V11.3.10. The Riemann surface of (3.9.1) actually has 168 = 84(3 — 1) automorphisms, which shows that the maximum number (of Hurwitz's theorem, V.1.3) is achieved. It is easy to exhibit an automorphism of period 7: (
(z,e w),
z,w)
e = exp(--). 7
To exhibit other automorphisms of M we must use some algebraic geometry. The abelian differentials of the first kind provide an embedding of M into P3 (111.10). Thus setting
z — 1 = +X 3 Y 2 ,
w=
we see from (3.9.1) or (3.9.2) that a projective equation for our curve is X3
Y + Y 3 Z Z 3 X = O.
Hence we see that we have another automorphism of M (of order 3) given by the permutation (X. Y,Z) ( Y,Z,X). We will not proceed with the above line of thought. (The interested reader should consult the work of A. Kuribayashi and K. Komiya for the complete classification of automorphism groups and Weierstrass points for surfaces of genus 3.) (The authors have a preprint of their forthcoming article, and hence cannot supply a reference to the literature.) If we are willing to use the fact that M has 168 automorphisms, we can conclude without calculation that each Weierstrass point is simple. We have seen in the proof of Hurwitz's theorem, that the maximum number of automorphisms occur only if M,'Aut M C i {on} and the canonical projection n:M —■ C1/4.)f)o} is branched over three points with ramification numbers 2, 3, 7. Thus these are the only possible orders of the stability subgroups of points. We conclude that each orbit under Aut M must contain at least = 24 points. In particular, the Weierstrass points must be the fixed points of the elements of order 7, and there must be 24 such points. Hence they must all be simple. VIL3.11. For our last example we consider the Riemann surface of W2
z6
zt
which is a hyperelliptic surface of genus 2. In addition to the hyperelliptic involution this surface permits an automorphism of period 5 (ez,e 3 w)
2rci
with e = exp(-5--).
Let us denote this automorphism by T and observe that the automorphism JT with J the hyperelliptic involution is of order 10. The automorphism
326
VII Examples
JT is given by (z,w)i— ■ fez, — e 3 w), so that the only possible fixed points of JT are P, = z -1 (0) or Qt, Q z the two points lying over co. Since T is of prime order 5, the Riemann-Hurwitz formula gives 2 = 5(2 4- — 2) + 4v(T), with v(T) as usual the number of fixed points of T. The only way this can be satisfied is with g = 0 and v(T) -- -, 3. Hence T fixes P 1 , Q 1 , and Q2. It is therefore obvious that JT fixes P 1 , but cannot fix either Q, or Q2, because JTQ, = JQ, = Q2 and JTQ 2 = = Q1. The reader should now recall Theorem V.2.11.
VII.4. Branch Points of Hyperelliptic Surfaces as Holomorphic Functions of the Periods In IV.7, we solved two elementary moduli problems. In this section we will describe one way of obtaining moduli for hyperelliptic surfaces; another elementary case. VII.4.1. We return once again to the concrete representation of a hyperelliptic surface M of genus g > 2. We will now assume that our function z of degree two is blanched over 0, 1, and :o, and hence represent M by 2g-1
(z
11,2 = z(:,
-
k= 1
)-o,
(4.1.1)
where A l , , z are distinct points in C \ {0,1 } . We are using a slight variation of (1.0.1). To fix notation, we set
= z -1 (0), Pj+ 2 =
Z
P2 =
1 (1),
j = 1, . , 2g — I,
P2 9 +2 = Z
We have seen in VII.1.2, that using P1 as a base point for 9: M J(M), the vector K of Riemann constants is given by -
1
K= —
g—1
2 1
+1
g 1 1
where 17 is the period matrix for the canonica homology basis constructed in VIL 1.1. We shall show that the Ai are holomorphic functions of the entries rcki of H. We will accomplish this by expressing the function z in terms of 0-functions. The function z e or(M) is characterized (up to a constant multiple) by the property that it has a double pole at P29+ 2, a double zero at 1' 1 , and is
VII.4. Branch
327
Points of Hyperelliptic Surfaces
holomorphic on Al\,/) 2, +2 ). We will produce such a function in terms of
0-functions. One main tool will be the Riemann vanishing theorem, and our explicit knowledge of the images in J(M) of the Weierstrass points on M. We shall see that there are many ways to proceed. We begin with the point of order 2
VII.4.2.
(P(P1P5P7 ' • P2g+ 1) + K = (P(P1P3) =
by virtue of (1.2.1) and the fact that ç(P2) = 0 for every Weierstrass point P on M. We have computed (p(P3) in =
Similarly,
Vie t) + et." 71- e(2)). K
(P(P20 +2P5P7 • • P29+1)
=
9(P29 . 2 P3) = lop +
e(2)),
We also observe that i(P1P5P7 • • • P29+ I) =
= i(P2 9 +2P5P7 • ' • P2g+t)
(compare with VII.1.4). We now consider the multiplicative function
o P i--■
[l
0 0 (4.2.1)
[1 0 0 0 0 1
According to Theorem VI.2.4, the numerator vanishes precisely (it does not vanish identically by Theorem VI.3.3) at P1, P5, P7,. . . , 1329+1 and the denominator at P29+2, P5, P7,. . . , P2g _, 1 . In particular, the function (4.2.1) vanishes to first order at P, and has a simple pole at P20. 2 . Examining the multiplicative behavior of the above function, we see that 02 [1
f(P)=
oo •
• •
O
• -
02 F1 o o
•
1
Lo
1
1 0 • •
is a meromorphic function on M with divisor PP.„. Hence f = cz,
c e C\(01.
The constant c is evaluated by f(P 2) = cz(P2) = c. Thus we see that 1 1 0 1 1 0 = f(P2) = 02[1 0 0 0 1 0
o2[
• • • •
•• • ••
01 (0p2),H)
0
2, OI