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Chapter 8 Geodesies and Complete Riemannian Manifolds
CHAPTER 8
Geodesics and Complete Rie m a n n ia n M Q n if0 Ids In the first section the local minimizing properties of geodesics are established with the help of the Gauss lemma, while in the second the Hopf-Rinow theorem on complete Riemannian manifolds is proved. In particular, it is shown that in the complete case geodesics realize global distances. T h e arc length of continuous curves is also discussed [24, 33, 50, 831.
8.1 Geodesics
In 6.3 we defined for any m E M , b = (m,e, , ..., ed) E B ( M ) , maps exp, : M , -+ M , +b : M , -+ B ( M ) , depending on a connexion on B ( M ) and such that was a rather natural lifting of exp, . We noticed that these maps in general were only defined on a neighborhood of the origin of M , , they were C" on this neighborhood, and exp,, was a diffeomorphism of a possibly smaller neighborhood of the origin. I n order for them to be globally defined we must assume that geodesics from m are infinitely extendible. We now assume M is a Riemannian manifold and we obtain Z p , as follows: Let b = (m,fi , ...,fd) E F ( M ) .Let be the Riemannian connexion on F ( M ) , and let p = C p i f i E M , , and let 6 be the unique integral curve of CpiEi with G(0) = b. Then we define G , ( p ) = c?(1). Ei is the restriction of the vector field Ei on B ( M ) (6.1.2) to F ( M ) . exp, is C" and is in fact the radially horizontal lifting of exp, to F ( M ) .I n particular, the image of M , under eWb is inF(M). If @ is the curvature form of the Riemannian connexion, w the
+
145
146
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
solder formonF(M) as defined in 1.3, then we define forms 0, 0, yh on M , as follows: -
0 = exp,*+
@=E$+@ $
~
exp,*
=
W,
and the real-valued forms Oii, O i i , yhi are defined accordingly. T h e structural equations become d#
=
-0#
3 P, 01 + 0.
de = -
(In terms of components, the structural equations become
2 d
d#i
=
eki$k
k=l
d
doij =
-
k=l
+
OikOkj
@ij.)
If P E M , , p = ray from m to p , u = exp,o p, q b p = (m,fit, ..., fdt) (so fit is the parallel translate of fi along a), then if s, t E (W7')P 9
(4 (b)
d expms
j
=
2 #?(4,c
( d e ~ ~ rdexprnt) n~, = <#(s), #(t)>
I I d expms I l2
=
I I #(s) I l2 =
=
2 #i(s)
+i(t)
2 #N.
[(b) is an immediate consequence of (a), which in turn follows easily from the definition of 4.1 Thus, if xl, ..., xd is the dual base to the f i , s = Zs,D,((p), then &(s) - si is a measure of the difference between parallel translation in M and M , . If p = Zpifi E M , , s = c ZppiD,,(p)E ( M m ) p, that is, s is tangent to the ray p above, then (c)
(4
= CPi
qS)= 0.
8.1. Geodesics
147
This says that the lengths of radial vectors are preserved under d exp,, and that tangents to horizontal curves in F ( M ) are horizontal tangents. A C" rectangle in M is a map Q of a rectangle [a, b] x [c, d ] in R2 into M which can be extended to a C" map of a neighborhood of the rectangle. Q(a, c) is the initial corner of Q , while the base of Q is the (a, d )
/b,d)
FIG. 28.
curve T defined by Q o j c , where j c ( t ) = ( t , c). More generally, the curves T , = Q o j , are called longitudinal, while the curves = Q o 'j, ('jz(t)= (x, t ) ) are transversal. T h e "vector field" V defined along the base T by V(s) = r S * ( c )is the associated vectorjeld of Q (it is actually a curve in the tangent bundle of M ) , and Q is an associated rectangle of V . Canonical lifting. If Q is a C" rectangle in a Riemannian manifold M , and if F ( M ) has the Riemannian connexion 4, then for every f~ n-I(Q(a, c)) there exists a unique C" rectangle in F ( M ) with initial corner f such that (a) Q = n o Q,that is, Q is a lifting of Q , (b) +(Qo j,*) = 0, that is, the longitudinal curves of Q are horizontal, and (c) +(Qo 'ja*)= 0, that is, the initial transversal curve is horizontal. Problem 1. Prove the existence, uniqueness, and differentiability of the canonical lifting of a C" rectangle.
If Q is a C" rectangle in M , Q : [a, b] x [c, d ] -+ M , whose longitudinal curves are geodesics and the tangents to these geodesics are all of the same length, then if V is the associated vector field, the function (T,* , V ) is a constant. In par( t V) ( t ) for all t E [a, b]. ticular, if TC*(a)1 V ( a ) , then ~ ~ * 1
Theorem 1 (Gauss' lemma).
148
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
Proof.
C$Q
= dwQ =
Let Q be the canonical lifting to f E F ( M ) of Q. Also, write Q*$, wQ = & * w . T h e first structural equation becomes
-4QwQ, which when applied to (Dl, D,)
QwQ(D2) - D,UQ(D1)- wQ([D1 9
gives
Del) = -+Q(D,) wQ(Dz)
+ $Q(Dz)wQ(D,).
Now [Dl ,D,]= 0, and C$O(Dl)= 0 since the longitudinal curves of are horizontal; so we have, taking inner products with wQ(D,), (D1wQ(D2),w Q ( W - (DzwQ(D1), wQ(D1)> =
Q
(CQ(D2)W Q ( D l ) > w Q ( W
=o
since dQ(D2)is skew-symmetric. Now wQ(D,)oj, is constant since the curve T, is a geodesic, so
also,
Dl (wQ(D,), w
D,( w Q ( 4 ) ,
Q ( W =
wQ(D1)) =
PQm, wQ(D1)>;
(D
2 (D2wQ(D1), wQ(D1)>.
But (wQ(Dl), wQ(Dl))is constant, since we assumed the tangents to the longitudinal curves all had the same length, so Dz(wQ(D1), w Q ( W = 0.
Hence, the above equation becomes Dl(wQ(D2),wQ(Dl)> = 0. But clearly, along the base curve (wQ(D2),wQ(Dl)) = ( V , T,*). QED Corollary. Let p E M , , p = ray from 0 to p , cr = exp,, o p, and s E (M& . Then s Ip (in the Euclidean inner product) implies that d exp,s 1a.
8.1. Geodesics
149
This follows by applying Gauss’ lemma to a rectangle Q whose initial transversal is the degenerate curve m E M and the longitudinal curves are the images under exp, of the rays from 0 in M,. I n particular, the base is u. T h e exact construction is left to the reader. (This rectangle may be described briefly as “a piece of pie.”) T h e following result expresses the fact that locally geodesics minimize arc length among broken C“ curves. Let B C M , be a ball about 0 on which exp, is a diffeomorphism, let p E B , p = ray from 0 to p, a = exp, o p, and let T be any broken C“ curve from m to exp,,p in M . Then I T 1 3 I a 1, and equality holds only if T is a broken C“ reparametrization of a.
Theorem 2.
Proof. Let xl, ..., xa be dual to f l , ...,fa on M , and define the following objects on either B or B‘ = exp,B: f = r o expi]
T
=
2
=
d exp,T o expm-l, defined on B’ - {m}.
Xi%,
, the radial unit vector field defined on B - (0).
+
If q E M , , s E (M,,Jq, then we write s = sT sN where sT is a multiple of T(q) and sN 1T(q).For t E M,, , b E B‘, we write similarly t = tT t N ,where t T is a multiple of T ( b ) and t N _L T(b). We have d exp,,s = d expmsT d explnsN, and from the corollary above, d expl,,sN 1d exp,sT . Hence,
+
+
. Also, from (b), p. 146, (ii) I I d expnis, I I = I I S T I I . (i) d exp,sT
=
(d exp,s),
Let [a, b] be the interval on which T is defined, and let c E [a,b] be the smallest number such that P ( T ( c ) ) = r(p) = I a I. Now define a curve 7 on [a, c] by 7)=ao---
7
IIPII
Or.
150
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
Noting that
1 1 u*
-
jj =
1
__
II P II
/ I p 11, we have
d a o dr o d exp,-'(T*)
hence,
this is true on [a, c]. Therefore, if
T'
= T jLa,cl
, we have
~ ~ ~ > ~ ~ ' ~ > ~ ~ l > l ~ l ,
FIG. 30.
as asserted. If I T I = I u 1, then we clearly must have T * ( u ) = 0 for u > c and T*,, = 0, from which it follows that T and u have the same image. QED T h e square of the distance to m, p(m, Z)z, is a C" function on B'. Corollary 1. Proof.
It equals Z ( x i o exp,-1)2. (Z is the identity map on M.)
8.1. Geodesics Corollary 2.
for c
< radius
Corollary 3.
1TI
=
151
If B ( m , c ) denotes the ball of radius c about m, then of B, exp,(B(O, c)) = B ( m , c). If
T
p(m, n ) , then
is a broken C" curve from m to n such that T is a broken C" reparametrization of a geodesic.
Proof. T minimizes arc length from m to n, so T locally minimizes arc length and, by the theorem, is locally a broken C" reparametrization of a geodesic. This is enough. Q E D
Lemma 1. Let m E M , 0 a ball C M,, , such that exp, is a diffeomorphism : 0 -+ U , y : (a, b) + U a C" curve in U. Suppose that r = p(m, y ) has an argument t such that r'(t) = 0. Then the geodesic in U from m to y ( t ) is perpendicular to y J t ) at y ( t ) . Proof. This is an immediate consequence of Gauss' lemma applied to the lifting of the curves to M7,,and the comparable fact for Euclidean space.
Let N, P be submanifolds of M , u a geodesic from n E N to p E P such that I u I = p(P, N ) . Then u is perpendicular to both N and P. Theorem 3.
Proof. It is obvious that a piece of u minimizes arc length from N o r P to any point on u. For such points which are sufficiently close to N or P lemma 1 then shows that u is perpendicular to curves in N or P which pass through n or p , respectively, and hence u is perpendicular to N, and P, . Q E D If N is a submanifold of a Riemannian manifold M , let I ( N ) , the normal bundle to N , be defined by
I ( N ) = {(n,t ) E T ( M ) 1 t E M , for some n E N and t 1N,} [cf. 3.3(4)]. Show that I ( N ) is a submanifold of T ( M ) and that Exp I I , N ) is nonsingular on the trivial cross section of I ( N ) . T h e tubular neighborhood, L r ( N ) , of N with radius r in I ( N ) is the open set of I ( N ) which intersects each fibre in the open ball of radius r about the origin of the fibre. Problem 2 .
If N is a compact submanifold of the Riemannian manifold M , then there exists an r > 0 such that exp maps I , ( N )
Theorem 4.
152
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
diffeomorphically. T h e image of 1r ( N ) ,called a tubular neighborhood of N in M , has the property that all its points are joined to N by unique geodesics which minimize arc length to N . T h e proof is left as an exercise. A local version of this gives us the existence of normal coordinates for N in the following sense. Let n E N . A coordinate system at n in M is normal for N if the points of N correspond to part of a linear subspace of dimension equal to that of N and the straight lines perpendicular to this subspace correspond to geodesics perpendicular to N . Problem 3 . If i : N -+M is an immersion (diis an isomorphism into for every point), define a normal bundle of the immersion and tubular neighborhoods in it. When N is compact show that Exp : T ( M )-+ M gives rise to an immersion of l , ( N )into M , some r . Give an example to show that the property of having unique minimizing geodesics to i ( N )need not hold for points of the immersion of L r ( N ) . Problem 4. Consider the flat 2-dimensional torus obtained by identifying opposite sides of the unit square in R2 having opposite corners (0, 0) and (1, 1). Sketch the locus of points at distance 2/3 from the corner point. If we obtain a noncomplete (see below) manifold from this torus by removing the closed line segment from (1/4, 0) to (1/4, 1/2), how does this change the locus ? Problem 5. Show that a Riemannian covering map is distancedecreasing.
8.2 Complete Riemannian Manifolds
Assume M is a Riemannian manifold. We have a map E : T ( M )+ M x M , given by: E(m, t ) = (m, exp,t). For E to be defined on all of T ( M ) we must assume that all geodesics are infinitely extendible. But in any case, E is defined on a neighborhood of the zero cross section of T ( M )and is in fact C" there. We also have Lemma 2. For each m E M , dE is an isomorphism on T(M)cm,ol , so by the inverse function theorem, E is a diffeomorphism of a neighborhood of (m,0) onto a neighborhood of (m, m).
8.2. Complete Riemannian Manifolds
153
Proof. It is sufficient to prove that dE maps T(M),,,,,, onto M x M(,,,,,, , since the dimensions are the same. Let ni : M x M + M be the projection onto the ith factor i = 1, 2. Now we know that , which is given by E I n - ~ ( , n ) ( m , t ) = (m , exp,t), maps E (MvJo onto the tangent space to n;l(m) at (m,m). We conclude by showing that dE maps the tangents to the zero cross section of T ( M ) onto the tangent space to the diagonal of M x M , which suffices since M x M(,,,,,, is clearly spanned by tangents to ncl(m) and tangents to the diagonal. Let D : M + M x M be the diagonal map, D(m) = (m , m). Then we have E I(zero cross section) = D 0 I(zero cross section) . dn is onto, and dD is onto the tangent space to the diagonal. QED
Let C be any compact subset of M . Then there exists is defined on B(O(m),c) and maps it diffeomorphically onto B( m , c), where O(m ) is the origin in Mm . Proof. We first note that if all geodesics are not infinitely extendible, exp, may only be defined in a neighborhood of O(m). However, by the theory of differential equations, it is clear that for each m E M exp, is defined on a ball whose radius is a continuous function of m, and hence we may take a c1 such that exp,n is defined on B (O (m ),cl) for every m E C. By corollary 2 to theorem 2, we need only show that there exists c > 0 such that, for each m E C, exp, is a diffeomorphism on B(O(m),c). Lemma 2 essentially says this locally, and hence by compactness the result follows. However, we must first translate lemma 2, which says precisely that for every m E C, there exists a neighborhood P, of (m, 0) which is mapped diffeomorphically onto a neighborhood of (m, m). Therefore, d exp, is regular (that is, 1 - 1 onto) and hence for every n E n (Pm), exp, is a diffeomorphism on the set {t E M , I ( n , t ) E PnL}= PnL,,;so we wish to show there is a c,, > 0 such that the ball, B(O(n),cm),in the Riemannian metric is contained in P,n,, , for all n E n(P,). Using the facts that T ( M ) has a local product structure with the topology of the fibre being given by the Euclidean metric, that the Riemannian and Euclidean metrics are equivalent on each tangent space (see lemma 7.1), and that the Riemannian metric is continuous, one finds that P,, contains a neighborhood of the form Lemma 3.
ac
> 0 such that, for each m E C, exp,
=
{(% t ) I n
E u r nI
II t II < C , J ,
154
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
where Urn is a neighborhood of m and c, > 0. Hence for n E U,,, , exp, maps B(O(n), ern) diffeomorphically onto B(n, cm). By compactness, C is covered by a finite number of U r n ,to each of which corresponds a c , ~ .Letting c = min c, gives the desired result. QED It now follows that the square of the distance function, p2, is C“ on a neighborhood of (m, m) in M x M . Theorem 5. (Hopf-Rinow [40, 77, 931). Consider the following
conditions on a connected Riemannian manifold M :
(a) M is complete. (b) All bounded closed subsets of M are compact. (c) For some point m of M , all geodesics from m are infinitely extendible. (d) All geodesics are infinitely extendible. (e) Any m, n E M can be joined by a geodesic whose arc length equals p(m, n). T h e conditions (a)-(d) are equivalent, and they imply (e). Problem 6. Find an example to show that (e) does not imply (a). Also find an example to show that a minimizing geodesic between two points need not be unique; in fact, there may be infinitely many. Notice that (c) can be stated “exp,, is defined on all of Mm”, and (d) can be stated “ T h e Riemannian connexion is complete’’ (6.3).
Proof. In any metric space it is true that (b) implies (a). That (d) implies (c) is trivial, and that (a) implies (d) follows from extendibility theorems in differential equations (see appendix). We therefore have only to prove the implications from (c) to (b) and from (b) to (e). We now fix m, E M , and define, for any real Y > 0,
8, = {m E &I I p(m, ,111)
< r } = B(m, , r ) ; -
E,. = {m E B, I m can be joined to m, by a geodesic of length p(m, m,)).
I t suffices, assuming (c) for m, , to prove (1) E, is compact and (2) E, = 8,;for any bounded subset is in a 8,, so by (1) and (2) a closed bounded subset is compact, which is (b); then since (b) implies (d), we can use any m for m, , so that (2) implies (e).
8.2. Complete Riemannian Manifolds
I55
which is clearly compact, since it is closed and bounded. Compactness of E , then follows from the fact that E , = exptnOErand the continuity of expm,. QED For the proof of (2) we need the following: Lemma 4. If E , = 8, for some r , and if p(m, , n) > r, then there p(m, n). exists m such that p(m, , m) = r and p(m, , n) = r
+
Proof. For K = 1, 2, ..., choose a, , a broken C" curve from mo to n with I a, I < p(m, , n) l/k (this is possible by the definition of p). Let mk be the last point on a, in B , , so p(m,, m,) = r . By com-
+
FIG. 31.
pactness of E , = 8,, the m k have a limit point m,and by passing to a subsequence we may assume {m,} converges to m. Now p(m0 7 n )
<
dm0
9
m)
+
f(H4
n)
=
r
+ p(m, 4.
On the other hand, p(m, , n) > 1 a , I - l/k = I part of ak from m, to m k I I part of ak from mk to n I - l / k r p(m, , a) - l/k, which has limit r p(m, n), so p(m, , n ) >, Y p(m, n), so equality holds. This proves lemma 4. We now prove (2) using the connectedness of the nonnegative real numbers; that is, we show:
+
(i) E, (ii) E , (iii) E,!
+
> + +
8,. = B , , r' < r , implies E,,! = 8,f . = 8,t for all r' < r implies E , = 8 , .
=
156
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
(iv) E , = 8, implies there exists c Er+c =
> 0 such that
Bwc-
(i) and (ii) are trivially true. Proof o f (iii). Let m E B , . If m E B,t for r' < r, then by hypothesis m E E,) C E , . Hence, we assume p(m, , m) = r . By lemma 4 we may choose a sequence (m,} which has limit m, where mk E 8, , r k < r . Then by hypothesis, mk E ErkC E, , so m E E , by compactness of E , , which proves (iii). Proof of (iv). By lemma 3, there exists c > 0 such that for each m E 8, , exp, maps B(O(m),2c) diffeomorphically onto B(m, 2c), since E, is compact. Let n E B,+, . We show n E E,+, . By lemma 4 there p(m, n), where p(m, , m) = r . exists m E 8,such that p(m, , n ) = r Therefore, p(m, n) c, so there exists a geodesic y from m to n with I y j = p(m, n). Let u be a geodesic from m, to m with I u 1 = p(m, , m) = r . Then u y is a broken C" curve from m, to n with I 0 y I = I 0 I I y I = r p(m,n) = p(m0, n), so by corollary 3 to theorem 2, u y can be reparametrized as a geodesic. Hence, n E E,,, . This completes the proof of the theorem. It has been shown by K. Nomizu and H. Ozeki [67] that every connected paracompact manifold admits a complete Riemannian metric; furthermore, they show that if every Riemannian metric is complete then the manifold is compact. T h e converse, that a compact metric space is complete, is well known.
+
<
+
++
+
+
Problem 7. Construct a noncomplete connected Riemannian manifold of infinite diameter such that no two points at distance greater than 1 from each other may be connected by a curve which minimizes arc length.
Let i : N + M be an isometric imbedding. Let p be the Riemannian distance on M , p' = p o i, p " the Riemannian distance on N . Give examples to show that all eight possibilities for these metrics to be complete or not may occur. Problem 8.
Problem 9. Show that for a connected Lie group which admits a left and a right invariant metric the exponential map is onto. Problem 10. Let M and N be complete Riemannian manifolds. Show that the Riemannian product M x N is complete. Completeness of affine connexions is also a property preserved under products.
8.2. Complete Riemannian Manifolds Problem 11.
157
Let M be a complete Riemannian manifold.
(a) Then any Riemannian covering of M is complete. (b) For every m E M there is a geodesic segment starting and ending at m in every homotopy class of loops at m. Problem 12. Give examples to show that the result in problem 1 l(b) may or may not hold if M is not complete. Problem 13. Let M be a complete Riemannian manifold which is not cannot be a C" function on all simply connected. Show that p(m, of M . Example. Consider a mechanical system with a finite number of degrees of freedom and no elements which dissipate energy. Then the conjiguration space is a manifold M which is a mathematical model of the totality of all positions of the system. A curve in M is generally thought of as occurring in time, so that a tangent vector is an assignment of velocities to the elements of the system consistent with the constraints of the system. T h e phase space of the system is T ( M ) . T h e kinetic energy given by an assignment of velocities is a positive definite quadratic form on each Mm , and thus gives a Riemannian metric on M . A force field on M is a 1-form 9; the integral of this 1-form on a curve is the amount of work done in traversing the curve. If it is a conservative force field then 9 = - d V , where V is the potential energy. If D is the covariant derivative symbol of the metric and X is the vector field defined by 2 ( X , Y ) = O( Y ) for every Y , then Newton's laws of motion for the system become
where y is a curve parametrized by time. I n particular, when the motion is free the path of the motion is a geodesic and the time is proportional to the length of the geodesic. T h e Riemannian manifold is thus complete if the system will coast indefinitely when given an arbitrary push. I n this case the system may be started in any one configuration with kinetic energy 1 in such a way that it attains another given configuration by free motion with an elapse of time equal to the distance between the two points on M . More specifically, the configuration space of a rigid body constrained
158
8. GEODESICS AND COMPLETE RIEMANNIANMANIFOLDS
to rotate about its center of gravity is homeomorphic to P3 = SO(3).
T h e kinetic energy metric is invariant if and only if the ellipsoid of inertia is spherical. 8.3 Continuous Curves
Let y be a continuous curve, y : [a, b] -+ M , and let a = to < t ,
< t , < .'. < t , < b
=
t,+,
be a sequence of numbers between a and b. Then we define the arc length of y by I y I = sup { p(y(ti), y(ti+,)) I all such sequences ti}. Proposition 1. If y : [a, b] -+ M is a continuous curve from m to n with I y I = p(m, n), then y is a continuous reparametrization of a geodesic. This implies that geodesics locally minimize arc length among continuous curves. Proof. Pick a c > 0 by lemma 3 with respect to the image of y . Take a < t, < t, < b such that p(y(tl), y(t,)) < c/2, so there exists a geodesic u from y ( t l ) to y(t,) with I u 1 = p(y(tl),y(t,)). We claim that for every t E (tl , t J , y ( t ) lies on u. For choose geodesics u1 , u, from r(t1) to y(t), y ( t ) to y(t,), respectively, with I 0 1 I = p(y(t,), y ( t ) ) and 1 u, 1 = p(y(t), y(tz)). Then since y clearly locally minimizes arc length, we have 1 0 1 0 2 1 = p(y(t1), y ( t ) ) p(y(t), y(t,)) = I y from r(t1) to y(t,) I I y from y ( t ) to y(t2) I = p(y(t,), y(t,)), so by the theorem 2, I u, u, I is a broken C" reparametrization of u, which proves y ( t ) lies on the image of u. Hence, y is locally a continuous reparametrization of a geodesic; so in particular it is a reparametrization of a broken C" curve. T h e result then follows from corollary 3 to theorem 2.
+ +
+
Proposition 2.
+
If y is a broken C" curve, then the two definitions of
arc length agree. Proof. Let { y } be the length of y as a continuous curve, retaining I y 1 as the notation for the integral of tangent lengths. Then it follows trivially from the definition of distance that { y } 1 y I. We
<
also have, easily, that both definitions are additive:
{r+d={r)S{4, I r + u l
=
IrlS
101.
8.3. Continuous Curves
159
+
Now suppose y is a curve for which { y } 12 = I y I, 12 > 0. Then by splitting y in half, we must have that the discrepancy on one of the halves is at least k / 2 . By repeatedly halving we get a nested sequence u, such that the discrepancy of y restricted of parameter values s, to [s,, u,] is at least k/2", and u, - s, = c/2,, where c = b - a, [a, b] is the interval of definition of y. Let t be the common limit of s, and u, , and let t , be a number in [s, , u,] such that
<
which exists by the mean value theorem. Then
Now reparametrize y so that t 2"/c, and take limits, getting
=
0, multiply this inequality by
Let xi be normal coordinates at y(O), and fi = xi o y . Then we have p(y(s), y(0)) = ( Zfi(s)2)1/2= ( Xfi(Ois)2)1/2 I s I, where 0 < Bi < 1, by the mean value theorem. Thus if we write g(4
=
P ( Y ( 4 , r(O))/ls I
and d u n ) un - As,) sn = g(Un)(un - s n )
+ ( d u n ) - d-4)sn,
we get
Since limn+mg(u,) = 1 1 y.+(O) 1 1 , this shows
II Y*(O> II 2 K i C + II Y*(O) 11, which contradicts 12
> 0. QED
Problem 14. Let y : [a, b] -+ M be a continuous curve with finite length in a Riemannian manifold. Show that y can be uniformly approximated by broken geodesics.
160
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
Problem 15. Let 4 : M ---t N be a map between Riemannian manifolds which is onto and preserves distances. Prove that 4 is an isometry, (See [50],p. 169.) This says that the topological metric of a Riemannian manifold determines both the Riemannian structure and the differential structure. Problem 16. Let 4 : M + N be a C" map of complete Riemannian manifold M onto Riemannian manifold N such that for every m E M , M,, decomposes orthogonally into subspaces V , and H,, V and H C" distributions, where V , = ker(d4,) and dr$, restricts to an isometry from H , onto NbC,,.Show that: (a) If y is a C" curve in N and +(m) = y(O), then there is a unique lift 7 o f y to M s u c h that Y(0) = m,y = 4 o 7, and I y I = I 7 I. (b) T h e lift, as in (a), of a geodesic is a geodesic.
Geodesics and Complete Rie m a n n ia n M Q n if0 Ids In the first section the local minimizing properties of geodesics are established with the help of the Gauss lemma, while in the second the Hopf-Rinow theorem on complete Riemannian manifolds is proved. In particular, it is shown that in the complete case geodesics realize global distances. T h e arc length of continuous curves is also discussed [24, 33, 50, 831.
8.1 Geodesics
In 6.3 we defined for any m E M , b = (m,e, , ..., ed) E B ( M ) , maps exp, : M , -+ M , +b : M , -+ B ( M ) , depending on a connexion on B ( M ) and such that was a rather natural lifting of exp, . We noticed that these maps in general were only defined on a neighborhood of the origin of M , , they were C" on this neighborhood, and exp,, was a diffeomorphism of a possibly smaller neighborhood of the origin. I n order for them to be globally defined we must assume that geodesics from m are infinitely extendible. We now assume M is a Riemannian manifold and we obtain Z p , as follows: Let b = (m,fi , ...,fd) E F ( M ) .Let be the Riemannian connexion on F ( M ) , and let p = C p i f i E M , , and let 6 be the unique integral curve of CpiEi with G(0) = b. Then we define G , ( p ) = c?(1). Ei is the restriction of the vector field Ei on B ( M ) (6.1.2) to F ( M ) . exp, is C" and is in fact the radially horizontal lifting of exp, to F ( M ) .I n particular, the image of M , under eWb is inF(M). If @ is the curvature form of the Riemannian connexion, w the
+
145
146
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
solder formonF(M) as defined in 1.3, then we define forms 0, 0, yh on M , as follows: -
0 = exp,*+
@=E$+@ $
~
exp,*
=
W,
and the real-valued forms Oii, O i i , yhi are defined accordingly. T h e structural equations become d#
=
-0#
3 P, 01 + 0.
de = -
(In terms of components, the structural equations become
2 d
d#i
=
eki$k
k=l
d
doij =
-
k=l
+
OikOkj
@ij.)
If P E M , , p = ray from m to p , u = exp,o p, q b p = (m,fit, ..., fdt) (so fit is the parallel translate of fi along a), then if s, t E (W7')P 9
(4 (b)
d expms
j
=
2 #?(4,c
( d e ~ ~ rdexprnt) n~, = <#(s), #(t)>
I I d expms I l2
=
I I #(s) I l2 =
=
2 #i(s)
+i(t)
2 #N.
[(b) is an immediate consequence of (a), which in turn follows easily from the definition of 4.1 Thus, if xl, ..., xd is the dual base to the f i , s = Zs,D,((p), then &(s) - si is a measure of the difference between parallel translation in M and M , . If p = Zpifi E M , , s = c ZppiD,,(p)E ( M m ) p, that is, s is tangent to the ray p above, then (c)
(4
= CPi
qS)= 0.
8.1. Geodesics
147
This says that the lengths of radial vectors are preserved under d exp,, and that tangents to horizontal curves in F ( M ) are horizontal tangents. A C" rectangle in M is a map Q of a rectangle [a, b] x [c, d ] in R2 into M which can be extended to a C" map of a neighborhood of the rectangle. Q(a, c) is the initial corner of Q , while the base of Q is the (a, d )
/b,d)
FIG. 28.
curve T defined by Q o j c , where j c ( t ) = ( t , c). More generally, the curves T , = Q o j , are called longitudinal, while the curves = Q o 'j, ('jz(t)= (x, t ) ) are transversal. T h e "vector field" V defined along the base T by V(s) = r S * ( c )is the associated vectorjeld of Q (it is actually a curve in the tangent bundle of M ) , and Q is an associated rectangle of V . Canonical lifting. If Q is a C" rectangle in a Riemannian manifold M , and if F ( M ) has the Riemannian connexion 4, then for every f~ n-I(Q(a, c)) there exists a unique C" rectangle in F ( M ) with initial corner f such that (a) Q = n o Q,that is, Q is a lifting of Q , (b) +(Qo j,*) = 0, that is, the longitudinal curves of Q are horizontal, and (c) +(Qo 'ja*)= 0, that is, the initial transversal curve is horizontal. Problem 1. Prove the existence, uniqueness, and differentiability of the canonical lifting of a C" rectangle.
If Q is a C" rectangle in M , Q : [a, b] x [c, d ] -+ M , whose longitudinal curves are geodesics and the tangents to these geodesics are all of the same length, then if V is the associated vector field, the function (T,* , V ) is a constant. In par( t V) ( t ) for all t E [a, b]. ticular, if TC*(a)1 V ( a ) , then ~ ~ * 1
Theorem 1 (Gauss' lemma).
148
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
Proof.
C$Q
= dwQ =
Let Q be the canonical lifting to f E F ( M ) of Q. Also, write Q*$, wQ = & * w . T h e first structural equation becomes
-4QwQ, which when applied to (Dl, D,)
QwQ(D2) - D,UQ(D1)- wQ([D1 9
gives
Del) = -+Q(D,) wQ(Dz)
+ $Q(Dz)wQ(D,).
Now [Dl ,D,]= 0, and C$O(Dl)= 0 since the longitudinal curves of are horizontal; so we have, taking inner products with wQ(D,), (D1wQ(D2),w Q ( W - (DzwQ(D1), wQ(D1)> =
Q
(CQ(D2)W Q ( D l ) > w Q ( W
=o
since dQ(D2)is skew-symmetric. Now wQ(D,)oj, is constant since the curve T, is a geodesic, so
also,
Dl (wQ(D,), w
D,( w Q ( 4 ) ,
Q ( W =
wQ(D1)) =
PQm, wQ(D1)>;
(D
2 (D2wQ(D1), wQ(D1)>.
But (wQ(Dl), wQ(Dl))is constant, since we assumed the tangents to the longitudinal curves all had the same length, so Dz(wQ(D1), w Q ( W = 0.
Hence, the above equation becomes Dl(wQ(D2),wQ(Dl)> = 0. But clearly, along the base curve (wQ(D2),wQ(Dl)) = ( V , T,*). QED Corollary. Let p E M , , p = ray from 0 to p , cr = exp,, o p, and s E (M& . Then s Ip (in the Euclidean inner product) implies that d exp,s 1a.
8.1. Geodesics
149
This follows by applying Gauss’ lemma to a rectangle Q whose initial transversal is the degenerate curve m E M and the longitudinal curves are the images under exp, of the rays from 0 in M,. I n particular, the base is u. T h e exact construction is left to the reader. (This rectangle may be described briefly as “a piece of pie.”) T h e following result expresses the fact that locally geodesics minimize arc length among broken C“ curves. Let B C M , be a ball about 0 on which exp, is a diffeomorphism, let p E B , p = ray from 0 to p, a = exp, o p, and let T be any broken C“ curve from m to exp,,p in M . Then I T 1 3 I a 1, and equality holds only if T is a broken C“ reparametrization of a.
Theorem 2.
Proof. Let xl, ..., xa be dual to f l , ...,fa on M , and define the following objects on either B or B‘ = exp,B: f = r o expi]
T
=
2
=
d exp,T o expm-l, defined on B’ - {m}.
Xi%,
, the radial unit vector field defined on B - (0).
+
If q E M , , s E (M,,Jq, then we write s = sT sN where sT is a multiple of T(q) and sN 1T(q).For t E M,, , b E B‘, we write similarly t = tT t N ,where t T is a multiple of T ( b ) and t N _L T(b). We have d exp,,s = d expmsT d explnsN, and from the corollary above, d expl,,sN 1d exp,sT . Hence,
+
+
. Also, from (b), p. 146, (ii) I I d expnis, I I = I I S T I I . (i) d exp,sT
=
(d exp,s),
Let [a, b] be the interval on which T is defined, and let c E [a,b] be the smallest number such that P ( T ( c ) ) = r(p) = I a I. Now define a curve 7 on [a, c] by 7)=ao---
7
IIPII
Or.
150
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
Noting that
1 1 u*
-
jj =
1
__
II P II
/ I p 11, we have
d a o dr o d exp,-'(T*)
hence,
this is true on [a, c]. Therefore, if
T'
= T jLa,cl
, we have
~ ~ ~ > ~ ~ ' ~ > ~ ~ l > l ~ l ,
FIG. 30.
as asserted. If I T I = I u 1, then we clearly must have T * ( u ) = 0 for u > c and T*,, = 0, from which it follows that T and u have the same image. QED T h e square of the distance to m, p(m, Z)z, is a C" function on B'. Corollary 1. Proof.
It equals Z ( x i o exp,-1)2. (Z is the identity map on M.)
8.1. Geodesics Corollary 2.
for c
< radius
Corollary 3.
1TI
=
151
If B ( m , c ) denotes the ball of radius c about m, then of B, exp,(B(O, c)) = B ( m , c). If
T
p(m, n ) , then
is a broken C" curve from m to n such that T is a broken C" reparametrization of a geodesic.
Proof. T minimizes arc length from m to n, so T locally minimizes arc length and, by the theorem, is locally a broken C" reparametrization of a geodesic. This is enough. Q E D
Lemma 1. Let m E M , 0 a ball C M,, , such that exp, is a diffeomorphism : 0 -+ U , y : (a, b) + U a C" curve in U. Suppose that r = p(m, y ) has an argument t such that r'(t) = 0. Then the geodesic in U from m to y ( t ) is perpendicular to y J t ) at y ( t ) . Proof. This is an immediate consequence of Gauss' lemma applied to the lifting of the curves to M7,,and the comparable fact for Euclidean space.
Let N, P be submanifolds of M , u a geodesic from n E N to p E P such that I u I = p(P, N ) . Then u is perpendicular to both N and P. Theorem 3.
Proof. It is obvious that a piece of u minimizes arc length from N o r P to any point on u. For such points which are sufficiently close to N or P lemma 1 then shows that u is perpendicular to curves in N or P which pass through n or p , respectively, and hence u is perpendicular to N, and P, . Q E D If N is a submanifold of a Riemannian manifold M , let I ( N ) , the normal bundle to N , be defined by
I ( N ) = {(n,t ) E T ( M ) 1 t E M , for some n E N and t 1N,} [cf. 3.3(4)]. Show that I ( N ) is a submanifold of T ( M ) and that Exp I I , N ) is nonsingular on the trivial cross section of I ( N ) . T h e tubular neighborhood, L r ( N ) , of N with radius r in I ( N ) is the open set of I ( N ) which intersects each fibre in the open ball of radius r about the origin of the fibre. Problem 2 .
If N is a compact submanifold of the Riemannian manifold M , then there exists an r > 0 such that exp maps I , ( N )
Theorem 4.
152
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
diffeomorphically. T h e image of 1r ( N ) ,called a tubular neighborhood of N in M , has the property that all its points are joined to N by unique geodesics which minimize arc length to N . T h e proof is left as an exercise. A local version of this gives us the existence of normal coordinates for N in the following sense. Let n E N . A coordinate system at n in M is normal for N if the points of N correspond to part of a linear subspace of dimension equal to that of N and the straight lines perpendicular to this subspace correspond to geodesics perpendicular to N . Problem 3 . If i : N -+M is an immersion (diis an isomorphism into for every point), define a normal bundle of the immersion and tubular neighborhoods in it. When N is compact show that Exp : T ( M )-+ M gives rise to an immersion of l , ( N )into M , some r . Give an example to show that the property of having unique minimizing geodesics to i ( N )need not hold for points of the immersion of L r ( N ) . Problem 4. Consider the flat 2-dimensional torus obtained by identifying opposite sides of the unit square in R2 having opposite corners (0, 0) and (1, 1). Sketch the locus of points at distance 2/3 from the corner point. If we obtain a noncomplete (see below) manifold from this torus by removing the closed line segment from (1/4, 0) to (1/4, 1/2), how does this change the locus ? Problem 5. Show that a Riemannian covering map is distancedecreasing.
8.2 Complete Riemannian Manifolds
Assume M is a Riemannian manifold. We have a map E : T ( M )+ M x M , given by: E(m, t ) = (m, exp,t). For E to be defined on all of T ( M ) we must assume that all geodesics are infinitely extendible. But in any case, E is defined on a neighborhood of the zero cross section of T ( M )and is in fact C" there. We also have Lemma 2. For each m E M , dE is an isomorphism on T(M)cm,ol , so by the inverse function theorem, E is a diffeomorphism of a neighborhood of (m,0) onto a neighborhood of (m, m).
8.2. Complete Riemannian Manifolds
153
Proof. It is sufficient to prove that dE maps T(M),,,,,, onto M x M(,,,,,, , since the dimensions are the same. Let ni : M x M + M be the projection onto the ith factor i = 1, 2. Now we know that , which is given by E I n - ~ ( , n ) ( m , t ) = (m , exp,t), maps E (MvJo onto the tangent space to n;l(m) at (m,m). We conclude by showing that dE maps the tangents to the zero cross section of T ( M ) onto the tangent space to the diagonal of M x M , which suffices since M x M(,,,,,, is clearly spanned by tangents to ncl(m) and tangents to the diagonal. Let D : M + M x M be the diagonal map, D(m) = (m , m). Then we have E I(zero cross section) = D 0 I(zero cross section) . dn is onto, and dD is onto the tangent space to the diagonal. QED
Let C be any compact subset of M . Then there exists is defined on B(O(m),c) and maps it diffeomorphically onto B( m , c), where O(m ) is the origin in Mm . Proof. We first note that if all geodesics are not infinitely extendible, exp, may only be defined in a neighborhood of O(m). However, by the theory of differential equations, it is clear that for each m E M exp, is defined on a ball whose radius is a continuous function of m, and hence we may take a c1 such that exp,n is defined on B (O (m ),cl) for every m E C. By corollary 2 to theorem 2, we need only show that there exists c > 0 such that, for each m E C, exp, is a diffeomorphism on B(O(m),c). Lemma 2 essentially says this locally, and hence by compactness the result follows. However, we must first translate lemma 2, which says precisely that for every m E C, there exists a neighborhood P, of (m, 0) which is mapped diffeomorphically onto a neighborhood of (m, m). Therefore, d exp, is regular (that is, 1 - 1 onto) and hence for every n E n (Pm), exp, is a diffeomorphism on the set {t E M , I ( n , t ) E PnL}= PnL,,;so we wish to show there is a c,, > 0 such that the ball, B(O(n),cm),in the Riemannian metric is contained in P,n,, , for all n E n(P,). Using the facts that T ( M ) has a local product structure with the topology of the fibre being given by the Euclidean metric, that the Riemannian and Euclidean metrics are equivalent on each tangent space (see lemma 7.1), and that the Riemannian metric is continuous, one finds that P,, contains a neighborhood of the form Lemma 3.
ac
> 0 such that, for each m E C, exp,
=
{(% t ) I n
E u r nI
II t II < C , J ,
154
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
where Urn is a neighborhood of m and c, > 0. Hence for n E U,,, , exp, maps B(O(n), ern) diffeomorphically onto B(n, cm). By compactness, C is covered by a finite number of U r n ,to each of which corresponds a c , ~ .Letting c = min c, gives the desired result. QED It now follows that the square of the distance function, p2, is C“ on a neighborhood of (m, m) in M x M . Theorem 5. (Hopf-Rinow [40, 77, 931). Consider the following
conditions on a connected Riemannian manifold M :
(a) M is complete. (b) All bounded closed subsets of M are compact. (c) For some point m of M , all geodesics from m are infinitely extendible. (d) All geodesics are infinitely extendible. (e) Any m, n E M can be joined by a geodesic whose arc length equals p(m, n). T h e conditions (a)-(d) are equivalent, and they imply (e). Problem 6. Find an example to show that (e) does not imply (a). Also find an example to show that a minimizing geodesic between two points need not be unique; in fact, there may be infinitely many. Notice that (c) can be stated “exp,, is defined on all of Mm”, and (d) can be stated “ T h e Riemannian connexion is complete’’ (6.3).
Proof. In any metric space it is true that (b) implies (a). That (d) implies (c) is trivial, and that (a) implies (d) follows from extendibility theorems in differential equations (see appendix). We therefore have only to prove the implications from (c) to (b) and from (b) to (e). We now fix m, E M , and define, for any real Y > 0,
8, = {m E &I I p(m, ,111)
< r } = B(m, , r ) ; -
E,. = {m E B, I m can be joined to m, by a geodesic of length p(m, m,)).
I t suffices, assuming (c) for m, , to prove (1) E, is compact and (2) E, = 8,;for any bounded subset is in a 8,, so by (1) and (2) a closed bounded subset is compact, which is (b); then since (b) implies (d), we can use any m for m, , so that (2) implies (e).
8.2. Complete Riemannian Manifolds
I55
which is clearly compact, since it is closed and bounded. Compactness of E , then follows from the fact that E , = exptnOErand the continuity of expm,. QED For the proof of (2) we need the following: Lemma 4. If E , = 8, for some r , and if p(m, , n) > r, then there p(m, n). exists m such that p(m, , m) = r and p(m, , n) = r
+
Proof. For K = 1, 2, ..., choose a, , a broken C" curve from mo to n with I a, I < p(m, , n) l/k (this is possible by the definition of p). Let mk be the last point on a, in B , , so p(m,, m,) = r . By com-
+
FIG. 31.
pactness of E , = 8,, the m k have a limit point m,and by passing to a subsequence we may assume {m,} converges to m. Now p(m0 7 n )
<
dm0
9
m)
+
f(H4
n)
=
r
+ p(m, 4.
On the other hand, p(m, , n) > 1 a , I - l/k = I part of ak from m, to m k I I part of ak from mk to n I - l / k r p(m, , a) - l/k, which has limit r p(m, n), so p(m, , n ) >, Y p(m, n), so equality holds. This proves lemma 4. We now prove (2) using the connectedness of the nonnegative real numbers; that is, we show:
+
(i) E, (ii) E , (iii) E,!
+
> + +
8,. = B , , r' < r , implies E,,! = 8,f . = 8,t for all r' < r implies E , = 8 , .
=
156
8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
(iv) E , = 8, implies there exists c Er+c =
> 0 such that
Bwc-
(i) and (ii) are trivially true. Proof o f (iii). Let m E B , . If m E B,t for r' < r, then by hypothesis m E E,) C E , . Hence, we assume p(m, , m) = r . By lemma 4 we may choose a sequence (m,} which has limit m, where mk E 8, , r k < r . Then by hypothesis, mk E ErkC E, , so m E E , by compactness of E , , which proves (iii). Proof of (iv). By lemma 3, there exists c > 0 such that for each m E 8, , exp, maps B(O(m),2c) diffeomorphically onto B(m, 2c), since E, is compact. Let n E B,+, . We show n E E,+, . By lemma 4 there p(m, n), where p(m, , m) = r . exists m E 8,such that p(m, , n ) = r Therefore, p(m, n) c, so there exists a geodesic y from m to n with I y j = p(m, n). Let u be a geodesic from m, to m with I u 1 = p(m, , m) = r . Then u y is a broken C" curve from m, to n with I 0 y I = I 0 I I y I = r p(m,n) = p(m0, n), so by corollary 3 to theorem 2, u y can be reparametrized as a geodesic. Hence, n E E,,, . This completes the proof of the theorem. It has been shown by K. Nomizu and H. Ozeki [67] that every connected paracompact manifold admits a complete Riemannian metric; furthermore, they show that if every Riemannian metric is complete then the manifold is compact. T h e converse, that a compact metric space is complete, is well known.
+
<
+
++
+
+
Problem 7. Construct a noncomplete connected Riemannian manifold of infinite diameter such that no two points at distance greater than 1 from each other may be connected by a curve which minimizes arc length.
Let i : N + M be an isometric imbedding. Let p be the Riemannian distance on M , p' = p o i, p " the Riemannian distance on N . Give examples to show that all eight possibilities for these metrics to be complete or not may occur. Problem 8.
Problem 9. Show that for a connected Lie group which admits a left and a right invariant metric the exponential map is onto. Problem 10. Let M and N be complete Riemannian manifolds. Show that the Riemannian product M x N is complete. Completeness of affine connexions is also a property preserved under products.
8.2. Complete Riemannian Manifolds Problem 11.
157
Let M be a complete Riemannian manifold.
(a) Then any Riemannian covering of M is complete. (b) For every m E M there is a geodesic segment starting and ending at m in every homotopy class of loops at m. Problem 12. Give examples to show that the result in problem 1 l(b) may or may not hold if M is not complete. Problem 13. Let M be a complete Riemannian manifold which is not cannot be a C" function on all simply connected. Show that p(m, of M . Example. Consider a mechanical system with a finite number of degrees of freedom and no elements which dissipate energy. Then the conjiguration space is a manifold M which is a mathematical model of the totality of all positions of the system. A curve in M is generally thought of as occurring in time, so that a tangent vector is an assignment of velocities to the elements of the system consistent with the constraints of the system. T h e phase space of the system is T ( M ) . T h e kinetic energy given by an assignment of velocities is a positive definite quadratic form on each Mm , and thus gives a Riemannian metric on M . A force field on M is a 1-form 9; the integral of this 1-form on a curve is the amount of work done in traversing the curve. If it is a conservative force field then 9 = - d V , where V is the potential energy. If D is the covariant derivative symbol of the metric and X is the vector field defined by 2 ( X , Y ) = O( Y ) for every Y , then Newton's laws of motion for the system become
where y is a curve parametrized by time. I n particular, when the motion is free the path of the motion is a geodesic and the time is proportional to the length of the geodesic. T h e Riemannian manifold is thus complete if the system will coast indefinitely when given an arbitrary push. I n this case the system may be started in any one configuration with kinetic energy 1 in such a way that it attains another given configuration by free motion with an elapse of time equal to the distance between the two points on M . More specifically, the configuration space of a rigid body constrained
158
8. GEODESICS AND COMPLETE RIEMANNIANMANIFOLDS
to rotate about its center of gravity is homeomorphic to P3 = SO(3).
T h e kinetic energy metric is invariant if and only if the ellipsoid of inertia is spherical. 8.3 Continuous Curves
Let y be a continuous curve, y : [a, b] -+ M , and let a = to < t ,
< t , < .'. < t , < b
=
t,+,
be a sequence of numbers between a and b. Then we define the arc length of y by I y I = sup { p(y(ti), y(ti+,)) I all such sequences ti}. Proposition 1. If y : [a, b] -+ M is a continuous curve from m to n with I y I = p(m, n), then y is a continuous reparametrization of a geodesic. This implies that geodesics locally minimize arc length among continuous curves. Proof. Pick a c > 0 by lemma 3 with respect to the image of y . Take a < t, < t, < b such that p(y(tl), y(t,)) < c/2, so there exists a geodesic u from y ( t l ) to y(t,) with I u 1 = p(y(tl),y(t,)). We claim that for every t E (tl , t J , y ( t ) lies on u. For choose geodesics u1 , u, from r(t1) to y(t), y ( t ) to y(t,), respectively, with I 0 1 I = p(y(t,), y ( t ) ) and 1 u, 1 = p(y(t), y(tz)). Then since y clearly locally minimizes arc length, we have 1 0 1 0 2 1 = p(y(t1), y ( t ) ) p(y(t), y(t,)) = I y from r(t1) to y(t,) I I y from y ( t ) to y(t2) I = p(y(t,), y(t,)), so by the theorem 2, I u, u, I is a broken C" reparametrization of u, which proves y ( t ) lies on the image of u. Hence, y is locally a continuous reparametrization of a geodesic; so in particular it is a reparametrization of a broken C" curve. T h e result then follows from corollary 3 to theorem 2.
+ +
+
Proposition 2.
+
If y is a broken C" curve, then the two definitions of
arc length agree. Proof. Let { y } be the length of y as a continuous curve, retaining I y 1 as the notation for the integral of tangent lengths. Then it follows trivially from the definition of distance that { y } 1 y I. We
<
also have, easily, that both definitions are additive:
{r+d={r)S{4, I r + u l
=
IrlS
101.
8.3. Continuous Curves
159
+
Now suppose y is a curve for which { y } 12 = I y I, 12 > 0. Then by splitting y in half, we must have that the discrepancy on one of the halves is at least k / 2 . By repeatedly halving we get a nested sequence u, such that the discrepancy of y restricted of parameter values s, to [s,, u,] is at least k/2", and u, - s, = c/2,, where c = b - a, [a, b] is the interval of definition of y. Let t be the common limit of s, and u, , and let t , be a number in [s, , u,] such that
<
which exists by the mean value theorem. Then
Now reparametrize y so that t 2"/c, and take limits, getting
=
0, multiply this inequality by
Let xi be normal coordinates at y(O), and fi = xi o y . Then we have p(y(s), y(0)) = ( Zfi(s)2)1/2= ( Xfi(Ois)2)1/2 I s I, where 0 < Bi < 1, by the mean value theorem. Thus if we write g(4
=
P ( Y ( 4 , r(O))/ls I
and d u n ) un - As,) sn = g(Un)(un - s n )
+ ( d u n ) - d-4)sn,
we get
Since limn+mg(u,) = 1 1 y.+(O) 1 1 , this shows
II Y*(O> II 2 K i C + II Y*(O) 11, which contradicts 12
> 0. QED
Problem 14. Let y : [a, b] -+ M be a continuous curve with finite length in a Riemannian manifold. Show that y can be uniformly approximated by broken geodesics.
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8. GEODESICS AND COMPLETE RIEMANNIAN MANIFOLDS
Problem 15. Let 4 : M ---t N be a map between Riemannian manifolds which is onto and preserves distances. Prove that 4 is an isometry, (See [50],p. 169.) This says that the topological metric of a Riemannian manifold determines both the Riemannian structure and the differential structure. Problem 16. Let 4 : M + N be a C" map of complete Riemannian manifold M onto Riemannian manifold N such that for every m E M , M,, decomposes orthogonally into subspaces V , and H,, V and H C" distributions, where V , = ker(d4,) and dr$, restricts to an isometry from H , onto NbC,,.Show that: (a) If y is a C" curve in N and +(m) = y(O), then there is a unique lift 7 o f y to M s u c h that Y(0) = m,y = 4 o 7, and I y I = I 7 I. (b) T h e lift, as in (a), of a geodesic is a geodesic.