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n
(6a)
The radius of convergence of exp with respect to I | p is p" 1 ^' 1 ^
As for the logarithm function, since I n | p < p"^logp'n' \ (6b)
The radius of convergence of log(l+x) with respect to | | p is 1.
Writing D(x,r) = {y e Fp \ | y-x | p < r}, we define exp : D(0,p"1/(p"1}) -> Opx and log : D(1,1) -» Op by the p-adically convergent power series (*). Let p be 4 if p = 2 and p = p if p > 2. Then for x e Z p with I x-1 | p < p \ log(x) e D(0,p"1/(p"1}), and we define xs = exp(s log(x)) for s e Z p . Using the formal identities in the power series ring, we can verify the following properties: (7a) (7b) (7c)
exp(x+y) = exp(x)exp(y) and log(xy) = log(x)+log(y), log(xs) = slog(x) for s e Zp, exp(log(x)) = x and log(exp(x)) = x,
Exercise 3. Give a proof of the above identities. If f(s) = 2T_ an(s-a)n is a power series converging around a e Op, then its formal derivative j-(s) = Z°°_1ann(s-a)n"1 also converges at a. This is clear from the inequality lim sup | an | ^ lim sup I (n+l)a n+ i | Writing f^n) for the formal n-th derivative of f, we have an = L^L
(8a)
, since
| n+11 p < 1.
for all n e N.
In particular, we see (8b)
^
=
Let Q be the p-adic completion of the algebraic closure Q p of Q p under | | p which is the unique extension of x
I l p on Q p . Let A = { x e Q
|x|p Gal(F/F). The kernel of p is called the inertia subgroup, which we write I. Note that all non-zero elements of F are roots of unity. Then, it is easy to check that p is surjective by considering the extension F(Q for various roots of unity £ ([Bourl, VI.8.5]). There is a canonical generator n in Gal(F/F): rc(x) = x q for q = #(F). The element % is called the Frobenius element of Gal(F/F). If K/F is a (finite or infinite) Galois extension of F (see [N, Chapter I] for the Galois theory of infinite extensions) and if K is fixed by I, K is called unramified over F. If this is the case, on the residue field F' of OK, p induces a canonical isomorphism Gal(K/F) = Gal(F'/F). Thus in this case, we have a naturally specified element p4(7c) = Frob in Gal(K/F), which we call the Frobenius element of Gal(K/F). Now we suppose F to be a number field in a fixed algebraic closure Q of Q. We pick a maximal ideal p of the integer ring O of F. Let K/F be a (finite or infinite) Galois extension of F inside Q. We pick a maximal ideal T of PK over p, where OK is the integer ring of K. Then a e Gal(K/F) naturally acts
24
1: Algebraic number theory
on maximal ideals of OK- We denote by D = D((P/p) the stabilizer of (s) = (2TI V z T)- 1 J | z | = t ^dz if | s | < t < r. Moreover we see that 36 3S
=
^ 4>(s+AsH)(s) As->0 As
26
2: L-functions and Eisenstein series
Here the interchange of the integral and the limit is possible because r
f(z-s-As)-1-(z-s)"1]
lim < As—»0 I
^g
converges uniformly on the circle ( Z G C I |z| =t}. In particular, we have —(0) = (2ft V—l) 3s Repeating the above process, we have
. , 0 and satisfies F(n) = (n-1)! for positive integers n. Therefore, essentially the value of £(s) can be defined by the infinite series when either Re(s) > 1 or
2.1. Euler's method of computing £-values
27
Re(s) < 0. Now there is an interesting way to compute the value of £(l-n) for positive integers n, which was invented by Euler in 1749. We consider instead of the alternating sum: We want to compute (l-2m+1)£(-m) for m > 0. Euler's idea is to introduce an auxiliary variable t and consider
Then Euler pretended that the above series were convergent at s = -m and concluded by replacing t by 1:
(l-2m+1)^(-m) = f ^ T Y — 1 | t=1 for every integer m>0.
(1)
Here the right-hand side is the value of a rational function at t = 1 and hence a rational number. Thus if the above argument is correct, we have (2)
£(-n)
E
Q for n > 0.
Of course the above argument needs justification, but the result and the formula (1) are actually true. Exercise 2. (a) Show that the rational function
t—I f
J does not have
(t-1) as a factor in its denominator. (b) Explain why the argument of Euler is a little problematic. Of course, Euler was fully aware of the shakiness of his argument. Here is how he justified it. First he replaced t by e x . By the chain rule, we see that t-jrf(t) I
(3)
= T~f(ex) I _Q. Thus if you believe the formula (1), we have
(l-2m+1)C(-m) = QL)m( T f- r | | _ n for each integer m > 0.
Instead of x, we put 27cV~lz and write e(z) = e consider the function F(z) = ,
(/ = V - I ) . We then
/ y By (believing) (3), the Taylor expansion of
F at z = 0 is given by
(4)
F(z) = X (F(n)(0)zn/n!) = X ((l-2n+1)C(-n)(27CV^zf/n!). n=0
n=0
28
2: L-functions and Eisenstein series
By another formula of Euler, e*e = cos0 + V-IsinG, we know that iz , -iz
iz
-iz
cos(z) = —~—, sin(z) = From this fact,
cot(z) = V ^ l —
Therefore we have e(z)+l On the other hand, the function cot has the following partial fraction expansion: (5) For the moment, let us believe this expansion (which is absolutely convergent if z ^ Z). By the expansion of geometric series, we know, if | z | < 1 and z * 0, that
Then we see that
(6)
ei(z) = ^ + f ; £ n=lr=0
ln=l,k=l
J
k=l
Here only the terms for odd r survive and then we have written r+1 = 2k. Exercise 3. (a) Suppose that z g Z. Show that absolute convergence of E
{—— + —:} and also show that Z — z +— n is not absolutely convergent n=i z+n z-n (b) In (6), justify the interchange of the summations with respect to k and n (i.e. show rigorously the equality marked by ? in (6)).
From (6), we know that ! ; 2(l-22k)C(2k)z2k"1 (=} (VzT7i)-1(ei(z)-2ei(2z)) k=l
e(2z)+l e(z) 2 -2e(z)+l (e(z)-l)2 (e(z)-l)(e(z)+l) " ' e(2z)-l " " (e( _ 1 _ e(z) e(-z) e(z) , v( ? z
(
= -£ k=l
2(l-22k)C(l-2k)(27cV::Tz)2k-1/(2k-l)!.
2.1. Euler's method of computing £-values This shows that
29
C(2k) =
By specializing the functional equation £(s) =
at s = 2k, this 2r(s)cos(ics/2) equality is in fact true, because cos(kTt) = (-l) k and F(2k) = (2k-l)!. At the time of Euler, the functional equation was not known and in this way, Euler predicted its form. To make sure of our logic, we summarize our argument. Introducing an auxiliary (d y^f ex ^ I variable t, Euler related the value I —I - — r | _0 with £(-m); so we write (l-2 m+1 )a m for this value, which is not yet proven to be equal to (l-2 m+1 )£(-m). Then by definition, the formula (4) read F(z)=n=0
On the other hand, by using the partial fraction expansion of the cotangent function, we computed the power series expansion in (6): ^
e(z)+l_l__
+i Since F(-z)-F(z) =
(
T+"iz
k=i
fl , - 2 n \ i > equating the power series coefficients of
the two sides using the above two formulas, we obtain Thus we know Proposition 1.
£(2k) e 7C2kQ for all 0 < k e Z.
On the other hand, by specializing the functional equation (which we have not proved yet) at s = 2k, we have C(2k)= Then we conclude, assuming the functional equation, that
C(l-2k) = a*.! = (l-2 2 Vfxf
"TTT>
I I x=0-
Here are some remarks, (a) By the formula (1), one can compute C(2k) or £(l-2k). Here are some examples: C(2) = 1 + 2 - 2 + 3 -
2
+-
= \ , ....
30
2: L-functions and Eisenstein series
For more examples, see the table given in [Wa, p.352]. (b) The primes appearing in the numerator of £(2k), for example 691, are called irregular primes and have arithmetic significance. In fact, the class number of the cyclotomic extension Q(CP) for a root of unity £p with £pp = 1 but £p * 1 is divisible by p if and only if the prime p appears in the numerator of C(l-2k) for some k with 2 k < p - l . This is the famous theorem of Kummer proved in the mid 19th century and immediately implies the impossibility of a non-zero integer solution to the Fermat equation xp + yp = zp if p is regular (i.e. not irregular). We refer to [Wa] for more details of this direction of research and to [Ri] for the approach using modular forms. (c) The nature of the value £(2k+l) for a positive odd integer is quite different from the even values £(2k) and they are supposed not to equal a rational number times a power of n. We insert here a sketch of the proof of (5) due to Eisenstein and Weil [W2, II]. We shall show that the right-hand side of (5) satisfies the differential equation y' = -y2-7i2. The solution of this equation which goes to °° at z = 0, as is easily shown by a standard argument, is unique and equals Ttcot(Ttz). We put l v-»~ r 1 1 ^ >* + + lf r = 1 if r > 1. de These series are absolutely convergent. Here note that - p = -re r+ i. Taking two
z
X
fe
z^
independent variables p and q and putting r = p+q, we get — = — + —. pq pr qr
a
a
Differentiating once by — and —, we get (keeping the fact that r = p+q in dp dq mind) _1 1_ J_ _2_ _2_ _1_ J _ _1 2_ _2_
pV
=
PV
+
qV
+
pr3
+
qr3
Or
p V " PV " q V
=
pr3
+
qr3 '
In this equality, we put p = z+n, q = w+m-n with integers m and n. Then r = z+w+m and summing up with respect to n keeping m constant, we have
r- 3 X n {jU 1-} = 2(z+w+m)-3{e1(z)+e1(w)}. Now we sum with respect to m: (7) e2(z)e2(w) - 82(z)e2(z+w) - e2(w)e2(z+w) = = 2{ei(z)+ei(w)}]Tm(z+w+m)-3 = 2e3(z+w){ei(z)+£i(w)}. Differentiating (6) with respect to z (noting the fact that —- = -e 2 ), we get QZ
e 2 (w) = w"2 + Y~=i (2r-l)2^(2r)w2r-2 = w"2+ 2£(2) + 6i;(4)w2 + —
2.1. Euler's method of computing £-values
31
Now expanding £2(z+w) into a power series in w at w = 0 regarding z as a constant, we have, by the formula —L = -re r + i, e 2 (z+w) = X ~ o (e 2 (r) (z)w r /r!) = £ ~
0
(-l)r(r+l)er+2(z)wr,
£ 3 (z+w) = X r °l 0 (-Dr(r+l)(r+2)er+.3(z)wV2. Then the constant term of the left-hand side of (7) in the power series expansion with respect to w is given by 2C(2)e 2 (z) - £ 2 (z) 2 - 3£4(z) - 2£(2)£ 2 (z) = - £ 2 ( z ) 2 - 3e 4 (z). The constant term of the right-hand side is (by using (6)) 2£3(z)£1(z)-6£4(z). Thus we have (8a)
3e4(z) = e 2 (z) 2 + 2e 3 (z)ei(z).
Similarly by expanding both sides of (7) into a power series in x = z+w at x = 0 regarding z as a constant and equating the constant term, we have (8b)
£ 2 (z) 2 = £4(z) + 4C(2)£2(z).
Eliminating £4 using (8a,b), we know that (8c)
£i£ 3 = £ 2 2 - 6£(2)£ 2 .
Differentiating this formula, we know from —L = -r£r+i that QZ
(8d)
£ 2 £ 3 - 4^(2)£ 3 = £i£ 4 .
Multiplying (8b) by £i and eliminating £i£4 using (8d), we have ei£22-4^(2)£i£2 = £2£3-4C(2)£3 or equivalently £2(£3-£i£2) = 4£(2)(£3-£i£2). Since £2 is not the constant 4^(2), we know that £3 = £i£2. In (8c), we replace £3 by £i£2 and then divide by £2 to obtain 2 = £ 2 - 6C(2). £l Then by the facts - p = -62 and 6£(2) = n2, we know that £1 satisfies the differential equation y1 = -y2-7C2. The fact 6£(2) = 7c2 will be shown independently of this argument. Now we give a generalization of the formula (1):
(^J^) t=i
(l-2m+1)C(-m) = ( t ^ J * ^ ) I t=i for each integer m>0, which was found by Katz [Kl]. Instead of 2, we fix an integer a > 2. We define a function \ : Z —> Z by 1 if n gfe 0 mod a, 11-a if n = 0 mod a.
32
2: L-functions and Eisenstein series
We note that
Y^l^{b)
= £j = 1 5(b) = 0. We consider, instead of j ^ , the
rational function O(t) = — 7 ~ a i — • Then we see that t—l t —1 e^y0"1. Thus writing a = Re(s), we see that
I r(s) I = I Jo~ e-V-xdy I * J~ I e-V 1 1 dy = J~ e V ' d y . In particular, we know that
I JEf e-VMy I 0. To find the analytic continuation of F(s), we consider eyys~l as a function of the complex variable y. Here note that the function y — i > ys is not well defined s slog because y = Q W and log is multivalued. To fix a branch, we write /G with 0 < 9 < 2TC and define log(y) = log \ y \ +/0 and y = | y | e s/ s y = e ^(y). When 0 = 0 or 2K (i.e. y is on the positive real axis), we write ys when 0 = 0 and y.s when 0 = 2n. Note that y11 = y.n for integers n. We fix a positive real number e and denote by 3D(e) the integral path which is the circle of radius e with center 0 starting from e = I e I e l0 and with counterclockwise orientation, by P+(e) the path on the real line from +°o to e and by P_(£) the path on the real line from e to +. We consider that 0 = 0 on P+(e) and 0 = 2% on P_(e). We write the total path as P(e) = P+(e)U3D(e)UP-(e). We similarly write 9D(e,ef) for the boundary of the annulus cut along the real axis: {P+(ef)-P+(e)}UaD(e)UaD.(eI)U{P-(e')-P-(e)} for e > e' > 0, where 3D.(e') = 3D(e') as a path but has the clockwise orientation: Then
j P + ( e )e" y y s " 1 dy
and
Jp (£)e~yy-s~1(ty converge for all s and give analytic functions on C. Note that
2.2. Analytic continuation and the functional equation e y S ld +(e) " y " y
35
= - Je~ e-V'dy and / p ^ e ' V ^ y = e2ltis J£" e ^ d y .
Changing variables by y = ee l9 (dy = *ee'ed9), we have, for a sufficiently small e,
I JaD(E)e"yyS"ldy I ^ Me a J^dG = 27iMe a if a > 0, where M is a constant independent of e. Thus if a = Re(s) > 0, we know that
^ o W ' V ^ d y = (e27t's-l)r(s). The function e^y**"1 has no singularity on 3D(e,e') and hence, by Cauchy's integral formula (see [Hor, 1.2]),
Zz:poles ta v^^-yf-1
f y y
= o.
Then we have
Thus we know that the holomorphic function Jp(8)e"yys"1dy of s is independent of £ and gives (e 27ns -l)r(s) if a > 0. Thus we have the meromorphic continuation of F(s) given by
(2)
r(s) = (e2ms-l)-1Jp(E)e-yys"1dy for all s e C.
Since (e271^-!)"1 has singularities only at integers s e Z, which are simple poles, F(s) has at most simple poles at non-positive integers. Through integration by parts, we have the well known functional equation
r(s+l) = Jo~ e'Vdy = [-e"yys]Q+s Jo~ e ' V ^ y = sr(s). The analytic continuation of F(s) can be also proven by using this functional equation, which shows that F(s) has in fact simple poles at integers m < 0. Exercise 1. Show that for each positive integer n
Now we go into the integral expression of £(s) using G(z). Expanding G into a geometric series, we know that
(3)
G ( y ) = - ^ = £ e"ny. X
This series is convergent when integrate G on R+:
"e
n=l
| e' y | < 1 ( Re(y) > 0). We first formally
36
2: L-functions and Eisenstein series
n=l
n=l
n=l oo
X n=l
/.OO
and I
made at the equality marked "V
by "?". For that, we look at the poles of G(y) =——. Since l-e' y has zeros x-e only at 2%4-irL = {27cV-Tn|nG Z}, G(y) can have a pole only at 0 on R. Exercise 2. Show that | yG(y) I is bounded (independently of y) in the unit disk of radius one with center 0 (hint: show lim yG(y) = 1 and deduce the result from this). On the other hand, since e"y -» 0 as y -> +©°, there is a constant M > 0 such that I yG(y) I < Me" y/2 by the above exercise. This shows first of all, for a = Re(s),
I J~ G(y)y s " 1 dy I < Jo°° I G(y)y I y°-2dy < M Jo°° e" y/2 y a ' 2 dy = 2°-iMr(a-l). Thus
I G(y)y
dy is convergent if a > 1. Since the domain of integration
R+ is not compact, the uniform convergence of (1) is not sufficient to get the in(•OO
terchange of Z°° n—1
and
. In order to assure the interchange, we shall use the JQ
following dominated convergence theorem in the integration theory (due to Lebesgue): If a sequence of continuous (actually integrable) functions fn(x) (on an interval [a,b] in R; a and b can be ± 1. (c) Show that the denominator of B n consists of primes p such that n is divisible by p-1 (use Exercise 6(b) in the previous section). Now we prove the functional equation. The idea is to relate the function of s given by Jp(e)G(y)ys"1dy with the integration of G(y)y's"x on the following integral path for each integer m > 0: (2m+l) 1. This zeta function is called the Hurwitz L-function (and was introduced by Hurwitz in the 188O's). We then have
In the same manner as in the case of the Riemann zeta function, we have the following integral expression of £(s,x): (2)
r(sK(s,x) = J~ G(y,x)ys"1dy if Re(s) > 1,
where G(y,x) = ^ y - ^ - = - ^ y
= £~=Q
e-
(n+x)y
.
Exercise 2. (a) Show that £(s,x) is absolutely convergent if Re(s)>l. (b) Show the formula (2) along the lines of the proof of Proposition 2.1. Note that the zeros of l-e' y are situated at 27iin for n € Z and are all simple. In particular, yG(y,x) is bounded in any small neighborhood of 0. Thus for sufficiently small e > 0 (actually, any e with 0 < e < 2n does the job), the integral Jp/8xG(y,x)ys"1dy is convergent for all s and gives an analytic function of s. The same computation as in the proof of Corollary 2.1 gives Proposition 1. (e27liS-l)F(s)^(s,x) can be continued to a holomorphic function on C and has an integral expression: (e 27U ' s -l)r(s)C(s,x) = J P(e) G(y,x)y s - 1 dy for 0 < e < 2TC.
42
2: L-functions and Eisenstein series
This combined with the formula (1) yields Corollary 1. The function (e27US-l)r(s)L(s,(|)) of s can be continued to a holomorphic function on C and has an integral expression: -i d y for
0 < e < 27C/N.
As a byproduct of the analytic continuation, we can compute the value of £(l-n,x) using the formula {(e27C/s-l)r(s) | s=1.n}Cd-n,x) = JaD(£)G(y,x)y-ndy = (2TC VzT)Resy=oG(y,x)y-n. By Exercise 2.1, we know that
(e 2 ™ D H s ) I i;i
(e
Wi
( 1)n 1(27cV=II)
'
"
n
Thus we compute Resy=oG(y,x)y" . Write F(y,x) = yG(y,l-x) and expand F(y,x) into a power series in y (regarding x as a constant):
We can interpret this argument in a manner similar to §1 as follows. Writing tx y 11
t = e , we have F(y,x) = yfy. Thus pr = - + Z°° / t n t y - This shows i~ x
i~ x
^ - a-f^ = £^(B n + 1 (x)-a^B n + i(x)) y n
(1 a
"
}
n+i
n^O \n~T" x) *
y
for a n y i n t e g e r a > 1? a n d t h u s
- I'dTj \t-rV-ij I '=!•
Then Bn(x) is a polynomial in x of degree n with rational coefficients. These are called "Bernoulli polynomials". We can make this more explicit as follows:
Therefore, equating the coefficients in y11, we see that Bn(x)
=
yn
n! ^ In other words, we have the formula
j=o
B
j
x n-j
j!(n-j)!
Bn(x) = ]T*=o (^Bjx 1 1 ^ G Q[x].
(3b) Thus we know that
Res y=0 G(y,x)y- n = Res y=0 F(y,l-x)y- n - 1 = the coefficient of y11 in F(y,l-x) = This shows that
Bn(1
"x). n!
2.3. Hurwitz and Dirichlet L-functions
(
)
;
j
43
j
Thus we conclude that £(l-n,x) = (-1)11"1 n^ (l-x)y
. Here we note that
e -xy
This equality yields B n (l-x) = (-l) n B n (x).
(4) Thus we obtain
Theorem 1. We have, for any integers a > 1 and 0 < b < N,
N»(l-,~«K(-m|) - ( . i y ^ - . J * , ) | „, for allra20, and
Then the formula (1), L(s,) = E^=1(|)(a)N"sC(s,^), combined with Theorem 1, implies Corollary 2. Let Q(%) denote the cyclotomic field generated by the values of the character %. Then we have Moreover, suppose %(-l) ^ (-l) n . ^/zen we have L(l-n,%) = 0 /or n > 0 if % is non-trivial, and £Q- n ) = 0 if n > 1. In general, for ty : Z/NZ --» C, we have, if SL is prime to N, L(-m,Hm+1W = fffffyWt" a f ^ f ] I t , /or a// m > 0, dtJ t=1
^
l&T^ ST^^J '
The vanishing of L(l-n,%) follows from (4) if %(-l) * (-l) n . The number Bn>% = Z asiXteJN^Bnfe) is called the generalized Bernoulli number. Using the notation
Bn>5c, the above formula takes the following
Examples of Bernoulli polynomials:
B0(x) = 1, Bi(x) = x-i , B2(x) = x 2 - x + \ , B3(x) = Using the above formula, we can get X
^ ^
if
X is non-trivial.
form:
44
2: L-functions and Eisenstein series
Since % : (Z/NZ) X -» C x is a group character and (-1)2 = 1, %(-l) 2 = 1. Thus % ( - l ) = ± L Now we prove the functional equation of L(s,%). We proceed in the same way as in the case of the Riemann zeta function. We consider the integral on the path A(m) for each positive integer m given in §2. By the Cauchy integral formula,
The function G(y,x) has a simple pole at 2ft V—In for integers n and the residues at 27iV—In can be computed as ,„,
NsK
Resy=27t(n(G(y)x)y
U
«
because as already computed, Resy=2mn c-2ninx+(s-l)ni/2
| 2n% | s-1
>
W
I2nn | ^
) = je.27tinx+3(s.1)7li/2, ^
Qr
z=l l-e~y
e-27imx+3(s-l)7rf/2
if n > 0 ,
,s ,
.{n C x be a character. If % is not identically equal to 1, then Xg^GX(g) = 0. Similarly if g ^ 1, X%%(S) = 0» where % runs over all characters of G. Proof. Since % has values in the group of roots of unity, which is cyclic, we may assume that G is cyclic and % is injective by replacing G by G/Ker(%). Let N be the order of G. Then % induces an isomorphism of G onto the group |IN of N-th roots of unity. Thus X*(g) = X s = 0 because n^e^ N (X-C) = X N - 1 . geG
;ey. N
Let G* be the set of all characters of G. Defining the multiplication on G* by %V(g) = %(g)V(g)» G* is a group. If G is cyclic of order N, then the character is determined by its value at a generator go. Thus G* s % h-> %(go) e (IN = {£ e C x I £ = 1} defines an injection. For any given £ e [i^, defining %(gom) = £m> % is a character having the value £ at go- Thus G* = (IN- Then assigning g the character of G* which sends % to %(g), we have a homomorphism: G -> G**. Since %(g) takes all the N-th roots of unity as its values at some %, this map is surjective. Then by counting the order of both sides, we conclude that G = G**. In general, decomposing G into the product of cyclic groups, G* will be decomposed into the product of that character groups of each cyclic component. Thus G = G** in general. Then replacing G by G* and applying the first assertion of the lemma, we get the second. Lemma 2. Define the Gauss sum G(%) = Z^=1%(a)e(a/N). Suppose that % is primitive modulo N. Then ^
%(a)e(na/N) = %~l(n)G(%) for all integers n.
Note here that %~l(n) = %(n)"1 is again a character of (Z/NZ)X, which is primitive. In particular, this implies Za=1%(a)e(na/N) = 0 if n g (Z/NZ) X because %"1(n)=0 by our way of extending %A outside (Z/NZ)X.
46
2: L-functions and Eisenstein series
Proof. Define \|/ : Z/NZ -> C x by \j/(t) = e(t/N). Then, \|/(t) does not depend on the choice of t. In fact, if t = s mod N, then t = s+Nn for an integer n and thus e(t/N) = e((s/N)+n) = e(s/N)e27l/n = e(s/N). Thus \|/ gives a homomorphism of the additive group Z/NZ into the multiplicative group C x . This fact is obvious because \j/(t+s) = e(t+s) = e 27lf(t+s) = e 27lft e 27C/s = \|f(t)\|f(s). Then S^=1%(a)e(na/N) = S ae(Z/NZ)X X(a)\|/(na). We first treat the case where n mod N e (Z/NZ)X. Then the multiplication of n induces a bijection x h-> nx on (Z/NZ)X. Thus we can make the variable change in the above summation; so, rewriting na as a, we have
£ %(a)\|/(na) = ae(Z/NZ)
x
^ ( n ^ a ^ a ) = %l(n) a€(Z/NZ)
x
£x(a)\|f(a) = X'l(n)G(x)-
ae(Z/NZ)x
Now assume that n £ (Z/NZ)X. In this case %"!(n) = 0 by our way of extending x"1 outside (Z/NZ)X. Thus we need to prove that
Let p be a prime which is a common divisor of N and n. Write N = pD and n = pnf. Then we have
£x(a)\j/(na) = ae(Z/NZ)x
]T%(a)e(n'a/D) = ae(Z/NZ)x
I^n' ae(Z/DZ)x
because e(n'a/D) only depends on the class of n'a modulo D (but not N). We shall show that S t e a m o d D%(b) = 0. We see that
X
E
1
X
D%(b)«
Xb=a mod D%(b) = Eb^a mod D % ( a b a ) = X( a )Xb.l mod
Let H = ( x 6 (Z/NZ) X | x = 1 mod D). Since H = Ker(p D ), it is a subgroup of (Z/NZ)X. If % is trivial on H, then we define %0 : (Z/DZ) X -> C x by %O(PD(C)) = %(c). Xo is well defined because if PD(C) = PD(C'), then c = c'h for h G H = Ker(p D ). Thus %(c) = %(c'h) = %(c')%(h) = %(c') because of the triviality of % on H. Then % = %O°PD> which contradicts the primitivity of %. Therefore we can conclude % is non-trivial on H. Thus the orthogonality relation of characters (Lemma 1) shows that
2.4. Shintani L-functions
47
Now by using this lemma, we finish the computation:
(e27tis-l)r(s)L(s,x)
We see easily that c2nis-l e37t;s/2_x-i(_1)e™s/2
if x ( - l ) = 1, if %(-l) = - 1 .
|2COS(TCS/2)
{2^pLsin(ns/2)
This shows Theorem 2. Suppose that % is primitive modulo N. Then we have 'G(%)(2TC/N) S L(1-S,%- 1 )
^>3"
2r(s)cos(7is/2) ^ G(x)(27r/N)sL(l-s,%-1) . 2Vzir(s)sin(7Cs/2)
.,
if
^
Exercise 4. Using the above functional equation, show that L(s,%) is a holomorphic function on the whole complex plane C if % is a primitive character modulo N > 1. (The main point is to show that L(s,%) is holomorphic at s = 1; use also Corollary 2.) Exercise 5. Suppose that % is primitive modulo N. By using the functional equation (and also the power series expansion of L(s,%) at s = -j), show G(x)G(%"1) = %(-l)N for general primitive %, and supposing that L(w,%) * 0, show
G(%) = ^3C(-1)N
if % has values in {±1}. (The fact that G(x) = has values in {±1} is true without the
assumption of the non-vanishing shing of L(s,x) at ^- Try to prove it without the non-vanishing assumption.) §2.4. Shintani L-functions In this section, we introduce the contour integral of several variables and Shintani L-functions [Stl-6] and later, we will relate them with Dedekind and Hecke L-functions of number fields. We now take another branch of log different from the one in the previous section; namely, for z e C, we write it as z = | z | tlQ
48
2: L-functions and Eisenstein series
with -K < 0 < n using the polar coordinate and define log(z) = log I z | +/0. Accordingly, we define the complex power zs = es/o^(z^ by this logarithm function. We put H ' = { z e C | Re(z) > 0}: the right half complex plane, R ± = j x e R | i x > 0 ) : the right or left real line, R ± = R ± U{0}, N = By our choice of log, we have the luxury of (ab) s = a V , as = a ? and (a 1 ) 8 = a"s for any two a,b
G
H' and s e C.
To define the Shintani L-function, we need the following data: (i) a complex rxm matrix A = (ay); (ii) % = (%i, ..., %r) e C r with |%il £ 1 for all i; and (iii) x = (xi, ..., xr) G R r such that 0 < Xi < 1 for all i but not all Xi are 0. We define linear forms Li on C m and Lj* on Cr by m
r
Li(z) = X aikzk> Lj*(w) = X a kj w k k=l
(z = (zi,...,z m ), w = (wi,...,w r )).
k=l
We suppose throughout this section that (1)
Re(aij) > 0 for all i and j .
This assumption guarantees that Li(z) and Lj*(w) for z e R + m -{0} and w G R+r-{0} stay in H', because H ' 3 R+H1 and H ' D H ' + H ' . Then we formally define the Shintani L-function by (2)
C(s,A,x,x) = X n G N rX n L*(n+x)" s for s = (si
sm) G C m ,
where we write L*(n+x) = (Li*(n+x), ..., L m *(n+x)) G C m and for w = (wi, ..., w m ) G H' m , we write ws = I I ^ w / J . When A is the scalar 1 and % = 1, then
£(s,A,x,%) = ^(s,x) = Z°° (n+x)"s and thus the Shintani n=0
L-function is a direct generalization of the Hurwitz L-function. We leave the proof of the following lemma to the reader as an exercise: Lemma 1. £(s,A,x,%) converges absolutely and uniformly on any compact subset in the region Re(si) > — for all i.
2.4. Shintani L-functions
49
Exercise 1. (a) Prove the above lemma. (Reduce the problem to the case where all entries of A, %, and x are 1 and use the fact that r 1 #{k e Z + | ki + —+k r = n} < Cn*" for a constant C > 0. Actually, £(s,A,x,%) converges if Re(si+---+sm) > r and Re(sj) > 0 for all j.) (b) When all the entries of % are equal to 1 and A is a real matrix, show that £(s,A,x,%) diverges at s = —(1,...,1) (actually it diverges if si+---+sm = r). Now we give another exercise which generalizes the fact that T(s) = P e ' V ^ d y if Re(s) > 0: Jo Exercise 2. If a e Hf and s e H', then
Jot
already explained, a' s = | a r s e r°°
f e ' ^ y ^ d y = a"T(s), where as Jo
writing a = | a | e
-1
ft
__
.7C
with — < a < — . First
1
1
ay
s
y
interpret the integral J e~ y dy as a" times the integral of e" y
on the line
in H' from 0 to + with argument a. Then relate this integral with the T integral J e^y^dy by using the following integral path: and show that the integral on the inner circle of radius e (resp. the outer circle of radius N) goes to 0 as e -^ 0 (resp. as N - > +°o). N
Now we want an integral expression of £(s,A,x,%) converging in the domain with sufficiently large real part. We consider the following function G(y) = G(y,A,x,%) with variable y in R+m given by c
G(y) = X n e N r Xnexp(- ^
(3)
L*j(n+x)Vj).
The convergence of this series can be shown as follows. First of all, we see that
This shows that G(y) =
50
2: L-functions and Eisenstein series
Since Li(y) e H' as already remarked, | %iexp(-Li(y)) | < 1 and the geometric series in the inside summation converges absolutely. We then have (4) Now writing
V( X V G(y,A,x,X) = Uft r1-Xiexp(-Li(y)) ' ff r\ ' y s = n™ = 1 y; s i
for
s e C m , 1 = (1,1,...,1) e
Cra,
dy = dyidy2"-dy m and F m (s) = n^jIXsi), we have, by Exercise 2,
(5) J J - J ~ G(y,A,x,z)y s - 1 dy = JJ...J" In€Nr%nexp(-I™1L*j(n+x)yj)ys-1dy
-s = rm(s)C(s,A,x,x). As in the case of the Riemann zeta function (Proposition 2.1), we can justify the interchange of the integral [)•••[
and the summation £
r
marked by "?" if
Re(si) > — for all i, thus (5) is valid (if Re(sO > — for all i). In fact, if A is a real matrix and % = 1, the convergence of E
N rexp(-Z^ 1 L*j(n+x)yj)
to
r
G(y,A,x,l) is monotone on R+ and hence we can interchange the integral and the summation at the equality marked by "?". We also know from this that G(y,A,x,l)y Re(s) " 1 is an integrable function if Re(sO > — for all i. In general, we know that I G(y,A,x,%)ys"11 < G(y,Re(A),x,l)y Re(s) " 1 and using the dominated convergence theorem of Lebesgue, If a sequence of continuous (actually integrable) functions fn(x) (on an interval [a,b] in R; a and b can be ±oo) is dominated by a continuous and integrable function and f(x) = lim fn(x) at every point x, then n—>o
rb
rb
lim fn(x)dx = lim Ja. n—>°°
fn(x)dx,
n—>ooJa
we can justify the interchange. Things have worked in exactly the same way so far, but we encounter a serious difficulty in converting the above integral into the contour integral convergent for all s G C m . Probably, many people before Shintani considered the zeta function of type (2) and tried to get its analytic continuation, but because of this difficulty, we had to wait until 1976 [Stl] to get the analytic continuation of £(s,A,x,%).
2.4. Shintani L-functions
51
First, we explain why the naive conversion to the contour integral does not work. For simplicity, we only treat the case where % = 1 and x = 1. As already seen, words,
e -z
1-e
— has a simple pole at z = 0 whose residue is equal to 1; in other
1-e
—£ is holomorphic at z = 0. Thus each factor of
has a simple pole at the hyperplane S[= {ye C m | Lj(y) = 0} if m > 2 (if m = 1, then Si = {0}). On the other hand, if we denote D(e) = { y e Cl | y | < e } , then D(e) m gives a neighborhood of 0 in C m and thus D(e) m riSi •*• 0 if m > 2. This implies that we cannot avert the singularity by taking the path 3D(e)m. Shintani's idea is to convert the integral (5) into a contour integral by means of an ingenious variable change. We divide R + m into the following m regions: R+ m = U £ = 1 D k ,
D k = {y = (yi, ..., y m ) | y k > yi for all i * k } .
This decomposition can be illustrated in the case of m = 2 as follows: Y2 On each D k , we shall make the following variable change:
(6) (0
) = M(f;,f2,...,rm) k) and tk = 1).
0 so that if 0 < | uLi(t) I < 5 for all i. We fix such a 8. Then we can find 0 < 8' so that if | u | < 8' and T
7^-2^
| ti | < 1 for all i, then
| uLj(t) | < 8 . We also
fix such a 8'. On the other hand, we know from (7) that
limRe(Li(t)) = R e ( a i k ) > 0.
Thus we can find
Put
a = Min{Re(a i k ) | i = 1,..., r } .
1 > 8" > 0 so that if
I til < 8" for all i * k, then
Re(Li(t)) > y- Thus the only possible zeros of U\=l{ l-%iexp(-uLi(t))} in the neighborhood U = {(u,ti, ..., t k _i,t k + i, ..., t m ) | | u | < 8', | til < 8M for i * k } are at Ufl{u = 0 } . On the other hand, if R s u > 5 ' and | ti | < 8 " , Re(uLi(t)) > a8'/2 (i.e. | %iexp(-uLj(t)) | < 1), and hence no poles are expected in this remaining case. Thus on the following integral path, we do not have any singularity of G k (u,t,A,x,%) for 0 < e < min(8',8 M ) independent of t and u: P(e):
I
^
P(£,l): U , -
+oo for u
1 for ti,
where the circle is of radius e centered at 0. Note that if t is on the real line, Re(Li(t)) > a always because ay e H' for all i and j . Thus I Gk(u,t,A,x,%) | decreases exponentially as u goes to infinity when t e P(e,l) m " 1 . Thus the integral of u on the real line from e to -h» always converges. On the other hand P(e,l) m " 1 is compact and therefore the integral on this path also converges always. Thus
2.4. Shintani L-functions
53
gives an analytic function of m variables on the whole complex space C m . Thus we have the following result: Theorem 1. £(s,A,x,%) can be continued to the whole space C m as a meromorphic function and has the following integral expression valid for all s e C m :
' "
X )
"
Here we insert a general formula. We assume that %i * 1 for all i. Then G(y,A,x,%) has no singularity on a neighborhood {y | lyil C x be a Hecke character. Then the Hecke L-function L(s,%) can be continued to a meromorphic function on the whole s-plane. Moreover, it has the following expression in terms of Shintani zeta functions: eO J,(Xl,X 2 ),l),
where 1 = (1,1) and e is a totally positive fundamental unit of F. The analytic continuation of L(s,%) was first shown by Hecke in 1917. Actually, one can show, by Hecke's method, that L(s,%) is entire if % is non-trivial and only has a simple pole at s = 1 even when % is trivial. Here the word "entire" means that the function is analytic everywhere on the s-plane. We will come back to this question later in Chapter 8. Corollary l(Siegel-Klingen). For a positive integer n, L(l-n,%) e Q(%). Proof. We here give a proof due to Shintani. We will come back later to this problem and give a proof due to Siegel and another proof due to Shimura (see Corollary 5.2.2 and Theorem 5.2.2). What we need to prove is that for a positive integer n, £((l-n,l-n),
o
,(xi,X2),l) e Q.
By the study of the Shintani L-function, we know that
(e 4 *<M)(e 2 ™-l)r(s) 2 | s=1 . n }
58
2: L-functions and Eisenstein series
where
exK-x.ud + t)) x ex P (-x 2 u(e + e«t)) 1 - exp(-u(l +1)) 1 - exp(-u(e + eat)) expC-x^Ct + l)) x exp(-x2u(et + £a)) 1 - exp(-u(t + 1)) 1 -exp(-u(et + e a ))'
0(Uft) =
G a
, t) ^ '
=
Writing (n!)"2B'n(xi,X2) for the coefficient of u^11'1^11"1 in the power series expansion of G(u,t), we see that B'n(xi,X2) is a polynomial with coefficients in F. Moreover if we denote by B'na(xi,X2) the polynomial obtained by applying a to all coefficients of B'n(xi,X2), then (n!)~2B'na(xi,X2) gives the coefficient of U2(n-i)tn-i i n G a (u,t). Thus by the Cauchy integral formula and the fact that
(e4*'s-l)(e2*/s-l)r(s)21 s=1 _ = 2 ? ^ ?
}
,
we have, noting the fact that
(n-1)! Xi
e Q, ^
^ x ^ U )
= r 1 n- 2 Tr F /Q(B l n (xi > x 2 )) e Q.
We now want to introduce Eisenstein series which will be studied in detail in Chapters 5 and 9 and which is one of the simplest examples of modular forms. First let us give a brief definition of modular forms. We write H = {z € C I Im(z) > 0} for the upper half complex plane. Then the group Gl4(R) = {oce M 2 ( R ) | d e t ( a ) > 0} acts on H via zh-> cc(z) = ^ ± 1 cz + d for a = , (cz+d * 0 because z is not real). We see easily that (a(3)(z) = a(p(z)). To see a(z) stays in H, we use the following identity: for z1 = oc(z), a bYz w\ fa(z) a(w)Ycz + d 0
c dj[l
lj
=
[ 1
1 A 0
cw + d
Then replacing w by the complex conjugate of z and taking the determinant, we see that det(oc)Im(z) = Im(a(z)) |cz+d| 2 and hence if det(oc) > 0 and Z G H , then a(z) e H. For any discrete subgroup T of GL^R), a modular form on T of weight (s,k) (s e C and k e Z) is a function f: H -> C such that f(Y(z))=j'(y,z) k lj(y,z)| 2s f(z) for all ye T, where j(y,z) = det(y)"1/2(cz+d) if y =
, . We see easily that
j(y8,z) = j(y,8(z))j(8,z) (a cocycle relation). Then we see that f(yS(z)) = j(y5,z)k | j(y5,z) 12sf(z) = j(y,8(z))k | j(y,8(z)) 12sj'(8,z)k | j(8,z) 12sf(z) = J(T,5(z))k | j(y,8(z)) 12sf(8(z)) = f(y(8(z)))
2.5. L-functions of real quadratic fields and Eisenstein series
59
and hence the definition is at least consistent. When f is holomorphic (resp. C°°-class), then f is called a holomorphic (resp. C°°-class) modular form. The importance of modular forms lies in the fact that it is a non-abelian replacement of Dirichlet and Hecke characters (see Chapter 9 for more details about this fact). This point will be clarified later in Chapters 8 and 9. We will study modular forms in detail in later chapters: Chapters 5 to 10. Here we introduce an example of modular forms: the Eisenstein series. Take a positive integer N, integers 0 < a < N, 0 < b < N (so that (a,b) * 0), and k > 0. Put as a function of z e H and s e C
I mZ+n I
E'k,N(z,S;a,b) = £ ( m , n ) G Z 2. { 0 } , (m,n).(a,b) mod N
One can easily verify that this series is absolutely and locally uniformly (with respect to s and z) convergent if Re(s) > l-y- We can write E' k ,N(z,S;a,b) = 2,(m,n)eZ2-{0}, (m,n)E(a,b)modN(mZ+
mz+n
I -2s
= X'(m,n)Ez2((a+Nm)z+(b+Nn))-k I (a+Nm)z+(b+Nn)
-2s -2s
where
Z'
means the summation over all
(m,n) e Z
for which
(Tr+m)z+(TT+n) * 0. To relate this series with Shintani L-functions, we consider, for 0 < u < 1 and 0 < v < 1, q>k(z,s;u,v) = X f (mn)eZ 2 ( u z + v + m z + n )~ k ' uz+v+mz+n | "2s. Then we have -k_ /_ _ a b x E' k , N (z,s;a,b) = (1) We now split the summation of 9k into four pieces:
Here we write u* = 1-u (0 < u* < 1) and v* = 1-v (0 < v* < 1). We consider the function summed on a cone: £(CDI,G)2;S;U,V) = X(m,n)€N2(ucoi+vco2+mcoi+no)2)"k I ucoi+vco2+mcoi+nco21
-2s
60
2: L-functions and Eisenstein series
for a>i,G)2e C and for u, v not both 0- Then it is easy to see from the above figure that, except when (u,v) = (0,1), q>k(z,s;u,v) = 5(zJ;s;u,v)+^ When (u,v) = (0,1), we have 9 k (z,s;0,l) = §(z,l;s;(U) + $(-z,l;s;l,0) + £(z,-l;s;l,0) + £(-z,-l;s;0,l). We can choose a,p G C X SO that cc,ocz e H' and P,Pz G H' for any given z E C -R. The choice of p = a works well, but we keep the freedom of choosing P differently. Then we see that uocz+va+naz+ma G Hf and upz+vp+npz+mp e H' for any (m,n) e N 2 . Recall that (xy)s = x s y s if x,y e H', where we define xs by I x | seI'8s for x = | x | e1"9 with - - < 0 < - . Thus we have (uccz+va+maz+na)'s(uP z +vp+mP z +n P )"s = {(uaz+va+maz+na)(uP z +vP+mp z +np)} "s = {aP | uz+v+mz+n 12}"s = (ap)" s | uz+v+mz+n | ~2s. From this, we conclude that
2(uaz+va+maz+na)" k " s (up z +vp+mp z +np)' faz )^ In the above argument, replacing (u,v) by (u*,v*), (z,l) by (-z,-l) and (a,p) by (-a,~P), we obtain faz FQz\
(-a)-k(ap)-s^(-z,-l;s;u*,v*) = C((k+s,s),^a
p
J,(u*,v*),l).
Similarly choosing a',p' e C x so that a'(-z),a' e H' and p'(-z),p f G H \ we have
and
(-aT T kk(oc'PT ( o c 's^z,-l;s;u,v*) P T s ^ l * ) = C((k+s,s),^, C ( ( k + ) ^ ZZ Thus we have a theorem of Shintani ([St6, §4]):
"p(ZZ))J,(u,v*),l).
2.5. L-functions of real quadratic fields and Eisenstein series
61
Theorem 2. The Eisenstein series E'k,N(z,s;a,b) can be continued to a meromorphic function of s to the whole plane and is real analytic with respect to z. Moreover writing u = (-^, ^ ) and supposing u * (0,N) ifN > 1 and u = (0,1) if N = 1, we have the following expression in terms of the Shintani L-function:
E'k,N(z,s;a)b) = N- 2 s - k {(ap) s a k C((k+)H
j
(a'p')salkC((k+s,s),r'
J,u*,l) faz
a,|3,a',p' are m/:en ^ r/zaf a z , a e Hf, Pz,P e H1, a'(-z), a ' € H' p'(-z),p'e H \
and
The analytic continuation of Eisenstein series was first proven by Hecke in the 1920's. Exercise 4. (a) Show that the following subset of SL2(Z) is a subgroup of finite index:
T(N) = {a e SL 2 (Z) | a = L
A mod NM 2 (Z)}.
(b) Show that E'k,N(z,s;a,b) is a modular form of weight (k,s) on the subgroup
r(N). Now we compute the residue of E'o,N(z»s;a,b) at s = 1, which is a key to proving the class number formula for imaginary quadratic fields. We first deal with the residue at s = 1 of the Shintani zeta function. For each matrix A =
(a. b {c d
with a,b,c,d € Hf, we have seen the following integral expression: (e 47lfs -l)(e 27lIS -l)r(s) 2 C((s,s),A,x,(l,l)) = Jp (e )Jp ( e,i) G i( u ' t )u 2s - 1 t s - 1 dtdu + J p(£) Jp (el) G 2 (u,t)u 2s - 1 t s - 1 dtdu, where Ullu tj
'
_ " _ -
fexp(-xiu(a+bt))l [l-exp(-u(a+bt))j fexp(-xiu(at+b))l j j
x
x
Jexp(-x2u(c+dt))] {l-exp(-u(c+dt))f Jexp(-x2u(ct+d))] {l-exp(-u(ct+d))J*
Even if t moves around the interval [0,1], Gj(u,t)u is finite provided that (Gj(u,t) has a pole of order 2 at u = 0). Thus if Re(s) > 0, we do not have to make the contour integral with respect to t; that is, we have if Re(s) > 0
62
2: L-functions and Eisenstein series
(e 47l/s -l)r(s) 2 C((s,s),A,x,(l,l)) G2(u,t)u2s-1ts-1dtdu. Note that the expansion of Gi(u,t) at u = t = 0 is, for the Bernoulli polynomial Bj(x) introduced in §3, ix v(u(c+dt))k-1 2/j=o v-lrBjCxi) j, 2/k=o ^ Bk(X2) k! In particular, the coefficient of u"2 is (a+bfT^c+dt)"1. Now we compute the value 0
1 dt ( a + b t ) ( c + dt)
A
A.tl{A) {-log(2i+b)+log{c+d)+log{&)-log{c)} if det(A) * 0, ^
. f d . . ( A , - 0 .
We can compute similarly the integral: J p(£) J
G2(u,t)u2s~1ts~1dtdu and have
Jp(e)Jo G 2 (u,t)u 2s - 1 t s - 1 dtdu f
1
d
)
-
Z
o
^
(
d
)
}
if det(A) / 0,
. f 4 . . ( A , - 0 . Exercise 5. Explain how one can compute J
. . y ^^dt.
Now we start the computation of the residue of E'ojvjCz.sja.b). We have (<xz
i(a'p')sC((s,s)^ a, =
p
J.Cx.x1),!)
-2-i(z-zy1{log(-a'z)-log(-pz)+log(p)-hg(a')}.
If ze IT, then z € H', and a and (3 can be taken in H1. Hence log(az) = log(a)+log(z) and log($z) = log($)+log(z). On the other hand, if we write a 1 = I a' I e m and z = | z | e10, then — < 0 < — 2 2 and — < 0 —7i + a C x such that %((a)) = a k a c m for a = 1 mod m, where c denotes complex conjugation, occm = (a c ) m , (a) = aO is the principal ideal generated by a and (k,m) is a pair of integers. Such a character is called a Hecke character (of ©o-type (k,m)). When (k,m) = (0,0), % gives a character of the ideal class group Cl(m) = I(m)/2(m). The Hecke L-function is then defined by
^X)
= yLn.I(m),o^X(n)N(nr
(s E C).
Note that %(ot) = a k oc c m = a k - m N ( a ) m = a c ( m " k ) N ( a ) k . Thus writing N : 1(0) —» C x for the Hecke character given by N(a) = NF/QU) for all ideals a, L(s,%) =L(s-m,Xi) = L(s-k,%2) for %i = %N'm and %2 = XN'k. Note that %i is of type (k-m,0) and yji is of type (0,m-k). Thus without loss of generality, we may assume that % is of type (k,0) or (0,k) for k > 0. Since the argument is the same in either case, hereafter we assume that % is of type (0,k) with k > 0. Exercise 1. Assume F = Q (V d) for a square-free integer d > 0. Let % : I(m) -> C x be a quasi-character such that %((cc)) = cc k a a m whenever a = 1 mod m for a pair of integers (k,m) and the unique non-trivial field automorphism a of F. Show that k = m. (Use the existence of non-trivial units.)
64
2: L-functions and Eisenstein series
By Exercise 1, we do not lose generality by assuming k = m = 0, i.e. % is of finite order, when F is real quadratic. As in the case of real quadratic fields F, we take a representative set {a\,...,%} of Cl(m) consisting of integral ideals. Then we see that
where [i(m) = {£ E ( f l ^ s 1 mod /rc}, which is a finite group and trivial if m is sufficiently small. To express S aGa .-i a H l moda -i m a ck A^(a)" s as a sum of special values of Eisenstein series, we pick a basis (0)1,0)2) = (0)14,0)2,1) of of1 and a positive integer N such that N e m. We may assume that a* is prime to N and Z{ = 0)1/0)2. By changing 0)2 to -0)2 if necessary, we may assume that Zi = 0)1/0)2 e H. Let R i = {(a,b) e Z 2 | 0 < a < N, 0 < b < N and ao)i+bco2 = 1 mod
a(lm\.
Then Ri is a finite set, and we have
ckx
(ao)i+bo)2+Nmo)i+Nno)2)"k (a,b)ERi(m,n)EZ2 I (ao)i+bco2+Nmo)i+Nno)2) 12s"2k k = £ (afb)eRi I ^ I -2s+2kE'k,N(0)i/0)2,s-k;a,b). ( a f b ) e RG>2iG>2 Thus we have
/(m), oz>nX(n)N(nys I ^ ( w ) I ' 1 ^(a,b)ER i c °2,i" k i co2,i I • 2 s + 2 k E l k f N (z i ,s-k;a > b). Theorem 1. L(s,%) ca« Z?e continued to a meromorphic function on the whole s-plane. In particular, when k = 0, and m= O, and % is trivial, then CF(S) = w-1X"1^(«i)"Ilo)2.il"2lEIo.i(zi.s;O,l), where H is the class number of F and w = \(f\. 1 2
Idetf" "
Note that
2.6. L-functions of imaginary quadratic fields
65
On the other hand, we see that fCGj det _
©2^ _ _ _ , 9 _ = O i CG2-CO2CO1 = 0)2 0)2(zi-Zi) = l o ) 2 r 2 V - H m ( z i ) .
By the residue formula of E'o,i(zi,s;O,l) (Corollary5.2):
we have Dirichlet's residue formula: Theorem 2. Let H be the class number of the imaginary quadratic field of discriminant -D. Then we have —T=.
wVD
Let R = Z [V^D] 3 Z. Thus R is a subring of the integer ring O and R = Z[X]/(X 2 +D). Let p be an odd prime in Z prime to D. Then we see easily that X +D mod p is reducible if and only if -D is a quadratic residue mod p.
In fact, if
—
= 1, then we can find
a e Z
such that
a 2 = -D mod p and R/pR = F[X]/(X-cc)(X+a) = F 0 F for F = Z/pZ. Thus pR = p\C\p2 m R fo r two distinct prime ideals p\ and p2> If D = 0 mod 2, then O = R and thus (1) If
pO = p\p2 for p\*p2 D
is odd, then for
co =
if and only if ,
—
= 1.
the minimal polynomial of
2
co: X -X+N(co) = 0 is reducible over F if and only if there is a e Z such that a 2 =-D mod p, because 2 is invertible in F. Thus (1) is still true. For example, if D is a prime, then D = 3 mod 4 (i.e. (=£\ = (-l){D'1)/2
= -l),
and we know from the quadratic reciprocity law that
D
Thus the map p H-> —
is a Dirichlet character modulo D. More generally, we
have the following fact: Exercise
2.
Using the quadratic reciprocity law, show that the
map: p n
—
is induced by a Dirichlet character %D : (Z/DZ)X -> {±1}.
Thus we see that pO = p\p2 if and only if %D(P) = 1- Now we look at the Euler factor of the Dedekind zeta function of F. We see
2: L-functions and Eisenstein series
66
XD(P) = ! XD(P) = - 1 XD(P)
=0
Euler factor at prime ideals dividing p (l-iV( Pl )- s )(l-N(^)- s ) = (l-p"s)(l-XD(p)p-s) l-N(pYs = (l+p- s )(l-p- s ) = (l-p- s )d-% D (p)p- s ) l-iV(p)-s = (l-p- s )(l-x D (p)p- s )
Thus we have CF(S) = Up Thus we know that
l
= C(s)L(s,%D).
L(l,x D ) = Ress=iC(s)L(s,%D) = Res s= iC F (s) = On the other hand, by the functional equation, we can relate value L(0,%D) = -D"1I^:i1xD(a)a (see §3). Thus we have
L(1,%D)
with the
Theorem 3 (Dirichlet's class number formula). H = -—X a =Ti %D(a)aExercise 3. Using the above class number formula, show (i) For a prime p > 3 with p = 3 mod 4, the number a of quadratic residues in [0, S] exceeds the number b of quadratic non-residues in the same interval; (ii) If p > 3 and p = 3 mod 8, then a-b = 0 mod 3.
§2.7. Hecke L-functions of number fields Let F be a general number field and let I be the set of all embeddings of F into C. Let I(R) be the subset of I consisting of real embeddings and put I(C) = I-I(R). Then the number of real places of F is given by r = #I(R) and the number of complex places is given by t = #I(C)/2. We start with the study of the fundamental domain of F+/E, where E = ( e e O x | £ ° > 0 for all o e I(R)}, F+ = {a e F x I a a > 0 for all a e I(R)}. Thus if I(R) = 0 , then we simply put F + = F x . The result we want to prove first is Theorem 1 (Shintani [Stl, St5]). Let E' be a subgroup of finite index in E. Then there are finitely many open simplicial cones Q = C(vii,...,Vimi) with vij G F+ such that C = UjCi and F + = U£GE-eC are both disjoint unions. We can take the Q's so that there exists ua>i e C x for each a e I and i such that Re(ua)iVija) > 0 for all j = 1, ..., mi.
2.7. Hecke L-functions of number fields
67
Here an open simplicial cone C(vi, ...,v m ) in an R-vector space or Q-vector space V with generators vi e V is by definition C(vi, ..., v m ) = {xivi+---+x m v m I Xi > 0 for all i}, where the m vectors v^ are supposed to be linearly independent. We divide I(C) = X(C)U£(C)c into a disjoint union of two subsets X(C) and its complex conjugate Z(C)c for complex conjugation c and write £ for I(R)LE(C). In the theorem, we regard Q as an open simplicial cone in the real vector space V = F ^ = F (oc°)ae £ G V. Then F is a Q-vector-subspace of V which is dense in V. We put V + = R + I < R >x(C x ) L(C) , where R + = ( x e R
| x > 0 ) . Then F + = V+flF.
Proof of Theorem 1. Since the proof is the same for any E', we simply treat only the case E' = E. Consider the hypersurface X in V+ defined by X = {(x a ) a G s | N(x) = IT
aG
i
|2 =
Then, for S = {x e C \ | x | = 1}, we have p : X = S^R1"""1. In fact the projection to S l can be given by x h ) (xa/1 x a | )aeZ(C) G Sl and the projection to R1**"1 is given by xi—> /(x) = (/a(xa))z-{T}» where we exclude one embedding x e Z and / a (x a ) = log{ I x a |) or 2log( I x a |) according as a is real or complex. By definition E acts on X by componentwise multiplication. The image of E in R1"1"*'1 is a lattice by Dirichlet's theorem (Theorem 1.2.3) and hence X/E is a compact set. Thus we can find a compact subset K of X such that X = U e G E eK. We can project V+ to X via x H> N(x)~ 1/d x for d = [F:Q], which will be denoted by n. This is obviously continuous and surjective and hence takes the dense subset F + to a dense subset of X. We can find a small neighborhood U of 1 in the multiplicative group X such that eUflU = 0 if 8 * 1 in E. We may assume that U = CoflX for an open simplicial cone Co in V+ with generators in F+. Thus UXGKfto(F ) x ^° -^ ^* Since K is compact, we can choose finitely many xi e 7i(F+)flK such that UjLjXiCo ^ K. We write F+flxiCo = C0,i. Note that eCo,iflCo,i = 0 if 6 * 1 because eUflU = 0 if 8 * 1 in E and Co is the R+-span of U. Moreover Co,i is a cone with generators in F + . In fact, taking yi e F + such that 7t(yi) = xi, then Co.i = XiCo = yiCo. Since Co = C(vi, ..., Vd) with Vi G F + , we see that Co,i = C(yjVi, ..., yiv^) is a cone generated by vectors in F+. Now admitting the following lemma, we finish the proof of the theorem:
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2: L-functions and Eisenstein series
Lemma 1. Let C and C be two polyhedral cones whose generators are in F (where a polyhedral cone means a disjoint union offinitely many open simplicial cones). Then CflC, CUC and C-C are all polyhedral cones whose generators are in F. We have F + = U" =1 U eeE eC 0 ,i and Co/leCo,! = 0 if e * l . If n = l , C0,i is the desired cone. When n > 1, we define C i j = Co,i and for i > 2, Ci,i = Co,i-U eeE eCo,i. Since Co,i ( i > l ) intersects with eCo,i for only finitely many e, Cij is a polyhedral cone by the lemma. We now have ( i > 2 ) and F + = U i = 1 U £ G E e C i , i and Ci,in{U£GEeCi,i} = 0 Ci,iDeCifi = 0 if e * 1. We now construct inductively (on j) polyhedral cones Cj,i with generators in F+ for each 0 < j < n by Cy = Cj.^i if i < j and Cjj = Cj_i,i-UeeEeCj_ij (Cy = CJ.IJ) for j < i. Then we see that Cj,inU £GE eCj, k = 0
for i > j > k, F + = Ur=1UeEEeCj,i,
and Cj/leCj.i = 0 if e * 1. Then Cn,i is a disjoint union of finitely many simplicial cones which give the desired simplicial cones. This proves the first assertion of the theorem. We can subdivide the cones Cj so that the last assertion follows. Exercise 1. Give a detailed proof of the last assertion of Theorem 1. Proof of Lemma 1. We may assume that C and C are open simplicial cones. Write C = C(vi, ..., v m ), where the vi's are linearly independent. By adding v m + i, ..., v 0 for i * j and Lj(w) = 0}-{0}, X = {w e W | Li(w) > 0 for i = 1, ..., /}-{0}. Since Xj is contained in Ker(Lj) which has dimension less than dim(W), by the induction hypothesis, Xj is a disjoint union of open simplicial cone. Moreover, Xj-Ui^jXi is a disjoint union of open simplicial cones, and hence it is easy to see that X-X = UjXj is also a disjoint union of open simplicial cones. Write these cones as UjXj = UkC(vki,...,Vkik). Let u be a point in XflF, which exists because X is open in W and Ff|W is dense as already remarked. Since v ki>---»vkik a r e i n a proper subspace of W and u is not in the subspace spanned by Vki,...,Vkik, u,Vki,...,Vkik are linearly independent. Write Ck(u) for C(u,Vki,..-,Vkik). Then we claim that X = UkCk(u)UR+u (disjoint). By definition, Li(x) > 0 for all i if x e X. Thus in particular, if x = Xu for X e R , then Li(x) = A,Li(u) > 0, Li(x) > 0 and Li(u) > 0. Thus, RuflX = R+u. Suppose that x e X is not a scalar multiple of u. Let s be the minimum of Lj(x)/Li(u) for i = 1,..., /. Then s > 0. There is an index i (maybe several) such that s = Li(x)/Li(u). Then Lj(x-su) = Lj(x)-sLj(u) > 0 and Li(x-su) = 0. Thus x-su e X-X. Therefore x-su G Ck for a unique k (because X-X is a disjoint union of the Ck's) and thus x G Ck(u). This shows the desired assertion. Exercise 2. Write down the proof of X-X = UjXj explaining every detail. Let Z[I] be the free module generated by embeddings of F into C. We can think of each element t, e Z[I] as a quasi-character of F x which takes a e F x to a^ = I I a G l a a ^ a G C x . A quasi-character X : I(m) -» C x for an ideal m of O is called an arithmetic Hecke character if there exists ^ G Z[I] such that X((a)) = oc^ for all a G P(W), where P(m) = {a G F+ I (3(a-l) G m for some (3 G O prime to m}.
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2: L-functions and Eisenstein series
Theorem 2 (Hecke). If X is an arithmetic Hecke character modulo m, then L(s,X) = £ne /(m), ODn X(n)N(n)~s can be continued to a meromorphic function on the whole complex s-plane C. Proof. Let {a\9 ...,%} be a representative set of ideal classes for //fP+ (see Exercise 1.2.1). We may assume that a\ are all prime to nu Since still gives a representative set, we can write
Now we take the fundamental domain C = U ^ C j for F+/E with disjoint open simplicial cones Cj. Here note that for a positive rational number u, Cj = uCj by definition. In particular, NCj = Cj for any positive integer N. Thus we may assume that Cj is generated by totally positive integers in 0. Fix a set of generators {vi,..., vb} of Cj in O and consider the Shintani L-function where v a = ( v i a , ..., v b a ) e C b , x = x 1 v 1 + «"+x b v b with Xi e (0,1], x a = viaxi+---+VbaXb and n«va = vi a ni+--+Vb a nb. By Theorem 1, we also have u a e C x such that all the entries of u a v a have positive real parts. Then
where we have to choose the branch of /