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0 be such that the $ 62 }
with centre (uo, vo) and radius 6 is contained in U (such a 6 exists because U is open). Then,
lim AN(Ro) = lKl, 0--+0 Aa(Ro) where K is the gaussian curvature of a at a(uo, vo).
167
7. Gaussian Curvature and the Gauss Map
Proof 7.1
By Definition 5.3,
ffR6 11 N u x N tl 11 dudv Aa(Ro) - ffR6 11 au x af} 11 dudv .
AN(Ro)
_--'--:--,'- _
(14)
By Proposition 6.4, N u x N tl = (aa u x bD'tI) x
(CO'u
x dati)
= (ad - bc)a u x a tl
== det(-.ri 1 .r11 )a u x a tl det(.rl1 ) = det(.r1) au x a tl L M E
M N F au x af}
G
F
=
(by the definition of .r1 and .r11)
LN-M 2 EG _ F2 au x a
= Ka u x af}
f}
(by Proposition 7.1(i)).
(15)
Sustituting in Eq. (14), we get AN(Ro)
_--'----'- _
Aa(Ro)
ffR 6 ]K]l1 au x a v 11 dudv ffR6 11 au x af} 11 dudv .
Let f be any positive number. Since K(u, v) is a continuous function of (u, v) (see Exercise 7.7), we can choose 6 > 0 so small that
]K(u, v) - K(uo, vo)] < f if (u, v) E Ro. Since, for any real numbers a, b, ]a - b] that 11K(u, v)] - ]K(uo, vo)11 < f if (u, v) E R o, Le.
~
]K(uo, vo)]- f < ]K(u, v)] < ]K(uo, vo)]
lla] - ]b11, it follows
+f ~
if (u, v) E R o. Multiplying through by 11 au x af} 11 and integrating over R o, ~ft get
(]K(uo,vo)]-
f)/
/11
au x a tl lldudv
To find the missing geodesics, we recall that the cylinder is isometric to the plane (see Example 5.5). In fact, the isometry takes the point (u, v, 0) of the xy-plane to the point (cos u, sin u, v) of the cylinder. By Corollary 8.2, this map takes geodesics on the plane (Le. straight lines) to geodesics on the cylinder, and vice versa. So to find all the geodesics on the cylinder, we have only to find the images under the isometry of all the straight lines in the plane. Any line not parallel to the y-axis has equation y = mx + c, where m and c are constants. Parametrising this line by x = U, Y = mu + c, we see that its image is the curve
")'(u)
= (cosu,sinu, mu + c)
on the cylinder. Comparing with Example 2.1, we see that this is a circular helix of radius one and pitch 271'"lml (adding c to the z-coordinate just translates the helix vertically). Note that if m = 0 we get the circular geodesics that we already know. Finally, any straight line in the xy-plane parallel to the y-axis is mapped by the isometry to a straight line on the cylinder parallel to the z-axis, giving the other family of geodesics that we already know.
EXERCISES 8.6 Show that, if P and Q are distinct points of a circular cylinder, there are either two Or infinitely many geodesics on the cylinder joining P and Q. Which pairs P, Q have the former property? 8.7 Use Corollary 8.2 to find all the geodesics on a circular cone. (Use Exercise 5.5.) 8.8
Use Corollary 8.2 to show that the geodesics on a generalised cylinder are exactly those constant-speed curves on the cylinder whose
181
8. Geodesics
tangent vector makes a constant angle with the rulings of the cylinder. 8.9
Find the geodesics on a circular cylinder by solving the geodesic equations.
8.10 Let ")'(t) be a unit-speed curve on the helicoid
a(u, v)
= (ucosv,usinv,v).
Show that
il,2
+ (1 + u 2)v 2 = 1
(a dot denotes d/dt). Show also that, if")' is a geodesic on a, then . a v = - - - , -2 1 +u
'
where a is a constant. Find the geodesics corresponding to a and a = l.
=0
8.11 Show that, if N is the standard unit normal of a surface patch a( u, v) with first fundamental form Edu 2 + 2Fdudv + Gdv 2, then N x a
Eav - Fa u = u "';EG-- F2 '
N
xa
tI
v - Ga u = Fa "';EG _ F2
---;::;:;::;;;;:=~
(Use Proposition 5.2.) Deduce that, if a(u(t), v(t)) is a unit-speed curve on a, its geodesic curvature "'g
= (vii, -
vu)JEG - F2
+ AiI,3 + BiI,2v + Cil,i;2 + Dv 3 ,
where A, B, C and D can be expressed in terms of E, F, G and their derivatives. (Use the method of proof of Theorem 8.1 to calculate dot products such as au.a vtI .) This gives another proof of Corollary 8.2. 8.12 Show directly that the parameter of any curve satisfying the geodesic equations (2) is proportional to arc-length.
8.3. Geodesics on Surfaces of Revolution It turns out that, although the geodesic equations for a surface of revolution cannot usually be solved explicitly, they can be used to get a good qualitative understanding of the geodesics on such a surface. We parametrise the surface of revolution in the usual way
a(u,v)
= (j(u)cosv,j(u)sinv,g(u)),
Elementary Differential Geometry
182
where we assume that
f > 0 and
(1u)
2
+ (*) 2 = 1 (see
Examples 4.13 and
6.2 - note that in these examples we used a dot to denote djdu, but now a dot is reserved for djdt, where t is the parameter along a geodesic). We found in Example 6.2 that the first fundamental form of a is du 2 + f(u)2dv 2. Referring to Eq. (2), we see that the geodesic equations are
dd (f(u)2i;) = O. t We might as well consider unit-speed geodesics, so that u2 + f(u)2 V2 = l. ii
= f(u) du df i;2 '
(7)
(8)
l.From this, we make the following easy deductions: Proposition
8.5
On the surface of revolution a(u,v) = (f(u)cosv,f(u)sinv,g(u)), (i) every meridian is a geodesic; (ii) a parallel u = Uo (say) is a geodesic if and only if df j du i.e. Uo is a stationary point of f·
= 0 when u = uo,
--geodesics
8. Geodesics
183
Proof 8.5
On a meridian, we have v = constant so the second equation in (7) is obviously satisfied. Equation (8) gives u = ±1, so u is constant and the first equation in (7) is also satisfied. For (ii), note that if u = Uo is constant, then by Eq. (8), v = ±1/ f(uo) is non-zero, so the first equation in (7) holds only if df /du = O. Conversely, if d// du 0 when u Uo, the first equation in (7) obviously holds, and the second holds because v = ±l/f(uo) and f(u) = f(uo) are constant. 0
=
=
Of course, this proposition only gives some of the geodesics on a surface of revolution. The following result is very helpful in understanding the remaining geodesics.
Proposition 8.6 (Clairaut's Theorem) Let 'Y be a geodesic on a surface of revolution 5, let p be the distance of a point of 5 from the axis of rotation, and let 'tj; be the angle between'Y and the meridians of 5. Then, psin'tj; is constant along 'Y. Co nversely, if P sin 'tj; is co nstan t along some curve 'Y in the surface, and if no part of'Y is part of some parallel of 5, then 'Y is a geodesic.
s
-~--------~.,----
-------
In the second paragraph of the proposition, by a 'part' of'Y we mean 'Y( J), where J is an open interval. The hypothesis there cannot be relaxed, for on a
184
Elementary Differential Geometry
parallel 'ljJ == 1r /2, so p sin 'ljJ is certainly constant. But parallels are not geodesics in general, as Proposition 8.5(ii) shows. Proof 8.6
Parametrising 5 as in Proposition 8.5, we have p = f(u). Note that aul II au II =au and avl II a v II = p-1a v are unit vectors tangent to the parallels and the meridians, respectively, and that they are perpendicular since F = O. Assuming that "Y( t) = a( u( t), v( t)) is unit-speed, we have 'Y
= cos 'tj; au + p-l sin 'tj; a v
(this equation actually serves to define the sign of 'tj;, which is left ambiguous in the statement of Clairaut's Theorem). Hence, au x'Y Since 'Y
= p-l sin'tj;a u
x avo
= uau + va v , this gives = p-l sin 'tj; au x a v , pi; = sin 'tj;.
vau x au
Hence, p sin 'tj;
= p2 iJ •
But the second equation in (7) shows that this is a constant, say n, along the geodesic. For the converse, if p sin 'tj; is a constant n along a unit-speed curve "Y in 5, the above argument shows that the second equation in (7) is satisfied, and we must show that the first equation in (7) is satisfied too. Since
v=
sin'tj; p
=n
(9)
p2'
Eq. (8) gives ·2
=
1
n2
(10)
-"2' P Differentiating both sides with respect to t gives
u
2n2 . 2n2 dp . 2uu= - p = - - u p3 p3 du ' . ..
. (-U
U
-
.2) =.
dp v Pdu
0
If the term in brackets does not vanish at some point of the curve, say at "Y(to) a(uo, vo), there will be a number e > 0 such that it does not vanish for It - tol < e. But then u 0 for It - tol < e, so "Y coincides with the parallel
=
=
8. Geodesics
185
u = Uo when It - tol < €, contrary to our assumption. Hence, the term in brackets must vanish everywhere on "(, Le. ., U
dp'2
= Pdu v
,
showing that the first equation in (7) is indeed satisfied.
0
Clairaut's Theorem has a simple mechanical interpretation. Recall that the geodesics on a surface S are the curves traced on S by a particle subject to no forces except a force normal to S that constrains it to move on S. When S is a surface of revolution, the force at a point P of S lies in the plane containing the axis of revolution and P, and so has no moment about the axis. It follows that the angular momentum n of the particle about the axis is constant. But, if the particle moves along a unit-speed geodesic, the component of its velocity along the parallel through Pis sin1/', so its angular momentum about the axis is proportional to p sin 1/'. Example 8.8
We use Clairaut's Theorem to determine the geodesics on the pseudosphere (Section 7.2): a( u, v)
= (e
U
cos v, e U sin v,
VI - e2u + cosh
-1 (e-
U
)).
We found there that its first fundamental form is du 2
+ e2u dv 2 .
It is convenient to reparametrise by setting w surface is j
v,w)
= (~ cosv
= e-
U
•
The reparametrised
~ siuv Vi ~2 + cosh-
1
w)
and its first fundamental form is dv 2
+ dw 2 w2
We must have w > 1 for ii to be well defined and smooth. If "((t) = ii(w(t), v(t)) is a unit-speed geodesic, the unit-speed condition gives (11) and Clairaut's Theorem gives 1 .
-w sm1/'
= -w12v. = n ,
(12)
AVV
L..H••••• ""II\.C.J
=
=
LJ
CI
'\,.I~U•• I~
J
=
where n is a constant, since p l/w. Thus, 1; f}w 2. If f} 0 we get a meridian v = constant. Assuming now that f} :/: 0 and substituting in Eq. (11) gives
Hence, along the geodesic, dv dw (v - vo) :.
= 1; = ± W
f}w v'1- f}2 w 2'
= =f ~ VI (v-vO)2 +w
f}2 w 2, 2
1 = f}2'
(13)
where Vo is a constant. So the geodesics are the images under jj of the parts of the circles in the vw-plane given by Eq. (13) and lying in the region w > 1. Note that these circles all have centre on the v-axis, and so intersect the v-axis perpendicularly. The meridians correspond to straight lines perpendicular to the v-axis. w
1
v
Since w > 1, any geodesic other than a meridian has a maximum value of w, which it attains, and a maximum and minimum value of v, which it approaches arbitrarily closely but does not attain (see the diagram below). This shows that the pseudosphere is 'incomplete', i.e. a geodesic on the pseudosphere cannot be continued indefinitely (in one direction if it is a meridian, in both directions otherwise) .
..lUI
Returning now to an arbitrary surface of revolution 5, we describe how Clairaut's Theorem allows us to describe the qualitative behaviour of the geodesics on 5. Note first that, in general, there are two geodesics passing through any given point P of 5 with a given angular momentum n, for v is determined by Eq. (9) and u up to sign by Eq. (10). In fact, one geodesic is obtained from the other by reflecting in the plane through P containing the axis of rotation (which changes n to -n) followed by changing the parameter t of the geodesic to -t (which changes the angular momentum back to n again). The discussion in the preceding paragraph shows that we may as well assume that n > 0, which we do from now on. Then, Eq. (10) shows that the geodesic is confined to the part of 5 which is at a distance > n from the axis. If all of 5 is a distance> n from the axis, the geodesic will cross every parallel of 5. For otherwise, U would be bounded above or below on 5, say the former. Let Uo be the least upper bound of u on the geodesic, and let n + 2e, where e > 0, be the radius of the parallel u = uo. If u is sufficiently close to uo, the radius of the corresponding parallel will be ;?: n + e, and on the part of the geodesic lying in this region we shall have
by Eq. (10). But this clearly implies that the geodesic will cross u = uo, con-
tradicting our assumption. Thus, the interesting case is that in which part of 5 is within a distance n of the axis. The discussion of this case will be clearer if we consider a concrete example whose geodesics nevertheless exhibit essentially all possible forms of behaviour. Example 8.9
We consider the hyperboloid of one sheet obtained by rotating the hyperbola x 2 - z2
= 1,
x> 0,
in the xz-plane around the z-axis. Since all of the surface is at a distance > 1 from the z-axis, we have seen above that, if 0 ::.; n < 1, a geodesic with angular momentum n crosses every parallel of the hyperboloid and so extends from z = -00 to z = 00.
Elementary Differential Geometry
188
0< Suppose now that regions
n> z
>
n 0 for all t E (to - T], to + T]), so the integral on the right-hand side of Eq. (20) is > O. This contradiction proves that we must have U(O, t) = 0 for all t E (a, b). One proves similarly that V(O, t) = 0 for all t E (a, b). Together, these results prove that "Y satisfies the geodesic equations.
o It is worth making several comments on Theorem 8.2 to be clear about what it says, and also what it does not say. Firstly, if"Y is a shortest path on a from p to q, then £(T) must have an absolute minimum when T = O. This implies that d~£(T) = 0 when T = 0, and hence by Theorem 8.2 that "Y is a geodesic. Secondly, if "Y is a geodesic on a passing through p and q, then £(T) has a stationary point (extremum) when T == 0, but this need not be an absolute minimum, or even a local minimum, so "Y need not be a shortest path from p
-------_ -----....
- - -.ep
-.e- q
to q. For example, if p and q are two nearby points on a sphere, the short great circle arc joining p and q is the shortest path from p to q (this is not quite obvious - see below), but the long great circle arc joining p and q is also a geodesic.
195
8. Geodesics
Thirdly, in general, a shortest path joining two points on a surface may not exist. For example, consider the surface 5 consisting of the xy-plane with the origin removed. This is a perfectly good surface, but there is no shortest path on the surface from the point p = (-1,0) to the point q = (1,0). Of course, the shortest path should be the straight line segment joining the two points, but this does not lie entirely on the surface, since it passes through the origin which is not part of the surface. For a 'real life' analogy, imagine trying to walk from p to q but finding that there is a deep hole in the ground at the origin. The solution might be to walk in a straight line as long as possible, and then skirt around the hole at the last minute, say taking something like the following route:
_ - - - t•
....-->-+----------l(\L..---->~.
P
---
q
This path consists of two straight line segments of length 1 - 1:, together with a semicircle of radius 1:, so its total length is
2(1-1:)+1r1:= 2+(1r-2)L Of course, this is greater than the straight line distance 2, but it Can be made as close as we like to 2 by taking I: sufficiently small. In the language of real analysis, the greatest lower bound of the lengths of curves on the surface joining p and q is 2, but there is no curve from p to q in the surface whose length is equal to this lower bound. Finally, it Can be proved that if a surface 5 is a closed subset of R 3 (i.e. if the set of points of R 3 that are not in 5 is an open subset of R 3), and if there is some path in 5 joining any two points of 5, then there is always a shortest path joining any two points of 5. For example, a plane is a closed subset of R 3 , so there is a shortest path joining any two points. This path must be a straight line, for by the first remark above it is a geodesic, and we know that the only geodesics on a plane are the straight lines. Similarly, a sphere is a closed subset of R 3 , and it follows that the short great circle arc joining two points on the sphere is the shortest path joining them. But the surface 5 considered above is not a closed subset of R 3 , for (0,0) is a point not in 5, but any open ball containing (0,0) must clearly contain points of 5, so the set of points not in 5 is not open.
196
Elementary Differential Geometry
Another property of surfaces that are closed subsets of R 3 (that we shall also not prove) is that geodesics on such surfaces can be extended indefinitely, Le. they can be defined on the whole of R. This is clear for straight lines in the plane, for example, and for great circles on the sphere (although in the latter case the geodesics 'close up' after an increment in the unit-speed parameter equal to the circumference of the sphere). But, for the straight line "Y(t) == (t-l,O) on the surface S defined above, which passes through p when t == 0, the largest interval containing t == 0 on which it is defined as a curve in the surface is (-00,1). We encountered a less artificial example of this 'incompleteness' in Example 8.8: the pseudosphere considered there fails to be a closed subset of R 3 because the points of its boundary circle in the xy-plane are not in the surface.
EXERCISES 8.19 The geodesics on a circular (half) cone were determined in Exercise 8.7. Interpreting 'line' as 'geodesic', which of the following (true) statements in plane euclidean geometry are true for the cone? (i) There is a line passing through any two points. (ii) There is a unique line pa.'3sing through any two points. (iii) Any two distinct lines intersect in at most one point. (iv) There are lines that do not intersect each other. (v) Any line can be continued indefinitely. (vi) A line defines the shortest distance between any two of its points. (vii) A line cannot intersect itself transversely (Le. with two nonparallel tangent vectors at the point of intersection). 8.20 Construct a smooth function with the properties in (19) in the following steps: 2 (i) Show that, for all integers n (positive and negative), ttl e - l j t tends to 0 as t tends to O. (Use L'Hopital's rUle.) (ii) Deduce from (i) that the function
B( t) == { e-
1jt2
o
if t if t
> 0, 0, and let b = cosh a. The surface S consisting of the part of the catenoid with Izl < a has boundary the two circles C± of radius b in the planes z ±a with centres on the z-axis. Another surface spanning the same two circles is, of course, the surface So consisting of the two discs x 2 + y2 < b2 in the planes z = ±a. The area of S is, by Proposition 5.2 1
=
{21f 0
1
fa -0
(EG _ F 2)1/2dudv
=
(21f 0
1
f
Q
cosh 2 u dudv
= 21r(a + sinh a cosh a).
-0
The area of So iS 1 of course, 21rb 2 = 21r cosh 2 a. So the minimal surface S will not minimise the area among all surfaces with boundary the two circles C± if cosh 2 a < a + sinh a cosha, Le. if 1 + e- 2a
< 2a.
(2)
2a
a
9. Minimal Surfaces
205
The graphs of 1 + e- 2a and 2a as functions of a clearly intersect in exactly one point a aO I saYI and the inequality (2) holds if a > ao. If this condition is satisfied, the catenoid is not area minimising. It can be shown that if a < ao the catenoid does have least area among all surfaces spanning the circles C+ and C-.
=
It is time to prove Theorem 9.1.
Proof 9.1 Let tp' = u T, so that tp0 = tp, and let NT be the standard unit normal of aT. There are smooth functions aT I j3T and "'( T of (u, V I 7) such that
= aTNT + 13' a~ + "'(T a~
tp'
=
I
so that a a O . To simplify the notation, we drop the superscript of the proof; at the end of the proof we put 7 O. We have
A(7) =
7
for the rest
=
J1.
int(1r)
so
.A ==
II au
x av
J1.
Int(1r)
II
dudv =
J1.
int(1r)
N.(au x a v ) dudv,
88 (N.(a u x a v )) dudv.
(3)
7
Now,
:7 (N.(a u
x a v ))
= N.(a u x a v ) + N.(uu x a v ) +-N.(au x u v ).
Since N is a unit vectorI
N.(au x a v ) = N.N II au x a v
II = o.
On the other hand, ) _ (au x av).(uu x a v) II au x a v II
N (.
. au x a v -
_ (au.uu)(av.a v ) - (au.av)(av.iT u ) II au x a v II _ G(au.u u ) - F(av.u u ) (EG - F2)l/2 using Proposition 5.2. SimilarlYI N (
. ) _ E(av.u v ) - F(au.uv)
. au x a v -
I
(EG _ F2)1/2
(4)
Elementary Differential Geometry
206
Substituting these results into Eq. (4) we get I
8 (N ( 8T . lTu
lTv
X
)) _ E{lTv.uv) - F{uu.lTv + lTu.uv) + G(lTu.uu) (EG _ F2)1/2
(5)
Now
= tpu = auN + j3u lTu + '"'(ulTv + aNu + j3lTuu + '"'(lT lTu.u u = Ej3u + F'"'(u + (uu.Nu)a + (lT u .lTuu )j3 + (lTu.lTuv)T Since lTu.N u = -lTuu.N = -L lTu.lT uu = ~Eu and lTu.lT uv = ~Ev, we get Uu
UVI
l
lTu.uu
= Ej3u + F'"'(u -
La + ~ Euj3 + ~ Ev'"'(.
Similarly,
= Fj3u + G'"'(u lTu.u v = Ej3v + F'"'(v -
Ma
+ (Fu - ~Ev)j3 + ~Gu'"'(,
Ma
+ !Ev j3 + (Fv - !Gu)'"'(,
= Fj3v + G'"'(v -
Na
+ ~Guj3 + ~Gv'"'(.
lTv.Uu
lTv.u v
2
2
Substituting these last four equations into the right-hand side of Eq. (5), simplifying, and using the formula for H in Proposition 7.1(ii), we find that
:T(N.(lT u x lTv))
= (j3(EG -
F 2)1/2)u + ('"'((EG - F 2)1/2)v
- 2aH(EG - F 2)1/2.
(6)
Comparing with Eq. (3), and reinstating the superscripts, we see that we must prove that
~
J
{(j30( EG - F 2)1/2)
Jint(fr)
+ ('"'(0 ( EG -
F 2)1/2) } dudv
u
= O.
(7)
v
But by Green's Theorem (see Section 3.1), this integral is equal to
1
(EG - F 2)1/2(j3odv - '"'(Odu) ,
and this obviously vanishes because 13° = '"'(0 This completes the proof of Theorem 9.1.
= 0 along the boundary curve fr. 0
Note that we did not quite use the full force of the assumptions in Theorem 9.1, since they imply that aO (= a) vanishes along the boundary curve, and this was not used in the proof. So Eq. (1) holds provided the surface variation tp is normal to the surface along the boundary curve. Note also that Theorem 9.1 is intuitively obvious for variations tp that are parallel to the surface, Le. those for which a = 0 everywhere on the surface,
9. Minimal Surfaces
207
since such a parallel variation causes the surface to slide along itself and will not change the shape, and in particular the area, of the surface. Thus, the main point is to prove Theorem 9.1 for normal variations, Le. those for which j3 "y 0 everywhere on the\. surface. Making this restriction simplifies the above proof considerably.
= =
EXERCISES 9.1 Show that any rigid motion of R 3 takes a minimal surface to another minimal surface, as does any dilation (x,y,z) 1-4 a(x,y,z), where a is a non-zero constant. 9.2 Show that z = f(x y), where f is a smooth function of two variables, is a minimal surface if and only if l
(1
+ f;)fxx
- 2fxf7l f x71
+ (1 + {;)f7l71 = o.
9.3 Show that every umbilic on a minimal surface is a planar point (see Proposition 6.3). 9.4 Show that the gaussian curvature of a minimal surface is < 0 everywhere, and that it is z"ero everywhere if and only if the surface is part of a plane. (Use Proposition 6.5.) We shall obtain a much more precise result in Corollary 9.2. 9.5 Show that there is no compact minimal surface. (Use Proposition 7.6 and Exercise 9.4.)
9.2. Examples of Minimal Surfaces The simplest minimal surface is, of course, the plane, for which both principal curvatures are zerO everywhere. Apart from this) the first minimal surfaces to be discovered were those in the following two examples.
Example 9.2 A catenoid is obtained by rotating a curve x = 1a cosh az in the xz-plane around the z-()Xis, where a > 0 is a constant. We showed in Example 9.1 that this is a minimal surface (we only dealt there with the case a = 1, but the general case followsfrom it by using Exercise 9.1).
Elementary Differential Geometry
208
The catenoid is a surface of revolution. In fact, apart from the plane it is the only minimal surface of revolution:
Proposition 9.1
Any minimal surface of revolution is either part of a plane or can be obtained by applying a rigid motion to part of a catenoid. Proof 9.1 By applying a rigid motion, we can assume that the axis of the surface S is the z-axis and the profile curve lies in the xz-plane. We parametrise S in the usual way (see Example 6.2):
a(u, v)
= (f(u) cosv, f(u) sin v, g(u)),
where the profile curve u t-+ (f(u), 0, g(u)) is assumed to be unit-speed and f > O. From Example 6.2, the first and second fundamental forms are du 2 + f(U)2dv 2 and (ig -liJ)du2 + fiJdv 2 , respectively, a dot denoting dJdu. By Proposition 7.1(ii), the mean curvature is
9. Minimal Surfaces
209
We suppose now that, for some value of u, say u == uo, we have g(uo) :/: O. We shall then have g(u) 1- 0 for u in some open interval containing uo. Let (a,;3) be the largest such interval. Supposing now that u E (a, ;3), the unitspeed condition j2 + il = 1 gives-(as in Example 7.2)
.
..
fg - fi;
1 = --;, 9
so we get
Since iJ2
= 1 - j2, S is minimal if and only if fi = 1- j2.
(8)
To solve the differential equation (8), put h ..
f
dh dhdj == dt == df dt
= j, and note that
dh = h df .
Hence, Eq. (8) becomes dh fh df Note that, since iJ follows:
-#
= 1- h2 .
0, we have h2 -# 1, so we can integrate this equation as hdh =Jdf f J 1 - h2 1
--;==~ -
./1- h 2 -
h=
af
'
,
vI'-a'2:""""f---:2- - -1 af
,
where a is a non·zero constant. (We have omitted a ±, but the sign can be changed by replacing u by -u if necessary.) Writing h dfJdu and integrating again,
=
. where b is a constant. By a change of parameter u b = O. So
t-+
u + b, we can assume that
Elementary Differential Geometry
210
To compute g, we have
il = 1 - /2 = 1 _ dg =
±
= _1_, 2 2
h2
a
1
/
VI + a2u 2
,
.,
du
..
= ±.!a sinh- 1 (au) + c au = ± sinh(a(g - c)), / = -a1 cosh(a(g - c)). 9
(where c is a constant),
Thus, the profile curve of S is 1 a
x = - cosh(a(z - c)). By a translation along the z·ax.is, we can assume that c = 0, so we have a catenoid. We are not quite finished, however. So far, we have only shown that the part of S corresponding to u E (a, /3) is part of the catenoid, for in the proof we used in an essential way that 9 -# O. This is why the proof has so far excluded the possibility that S is a plane. To complete the proof, we argue as follows. Suppose that /3 < 00. Then, if the profile curve is defined for values of u > /3, we must have 9(/3) = 0, for otherwise iJ would be non-zero on an open interval containing /3, which would contradict our assumption that (a,/3) is the largest open interval containing Uo on which iJ -# O. But the formulas above show that 2
9
= 1 + ~2U2
if u E (a,/3),
so, since iJ is a continuous function of u, 9(/3) = ±(1 + a2 f32)-1/2 -# O. This contradiction shows that the profile curve is not defined for values of u 2: /3. Of course, this also holds trivially if /3 = 00. A similar argument applies to a, and shows that (a, /3) is the entire domain of definition of the profile curve. Hence, the whole of S is part of a catenoid. The only remaining case to consider is that in which iJ(u) = 0 for all values of u for which the profile curve is defined. But then g( u) is a constant, say d, and S is part of the plane z = d. 0 Example 9.9
A helicoid is a ruled surface swept out by a straight line that rotates at constant speed about an axis perpendicular to the line while simultaneously moving at constant speed along the axis. We can take the axis to be the z-axis. Let w be the angular velocity of the rotating line and a its speed along the z-axis. If the
9. Minimal Surfaces
211
line starts along the x-axis l at time v the centre of the line is at (0,0, av) and it has rotated by an angle wv. Hence, the point of the line initially at (u, 0,0) is now at the point with position vector
a(u, v)
= (u'coswv, u sinwv, av).
We leave it to Exercise 9.6 to check that this is a minimal surface.
We have the following analogue of Proposition 9.1.
Proposition 9.2
•
Any ruled minimal surface is part of a plane or part of a helicoid. Proof 9.2 We take the usual parametrisation
a(u, v)
=,,),(u) + v6(u)
(see Example 4.12)l where")' is a curve that meets each of the rulings and 6(u) is a vector parallel to the ruling through ,,),(u). We begin the proof by making some simplifications to the parametrisation. First, ~e can certainly assume that 116(u) 11= 1 for all values of u. We assume also that 6 is never zero, where the dot denotes dJdu. (We shall consider later what happens if 6(u) = 0 for some values of u.) We can then assume that 6 is a unit-speed curve (we do not assume that")' is unit-speed). These assumptions imply that 6.6 = 6.6 = O. Now we consider the curve
.y(u)
= ")'(u) -
("(.6)6(u).
If v = v + "(.6, the surface can be reparametrised using u and v, namely
a(u,v) = .y(u) + v6(u),
.y and
the parameters
212
Elementary Differential Geometry
but now we also have
1·6 = ("t - d~ ("1.6)6 - ("t. 6)6).6 =
0,
=
since 6.6 0 and 6.6 = 1. This means that we could have assumed that at the beginning) and we make this assumption from now on. We have au ..
Let A
= (EG -
E
= -y + v6)
=II -y + v6 11
2
,)
a 11
=6,
F
= (-y + v6).6 = -y.6)
G
-y.6 = 0
= 1.
F 2)1/2. Then) N
= A- 1 (-y + v6) x 6.
Next, we have
= .:y + v6, a uv = 6) a V11 = 0) L = A-I (.:y + v6). ((-y + v6) x 6)) M = A- 16.((-Y + v6) x 6) = A- 1 6.(-Y X 6))
aU'll.
N=O. Hence, the minimal surface condition H
= LG -
2M F
+N E
2A2
=0
gives
(.:y + v6).((-y + v6) x 6)
= 2(6.-y)(6.(-y x 6)).
This equation must hold for all values of (u, v). Equating coefficients of powers of v gives
.:y.(-y x 6) = 2(6.-y)(6.(-y
X
6)))
= 0, 6.(6 x 6) = O.
.:y.(6 x 6) + 6.(-y x 6)
(9) (10) (11)
Equation (11) shows that 6,6 and 6 are linearly dependent. Since 6 and 6 are perpendicular unit vectors, there are smooth functions a(u) and j3(u) such that
6=
a6 + 136.
B\lt! since 6 is unit-speed, 6.6 = O. Also) differentiating 6.6 -6.6 = -1. Hence, a = -1 and 13 = 0) so
6= -6.
= 0 gives 6.6 = (12)
9. Minimal Surfaces
213
Equation (12) shows that the curvatu~e of the curve 6 is 1, and that its principal normal is -6. Hence, its binormal is 6 x (-6), and since .. d . " -d (6 x 6) 6 x 6 + 6 x 6 -6 x 6 0, u _
=
=
=
it follows that the torsion of 6 is zero. Hence, 6 parametrises a circle of radius 1 (see Proposition 1.5). By applying a rigid motion, we can assume that 6 is the circle with radius 1 and centre the origin in the xy-plane, so that
= (cos u, sin u, 0). l.From Eq. (12), we get 6.("'" x 6) = -6.(...,. x 6) = 0, so by Eq. (10), 6(u)
-)'.(6 x 6)
= O.
It follows that -)' is parallel to the xy-plane, and hence that ",((u)
= (f(u), g(u), au ,. + b),
where f and 9 are smooth functions and a and b are constants. If a surface is part of the plane z = b. Otherwise, Eq. (9) gives
= 2 (1 cos u + 9 sin u) . We finally make use of the condition ...,..6 = 0, which gives 1sin u = iJ cos u. 9 cos u -
/ sin u
= 0, the (13)
(14)
Differentiating this gives
j sin u + i cosu = 9 cosu -; gsin u.
(15)
Equations (13) and (15) together give
1cos u + 9sin u = O. and using Eq. (14) we get j = 9 = O. Thus, f and 9 are constants. By a translation of the surface, we can assume that the constants f, 9 and b are zero, so that ",((u)
= (0, 0, au)
and a(u, v)
= (v cos u, vsin u, au),
which is a helicoid. We assumed at the beginning that 6 is never zero. If 6 is always zero, then 6 is a constant vector and" the surface is a generalised cylinder. But in fact a generalised cylinder is a minimal surface only if the cylinder is part of a plane (Exercise 9.8). The proof is now completed by an argument similar to that used
Elementary Differential Geometry
214
at the end of the proof of Proposition 9.1, which shows that the whole surface is either part of a plane or part of a helicoid. 0 After the catenoid and helicoid, the next minimal surfaces to be discovered were the following two. Example 9.4 E'imeper's minimal surface is a(u, v)
= (u - ~u3 + uv2 , V - ~v3 + vu" u 2 -
v
2
)
•
It was shown in Exercise 7.15 that this is a minimal surface.
I
Strictly speaking, this is not a surface patch in the Sense used in this book as it is not injective. The self-intersections are clearly visible in the picture above. However, if we restrict (u, v) to lie in sufficiently small open sets, a will be injective by the inverse function theorem. Example 9.5 Scherk's minimal surface is the surface with cartesian equation z
= In (COS Y) .
cos x It was shown in Exercise 7.16 that this is a minimal surface. Note that the surface exists only when cos x and cos y are both > 0 or both < 0, in other words in the interiors of the white squares of the following chess board pattern, in which the squares have vertices at the points (1r/2 + m1r, 1r/2 + n1r), where m and n are integers, no two squares with a common edge have the same colo'l!-r, and the square containing the origin is white:
9. Minimal Surfaces
215
The white squares have centres of the form (m7r l n7r)l where m and n are integers with m + n even. Since, for such m l n l
+ n7r) cos(x + m7r) cos(y
cosy
=--l cos x
it follows that the part of the surface over the square with centre (m7r l n7r) is obtained from the part over the square with centre (OlO) by the translation (Xl Yl Z).I-+ (x + m7r, Y + n7r, z). So it suffices to eX~bit the part of the surface over a smgle s q u a r e : -
216
Elementary Differential Geometry
EXERCISES 9.6
Show that the helicoid is a minimal surface.
9.7
Show that the surfaces at in the isometric deformation of the helicoid into th'e catenoid given in Exercise 5.8 are minimal surfaces. I
9.8 Show that a generalised cylinder is a minimal surface only when the cylinder is part of a plane. 9.9
A translation surface is a surface of the form z
= f(x) + g(y),
where f and 9 are smooth functions. (It is obtained by moving the curve U t-+ (u, 0, f(u)) parallel to itself along the curve v t-+ (0, v, g( v) ).) Use Exercise 9.2 to show that this is a minimal surface if and only if tP f jdx2 tPgjdy2 -_""--:-~"'::'"
1 + (df jdX)2
=- 1 + (dgjdy)2' --..:.........:-~~
Deduce that any minimal translation surface is part of a plane or can be transformed into part of Scherk's surface in Example 9.5 by a translation and a dilation (x, y, z) t-+ a(x, y, z) for some non-zero constant a. 9.10 Verify that Catalan's surface a( u, v) = (u - sin u cosh v, 1 - cos u cosh v, - 4 sin ~ sinh
¥)
is a conformally parametrised minimal surface. (As in the case of Enneper's surface, Catalan's surface has self-intersections, so it is only a surface if we restrict (u, v) to sufficiently small open sets.)
Show that (i) the parameter curve on the surface given by u ::. 0 is a straight line;
217
9. Minimal Surfaces
(ii) the parameter curve u (iii) the parameter curve v
is a parabola; = 0 is a cycloid (see Exercise 1.7). Show also that each of these curves, when suitably parametrised, is a geodesic on Catalan's surface. =
1r
9.3. Gauss Map of a Minimal Surface Recall from Section 7.3 that the Gauss map of a surface patch a : U -+ R 3 associates to a point a(u, v) of the surface the standard unit normal N(u, v) regarded as a point of the unit sphere S2. By Eq. (15) in Chapter 7,
where K is the gaussian curvature of a, so N will be regular provided K is nowhere zero, and we assume this for the remainder of this section.
Proposition 9.3 Let a( u, v) be a minimal surface patch with nowhere vanishing gaussian cur· vature. Then, the Gauss map is a conformal map from a to part of the unit sphere.
We should really be a little more careful in the statement of this proposition, since for us conformal maps are always diffeomorphisms (see Section 5.3). However, even if N u x N v is never zero, it does not follow that the map a(u, v) 1-4 N(u, v) is injective (see Exercise 9.12(ii)). Nevertheless, the inverse function theorem tells us that, if (uo, vo) E R 2 is a point where a (and hence N) is defined, there is an open set U containing (uo, va) on which a is defined and on which N is injective. Then, N : U --t S2 is an allowable surface patch on the unit sphere S2, and the Gauss map is a diffeomorphism from a(U) to N(U). Proof 9.3
By Theorem 5.2, we have to show that the first fundamental form (N u .Nu )du 2
+ 2(Nu .Nv )dudv + (N v .Nv )dv 2
of N is proportional to that of a. Form the symmetric 2 x 2 matrix Nu.Nu :F111= ( Nu.N v
Nu.Nv) Nv.N v
218
Elementary DifFerential Geometry
in the same way as we associated symmetric 2 x 2 matrices :Fl and Fl I to the first and second fundamental forms of a in Section 6.3. Then, we have to show that (16) 'I
for some scalar A. By Proposition 6.4, I
= a2 au.a u + 2abau .av + b2 a'l).av = a2 E + 2abF + b2 0, ~) = -Wand W =:FIl:FII is the Weingarten matrix. Computing
Nu.N u
where (:
Nu.N v and Nv.N v in the same way gives F
_ ( HI -
+ 2abF +';0 acE + (ad + bc)F + bdG 2
a E
acE + (ad + be)F + bdO) c2 E + 2cdF + ~G
= (: :)(; ~)(: ~) =(-W)t:Fl(-W) = (-:F1 1 :FI1)t:Fl( -:Fi 1:F11) =:F11T]1 :FIT] I :F11 =:F11:F1 1 :FH. Hence, Eq. (16) is equivalent to :F1 :FI1:Fi 1:F11 = AI, 1
i.e.
= AI.
W2
But,
W2
= (a
b
dC )
2
2 _
(
a + be b(a + d)
c(a + d)) ~ + be .
Now, recall from Section 6.3 that the principal curvatures Kl and K2 are the eigenvalues of W. Since the sum of the eigenvalues of a matrix is equal to the sum of its diagonal entries,
= -(a+d). If a is minimal, the mean curvature H = t (K} + K2) vanishes, so a + d :::::: 0 and K1 +K2
hence
as we want.
o
219
9. Minimal Surfaces
We saw in Exercise 5.14 that a conformal parametrisation of the plane is necessarily holomorphic or anti-holomorphic, so this proposition strongly sug· gests a connection between minimal surfaces and holomorphic functions. This connection turns out to be very extensive, and we shall give an introduction to it in the next section.
EXERCISES 9.11 Show that the scalar A appearing in the proof of Proposition 9.3 is equal to -K, where K is the gaussian curvature of the surface. 9.12 Show that (i) the Gauss map of the catenoid is injective and its image is the whole of the unit fiphere except for the north and south poles; (ii) the image of the Gauss map of the helicoid is the same as that of the catenoid, but that infinitely many points on the helicoid are sent by the Gauss map to any given point in its image.
9.4. Minimal Surfaces and Holomorphic Functions In this section, we shall make use of certain elementary properties of holomor· phic functions. Readers without the necessary background in complex analysis may safely omit this section, whose results are not used anywhere else in the book. We shall need to make use of special surface patches on a minimal surface. Recall from Section 5.3 that a surface patch a : U --t R 3 is called conformal if its first fundamental form is equal to E(du 2 + dv 2 ) for some positive smooth function E on U.
Proposition 9.4 Every surface has an atlas consisting of conformal surface patches.
We shall accept this result without proof (the proof is non-trivial). Let a : tJ --t R 3 be a conformal surface patch. We introduce complex coordinates in the plane in which U lies by setting ( = u
+ iv
for (u, v)
E U,
220
Elementary Differential Geometry
and we define
(17) Thus, tp = ('PI, 'P2, 'P3) has three components, each of which is a complex·valued function of (u, V),I i.e. of (. The basic result which establishes the connection between minimal surfaces and holomorphic functions is
Proposition 9.5 Let a : U -+ R 3 be a conformal surface patch. Then a is minimal if and only if the function tp defined in Eq. (17) is holomorphic on U.
Saying that tp is holomorphic means that each of its components 'PI, 'P2 and 'P3 is holomorphic. Proof 9.5
Let 'P(u, v) be a complex·valued smooth function, and let a and j3 be its real and imaginary parts, so that 'P = a + ij3. The Cauchy-Riemann equations au
= j3v
and
a v = -j3u
are the necessary and sufficient conditions for 'P to be holomorphic. Applying this to each of the components of tp, we see that tp is holomorphic if and only if
(au)u = (-av)v
and
(au)v = -( -av)u.
The second equation imposes no condition on a, and the first is equivalent to a uu +a vv = O. SO we have to show that a is minimal if and only if the laplacian L1a = a uu + a vv is zero. By Proposition 7.2(ii) and the fact that a is conformal, the mean curvature of a is given by
H=L+N 2E
I
+ N = 0, Le. (a uu + avv).N = 0.
so a is minimal if and only if L
(18)
Obviously, then, a is minimal if L1a = O. For the converse, we have ~o show that L1a = 0 if Eq. (18) holds. It is enough to prove that L1a.a u = L1a.av = 0, since {au, a v , N} is a basis of R 3 •
9. Minimal Surfaces
221
We compute .d.O'·O'u
= O'uu.O'u + O'vv.O'u =
1
"2 (O'u.O'u)u + (O'v'O'u)v
=1
'2(O'u.O'u -
But, since 0' is conformal) Similar ly, .d.O'.O'v = O.
O'U'O'u
O'v.O'v)u
= O'V'O'v
and
-
(O'v.O'uv)
+ (O'tI.O'u)v' O'u.O'v
= O.
Hence,
.d.O'.O'u
= O. 0
The holomorphic function tp associated to a minimal surface trary, however:
0'
is not arbi·
Theorem 9.2 If 0' : U --+ R 3 is ti conformally parametrised minimal surface, the vector-valued holomorphic function tp ( 1 at (0 E U, and f has a zero of order n > 2m at (0, then the Laurent expansions of f and 9 about (0 are of the form
f (()
= a (( -
(0
t
+ ...
b
and
9 (()
= (( _ (0) m +"',
where a and b are non·zero complex numbers and the··· indicates terms in· volving higher powers of ( - (0. Then, f(l ± g2) ±ab2(( - (ot- 2m + ... and fg ab(( - (ot- m + ...
=
=
involve only non·negative powers of ( - (0, so tp is holomorphic near (0. Since it is clear that tp is holomorphic wherever 9 is holomorphic, it follows that the function tp defined by Eq. (21) is holomorphic everywhere 0t.'l u. It is clear that tp is identically zero only if f is identically zero, and simple algebra shows that tp satisfies condition (i) in Theorem 9.2. Conversely, suppose that tp = ('PI, 'P2, 'P3) is a holomorphic function satis· fying conditions (i) and (ii) in Theorem 9.2. Define
f
= 'Pl - i'P2,
9
= 'Pl 'P3 . - 1,'P2
(22)
This definition makes sense, for if 'PI and 'P2 were both zero, then 'P3 would be zero by condition (i), and this would violate condition (ii). Since tp is holomorphic, f is holomorphic and 9 is meromorphic. Condition (i) implies that ('PI + i'P2)( 'PI - i'P2) = -'P~, and hence that 'PI
+ i'P2 =- fl·
(23)
225
9, Minimal Surfaces
Simple algebra shows that Eqs. (22) and (23) imply Eq. (21). Finally, Eq. (23) implies that f g2 is holomorphic, and the argument with Laurent expansions in the first part of the proof now gives the condition on the zeros and poles of f and g. 0 We give only one application of Weierstrass's representation.
Proposition 9.7 The gaussian curvature of the minimal surface corresponding to the functions f and 9 in Weierstrass's representation is 2 - 16 Idgjd(1 K = lfl2 (1
+ IgI2)4'
Proof 9.7 This is a straightforward, if tedious, computation, and We shall omit many of the details. Define tp by taking complex·conjugates of each component of tp. Then, tp+tp ~-tp au = 2 ' a v = 2i . Hence, the coefficients of the first fundamental form are given by E
= G = ~ II tp + tp W= ~tp.tp,
F
=0,
since tp.tp = tp.tp = O. Substituting the formula for tp into Eq. (14) and simpli· fying, we find that the first fundamental form is 1
4lfl2 (1 + Ig1 2)2 (du 2 + dv 2).
(24)
Next,
= ~i (cp + tp) x (tp - cp) = ;iCP x 'ip, 2 II au x a1J 11 = -~(CP X cp).(cp x tp), = _~((CP.CP)(tp.tp) _ (cp.tp)2), au x a v
..
= 41 (cp.cp-)2 , .
.. In terms of
f and N
g,
N=i~xCP. cp.cp
this becomes
= 1 +11g12 (g + g, -i(g -
g), Igl 2 -1) .
(25)
clementary UlHerentlal ueometry
...... v
Using the remarks preceding the proof of Theorem 9.2 and the formulas
(which follow by differentiating au.N = av.N = 0), we find that the second fundamental form is I
-~ ((Vg' + fg')(du 2 + dv 2 ) + 2i(fg' -
fg')dudv).
(26)
Combining Eqs. (24), (25) and (26), and using the formula for the gaussian curvature K in Proposition 7.1(i), we finally obtain the formula in the statement of the proposition. 0
Corollary 9.2 Let 5 be a minimal surface that is not part of a plane. Then, the zeros of the gaussian curvature of 5 are isolated. This means that, if the gaussian curvature K vanishes at a point P of 5, then K does not vanish at any other point of 5 sufficiently near to P. More precisely, if P lies in a surface patch a of 5, say P = a(uo, vo), there is a number € > 0 such that K does not vanish at the point a(u, v) of 5 if
0
.. : a(u, v)
1-4
a(u, v
+ A).
This proves the proposition (the isometry Ry oS>.. arises because R y = R z oR z .)
o EXERCISES 10.5 Show that there is no isometry between any region of a sphere and any region of a (generalised) cylinder or a (generalised) cone. (Use Proposition 10.1 and Exercise 5.7.) 10.6 Show that the gaussian curvature of the Mobius band in Example 4.9 is equal to -1/4 everywhere along its median circle. Deduce that this Mobius band cannot be constructed by taking a strip of
240
Elementary Differential Geometry
paper and joining the ends together with a half-twist. (The analytic description of the 'cut and paste' Mobius band is more complicated than the version in Example 4.9.) 10.7 Consider the surf~ce patches
O'(u, v)
= (u cos v, u sin v, In u),
&(u, v) = (u cos v, u sin v, v).
Prove that the gaussian curvature of 0' at O'(u, v) is the same as that of iT at &(u,v), but that the map from 0' to & which takes O'(u,v) to &(u, v) is not an isometry. Prove that, in fact, there is no isometry from 0' to &. 10.8 Show that the only isometries from the catenoid to itself are products of rotations around its axis, reflections in planes containing the axis, and reflection in the plane containing the waist of the catenoid.
10.3. The Codazzi-Mainardi Equations Gaussls Theorema Egregium shows that the coefficients of the first and second fundamental forms of a surface cannot be arbitrary smooth functions, for it shows that LN - M 2 can be expressed in terms of E, F and G. It is natural to ask if there are any further relations between these coefficients. In this section, we find that there are indeed some additional relations, and we show that, in a sense we shall explain, there are no others. We begin with a computation similar to that in Lemma 10.1.
Proposition 10.3 (Gauss Equations) Let O'(u, v) be a surface patch. Then, O'uu O'uv
= rflO'u + rflO'v + LN, = rf2O'u + rf2O'v + MN,
0'1.11.1
= ri2O'u + ri2O'v
+ NN,
where
rl 11
r. l
_ GEu - 2FFu + FE v r 2 11 2(EG - F2) 2 1 GEv -FGu l2 12 r = 2(EG _ F2)' r _
22 -
2GFv - GGu - FG u 2(EG - F2) ,
r.2 22
2EFu - EE v - FEu 2(EG - F2) , EG u - FEu = 2(EG - F2) , _ EG v - 2FFv + FG u 2(EG - F2) _
10. Gauss's Theorema Egregium
241
r coefficients in these formulas are called Christoffel symbols.
The six Proof 10.9
Since {u u, U v, N} is a basis of R 3 , scalar functions
ai, '.' • ,1'3
satisfying
= alu u + a2Uv + a3 N , Uuv = 131uu + 132 u v + 133 N , Uvv = 1'lUu + 1'2 Uv + 1'3 N ,
Uuu
(15)
'. certainly exist. Taking the dot product of each equation with N gives a3
= L,
133
= M,
1'3
= N.
Now we take the dot product of each equation in (15) with Uu and Uv - This gives six scalar equations from which we determine the remaining six coefficients. For example, taking the dot product of the first equation in (15) with Uu and Uv gives the two equations 1
Eal + Fa 2 =
Uuu·Uu
= 2Eu, 1
~
Fal
+ Ga 2 = Uuu'UV = (uu·uv)u
Solving these equations gives coefficients in Eqs. (15).
al
= rtl' a2
- Uu·Uuv =. Fu - 2Ev.
rfl j
:::
similarly for the other four 0
The new relations between the coefficients of the first and second fundamental forms of a surface patch are contained in the following result.
Proposition 10.4 (Codazzi-M aina rd i Eq uations) Define the Christoffel symbols of a surface patch u(u, v) as above. Then,
Lv - M u = Lrl2 + M(rr2 - rll) - N rfl' M v - N u = Lri2
+ M(ri2
- rl2) - N rf2'
Proof 10·4
We write down the equation (uuu)v Uuu and Uuv: (rl\uu
(
8r1l
av
_
= (uuv)u,
using the Gauss equations for
+ ri\uv+LN)v = (rl2u u + rf2Uv + MN)u,
8rl 2 )
au
= r12u uu
uu +
(8r avt1
+ (r122 -
_
8r auf2 )
uv
+
(L - M )N v
u
r1\)uuv - r121UVV - LNv + MN u
= rl2(rlluU+rrlUV + LN) + (rf2 - rfl)(rl12u u + rf2Uv + MN) - r 121 (ri2Uu + ri2UV + NN) - LNv + MN u, (16)
242
Elementary Differential Geometry
using the Ga'uss equations again. Now, N u and N v are perpendicular to N, and so are linear combinations of au and avo Hence, equating N components on both sides of the last equation gives 1
Lv - M u ="'LTl 2
+ M(Ti2
- TA) - NT;l'
which is the first of the Codazzi-Mainardi equations. The other equation follows in a similar way from (auv)v = (avv)u. 0 At first sight, it seems that we could get four other identities like those in Theorem 10.2 by equating the coefficients of au and a v in Eq. (16) and in its analogue coming from (auv)v = (avv)u. It turns out, however, that these identites are all equivalent to the formula in Corollary 10.1 (and so, in partie· ular, they give another proof of the Theorema Egregium). In fact, there are no further identites to be discovered, as the following theorem shows. Theorem
10.3
Let a : U --t R 3 and jj : U --t R 3 be surface patches with the same first and second fundamental forms. Then, there is a rigid motion M of R 3 such that
i1=Moa. Moreover, let V be an open subset ofR 3 and let E,F,G,L,M and N be smooth functions on V. Assume that E > 0, G > 0, EG - F 2 > 0 and that the equations in Corollary 10.1 and Proposition 10.4 hold, with K = 1:~:'¥; and the Christoffel symbols defined as in Proposition 10.9. Then, if (uo, va) E V, there is an open set U contained in V and containing (uo, va), and a surface patch a : U --t R 3 , such that E du 2+ 2F dudv+ Gdv 2 and Ldu2+ 2M dudv +N dv 2 are the first and second fundamental forms of a, respectively.
This theorem is the analogue for surfaces of Theorem 2.3, which shows that unit·speed plane curves are determined up to a rigid motion by their signed curvature. We shall not prove Theorem 10.3 here. The first part depends on uniqueness theorems for the solution of systems of ordinary differential equa· tions, and is not particularly difficult. The second part is more sophisticated and depends on existence theorems for the solution of certain partial differential equations. The following example will illustrate what is involved. Example 10.2
Consider the first and second fundamental forms du 2 + dif and -du2 , respec· tively. Let us first see whether a surface patch with these first and second fundamental forms exists. Since all the coefficients of these forms are constant,
243
10. Gauss's Theorema Egregium
all the Christoffel symbols are zero and the Codazzi-Mainardi equations are obviously satisfied. The formula in Corollary 10.1 gives K = 0, so the only other condition to be checked. is LN - M2 = 0, and this clearly holds since M = N :::::: o. Theorem 10.3 therefore tells uS that a surface patch with the given first and second fundamental forms exists. To find it, we note that the Gauss equations give D'uu ::::::
-N,
The last two equations tell us that
D'uv D'v
= 0,
D'vv
=
o.
is a constant vector, say a, so
D'(u, v) :::::: b(u)
+ av,
(17)
where b is a function of u only. The first equation then gives N = -b" (a dash denoting d/dv). We now need to use the expressions for N u and N v in terms of D'u and D'v in Proposition 6.4. The Weingarten matrix is
W
= :Fi':FII = (~
0)-1(-10)=(-1 1 0 0 0
0) 0 '
so Proposition 6.4. gives Nu
= D'u,
Nv =
o.
The second equation tells us nothing-new, since we already knew that N depends only on u. The first equation gives bin
= -b"
+ b ' = O.
Hence, b" + b is a constant vector, which we can take to be zero by applying a translation to D' (see Eq. (17)). Then, b (u) :::::: C cos U
+ d sin u,
where c and d are constant vectors, and N :::::: - b" :::::: b. This must be a unit vector for all values of u. It is easy to see that this is possible only if c and d are perpendicular unit vectors, in which case we can arrange that c :::::: (1,0,0) and d :::::: (0,1,0) by applying a rigid motion, giving b(u) :::::: (cosu,sin u, 0). Finally, D'u x D'u :::::: AN for some non·~ero scalar A, so b l x a:::::: Ab. This forces a:= (0,0, A), and the patch is given by D'(u,v)
= (cosu,sinu,Av),
a parametrisation of a circular cylinder of radius 1 (which the reader had prob· ably guessed some time ago).
Elementary Differential Geometry
244
EXERCISES 10.9
A surface patch has first and second fundamental forms cos2 v di.i 2 , + dv 2
and
- co~ v du 2
-
dv 2 ,
respectively. Show that the surface is part of a sphere of radius one. (Compute the Weingarten matrix.) Write down a parametrisation of the unit sphere with these first and second fundamental forms. 10.10 Show that there is no surface patch whose first and second funda· mental forms are du2
+ cos 2 U dv 2
and
cos2 u du2
+ dv 2 ,
respectively. 10.11 Suppose that the first and second fundamental forms of a surface patch are Edu2 + Gdv2 and Ldu2 + Ndv 2 , respectively (d. Propo· sition 7.2). Show that the Codazzi-Mainardi equations reduce to
Lv=~E.(~+~), Nu=~Gu(~+~). Deduce that the principal curvatures satisfy the equations (Kl)V
= : ; (K2 - KI),
Kl ::::::
(K2)u ::::::
L/ E and
~G (Kl -
K2 -
N /G
K2).
10.4. Compact Surfaces of Constant Gaussian
Curvature We conclude this chapter with a beautiful theorem that is the analogue for surfaces of the characterisation given in Example 2.2 of circles as the plane curves with constant curvature.
Theorem 10.4 Every compact surface whose gaussian curvature is constant is a sphere.
Note that, by Proposition 7.6, the value of the constant gaussian curvature in this theorem must be > o. The proof of this theorem depends on the following lemma.
245
10. Gauss's Theorema Egregium
Lemma 10.2 Let a : U --t R 3 be a surface patch containing a point P '= a(uo l vo) that is not an umbilic. Let Kl > K2 be the principal curvatures of a and suppose that Kl has a local maximum at P and K2 has a local minimum there. Then, the gaussian curvature of a at P is < O. Proof 10.2
Since P is not an umbUic, Kl > K2 at P, so by shrinking U if necessary, we may assume that Kl > K2 everywhere. By Proposition 7.2, we can assume that the first and second funqamental forms of a are E du 2
+ Gdv2
and
Ldu2
+ N dv 2 ,
respectively. By Exercise 10.8, Ev
= - Kl 2E (Kl)v, - K2
Gu
= Kl 2G (K2)u, - K2
and by Corollary 10.2(ii), the gaussian Curvature K
=-
2~ (:u (Jia) + :v (!ia)) .
Since P is a stationary point of Kl and hence E v = Gu = 0, at P. Hence, at P, 1
K = -2EG(G uu
1 + E vv ) = -2EG
(
K2,
we have
2G Kl - K2
(Kl)V
(K2)uu -
= (K2)tt
= 0, and
2E) Kl - K2
(KI)vv
(again dropping terms involving E v , Gu and the first derivatives of Kl and K2). Since Kl has a local maximum at P, (KI)vv < 0 there, and since K2 has a local minimum at P, (K2)uu > 0 there. Hence, the last equation shows that K < 0 at P. 0 Proof 10·4
a
The proof of Theorem 10.4 will use little point set topology. We consider the continuous function on the surface S given by J = (Kl _1\.2)2, where Kl and K2 are the principal curvatures. Note that this function is well defined even though Kl and K2 are not, partly because we do not know which principal curvature is to be called Kl and which 1\.2, and partly because the sign of the principal curvatures depends on the choice of parametrisation of S. We shall prove that this function is identically zero on S, so that every point of S is an umbilic. Since the gaussian curvature K > 0, it follows from Proposition 6.5 that S is part of a sphere, say S. In fact, S must be the whole of S. For, any point P of
246
Elementary Differential Geometry
5 is contained in a patch a: U --t R 3 of 5, and a(U) = 5 n W, where W is an open subset of R 3 ; it follows that 5 is an open subset of S. On the other hand, since 5 is compact, it is necessarily a closed subset of R 3 , and hence a closed subset of S. But since S is connected, the only non-empty subset of S that is both open and closed is S its~lf. Suppose then, to get a contradiction, that J is not identically zero on 5. Since 5 is compact, J must attain its maximum value at some point P of 5, and this maximum value is > O. Choose a patch a: U --t R 3 of 5 containing P, and let Itl and K2 be its principal curvatures. Since KIK2 > 0, by reparametrising if necessary, we can assume that Kl and K2 are both> 0 (see Exercise 6.17). Suppose that Kl > K2 at Pi then by shrinking U if necessary, we can assume that Kl > K2 everywhere on U. Since K is a constant> 0, the function (x _ ~) 2 increases with x provided that x > K/x > O. Since Kl > K/ Kl = K2 > 0, this function is increasing at x = Kl' so Kl must have a local maximum at P, and then K2 = K/Kl must have a local minimum there. By Lemma 10.2, K < 0 at P. This contradicts the assumption that K > O. 0
EXERCISES 10.12 Show that a compact surface with gaussian curvature> 0 everywhere and constant mean curvature is a sphere. (As in the proof of Theorem 10.4, if Kl has a local maximum at a point P of the surface, then K2 = 2H - Kl has a local minimum there.)
The Gauss-Bonnet Theorem
The Gauss-Bonnet theorem is the most beautiful and profound result in the theory of surfaces. Its most important version relates the average over a surface of its gaussian curvature to a property of the surface called its 'Euler number' which is 'topological', Le. it is unchanged by any continuous deformation of the surface. Such deformations will in general change the value of the gaussian curvature, but the theorem says that its average over the surface does not change. The real importance of the Gauss-Bonnet theorem is as a prototype of analogous results which apply in higher dimensional situations, and which relate geometrical properties to topological ones. The study of such relations is one of the most important themes of 20th century Mathematics.
11.1. Gauss-Bonnet for Simple Closed Curves The simplest version of the Gauss-Bonnet Theorem involves simple closed curves on a surface. In the special case when the surface is a plane, these curves have been discussed in Section 3.1. For a general surface, we make
Definition 11.1 A curve "Y(t) = a(u(t), v(t)) on a surface patch a : U --t R 3 is called a simple closed curve with period a if 1r(t) = (u(t), v(t)) is a simple closed curve in R 2 with period a such that the region int(1r) of R 2 enclosed by 1r is entirely 247
Elementary Differential Geometry
248
contained in U (see the diagrams below). The curve "y is said to be positivelyoriented if 1r is positively-oriented. Finally, the image of int(1r) under the map a is defined to be the interior int("Y) of "Y.
not allowed
allowed
We can now state the first version of the Gauss-Bonnet Theorem.
Theorem 11.1 Let "Y(s) be a unit-speed simple closed curve on a surface a of length f("Y)} and assume that "Y is positively-oriented. Then} 1
1
("Y) Kgds
o
= 271'" -
J.l.
KdAal
int("Y)
where Kg is the geodesic curvature of "Yl K is the gaussian curvature of a and dAa = (EG - F 2 )1/2dudv is the area element on a (see Section 5.4).
We use s to denote the parameter of"Y to emphasize that "Y is unit-speed. Proof 11.1
As in the proof of Theorem 10.1 1choose a smooth orthonormal basis {e' Ie"} of the tangent plane of a at each point such that {e' I elf I N} is a right-handed orthonormal basis of R 3 I where N is the unit normal to a. Consider the following
249
11. The Gauss-Bonnet Theorem
integral: f"("()
I =
J
e'.e" ds
o
fl("()
=
Jo
=
L(e'.e~)du + (e'.e~)dv.
e'.(e~u
+ e~iJ)ds
By Green's theorem (see Section 3.1), this can be rewritten as a double integral: / I =
i' f i' f
Jint('1r)
=
Jint('1r)
=
i
f
Ji t('1r) n
i' =i' f
{(e'.e~)u - (e/.e~)v}dudv {(e~.e~) - (e~.e~)}dudv LN-M2 (EG _ F2)1/2 dudv
(by Lemma 10.1)
2
= f Jint('1r)
LN - M (EG - F 2 )1/2dudv EG - F2
KdAa.
J int('1r)
(1) .
Now let B(8) be the angle between the unit tangent vector Ar of"( at ,,((s) and the unit vector e' at the same point. More precisely, B is the angle, uniquely determined up to a multiple of 271'", such that
'Y =
+ sin Be" .
cos Be'
(2)
Then, N x
'Y : : : -
sin Be' N
,
I
u
,...
...
+ cos Be" .
(3)
250
Elementary Differential Geometry
Now, by Eq. (2L
.:y = cos Be'
+ sin Be" + B(-
sin Be'
+ cos Be"),
(4)
so by Eqs. (3) and (4) the geodesic curvature of"Y is Kg
= (N x
t).;y (see Section 6.2)
sin Be' + cos Be"). (- sin Be'
=
B(-
=
iJ + cos 2 B(e'.e") - sin 2 B(e".e') + sin BcosB(e".e" - e'.e')
+ cos Be") + (- sin Be' + cos Be"). (cos Be' + sin Be") (by Eqs. (2) and (3)).
Since e' and elf are perpendicular unit vectors,
e'.e' = e".e"
= 0,
e'.e"
= B· -
, ." , e.e
=
-e'.e".
Hence, Kg
and by the definition of I,
{"("Y)
I =
10
(iJ -
Kg
)ds.
Thus, to complete the proof of Theorem 11.1, we must show that
{"("Y)
1
Bds = 27r.
(5)
0
Equation (5) is called 'Hopf's Umlaufsatz' - literally 'rotation theorem' in German. We cannot give a fully satisfactory proof of it here because the proof would take us too far into the realm of topology. Instead, we shall justify Eq. (5) by means of the following heuristic argument. The main observation is that, if.y is any other simple closed curve contained in the interior of "Y, there is a smooth family of simple closed Curves "Y T , defined for 0 < T < 1, say, with "Yo = "Y and "Y 1 =.y (see Section 8.4 for the notion of a smooth family of curves). The existence of such a family is supposed to be 'intuitively obvious'.
251
11. The Gauss-Bonnet Theorem
Note, however, that it is crucial that the interior of 1(" is entirely contained in U, otherwise such a family will not exist, in general:
Observe next that the integral fOl('Y~) iJds should depend continuously on T. FUrther, since 'Y'" and e' return to their original values as one goes once round 'Y"', the integral is always an integer multiple of 271'". These two facts imply that the integral must be independent of T - for by the Intermediate Value Theorem a continuous variable cannot change from one integer to a different integer without passing through some non-integer value. To compute fol('Y) iJ ds, we can therefore replace 'Y by any other simple closed. curve 'Y in the interior of 'Y, since this will not change the value of the integral. We take .:y to be the image under a of a small circle in the interior of 1r. It is 'intutively clear' that
(l(i)
1
iJ ds
= 271'",
0
because (i) e' is essentially constant at all points of.:y (because the circle is very small), and
(ii) the tangent vector to .:y rotates by 271'" on going once round .:y because the interior of.:y can be considered to be essentially part of a plane, and it is 'intutively clear' that the tangent vector of a simple closed curve in the plane rotates by 271'" on going once round the curve. This completes the 'proof' of Hopf's Umlaufsatz, and hence that of Theorem 11.1.
Elementary Differential Geometry
252
EXERCISES 11.1 A surface patch u has gaussian curvature < 0 everywhere. Prove that there are no simpl~closed geodesics in u. How do you reconcile this with the fact that the parallels of a circular cylinder are geodesics? 11.2 Let "Y(s) be a simple closed curve in R Z , parametrised by arc-length and of total length l("Y). Deduce from HopPs Umlaufsatz that, if Ks(S) is the signed curvature of "Y, then
("("Y)
1
0
K8 (S) ds = 271'".
(Use Proposition 2.2.)
11.2. Gauss-Bonnet for Curvilinear Polygons For the next version of Gauss-Bonnet, we shall have to generalise our notion of a curve by allowing the possibility of 'corners'. More precisely, we make the following definition.
Definition 11.2 A curvilinear polygon in R Z is a continuous map 1r : R --+ R Z such that, for some real number a and some points 0 = to < h < ... < t n = a, (i) 1r(t) = 1r(t') if and only if t' - t is an integer multiple of a; (ii) 1r is smooth on each of the open intervals (to, tz), (til tz), ... , (tn-I, t n ); (iii) the one-sided derivatives (6)
exist for i = 1, ... ,n and are non-zerO and not parallel. The points "Y(ti) for i = 1, ... , n are called the vertices of the curvilinear polygon 1r, and the segments of it corresponding to the open intervals (ti-l, t i ) are called its edges. It makes sense to say that a curvilinear polygon 'R' is positively-oriented: for all t such that 1r(t) is not a vertex, the vector n s obtained by rotating ir anticlockwise by 71'"/2 should point into int(1r). (The region int(1r) enclosed by 1r makes sense because the Jordan Curve Theorem applies to curvilinear polygons in the plane.)
253
lL The Gauss-Bonnet Theorem
Now let a : U --+ R 3 be a surface patch and let 1r : R --+ U be a curvilinear polygon in U, as in Definition 11.2. Then, 'Y = a 0 1r is called a curvilinear polygon on the'surface patch a, int('Y) is the image under a of int(1r), the vertices of 'Y are the points 'Y(ti) for i = 1, ... , n, and the edges of a are the segments of it corresponding to the open intervals (ti-l, t i ). Since a is allowable, the one-sided derivatives -1' 'Y(t) - 'Y(ti) , 'Y. -(t.) I 1m ttt. t - ti
exist and are not parallel. Let be the angles between ...,.± (ti) and e', defined as in Eq. (2), let 6i = 0i be the external angle at the vertex 'Y(td, and let ai = 1r - 6i be the internal angle. Since the tangent vectors ""'+(ti) and ...,.- (ti) are not parallel, the angle 6i is not a multiple of 1r. Note that all of these angles are well defined only up to multiples of 21r. We assume from now on that 0 < ai < 21r for i = 1, .. . ,n. A curvilinear polygon 'Y is said to be unit-speed if II . .,. II = 1 whenever...,. is defined, i.e. for all t such that 'Y(t) is not a vertex of 'Y. We denote the parameter of 'Y by s if 'Y is unit-speed. The period of 'Y is then equal to its length l('Y), which is the sum of the lengths of the edges of 'Y.
ot
ot -
Theorem 11.2 Let'Y be a positively-oriented unit-speed curvilinear polygon with n edges on a surface a, and let aI, a2, ... ,an be th e i nterior angles at its vertices. The n,
l
l
o
('Y) Kgds =
Ln i= 1
ai -
(n - 2)1r -
Jl
int('Y)
KdAa-
Elementary Differential Geometry
254
Proof 11.2
Exactly the same argument as in the proof of Theorem 11.1 shows that 1 (")')
1
1 1
"'gds. .,....."
o
(")')
8ds -
J1
0
KdAa.
int("Y)
We shall prove that (l(")')
10
8ds
o
= 27r -
1
L ai.
(7)
i=l
Assuming this, we get 1 (")')
n
Kgds = 27r -
o
J1
n Lai -
KdAa
int(")')
i=l
n
= 27r - L(7r - Oi) -
J1. J1.
KdAa
mt(")')
i=l
n
= LOi - (n -
2)7r -
KdAa·
Int(")')
i=l
To establish Eq. (7), we imagine 'smoothing' each vertex of")' as shown in the following diagram.
If the 'smoothed' curve
.y is smooth
(!), then, in an obvious notation,
(l(~) ~
1
Ods
0
Since")' and
= 27r.
.y are the same except near the
1
1(~)
o
:.
Ods -
1
(8)
vertices of ")', the difference
1(")') .
Ods
(9)
0
is a sum of n contributions, one from near each vertex. Near ")'( Si), the picture is
11.
255
The Gauss-Bonnet Theorem
Le. "( and .:y agree except when s belongs to a small interval taining Si, so the contribution from the ith vertex is II
(s~, s~'),
say, con-
II
J.~i Ods - J.~i iJds - J.~i iJds. Si
s,
S,
The first integr3J. is the angle between si) and 4( si), which as s~ and s~' tend to Si becomes the angle between ""'+(Si) and ""'-(Si), Le. 6i. On the other hand, since "((s) is smooth on each of the intervals (si, Si) and (Si' si'), the last two integrals go to zero as si and si' tend to Si. Thus, the contribution to the expression (9) from the ith vertex tends to 6i as si and s~' tend to Si . Summing over all the vertices, we get
.y(
1(,),) .
1 D
Ods -
1
1 ("()
n
iJds =
0
L6
i.
i:;;;;l
o
Equation (7) now follows from this and Eq. (8).
Corollary 11.1 If"( is a curvilinear polygon with n edges each of which is an arc of a geodesic, then the internal angles aI, az, ... ,an of the polygon satisfy the equation n
Lai = (n - 2)7r + i=l
J1
KdAa.
mt("()
Proof 11.1
This is immediate from Theorem 11.2, since
Kg
= 0 along a geodesic.
0
As a special case of Corollary 11.1, consider an n-gon in the plane with
Elementary Differential Geometry
256
straight edges. Since K = 0 for the plane, Corollary 11.1 gives n
L
D:i
= (n - 2)'71'",
i=l ,
a well known result of element~ry geometry.
For a curvilinear n-gon on the unit sphere whose sides are arcs of great circles, we have K = 1 so 2: D:i exceeds the plane value (n - 2)'71'" by the area II dAD- of the polygon. Taking n = 3, we get for a spherical triangle ABC whose edges are arcs of great circles,
A(ABC) = LA + LB + LC -
1r.
This is just Theorem 5.5, which is therefore a special case of Gauss-Bonnet.
U. The Gauss-Bonnet Theorem
257
Finally, for a geodesic n-gon on the pseudosphere (see Section 7.2), for which K = -1, we see that 2: D:i is less than (n - 2)71'" by the area of the polygon. In particular, for a geodesic triangle ABC on the pseudosphere,
A(ABC)
= 71'" -
LA - LB - LC.
EXERCISES 11.3 Suppose that the gaussian curvature K of a surface patch a satisfies K ~ -1 everywhere and that 'Y is a curvilinear n-gon on a whose sides are geodesics. Show that n > 3, and that, if n = 3, the area enclosed by 'Y cannot exceed 71'". 11.4 Consider the surface of revolution
a(u,v)
= (f(u)cosv,f(u)sinv,g(u)),
where 'Y(u) = (f(u),O,g(u)) is a unit-speed curve in the xz-plane. Let U1 < U2 be constants, let 'Y1 and 'Y2 be the two parallels u = Ul and u = U2 on a, and let R be the region of the uv-plane given by Ul$U 0) for all non-zero 2 x 1 matrices v.) 11.17 For which of the following functions on the plane is the origin a nondegenerate critical point? In the non-degenerate case(s), classify the origin as a local maximum, local minimum or saddle point. (i) x 2 - 2xy + 4y 2 j (ii) x 2 + 4xyj (iii) x 3 - 3xy 2. 11.18 Let 5 be the torus obtained by rotating the circle (x _2)2 +Z2 = 1 in the xz-plane around the z-axis, and let F : 5 --+ R be the distance from the plane x ::: - 3. Show that F has four critical points, all non-degenerate, and classify them as local maxima, saddle points, or local minima. (See Exercise 4.10 for a parametrisation of 5.)
Solutions
Chapter 1 It is a parametrisation of the part of the parabola with x > o. (i) "Y(t) ::: (sec t, tan t) with -1r/2 < t < 1r/2 and 1r/2 < t < 31r/2. (ii) "Y(t)::: (2cost,3sint). 1.3 (i) x + y ::: 1. (ii) y (In X)2 . 1.4 (i) "'r(t) sin 2t( -I, 1). (ii) "'r(t) ::: (e t ,2t). 1.5 "'r(t) 3 sin t cos t( - cos t, sin t) vanishes where sin t ::: 0 or cos t ::: 0, Le. t ::: n1r /2 where n is any integer. These points correspond to the four cusps of the astroid. 1.1 1.2
=
=
=
281
Elementary Differential Geometry
282
(i) Let OP make an angle I) with the positive x-axis. Then R has coordinates 'Y( I)) ::: (2a cot I), a (1 - cos 21)) ) . (ii) From x ::: 2acotl), Y ::: a(l - cos 21)), we get sin2 I) ::: y/2a, cos2 () = cot2 I) sin2 I) ::: x 2 y/ Sa 3 , so the caitesian equation is y /2a + x 2 y/ 8a 3 ::: 1. 1.7 When the circle has rotated through an angle t, its centre has moved to (at, a), so the point on the circle initially at the origin is now at the point (a(t - sint),a(l- cost)). 1.6
----
......
~~------------- -~~---::;...-
a
o
at
1.8 Let the fixed circle have radius a, and the moving circle radius b (so that b < a in the case of the hypocycloid), and let the point P of the moving circle be initially in contact with the fixed circle at (a, 0). When the moving circle has rotated through an angle O. Since t = 'Y / II 'Y II::: vk t+l (k cos t - sin t, k sin t + cos t), we have ns vk!+1 (-ksint -cost,kcost - sint). So dt/ds (dt/dt)/(ds/dt)::: kt e- +1 n so K = 1/ ks. B y Th eorem k2-lot +1 ( -k sin t - cos t, k cos t - sin t ) ::: jk s s 2 2.1, any other curve with the same signed curvature is obtained from the logarithmic spiral by applying a rigid motion. (i) Differentiating "Y ::: rt gives t ::: ft + Ksrn s . Since t and n s are perpendicular unit vectors, it follows that K s ::: 0 and "Y is part of a straight line. (ii) Differentiating "Y ::: rns gives t ::: rns + rns = fns - Kart (Exercise 2.3). Hence, f = 0, so r is constant, and K s ::: -1/r, hence K a is constant. So "Y is part of a circle. (iii) Write"Y ::: r(t cos O+n s sin 0). Differentiating and equating coefficients of t and n s gives f cos 0 - Ksr sin 0 ::: 1, f sin 0 + Ksr cos 0 = 0, from which f = cosO and KsT = - sinO. From the first equation, r ::: scosO (we can assume the arbitrary constant is zero by adding a suitable constant to s) so Ks = -1/scotO. By Exercise 2.5, "Y is obtained by applying a rigid motion to the logarithmic spiral defined there with k ::: - cot O. We can assume that "Y is unit-speed. Then Ar>" ::: (1 - AKs)t, and this is non-zero since 1 - AKs > O. The unit tangent vector of "Y). is t, and the
=
2.6
II
(0,0, -9 sin2 t cos2 t) II (-3cos2 tsint,3sin2 tcost,O)
=
Elementary Differential GeometrY
286
arc~length s of 'Y A satisfies ds / dt ~ 1 - AKa. Hence, the curvature of 'Y). is
II dt/ds II ~ II i II /(1 - AKa) ~ Ka/(l- AKa). 2.8 The circle passes through 'Y(s) because II f - 'Y Il'= II K.10. II ~ l/IKs l, which is the radius of the circle. It is tangent to 'Y at this point because f - 'Y = ...!.. fi s is perpendicular to the tangent t of 'Y. The curvature of the K. circle is the reciprocal of its radius, Le. IKsl, which is also the curvature of 'Y. 2.9 The tangent vector of f is t +...!..( -Kst) - ~ns ~ -~ns so its arc-length K. K. K. I
is u == f II f II ds = f ~ds = Uo - ...!.., where Uo is a constant. Hence, K. K. the unit tangent vector of f is -ns and its signed unit normal is t. Since :3 -dnlJ/du = Kst/(du/ds) = the signed curvature of f is K;/k s. Denoting d/dt by a dash, 'Y' ~ a( 1 - cos t, sin t) so the arc-length s of 'Y is given by ds / dt = 2a sinet /2) and its unit tangent vector is t = 'Y (sin(t/2), cos(t/2)). So n s cos(t/2), sin(t/2)) and i = (dt/dt)/(ds/dt) = 4a8i~(t72)(cOs(t/2), - sin(t/2)) ~ -1/4asin(t/2)n lJ , so the signed curvature of 'Y is -1/4a sinet /2). Its evolute is therefore f = aCt -sin t, I-cos t) -4a sin(t/2)( - cos(t/2), sin(t/2)) = a(t+sin t, -1+ cos t). Reparametrising by f = 7I'"+t, we get aCt-sin f, I-cos f)+a( -71'", -2), which is obtained from a reparametrisation of 'Y by translating by the vector a(-7I'",-2). 2.10 The free part of the string is tangent to 'Y at 'Y( s) and has length f- s, hence the stated formula for ..(s). The tangent vector of .. is 'Y - 'Y + (f- s).:y = Ks(f - s)n s (a dot denotes d/ds). The arc-length v of .. is given by dv/ds = Ks(f - s) so its unit tangent vector is n s and its signed unit normal is -to Now dns/dv = K.(;_S)J1 s = {_Is t, so the signed curvature of .. is l/(f - s). 2.11 (i) With the notation in Exercise 2.9, the involute of f is
i:"t,
=
..(u)
~ f
= (-
~
1
+ (f - u) -d = 'Y + -n s - (f - u) n s
~
.
'Y - (f - Uo) n s ,
u KIJ since u = Uo - ...!.., SO" is the parallel curve 'Y-(l-uo). K. (ii) Using the results of Exercise 2.10, the evolute of .. is
.. + (f- s)( -t) = 'Y + (f- s)t -
(f - s)t ~ 'Y.
2.12 If we take the fixed vector in Proposition 2.2 to be parallel to the line of reflection, the effect of the reflection is to change
0). For the part with z < 0, we can take a(r,O) ::: (r sinh 0, r cosh 0, _r 2) defined on the open set Rj and a( u, v) = (u, v, u 2 - v 2 ) defined On the open set V ::: {(u, v) E R 2 I u 2 - v 2 < OJ. This is similar to Example 4.5, but using the 'latitude longitude' patch a(O, ' t cos t, e>' t sin t, e >'t) and au:::: (cos t, sin t, 1) at "y( t ), we get cos B :::: J2A 2 / (2A2 + 1), which is independent of t. The first fundamental form is sech2 u(du 2 + dv 2 ). The first fundamental form is (1 + j2)du 2 + u 2 dv 2, where a dot denotes d/du. So a is conformal if and only if j :::: ±vu2 - 1, i.e. if and only if I(u) = ±(!uvu2 -1- cosh- 1 u) + c, where c is a constant. The first fundamental form is (1 + 2v"{.6 +v 26.6)du 2 + 2"{.6 dudv + dv 2. So a is conformal if and only if 1 + 2v"{.6 + v 26.6 = 1 and "(.6 = for all u, v j the first condition gives 6 :::: 0, so 6 is constant, and the second condition then says that "Y.6 is constant, say equal to d. Thus, a is conformal if and only if 6 is constant and "y is contained in a plane r.6 = d. In this case, a is a generalised cylinder. a is conformal if and only if I~ + y~ = f; + y~ and lulv + YuYv = 0. Let z = lu +iyu, W = Iv +iYvj then a is conformal if and only if zz:::: ww and zw + zw = 0, where the bar denotes complex conjugate; if z = 0, then w = and all four equations are certainly satisfied; if z :/: 0, the equations give Z2 == _w 2 , so Z :::: ±iwj these are easily seen to be equivalent to the Cauchy-Riemann equations if the sign is +, and to the 'anti-CauchyRiemann' equations if the sign is -. Parametrise the paraboloid by a(u, v) = (u, v, u 2+v 2 ), its first fundamental form is (1 + 4u2)du2 + 8uv dudv + (1 + 4v 2)dv 2. Hence, the required area is II J1 + 4(u2 + v 2)dudv, taken over the disc u 2 + v 2 < 1. Let u = TsinB,V = TcosB; then the area is 21r Io1 VI + 4T 2TdT:::: ~(53/2 -1).
UVZ),
5.8
5.9
5.10
5.11 5.12
4
5.13
5.14
°
°
5.15
297
Solutions
5.16
5.17
This is less than the area 271'" of the hemisphere. Parametrize the surface by O'(u,v) (p(U) cosv, p(u) sinv, a(u)), where ...,.(u) = (p(u), 0, u(u)). By Example 6.2, the first fundamental form is du 2 + p(U)2 dv 2, so the area is II p(u) dudv = 271'" I p(u) duo (i) Take p(u) = cosu, u(u) = sinu, with -1r/2 ::; u ::; 71'"/2; so 271'" I~~~2 cos u du :::: 471'" is the area. (ii) For the torus, the profile curve is "y( B) :::: (a + b cos B, 0, b sin B), but this is not ,unit-speed; a unit-speed reparametrisation is .:y(u) :::: (a+bcos!,O,bsin~) with 0::; u::; 21rb. So 271'"I0211"b(a+bcos~) du = 471'"2 ab is the area. 0' is the tube swept out by a circle of radius a in a plane perpendicular to "y as its centre moves along "Y. D's = (1 - Ka cos B)t - ra sin Bn + ra cos Bb, 0'1) = - a sin Bn+a cos Bb, giving 0'8 XO'I) == -a (l-Ka cos B)( cos Bn+sin Bb); this is never zero since Ka < 1 implies that 1- Ka cosB > 0 for all B. The first fundamental form is ((1 - Ka COSB)2 + T 2a2) ds 2 + 2ra 2 dsdB + 1 a 2 dB 2, SO the area is 0 I;1I" a( 1 - Ka cos B)dsdB = 271'"a (s 1 - so). If E 1 = E 2, F1 :::: F2, G1 = G2, then E 1G 1 - Ff = E 2G2 - Fi, so any isometry is equiareal. The map f in Archimedes's theorem is equiareal but not an isometry (as E 1 :/: E 2 , for example). If E 1 :::: AE2,F1 = AF2,G 1 ::::AG 2 , and if E 1G1 - Ff = E 2G2 - Fi, then A2 :::: 1 so A = 1 (since A > 0). By Theorem 5.5, the sum of the angles of the triangle is 71'" + A/R 2 , where A is its area and R is the radius of the earth, and so is 2: 71'" + (7500000)/(6500)2 = 1r + 136°9 radians. Hence, at least one angle of the triangle must be at least one third of this, Le. 71'" + 116°9 radians. 3F 2E because every face has three edges and every edge is an edge of exactly two faces. The sum of the angles around any vertex is 271'", so the sum of the angles of all the triangles is 271'"Vi on the other hand, by Theorem 5.5, the sum of the angles of any triangle is 71'" plus its area, so since there are F triangles and the sum of all their areas is 471'" (the area of the sphere), the sum of all the angles is 71'" F + 471'". Hence, 2V = F + 4. Then, V - E + F :::: 2 + !F - E + F = 2 + ~(3F - 2E) = 2. Let 0' : U --t R 3 . Then, f is equiareal if and only if
=
I:
5.18
5.19 5.20
5.21
5.22
=
J1
(E 1G 1 - Ff)1/2 dudv
=
J1
(E 2G2 - Fi)1/2 dudv
for all regions R S; U. This holds if and only if the two integrands are equal everywhere, Le. if and only if E 1 G 1 - Ff = E 2 G2 - F:j.
298
Elementary Differential Geometry
Chapter 6 au:::: (1,0, 2u), au = (0 11, 2V)1 sej N = A( -2u, -2v, 1), where A = (1 + 4u2 +4v 2 ) -1/2 j a uu :::: (0,0,2), a ~~ :::: 0, a vv :::: (0,0,2), so L = 2A, M = 0, N :::: 2A 1 and the second fundamental form is 2A(du 2 + dv 2). 6.2 au.N u :::: -auu.N (since au.N :::: 0), so Nu.a u =: OJ similarly, Nu.a v :::: Nv.a u = Nv.a v :::: OJ hence, N u and N u are perpendicular to both au and au, and so are parallel to N. On the other hand, N u and N v are perpendicular to N since N is a unit vector. Thus, N u :::: N v = 0, and hence N is constant. Then, (a.N)u :::: au.N :::: 0, and similarly (a.N)v :::: 0, so a.N is constant, say equal to d, and then a is part of the plane r.N :::: d. 6.3 From Section 4.3, N :::: ±N, the sign being that of det(J). From au 8u + 8v - _ + au 8ii' 8v au 8u a v 8u' a:v - au 8u 8:V we get 2 2 _ 8 u 8 v (8U)2 8u 8v (8v)2 a uu :::: au 8u2 + a v 8fJ.2 + a uu 8u + 2auu 8u au + a vv au
6.1
So _ L ::::
(8v)2) ± ((8U)2 L 8u + 2M 8u8v 8u 8u + N 8fJ.
since au.N :::: av.N :::: 0. This, together with similar formulas for M and N, are equivalent to the matrix equation in the question. 6.4 Applying a translation to a surface patch a does not change au and a V1 and hence does not change N,auu,a uv or a vv , and hence does not change the second fundamental form. A rotation A about the origin has the following effect: au --t A(a u ), au --t A(a v ) and hence N --t A(N), a uu --t A(a uu ), a uv --t A(a uv ), a vv --t A(avv)j since A(p).A(q) :::: p.q for any vectors p, q E R 3 , the second fundamental form is again unchanged. 6.5 By Exercise 6.1, the second fundamental form of the paraboloid is 2 (du 2 + dv 2 )/J1 + 4u 2 + 4v 2; so
"'n :::: 2(( -
sin t)2
+ cos2 t)/V1 + 4 cos2 t + 4 sin2 t :::: 2/VS.
°
= "'; + "'; :::: 0, so '" :::: and "( is part of a straight line. 6.7 Let "( be a unit-speed curve on the sphere of centre a and radius T. Then, ("( - a).("( - a) :::: T 2 j differentiating gives Ar.("( - a) :::: 0, so .y.("( - a) :::: -"(."( =: -1. At the point "((t), the unit normal of the sphere is N ±~("((t) - a), so "'n :::: .:y.N :::: ±~.:y.("( - a) :::: =f~. 6.8 If the sphere has radius R, the parallel with latitude I) has radius T Rcosl); if P is a point of this circle, its principal normal at P is parallel to the line through P perpendicular to the z-axis, while the unit normal to the sphere is parallel to the line through P and the centre of the sphere.
6.6
",2
Solutions
299
The angle 't/J in Eq. (8) is therefore equal to f) or 1r - f) so K 9 :::: ± ~ sin f) :::: ± ~ tan f). Note that this is zero if and only if the parallel is a great circle. 6.9 The unit normal is N :::: (-g cos v, -g sinv, j), where a dot denotes d/du. On a meridian v :::: constant, we can use u as the parameter; since au:::: (j cosv, J sin v, 9) is a unit vector, u is a unit-speed parameter on the meridian and !sinv 9 /cosv -gsinv j -0 Kg :::: auu.(N x au) :::: -9 cos v - . jcosv fsinv 9 On a parallel u :::: constant, we can use vasa parameter, but a (- f sin v, f cos v, 0) is not a unit vector; the arc-length s is given by ds/dv :::: II a v II:::: f(u), so s :::: f(u)v (we can take the arbitrary constant to be zero). Then, 1)
::::
- f cos v -fsinv -gcosv - f sinv
-g sin v fcosv
6.10 Kl :::: KN 1 .n, K2 :::: KN 2.n, so Kl N 2 - K2Nl :::: K((N 1 .n)N 2 - (N 2.n)N d
:::: K(N l
X
N 2 ) x n.
Taking the squared length of each side, we get KI
+ K~ -
2 2KlK2Nl.N2 :::: K2 II (N 1 x N 2) x n 11 .
Now, N I.N2 :::: cos a; 'Y is perpendicular to N 1 and N 2, so N 1 parallel to 'Y, hence perpendicular to nj hence,
X
N 2 is
II (N 1 x N 2) x n II:::: II N 1 x N 2 1111 n II:::: sina. 6.11 N. n :::: cos 't/J, N. t :::: 0, so N.b :::: sin 't/J j hence, N :::: n cos 't/J B :::: t x (n cos 't/J + b sin 't/J) :::: b cos 't/J - n sin 't/J. Hence,
+ b sin 't/J
and
N :::: Ii cos 't/J + b sin 't/J + ~ (- n sin 't/J + b cos 't/J ) :::: ( - K t + T b) cos 't/J - n 7 sin 't/J + ~ (- n sin 't/J + b cos 't/J) :::: - K cos 't/Jt + (T + ~) (b cos 't/J - n sin 't/J) :::: -Knt + 7 g B. The formula for :B is proved similar Iy. Since {t, N, B} is a right-handed orthonormal basis of R 3 , Exercise 2.22 shows that the matrix expressing t, N,::B in terms of t, N, B is skew-symmetric, hence the formula for i. 6.12 A straight line has a unit-speed parametrisation ")'(t) :::: p + qt (with q a unit vector), so .y :::: 0 and hence K n :::: .y.N :::: O. In general, K n :::: 0 .:y is perpendicular to N N is perpendicular to n N is parallel to b (since N is perpendicular to t).
Elementary Differential Geometry
300
vI
6.13 The second fundamental form is (- du 2 + U 2 dv 2)/ u + u 2, so a curve "Y(t) :::: O'(u(t), v(t)) is asymptotic if and only if -iJ? + u 2iP = 0, i.e. dv/du:::: v/iI, == ±l/u, so lnu "': ±(v + c), where c is a constant. 6.14 By Exercise 6.12, b is parallel to N, so b:::: ±Nj then, B :::: t x N:::: =fn. Hence, :B = =fn :::: =f( - Id + T b) = ±Kt - TN j comparing with the formula for :B in Exercise 6.11 shows that 7 9 :::: 7 (and Kn :::: ±K). 6.15 For the helicoid O'(u,v) = (vcosu,vsinu,Au), the first and second funda, mental forms are (A2 + v 2)du2 + dv 2 and 2Adudv/VA2 + v 2, respectively. Hence, the principal curvatures are the roots of -K(A
2
+ v2 )
>.
.J>.2+ v2
.J>.;+v2 -
:::: 0,
K
i.e. ±A/(A + v ). For the catenoid 0'( u, v) :::: (cosh u cos v, cosh u sin v, u), the first and second fundamental forms are cosh 2 u(du2+dv 2) and -du2+dv 2, respectively. Hence, the principal curvatures are the roots of 2
2
-1 - Kcosh2 U
o
0 1 - Kcosh 2 U
'
i.e. K :::: ±sech 2 u. 6.16 Let s be arc-length along "Y, and denote d/ds by a dash. Then, by Proposition 6.1, ,2 Kn
==Lu
,
,2
I
+ 2Mu v + Nv::::
LiI,2
+ 2MiI,v + Nv 2 LiJ? + 2Muv + Ni} (ds/dt)2 - Eu 2 + 2Fuv + GV2
6.17 By Exercises 5.4 and 6.3, we have (in an obvious notation), PI == Jt :FIJ, PII = ±Jt:FIIJ, where the sign is that of det( J). The principal curvatures of ii are the roots of det( PI1- kPI) ~ 0, Le. det( ±Jt:FII J - kJ t :FII J) = 0, whkh (since J is invertible) are the same as the roots of det(±:FII-k:FI ) = O. Hence, the principal curvatures of ii are ± those o{ 0'. Let ~au + ijuii be a principal vector for ii corresponding to the principal curvature k. Then, (FIJ - i (FIJ -
~) = (~). then ~11
u
+1)11 v
"F/)J
(~) = (~) ,
is a principal vectorfor 11 correspond-
ing to the principal curvature K. But, since ~ = aa~ ~ + ~~ry, 1] = ~~ ~ + ~~ ry, - a a &,v J,.'U v we have ~O'u + 1]O'v_:::: ~ (a~O'u + a~O'v) + ij (~~O'u + avO'v) = ~uu + ryiiv, which shows that ~iiu + ijiT v is a principal vector for 0' corresponding to the principal curvature K. The second part also follows from Corollary 6.2.
Solutions 6.18 ..y =
301
uau + va v
is a principal vector corresponding to the principal curva-
ture"=(;:I1-";:I)(~) = (~) =;:il;:I1(~) =,,(~) = (~ ~) (~) = -" (~) = au + ciJ = -"u and bu + dv = -"v. But, N :::: uNu+vN v = u(aau+bav)+v(CO'u+dav) = (au+cV)au+(bu+dv)a v . Hence, Ar is principal N = -K(W u + va v ) :::: -K'Y. From Example 6.2, the first and second fundamental forms of a surface of revolution are du 2 + f(U)2dv 2 and (jg -liJ)du2 + fiJdv 2 , respectively. Since the terms dudv are absent, the vectors au and a'll are principal; but these are tangent to the meridians and parallels, respectively. 6.19 By Exercise 6.11, N = -Knt + 7 g B, so by Exercise 6.18 'Y is a line of curvature if and only if 7 9 :::: 0 (in which case A -= K n ). 6.20 Let N 1 and N 2 be unit normals of the two surfaces; if 'Y is a unit-speed parametrisation of C, then N l = -AlAr for some scalar Al by Exercise 6.18. If C is a line of curvature of 52, then N 2 = -A2Ar for some scalar A2, and then (NI.N2)" -= -AI 'Y.N 2 - A2'Y.N1 = 0, so N I.N2 is constant along 'Y, showing that the angle between 51 and 52 is constant. Conversely, if N l .N 2 is constant, then N l .N 2 = 0 since Nl .N 2 == -Al'Y.N2 :::: 0; thus, N2 is perpendicular to N I, and is also perpendicular to N 2 as N 2 is a unit vector; but i' is also perpendicular to Nl and N 2; hence, N2 must be parallel to 'Y, so there is a scalar A2 (say) such that N2 :::: -A2'Y. 6.21 (i) Differentiate the three equations in (21) with respect to w, u and v, respectively; this gives
Subtracting the second equation from the sum of the other two gives au.a vw = 0, and similarly av.a uw = aw.a uv :::: O. (ii) Since av.a w :::: 0, it follows that the matrix :F1 for the u = Uo surface is diagonal (and similarly for the others). Let N be the unit normal of the u =:. Uo surface; N is parallel to a v xaw by definition, and hence to au since au, a v and aware perpendicular; by (i), avw.au :::: 0, hence avw.N = 0, proving that the matrix :FIl for the u :::: Uo surface is diagonal. (iii) By part (ii), the parameter curves of each surface u :::: 'Uo are lines of curvature. But the parameter curve v -= VOl say, on this surface is the curve of intersection of the u:::: Uo surface with the v = Va surface.
Elementary Differential Geometry
302
6.22 We have N u = at1 u
+ ba v , N v
Nu.Nu :FIII = ( N N 1.1' v
= C4 u
+ da v , so
Nu.Nv, ),.
N v·Nv·./
_ ( Ea 2 + 2Fab + Gb 2 Eae + F(ad + be) + Gbd
Eae + F(ad + be) + Gbd) Ec2 + 2Fcd + Gcf2
=( : :) ( ~ ~) (~ ~ ) = (- Fi 1FII ) t F 1 1 == :Fu:Fi :F/:F 1:Fu =:Fu:Fi :Fu ·
j ( -
Fi 1 F II )
1-
6.23 By Example 6.2, the principal curvatures are jg - jiJ and iJII. If iJ = 0, the surface is part of a plane z == constant and no point is parabolic. Thus, iJ :/: 0, and every point is parabolic if and only if jg - jiJ = O. Multiplying through by iJ and using j2 + iJ2 = 1 (which implies that j / + gg = 0), we get j = O. Hence, I(u) = au + b, where a and b are constants. If a = 0, we have a circular cylinder; if a :/: 0, we have a circular cone. 6.24 On the part of the ellipsoid with z :/: 0, we can use the parametrisation
t1(x,y) (x,y,z), where z ::: ±TJ1- ~ - ~. The first and second fundamental forms are (1 + z; )dx 2 + 2z x zydxdy + (1 + Z~)dy2 and (z xx dx 2+
=
2ZZydxdy+Zyydy2)1 J1 + z; + z~, respectively. By Proposition 5.3(ii), the condition for an umbilic is that :Fu = K:F/ for some scalar K. This leads to the equations Zzz
= ,\(1 + z;),
Zxy
= '\zzZy,
Zyy
= ,\(1 + z~),
= KJ1 + z; + z;. If x and yare both non-zero, the middle equation gives ,\ = liz, and substituting into the first equation gives the contradiction p2 = T'l. Hence, either x = 0 or y = O. IT x = 0, the equations where ,\
have the four solutions
x
= 0,
y
= ±q
~ 2_P2 2
q
-T
2'
Z
= ±T
~2_P2 T
2
-q 2'
Similarly, one finds the following eight other candidates for umbilics:
~
X = ±p P2
_q2
-T
2'
Y = 0, z
= ±T
~2_q2 T
2
-p
2 '
q x=±pV~-T:, y=±qvq: T:, z=O. -q p
-p
Of these 12 points, exactly 4 are real, depending on the relative sizes of p2, q2 and T2.
Solutions
303
Chapter 7 7.1
au:::: (1,I,v), a v :::: (1,-I,u),a uu :::: a vv :::: 0, a uv :::: (0,0,1). When u :::: V :::: 1, we find from this that E :::: 3" F :::: 1, G :::: 3 and L :::: N :::: 0, M :::: -1/0. Hence, K :::: (LN - M 2 )/(EG - F 2 ) :::: -1/16,
=
7.2
H -:.. (LG - 2MF + NG)/2(EG - F 2 ) 1/80. For the helicoid (7(u, v) :::: (vcosu,vstnu,~u), au:::: (-vsinu,vcosu,A), av (cos u, sin u, 0), N :::: (A 2 + V 2)-1/2( -A sin u, Acos u, -v), a uu :::: '( -v cos u, -v sin u, 0), (7 uv sin u, cos u, 0), a vv :::: o. This gives 2 2 E :::: A + v " F :::: 0, G = land L = N = 0, M :::: A/JA 2 + v2 -. Hence, K:::: (LN - M 2 )/(EG - F 2 ) :::: _A 2 /(A 2 + v 2 ? For the catenoid 'a( u, v) :::: (cosh u cos v, cosh u sin v, u), we have au:::: (sinh u cos v, sinh u sin v, 1), a v :::: (- cosh u sin v, cosh u cos v, 0), N :::: sech u( - cosv,- sin v, sinh u), a uu (cosh u cos v, cosh usinv, 0), a uv :::: (- sinh u sin v, sinh u cos v, lT vv :::: (- cosh ucos v, - cosh u sin v, 0). 2 This gives E ::::G ::::cosh u, F and L :::: -1, M :::: 0, N :::: 1. Hence, 2 2 K :::: (LN -M )/(EO - F ) :::: -sech 4 u. Altetnatively, use the results of Exercise 6.15. Parametris'e the su'dace by a(u,v) (u,v,!(u,v)). Then, 'au:::: (I,O,!u), a v :::: (O,I,!v), N :::: (I +!~ + J';)-1/2{-!U'-!V' 1), (7uu :::: (O,O,!uu), a uv :::: (0,0, !uv), vv :::: (O, 0, !vv). This gives E :::: 1 + !~, F :::: !u!v,O :::: 1 + /; and L :::: (1 + !~ + J';)-1/2 !uu, M :::: (1 + i~ + !;)-1/2!uv, N = (1 + !~ + !;)-1/2!vv. Hence,
=
= (-
°), =°
7.3
=
=
u
K::::
! uulvv - / 2uv(I +!~ + !;)2'
°
7.4 (i) ;F'rom Example 7.3, K :::: 0 6.N :::: 0 6.((t + v6) x 6) :::: 6.(t x 6) :::: 0. If 6 :::: 'n, 6 = -Kt + Tb, t x 6 :::: b, so K =0 T ::::0 "f Is planar (by Proposition 2.4). If 6 :::: b, 6:::: -Tn, t x 6 :::: -n, so again K =:0 T :::: 0. (ii)Le't N 1-be a unit 'normal of S. Then,K :::: N l.(t xN I) :::: 0. Since N1 is perpendicular to N1 and N l is perpendicular to t, this condition holds N'l is parallel to t, Le. N 1 :::: -Ai' 'for some scalar A. Now use Exercise 6.18. 7.5 Using the parametrisation a in 'Exercise 4.10, we find that E :::: b2 , F :::: 0, G :::: (a + b cosu)2and L b, M :::: 0, N :::: (a + bcosu) cos u. This gives K ::::cosu/b(a + bcosu), dAa :::: (EG - F 2)1/2dudv:::: b{a + bcosu)dudv. Hence,
°
=
ff
1'1" 2
KdAa
=
cosududv
= o.
304
Elementary Differential Geometry
7.6
The first part follows from Exercises 5.3 and 6.4. The dilation multiplies E, F, G by a 2 and L, M, N by a, hence H by a-I and K by a- 2 (using Proposition 7.1). 7.7 Since a is smooth and au x a v is never zero, N = au x avl II au x a v II is smooth. Hence, E, F, G, L, M and N are smooth. Since EG - F2 > 0 (by the remark following Proposition 5.2), the formulas in Proposition 7.1(i) and (ii) show that H and K are smooth. By Proposition 7.1(iii), the principal curvatures are smooth provided H 2 > K, Le. provided there are no umbilics. 7.8 At a point P of an asymptotic curve, the normal curvature is zerO. By Corollary 6.2, one principal curvature Kl 2: 0 and the other K2 :$ O. Hence K = Kl K2 :$ O. On a ruled surface, there is an asymptotic curve, namely a straight line, passing through every point (see Exercise 6.12). 7.9 By Exercise 6.22, :FIll = :FIl:Fi! :FIl. Multiplying on the left by :Fi1, the given equation is equivalent to
A 2 +2HA+KI=0, where A = -:Fi1:FII = (:
~). By the
remarks following Definition
6.1, the principal curvatures are the eigenvalues of -A. Hence, 2H = sum of eigenvalues of -A = -(a + d), K = product of eigenvalues of -A = ad - be. Now use the fact stated in the question. 7.10 By Eq. (9) in Chapter 6, 'Y.'Y = Tt:FIT; by Eq. (10) in Chapter 6, N.'Y = -N.':';' (since N.'Y = 0) = -K n = -Tt:FIlT. Now, N.N = (uN u + vNv).(uNu+vN v ) (Nu.Nu)u2+2(Nu.Nv)uv+(Nv.Nv)V2 Tt:FIlIT. Hence, multiplying the equation in Exercise 7.9 on the left by Tt and on the right by T gives N.N+2HN.'Y+K'Y.'Y;;:: O.J£.')ds an asymptotic curve, K n = 0 so N.'Y = O. So, assuming that 'Y is unit-speed, we get N.N = -K. But Exercise 6.12 gives N = ±b, so N = =fro and N.N = r 2 . 7.11 The parametrisation is a(u,v) = (f(u)cosv,f(u)sinv,g(u)), f(u) = e U , g(u) ;;:: VI - e 2u - cosh- 1(e- u ), -00 < u < O. (i) A parallel u = constant is a circle of radius f(u) = e U , so has length 27re u • (ii) From Example 7.2, E I, F = 0, G = f(U)2, so dAD- ;;:: f(u)dudv and Jr the area is f~oo eUdudv = 27r. (iii) From Example 7.2, the principal curvatures are Kl = jg - /f; -Jlh;;:: _(e- 2u _1)-1/2, K2;;:: fhlf 2 ;;:: hlf;;:: (e- 2u _1)1/2. (iv) Kl < 0, K2 > O. 7.12 (i) Setting u = v,v = w;;:: e- u , we have u = -lnv,V;;:: u so, in the
=
f:
=
=
notation of Exercise 5.4, J =
(~ -o~).
Since J is invertible, (u, v)
f4
Solutions
305
(v ,w) is a reparametrisation map. The first fundamental form in terms of v, w is given by
(~ ~) =
Jt ( ;
~) J
0 )(01 _4:) (~ : : (-~0 1)(1 0 0 j(U)2 OV :::: 0
so the first fundamental form is (dv 2 + dw 2)/w 2. (ii) We find that the matrix
8V) _ ( v(w + 1) 2 - gil ~ - -t(v -(w+1)2)
j _ (8;;
~(V2 - (w + 1)2)) v(w+1) ,
so the first fundamental form in terms of U and V is given by
0) - (v JOb J:::: -t (
~
2
+ (w + 1)2)2 4w2
1
I :::: (1 _ U2 _ V2)2 11
after some tedious algebra. In (i), U < 0 and -1r < V < 1r corresponds to -1r < V < semi-infinite rectangle in the upper half of the vw-plane.
o
1r
and w
> 1, a
v
To find the corresponding region in (ii), it is convenient to introduce the complex numbers z :::: V + iw 1 Z :::: U + iV. Then, the equations in (ii) are equivalent to Z :::: ~+~ 1 Z :::: i(~!.ll)' The line v :::: 1r in the vw-plane corresponds to z + :2 :::: 21r (the bar denoting complex conjugate), Le. i(~~ll) - i(1!.11) ::= 21r I which simplifies to 1Z - (1 - ~) 12 :::: ~; so V :::: 1r corresponds to the circle in the U~'-plane with centre 1- i11" and radius 1.. 11" Similar lY1 V :::: -1r corresponds to the circle with centre 1 + i11" and radius 1 F'mall Y, w:::: 1 correspon d s t 0 z - z:::: 2"" l.e. --.llL 2'+1 -- 2't. ;;. i(Z-I) + i(Z-I)
i;
This simplifies to IZ - ~ 12 :::: so w :::: 1 corresponds to the circle with centre 1/2 and radil;ls 1/2 in the UV-plane. The required region in the UV-plane is that bounded by these three circles:
Elementary Differential Geometry
306
v
- - - - + - - - -.....~~~",..E---~u o
7.13 Let "'((u) :::: (/(u), O,g(u)) and denote d/du by a dot; by Eq. (2), I+KI :::: O. If K < 0, the general solution is 1 :::: ae- FKu + be FKu where a, b are constants; the condition 1(1r /2) :::: I( -1r/2) == 0 forces a == b :::: 0, so "'( coincides with the z-axis, contradicting the assumptions. If K :::: 0, 1 :::: a + bu and again a :::: b :::: 0 is forced. So we must have K > 0 and 1 :::: a cos..jKu + b sin..jKu. This time, 1 (1r /2) :::: I( -1r/2) :::: 0 and a, b not both zero implies that the determinant
cos..jK1r/2 cos..jK1r /2
sin..jK1r/2 - sin..jK1r /2 == O.
This gives sin..jK1r = 0, so K = n 2 for some integer n :/: O. If n == 2k is even, 1 = b sin 2ku, but then 1(0) = 0, contradicting the assumptions. If n :::: 2k + 1 is odd, 1 :::: a cos(2k + l)u and 1(1r/2(2k + 1)) = 0, which contradicts the assumptions unless k == 0 Or -I, Le. unless K :::: (2k+ 1) 2 == 1. Thus, 1
= acosu, iJ ==
V1- j2 :::: V\ - a sin u. Now, ~:::: (j,O,g) is 2
2
perpendicular to the z-axis !J = o. So the assumptions give VI - a 2 == 0, i.e. a :::: ±1. Then, "'((u) = (± cos u, 0, ± sin u) (up to a translation along the z-axis) and S is the unit sphere. 7.14 Let &(u, v) be a patch of S containing P == O'(uo, vo). The gaussian curVature K of S is < 0 at P; since K is a smooth function of (u, v) (Exercise 7.7), K(u, v) < 0 !or (u,v) in Some open set (; containing (uo,vo); then every point of O'(U) is hyperbolic. Let K}, K2 be the principal curvatures of &, let 0 < f) < 1r /2 be such that tan f) == -Kr / K2, and let el and e2 be the unit tangent vectors of & making angles f) and -f), respectively, with the principal vector corresponding to K} (see Corollary 6.1). Applying
vi
307
Solutions
Proposition 7.4 gives the result. 7,15 See the proof of Proposition 9.5 for the first part. The first fundamental form of the given surface patch is (l+u 2 +v 2)2(du 2 + dv 2), so it is conformal, and a uu + a vv (-2u, 2v, 2) + (2u, -2v, -2) O. 7.16 Parametrize the surface by a(u,v) = (u,v,/(u,v)). By Exercise 7.3, E = 1 + t;,F = lu/v,G == 1 + I;,L = (1 + I~ + 1;)-1/2/uu,M = (1 + I~ + {;)-1/2 luv, N = (1 + I~ + 1;)-1/2 Ivv. Hence,
=
H _ LG - 2M F + N E _ luu(1 2(EG - F2) Taking I( u, v)
= ln (~~::)
=
+ I;) - 2/u1J/u/v + l11v(1 + I~) 2(1 + I~ + 1;)3/2
gives
H _ sec2 u(1 + tan 2 v) - sec 2 v(1 + tan2 u) _ 0 2(1 + tan2 u + tan 2 V)3/2 - .
7.17 E u = au + wN u , E v = a v + wN v , E w = N. Eu.E w = 0 since au.N == Nu.N = 0, and similarly Ev.Ew ::= O. Finally, Eu.E v = au.a v + w(au.N v + av.N u ) +w 2 N u .N v == F - 2wM +w 2 N u.Nv == w 2 N u.Nv ; by Proposition 6.4, N u == - iau, N v = v , so Nu.N v ::= iZ F == O. Every surface u = Uo (a constant) is ruled as it is the union of the straight lines given by v ::= constant; by Exercise 7 .4(ii) , this surface is flat provided the curve "y( v) = a(Uo, v) is a line of curvature of S, Le. if a v is a principal vector; but this is true since the matrices :F1 and :FII are diagonal. Similarly for the surfaces v = constant. 7.18 By Eq. (15), the area of a(R) is
Za
Jh ll
Nu x Nv
J
II dudv = hlKll1 au x a v II dudv ==
J
hlKldAa.
7.18 From the formula for K in the solution of Exercise 7.5, it follows that s+ and S- are the annular regions on the torus given by -1r /2 ::; u ~ 1r /2 and 1r /2 ::; u ::; 31r /2, respectively.
S+
S-
Elementary Differential Geometry
308
It is clear that as a point P moves over S+ (resp. S-), the unit normal at P covers the whole of the unit sphere. Hence, ffs + IKldA = ffs - IKldA = 471'" by Exercise 7.18j since IKI = ±K on S±, this gives the result.
Chapter 8 8.1
By Exercise 4.4, there are two straight lines on the hyperboloid passing through (1,0,0); by Proposition 8.3, they are geodesics. The circle given by z = 0, x 2 + y2 = 1 and the hyperbola given by y = 0, x 2 - z2 = 1 are both normal sections, hence geodesics by Proposition 8.4 (see also Proposition 8.5). 8.2 Let lIs be the plane through ,,),(s) perpendicular to t(s); the parameter curve s = constant is the intersection of the surface with lIs. From the solution to Exercise 5.17, the standard unit normal of a is N = -(cos f) n+ sin f) b). Since this is perpendicular to t, the circles in question are normal sections. 2 2 2 . 8.3 Take the ellipsoid to be } + ~ + ~ = 1; the vector (?' ~) is normal to the ellipsoid by Exercise 4.16. If ,,),(t) = (/(t),g(t),h(t)) is a curve on j2 ·2 .2 /2 2 2 / the ellipsoid, R = (~+ ~ + ~ )-1/2, S = (7 + ~ + ~ )-1 2. Now, ")' is a ~, -;;, for geodesic .y is parallel to the normal (I, jj, it) = A( . p q
?'
-!i)
& = 1 we get ft = 0, hence p +~ q + ~ r p + ~ q + ~ r j2 it fl" hh • j2 it h ~ + ~ + TT + ~ +,. +;:T = O,l.e. ~ + ~ + ~ + A 7 + ~ + 1:4 =0, which gives A = _S2 / R2. The curvature of")' is
some scalar A(t). From ·2
2
·2
/
2
2
. ( /2
h2 )1/2
2
II ;Y II = (/2 + jj2 + it 2)1/2 = IAI ( -p4 + !L. + -r 4 q4 Finally,
1d( 1) (Ii + "2 8 dt
R2 2
=
p4
gg' hh) q4 + ~
+
(i + 2
p2
2
9
q2
(1p42 + g2q4 + h2 ) r4
2
h
+ r2
-
2 )
IAI
S R2'
S
)
(i/+ p2
2
gjj q2
+
hit) r2
99 hh) A (Ii 99 hh) = R21 (Ii p4 + Q4 + r4 + 82 p4 + Q4 + r 4 = 0, since A = -82 / R 2 . Hence) R8 is constant. 8.4 If ")' is a geodesic, .:y = Kn is parallel to N (in the usual notation), so n = ±N. In the notation of Exercise 6.11, B = t x N = ±b, so ::B =
~olutions
Kgt -
TgN
== ±b ==
=fTn
==
-TN.
Hence,
Tg
==
T
(and
Kg
== 0, which we
knew already). 8.5 .:y is parallel to II since 'Y lies in II, and .:y is parallel to N since 'Y is a geodesic; so N is parallel to II. It follows that N is also parallel to II. Since N is perpendicular to N, and 'Y is also perpendicular to N and parallel to II) N is parallel to 'Y. By Exercise 6.18, 'Y is a line of curvature of S. 8.6 If P and Q lie on the same parallel of the cylinder, there are exactly two geodesics joining them, namely the two circular arcs of the parallel of which P and Q are the endpoints. If P and Q are not on the same parallel, there are infinitely many circular helices joining P and Q (see Example 8.7). 8.7 Take the COne to be O'(u,v) == (ucosv,usinv,u). By Exercise 5.5,0' is isometric to part of the xy-plane by 0'(U, v) 1-4 (u V'i cos .:12, uV'i sin .:12, 0). By Corollary 8.2, the geodesics on the COne correspond to the straight lines in the xy-plane. Any such line) other than the axes x == 0 and y == 0, has equation ax + by == 1, where a, b are constants; this line corresponds to the curve v 1-4 (
cos v J2(acos
sin
V I ). J2(acos :72+ bsin :72) ,
:72+ bsin :72) ' ..;2(acos 72+ bsin 72))
the x and y-axes correspond to straight lines on the cone. 8.8 From Example 5.3, O'(u, v) == 'Y(u) + va) where 'Y is unit-speed, II a 11== 1, and 'Y is contained in a plane perpendicular to aj the map O'(u, v) 1-4 (u, v, 0) is an isometry from the cylinder to the xy-plane. A curve t 1-4 O'(u(t), v(t)) is a geodesic on the cylinder t t-t (u(t), v(t), 0) is a constant-speed parametrisation of a straight line in the plane iI, and v are constant iJ is constant (since il,2 + v2 is constant). Since a. JtO'(u(t), v(t)) == v, iJ == constant the tangent vector JtO'(u, v) makes a fixed angle with the unit vector a parallel to the axis of the cylinder. 8.9 Take the cylinder to be O'(u, v) = (cos u, sin u, v). Then, E == G == 1, F == 0, so the geodesic equations are u == v == O. Hence) u == a + bt, v == c + dt, where a) b, C, d are constants. If b = 0 this is a straight line on the cylinder; otherwise, it is a circular helix. 8.10 E == 1, F == 0, G == 1 + u 2 , so 'Y is unit-speed i} + (1 + U 2 )V 2 == 1. The second equation in (2) gives Jt ((1 + u 2 )v) == 0, Le. iJ == l':U:l' where a is a constant. So
il,2
== 1 -
dv == ~ == du U
2
(1~U2) and, along the geodesic,
±
a -/(1 - a 2 + u 2 )(1
+ u2 )
If a = 0, then v = constant and we have a ruling. If a == 1, then dv jdu = 1 ±ljuvl + u 2 , which can be integrated to give v == Vo =f sinh- ~, where
310
Elementary Differential Geometry Va is a constant.
8.11 We have N x
a ::;; (au x a v) x au ::;; (au.au}t!v - (au.av)au ::;; Eav - Fa u u II au x l1 v II JEG - F2 JEG _ F2 l
'Y ::;; ita u + va v , so u(Eav - Fa u) + v(F&v - Gau )
and similarly for N x avo Now, N x . ::;;
"(
JEG- F2 .:y = iUru + va v + u 2a uu + 2uvauv
' + iJ2 avv . Hence, "'g ::;; .y.(N X'Y) ::;; (uv-iJ'u)JEG - F2 +Au 3 +Bu 2v+CuV 2+Dv 3 , where A ::;; auu.(Eav - Fa u ) ::;; E((a u .l1 v )'U - au.a uv ) - !F(au.au)u ::;; E(Fu - !Ev ) - !FEu , with similar expressions for B,C,D. 8.12 We have
(Eu 2+2Fuv + Gv 2)' ::;; (Euu + Evv) u 2 + 2(Fuu + Fvv)uv + (Guu + Gvv)v 2 + 2Euu + 2F(uv + uv) + 2Gvv ::;; u{Euu 2 + 2Fuuv + G uv 2) + v(Ev u 2 + 2Fvuv +'G vv 2) + 2EufL + 2F(uv + iiv) + 2Gvv ::;; 2(Euu + Fv)'u + 2(Fit + Gvrv + 2(Eu + Fv)u + 2(Fu + Gv)v (by the geodesic equations ::;; 2[(Eu + Fv)u)' + 2[(Fu + Gv)v)' ::;; 2(Eu 2 + 2Fuv + Gv 2)'. Hence, (Eu 2 +2Fuv + Gv 2)' ::;; 0 and so II 'Y11 2 ::;; Eu 2 + 2Fuv + GiJ 2 is constant. 8.13 They are normal sections. 8.14 Every parallel is a geodesic every value of u is a stationary point of f (u) (in the notation of Proposition 8.5 ) f ::;; constant the surface isa circular cylinder. 8.15 The two solutions of Eq. (13) are
V ::;; Vo
for a self-intersection is that, for some w
±
J~ -
w 2 , so the condition
> 1, 2J ~ - w2 = 2k1r for some
--b -
1 > 21r, ie. n < (1 + 1r 2)-1/2. In integer k > O. This holds 2J this case, there are k self·intersections, where k is the largest integer such
--b -
that 2k1r < 2 J 1. 8.16 (i)If"((t)isageodesic,soisf("((t)),andif"(isdefinedforall-oo < t < 00, so is f("((t)). So f takes meridians to meridians, i.e. if v is constant, so is v. Hence, v does not depend 'on w.
Solutions·
311
(ii) f preserves angles and takes meridians to meridians, so must take parallels to parallels. Hence, tV does not depend on v. (iii) The parallel w constant has length 21r jw by Exercise 7.II(i) (w = U e- ). As f preserves lengths, part (ii) implies that 21r jw 21r jtV, so
=
=
w=tV. (iv) We now know that f(a(u, v)) = a(F(v), w) for some smooth function
F(v). The first fundamental form of a(F(v), w) is w- 2 ( (~~) 2 dv 2 + dw 2 ) j since f is an isometry, this is equal to w- 2 (dv 2 +dw 2 ), hence dF jdv = ±I, so f(v) = ±v + a, where a is a constant. If the sign is +, f is rotation by a around the z-axis; if the sign is -, f is refiection in the plane containing the z-axis making an angle aj2 with the xz-plane. U + iV ~~~, where z v+ 8.17 From the solution to Exercise 7.12, Z iw. Since the geodesics on the pseudosphere correspond to straight lines and circles in the vw-plane which are perpendicular to the v-axis, they correspond in the UV -plane to straight lines and circles perpendicular to the image of the v·axis under the transformation z I-t z-~, Le. the unit z+t circle U2 + V2 1. 1 around 8.18 (i) Let the spheroid be obtained by rotating the ellipse' ~ + ~ the z-axis, where a, b > O. Then, a is the maximum distance of a point of the spheroid from the z-axis, ~o the angular momentum n of a geodesic must be < a (we can assume that n ~ 0). If n 0, the geodesic is a meridian. If 0 < n < a, the geodesic is confined to the annular region
=
=
=
=
=
=
IF-,
on the spheroid contained between the circles z == ±bVI and the discussion in Example 8.9 shows that the geodesic 'bounces' between these two circles:
Elementary Differential Geometry
312
If n :::: a, Eq. (10) shows that the geodesic must be the parallel z :::: o. (ii) Let the torus be as in Exercise 4.10. If n :::: 0, the geodesic is a meridian (a circle). If 0 < n < a - b, the g~0desic spirals around the torus:
If n = a - b, the geodesic is either the parallel of radius a - b or spirals around the torus approaching this parallel asymptotically (but never crossing it):
If a - b < n < a + b, the geodesic is confined to the annular region consisting of the part of the torus a distance > n from the axis, and bounces between the two parallels which bound this region:
If
n :::: a + b,
the geodesic must be the parallel of radius a
+ b.
:)OlutlOns
8.19 From Exercise 5.5, the cone is isometric to the 'sector'S of the plane with vertex at the origin and angle 7rV2:
Geodesics on the cone correspond to possibly broken line segments in S: if a line segment meets the boundary of S at a point A, say, it may continue from the point B on the other boundary line at the same distance as A from the origin and with the indicated angles being equal:
o
B
A
(i) TRUE: if two points P and Q can be joined by a line segment in S there is no problem; otherwise, P and Q can be joined by a broken line segment satisfying the conditions above:
o
p
s
314
Elementary Differential Geometry
To see that this is always possible, let PI, 112, ql and q2 be the indicated distances, and let R and S be the points on the boundary of the sector at a distance CP2ql + Plq2)/~ + q~) from the origin. Then, the broken line segment P R followed by SQ is' the desired geodesic. (ii) FALSE: Q
(iii) FALSE: many meet in two points, such as the two geodesics joining P and Q in the diagram in (ii). (iv) TRUE: the meridians do not intersect (remember that the vertex of the cone has been removed), and parallel straight lines that are entirely contained in S do not intersect. (v) TRUE: since (broken) line segments in S can clearly be continued indefinitely in both directions. (vi) TRUE: a situation of the form
o
in which the indicated angles are equal is clearly impossible. But the answer to this part of the question depends On the angle of the COne: if the angle is 0:, instead of 1r/4, lines can self-intersect if 0: < 1r/6, for then the corresponding sector in the plane has angle < 1r:
j15
:)OlutlOns
o
8.20 (i) This is obvious if n > 0 since e- 1jt2 ~ 0 as t ~ O. We prove that t-ne-ljt2 ~ 0 as t ~ 0 by induction on n > O. We know the result if n = 0) and if n > 0 we can apply L'Hopital's rule: t- n t- n - l t-(n-2) r n l' n 1 jt2 = un 2 1 j t2 = 1m - -l-jt"'?~- ) t-.o e t-.o tye t-.o 2 e
r1m
which vanishes by the induction hypothesis. (ii) We prove by induction on n that f) is n-times differentiable with dnf) = { ~3~t) e- 1jt2 dt n 0
if t :/: 0, if t = 0 ,
where P n is a polynomial in t. For n = 0, the assertion holds with Po = 1. Assuming the result for some n > 0, a,n+lf) _ (-3nPn dtn+l t 3n + l
P~
2Pn ) e-ljt2
+ t 3n + t 3n+3
if t :/: 0) so we take P n + 1 = (2 - 3nt 2 )Pn dn+1f) = lim Pn(t) e- 1jt2 = dt n + 1 t-.O t 3n + l
+ t 3 P~.
If t
p. (0) lim en
=0) ljt2
t-.o t 3n + l
= 0
by part (i). Parts (iii) and (iv) are obvious. 8.21 Since 'Y8 is unit-speed, UT,U T = 1) so foR Ur.U r dr = R. Differentiating with respect to f) gives foR Ur .Ur 8 dr = 0, and then integrating by parts gives r:::R
U8, U r[r:::O -
i
0
R U8,U rr
dr = O.
Now u(O, f)) = P for all f), so U8 = 0 when r = O. So we must show that the integral in the last equation vanishes. But, Urr = ~8' the dot denoting the derivative with respect to the parameter r of the geodesic 'Y8, so Urr is parallel to the unit normal N of u; since U8.N = 0, it follows that U8,Urr = O. The first fundamental form is as indicated since ur.u r = 1 and U r .U8 = O.
316
Elementary Differential Geometry
Chapter 9 9.1 This follows from Exercise 7.6. 9.2 This follows from Exercise 7.3. 9.3 Kl + K2 = 0 and Kl = K2 ==> Kl = K2 == O. 9.4 Kl + K2 = 0 ==> K2 = -Kl :::::=:> K = KI K 2 = -KI ::; O. K = 0 KI ::: o Kl = K2 = 0 the surface is part of a plane (by Proposition 6.5). 9.5 By Proposition 7.6, a compact minimal surface would have K > 0 at some point, contradicting Exercise 9.4. 9.6 By the solution of Exercise 7.2, the helicoid a(u, v) = (v cos u, v sin u, AU) has E = A2 + v 2, F = 0, G == 1" L = 0 M = A/(A 2 + V2)1/2 " N = 0 so LG - 2MF+NG H = 2(EG _ F2) = O.
9.7
A straightforward calculation shows that the first and second fundamental forms of at are cosh2 u(du 2 + dv 2 ) and - cos t du 2 - 2 sin t dudv + cos t dv 2I
respectively, so 2 2 H = - costcosh U + cost cosh U 2cosh4 U
-
0 •
From Example 4.10, the cylinder can be parametrised by a(u, v) = "y( u) + va, where "y is unit-speed, II a II = 1 and "y is contained in a plane II perpendicular to a. We have au = l' = t (a dot denoting d/ du), a v = a l so E = 1, F = 0, G = 1; N = t x a, a uu = t = KD, a uv = a uu = 0, so L = KD.(t x a), M = N = O. Now t x a is a unit vector parallel to II and perpendicular to t, hence parallel to Dj so L = ±K and H = ±K/2. SO H = 0 K = 0 "y is part of a straight line the cylinder is part of a plane. 9.9 Using Exercise 9.2, the surface is minimal 9.8
(1
+ gI2)/ + (1 + j2)gll
= 0,
where a dot denotes d/ dx and a dash denotes d/ dy; hence the stated equation. Since the left-hand side of this equation depends only on x and the right-hand side only on y, we must have
1
---,..- - a 1+
j2 - ,
gil
-
1 + g1 2 -
- a,
for some constant a. Suppose that a:/: O. Let r = jj then! = rdr/df and the first equation is rdr/df = a(1 + r 2 ), which can be integrated to give af = tln(1 + r 2 ), up to adding an arbitary constant (which corresponds to translating the surface parallel to the z-axis). So df /dx = ±Je2af - 1,
317
Solutions
which integrates to give f = -~ In cos a(x + b), where b is a constant; we can assume that b = 0 by translating the surface parallel to the xaxis. Similarly, 9 = ~ In cos ay, after translating the surface parallel to the y-axis. So, up to a translation, we have z = ~ In (COS ay ) , a cos ax which is obtained from 5cherk's surface by the dilation (x, y, z) I-t a(x, y, z). IT a = 0, then j = gil = 0 so f = b + ex, 9 = d + ey, for some constants b, c, d, e, and we have the plane z = b + d + cx + ey. 9.10 The first fundamental form is (cosh v + l)(cosh v - cos u)(du 2 + dv 2 ), so a is conformal. By Exercise 7.15, to show that a is minimal we must show that a uu + a vv = 0; but this is so, since
= (sin u cosh VI cos u cosh v, sin ~ sinh ~), a vv = (- sin u eoshv, - cos ucosh v, - sin u sinh ~). 2 2
a uu
(i) a(O, v) = (0, 1 - cosh v, 0), which is the y-axis. Any straight line is a geodesic. (ii) a ('71'" , v) = ('71'", 1 + cosh v, -4 sinh ~), which is a curve in the plane x = '71'" such that Z2
= 16 sinh
2
~ == 8(cosh v-I) ::: 8(y - 2),
Le. a parabola. The geodesic equations are d I2d 1 2 2 dt (Eu) = 2Eu (u + v ), dt (Ev) = 2Ev (u + i?),
where a dot denotes the derivative with respect to the parameter t of the geodesic and E = (cosh v + 1)(cosh v - cos u). When u = '71'", the unit-speed condition is Ev 2 = 1, so v = l/(coshv + 1). Hence, the first geodesic equation is 0 = tEuv2, which holds because Eu = sinu(coshv + 1) :::;- 0 when u = '71'"; and the second geodesic equation is d dt (cosh v
+ 1) = (cosh v + 1) sinh v '11 2 = sinh v V,
which obviously holds. (iii) a (u, 0) = (u - sin u, 1- cos u, 0), which is the cycloid of Exercise 1.7 (in the xy-plane, with a = 1 and with t replaced by u). The second geodesic equation is satisfied because E v = sinh v(2 cosh v + 1 - cos u) = 0 when v = O. The unit-speed condition is 2(1-cos u)u 2 ::: 1, so U = 1/2 sin ¥. The first geodesic equation is (4 sin2 ~ u) = sin uu 2 , Le. (2 sin ~) = cos ¥U, which obviously holds. 9.11 A = a2 + be = -(ad - be) (since d = -a) = - det W = -K.
-it
tt
Elementary Differential Geometry
318
9.12 (i) From Example 9.1, N = (-sech u cos v, -sech u sin v, tanh u). Hence, if N(u) v) = N(u l , Vi), then u = u l since u I-t tanh u is injective, so cos v = cos Vi and sin v = sin Vi, hence v = v'; thus, N is injective. If N = (x, y, z)) then x 2 + y2 = sech 2 u :/: 0, so the image of N does not contain the poles. Given a point (x, y, z) on the unit sphere with x 2 + y2 :/: 0, let u = ±sech-1vx2 + y2, the sign being that of z, and let v be such that cos v ~ -xl vx2 + y2, sin v -y/ vx2 + y2; then) N(u) v) (x, y, z). 2 2 (ii) By the solution of Exercise 7.2, N (A +V )-1/2( -A sin u, Acos u, -v). Since N(u) v) == N (u+ 2k7r, v) for all integers k, the infinitely many points O'(u+ 2k7r, v) = 0'( u) v) + (0,0, 2k7r) of the helicoid all have the same image under the Gauss map. If N (x)y)z), then x 2 + y2 = A2/(A2 + v 2 ):/: 0, so the image of N does not contain the poles. If (x, y, z) is on the unit sphere and x 2 + y2 :/: 0, let v = - AZ/ x 2 + y2 and let u be such that sinu = -x/vx2 + y2) cosu -y/vx2 + y2; then N(u,v) = (x)y)z). 9.13 The plane can be parametrised by 0' (u, v) ub + vc) where {8, b, c} is a 3 right-handed orthonormal basis of R . Then) tp = O'u --,- iO'v = b --,- ic. The conjugate surface corresponds to ilp = c + ib; since {a, c, - b} is also a right-handed orthonormal basis of R 3 ) the plane is self-conjugate (up to a translation). 9.14 tp = (~/(I- 92), !/(1 + 92)) 19) => itp = (~i/(l- 92)) ~i/(1 + 9 2 )) iI9)) which corresponds to the pair il and 9. 9.15 By Example 9.6) tp(() = (sinh () -i cosh (, 1). From the proof of Proposition 9.7,1 = OJ since K < 0, we have (it - 2)71'" > 0 and hence n ~ 3; and if n = 3, then fhnt('Y) (-K)dAa < 71'" so
J1.w'()
=
dA.,.
1, so r < 6. Triangulations for r ::= 3,4 and 5 can be obtained by 'inflating' a regular tetrahedron, octahedron and icosahedron, respectively.
!
tetrahedron
octahedron
icosahedron
325
Solutions
11.7 If such curves exist they would give a triangulation of the sphere with 5 vertices and 5 x 4/2 = 10 edges, hence 2 + 10 - 5 = 7 polygons. Since each edge is an edge of two polygons and each polygon has at least 3 edges, 3F < 2E; but 3 x 7 > 2 x 10. If curves satisfying the same conditions exist in the plane, applying the inverse of the stereographic projection map of Example 5.7 would give curves satisfying the conditions on the sphere, which we have shown is impossible. 11.8 Such a collection of curves would give a triangulation of the sphere with V = 6, E = 9, and hence F = 5. The total number of edges of all the polygons in the triangulation is 2E = 18. Since exactly 3 edges meet at each vertex, going around each polygon once counts each edge 3 times, so there should be 18/3 6 polygons, not 5. 11.9 By Corollary 11.3, IIs KdA = 471'"(1 - g), and by Theorem 11.6, 9 = 1 since S is diffeomorphic to T1 • By Proposition 7.6, K > 0 at some point of S. 11.10 The ellipsoid is diffeomorphic to the unit sphere by the map (x, y, x) I-t (x/a, y/a, z/b), sO the genus of the ellipsoid is zero. Hence, Corollary 11.3 gives IIs K dA = 471'"(1 - 0) = 471'". Parametrising the ellipsoid by a((), 0, < O.
By de Moivre's theorem, 0 = cos ks, j3 = sin ks in both cases. Hence, the angle 't/J between V and is equal to ks, and Definition 11.6 shows that the multiplicity is k. 11.14 If D'(u,v) = a(u,v), where (u, v) I--t (u,v) is a reparametrisation map, u then V = OD'u _+ j3D'v = aail + j3a v ==> a = 0 aau u + j3 aav u , j3 = 0 aa v + j3 aaii v• Hence, a and j3 are smooth if 0 and j3 are smooth. Since the components of the vectors D' u and D'v are smooth, if V is smooth sO are its components. If the components of V = OD'v + j3D'v are smooth, then V.D'u and V.D'v are smooth functions, hence
e
0=
G(V.D'u) - F(V.D'v) EG _ F2 '
j3 = E(V.D'v) - F(V.D'u) EG-F2
are_smooth functions, sO V is sm~oth. _ 11.15 If't/J is the angle between V and we have 't/J - 't/J = f) (up to multiples l of 271'-)j so we must show that fo (")') iJ ds = 0 (a dot denotes d/ ds). This is not obvious since f) is not a well defined smooth function of s (although df)/ ds is well defined). However, p = cos f) is well defined and smooth, since p = II llll II· Now, p = -0 sin f), so we must prove that
e,
e.e/ e e
327
Solutions
fo1(1') ~ds yl_p2
= O. Using Green's theorem, this integral is equal to
Pudu + Pv dv 11' ..)1 - p2
{
1.
a (
Pv
where 11' is the curve in U such that ")'(s) vanishes because
a ( Pv ) au ..)1 _ p2
)
a (
Pu
)
= Jint(1I') au ..)1 - t:? - av ..)1 _ p2 '
=0'(11'(8)); and this line integral
a ( Pu ) = av ..)1 _ p2
11.16 Let F : S ~ R be a smooth function on a surface S, let P be a point of S, let 0' and jj be patches of S containing P, say 0'(110, Vo) = a(Uo"iio) = P, and let I = F 0 0 ' and I = F 0 iT. Then, Iv. _ lu$~ + I"g~, = lu g~ + I" g~, so if lu = I" = 0 at (Vo, Vo), then Iv. = Iv = 0 at (uo, iio). Since lu = I" = 0 at P, we have
Iv
luv.
au av (av)2 = luu (au)2 au + 2/11.v au au + Iv" au '
with similar expressions for tation,
it.
=
Jt 1l.J,
where
Iv.v J =
Ivv. This gives, in an obvious no(~; ft) is the jacobian matrix -of
and
8il
8v
the reparametrisation map (u, v) I-t (u, v). Since J is invertible, it is invertible if 11. is invertible. Since the matrix 11. is real and symmetric, it has eigenvectorsv1,v2, with eigenvalues AI, A2', say, such that V~Vj = 1 if i = j and (J if 'i# j. Then, if v = a1 VI + a2V2 is any vector, where aI, a2 are :scalars, v't1iv = Alai + A2a~jhence, vt 1l.v > 0 (resp. < 0) for all V :/: '0 ),11 :and A2 are both> 0 (resp. both < 0) P is a local minimum '(r~. 1aca1 maximum)jand hence P is a saddle point vt 1l.v :can be!bmth '> I(j) and < 0, depending on the choice of v. Since J is invertible, :a vectar v :/: 0 v = J fj :/: 0 j and fj tit.v = vt Jt 11. Jv = vt 1l.v. iDhe .asser.tioos in the last sentence of the exercise follow from this. 11.17 (i) Ix = 2x - 2y, I y = -2x +8y, sO Ix = I y = 0 ·at the 0i~gin. .!:xz= I = '8 ,so '1J 2 -82 ) . '1J" 'bl ',j..)h '. ,. ,2, I ;cy = - '2 ,'yy n = ,( -2 n IS mverh,'e-S01L:e:0ngm is non-degeneratejand the eigenvalues 5 ± is a local minimum.
(ii)
I. = Iv = 0 and
Ji
=
n
~)
v'i3 of 11. are ibdth > !,@"
at the origin; detJi
'-sa it
= -16 < i0, 'S0
the eigenvalues of 11. are of opposite sign and the origin is a:saddle \}Joint. (iii) Ix = I y = 0 and 11. =. 0 at the origin, which is therefot:e;a-,aegenerate critical point.
Elementary Differential Geometry
328
11.18 Using the parametrisation a in Exercise 4.10 (with a = 2, b = 1) gives I((),
