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8. Symmetry and Constant Curvature
8
SYMMETRY AND CONSTANT CU RVATURE
A natural condition to impose on a semi-Riemannian manifold is that its curvature tensor R be parallel, that is, have vanishing covariant differential, DR = 0. Such a manifold is said to be locally symmetric. In particular, manifolds of constant curvature turn out to be locally symmetric. A fundamental property of curvature is its control over the relative behavior of nearby geodesics. Because a normal neighborhood 42 is filled with radial geodesics, curvature thereby gives a description of the geometry of 42. Considering only the locally symmetric case, we show that this description is so accurate that, if 42 and 42' are normal neighborhoods with the same description (and same dimension and index), then 42 and 42' are isometric. Using covering techniques this local result is given a global formulation that as a first consequence provides a list of all complete, simply connected manifolds of constant curvature.
JACOB1 FIELDS
A curve can be compared with nearby curves using the following notion. 1. Definition. A variation of a curve segment parameter mapping
x:[a,b] x(-d,d)+M, such that ~ ( u = ) x(u, 0) for all a I u I b. 21 5
ci:
[a, b] -+ M is a two-
216 8
Symmetry and Constant Curvature
The u-parameter curves of a variation are called longitudinal and the u-parameter curves transuerse. The base curve of x is ci. Typically we are interested in the longitudinal curves and the number 6 > 0 is not important. The vector field V on CI given by V ( u ) = xv(u, 0) is called the variation uectorjeld of x. Each V ( u ) is the initial velocity of the transverse curve u -+ x(u, u ) ; thus, for 6 > 0 sufficiently small, the vector field V is an infinitesimal model of the variation x. If every longitudinal curve of x is geodesic, x is called a geodesic variation or one-parameterfamily of geodesics. 2. Definition. If y is a geodesic, a vector field Y on y that satisfies the Jacobi diflerential equation Y" = R y J y ' ) is called a Jacobi vector field.
3. Lemma. The variation vector field of a geodesic variation is a Jacobi field. Proof: Since each longitudinal curve is geodesic, xu, = 0. Thus b y Proposition 4.44, xu,, = xu,, = xu,,
+ Nx,,
x,)x, = R(x,, x,)x,.
Hence x, satisfies the Jacobi equation on every longitudinal curve, in parD ticular on the base curve, where xu is the variation vector field. Because of this result the Jacobi equation is also called the equation of geodesic deviation. There is a far-reaching heuristic interpretation. If we think of a geodesic variation x of y as a one-parameter family of freely falling particles, then the variation vector field V gives the position, relative to y, of arbitrarily nearby particles. Thus the derivative V' gives relative velocity, and V" relative acceleration. Assigning these particles unit mass we can read the Jacobi equation I/" = RV,.y' as Newton's second law with the curvature vector R,,.y' in the role of force, the so-called tidal force. We shall see in Chapter 12 that this is the key to the interpretation of curvature as gravitation in general relativity. In the following example, attracting tidal forces pull radiating geodesics back together again. 4.
Example. In the sphere S 2 ( r ) consider the variation x(u, u ) =
r(cos u cos v, cos u sin u, sin u )
for - n/2 I u I742 and I u I small. The curves u constant are indeed longitudinal, since they parametrize semicircles joining the north and south poles (0, 0, ? r ) . Thus x is a geodesic variation of y(u) = r(cos u, 0, sin u). The variation vector field is V ( u ) = x,(u, 0) = r cos u a,,. Since dYis parallel in R3
Jacobi Fields
217
and tangent to the sphere it is parallel in the geometry of the sphere. Hence V” = - r cos u a , Since S2(r)has constant curvature K we compute
Rv&’)
=
=
-V.
l/r2 and y’ =
= (1/r2)[Y’-
-r
(y’, Y’)VI
sin u 8,
+ r cos u a,,
= - V.
Thus, as predicted by the lemma, V is a Jacobi field. 5. Lemma. Let y be a geodesic with y(0) = p , and let u, w E Tp(M).Then there is a unique Jacobi field Y on y such that Y(0) = u and Y’(0) = w.
Proof: Let El, . . . , En be a parallel frame field on y, and write Y
=
1y i E i . Let ui and wi (1 Ii In) be the coordinates of u and w relative to
E,(O), . .. ,E,(O). Then
Y(0) = u 0~ ‘ ( 0=) 0 ’ ;
Y’(0) = w c> (d$/ds)(O)
=
wi.
1
Because 7’ is parallel, y’ = aiEi with constant coefficients. Thus the Jacobi is equivalent to the linear system equation Y ” = Ryy,(y‘)
where the smooth coefficient functions are uniquely determined by R E , E ~ ( E=~C ) R;kE,
(1 I i , j , k I n).
Such linear systems have smooth solutions (on the entire domain of y) that are uniquely determined by initial conditions as above. Because the Jacobi equation is linear, the set of all Jacobi fields on y forms a real vector space. The lemma shows that its dimension is 2n, since u and w can be chosen arbitrarily in the n-dimensional space Tp(M). Exponential maps are defined in terms of radial geodesics; thus the Jacobi influence of curvature on geodesics leads to a Jacobi description of exponential maps.
TX(
6. Proposition. Let o be a point of M and let x E T,(M). For u, E T,M I , d expAu,) =
W),
where V is the unique Jacobi field on the geodesic y x such that V ( 0 )= 0
and
V’(0)= u E T,(M).
218 8
Symmetry and Constant Curvature
Proof: As in the proof of the Gauss lemma (5.1) consider the twot I1 and I s I small. Its parameter map t ( r , s) = t(x su) in T,(M),with 0 I exponential image in M ,
+
x ( t , s) = exp,(t(x + su>> = ~,+,,(t), is a geodesic variation of y x l [O, 11. Thus the variation vector field V ( u ) = x,(u, 0) = d exp,(i,(u, 0))
is a Jacobi field on y x . Since k(1, s) = x + su, V ( 1) = d exp,(u,). The curve s + x(0, s) is constant at 0,hence V ( 0 ) = x,(O, 0) = 0. By Proposition 4.44 V’(0) = X,,(O, 0) = X,,(O, 0).
Now s + x,(O, s) = x + su is a vector field on the constant curve at o. Hence, by Proposition 3.18, x,,(O, s) = v for all s. In particular, V’(0)= u. This result provides a strong link between M and its curvature tensor. TIDAL FORCES
A vector field Y on a curve a : 1 -+ M is tangent to a if Y =fa‘ for some % ( I ) and perpendicular to u if ( Y , a’) = 0. We consider how these notions relate to Jacobi fields. If la’] > 0, then each tangent space x ( , ) ( M )has a direct sum decomposition Ru‘ + a‘l. Hence each vector field Y on ci has a unique expression Y = Y T + Y’, where YT is tangent to u and Y’ is perpendicular to a. If y is a geodesic, then Y I y implies Y’ I y. since d/ds( Y , y’} = (Y’, y’}; similarly for tangency. Then, if y is also nonnull, it follows that (Y‘)’ = ( YT)r and (Y‘)’ = (Y’)’. {E
7. Lemma. Let Y be a vector field on a geodesic y. (1) If Y is tangent to y then it is a Jacobi field o Y” = 0 0 Y(s) = + b)y’(s)for all s. (2) If Y is a Jacobi field, then Y I y o there exist a # b such that Y ( a )Iy, Y ( b )I y 9 there exists a such that Y ( a ) I y, Y’(a) I y. (3) If y is nonnull, then Y is a Jacobi field o both Y T and Y’ are Jacobi fields.
(as
Proof: (1) Since R(u, u ) = 0, the Jacobi equation forfy’ is equivalent to d2f/ds2= 0. (2) Again since R(u, u ) = 0, the second derivative of ( Y , y‘) is zero, hence ( Y ( s ) ,y’(s)} = A sB, and the result follows. ( 3 ) R( YT, 7’) = 0, hence R( Y’, y‘) = R( Y , y’). Also, since R( Y , 7’) is skew-adjoint, R( Y, y’)y’ Iy.
+
Locally Symmetric Manifolds 21 9
Thus the Jacobi equation Y ” = R ( Y , y‘)y’ splits into the two equations (YT)”= 0 and (Y’)’’ = R ( Y 1 , y’)y’. The result follows using (1).
A Jacobi field tangent to a geodesic y is of scant importance since it is the infinitesimal model for a family of geodesics that merely reparametrize y. Thus in considering the Jacobi equation Y ” = R,,,. y’ as relatiue acceleration produced by tidalforce we shall emphasize the case Y Iy. 8. Definition. For a vector 0 # u E T,(M) the tidal force operator F,: u’- + u’ is given by F,(y) = Ryvu. -
9. Lemma. F , is a self-adjoint linear operator on u’, and trace F , Ric(o, 0).
=
Proof. Clearly F , is linear (and carries u’ to itself). The pair symmetry of curvature implies that F , is self-adjoint, but note that u’ is a degenerate subspace if u is a null vector. If u is nonnull, let e 2 , .. . , en be an orthonormal basis for u’-. Then Ric(u, v ) = ci(RUe,v,ei) = -1 Ei(Fv(ei),e i ) = -trace F,. If u is null, let w be a null vector such that ( u , w ) = - 1; these two vectors span a Lorentz plane IZ. Then e , = ( u + w)/,,h and e2 = ( v - w)/$ form an orthonormal basis for IZ with e l timelike and e2 spacelike. Let e 3 , .. . , en be an orthonormal basis for Il’ c u‘. Then
c
Ric(v, u> = - + ( R o e ,u, e2> +
C Ej(Rve,U, ej>.
j>2
The first two terms on the right-hand side cancel since (RuelV, e l > = HRuwv, W > = (Rve2v, e l > . Now u, e 3 , .. . , en is a basis for d , and F,(v) = 0. Hence trace F , = ej(F,(ej), e j ) = - E~(R,,,u,e j ) . j> 2
1
j>2
In the nonnull case tidal forces are often normalized by taking u to be a unit vector. For example, if M has constant curvature C , then Corollary 3.43 gives F,(y) = - &Cy,where E = ( v , u ) = f 1. LOCALLY SYMMETRIC MANIFOLDS
10. Proposition. The following conditions on a semi-Riemannian manifold M are equivalent: (1) DR = 0, that is, M is locally symmetric. (2) If X , Y , 2 are parallel vector fields on a curve a, then the vector field R,,Z on a is also parallel.
220 8
Symmetry and Constant Curvature
( 3 ) Sectional curvature is invariant under parallel translation, that is, the sectional curvature of a nondegenerate tangent plane I7 remains constant as n is parallel translated along any curve a. Proof: For simplicity assume that the curve a is regular; this will be sufficient for later work. (2). Fix an arbitrary point on a,say a(0). By Exercise 3.12 there (1) exist vector fields V , X , F, Z on a neighborhood of a(0) in M such that K,f) = a'(t), = X ( t ) , and similarly for P and Z . Since D R = 0, - - --0 = (D,R),?;Z = D,(R(X, Y ) Z ) - R ( D V X , Y ) Z - R ( X , D v F ) Z - R ( X , P)(D,Z). Now evaluate at 0. There, for example,
(4m
( O ) =
( E m = Y'(0) = 0,
since Y is parallel. Thus the equation above reduces to
(RXYZ)'(O) = Dd(O,(RXYZ)= 0. (2) (1). If u, x, y , z E T,(M), let X , Y, Z be the parallel vector fields on yu gotten by parallel translation of x, y, z, respectively. Then
-
(D,R),,z
=
( R x y Z ) ' ( 0 )= 0.
(2) (3). Let n(0)be a nondegenerate tangent plane at a(O), and let X , Y be parallel vector fields on a such that X(O), Y ( 0 ) is a basis for n(0). Thus X(r), Y ( t ) is a basis for the parallel translate n(t)of n(0)along a. Then both ( R , , X , Y) and Q ( X , Y ) are constant along a, hence the sectional curvature of n(t)is constant. ( 3 ) (2). Suppose a : I -+ M starts at p . By orthonormal expansion it suffices to show that, if X , Y , Z , W are parallel vector fields on a, then ( R x y Z , W ) is constant. Fix t E I and define a function A : T,(M)4 -+ R by A(x, y , Z, W ) = (RxrZ, w)(t>,
where X , Y, Z , W are parallel vector fields on a extending x, y, z , w. Since Q ( X , Y ) is constant, ( 3 ) gives
for all x, y e T,(M) spanning a nondegenerate tangent plane. It is easy to check that the function A is curvaturelike (page 79); hence by Corollary 3.42
&, y , z, 4 = (RxyZ,w ) . Thus ( R , , Z , W ) ( t ) is independent oft.
lsometries of Normal Neighborhoods
221
11. Corollary. A semi-Riemannian manifold of constant sectional curvature is locally symmetric.
This is obvious from criterion (3) above. Semi-Riemannian products of locally symmetric manifolds are again locally symmetric but (if nonflat) do not have constant curvature. ISOMETRIES
OF NORMAL NEIGHBORHOODS
Let M and R be semi-Riemannian manifolds of the same dimension and index and let o E M and A E M . Our goal is to find conditions under which a given linear isometry T,(M) + &(M) is the differential map of an isometry defined on some normal neighborhood of 0. 12. Definition. Let L: & ( M ) -+ T,(M) be a linear isometry, and let Q be a normal neighborhood of o in M such that exp, is defined on the set L(exp,- '(42)). Then the mapping
rbL = exp50Loexp;':%
+
M
is called the polar map of L on 42. In short, +L sends exp,(u) to exp,(Lu) for all u E Q c 7',,(M). Polar maps always exist for Q sufficiently small, and the first two properties below show that if the isometry we seek exists it must be a polar map of L. 13. Lemma. With notation as above, ( 1 ) 4Lcarries radial geodesics to radial geodesics; explicitly, if u E T,(M), then & o y u = yLu, where both sides are defined. (2) The differential map of 4 at o is L. (3) If 42 is sufficiently small, then 4Lis a diffeomorphism onto a normal neighborhood of o in (4) If M is complete, +L is defined on every normal neighborhood of 0.
a.
Proof. (1) Since y,(t)
=
exp,(tu),
4L(YU(t))= exPo(L(t4) = exPo(tL(u))
=
YLdt)
for all t such that y,(t) remains in Q. (2) If u E T,(M), then by (1) &L(U)
=
ddL(YL(0))
= (rbL
O
Y,)'(O)
=
YkU(0)
= LU.
222 8
Symmetry and Constant Curvature
(3) If @ is sufficiently small, $L(%) is contained in a normal neighborhood ofo. Then exp, is a diffeomorphism of P = L(exp; ' %) onto a (normal) neighborhood Y" of 6 Thus 4L:% -+ V is a composition of diffeomorphisms. Finally, (4) is clear since exp, is defined on all of %(AT). a If L is to be the differential map at o of an isometry, then, as we saw in Chapter 3, it must preserve curuature at 0 ;that is, L(RxyZ)= RLX,&z)
for x, y ,
E
T(W.
By Corollary 3.42 it is equivalent that L preserve sectional curvature: K(ZZ) = K ( ~ 1 7 for ) nondegenerate planes in To(M). This necessary condition is sufficient. 14. Theorem. Let M and
manifolds, and let L: curvature. Then
T(M)
-+
be locally symmetric semi-Riemannian &(AT) be a linear isometry that preserves
(1) If 2% is a sufficiently small normal neighborhood of o, there is a unique isometry 4 of -& onto a normal neighborhood V o f 0 such that d#lo = L. (2) If M is complete then for any normal neighborhood 42 of 0,there is a unique local isometry 4: 1 -+ M such that d40 = L.
Proof: In both cases uniqueness follows from Proposition 3.62. By the preceding lemma, existence in both cases follows if every polar map 4 = 4L:42 -+ is a local isometry. Thus for u E T J M ) , p E %, we must prove that (dqh, d q h ) = (0, u ) . The idea is to use Proposition 6 and show that corresponding Jacobi fields grow at the same rate in both manifolds. (1) Let 02 be the neighborhood in T'(M) corresponding to the normal neighborhood 42. There is a unique x E 4 and y X g T,.(T'M) such that d exp,(y,) = u. By Proposition 6, (0, o) = (Y(1), Y(1)), where Y is the Jacobi field on yx such that Y(0) = 0 and Y'(0) = Y E T,(M). (2) Now look at the corresponding situation in AT. Since L is linear, d ~ ( y ,= ) ( L Y ) L ~ . Hence by the definition of the polar map 4,
d4(u)
=
d exPo((Ly)Lx).
Thus, as above, (d&, d 4 v ) = ( P ( 1 ) , P(1)), where P is the unique Jacobi field on yLx such that P(0) = 0 and P'(0) = LY E G(R). (3) Let E , , . . . , En and El, . . . , Enbe parallel frame fields on y x and yLx respectively, such that L(E~(O))= E,(O) for all i. Since L is a linear isometry, the coordinates of x and y relative to the Ei(0)s are the same as the coordinates of LX and ~y relative to the E,(O)s.
lsometries of Normal Neighborhoods
223
If we write Y = y i E i , then the functions y ' , . . . ,y" satisfy the system of differential equations in the proof of Lemma 5 as well as the initial conditions corresponding to Y(0) = 0 and Y'(O)'= y . Over in El, writing P = y'Ei gives the corresponding differential equations
1
Furthermore, by a remark above, the functions y ' , . . ., y" and j ' , . . .,y" satisfy exactly the same initial conditions. (4) We assert that R $ k = RZk on common domain I of y x and y L x (keeping the former in 42). In fact, since the linear isometry L preserves curvature, it follows from our choice of corresponding frame fields that j ? Z k ( O ) = R&(O). Since M is locally symmetric, Proposition 1 shows that the functions ( R E j E k E iE, m ) and hence RZk are constant on I . Similarly REk is constant, giving the result. i I n) satisfy the same system of dif( 5 ) The functions y' and y i (1 I ferential equations and the same initial conditions. By the uniqueness of such solutions, y' = y' for all i; hence ( d 4 u , d$u) = (P(l), Y(1)) = ~ ~ ( y ' ( 1=) )(Y(1), ~ Y ( 1 ) ) = (0, u ) .
C
A first consequence of the theorem is that constant curvature determines local geometry: 15. Corollary. Let M and R be semi-Riemannian manifolds with the same dimension and index and the same constant curvature C . Then Any points o E M and B E R have isometric neighborhoods.
This is clear since a linear isometry from T J M ) to &(R) exists (by 2.27) and must preserve curvature. Another application will explain the term locally symmetric. For p E M let (, be the polar map of the linear isometry u + - u of T,(M). If 42 is a suitably chosen normal neighborhood, then [,: (42 + 42 is a diffeomorphism. Evidently (, reverses geodesics through p: if y(0) = p , then (,(y(s)) = (y -s). In fact this property uniquely determines (, (once is specified), hence [, is called the local geodesic symmetry of M at p . 16. Corollary. The following conditions on a semi-Riemannian manifold are equivalent:
(1) M is locally symmetric. (2) If L: T,(M) + T , ( M ) is a local isometry that preserves curvature, then there is an isometry 4 of normal neighborhoods of p and q such that dcjp = L. (3) At each point p of M the local geodesic symmetry (, is an isometry.
224 8
Symmetry and Constant Curvature
Proof: (1) (2). By the theorem. (2) (3). The linear isometry -id on T,(M) carries each nondegenerate plane Il c T,(M) to itself; thus -id preserves curvature. If 4-idis the isometry given by (2), then reverses geodesics through p , hence is the local geodesic symmetry at p . (3) * (1). We show that the tensor field D R vanishes at an arbitrary point p . Since [, is an isometry it preserves curvature and covariant derivatives. The differential map of l pat p is -id, hence for all u, x, y , z E T'(M)
-(D,R),,z Hence DR = 0 at p .
=
u - " W - , ,-,(-4 = (D,R),,z.
SYMMETRIC SPACES
The local result in Theorem 14 can be generalized as follows. 17. Theorem. Let M and hl be complete, connected, locally symmetric semi-Riemannian manifolds, with M simply connected. If L: T,(M) -+ %(a) is a linear isometry that preserves curvature, then there is a unique semi-Riemannian covering map 4: M -+ R such that d 4 0 = L. We postpone the proof to look first at some consequences. Roughly speaking, the theorem asserts that a complete connected locally symmetric manifold R is determined-up to semi-Riemannian covering-by its curvature at one point. The result is due in principle to E. Cartan (1869-1951), who singlehandedly created the theory of symmetric spaces. (Also see [A] and Theorem 1.36 of [CE].)
18. Definition. A semi-Riemannian symmetric space is a connected semi-Riemannian manifold M such that for each p E M there is a (unique) isometry 5,: M -+ M with differential map -id on T,(M). The isometry i,is called the global symmetry of M at p . Since it reverses the geodesics through p , i,is the unique extension to all of M of the local geodesic symmetry at p . Thus the latter is an isometry; so by Corollary 16, symmetric implies locally symmetric. 19. Examples. ( 1 ) R" is symmetric, since for each point p the map p x -+ p - x is an isometry. (2) The sphere S" is symmetric, since for each p it is symmetric in the usual Euclidean sense about the line in R"" through p and - p . (3) In fact, by Proposition 4.30, euery connected hyperquadric is to be -el,. . . , --en). symmetric (take the framef,, . . . ,,L
+
Symmetric Spaces 225
Chapter 11 will exhibit a variety of symmetric spaces with nonconstant curvature. A locally symmetric manifold need not be complete, since any open submanifold of one is again one; however,
20. Lemma. A semi-Riemannian symmetric space is complete. Proof: To show that a geodesic y: [0, b) + M is extendible, choose c near b in the interval, and let [ be the global symmetry at y(c). Since [reverses geodesics through y(c), a reparametrization of [ 0 y provides the required rn extension of y.
21. Corollary. A complete, simply connected, locally symmetric semiRiemannian manifold is symmetric. Proof: At any point p E M the linear isometry -id on T’(M) preserves curvature. Thus we can apply Theorem 17, with R = M , to obtain a semiRiemannian covering map 4: M + M such that ddP = -id. Since M is rn simply connected, 4 is an isometry, by Corollary 7.27.
It follows that, if M is a complete, connected, locally symmetric manifold, then its simply connected semi-Riemannian covering manifold fi is symmetric. In fact, by Corollary 7.29, fi is complete and (since the covering map is in particular a local isometry) fi is locally symmetric. Thus the preceding corollary shows that fi is symmetric. Now we return to the proof of Theorem 17. The scheme is as follows. By Theorem 14 there is an isometry on some neighborhood of o E M whose differential map is L. Any point p E M can be connected to o by a broken geodesic /?, so we can parallel translate L along p to p and there get another isometry on a neighborhood of p . Because M is simply connected, these locally defined isometries fit together to give the required map 4. Some notation is required to cope with broken geodesics. For any k 2 1 let u be a k-tuple ( u l , . . . , u k ) of vectors in T,(M). Then the broken geodesic 8,: [0, k ] + M is defined inductively as follows (using the fact that M is complete). Let = y,, I[O, 11. If pj has been defined on [ O , j ] , for 1 Ij < k , let w = P ( U ~ +be~ the ) parallel translate of u j + l along pj to its endpoint bj(j).Then attaching t
to
pi gives pi+ 1.
-,y,(r
-
j)
( j IE Ij
+ I)
(1) For u as above, let P,: T,,(M) + TB&)(M) be parallel translation along Similarly in R,for LU = ( L u l , . . . ,L u k ) let PLv:T(M)+ q , , ( k ) ( R ) be parallel translation along I j L v . Then the parallel translate of L along Ijk is
8,.
L” = P L ,
0
L
0
Pv- : T,(M)
+
T(@),
226 8
Symmetry and Constant Curvature
where p = /l,(k) and q = fiL,(k). Since M and R are locally symmetric the linear isometries P , and PLupreserve curvature, hence so does the linear isometry L,. By Theorem 14, there is a neighborhood 02, of each fl,(k) and a local isometry 4u:4Y, + h? whose differential map at &(k) is L,. In view of Lemma 5.10, we can suppose that each J2,is convex and that if $2, and $2w meet, then their intersection is convex. hence connected. (2) We now construct a matched covering of M . Let A be the set of ktuples in T , ( M ) for all k 2 1. Let M be the collection of convex open sets 42, as above for all v E A. Since M is connected, 53 is a covering of M . If u, w E A , define t' w to mean that 4u= 4won $2, n aW. Evidently the relation is reflexive and symmetric, so it remains to prove the quasi-transitive condition (3) of Definition 7.30. Suppose u v, v w, and p E 4Yu n $2, n $2w. It follows from the two relations that the differential maps d4", d&, d'$, all agree at the point p . Hence by Proposition 3.62 4u = bwon $2,, n 4Yw. Thus u w, and (A, ) is a matched covering. (3) The essential step of the proof is to show that this matched covering is chainable over any geodesic segment a: [O, I] + M , since then Proposition 7.32 applies. Choose u = ( u l , . . . , u k ) E A so that fiu runs from o to a(0). Let uk+ be the ) to o along flu. For each s E [0, 11 let w(s) = parallel translate of ~ ' ( 0 back (vlr . . . , u k , suk+ A . Thus the last segment of flw(s)is a reparametrization of CJI[O, s]. Then there is a partition 0 = so < s1 < . . . < s, = 1 such that
-
-
-
- -
-
for
a([si- 1, s i ] ) c $2w(sI)
1 I i I m.
-
See Figure 1. It remains to check that w(siw(si). Abbreviate 4w(si) to 4i. Let P be parallel translation along CJ from a(sito o(si),and correspondingly let P be parallel translation along 4i0 a from c$~cJ(s~- to 4ia(si).It follows from the definition of L, that LW(S,. I)
=
p-
O LW(Si)
p.
As a local isometry, 4i preserves parallel translation ; hence, ( d 4 i ) o , s z- I )
=
P-
0
(d4i)ofsi) 0 P .
Figure 1
Simply Connected Space Forms
By construction, for each i, L , , , ( ~ ~is) the differential map of q5i at b,,,(& Thus the two preceding formulas yield
= a(s,).
227
+ 1)
-
= 4i on n a,+); that is, w(siPI) w(si). By Proposition 7.32, the matched covering (9, -) gives a semiRiemannian covering map &: M* + M . As with any matched covering, for meets A,,,(%,,,) if each v E A there is local section A":%" + M * , and and only if u w. It follows that the maps 4" A : &(%J + R for all u E A agree on overlaps, hence define a single map 4*: M* -+ R.This is a local isometry since A and every 4" are. Since M is simply connected, Corollary A.14 applies, showing that there is a (locally isometric) global section A: M + M * such that 111q0= Lo for (0) E A . Thus 4 = 4* 0 A: M -+ R is a local isometry with 4la0= $o; hence d$o = L. B
Hence,
$i-l
(4)
-
0
SIMPLY CONNECTED SPACE FORMS 22. Definition. A space,forrn is a complete connected semi-Riemannian manifold of constant curvature.
Space forms must be regarded as the simplest important class of semiRiemannian manifolds. In the simply connected case, Theorem 17 is decisive:
23. Proposition. Simply connected space forms are isometric if and only if they have the same dimension, index, and curvature C. ProoJ: Obviously the conditions are necessary. For the converse, note that, since constant curvature implies local symmetry, Corollary 2 1 shows that simply connected space forms are symmetric. Suppose M and N are simply connected space forms of the same dimension and index. The latter show that linear isometries T,(M) -+ T,(N) exist. Then Theorem 17 gives a semi-Riemannian covering map M -+ N which, since M is simply connected, is an isometry. Thus up to isometry there is at most one simply connected space form M(n, v, C) of dimension n, index v, and curvature C. Existence for all (n, v, C) is settled for C = 0 by the semi-Euclidean spaces R:, and, with suitable modifications, for C > 0 by pseudospheres and for C < 0 by pseudohyperbolic spaces.
228 8
24.
Symmetry and Constant Curvature
Corollary. For n 2 2,
i-
S:(r) if C = l/r2 and 0 I v I n - 2, if C = 0, M(n, v, C ) = R: Ht(r) if C = - l/rz and 2 I v I n. H;-,,(r), these space forms are all In fact, since St(r) s"-' x R simply connected. The remaining cases come in anti-isometric pairs. One-dimensional semi-Riemannian manifolds are trivially flat ; hence by Exercise 8, the only simply connected one-dimensional space forms are R' and R : . For n 2 2 we have .V
M(n, n, l / r 2 ) = one component cS#) of S:(r), and by metric reversal, M(n, 0, - l/r2) = hyperbolic n-space H"(r), one component of H$(r). M ( n , n - 1, l/r2) = S:- l(r), the simply connected semi-Riemannian covering manifold of S:- l(r), and M ( n , 1, - l/r2) = I?y(r), the simply connected semi-Riemannian covering manifold of H l ( r ) .
-
Since R" is the simply connected covering manifold of S' x R n - ' , these four types are diffeomorphic to R". 25. Corollary (Hopf). A complete, simply connected, n-dimensional Riemannian manifold of constant curvature C is isometric to
the sphere S"(r) Euclidean space R" hyperbolic space H"(r)
if if if
c = 1/r2,
C = 0, C = - l/r2.
This result is particularly satisfying from a historical viewpoint, since all three of these constant curvature geometries were well known, at least in low dimensions, before Riemannian geometry was invented (1 845) and long before a rigorous proof of this corollary (1926). Euclidean geometry is of course the oldest, and hyperbolic geometry, dating from the early 1800s, the newest. Writing fi for the simply connected semi-Riemannian covering manifold of M , the Lorentz analogue of Hopf's result is
26. Corollary. A complete, simply connected, n-dimensional Lorentz manifold of constant curvature C is isometric to the Lorentz sphere S ; ( r ) s:(r) Minkowski space RY @(r>
if C = l/r2 and if C = l/r2 and if C = 0, if c = -l/r2.
n 2 3, n = 2;
Simply Connected Space Forms
229
In relativity theory S:(r) is called de Sitter spacetime, and fl:(r) is universal anti-de Sitter spacetime. 27. Example. A model for pl.As Figure 2 suggests, H t can be regarded as a hypersurface of revolution in R;i ' = R: x R"- Unwrapping the circles of revolution gives a covering map R : R' x R"-' + H;, where
'.
R(t, XI
=
(JEG 'cos t, J
sin t, XI.
Furnished with the pulled-back metric, R" becomes the simply connected semi-Riemannian covering manifold of H ; -and hence can be denoted by
A;.
R: (coordinates uo, u ' )
' (coo1-dinates u 2 , . . . ,u")
Figure 2. H; in R;+'
The geometry of the model can be derived from that of H ; as modified by the map A. For example, if the coordinate slice uo = 0 in RI' is regarded as R;, then R; n H ; is just H",'. The latter's two components H"-' are totally geodesic in H:. By rotational symmetry the same is true for all slices of H ; by hyperplanes through R'- '. Thus in plthe slices t constant are totally geodesic spacelike hypersurfaces, isometric to H"- (or R' if n = 2).
In Riemannian geometry, curvature inequalities such as K 2 c and aI K I b have been intensively studied. For indefinite metrics we shall now see that such conditions imply constant curvature. Sectional curvature K is undefined on a degenerate plane Z l in T'(M), but a formula in the proof of Lemma 3.39 shows that sgn(R,,X, Y ) is the same for all bases X, Y for n. This provides a well-defined function JV from the set of degenerate planes in T,(M) to the set (+ 1,0, - 1 >. 28. Proposition (Kulkarni et a/.). Let p be a point of a semi-Riemannian manifold of indefinite metric. The following conditions on T,(M) are equivalent : (1) K is constant, (2) N = 0,
230 8
Symmetry and Constant Curvature
(3) a I K or K I b, where a, b E R, (4) a I K 5 b on indefinite planes, ( 5 ) a I K < b on definite planes. Proof (S. Harris et al.). Evidently we can assume dim M 2 3. Obviously (1) implies (3, 4, 5), and by Corollary 3.43 it implies (2). The reverse implications will use this consequence of .Ar = 0: (*) If X , U , Z are orthonormal vectors and X and U have opposite causal character, then K ( X , Z ) = K ( U , Z ) .
To prove this note that U are degenerate. Hence
(Rx+u , A X
k X are null vectors and the planes (U i-X , 2)
+ u),Z > = 0 = < R x - u , z ( X
-
Expansion yields first ( R x z U , Z ) = 0, then ( R x z X , Z ) Since Q ( X , Z ) + Q ( U , Z ) = 0, the result follows.
W, Z > .
+ (RuZ U , Z ) = 0.
( 2 3 1) Case I v = 1 or n - 1. By metric reversal we can assume U , X , Y is orthonormal with U timelike, then (*)implies K ( U , X ) = K ( X , Y ) . It follows that every nondegenerate plane that either contains or is orthogonal to U has the same sectional curvature k(U). If V is an independent timelike vector, the plane spanned by U and V is nondegenerate, hence k(U) = K ( U , V ) = k ( V ) . v = 1. If
Case I I 1 < v < n - 1. Let U be a positive definite plane, U’negative definite, and let X E ZZ and U E U’ be unit vectors. Now Q ( X , U ) < 0, and using Lemma 2.19 we easily check that ZZ + RU and U‘ + R X are nondegenerate. Hence by Case I, K(ZZ) = K ( U , X ) = K ( n ’ ) . This suffices because, if X , U is an orthonormal basis for any indefinite plane, then extension to an orthonormal basis for T’JM) will provide planes U , U’ as above.
(332). Suppose a I K (reversing the metric gives K I b). By Exercise 9 every degenerate plane n is a limit of definite planes and a limit of indefinite + ZZ = n ( X , Y ) , where X i + X and +Y. planes. Consider U(Xi, Then ( R ( X t , V X i , X) a.
x)
x
Q(xi,V)
Here the numerator approaches ( R ( X , Y ) X , Y ) and the denominator approaches zero. Thus, if all Q ( X i , > 0, then ( R ( X , Y ) X , Y ) 2 0, hence N ( U ) 2 0. Similarly, if all Q ( X , , < 0, then M ( n ) 5 0; hence N ( U ) = 0.
x) x)
x)l/
( 4 a 2) and (5 * 2 ) . With notation as above, I ( R ( X i , x ) X i , I Q ( X i , 6)l is bounded, where Q ( X i , is negative for (4), positive for (5). Hence ( R ( X i , x ) X i , approaches zero, so N(n)= 0. rn
x)
x)
Transvections 231
If at each point of M, connected and dim 2 3, one of the conditions in the proposition holds, then, by Schur’s theorem (Exercise 3.21), M has constant curvature. TRANSVECTIONS 29. Definition. An isometry geodesic y : R + M provided
4: M
+M
is a transvection along a
(1) 4 translates y; that is, &(s) = y(s + c) for all s E R and some c. (2) d 4 gives parallel translation along y ; that is, if x E q(s)(M),then d4(x) E T ( s + c l ( Mis) the parallel translate of x along y. For example, if 4 is a rotation of the sphere S2about the z-axis in R3,then 4 is a transvection along the (geodesic) equatorial circle z = 0. In a symmetric space there are transvections along every geodesic.
be the global 30. Lemma. If y is a geodesic in a symmetric space, let is symmetry of M at y(s). Then for any c, the isometry [cizio is a transvection along y that translates it by c. Proof. For all s, [o(y(s))
=
y( - s). But c/2 is the midpoint of [ -s, s
+ c],
+ c). so i c i 2 i o Y ( S ) = If X is a parallel vector field on y , then for any s, d [ , ( X ) is a parallel vector 0 y, which is a reparametrization of y. (Since parallel translation field on is is independent of parametrization, we can ignore such reparametrizations.) If x E TCs, M , then x is parallel along y to a tangent vector y at y(0). Hence dio(x) is parallel to diO(y) = - y , and also to a vector z at y(c/2). Thus di,,, dcO(x)is parallel to - z , hence to y, hence to x.
31. Corollary. Every nonconstant geodesic in a symmetric space is either one-to-one or simply periodic.
Proof. It suffices to show that, if y is a geodesic with y(b) = y(0), then y(s b) = y(s) for all s. For each s let 4sbe the transvection that translates y by s. Then
+
Y(S
+ b) = qbs(Y(b)) = q b S ( Y ( 0 ) )
=
’
Exercises
1. An isometry iof M , connected, is a global symmetry at p E M if and only if [ is involutive (that is, 1’ = id) and p is an isolated fixed point of [.
232 8
Symmetry and Constant Curvature
2. (a) Let M be a semi-Riemannian submanifold of a symmetric space M.If
M is connected, closed in M,and totally geodesic, then M is symmetric. (b) A semi-Riemannian product M x N is symmetric if and only if both M and N are symmetric. (c) Let k : A -+ M be a simply connected semi-Riemannian covering. If M is symmetric, then fi is symmetric (but not conversely). 3. (a) If M is a complete Riemannian manifold, then every element of nl(M, p ) contains a geodesic loop. (Hint: Use the Hopf-Rinow theorem in the simply connected covering manifold.) (b) The fundamental group of a Riemannian symmetric space is abelian. (Hint: Show that the geodesic symmetry at p induces the homomorphism g -+ g - l on q ( M , p).) 4. A model of hence (reversing the metric) R:. Consider the smooth map x: R2 + R: given by x(t, 9) = (sinh t, cosh t cos 9, cosh t sin 9). (a) Prove that x is a covering map of R2 onto the unit pseudosphere S: in R:. (b) Prove that R' furnished with the pulled-back metric is $ and the line element is -dt2 + cosh' t d 9 2 . (c) Sketch some null geodesics of this model and determine which points can be reached by geodesics starting at the origin. 5. (a) The model of in Example 27 has line element - S2 dt2 + do2,where S = (1 + x x ) ~ and ' ~ do2 = dx dx - ( ( x d x ) 2 / S 2 ) on R"- l . (Hint: S d S = x - d x . ) (b) do2 on R"-' gives H"-'.(For other models of hyperbolic space see [Wo].) 6. Let y be a geodesic with (y', y ' ) = E = f 1 in a semi-Riemannian surface. If Y is a Jacobi field on y perpendicular to y, write Y = Y E , where E is a parallel unit vector field on a. Show that the Jacobi equation for Y reduces to y" + EKY = 0. 7. Let y be a geodesic in a manifold of constant curvature C , and let Y be a Jacobi field on y that is ly. (a) the Jacobi equation for Y is Y" + C(y', y')Y = 0. (b) Let c = 1 C(y', y ' ) I '". On y there exist parallel vector fields A , B 1y such that Y(s) = cos(cs) A(s) + sin(cs) B(s) if C(y', y ' ) > 0. Y ( s ) = A(s) + sB(s) if C(y', y ' ) = 0. Y ( s ) = cosh(cs) A(s) + sinh(cs) B(s) if C(y', y ' ) < 0. 8. (a) Let M be a flat connected semi-Riemannian manifold complete at a E M (that is, expo is defined on all of T,(M)).Prove that exp,: T,(M) -+ M is a semi-Riemannian covering map. (Hint: Use Proposition 6.) (b) Give an example of a connected semi-Riemannian manifold that is complete at one point but not complete. 9. In a vector space with indefinite scalar product, every degenerate 2-plane is a limit of (a) indefinite planes (b) definite planes. (Hint: See proof of Lemma 3.40, but for (b), if gln = 0, consider n(u + 6x, w + dy).)
s:,
-
-
.
SYMMETRY AND CONSTANT CU RVATURE
A natural condition to impose on a semi-Riemannian manifold is that its curvature tensor R be parallel, that is, have vanishing covariant differential, DR = 0. Such a manifold is said to be locally symmetric. In particular, manifolds of constant curvature turn out to be locally symmetric. A fundamental property of curvature is its control over the relative behavior of nearby geodesics. Because a normal neighborhood 42 is filled with radial geodesics, curvature thereby gives a description of the geometry of 42. Considering only the locally symmetric case, we show that this description is so accurate that, if 42 and 42' are normal neighborhoods with the same description (and same dimension and index), then 42 and 42' are isometric. Using covering techniques this local result is given a global formulation that as a first consequence provides a list of all complete, simply connected manifolds of constant curvature.
JACOB1 FIELDS
A curve can be compared with nearby curves using the following notion. 1. Definition. A variation of a curve segment parameter mapping
x:[a,b] x(-d,d)+M, such that ~ ( u = ) x(u, 0) for all a I u I b. 21 5
ci:
[a, b] -+ M is a two-
216 8
Symmetry and Constant Curvature
The u-parameter curves of a variation are called longitudinal and the u-parameter curves transuerse. The base curve of x is ci. Typically we are interested in the longitudinal curves and the number 6 > 0 is not important. The vector field V on CI given by V ( u ) = xv(u, 0) is called the variation uectorjeld of x. Each V ( u ) is the initial velocity of the transverse curve u -+ x(u, u ) ; thus, for 6 > 0 sufficiently small, the vector field V is an infinitesimal model of the variation x. If every longitudinal curve of x is geodesic, x is called a geodesic variation or one-parameterfamily of geodesics. 2. Definition. If y is a geodesic, a vector field Y on y that satisfies the Jacobi diflerential equation Y" = R y J y ' ) is called a Jacobi vector field.
3. Lemma. The variation vector field of a geodesic variation is a Jacobi field. Proof: Since each longitudinal curve is geodesic, xu, = 0. Thus b y Proposition 4.44, xu,, = xu,, = xu,,
+ Nx,,
x,)x, = R(x,, x,)x,.
Hence x, satisfies the Jacobi equation on every longitudinal curve, in parD ticular on the base curve, where xu is the variation vector field. Because of this result the Jacobi equation is also called the equation of geodesic deviation. There is a far-reaching heuristic interpretation. If we think of a geodesic variation x of y as a one-parameter family of freely falling particles, then the variation vector field V gives the position, relative to y, of arbitrarily nearby particles. Thus the derivative V' gives relative velocity, and V" relative acceleration. Assigning these particles unit mass we can read the Jacobi equation I/" = RV,.y' as Newton's second law with the curvature vector R,,.y' in the role of force, the so-called tidal force. We shall see in Chapter 12 that this is the key to the interpretation of curvature as gravitation in general relativity. In the following example, attracting tidal forces pull radiating geodesics back together again. 4.
Example. In the sphere S 2 ( r ) consider the variation x(u, u ) =
r(cos u cos v, cos u sin u, sin u )
for - n/2 I u I742 and I u I small. The curves u constant are indeed longitudinal, since they parametrize semicircles joining the north and south poles (0, 0, ? r ) . Thus x is a geodesic variation of y(u) = r(cos u, 0, sin u). The variation vector field is V ( u ) = x,(u, 0) = r cos u a,,. Since dYis parallel in R3
Jacobi Fields
217
and tangent to the sphere it is parallel in the geometry of the sphere. Hence V” = - r cos u a , Since S2(r)has constant curvature K we compute
Rv&’)
=
=
-V.
l/r2 and y’ =
= (1/r2)[
-r
(y’, Y’)VI
sin u 8,
+ r cos u a,,
= - V.
Thus, as predicted by the lemma, V is a Jacobi field. 5. Lemma. Let y be a geodesic with y(0) = p , and let u, w E Tp(M).Then there is a unique Jacobi field Y on y such that Y(0) = u and Y’(0) = w.
Proof: Let El, . . . , En be a parallel frame field on y, and write Y
=
1y i E i . Let ui and wi (1 Ii In) be the coordinates of u and w relative to
E,(O), . .. ,E,(O). Then
Y(0) = u 0~ ‘ ( 0=) 0 ’ ;
Y’(0) = w c> (d$/ds)(O)
=
wi.
1
Because 7’ is parallel, y’ = aiEi with constant coefficients. Thus the Jacobi is equivalent to the linear system equation Y ” = Ryy,(y‘)
where the smooth coefficient functions are uniquely determined by R E , E ~ ( E=~C ) R;kE,
(1 I i , j , k I n).
Such linear systems have smooth solutions (on the entire domain of y) that are uniquely determined by initial conditions as above. Because the Jacobi equation is linear, the set of all Jacobi fields on y forms a real vector space. The lemma shows that its dimension is 2n, since u and w can be chosen arbitrarily in the n-dimensional space Tp(M). Exponential maps are defined in terms of radial geodesics; thus the Jacobi influence of curvature on geodesics leads to a Jacobi description of exponential maps.
TX(
6. Proposition. Let o be a point of M and let x E T,(M). For u, E T,M I , d expAu,) =
W),
where V is the unique Jacobi field on the geodesic y x such that V ( 0 )= 0
and
V’(0)= u E T,(M).
218 8
Symmetry and Constant Curvature
Proof: As in the proof of the Gauss lemma (5.1) consider the twot I1 and I s I small. Its parameter map t ( r , s) = t(x su) in T,(M),with 0 I exponential image in M ,
+
x ( t , s) = exp,(t(x + su>> = ~,+,,(t), is a geodesic variation of y x l [O, 11. Thus the variation vector field V ( u ) = x,(u, 0) = d exp,(i,(u, 0))
is a Jacobi field on y x . Since k(1, s) = x + su, V ( 1) = d exp,(u,). The curve s + x(0, s) is constant at 0,hence V ( 0 ) = x,(O, 0) = 0. By Proposition 4.44 V’(0) = X,,(O, 0) = X,,(O, 0).
Now s + x,(O, s) = x + su is a vector field on the constant curve at o. Hence, by Proposition 3.18, x,,(O, s) = v for all s. In particular, V’(0)= u. This result provides a strong link between M and its curvature tensor. TIDAL FORCES
A vector field Y on a curve a : 1 -+ M is tangent to a if Y =fa‘ for some % ( I ) and perpendicular to u if ( Y , a’) = 0. We consider how these notions relate to Jacobi fields. If la’] > 0, then each tangent space x ( , ) ( M )has a direct sum decomposition Ru‘ + a‘l. Hence each vector field Y on ci has a unique expression Y = Y T + Y’, where YT is tangent to u and Y’ is perpendicular to a. If y is a geodesic, then Y I y implies Y’ I y. since d/ds( Y , y’} = (Y’, y’}; similarly for tangency. Then, if y is also nonnull, it follows that (Y‘)’ = ( YT)r and (Y‘)’ = (Y’)’. {E
7. Lemma. Let Y be a vector field on a geodesic y. (1) If Y is tangent to y then it is a Jacobi field o Y” = 0 0 Y(s) = + b)y’(s)for all s. (2) If Y is a Jacobi field, then Y I y o there exist a # b such that Y ( a )Iy, Y ( b )I y 9 there exists a such that Y ( a ) I y, Y’(a) I y. (3) If y is nonnull, then Y is a Jacobi field o both Y T and Y’ are Jacobi fields.
(as
Proof: (1) Since R(u, u ) = 0, the Jacobi equation forfy’ is equivalent to d2f/ds2= 0. (2) Again since R(u, u ) = 0, the second derivative of ( Y , y‘) is zero, hence ( Y ( s ) ,y’(s)} = A sB, and the result follows. ( 3 ) R( YT, 7’) = 0, hence R( Y’, y‘) = R( Y , y’). Also, since R( Y , 7’) is skew-adjoint, R( Y, y’)y’ Iy.
+
Locally Symmetric Manifolds 21 9
Thus the Jacobi equation Y ” = R ( Y , y‘)y’ splits into the two equations (YT)”= 0 and (Y’)’’ = R ( Y 1 , y’)y’. The result follows using (1).
A Jacobi field tangent to a geodesic y is of scant importance since it is the infinitesimal model for a family of geodesics that merely reparametrize y. Thus in considering the Jacobi equation Y ” = R,,,. y’ as relatiue acceleration produced by tidalforce we shall emphasize the case Y Iy. 8. Definition. For a vector 0 # u E T,(M) the tidal force operator F,: u’- + u’ is given by F,(y) = Ryvu. -
9. Lemma. F , is a self-adjoint linear operator on u’, and trace F , Ric(o, 0).
=
Proof. Clearly F , is linear (and carries u’ to itself). The pair symmetry of curvature implies that F , is self-adjoint, but note that u’ is a degenerate subspace if u is a null vector. If u is nonnull, let e 2 , .. . , en be an orthonormal basis for u’-. Then Ric(u, v ) = ci(RUe,v,ei) = -1 Ei(Fv(ei),e i ) = -trace F,. If u is null, let w be a null vector such that ( u , w ) = - 1; these two vectors span a Lorentz plane IZ. Then e , = ( u + w)/,,h and e2 = ( v - w)/$ form an orthonormal basis for IZ with e l timelike and e2 spacelike. Let e 3 , .. . , en be an orthonormal basis for Il’ c u‘. Then
c
Ric(v, u> = -
C Ej(Rve,U, ej>.
j>2
The first two terms on the right-hand side cancel since (RuelV, e l > = HRuwv, W > = (Rve2v, e l > . Now u, e 3 , .. . , en is a basis for d , and F,(v) = 0. Hence trace F , = ej(F,(ej), e j ) = - E~(R,,,u,e j ) . j> 2
1
j>2
In the nonnull case tidal forces are often normalized by taking u to be a unit vector. For example, if M has constant curvature C , then Corollary 3.43 gives F,(y) = - &Cy,where E = ( v , u ) = f 1. LOCALLY SYMMETRIC MANIFOLDS
10. Proposition. The following conditions on a semi-Riemannian manifold M are equivalent: (1) DR = 0, that is, M is locally symmetric. (2) If X , Y , 2 are parallel vector fields on a curve a, then the vector field R,,Z on a is also parallel.
220 8
Symmetry and Constant Curvature
( 3 ) Sectional curvature is invariant under parallel translation, that is, the sectional curvature of a nondegenerate tangent plane I7 remains constant as n is parallel translated along any curve a. Proof: For simplicity assume that the curve a is regular; this will be sufficient for later work. (2). Fix an arbitrary point on a,say a(0). By Exercise 3.12 there (1) exist vector fields V , X , F, Z on a neighborhood of a(0) in M such that K,f) = a'(t), = X ( t ) , and similarly for P and Z . Since D R = 0, - - --0 = (D,R),?;Z = D,(R(X, Y ) Z ) - R ( D V X , Y ) Z - R ( X , D v F ) Z - R ( X , P)(D,Z). Now evaluate at 0. There, for example,
(4m
( O ) =
( E m = Y'(0) = 0,
since Y is parallel. Thus the equation above reduces to
(RXYZ)'(O) = Dd(O,(RXYZ)= 0. (2) (1). If u, x, y , z E T,(M), let X , Y, Z be the parallel vector fields on yu gotten by parallel translation of x, y, z, respectively. Then
-
(D,R),,z
=
( R x y Z ) ' ( 0 )= 0.
(2) (3). Let n(0)be a nondegenerate tangent plane at a(O), and let X , Y be parallel vector fields on a such that X(O), Y ( 0 ) is a basis for n(0). Thus X(r), Y ( t ) is a basis for the parallel translate n(t)of n(0)along a. Then both ( R , , X , Y) and Q ( X , Y ) are constant along a, hence the sectional curvature of n(t)is constant. ( 3 ) (2). Suppose a : I -+ M starts at p . By orthonormal expansion it suffices to show that, if X , Y , Z , W are parallel vector fields on a, then ( R x y Z , W ) is constant. Fix t E I and define a function A : T,(M)4 -+ R by A(x, y , Z, W ) = (RxrZ, w)(t>,
where X , Y, Z , W are parallel vector fields on a extending x, y, z , w. Since Q ( X , Y ) is constant, ( 3 ) gives
for all x, y e T,(M) spanning a nondegenerate tangent plane. It is easy to check that the function A is curvaturelike (page 79); hence by Corollary 3.42
&, y , z, 4 = (RxyZ,w ) . Thus ( R , , Z , W ) ( t ) is independent oft.
lsometries of Normal Neighborhoods
221
11. Corollary. A semi-Riemannian manifold of constant sectional curvature is locally symmetric.
This is obvious from criterion (3) above. Semi-Riemannian products of locally symmetric manifolds are again locally symmetric but (if nonflat) do not have constant curvature. ISOMETRIES
OF NORMAL NEIGHBORHOODS
Let M and R be semi-Riemannian manifolds of the same dimension and index and let o E M and A E M . Our goal is to find conditions under which a given linear isometry T,(M) + &(M) is the differential map of an isometry defined on some normal neighborhood of 0. 12. Definition. Let L: & ( M ) -+ T,(M) be a linear isometry, and let Q be a normal neighborhood of o in M such that exp, is defined on the set L(exp,- '(42)). Then the mapping
rbL = exp50Loexp;':%
+
M
is called the polar map of L on 42. In short, +L sends exp,(u) to exp,(Lu) for all u E Q c 7',,(M). Polar maps always exist for Q sufficiently small, and the first two properties below show that if the isometry we seek exists it must be a polar map of L. 13. Lemma. With notation as above, ( 1 ) 4Lcarries radial geodesics to radial geodesics; explicitly, if u E T,(M), then & o y u = yLu, where both sides are defined. (2) The differential map of 4 at o is L. (3) If 42 is sufficiently small, then 4Lis a diffeomorphism onto a normal neighborhood of o in (4) If M is complete, +L is defined on every normal neighborhood of 0.
a.
Proof. (1) Since y,(t)
=
exp,(tu),
4L(YU(t))= exPo(L(t4) = exPo(tL(u))
=
YLdt)
for all t such that y,(t) remains in Q. (2) If u E T,(M), then by (1) &L(U)
=
ddL(YL(0))
= (rbL
O
Y,)'(O)
=
YkU(0)
= LU.
222 8
Symmetry and Constant Curvature
(3) If @ is sufficiently small, $L(%) is contained in a normal neighborhood ofo. Then exp, is a diffeomorphism of P = L(exp; ' %) onto a (normal) neighborhood Y" of 6 Thus 4L:% -+ V is a composition of diffeomorphisms. Finally, (4) is clear since exp, is defined on all of %(AT). a If L is to be the differential map at o of an isometry, then, as we saw in Chapter 3, it must preserve curuature at 0 ;that is, L(RxyZ)= RLX,&z)
for x, y ,
E
T(W.
By Corollary 3.42 it is equivalent that L preserve sectional curvature: K(ZZ) = K ( ~ 1 7 for ) nondegenerate planes in To(M). This necessary condition is sufficient. 14. Theorem. Let M and
manifolds, and let L: curvature. Then
T(M)
-+
be locally symmetric semi-Riemannian &(AT) be a linear isometry that preserves
(1) If 2% is a sufficiently small normal neighborhood of o, there is a unique isometry 4 of -& onto a normal neighborhood V o f 0 such that d#lo = L. (2) If M is complete then for any normal neighborhood 42 of 0,there is a unique local isometry 4: 1 -+ M such that d40 = L.
Proof: In both cases uniqueness follows from Proposition 3.62. By the preceding lemma, existence in both cases follows if every polar map 4 = 4L:42 -+ is a local isometry. Thus for u E T J M ) , p E %, we must prove that (dqh, d q h ) = (0, u ) . The idea is to use Proposition 6 and show that corresponding Jacobi fields grow at the same rate in both manifolds. (1) Let 02 be the neighborhood in T'(M) corresponding to the normal neighborhood 42. There is a unique x E 4 and y X g T,.(T'M) such that d exp,(y,) = u. By Proposition 6, (0, o) = (Y(1), Y(1)), where Y is the Jacobi field on yx such that Y(0) = 0 and Y'(0) = Y E T,(M). (2) Now look at the corresponding situation in AT. Since L is linear, d ~ ( y ,= ) ( L Y ) L ~ . Hence by the definition of the polar map 4,
d4(u)
=
d exPo((Ly)Lx).
Thus, as above, (d&, d 4 v ) = ( P ( 1 ) , P(1)), where P is the unique Jacobi field on yLx such that P(0) = 0 and P'(0) = LY E G(R). (3) Let E , , . . . , En and El, . . . , Enbe parallel frame fields on y x and yLx respectively, such that L(E~(O))= E,(O) for all i. Since L is a linear isometry, the coordinates of x and y relative to the Ei(0)s are the same as the coordinates of LX and ~y relative to the E,(O)s.
lsometries of Normal Neighborhoods
223
If we write Y = y i E i , then the functions y ' , . . . ,y" satisfy the system of differential equations in the proof of Lemma 5 as well as the initial conditions corresponding to Y(0) = 0 and Y'(O)'= y . Over in El, writing P = y'Ei gives the corresponding differential equations
1
Furthermore, by a remark above, the functions y ' , . . ., y" and j ' , . . .,y" satisfy exactly the same initial conditions. (4) We assert that R $ k = RZk on common domain I of y x and y L x (keeping the former in 42). In fact, since the linear isometry L preserves curvature, it follows from our choice of corresponding frame fields that j ? Z k ( O ) = R&(O). Since M is locally symmetric, Proposition 1 shows that the functions ( R E j E k E iE, m ) and hence RZk are constant on I . Similarly REk is constant, giving the result. i I n) satisfy the same system of dif( 5 ) The functions y' and y i (1 I ferential equations and the same initial conditions. By the uniqueness of such solutions, y' = y' for all i; hence ( d 4 u , d$u) = (P(l), Y(1)) = ~ ~ ( y ' ( 1=) )(Y(1), ~ Y ( 1 ) ) = (0, u ) .
C
A first consequence of the theorem is that constant curvature determines local geometry: 15. Corollary. Let M and R be semi-Riemannian manifolds with the same dimension and index and the same constant curvature C . Then Any points o E M and B E R have isometric neighborhoods.
This is clear since a linear isometry from T J M ) to &(R) exists (by 2.27) and must preserve curvature. Another application will explain the term locally symmetric. For p E M let (, be the polar map of the linear isometry u + - u of T,(M). If 42 is a suitably chosen normal neighborhood, then [,: (42 + 42 is a diffeomorphism. Evidently (, reverses geodesics through p: if y(0) = p , then (,(y(s)) = (y -s). In fact this property uniquely determines (, (once is specified), hence [, is called the local geodesic symmetry of M at p . 16. Corollary. The following conditions on a semi-Riemannian manifold are equivalent:
(1) M is locally symmetric. (2) If L: T,(M) + T , ( M ) is a local isometry that preserves curvature, then there is an isometry 4 of normal neighborhoods of p and q such that dcjp = L. (3) At each point p of M the local geodesic symmetry (, is an isometry.
224 8
Symmetry and Constant Curvature
Proof: (1) (2). By the theorem. (2) (3). The linear isometry -id on T,(M) carries each nondegenerate plane Il c T,(M) to itself; thus -id preserves curvature. If 4-idis the isometry given by (2), then reverses geodesics through p , hence is the local geodesic symmetry at p . (3) * (1). We show that the tensor field D R vanishes at an arbitrary point p . Since [, is an isometry it preserves curvature and covariant derivatives. The differential map of l pat p is -id, hence for all u, x, y , z E T'(M)
-(D,R),,z Hence DR = 0 at p .
=
u - " W - , ,-,(-4 = (D,R),,z.
SYMMETRIC SPACES
The local result in Theorem 14 can be generalized as follows. 17. Theorem. Let M and hl be complete, connected, locally symmetric semi-Riemannian manifolds, with M simply connected. If L: T,(M) -+ %(a) is a linear isometry that preserves curvature, then there is a unique semi-Riemannian covering map 4: M -+ R such that d 4 0 = L. We postpone the proof to look first at some consequences. Roughly speaking, the theorem asserts that a complete connected locally symmetric manifold R is determined-up to semi-Riemannian covering-by its curvature at one point. The result is due in principle to E. Cartan (1869-1951), who singlehandedly created the theory of symmetric spaces. (Also see [A] and Theorem 1.36 of [CE].)
18. Definition. A semi-Riemannian symmetric space is a connected semi-Riemannian manifold M such that for each p E M there is a (unique) isometry 5,: M -+ M with differential map -id on T,(M). The isometry i,is called the global symmetry of M at p . Since it reverses the geodesics through p , i,is the unique extension to all of M of the local geodesic symmetry at p . Thus the latter is an isometry; so by Corollary 16, symmetric implies locally symmetric. 19. Examples. ( 1 ) R" is symmetric, since for each point p the map p x -+ p - x is an isometry. (2) The sphere S" is symmetric, since for each p it is symmetric in the usual Euclidean sense about the line in R"" through p and - p . (3) In fact, by Proposition 4.30, euery connected hyperquadric is to be -el,. . . , --en). symmetric (take the framef,, . . . ,,L
+
Symmetric Spaces 225
Chapter 11 will exhibit a variety of symmetric spaces with nonconstant curvature. A locally symmetric manifold need not be complete, since any open submanifold of one is again one; however,
20. Lemma. A semi-Riemannian symmetric space is complete. Proof: To show that a geodesic y: [0, b) + M is extendible, choose c near b in the interval, and let [ be the global symmetry at y(c). Since [reverses geodesics through y(c), a reparametrization of [ 0 y provides the required rn extension of y.
21. Corollary. A complete, simply connected, locally symmetric semiRiemannian manifold is symmetric. Proof: At any point p E M the linear isometry -id on T’(M) preserves curvature. Thus we can apply Theorem 17, with R = M , to obtain a semiRiemannian covering map 4: M + M such that ddP = -id. Since M is rn simply connected, 4 is an isometry, by Corollary 7.27.
It follows that, if M is a complete, connected, locally symmetric manifold, then its simply connected semi-Riemannian covering manifold fi is symmetric. In fact, by Corollary 7.29, fi is complete and (since the covering map is in particular a local isometry) fi is locally symmetric. Thus the preceding corollary shows that fi is symmetric. Now we return to the proof of Theorem 17. The scheme is as follows. By Theorem 14 there is an isometry on some neighborhood of o E M whose differential map is L. Any point p E M can be connected to o by a broken geodesic /?, so we can parallel translate L along p to p and there get another isometry on a neighborhood of p . Because M is simply connected, these locally defined isometries fit together to give the required map 4. Some notation is required to cope with broken geodesics. For any k 2 1 let u be a k-tuple ( u l , . . . , u k ) of vectors in T,(M). Then the broken geodesic 8,: [0, k ] + M is defined inductively as follows (using the fact that M is complete). Let = y,, I[O, 11. If pj has been defined on [ O , j ] , for 1 Ij < k , let w = P ( U ~ +be~ the ) parallel translate of u j + l along pj to its endpoint bj(j).Then attaching t
to
pi gives pi+ 1.
-,y,(r
-
j)
( j IE Ij
+ I)
(1) For u as above, let P,: T,,(M) + TB&)(M) be parallel translation along Similarly in R,for LU = ( L u l , . . . ,L u k ) let PLv:T(M)+ q , , ( k ) ( R ) be parallel translation along I j L v . Then the parallel translate of L along Ijk is
8,.
L” = P L ,
0
L
0
Pv- : T,(M)
+
T(@),
226 8
Symmetry and Constant Curvature
where p = /l,(k) and q = fiL,(k). Since M and R are locally symmetric the linear isometries P , and PLupreserve curvature, hence so does the linear isometry L,. By Theorem 14, there is a neighborhood 02, of each fl,(k) and a local isometry 4u:4Y, + h? whose differential map at &(k) is L,. In view of Lemma 5.10, we can suppose that each J2,is convex and that if $2, and $2w meet, then their intersection is convex. hence connected. (2) We now construct a matched covering of M . Let A be the set of ktuples in T , ( M ) for all k 2 1. Let M be the collection of convex open sets 42, as above for all v E A. Since M is connected, 53 is a covering of M . If u, w E A , define t' w to mean that 4u= 4won $2, n aW. Evidently the relation is reflexive and symmetric, so it remains to prove the quasi-transitive condition (3) of Definition 7.30. Suppose u v, v w, and p E 4Yu n $2, n $2w. It follows from the two relations that the differential maps d4", d&, d'$, all agree at the point p . Hence by Proposition 3.62 4u = bwon $2,, n 4Yw. Thus u w, and (A, ) is a matched covering. (3) The essential step of the proof is to show that this matched covering is chainable over any geodesic segment a: [O, I] + M , since then Proposition 7.32 applies. Choose u = ( u l , . . . , u k ) E A so that fiu runs from o to a(0). Let uk+ be the ) to o along flu. For each s E [0, 11 let w(s) = parallel translate of ~ ' ( 0 back (vlr . . . , u k , suk+ A . Thus the last segment of flw(s)is a reparametrization of CJI[O, s]. Then there is a partition 0 = so < s1 < . . . < s, = 1 such that
-
-
-
- -
-
for
a([si- 1, s i ] ) c $2w(sI)
1 I i I m.
-
See Figure 1. It remains to check that w(siw(si). Abbreviate 4w(si) to 4i. Let P be parallel translation along CJ from a(sito o(si),and correspondingly let P be parallel translation along 4i0 a from c$~cJ(s~- to 4ia(si).It follows from the definition of L, that LW(S,. I)
=
p-
O LW(Si)
p.
As a local isometry, 4i preserves parallel translation ; hence, ( d 4 i ) o , s z- I )
=
P-
0
(d4i)ofsi) 0 P .
Figure 1
Simply Connected Space Forms
By construction, for each i, L , , , ( ~ ~is) the differential map of q5i at b,,,(& Thus the two preceding formulas yield
= a(s,).
227
+ 1)
-
= 4i on n a,+); that is, w(siPI) w(si). By Proposition 7.32, the matched covering (9, -) gives a semiRiemannian covering map &: M* + M . As with any matched covering, for meets A,,,(%,,,) if each v E A there is local section A":%" + M * , and and only if u w. It follows that the maps 4" A : &(%J + R for all u E A agree on overlaps, hence define a single map 4*: M* -+ R.This is a local isometry since A and every 4" are. Since M is simply connected, Corollary A.14 applies, showing that there is a (locally isometric) global section A: M + M * such that 111q0= Lo for (0) E A . Thus 4 = 4* 0 A: M -+ R is a local isometry with 4la0= $o; hence d$o = L. B
Hence,
$i-l
(4)
-
0
SIMPLY CONNECTED SPACE FORMS 22. Definition. A space,forrn is a complete connected semi-Riemannian manifold of constant curvature.
Space forms must be regarded as the simplest important class of semiRiemannian manifolds. In the simply connected case, Theorem 17 is decisive:
23. Proposition. Simply connected space forms are isometric if and only if they have the same dimension, index, and curvature C. ProoJ: Obviously the conditions are necessary. For the converse, note that, since constant curvature implies local symmetry, Corollary 2 1 shows that simply connected space forms are symmetric. Suppose M and N are simply connected space forms of the same dimension and index. The latter show that linear isometries T,(M) -+ T,(N) exist. Then Theorem 17 gives a semi-Riemannian covering map M -+ N which, since M is simply connected, is an isometry. Thus up to isometry there is at most one simply connected space form M(n, v, C) of dimension n, index v, and curvature C. Existence for all (n, v, C) is settled for C = 0 by the semi-Euclidean spaces R:, and, with suitable modifications, for C > 0 by pseudospheres and for C < 0 by pseudohyperbolic spaces.
228 8
24.
Symmetry and Constant Curvature
Corollary. For n 2 2,
i-
S:(r) if C = l/r2 and 0 I v I n - 2, if C = 0, M(n, v, C ) = R: Ht(r) if C = - l/rz and 2 I v I n. H;-,,(r), these space forms are all In fact, since St(r) s"-' x R simply connected. The remaining cases come in anti-isometric pairs. One-dimensional semi-Riemannian manifolds are trivially flat ; hence by Exercise 8, the only simply connected one-dimensional space forms are R' and R : . For n 2 2 we have .V
M(n, n, l / r 2 ) = one component cS#) of S:(r), and by metric reversal, M(n, 0, - l/r2) = hyperbolic n-space H"(r), one component of H$(r). M ( n , n - 1, l/r2) = S:- l(r), the simply connected semi-Riemannian covering manifold of S:- l(r), and M ( n , 1, - l/r2) = I?y(r), the simply connected semi-Riemannian covering manifold of H l ( r ) .
-
Since R" is the simply connected covering manifold of S' x R n - ' , these four types are diffeomorphic to R". 25. Corollary (Hopf). A complete, simply connected, n-dimensional Riemannian manifold of constant curvature C is isometric to
the sphere S"(r) Euclidean space R" hyperbolic space H"(r)
if if if
c = 1/r2,
C = 0, C = - l/r2.
This result is particularly satisfying from a historical viewpoint, since all three of these constant curvature geometries were well known, at least in low dimensions, before Riemannian geometry was invented (1 845) and long before a rigorous proof of this corollary (1926). Euclidean geometry is of course the oldest, and hyperbolic geometry, dating from the early 1800s, the newest. Writing fi for the simply connected semi-Riemannian covering manifold of M , the Lorentz analogue of Hopf's result is
26. Corollary. A complete, simply connected, n-dimensional Lorentz manifold of constant curvature C is isometric to the Lorentz sphere S ; ( r ) s:(r) Minkowski space RY @(r>
if C = l/r2 and if C = l/r2 and if C = 0, if c = -l/r2.
n 2 3, n = 2;
Simply Connected Space Forms
229
In relativity theory S:(r) is called de Sitter spacetime, and fl:(r) is universal anti-de Sitter spacetime. 27. Example. A model for pl.As Figure 2 suggests, H t can be regarded as a hypersurface of revolution in R;i ' = R: x R"- Unwrapping the circles of revolution gives a covering map R : R' x R"-' + H;, where
'.
R(t, XI
=
(JEG 'cos t, J
sin t, XI.
Furnished with the pulled-back metric, R" becomes the simply connected semi-Riemannian covering manifold of H ; -and hence can be denoted by
A;.
R: (coordinates uo, u ' )
' (coo1-dinates u 2 , . . . ,u")
Figure 2. H; in R;+'
The geometry of the model can be derived from that of H ; as modified by the map A. For example, if the coordinate slice uo = 0 in RI' is regarded as R;, then R; n H ; is just H",'. The latter's two components H"-' are totally geodesic in H:. By rotational symmetry the same is true for all slices of H ; by hyperplanes through R'- '. Thus in plthe slices t constant are totally geodesic spacelike hypersurfaces, isometric to H"- (or R' if n = 2).
In Riemannian geometry, curvature inequalities such as K 2 c and aI K I b have been intensively studied. For indefinite metrics we shall now see that such conditions imply constant curvature. Sectional curvature K is undefined on a degenerate plane Z l in T'(M), but a formula in the proof of Lemma 3.39 shows that sgn(R,,X, Y ) is the same for all bases X, Y for n. This provides a well-defined function JV from the set of degenerate planes in T,(M) to the set (+ 1,0, - 1 >. 28. Proposition (Kulkarni et a/.). Let p be a point of a semi-Riemannian manifold of indefinite metric. The following conditions on T,(M) are equivalent : (1) K is constant, (2) N = 0,
230 8
Symmetry and Constant Curvature
(3) a I K or K I b, where a, b E R, (4) a I K 5 b on indefinite planes, ( 5 ) a I K < b on definite planes. Proof (S. Harris et al.). Evidently we can assume dim M 2 3. Obviously (1) implies (3, 4, 5), and by Corollary 3.43 it implies (2). The reverse implications will use this consequence of .Ar = 0: (*) If X , U , Z are orthonormal vectors and X and U have opposite causal character, then K ( X , Z ) = K ( U , Z ) .
To prove this note that U are degenerate. Hence
(Rx+u , A X
k X are null vectors and the planes (U i-X , 2)
+ u),Z > = 0 = < R x - u , z ( X
-
Expansion yields first ( R x z U , Z ) = 0, then ( R x z X , Z ) Since Q ( X , Z ) + Q ( U , Z ) = 0, the result follows.
W, Z > .
+ (RuZ U , Z ) = 0.
( 2 3 1) Case I v = 1 or n - 1. By metric reversal we can assume U , X , Y is orthonormal with U timelike, then (*)implies K ( U , X ) = K ( X , Y ) . It follows that every nondegenerate plane that either contains or is orthogonal to U has the same sectional curvature k(U). If V is an independent timelike vector, the plane spanned by U and V is nondegenerate, hence k(U) = K ( U , V ) = k ( V ) . v = 1. If
Case I I 1 < v < n - 1. Let U be a positive definite plane, U’negative definite, and let X E ZZ and U E U’ be unit vectors. Now Q ( X , U ) < 0, and using Lemma 2.19 we easily check that ZZ + RU and U‘ + R X are nondegenerate. Hence by Case I, K(ZZ) = K ( U , X ) = K ( n ’ ) . This suffices because, if X , U is an orthonormal basis for any indefinite plane, then extension to an orthonormal basis for T’JM) will provide planes U , U’ as above.
(332). Suppose a I K (reversing the metric gives K I b). By Exercise 9 every degenerate plane n is a limit of definite planes and a limit of indefinite + ZZ = n ( X , Y ) , where X i + X and +Y. planes. Consider U(Xi, Then ( R ( X t , V X i , X) a.
x)
x
Q(xi,V)
Here the numerator approaches ( R ( X , Y ) X , Y ) and the denominator approaches zero. Thus, if all Q ( X i , > 0, then ( R ( X , Y ) X , Y ) 2 0, hence N ( U ) 2 0. Similarly, if all Q ( X , , < 0, then M ( n ) 5 0; hence N ( U ) = 0.
x) x)
x)l/
( 4 a 2) and (5 * 2 ) . With notation as above, I ( R ( X i , x ) X i , I Q ( X i , 6)l is bounded, where Q ( X i , is negative for (4), positive for (5). Hence ( R ( X i , x ) X i , approaches zero, so N(n)= 0. rn
x)
x)
Transvections 231
If at each point of M, connected and dim 2 3, one of the conditions in the proposition holds, then, by Schur’s theorem (Exercise 3.21), M has constant curvature. TRANSVECTIONS 29. Definition. An isometry geodesic y : R + M provided
4: M
+M
is a transvection along a
(1) 4 translates y; that is, &(s) = y(s + c) for all s E R and some c. (2) d 4 gives parallel translation along y ; that is, if x E q(s)(M),then d4(x) E T ( s + c l ( Mis) the parallel translate of x along y. For example, if 4 is a rotation of the sphere S2about the z-axis in R3,then 4 is a transvection along the (geodesic) equatorial circle z = 0. In a symmetric space there are transvections along every geodesic.
be the global 30. Lemma. If y is a geodesic in a symmetric space, let is symmetry of M at y(s). Then for any c, the isometry [cizio is a transvection along y that translates it by c. Proof. For all s, [o(y(s))
=
y( - s). But c/2 is the midpoint of [ -s, s
+ c],
+ c). so i c i 2 i o Y ( S ) = If X is a parallel vector field on y , then for any s, d [ , ( X ) is a parallel vector 0 y, which is a reparametrization of y. (Since parallel translation field on is is independent of parametrization, we can ignore such reparametrizations.) If x E TCs, M , then x is parallel along y to a tangent vector y at y(0). Hence dio(x) is parallel to diO(y) = - y , and also to a vector z at y(c/2). Thus di,,, dcO(x)is parallel to - z , hence to y, hence to x.
31. Corollary. Every nonconstant geodesic in a symmetric space is either one-to-one or simply periodic.
Proof. It suffices to show that, if y is a geodesic with y(b) = y(0), then y(s b) = y(s) for all s. For each s let 4sbe the transvection that translates y by s. Then
+
Y(S
+ b) = qbs(Y(b)) = q b S ( Y ( 0 ) )
=
’
Exercises
1. An isometry iof M , connected, is a global symmetry at p E M if and only if [ is involutive (that is, 1’ = id) and p is an isolated fixed point of [.
232 8
Symmetry and Constant Curvature
2. (a) Let M be a semi-Riemannian submanifold of a symmetric space M.If
M is connected, closed in M,and totally geodesic, then M is symmetric. (b) A semi-Riemannian product M x N is symmetric if and only if both M and N are symmetric. (c) Let k : A -+ M be a simply connected semi-Riemannian covering. If M is symmetric, then fi is symmetric (but not conversely). 3. (a) If M is a complete Riemannian manifold, then every element of nl(M, p ) contains a geodesic loop. (Hint: Use the Hopf-Rinow theorem in the simply connected covering manifold.) (b) The fundamental group of a Riemannian symmetric space is abelian. (Hint: Show that the geodesic symmetry at p induces the homomorphism g -+ g - l on q ( M , p).) 4. A model of hence (reversing the metric) R:. Consider the smooth map x: R2 + R: given by x(t, 9) = (sinh t, cosh t cos 9, cosh t sin 9). (a) Prove that x is a covering map of R2 onto the unit pseudosphere S: in R:. (b) Prove that R' furnished with the pulled-back metric is $ and the line element is -dt2 + cosh' t d 9 2 . (c) Sketch some null geodesics of this model and determine which points can be reached by geodesics starting at the origin. 5. (a) The model of in Example 27 has line element - S2 dt2 + do2,where S = (1 + x x ) ~ and ' ~ do2 = dx dx - ( ( x d x ) 2 / S 2 ) on R"- l . (Hint: S d S = x - d x . ) (b) do2 on R"-' gives H"-'.(For other models of hyperbolic space see [Wo].) 6. Let y be a geodesic with (y', y ' ) = E = f 1 in a semi-Riemannian surface. If Y is a Jacobi field on y perpendicular to y, write Y = Y E , where E is a parallel unit vector field on a. Show that the Jacobi equation for Y reduces to y" + EKY = 0. 7. Let y be a geodesic in a manifold of constant curvature C , and let Y be a Jacobi field on y that is ly. (a) the Jacobi equation for Y is Y" + C(y', y')Y = 0. (b) Let c = 1 C(y', y ' ) I '". On y there exist parallel vector fields A , B 1y such that Y(s) = cos(cs) A(s) + sin(cs) B(s) if C(y', y ' ) > 0. Y ( s ) = A(s) + sB(s) if C(y', y ' ) = 0. Y ( s ) = cosh(cs) A(s) + sinh(cs) B(s) if C(y', y ' ) < 0. 8. (a) Let M be a flat connected semi-Riemannian manifold complete at a E M (that is, expo is defined on all of T,(M)).Prove that exp,: T,(M) -+ M is a semi-Riemannian covering map. (Hint: Use Proposition 6.) (b) Give an example of a connected semi-Riemannian manifold that is complete at one point but not complete. 9. In a vector space with indefinite scalar product, every degenerate 2-plane is a limit of (a) indefinite planes (b) definite planes. (Hint: See proof of Lemma 3.40, but for (b), if gln = 0, consider n(u + 6x, w + dy).)
s:,
-
-
.