On Geodesics in Riemannian Geometry

Home , Geodesic, Manifold, Riemannian manifold, Spray (mathematics)

Lectures on Geodesics Riemannian Geometry By M. Berger Lectures on Geodesics Riemannian Geometry

By M. Berger

No part of this book may be reproduced in any form by print, microﬁlm or any other means without written permission from the Tata Institute of Fundamental Research, Colaba, Bombay 5

Tata Institute of Fundamental Research Bombay 1965 Introduction

The main topic of these notes is geodesics. Our aim is twofold. The first is to give a fairly complete treatment of the foundations of riemannian geometry through the tangent bundle and the geodesic flow on it, following the path sketched in [2] and [19]. We construct the canonical spray of a riemannian manifold (M, g) as the vector field G on T(M) defined by the equation 1 i(G) dα = dE, · −2 where i(G) dα denotes the interior product by G of the exterior deriva- · tive dα of the canonical form α on (M, g) (see (III.6)) and E the energy (square of the norm) on T(M). Then the canonical connection is introduced as the unique symmetric connection whose associated spray is G. The second is to give global results for riemannian manifolds which are subject to geometric conditions of various types; these conditions involve essentially geodesics. These global results are contained in Chapters IV, VII and VIII. Chapter IV contains first the description of the geodesics in a symmetric compact space of rank one (called here an S.C.-manifold) and the description of Zoll’s surface (a riemannian manifold, homeomorphic to the two-dimensional sphere, non isometric to it and all of whose geodesics through every point are closed). Then we sketch results of Samelson and Bott to the effect that a riemannian manifold all of whose geodesics are closed has a cohomology ring close to that of an S.C.-manifold. In Chapter VII are contained the Hopf-Rinow theorem, the existence of a closed geodesic in a non-zero free homotopy class of a compact rie-

iii iv Introduction mannian manifold, and the isometry between two simply connected and complete riemannian manifolds of the same constant sectional curvature. In Chapter VIII one will find theorems of Myers and Synge, the Gauss-Bonnet formula and a result on complete riemannian manifolds of non-positive curvature. Then come the theorem of L.W. Green, which asserts that, on the two-dimensional real-projective space, a riemannian structure all of whose geodesics are closed has to be isometric to the standard one, and the theorem of E. Hopf: on the two-dimensional torus, a riemannian structure all of whose geodesics are without conjugate points has to be a flat one. As a counterpoint we have quoted the work of Busemann which shows that the theorems of Green and Hopf pertain to the realm of riemannian geometry, for they no longer hold good in G-spaces (see (VIII.10)). However, the result on complete manifolds with non-positive curvature is still valid in G-spaces. We have included in Chapter VIII theorems of Loewner and Pu, which are “isoperimetric” inequalities on the two dimensional torus and the two-dimensional real projective space (equality implying isometry with the standard riemannian structures on these manifolds). These results do not involve geodesics explicitly, but have been included for their great geometric interest. One should also note that the results of Green, Hopf, Loewner and Pu are two-dimensional and so lead to interesting problems in higher dimensions. The tools needed for these results are developed in various chapters: Jacobi fields, sectional curvature, the second variation formula play an important role; see also the formulas in (VIII.4) and (VIII.8). The reference for calculus is [36]; references for differential and riemannian geometry are [14]; [16], [17], [18], [19], [21], [33], [35], and lectures notes of I.M.Singer and a seminar held at Strasbourg University. We have used some of these references without detailed acknowledge- ment. I am greatly indebted to N. Bourbaki, P. Cartier and J.L. Koszul for communication and permission to use ideas and results of theirs on connections. For most valuable help I am glad to thank here P. Cartier, J.L. Koszul, N. Kuiper and D. Lehmann. Contents

Introduction iii

0 Preliminaries 1 1 ...... 1 2 Forms...... 6 3 Integration...... 9 4 ...... 14 5 Curves...... 21 6 Flows ...... 24

1 Sprays 27 1 Deﬁnition...... 27 2 Geodesics ...... 29 3 Expressions for the spray in local coordinates ...... 31 4 The exponential map ...... 33

2 Linear connections 37 1 Linear connection ...... 37 2 Connection in terms of the local coordinates ...... 41 3 Covariant derivation ...... 43 4 The derivation law ...... 48 5 Curvature ...... 53 6 Convexity ...... 60 7 Parallel transport ...... 65 8 Jacobiﬁelds...... 69

v vi Contents

3 Riemannian manifolds 81 2 Examples ...... 85 3 Symmetricpairs...... 87 4 TheS.C.-manifolds ...... 94 5 Volumes...... 100 6 The canonical forms α and dα ...... 104 7 Theunitbundle ...... 109 8 Expressions in local coordinates ...... 110

4 Geodesics 113 1 The ﬁrst variation ...... 113 2 The canonical spray on an r.m.(M, g)...... 117 3 First consequences of the deﬁnition ...... 122 4 Geodesics in a symmetric pair ...... 124 5 Geodesics in S.C.-manifolds ...... 127 6 Results of Samelson and Bott ...... 130 7 Expressions for G in local coordinates ...... 139 8 Zoll’ssurface ...... 142

5 Canonical connection 151 2 Riemannian structures on T(M) and U(M)...... 156 3 Dg = 0...... 162 4 Consequences of Dg = 0 ...... 165 5 Curvature ...... 167 6 Jacobi ﬁelds in an r.m...... 169

6 Sectional Curvature 179 2 Examples ...... 181 3 Geometric interpretation ...... 184 4 A criterion for local isometry ...... 187 5 Jacobi ﬁelds in symmetric pairs ...... 195 6 Sectional curvature of S.C.-manifolds ...... 197 7 Volumes of S.C.manifolds ...... 199 8 Ricci and Scalar curvature ...... 206 Contents vii

7 The metric structure 213 2 The metric structure ...... 221 3 Niceballs ...... 223 4 Hopf-Rinow theorem ...... 230 5 A covering criterion ...... 235 6 Closed geodesics ...... 239 7 Manifolds with constant sectional curvature ...... 246

8 Some formulas and applications 249 1 The second variation formula ...... 249 2 Second variation versus Jacobi ﬁelds ...... 254 3 The theorems of Synge and Myers ...... 261 4 Aformula...... 266 5 Index of a vector ﬁeld ...... 272 6 Gauss-Bonnet formula ...... 274 7 E.Hops’stheorem ...... 280 8 Anotherformula...... 282 9 L.W.Green’s theorem ...... 288 10 Concerning G-spaces ...... 292 11 Conformal representation ...... 294 12 The theorems of Loewner and Pu ...... 299

Chapter 0 Preliminaries

1 1

In this chapter we formulate certain notions connected with the notion of a manifold in a form we need later on and fix some notations and conventions. By means of a basis we identify a d-dimensional real vector space with set Rd of all d-tuples of real numbers with the standard vector V space structure and denote the element with 1 in the ith place and zeros elsewhere by e . Then e form a basis of Rd, we call it the canonical i { i} basis of Rd, and its dual basis (u1,..., ud) the canonical coordinate system in Rd. (For R(= R1) we set u1 = t). With this identification any real vector space becomes a manifold and this structure is independent of the basis chosen. A manifold will always be a C∞-manifold which is Hausdorff, para compact and of constant dimension d. Generally we denote it by M, a typical chart of it by (U, r) and local coordinates with respect to (U, r) by xi = ui r. Let us note that because of para compactness partitions ◦ of unity for M exist. We denote the tangent space of M at a by Ta(M) and define the tangent bundle T(M) of M to be (0.1.1) x = T(M) { } a[M x [Ta(M) ∈ ∈ and call elements of T(M) vectors of M. Then we have the natural

1 2 Preliminaries

projection map pM (or simply p) from T(M) onto M which takes every vector of M at a onto a. 2 We denote the set of all maps (C∞-maps, differentiable maps) of a manifold M into a manifold N byD(M, N) and when N = R we write F(M) for D(M, R). Every f in D(M, N) induces a map of F(N) into F(M) defined by g g f → ◦ and this in turn a map f T of T(M) into T(N) defined by the equation

(0.1.2) ( f T (x))(g) = x(g f ), g F(N). ◦ ∈ Then we have the following commutative diagram

f T T(M) / T(N)

(0.1.3) pM pN

f M / N

1.4

T Let us note that if f restricted to Tm(M) is one-one then we can choose a local coordinate system (x1,..., xe) for N at n = f (m) such that xi f { ◦ } (i = 1,..., d) form a local coordinate system at m for M. Also if f T restricted to Tm(M) is onto Tn(N) then we can choose a local coordinate system (x1,..., xd) for M at m and a local coordinate system (y1,..., ye) for N at n such that xi = yi f , i = 1,..., 0. ◦

1.5 We call a manifold N a sub manifold of M if

3 1) N M, ⊂ 2) the topology on N is induced by that on M, 1. 3

3) to each point p of N there is a chart (Up, rp) in the atlas of M such that for some positive integer k

r (N U ) = (x ) r (U ) x + = ... = x = 0 . p ∩ p i ∈ p p k 1 d n o

1.6

T Under these conditions a vector x of Tm(M) will be in i (Tm(N)) (where i : N M denotes the injection) if and only if x(ϕ i) = 0 ϕ F(N) → ◦ ∀ ∈ implies that x = 0. In case the manifold M is an open sub manifold A of a ﬁnite dimensional real vector space V, we identify T(A) with A V as follows: × 1.7 For y V, f F(A), a A set ∈ ∈ ∈ + 1 = f (a ty) f (a) ζa− (y) ( f ) lim − . · t 0 t ! → We see that ζ 1(y) T (M) and that the map ζ 1 is an isomorphism a− ∈ a a− of V with Ta(M). We denote its inverse by ζa and deﬁne a map from T(A) onto V by setting = (0.1.8) ζ(x) ζpA(x)(x).

Then the identiﬁcation is given by the map

(0.1.9) T(A) x (p (x), ζ(x)) A V. ∋ → A ∈ × Given (U, r) the maps ζ rT and (p , ζ) rT are called the principal ◦ r(U) ◦ part and the trivialising map respectively. (See Lang [19] : p. 49). 4 The use of these maps instead of the explicit use of the local coordinates has the advantages of the latter without its tediousness. So, more explicitly, given (U, r) we write

x =(a, b) ℧ 4 Preliminaries

a = (p rT )(x) and b = (ζ rT )(x). r(U) ◦ ◦ Generally we take (a1,..., ad), (b1,..., bd) as the coordinate represen- tations of a, b. Now, we can consider the collection (T(U), (p , ζ) rT ) corre- r(U) ◦ sponding to the charts (U, r) of M as an atlas on T(M) and thus deﬁne a structure of a manifold on T(M). With this structure on (T(M)) we have p D(T(M), M) and furthermore if N is a manifold and f D(M, N) M ∈ ∈ then f T D(T(M), T(N)). ∈

1.10 Since M is Hausdorﬀ and hence there are functions with arbitrarily small support we see that the class of all sections of T(M), i.e.,

X D(M, T(M)) p X = id ∈ ◦ M n o can be identiﬁed with C (M), the set of all derivations of the R-algebra F(M) into itself. Elements of C (M) are called vector ﬁelds on M.

1.11 Now let us take (U, r). Suppose that a vector (a, b) is given in U. In (a, b) we vary the first coordinate a over U keeping b fixed. Thus, clearly, we 5 get a vector field over U. Now we take functions ϕ on M such that

(i) 0 ϕ 1, and ≤ ≤ (ii) ϕ is zero outside U and is 1 in a neighbourhood of a in U.

Then we set

= ϕ(x) X(x) for x U Xℓc  · ∈ 0x for x < U   This vector ﬁeld so obtained is called a locally constant vector ﬁeld at a. 1. 5

If f D(M, V) is a map of a manifold M into a finite dimensional ∈ real vector space, we define, besides f T , a map d f , called the differential of f , from T(M) to V by setting

(0.1.12) d f = ζ f T . ◦

Further if M is a vector space V′, then denoting the Jacobian of f by Df we have (with the Df of [36]):

1.13

D f ζ = d f. ◦ Let us write down explicitly the expression for d f and D f . Let V be n-dimensional with a basis e ,..., e and corresponding coordinate { 1 n} system (x1,..., xn) and, as usual, let (U, r) be a chart of M with local 1 d coordinate system (x ,..., x ). Then f on U is given by nC∞-functions i f = x f (i = 1,..., n) on U. The canonical local bases for C (U) and i ◦ C (V) are ∂ ∂ and respectively. ( ∂xi !) ( ∂xi !)

Then we have 6

j ∂ ∂ f ∂ (0.1.14) f T = , i = 1,..., d. ∂xi ! ∂xi j Xj ∂x

Now is customary let us write the contra variant (tangent) vector as a row vector. Now suppose that d f takes a contra variant vector with coordinates (x1,..., xd) to one with (x1,..., xn), i.e.

∂ ∂ f (0.1.15) (d f ) x = x .  i ∂xi  i i Xi Xi ∂x     6 Preliminaries

Then the coordinates are related by the matrix equation (0.1.16) 1 2 n ∂ f ∂ f ∂ f ...  ∂x1 ∂x1 ∂x1     . . .   . . .     1 2 1  = ∂ f ∂ f ∂ f  = i.e. (x1,..., xn)  ...  (x1,..., xn); X J X  i i i  · ·  ∂x ∂x ∂x   . . .   . . .     1 2 n  ∂ f ∂ f ∂ f   ...   ∂xd ∂xd ∂xd    Hence the Jacobian isJ. 

2 Forms

We consider C (M) as an F(M) module and denote its dual by C ∗(M). 7 The elements of C ∗(M) are called the diﬀerential forms on M.

2.1

Now we deﬁne the cotangent bundle T ∗(M) to be

T ∗(M) = x∗ , { } a[M x [Ta(M) ∈ ∗∈ and deﬁne a manifold structure on T ∗(M) in a way analogous to that on T(M). For ω T (M), X C (M) by evaluation at a in M we see that ∈ ∗ ∈ ω(X)(a) depends only on X(a) and that we can identify C ∗(M) with

(0.2.2) ω D(M, T ∗(M)) p ω = id = C ∗(M) ∈ | M ◦ M More generally, M being any unitary module over a commutative ring A we write Ls(M) (resp. Es(M)) for the A-module of all multilinear (resp. alternating multilinear) forms of degree s on M. There + is a map, called multiplication, from Ls(M) Ls′ (M) into Ls s′ (M) de- × ﬁned by (ω,σ) ω σ, where (ω σ)(X ,..., X , X + ..., X + ) = → · · 1 s s 1 s s′ 2. FORMS 7

ω(X1,..., Xs) σ(Xs+1,..., Xs+s ), and a map, called exterior multipli- · ′ + cation, from Es(M) Es′ (M) into Es s′ (M), obtained by alternating the × latter one. Now we set

L s(M) = Ls(C (M)) and E s(M)Es(C (M)) considering C (M) as an F(M)-module, and

Ls(T(M)) = x s { } a[M x L [(Ta(M)) ∈ ∈ Es(T(M)) = x s { } a[M x E [(Ta(M)) ∈ ∈ considering the Ta(M) as R-modules. We can deﬁne a manifold struc- 8 ture on Ls(T(M)) and also on Es(T(M)) in a way analogous to that on T (M). By evaluation at a of M we can show that for ω L s(M) ∗ ∈

ω(X1,..., Xs)(a) depends only on (X1(a),..., Xs(a)) and hence we can identify L s(M) and E s(M) with

(0.2.3) ω D(M, Ls(T(M)) p ω = id ∈ | M ◦ M and ω D(M, Es(T(M)) p ω = id ∈ | M ◦ M respectively, and call them s-forms on M and s-exterior forms on M respectively. Now for ω L s(M) and a M, ω(a) makes sense and we ∈ ∈ often denote it by ωa, and we have a similar convention for the elements s of E (M)′ s.

2.4 Given a chart (U, r), with the notation

[i] = i <...< i , dx = dx ... dx 1 s [i] i1 ∧ ∧ is 8 Preliminaries

locally we can write any s-exterior form as

(0.2.5) ω dx , where ω F(r(U)). [i] [i] [i] ∈ X[i]

Every f D(M, N) induces a map f of multilinear forms on N into ∈ ∗ those on M. We have, by deﬁnition

T T s (0.2.6) ( f ∗ω)(x ,..., x ) = ω( f (x ),..., f (x )) for ω L (N) 1 s 1 s ∈ and x ,..., x T (M). Thus f takes L s(N) into L s(M) and further- 1 s ∈ a more it takes elements of E s(N) into those of E s(M) and hence can be s s 9 considered as a map of E (N) into E (M). f ∗ is a homomorphism of R-modules having the following properties: i) f (ω σ) = f (ω) f (σ) for ω L s(N), σ L s′ (N), ∗ · ∗ · ∗ ∈ ∈ (0.2.7) ii) f (ω σ) = f (ω) f (σ), for ω E s(N), σ E s′ (N) ∗ ∧ ∗ ∧ ∗ ∈ ∈ iii) for g D(N, L), L a manifold, (g f )∗ = f ∗ g∗. We deﬁne a map∈ d, called exterior diﬀerentiation◦ , from◦ E s(M) into E s+1(M) by setting

s (0.2.8) (dω)(X ,..., X ) = ( 1)iX (ω(X ,..., X ,..., X )) + 0 s − i 0 i s Xi=0 b ( 1)i+ jω([X , X ], X ,..., X ,..., X ,..., X ) − i j 0 i j s 0 Xi< j s ≤ ≤ b b where [X, Y] = X Y Y X for X, Y C (M), X ,..., X C (M) ◦ − ◦ ∈ 0 s ∈ and the element under is to be deleted from the sequence of elements. ∧ This map has the following properties: i) it is R-linear (0.2.9) ii) d d = 0 ◦ iii) d(ω σ) = (dω) σ + ( 1)degree ofωω dσ ∧ ∧ − ∧ iv) d f ∗ = f ∗ d for f D(M, N), N a manifold. As an example◦ we note◦ ∈

(0.2.10) (dω)(X, Y) = X(ω(Y)) Y(ω(X)) ω([X, Y]) − − for ω C (M); X, Y C (M). ∈ ∗ ∈ 3. INTEGRATION 9

For X C (M), ω L s(M) we deﬁne the interior product i(x)ω s 1 ∈ ∈ ∈ L − (M) by the equation

(0.2.11) (i(X)ω)(X1,..., Xs 1) = ω(X, X1,..., Xs 1) X1,..., Xs 1 C (M). − − ∀ − ∈ 2.12 10 s t We call multilinear forms on (C (M)) (C ∗(M)) tensors of type (s, t) × L s and denote the F(M)-module of such tensors by t (M).

3 Integration

3.1 a.1. Let V be a real vector space of ﬁnite dimension d. We call any nonzero element of Ed(V) an orientation on V A basis x ,..., x , in { 1 d} that order, is positive with respect to the orientation S on V is

(0.3.2) S (x1,..., xd) > 0.

3.3 Let V be a vector space. V together with a symmetric positive deﬁnite bilinear form g on V, i.e. a bilinear form such that

g(x, y) = g(y, x), and g(x, x) > 0 x , 0 ∀ is called a g-euclidean (or simply euclidean if there is no possible confusion) space. Sometimes we describe this situation as “V is provided with a euclidean structure g”. Two vectors x and y are said to be orthogonal relative to g if g(xy) = 0. A basis e ,..., e of V is called { 1 d} an orthogonal basis relative to g or simply an orthogonal basis of (V, g) if e1,..., ed are orthogonal i.e. g(ei, e j) = 0 if i , j. It is called an orthonormal basis relative to g if, further,

g(ei, ei) = 1. 10 Preliminaries

3.4 Example. Let V be provided with a euclidean structure and be oriented 11 by S (i.e. S is an orientation on V). Then it admits a canonical orientation S V deﬁned by the equation

S V (x1,..., xd) = 1

for every orthonormal basis x ,..., x , positive relative to S . { 1 d} 3.5 2. A volume t on V is a map Vd R such that it is the absolute value of → an orientation S : t = S , i.e., | | t(x ,..., x ) = S (x ,..., x ) for x ,..., x in V,. 1 d | 1 d | 1 d 3.6 Example 3.2. Let V be g-euclidean. Then it admits a canonical volume t deﬁned by t (x ,..., x ) = (det(g(x , x )))1/2 x ,..., x V. V V 1 d i j ∀ 1 d ∈ b. Let M be a manifold. We call an element σ E d(M) an orienting ∈ or volume form if m M : σ is an orientation of T (M). In that case ∀ ∈ m m we also say that σ orients M or that M is oriented by σ.

3.7 Example. On Rd there is an orientation τ, called the canonical orientation, given by

(0.3.8) τ = du1 ... dud. ∧ ∧ We consider Rd always with this orientation.

Now there is the notion of a diﬀeomorphism between oriented manifolds preserving orientation. If M is oriented then every open sub- manifold of M is oriented in a natural way. Now let (U, r) be a chart of an oriented manifold M. Then on U there is an induced orientation 3. INTEGRATION 11

and on r(U) Rd there is an induced orientation. We say that (U, r) is ⊂ positive if r preserves orientation between U and r(U). 12 If M is oriented, using partition of unity, we can deﬁne a notion of integration for elements of E d, denoted by ω where ω E d(M), M ∈ with some good properties some of which weR proceed to state. The

integral M ω exists if ω has compact support. Further if M is an open sub manifoldR of an oriented manifold N and M is compact and ω is the E d ω restriction of an element of (N) then also M exists. Let M, N be oriented manifolds and f be a diﬀeomorphismR of N with M preserving orientation, and let ω E d(M). Then if ω exists so does f ω, and ∈ M N ∗ R R

(0.3.9) f ∗ω = ω. Z Z N M

3.10 Lemma. Let f D(N, M) where N, M are oriented and let ω E d(M) ∈ ∈ be an orienting form for M such that M ω exists. Suppose that f is subjective, preserves orientation and thatR f T is an isomorphism n N. n ∀ ∈ Then:

f ∗ω ω. Z ≥ Z N M Proof. Under the assumptions on f T it follows that f is a local diffeomorphism. Now let us take an open covering V of N such that f re- { i} stricted to V is a diffeomorphism. Then U = f (V ) form a covering of i { i i } M since f is onto. Let us take a partition of unity ϕ on N subordinate { i} to the covering V . Let us define ϕ on M by { i} i 1 ϕ = ϕ f − on U i i ◦ i = 0 otherwise.

Then ϕi 1 since a point in M has at least one inverse. Now 13 i ≥ P

f ∗ω = ϕi f ∗ω = ϕ ω by (0.3.9) Z Z · Z i N Xi N Xi M 12 Preliminaries

ω since ϕ 1. ≥ Z i ≥ M X

3.11

c. We call a domain D in M a nice domain if m b(D) = D D (D ∀ ∈ − being the closure of D), there is an open neighbourhood U of m in M and a ϕ F(M) such that ∈ 1 dϕ , 0 and U D = ϕ− (] , 0[). m ∩ −∞

3.12

Then it follows that b(D) is a sub manifold of M of dimension d 1. The − notion of orientation can be extended to manifolds with good boundary. Then one sees that b(D) is oriented in a natural way if D is. Now let ω be a d 1 form on M and let us denote by i the injection of b(D) into M. − Then Stokes theorem can be stated as follows:

3.13

If D is a nice domain of M such that D is compact then the Stokes’ formula holds:

dω = i∗ω. Z Z D b(D)

ω = Let us note that if b(D) is empty, then D d 0. d. On a manifold M a positive oddR d-form is a map ω from C d with values in the set of functions on M such that it is everywhere, locally, the absolute value of a local d-exterior form. A volume element θ on M 14 is a positive odd d-form such that, for every m, θm is a volume of Tm(V). The volume element τ the absolute value of the canonical orientation on Rd, is called the canonical volume element of Rd. 3. INTEGRATION 13

3.14 On a manifold there is a notion of integration for positive odd d-forms and this allows us to deﬁne the integral of a positive odd d-form ω, which we denote by M ω. R 3.15 If f D(N, M) is a diﬀeomorphism, ω a positive odd d-form of M then ∈ we can, in a natural way, associate to ω a positive odd d-form f ∗ω on ω ω N. Then if M exists so does N f ∗ and further we have R R

ω = f ∗ω. Z Z M N

3.16 Remark . if M is a manifold oriented by ω then ω is a positive odd | | d-form and if ω exists then ω exists and M M | | R R ω = ω Z Z | | M M e. Let E be a diﬀerentiable ﬁbre bundle over M

p : E M, → with E and M oriented, so that the fibres are oriented in a natural way. Let ω E d(M), ϕ E f (E) ( f being the dimension of a fibre) and ∈ ∈ i : p 1(m) E be the injection of the fibre p 1(m) into E. Using a m − → − partition of unity and Fubini’s theorem we have

3.17 Integration along ﬁbres

ϕ (p∗ω) = ( i∗ (ϕ))ω Z ∧ Z Z m E m M p 1(m) ∈ − provided that M and the ﬁbres are compact. 15 14 Preliminaries

a) Double tangent bundle.

4.1 We have the following commutative diagram;

pT T(T(M)) / T(M)

pT(M)=p′ p

p T(M) / M In the case of (U, r), a chart of M, the situation in detail is given by the following commutative diagram:

((p,ζ) rT )T T(T(U)) ◦ / T(r(U) Rd) : × K (p ,ζ ) uu KK 1 1 u KK uu KK uu KK zuu KK Rd Rd Rd % (r(U) )( ) T(r(U)) T(Rd) ×, , , × ((a b) (c d)) ((p, ζ),×(p, ζ)) (a, b, c, d) (0.4.2) 99 99 99 99 9 d d d p′ 9 (r(U) R ) (R R ) 9 × × × 99 ooo 99 ooo 9 o T 9 ooo p (p,ζ) rT wo T(U) ◦ / r(U) Rd × 16 where the arrows with two heads denote isomorphisms and we identify corresponding spaces.

4.3

Generally we take (a1,..., ad), (b1,..., bd), (c1,..., cd), (d1,..., dd) as coordinate representatives of a, b, c, d if z=(a, b, c, d). ∪ 4. 15

4.4 Let Φ be a C function on T(U). Then let us denote Φ (r, ζ rT ) 1 on ∞ ◦ ◦ − T(r(U)) by Φ and the canonical coordinates in T(r(U)) by (x1,..., xd, y1,..., yd). Then

(0.4.5) z(dΦ) = (a, b, c, d)(dΦ) = d ∂Φ d ∂Φ = ((a); (b))c + ((a); (b))d . ∂xi i ∂yi i Xi=1 Xi=1

As a particular case when ϕ = dΦ, ϕ C (U) we have ∈ ∞ z(dϕ) = (a, b, c, d)(dϕ) d d ∂2Φ ∂ϕ = (a)b c + (a)d . ∂x j∂xi i j ∂xi i Xi=1 Xj=1 X

Now let T(T(U)) z =(a, b, c, d). ∋ ∪ Then we have

T (0.4.6) p′(z) =(a, b) and p (z) =(a, c). ∪ ∪

4.7 Example. Let φ F(M) and let φ r 1 = φ. ∈ ◦ − Then φ F(r(U)) and we have ∈ z(dφ) = (a, b, c, d)(dφ) o 2 o = + (0.4.8) Dφa(d) Dφa(b, c) by (0.4.5)

o o 2 o where Dφ is the Jacobian of φ. Note that Dφ is symmetric 17 (see Dieudonne [36] : p. 174). 16 Preliminaries

4.9 b) Vertical vectors. For x in T(M) we deﬁne the space of vertical T 1 vectors Vx of T(M) at x to be (px )− (0). (From (0.4.6) it follows that Y=(a, b, c, d) is vertical if and only if c = 0). From the injection ∪ i : T (M) T(M) at m = p(x) m m →

and the fact that p(Tm(M)) = m we conclude that

(pT iT )(T (T (M))) = 0 ◦ m x m and hence that iT (T (M)) V . m p(x) ⊂ x T T But since px is surjective the kernel of px has dimension d and since T the dimension of i (Tp(x)(M)), is d we have = T (0.4.10) Vx im(Tp(x)(M)). For y T(T(M)) set x = p (y) and for y vertical set ∈ T(M) T 1 (0.4.11) ξ(y) = (ζ (i )− )(y). ◦ m

ζx Tx(Tp(x)(M)) / Tp(x)(M) nn7 nnn nnn (0.4.12) iT nn m nnn nnn nnn nnn Vx

4.13 18 If y=(a, b, c, d) then since y is vertical c = 0, and it follows that ξ(y)=∪(a, d). ∪ Note that ξ is an isomorphism between Vx and Tm(M).

On T(M) T(M) = (x, y) T(M) T(M) p(x) = p(y) ×M { ∈ × | } 4. 17 there is canonical manifold structure. Now we deﬁne a map

1 ξ− : T(M) T(M) T(T(M)) × → by the equation

1 = 1 1 (0.4.14) ξ− (x, y) ξx− (y) where ξx− (y) is uniquely deﬁned by the conditions

1 = 1) ξ(ξx− (y)) y 2) ξ 1(y) V . −x ∈ x 4.15 If x=(a, b), and y=(a, c) then it follows that ∪ ∪ 1 = ξx− (y) (a, b, 0, c) ∪ 4.16 Lemma. If ω F(T(M)), restricted to T (M) is linear for every m of ∈ m M, and z is a vertical vector of T(M), then

z(ω) = ω(ξ(z)).

Proof. The proof follows from the deﬁnitions and that, for a linear map f in a vector space, one has D f = f .

4.17 Lemma. If z T(T(M)) and ∈ z(dϕ) = 0 ϕ F(M) ∀ ∈ then 19 z = 0 18 Preliminaries

Proof. With the notation of 4.1 a) let z=(a, b, c, d). Using 4.3 we get, for every linear function ∪

z(dφ) = Dφa(d) = φa(d) = 0 and hence d is zero; and hence for every quadratic function

2 z(dφ) = (D φa)(b, c) = 0 and hence c is zero.

4.18

Let hθ denote the map of T(M) which takes every vector x into θx. Then, with the usual notation relative to (U, r) we have

hθ(a, b) =(a,θb) ∪ and T = hθ (a, b, c, d) (a,θ b, c,θ d). ∪ · · c. The canonical involution on T(T(M)).

4.19 Theorem. There is an involution z zofT(T(M)) with the following → properties:

T 1) p (z) = p′(z),

T 2) p′(z) = p (z) 3) z(dφ) = z(dφ) F(M) ∀ ∈ and is uniquely determined by these conditions.

Proof. In the case of (U, r) the map

z =(a, b, c, d) (a, c, b, d) = z ∪ → ∪ 4. 19

20 has the properties 1, 2 and 3 thanks to formula (0.4.2) and the symmetry 2 of D φ. Conversely, if z=(a, b, c, d) and z′=(a′, b′, c′, d′), ∪ ∪ T p (z′) = p′(z) a = a′ and b = c′ ⇔ T p′(z′) = p (z) a = a′ and b′ = c ⇔ and z (dφ) = z(dφ) φ F(M) together with a = a , b = c , b = c ′ ∀ ∈ ′ ′ ′ gives d = d′. Because of this uniqueness the problem becomes local which has a solution. The deﬁnition being intrinsic the involution is canonical.

4.20 Remark. z = z p (z) = pT (z). ⇔ ′ d. Another vector bundle structure on T(T(M)).

4.21 Deﬁnition. If θ R, z, z T(T(M)) and pT (z) = pT (z ) ∈ ′ ∈ ′ = + set z z′ z z′ ⊕ and θ z = θ z ⊙ ·

Relative to (U, r) if z=(a, b, c, d) and z′=(a′, b′, c′, d′) the formula are ∪ ∪ z z′ = (a, b + b′, c, d + d′) ⊕ ∪ and θ z =(a,θb, c,θd). ⊙ ∪ With this deﬁnition if f , g are curves (for a deﬁnition of a curve and related notions see 5) in T(M) such that p f = p g we have § ◦ ◦

( f + g)′ = f ′ g′ and (θ f )′ = θ f ′. ⊕ · ⊙ p This last deﬁnition of and holds good for any vector bundle E M. ⊕ ⊙ −→ 20 Preliminaries

4.22

T 21 Lemma. For φ F(M), θ R and z, z′ T(T(M)) such that p (Z) = T ∈ ∈ ∈ p (z′), we have (z z′)(dφ) = z(dφ) + z′(dφ) ⊕ and (θ z)(dφ) = (θ z)(dφ). ⊙ · Proof. The ﬁrst part follows from the deﬁnition of and property 3) of ⊕ the involution , and similarly the latter. − e. The canonical forms µ and dµ. We have the following commutative diagram:

T (p∗) T(T ∗(M)) / T(M)

(0.4.23) pT(M)=p′′ p

T ∗(M) / M p∗

where p∗ is the natural projection from T ∗(M) on M. For z T(T ∗(M)) T ∈ we denote p′′(z)(p (z)) by µ(z). Then

1 (0.4.24) µ ξ (T ∗(M)) = C ∗(T ∗(M)). ∈ To describe locally the situation above we have a diagram similar to the one in (0.4.2). With a similar notation, if z = (a, β, c, δ) then p′′(z) = (a, β), p∗(z) = (a, c) and hence

(0.4.25) µ(z) = β(c) = βici Xi

22 To compute dµ, let z=(a, β, c, δ) and z′=(a, β′, c, δ′), and let Z, Z′ be local vector ﬁelds on T with∪ constant principal∪ parts such that

i) at (a, β) they are equal to z and z′ respectively and 5. CURVES 21

ii) [Z, Z′] = 0 in a neighbourhood of (a, β). Then we have (0.2.10):

dµ(Z, Z′) = Zµ(Z′) Z′µ(Z) − around (a, β). Hence

(0.4.26) dµ(z, z′) = δ(c′) δ′(c). − From this follows the

4.27

2 Lemma. dµ is a non degenerate element of E (T ∗(M)).

5 Curves

5.1 Deﬁnition . A curve in M is an open interval I together with an f ∈ D(I, M).

We denote, generally, a curve by (I, f ) and when no confusion is possible we omit I. Generally, whenever we consider a curve we assume that 0 I. A curve can also be viewed as a point set E obtained as the ∈ image under f of an interval I. Hence we sometimes say “the curve E is parametrised by f ”, “the curve f is parametrised by t I” and by these ∈ we simply mean that the curve under consideration is (I, f ).

5.2 23 When we take a closed interval [a, b] and say f is a curve from [a, b] to M we mean that there is an open interval I [a, b] such that f ⊃ ∈ D(I, M). We denote by P the element of C (R) such that ζ P = 1, so d ◦ that P = is a basis of C (R), dual to the basis dt of C (R). Let us ( dt) ∗ agree to denote the restriction of P to an interval by P itself. 22 Preliminaries

5.3 The curve f 1(t) = f (b (t a)) is called the inverse of the curve f or − − − the curve f described in the opposite way.

5.4 Deﬁnition. If f is a curve in M we deﬁne the speed or the tangent vector f to f by f = f T P. ′ ′ ◦ T(T(M)) oo7 ~> ooo ~~ ooo ~~ ( f )T oo ~ ′ oo ~ T oo ~~ p p′ oo f ′′ ~ ooo ~ ooo ~~ ooo ~~ ooo ~~ T(I) ~~ / T(M) (0.5.5) ~ T O ~ f n7 ~~ nn ~ nnn ~~ nnn ~~ nn P ~ nnn p ~~ nn f ~ nnn ′ ~~ nnn ~~ nnn ~~nnn I n / M f

5.7 Remarks. If N is a manifold and g D(M, N) and f is a curve in M ∈ then g f is a curve in N and ◦ T (g f )′ = g f ′. ◦ ◦ Sometimes this will be used as a geometric device for computation.

5.8 24 To compute x(φ) for x T(M) and φ F(M) one can start with a curve ∈ ∈ f in M such that f ′(0) = x and observe d x(φ) = f (0)(φ) = ( f T P) (φ) = P(φ f ) = (φ f ) ◦ 0 ◦ 0 dt ◦ 0

5. CURVES 23

b) We call the speed of the tangent vector f ′ of f acceleration f ′′ of f and thus T T T T f ′′ = ( f ′)′ = ( f ′) P = ( f ) P P. ◦ ◦ ◦ 5.9 = Remarks. We have f ′′ f ′′, for T T T T p f ′′ = (p f P) P = f P = p′ f ′′ ◦ ◦ ◦ ◦ ◦ ◦ and (4.20) now gives the result.

5.10 d2(φ f ) We have ◦ = f ′′(0)(dφ). In fact dt2 T f ′′(0)(dφ) = (( f ′) P)(0)(dφ) = P(dφ( f ′)) ◦ 0 d d d = (dφ)( f T P) = (φ f ) dt ◦ 0 dt dt ◦ ! 0

d2 = (φ f ) dt2 ◦ 0

5.11 Change of parameter.

For a, θ R, we deﬁne maps τ and kθ by setting ∈ a τ (t) = a + t and kθ(t) = θ t, a · for any t in R. It follows, directly, from the deﬁnition that

( f τ )′ = f ′ τ , ( f kθ)′ = θ( f ′ kθ) ◦ k ◦ k ◦ ◦ and 25 T ( f kθ)′′ = (h f ′′ kθ). ◦ θ ◦ ◦ If φ is a diﬀeomorphism of an interval I with I then f φ is a curve. ′ ◦ This new curve is called the curve obtained from the curve (I, f ) by re-parametrisation by φ. This situation is sometimes described as “φ re-parametrises f .” 24 Preliminaries

6 Flows

In this article we ﬁx, once for all, a manifold L and a vector ﬁeld X ∈ C (L) of L. a. Integral curves.

6.1

Deﬁnition. An integral or integral curve of X is a curve f D(I, L) such ∈ that f ′ = X f. ◦ From the fundamental existence theorem in the theory of diﬀerential equations we know that through any point of L there is an integral curve of X. Set

(0.6.2) ψ = (t, m) R L I [0, t] and an f D(I, L) { ∈ × |∃ ⊃ ∈ such that i) f is an integral of X, ii) f (0) = m . For m L we deﬁne + } ∈ t (m) by the equation

(0.6.3) t+(m) = sup t R (t, m) ψ , { ∈ | ∈ } + and similarly t−(m). We can see that t (resp. t−) is lower (resp. upper) semi continuous on L. Since L is Hausdorﬀ we can see that there exists a + unique integral curve fm of X deﬁned over ]t−(m), t (m)[ with f (0) = m, and that it is maximal.

6.4 26 b. Flow. For t ]t (m), t+(m)[ ∈ −

6.5 set

γ(t, m) = fm(t) = γt(m). 6. FLOWS 25

Then γ D(ψ, L) and in open sets of L where γ , γ and γ + are deﬁned ∈ t s t s we have

(0.6.6) γ γ = γ + t ◦ s t s and, in particular, γt is a local diﬀeomorphism.

6.7 The family γ is also called the local one parameter group of transfor- { t} mations generated by X. We call γ the ﬂow of X.

6.8 Now let us suppose that a map f from a manifold M onto N is a diffeomorphism and X and Y are vector fields in M and N respectively such that f T X = Y. ◦ Then if γM and γN are flows of X and Y respectively we have

(id , f ) γ = γ . R ◦ M N c. Lie derivative. Let ω L r(L), m L and t R be suﬃ- ∈ ∈ ∈ ciently small. Then

r (γ )∗ωγ , L (T (L)) t (t m) ∈ m for every t and depends diﬀerentiably on t. So it makes sense to set

d (0.6.9) (θ(X) ω)(m) = ((γ )∗ωγ , ) = . · dt t (t m) |t 0

Note that 27 θ(X) ω L r(L), · ∈ and that θ(X) ω is called the Lie derivative of ω with respect to X. If · ω E r(L), it is easy to see that θ ω E r(L). ∈ X ∈ 26 Preliminaries

6.10 We can see that

r θ ω(X ,..., X ) = X(ω(X ,..., X )) ω(X ,..., [X, X ],..., X ) X 1 r 1 r − 1 i r Xi=1 X ,..., X C (L). ∀ 1 r ∈ 6.11 Further we recall that

θ(X)ω = 0 γ∗(ω) = ω t. ⇔ t ∀ In this case we say that ω is invariant by X or under the ﬂow of X. We also have

(0.6.12) θ(X) = i(X) d + d i(X). ◦ ◦ Chapter 1 Sprays

Throughout this chapter M stands for a manifold and (U, r) for a typical 28 chart of it.

1 Deﬁnition.

27 28 Sprays

Our aim now is to define a “geodesic” so as to generalise a curve which is intuitively the shortest distance between two points on it, which are sufficiently close. Such a definition must have the following natural properties: 1) A geodesic curve is uniquely determined by its initial position and speed. 2) A change of parameter by a homothesy or a translation leaves 29 it a geodesic. So the family of geodesics determines a family of curves on T(M) one passing through each point x of T(M). Associating to each point x of T(M) the tangent vector at x of the unique curve of the family passing through x we get a map (set theoretic) : G : T(M) T(T(M)). → We have by the very definition and from 5.9 together with 4.20

(1) f ′′ = G f ′ ◦ and

T (2) p′ G = p G = id ◦ ◦ T(M) for any geodesic curve f . Further we have f kθ is a geodesic curve, ◦ and hence

(3) ( f kθ)′′ = G ( f kθ)′. ◦ ◦ ◦ But we have for x = f ′(0) and by 5.11:

( f kθ)′ = θ f ′(0) = θ x ◦ · and T T ( f kθ)′′(0) = θ (h f ′′(0)) = θ (h G(x)) ◦ · θ ◦ · θ ◦ and so

T (4) G hθ = θ(h G) ◦ θ ◦ Thus from the concept of a geodesic we are led to a G with the properties (2) and (4). Keeping this in mind we reverse the process and 30 ﬁrst deﬁne a spray and then a geodesic as its integral curve. 2. GEODESICS 29

1.1 Deﬁnition. A spray G on M is an element of C (T(M)) such that 1) pT G = id and ◦ T(M) T 2) G hθ = θ(h G) θ R ◦ θ ◦ ∀ ∈ 1.2 Note. From 2) it follows that

G(0 ) = 0 m M m ∀ ∈ Generally we denote a spray by G.

1.3 If N is an open sub manifold of M, G induces a spray on N and we denote it by NG.

2 Geodesics

Now we deﬁne a geodesic curve.

2.1 Deﬁnition. A geodesic of M relative to a spray G is a curve f in M such that f ′ is an integral of G, i.e. f is such that

f ′′ = G f ′. ◦ 2.2 Remark. If f : [0, 1] M is a geodesic, then the curve g : [0, 1] M → → deﬁned by g(t) = f (1 t) is a geodesic. − When no confusion is possible, we shall speak simply of a geodesic and omit the reference to the spray. 30 Sprays

2.3 Proposition. f is a geodesic if and only if there exists an integral g of G such that f = p g. ◦ 31 Proof. If g is an integral of G then g = G g and then if f = p g, we ′ ◦ ◦ have T T T f ′ = p g = p G g = g since p G = id . ◦ ◦ ◦ ◦ T(M) Hence f is a geodesic. For the other part, if f is a geodesic we can take g = f ′ for the integral of G.

2.4 Remarks. From (0.6.2) one knows that given x T(M) there is, locally, ∈ a geodesic f of M such that f ′(0) = x, and that it is unique.

2.5 From the fact that G(0 ) = 0 we see that m M f (R) = m is a m ∀ ∈ geodesic, called the trivial geodesic at m.

2.6 Proposition. If f D(I, M) is a geodesic so are ∈ 1 f τ D(τ− (I), M) and ◦ a ∈ a 1 f kθ D(k− (I), M) a R, θ R,θ , 0. ◦ ∈ θ ∀ ∈ ∀ ∈ This proposition is a direct consequence of the deﬁnitions.

2.7 Corollary. For x T(M), θ R and t R we have ∈ ∈ ∈ γ(t,θx) = θ γ(tθ, x), · i.e. whenever one of these terms is deﬁned so is the other and the equality holds. 3. EXPRESSIONS FOR THE SPRAY IN LOCAL COORDINATES31

Proof. Let f be the geodesic with f (0) = x and let g = f kθ. Then g ′ ◦ is the geodesic with g (0) = θ x and we have by the deﬁnition of γ 32 ′ ·

f ′(t) = γ(t, x) and g′(t) = γ(t,θx) But then γ(t,θx) = g′(t) = θ f ′(θ t) = θ γ(θ t, x). · · · ·

2.8 Corollary. For x T(M) and θ R and , 0, we have ∈ ∈ + 1 + t (θ x) = θ− (t (x)) · 1 and t−(θ x) = θ− (t−(x)) · + Proof. If f is the geodesic of M on ]t−(x), t (x)[ then

1 1 + f kθ on ]θ− t−(x),θ− t (x)[ ◦ is a geodesic of M with speed θ x. Hence by the deﬁnitions of t− and + · t (see 6.4)

1 t−(θ x) θ− t−(x) · ≤ · + 1 + and t (θ x) θ− t (x). · ≥ · The other inequalities follow if we interchange the roles of f and f ◦ kθ.

3 Expressions for the spray in local coordinates

Let (U, r) be a chart and let x=(a, b), G(x)=(a, b, c, d) (see 4) ∪ ∪ 32 Sprays

Then from the condition 1) for a spray and from (0.4.6) we have b = c. Setting d = ψ(a, b) we have

ψ D((r(U) Rd), Rd), and ∈ × G(θ x) = (a,θb,θb, ψ(a,θb)), · hT (G(x)) = (a,θb, b,θ ψ(a, b)), θ · 33 and θhT (G(x)) = (a,θb,θb,θ2 ψ(a, b)). θ · Now from the second condition for a spray we get

G(θ x) = θ(hT (G(x))) · θ and hence

(1.3.1) ψ(a,θb) = θ2 ψ(a, b) θ R · ∀ ∈ Using Euler’s theorem on homogeneous functions we see that ψ is, for a ﬁxed a, a quadratic form in b so that

(1.3.2) ψ(a, b) = (D2ψ )(b b) a · with D2ψ symmetric. Hence we have the following:

3.3 Proposition . If G is a spray then there exists, locally, a unique δ ∈ D(r(U) Rd Rd, R) such that, for a r(U), its restriction to a Rd Rd × × ∈ { }× × is bilinear, symmetric and

ψ(a, b) = δ(a, b, b).

3.4 Setting Γ j (a) = U j(δ(a, e , e )) for a r(U), we have ik − i k ∈ Γ j F(r(U)). ik ∈ 4. THE EXPONENTIAL MAP 33

Let f D(I, M) be a geodesic and let a = r ( f U), be its image in ∈ ◦ | r(U), then we have

da(t) f (t) =(a(t), ) ′ dt ∪ da(t) da(t) d2a(t) and f (t) =(a(t), , , ). ′′ dt dt 2 ∪ dt So if we write i a(t) = x (t)ei X then we see that f is a geodesic if and only if

d2 x j dxi dxk (1.3.5) + Γ j = 0 j. 2 ik dt · dt ∀ dt Xi,k 34

4 The exponential map

Let us consider G and its ﬂow. Setting

+ 1 (1.4.1) Ω= (t )− (]1, + [) (see 6a) ∞ we see, since t+ is lower semi continuous, that Ω is open in T(M). For every m M, 0 the zero vector at m clearly belongs to Ω. Hence a ∈ m neighbourhood of 0m is in Ω.

4.2

If N is an open subset of M then by 1.3 there is a natural spray NG on N and we can deﬁne t+ and t and Ω relative to N and N. We denote NG N−G G this open subset by NΩ. We identify T(N) with an open sub manifold of T(M) and then, clearly NΩ Ω T(N). Now let x be in Ω T(M). ⊂ ∩ ∩ Then by (0.6.a) it follows that x belongs to NΩ if the image of ]0, 1[ under p γ(t, G(x)) is in N. ◦ 34 Sprays

4.3 Deﬁnition. The map p γ from Ω into M is called the exponential map ◦ 1 (for the spray G) of T(M) and is denoted by exp.

4.4 Note . exp D(Ω, M) because p, γ are diﬀerentiable. We denote the ∈ restriction of exp to T (M) Ω by exp . m ∩ m 4.5 Lemma. For x in Ω the map

S : t exp(tx) →

deﬁned for suﬃciently small t is a geodesic of M and S ′(0) = x. In particular exp(x) is the point S (1) of this geodesic.

Proof. We have

S (t) = p γ (tx) = p γ(1, tx) ◦ 1 ◦ = p(t, γ(t, x)) by (2.7) = p(γ(t, x))

35 Hence S is a geodesic by (2.3). The fact that S ′(0) = x follows from the deﬁnition of γ.

4.6 We have T expm T0 (Tm(M)) / Tm(M) m 7 ooo ooo ooo ζ ooo ooo idTm(M) ooo oo Tm(M) 4. THE EXPONENTIAL MAP 35

Proof. For x in T (M) and θ R set U with: m ∈ U (θ) = θ x · so that U = ζ0m ( ′(0)) x. Now we have T 1 T (exp ζ− )(x) = exp (U ′(0)) = (exp U )′(0) = S ′(0). ◦ 0m ◦ and by the preceding lemma we have

S ′(0) = x.

4.7 Hence it follows by the inverse function theorem that exp is a diﬀeo- morphism between suitable neighbourhoods of 0 T (M) and m in 36 m ∈ m M.

4.8 Corollary. The map (p, exp) of Ω into M M has maximal rank at 0 × m for every m in M. Proof. Since the dimensions of Ω and M M are the same it is enough × to show that the map (p, exp)T is injective. Now let z T (T(M)) be ∈ 0m such that (p, exp .)T (z) = 0. Then since (1) pT (z) = 0 we have z V . ∈ 0m By the previous proposition we have (2) expT (z) = ζ(z) and hence ζ(z) = 0. The corollary follows from (1) and (2).

Chapter 2 Linear connections

1 Linear connection 37 1.1

Deﬁnition. From 4.9 we see that, for x in T(M), if Hx is transverse to Vx (i.e. a supplement to Vx in Tx(T(M)) then the restriction pT H : H T (x)(M) | x x → p is an isomorphism. So a natural question is whether it is possible to choose an interesting map x H where x runs through M. For an → x open subset A of Rd we see that the choice = T 1 (2.1.2) Hx (ζx )− (0) gives us such a distribution of Hx. We look upon such a choice as a map (2.1.3) : T(M) T(M) (x, y) (x, y) T(T(M)). C ×M ∋ →C ∈ where x determines the ﬁbre containing Hx and y, by means of the inverse of the isomorphism (pT H ) : H T (M), the element (x, y) | x x → p(y) C in H such that pT ( (x, y)) = y. x C Relative to (U, r), with the notation of 4 we have, by 1.1,

(2.1.4) ((a, b), (a, b′)) =(a, b, b′,.). C ∪ Keeping this in mind we deﬁne a connection.

37 38 Linear connections

1.5

38 Deﬁnition. A connection on T(M) (or a linear connection on M) is a

0) D(T(M) T(M), T(T(M))) C C∈ ×M such that

T 1) (p, p ) = id T(M) T(M) C ◦C | ×M ) (θ x + η x , y) = θ (x, y) η (x , y) C2 · · ′ ⊙C ⊕ ⊙C ′ ) (x,θ y + η y ) = θ (x, y) + η (x, y ). C3 · · ′ ·C ·C ′ (Note that this deﬁnition would make sense for any vector bundle.) Now let us consider the image set H of x T (M) under : x { }× p(x) C H = (x, T (M)). x C p(x) By ) it follows that H is a vector subspace of T (T(M)); and from C3 x x ) that C1 pT ( (x, y)) = y C and hence that

i) restricted to x T (M) is a monomorphism, and C { }× p(x)

ii) the only element common to Vx and Hx is zero.

Since the dimension of Tx(T(M)) is 2d and that of Vx is d it follows, now, that

(2.1.6) Tx(T(M)) = Hx + Vx (direct sum)).

Now we give the following deﬁnitions.

1.7 Deﬁnition. For x T(M),H is called the horizontal subspace or the ∈ x horizontal component of T (T(M)) relative to the connection . Gener- x C ally we omit the reference to . C 1. LINEAR CONNECTION 39

1.8

39 Deﬁnition. For x T(M) the map ∈

1.9

T (T(M)) z z (x, pT (z)) T (T(M)) is the projection onto V x ∋ → −C ∈ x x parallel to Hx.

1.10

Deﬁnition. For z in T(T(M)) the element

1.11

U (z) = ξ(z (p (z), pT (z))) is called the vertical component of z. −C ′

1.12

Note. From the very deﬁnition we have

p U = p p′. ◦ ◦

1.13

Remark. By ) we have the following equations: C1 For (x, y) T(M) T(M), ∈ ×M

T p ( (x, y)) = y = p′( (y, x)) C C T and p ( (y, x)) = x = p′( (x, y)). C C They show that the condition (x, y) = (y, x) is compatible with ), C C C1 ), ). C2 C3 In view of the above, the following deﬁnition makes sense: 40 Linear connections

1.14

Deﬁnition. A connection is called symmetric if C

1.15

4) (x, y) = (y, x) (x, y) T(M) T(M) C C C ∀ ∈ ×M Examples.

1.16

The canonical connection on an open subset A Rd is deﬁned by ⊂ = T 1 (2.1.17) Hx (ζx )− (0)

i.e.: C((a, b), (a, c)) = (a, b, c, 0) a A, b, c Rd. ∀ ∈ ∀ ∈ 40 From 4 one sees ), ), ) and ) are fulﬁlled so that this canonical C1 C2 C3 C4 connection is moreover symmetric.

1.18

If f : M N is a diﬀeomorphism and C a connection on N, then → (x, y) (( f 1)T )T ( f T (x), f T (y)) deﬁnes a connection on M. → − C Exercise. Use 1.16, 1.18 and partitions of unity to build up a connection on any manifold (a proof of that would follow also from 1.1.

1.19

Remark. If is a connection on T(M) then G(x) = (x, x) for every x C C in T(M) is a spray, for

i) p (G(x)) = p ( (x, x)) = x = pT ( (x, x)) = pT (G(x)) and ′ ′ C C 2. CONNECTION IN TERMS OF THE LOCAL COORDINATES 41

ii) (G hθ)(x) = G(θ x) = (θx,θx) = by 4.21 ◦ · C = θ (x,θ, x) = hT (θ (x, x)) = ⊙C θ ·C = θ hT ( (x, x)) = (θ (hT G))(x). · θ C · θ ◦ This spray is called the spray associated to the given connection.

1.20 Claim. For an open subset A Rd the geodesics relative to the spray ⊂ associated to the canonical connection are open segments of straight lines. With the above notation, the deﬁnition shows that f is a geodesic if and only if f ′′ G f ′ = 0, − ◦ i.e. if and only if f (p ( f ), pT ( f )) = 0. By the deﬁnition of 41 ′′ − C ′ ′′ ′′ canonical connection C and being horizontal we have T T (2.1.21) ζ (C(p′( f ′′), p ( f ′′)) = 0. Hence, by (2.1.21), (4.20) and (5.9) f is a geodesic if and only if T (2.1.22) ζ ( f ′′) = 0. But T T T T d ζ ( f ′′) = ζ f ′ P = (ζ f ′) P = (ζ f ′) ◦ ◦ ◦ ◦ dt ◦ d(ζ f ) Hence f is a geodesic if and only if ◦ = 0, i.e. if and only if dt ζ f = x, a constant vector. ◦ ′ But then f (t) = tx + y (for suitable values of t).

2 Connection in terms of the local coordinates

With the notation of 4 let x=(a, b) and y=(a, c). Then (x, y) ∈ T(M) T(M). Now given a connection∪ on T∪(M) we have, by (1.5). ×M C 42 Linear connections

2.1 (x, y) = (a, b, c, δ(a, b, c)) where δ depends on , and then ) becomes C C C0 (2.2.2) δ D(r(U) Rd Rd, Rd). ∈ × × Now ) and ) together mean that the restriction of δ to m Rd Rd C2 C3 { }× × for m in r(U) is bilinear. Further is symmetric if and only if δ restricted C to a Rd Rd is (see (4.19)). Now we prove a theorem which completes { }× × the remark (1.19).

2.3

42 Theorem [2], th. 2, p. 173. Given a spray G on M there exists a unique symmetric connection on T(M) such that C (x, x) = G(x). C Proof. In view of the first paragraph of this article, locally, the problem can be stated as follows: Given a ψ D(r(U) Rd, Rd) such that for a fixed m in r(U), ψ is ∈ × a quadratic form in the other variable to construct a δ D(r(U) Rd ∈ × × Rd, Rd) which, for fixed m in r(U), is bilinear and symmetric in other variables and to prove its uniqueness. This has been done in (3.3). The uniqueness of δ shows that the problem is local.

2.4 Remark. Let us write down explicitly in terms of G. By (4.17) it is C enough to express in terms of G on dφ for φ F(M). We have C ∈ G(x + y)(dφ) = (x + y, x + y)(dφ) = (x + y, x)(dφ)+ C C + (x + y, y)(dφ) = C = (C(x, y) (x, x))(dφ) + (C(x, y) C(y, y))(dφ) = ⊕C ⊕ = C(x, y)(dφ) + C(y, x)(dφ) + C(x, x)(dφ) + (y, y)(dφ) = C by (4.19), (3) = C(x, y)(dφ) + C(x, y)(dφ) + G(x)(dφ) + G(y)(dφ) = 3. COVARIANT DERIVATION 43

since C is symmetric = 2 (x, y)(dφ) + G(x)(dφ) + G(y)(dφ) by (4.22). C

Hence 43

1 (2.2.5) (x, y)(dφ) = G(x + y) G(x) G(y) (dφ) C 2 − − which shows, once again, the uniqueness (4.17).

3 Covariant derivation

For the rest of this chapter we suppose given a manifold M with a connection . C

3.1

Deﬁnition. Given a manifold N, an X C (N) and g D(N, T(M)), the ∈ ∈ map N T(M) deﬁned by →

(2.3.2) D g = U gT X X ◦ ◦ is called the covariant derivative of g with respect to X. If x T(N) we ∈ write D g = (gT (x)). x U 44 Linear connections

T(T(M)) x; xx xx T xx g xx xx p′ U xx xx xx xx T(N) T(M) O w; wwww; wwww D g wwww × wwww X wwww p wwww g wwww wwww wwww www N p g / M ◦

3.3

Remarks. Since X, gT and are diﬀerentiable we have U D g D(N, T(M)). X ∈

3.4

44 We have 3. COVARIANT DERIVATION 45

p D g = p U gT X = ◦ X ◦ ◦ ◦ T = p p′ g X by (1.12) = ◦ ◦ ◦ = p g ◦ and hence if g is a lift of f then DXg is again a lift of f .

3.5 Let f D(I, M). Then f T can be considered as a map from I to T(M). ∈ Now we have: f is a geodesic for the spray G associated to the connection if and only if C DP f ′ = 0. For we have

T D f ′ = U f ′ T = U f ′′ = ξ( f ′′ ( f ′, f ′)) = P ◦ ◦ P ◦ −C = ξ( f ′′ G f ′), and ξ is an isomorphism − ◦ between the vertical vectors at a point and the tangent space containing that point. Writing f = p g we have ◦ T T T T (2.3.6) D g = ξ(g X C(p′ g X, p g X)) X ◦ − ◦ ◦ ◦ ◦ = ξ(gT X (g, f T X)). ◦ −C ◦ Now considering an ω in 1(M) as a linear function on T(M) and ob- E serving that gT X (g, f T X) is vertical we obtain, by (4.16) 45 ◦ −C ◦ ω(D g) = (gT X (g, f T X))(ω) = X ◦ −C ◦ (2.3.7) = X(ω g) (g, f T X)(ω). ◦ −C ◦ In particular, for ω = dφ, where φ F(M), ∈ (2.3.8) dφ(D g) = X(dφ g) (g, f T X)(dφ). X ◦ − ◦ Sometimes we write D g(φ) for dφ(D g) and X(g(φ)) for X(dφ g). X X ◦ 46 Linear connections

3.9 Example . For an open subset A Rd with respect to the canonical ⊂ connection we have

ν = ζT . (2.3.10) For ν(z) = ν(a, b, c, d) T = ξ(z (p′(z), p (z))) −C = ξ((a, b, c, d) (a, b, c, 0)) = ξ(a, b, o, d) = d. − and ζT (a, b, c, d) = d. Hence

D g = ν gT X = ζT gT X = (ζ g)T X X ◦ ◦ ◦ ◦ ◦ where g D(A, T(A)). ∈ 3.11 In case f = p g is a constant map, i.e. in case f (y) = m for every y ◦ in N, g can be considered as a map from N to Tm(M), and in this case ν = ζ. Hence we have

D g = ν gT X = ζ gT X = dg X X ◦ ◦ ◦ ◦

46 and so DXg is nothing but the restriction of the diﬀerential map of g to the subset X(y) of T(N) In particular if g is a curve in T (M), and { } m X = P on R we have dg D g = P dt

and hence DPg is the ordinary derivative of a vector valued function.

3.12 If f D(N, M) and X, Y C (N) then f T Y D(N, T(M)) and by ∈ ∈ ◦ ∈ (2.3.8) we have

(2.3.13) D ( f T Y)(φ) = X(Y(φ f )) ( f T Y, f T X)(dφ) φ F(M). x ◦ ◦ −C ◦ ◦ ∀ ∈ 3. COVARIANT DERIVATION 47

3.14 For f D(N, M), X C (N), Y C (M) and n N we have ∈ ∈ ∈ ∈ T T D (Y f ) = v(Y ( f (X(n))) = D T Y. X(n) ◦ f (X(n)) In particular if f = id and X, Y C (M), then D Y C (M) and M ∈ X ∈ so D can be considered as a mapping C (M) C (M) C (M). For × → f D(N, M), ψ F(N), X, Y C (N) and g, g D(N, T(M)) such that ∈ ∈ ∈ ′ ∈ p g = p g we have ◦ ◦ ′ C.D. 1. D g is F(N) linear in X, X − C.D. 2. DX(g + g′) = DXg + DXg′

(2.3.15) C.D. 3. DX(ψg) = X(ψ)g + ψ(DXg) C.D. 4. D ( f T Y) D ( f T X) = f T [X, Y] X ◦ − Y ◦ ◦ if C is symmetric.

Proof.

C.D. 1. follows directly from the deﬁnition. To prove the others, and for 47 simplicity let us use the convention following (2.3.8). C.D. 2. For φ F(M) we have, setting f = p g, ∈ ◦ T D (g + g′)(φ) = X(g + g′)φ (g + g′, f X)(dφ) (by (2.3.8)) X − C ◦ T T = X(g(φ) + g′(φ)) ( (g, f X) C(g′, f X))(dφ) − C ◦ ⊕ ◦ T T = X(g(φ)) + X(g′(φ)) (g, f X))(dφ) (g′, f X)(dφ) − C ◦ − C ◦ by 4.22.

C.D. 3. D (ψg)(φ) = X(ψg)(φ) C(ψ g, f T x)(dφ) = X − · ◦ = X(ψ g(φ)) (ψ (g, f T X))(dφ) = · − ⊙C ◦ = X(ψ)g(φ) + ψX(g(φ)) ψ C(g, f T X)(dφ), by (4.22) − · ◦ C.D.4. (D ( f T Y) D ( f T X))(φ) = X ◦ − Y ◦ = X(Y(φ f )) ( f T Y, f T X)(dφ) ◦ − ◦ ◦ 48 Linear connections

Y(X(φ f )) ( f T X, f T Y)(dφ) = − ◦ − ◦ ◦ = [X, Y](φ f ) because is symmetric, and ◦ C property 3) of the involution , in (4.19) − = ( f T [X, Y])(φ) = ( f T [X, Y])(φ). ◦ ◦

3.16 Lemma. For f D(N, T(M)),X,Y C (N) the map ∈ ∈ (X, Y, f ) ([D , D ] D , )( f ) → X Y − [X Y] if F(N) - trilinear. Proof. We have for ψ F(N): ∈ (([Dψ , D ] D ψ , )( f ) X Y − [ X Y] = ψD D f D (ψD f ) D ψ , f and by C.D. 1., X Y − Y X − [ X Y] = ψD D f D (ψD f ) ψD , f + Y(ψ)D f = X Y − Y x − [X Y] X = ψD D f ψD D f Y(ψ) D f X Y − Y X − · X D , f + Y(ψ) D f by C.D. 3., − [X Y] · X = ψ([D , D ] D , ) f. X Y − [X Y] 48 Therefore the mapping is F(N)-linear in X and since the mapping is antisymmetric in X, Y it is F(N) - linear in Y. Applying C.D. 3. twice we obtain, by straightforward calculation

([D , D ] D , )(ψ f ) X Y − [X Y] · = ψ([D , D ] D , f ). X Y − [X Y] 4 The derivation law

4.1 In case X, Y C (M) and φ F(M), with N = M, C.D.’s can be written, ∈ ∈ respectively, as 4. THE DERIVATION LAW 49

D.L.1. DXY is F(M)-linear in X,

D.L.2. DX(Y + Y′) = DXY + DXY′ D.L.3. D (φY) = X(φ)Y + φ D Y X · X D.L.4. D Y D X = [X, Y] if C is symmetric. X − Y 4.2

Thus from a connection we obtain a mapping from C (M) C (M) into 49 × C (M) satisfying D.L. 1, 2, 3. Any such map is called a Derivation Law in M. It is said to be symmetric (or without torsion) if D.L.4 holds for it. Conversely, given a derivation law in M there is a connection which induces the given derivation law (see Koszul [18]: Th. 4. p. 94).

4.3 In local coordinates relative to (U, r), let X denote the dual basis of { i} dxi = d( i r) and now let U ◦ 4.4 = j DXi Xk λik X j. Then we have by 3.12 j P (D X )(x j) = X (X (x j)) (X , X )(dx j). Xi k i k −C k i j = j j = But Xk(x ) δk a constant and hence Xi(Xk(x ))) 0. Hence (D X )(x j) = (X , X )(dx j) Xi k −C k i = j(δ(a, e , e )) =Γ j (see (3.4)). −U k i ki 4.5 j =Γ j Thus λik ki. Given ω L r(M) and X C (M) we deﬁne a map ∈ ∈ Dω : C r+1(M) F(M) by setting → 50 Linear connections

4.6

(Dω)(X, X ,...... , X ) = X(ω(X ,...... , X )) 1 r 1 r − r ω(X1,..., Xi 1, DXXi, Xi+1,..., Xr) − − Xi=1

X ,..., X C (M). ∀ 1 r ∈ 50 Then we have

i) Dω is (r + 1)-additive

ii) (Dω)(φX, X1,..., Xr) =

r φ X(ω(X1,..., Xr)) ω(X1,..., Xi 1, φ DXXi, Xi+1,..., Xr) = · − − · Xi=1

by 4.1 D.L.1.

= φ(Dω)(X, X1,..., Xr)

and

iii) D(ω)(X, φX1,..., Xr) = X(ω(φX1,..., Xr))

r (D (φX ), X ,..., X ) ω(φX ,..., D X ,..., X ) = − X 1 2 r − 1 X i r Xi=2

= φX(ω(X1,..., Xr)) + (X(φ)ω(X1, X2,..., Xr) ω(X(φ), X , X ,..., X ) ω(φD X , X ,..., X ) − 1 2 r − X 1 2 r r φω(X1,..., Xi 1, DXXi, Xi+1,..., Xr) = − − Xi=2

= φ(Dω)(X, X1,..., X2)

and similarly for i = 2,..., r. 4. THE DERIVATION LAW 51

4.7 Hence D L r+1(M). However, in general, if ω r(M), Dω may ∈ + ∈ E not belong to r 1(M)). This form Dω is called the covariant derivative E of ω with the respect to D or C (the derivation law or the connection). More generally:

Proposition 2.4.7 bis. Let X C (N), g1,..., gr D(N, T(M)), f ∈ ∈ r ∈ D(N, M) maps such that p gi = f i = 1,..., r; and ω E (M). ◦ ∀ r ∈ T Then (Dω)( f X, g1,..., gr) = X(ω(g1,..., gr)) ω(g1,..., gi 1, ◦ − − Xi=1 DXgi, gi+1,..., gr).

Proof. This follows from the local expression of f X given by (5.14) 51 T ◦ and linearity.

Now we set

(2.4.8) D ω = i(X) Dω X ◦ 4.9 Proposition. For ω E 1(M) and (x, y) T(M)=T(M) we have ∈ ∈ M (Dω)(x, y) = (y, x)(ω). C Proof. Let X, Y C (M) be such that ∈ X(p(x)) = x and Y(p(y)) = y.

Then by (2.3.7) we have

ω(D Y) = X(ω Y) (Y, X)(ω) X ◦ −C and so

C(Y, X)(ω) = X(ω(Y)) ω(D Y) = − X = (Dω)(X, Y) by the deﬁnition of Dω. 52 Linear connections

4.10

Lemma. With the above notation

(dω)(x, y) = (Dω)(x, y) (Dω)(y, x) − provided that is symmetric. C Proof. With the same notation, we have

(Dω)(X, Y) (Dω)(Y, X) − = X(ω(Y)) ω(D Y) Y(ω(X)) + ω(D (X)) − X − Y = X(ω(Y)) Y(ω(X)) ω([X, Y]) − −

52 since C is symmetric = (dω)(X, Y) (by the deﬁnition of dω: see (0.2.10)).

4.11

Deﬁnition. For φ F(M) Ddφ E 2(M) is called the Hessian of φ with ∈ ∈ respect to the connection C.

4.12

Corollary. If C is symmetric then

Ddφ is symmetric φ F(M). ∀ ∈ Proof. By the above proposition we have, with the same notation,

(Ddφ)(x, y) = C(y, x)(dφ) = C(y, x)(dφ) = C(x, y)(dφ) (see 4.19 & 1.15) = = (Ddφ)(y, x).

5. CURVATURE 53

4.13 Proposition. If f is a geodesic of M with respect to the spray associated to C, then for φ F(M) we have ∈ d2(φ f ) ◦ = (Ddφ)( f ′, f ′). dt2 Proof. The fact that f is a geodesic implies that

f ′′ = C( f ′, f ′) and hence

f ′′(dφ) = C( f ′, f ′)(dφ). d2(φ f ) But f ′′(dφ) = ◦ (see 5.10) dt2 and by the proposition above

C( f ′, f ′)(dφ) = (Ddφ)( f ′, f ′).

5 Curvature

5.1 Let X, Y C (M) and let us write ∈

R(X, Y) = [D , D ] D , (defect of brackets). X Y − [X Y] Then by (3.16) it follows that R L 3(M). This map R is called the 53 ∈ 1 curvature tensor of M for D. R has the following properties:

5.2 C.T. 1. R(X, Y) = R(Y, X), − C.T. 2. R(X, Y)Z + R(Y, Z)X + R(Z, X)Y = 0 if C is symmetric. 54 Linear connections

The equality C.T. 1. follows from the deﬁnition of R, so let us consider C.T. 2. We have in succession, applying the deﬁnition of R and D.L. 4., 4.1, six times. R(X, Y)Z + R(Y, Z)X + R(Z, X)Y

= D D Z D D Z D , Z X Y − Y X − [X Y] + D D X D D X D , X Y Z − Z Y − [Y Z] + D D Y D D Y D , Y = Z X − X Z − [Z X] = D (D Z D Y) + D (D X D Z) + D (D X D Y) X Y − Z Y Z − X Z Y − X D , X D , Y D , Z − [Y Z] − [Z X] − [X Y] = D [Y, Z] D , X + D [Z, X] D , Y + D [Y, X] D , Z = X − [Y Z] Y − [Z X] Z − [Y X] = [X, [Y, Z]] + [Y, [Z, X]] + [Z, [XY]] which is zero by Jacobi identity in C (M). Examples.

5.3 1. If the dimension of M is 1 then R = 0 by C.T. 1.

5.4 2. For an open subset A of Rd we have R = 0

54 the connection being the canonical connection. To see this let x, y, z be vectors and X, Y, Z be vector ﬁelds such that X(m) = x, Y(m) = y, and ζ Z = ζ(Z) . Then we have ◦ { } D Z = ν ZT X = ζT ZT X = (ζ Z)T X. X ◦ ◦ ◦ ◦ ◦ ◦ But ζ Z is constant and hence ◦ (ζ Z)T = 0. ◦ This argument actually shows that

DXDY Z = 0, DY DXZ = 0, D[X,Y]Z = 0. 5. CURVATURE 55

5.5 Note. We can deﬁne the covariant derivative DΩ for any Ω L r(M) ∈ s and then for R we can prove the Bianchi identity.

(2.5.6) DXR(Y, Z) + DY R(Z, X) + DZR(X, Y) = 0 (see [18]).

5.7 Lemma. For ω E 1(M),X,Y,Z C (M): ∈ ∈

([D , D ] D , )ω = R(X, Y) ω X Y − [X Y] − · Proof. For any Z C (M) the left hand side, by deﬁnition 4.6, is equal ∈ to

(D D ω)(Z) (D D ω)(Z) (D , ω)(Z) X Y − Y X − [X Y] = X(D ω)(Z)) (D ω)(D Z) Y((D ω))(Z)) Y − Y X − X + (D ω)(D Z) [X, Y](ω(Z)) + ω(D , Z) X Y − [X Y] = X(Y(ω(Z))) X(ω(D Z)) Y(ω(D Z)) − Y − X + ω(D D Z) Y(X(ω(Z))) + Y(ω(D Z)) Y X − X + X(ω(D Z)) ω(D D Z) [X, Y](ω(Z)) + ω(D , Z) Y − X Y − [X Y] = ω(R(Y, X)Z) = (R(Y, X) ω)(Z). · 55

5.8 Deﬁnition. Given an X C (M) by the horizontal lift of X, denoted by ∈ XH, we mean the mapping T(M) to T(M) deﬁned by

XH(x) = C(x, X(p(x))), x T(M). ∈ Since C is diﬀerentiable it follows that

XH C (T(M)). ∈ 56 Linear connections

5.9 Lemma. For X C (M) and ω E 1(M) ∈ ∈ H X (ω) = DXω

Proof. By (4.9) and the deﬁnition of XH, we have

XH(ω)(x) = (x, X(p(x)))ω = (Dω)(X(p(x)), x) = = (D ω)(x) x T(M). X ∀ ∈

5.10 Proposition. For X, Y C (M) and ω E 1(M) ∈ ∈ [XH, Y H] [X, Y]H is vertical − and ω(ξ([XH, Y H] [X, Y]H)) = R(X, Y) ω. − − · Proof. Since XH and Y H are projectable by pT , i.e. images under pT are again vector ﬁelds, we have

pT [XH, Y H] = [pT XH, pT Y H] = [X, Y] ◦ = pT [X, Y]H. ◦

56 Further we have (4.16):

ω(ξ([XH, Y H] [X, Y]H)) = ([XH, Y H] [X, Y]H) ω − − · = XH(Y H(ω)) Y H(XH(ω)) [X, Y]H(ω) = − − H H = X (D ω) Y (D ω) D , ω = by (5.9) Y − X − [X Y] = D D ω D D ω D , ω = by (5.9) X Y − Y X − [X Y] = R(X, Y)ω by (5.7). − 5. CURVATURE 57

5.11

Corollary. x T(M) ∀ ∈ ξ([XH, Y H] [X, Y]H) − x = R(X(p(x)), Y(p(x)))x − For the above proposition holds ω E 1(M) and the elements of both ∀ ∈ sides of the equation are in T(M).

5.12 C.D. 5.

If N is a manifold, X, Y C (N) and g D(N, T(M)), then ∈ ∈ T T ([D , D ] D , )(g) = R( f X, f Y)(g) X Y − [X Y] ◦ ◦ where f = p g. ◦ Proof. We ﬁrst prove two lemmas.

5.13

Lemma. If g D(N, T(M)) and f = p g then there exist, locally on N, ∈ ◦ functions ψi and, locally on M, vector ﬁelds Ui such that, locally

g = ψ (U f ). i i ◦ Xi 57

Proof. Relative to a chart (U, r) of M let Ui be a basis for C (U). Then for n f 1(U) there exists ψ (n) such that ∈ − i

g(n) = ψi(n)Ui( f (n)). Xi

The ψi are the required functions. 58 Linear connections

5.14 Lemma. For f D(N, M),X,Y C (N) locally let ∈ ∈ f T X = ψ (U f ) and f T Y = θ (U f ). ◦ i · i ◦ ◦ i · i ◦ Xi Xi Then, locally

f T [X, Y] = (X(θ ) Y(ψ ))(U f ) ◦ i − i i ◦ Xi + θ ψ ([U , U ] f ). j i i j ◦ Xi, j Proof. φ F(M): ∀ ∈ ( f T [X, Y])(φ) = [X, Y](φ f ) ◦ ◦ = X(Y(φ f )) Y(X(φ f ) = ◦ − ◦ = X(( f T Y)φ) Y(( f T X)φ) = ◦ − ◦ = X( θ (U f )φ) Y( ψ (U f )φ) = j j ◦ − i i ◦ Xj Xi = X( θ (U (φ) f )) Y( ψ (U (φ) f )) = j j ◦ − i i ◦ Xj Xi = X(θ )(U f )(φ) + θ X(U (φ) f ) j j ◦ j ◦ j ◦ Xj Xj Y(ψ )(U f )(φ) ψ X(U (φ) f ) − i i ◦ − i i ◦ Xi Xi 58 But

X(θ )(U f )(φ) Y(ψ )(U f )(φ) j j ◦ − j j ◦ Xj Xj = (X(θ ) Y(ψ ))(U f )(φ) j − j j ◦ Xj and

θ X(U (φ) f ) ψ Y(U (φ) f ) = j j ◦ − i i ◦ Xj Xi 5. CURVATURE 59

= θ (( f T X)U (φ)) ψ (( f T Y)U (φ)) j ◦ j − i ◦ i Xj Xi = θ ( ψ (U f )U (φ)) ψ ( θ (U f )U (φ)) j i i ◦ j − i j j ◦ i Xj Xi Xi Xj = ψ θ (U (U (φ))) f ψ θ (U (U (φ)) f ) i · j · i j ◦ − i j j i ◦ Xi, j Xi, j = ψ θ ([U , U ] f )(φ). i · j · i j ◦ Xi, j

Hence the result. Now by (3.16) and (5.13) it is enough to prove the equation for g = Z f where Z C (M). Using the notations of (5.15) we see that ◦ ∈

5.15

(D D )(Z f ) (D D )(Z f ) D , (Z f ) X Y ◦ − Y X ◦ − [X Y] ◦ = DX(D f T Y Z) DY (D f T XZ) D f T [X,Y]Z by (5.14) ◦ − ◦ − ◦ = (DXD θ j(U j f )Z DY D ψi(Ui f )Z. j ◦ − i ◦ P P

(D(X(θi) Y(ψi))(Ui f )+ θ jψi([Ui,U j] f )Z) − − ◦ ij ◦ = P DX( θ jDU j f Z) DY ( ψiDUi f Z) ◦ − ◦ Xj Xi

(X(θi) Y(ψi))DUi f Z ψi θ j D[Ui,U j] f Z − − ◦ − ◦ ◦ ◦ Xi, j by 4.1 (D.L.1.) = A B C D say. − − − The, by D.L.2 and D.L.3 59

= + A B X(θ j)(DU j f Z θ jDXDU j f Z − ◦ ◦ Xj Xj

Y(ψi)DUi f Z ψiDY DUi f Z. − ◦ − ◦ Xi Xi 60 Linear connections

But the ﬁrst and the third terms together equal C. Hence, by (3.14):

= A B C D θ jD f T X(DU j Z) ψiD f T Y (DUi Z) − − − ◦ − ◦ Xj Xi

ψ θ (D , Z f ). − i · j [Ui U j] ◦ Xi, j = θ jD ψi(Ui f )(DU j Z) ψiD θ j(U j f )(DU j Z) ◦ − j ◦ Xj P Xi P

ψ θ (D , Z f ) by (D.L.1) − i · j [Ui U j] ◦ Xi, j = θ ψ (D D Z f ) j · i · Ui U j ◦ Xj,i θ ψ (D D Z f ) − j · i · U j Ui ◦ Xi, j

ψ θ D , Z f − i · j [Ui U j] ◦ Xi, j

= ψ θ (D D Z D D Z D , Z) f i · j · Ui U j − U j Ui − [Ui U j] ◦ Xi, j n o = ψ θ R(U , U )Z f i · j · i j ◦ Xi, j n o = ψ θ R(U f, U f )(Z f ) i · j · i ◦ j ◦ ◦ Xi, j = R( ψ (U f ), θ(U f )(Z f )) i i ◦ j ◦ ◦ Xi Xj = R( f T X, f T Y)(Z f ) ... ◦ ◦ ◦

6 Convexity

Throughout this article let us denote by M a manifold with a spray G and by Ω the open set of T(M) on which the function exp is deﬁned. 6. CONVEXITY 61

6.1 Deﬁnition . We say that M is convex if there exists s D(M M, Ω) ∈ × such that (p, exp) s = idM M . ◦ × 61

6.2 Since M M as well as Ω is of dimension 2d the above equation gives T × that s on T , (M M) is one-one and hence onto T , (Ω). Hence s (m n) × s(m n) is a local diffeomrophism. Supposing that M is convex, let m, n M. Then, for sufficiently ∈ small positive number ǫ, consider the map f :] ǫ, 1 + ǫ[ M defined − → by the equation f (t) = exp .(t s(m, n)). · From (4.5) we conclude that f is a geodesic and, from the fact that (p, exp .) s = idM M that, ◦ × i) f (0) = exp .(0m), since p(s(m, n)) = m and ii) f (1) = exp((s(m, n)) = n. Hence f is a geodesic connecting m and n. In the case of an open subset A of Rd, with the canonical connection, we know that the geodesics are segments of straight lines. Hence it follows that our convexity implies the usual one. Conversely, if we set 1 (2.6.3) s(m, n) = ζ− (n m) m − then, the convexity of A in the usual sense implies that s(m, n) Ω ∈ and hence it follows that convexity of A in the usual sense implies the convexity in our sense. This justifies our terminology.

6.4 Proposition. If M is convex, then

s (p, exp) = idΩ . ◦ 62 Linear connections

6.5

62 Remark. From this it follows, if M is convex, that

(p, exp) : Ω M M → × is a diﬀeomorphism and, in particular, that

exp : Ω T (M) M m ∩ m → is a diffeomorphism for every m M, and in particular given m, n M ∈ ∈ there is one and only one geodesic from m to n. In view of 6.2, the proposition follows from the following lemma. Lemma 6.5 (a). Let E and F be connected differentiable manifolds and f : E F and g : F E be differentiable maps such that g is a local → → diffeomorphism and f g = Id . Then g f = Id . (In other words, f ◦ F ◦ E is bijective.) Proof. Let A = x x E, g f (x) = x . It is sufficient to prove that { | ∈ ◦ } A = E. Since the maps f and g are continuous, A is closed. Evidently A g(F). On the other hand, if x = g(y), y F, ⊂ ∈ g f (x) = g ( f g)(y) = g(y) = x. ◦ ◦ ◦ Hence A = g(F); in particular A is non-empty. Finally, g being a local diffeomorphism, A = g(F) is open in E. Hence A is a non-empty open and closed subset of the connected space E. It follows that A = E.

As an immediate consequence of (6.4), we have

6.6

63 Corollary. Let f D(] ǫ, 1 + ǫ[, M) be a geodesic in M. Then, if M ∈ − is convex, s( f (0), f (1)) = f ′(0).

Proof. It is suﬃcient to check that exp f ′(0) = f (1); (4.5) asserts precisely this. 6. CONVEXITY 63

6.7 Deﬁnition. We say that an open subset U of M is convex if U, as a sub manifold of M, with the induced spray, is convex.

6.8 Theorem . For every point m of a manifold M there is an open neighbourhood of m which is convex. Proof. First let us select a function φ F(M) such that ∈

φ(m) = 0, dφm = 0 and Ddφm is positive deﬁnite, i.e. (Ddφ)(x, x) > 0 for every non-zero x in Tm(M). For example let us take any euclidean structure on T (M), and, locally, deﬁne φ to be || || m || 2 exp 1. Then clearly || ◦ −

φ(m) = 0 and dφm = 0.

To prove that (Ddφ)(x, x) > 0 for non-zero x T (M), let r > 0 be such ∈ m that x T (M) x < r Ω. { ∈ m | || || }⊂ Set: u : t tx(on ] r, r[) and f = exp u. → − x ◦ Then f (0) = x and φ f = 2 u and hence, since f is a geodesic, by x′ ◦ x || || ◦ (4.13) d2(φ f ) (Ddφ)(x, x) = x = 2 + x > 0. ◦2 = dt t 0 || ||

Thanks to the continuity of Ddφ, it is possible to find r′ such that 0 < 64 r < r and Ddφ is positive definite on W = exp( x T (M) x < r ). ′ { ∈ m || || ′} Since M is Hausdorff and W compact in exp( x T (M) x < r ) then { ∈ m || || } W is compact in M. Now let WG be the spray on W induced by G and

W be the open set in T(W) on which the corresponding exp . is deﬁned. Now let us consider the map

(p, exp .) : W Ω W W. → × 64 Linear connections

From (4.8) and inverse function theorem it follows that there is an 0 < r′′ < r′ and an open neighbourhood V of m in M such that for

V = exp .( x (x) < r′′ ) the map { || || } 1 (p, exp .) : (p , exp .)− (V V) V V is × → × a diﬀeomorphism. Now let us set

1 1 s = (p, exp .)− : V V (p, exp .)− (V V). 1 × → × Since φ is continuous and is positive in V except at m and V V is − compact, being a closed subset of the compact set W, it follows that

ǫ = inf φ(x) > 0. x V V ∈ − Then by the selection of V we have ǫ < r . Now let us set U = V ′′ ∩ φ 1(] , ǫ [) where 0 < ǫ < ǫ. Note also that: U = W φ 1(] , − −∞ 1 1 ∩ − −∞ ǫ [) = exp( x T (M) x < ǫ ). Then m U, U is open and we have 1 { ∈ m || || 1} ∈

(p, exp) s1 = idU U ◦ × 65 since this equation holds in V V. Now if we show that s (U U) Ω × 1 × ⊂ then it follows, by deﬁnition, that U is convex. Let (x, y) U U. Then ∈ × 6.2 there is a geodesic f in V such that:

f1(t0) = x and f1(t1) = y.

But since f1 is a geodesic we have (by our choice of W and by (4.13)): 2 d (φ f1) ◦ = (Ddφ)( f ′, f ′) > 0 dt2 1 1 and hence φ f is a convex function. Further since x, y U we have ◦ 1 ∈ φ(x) < ǫ1, φ(y) < ǫ1. Hence by the convexity of φ f it follows that ◦ 1 1 f [t , t ] W φ− (] , ǫ [) = U. 1 0 1 ⊂ ∩ −∞ 1

Now the result follows from (4.2). 7. PARALLEL TRANSPORT 65

7 Parallel transport

For the rest of this chapter let us suppose that a manifold M with a symmetric connection C is given.

7.1 Example. Let us consider Rd with its canonical connection. Let f be a curve in Rd. If g is a lift of f into T(Rd) such that

DPg = 0. then, since (see (3.9)) ν = ζT . 66

0 = D g = ν gT P = ζT gT P = (ζ g)T P, P ◦ ◦ ◦ ◦ ◦ ◦ and hence ζ g is a constant, i.e. g is the ﬁeld of vectors obtained by ◦ translating a constant vector along f . Keeping this in mind, we deﬁne a parallel lift.

7.2 Deﬁnition . A lift g of a curve f D(I, M) into T(M) is said to be a ∈ parallel lift of f if DPg = 0.

7.3

Example. A curve f is a geodesic, if and only if f ′ is a parallel lift of f (see 3.5). Now we shall prove the existence and uniqueness of parallel lifts.

7.4 Proposition. Let f D(I, M). Then corresponding to every t I and ∈ 0 ∈ x T (M) there exists a unique parallel lift g of f such that 0 ∈ f (t0) g(t0) = x0. 66 Linear connections

Proof. Let t I and (U, r) be a chart of M such that f (t) U. Then, ∈ ∈ by (5.13) relative to (U, r) the local existence of such a g is equivalent to 1 the existence of functions ψi on f − (U) satisfying the equations

ψ (t)(U f )(t ) = x , i i ◦ 0 0 D ( ψ (U f )) = 0, P i i ◦ Xi

67 where x T (M) and U is a basis for C (U). By (4.1) C.D.2. and ∈ f (t) { i} C.D.3. this is the same as

ψ (t)(U f )(t ) = x i i ◦ 0 0 dXψ and i (U f ) + ψ D (U f ) = 0. dt i ◦ i P i ◦ Xi Xi Setting D (U f ) = a (U f ) i P i ◦ ji j ◦ ∀ Xj and x0 = biUi( f (t0)) Xi

we see that the above equations are equivalent to ψi(t0) = bi dψ i + a ψ = 0. dt i j j Xj

But we know that this system being linear admits a unique solution in

the whole domain of deﬁnition of the ψi′ s. Hence locally the proposition is true. But since the deﬁnition of parallel lift is intrinsic and because we do have local solutions, we are done.

7.5 Deﬁnition. A path C(t) in M consists of

i) a family f D(I , M), i = 1,..., k and i ∈ i 7. PARALLEL TRANSPORT 67

ii) a family [a , b ] I of intervals such that i i ⊂ i f (b ) = f + (a + ) for i = 1,..., k 1. i i i 1 i 1 − f1(a1) is called the origin of the path C and fk(bk) the end. Let C = f D(I , M), [a , b ] I be a path in M. Then for { i ∈ i i i ⊂ i} every x T (M) we can consider the end point of the parallel lift ∈ f1(a) g1 of f1 with g1(a1) = x; then the end point of the parallel lift g2 of f2 68 with g (a ) = g (b ), and so on; and thus arrive at the point g (b ) 2 2 1 1 k k ∈ T fk (bk)(M).

7.6

The vector gk(bk) is called the parallel transport of x along C and is denoted by τ(C)(x).

7.7 Proposition. With the above notation τ(C) is an isomorphism between

T f1(a1)(M) and T fk(bk)(M). Proof. By (4.1) C.D.2.-3, it follows that the mapping τ(C) is linear. Consider the path described in the opposite direction, i.e., the inverse 1 path C− of C which consists of 1 1 = fi− D(Ii, M) fk− i(t) fi(bi (t ai)) . { ∈ | − − − }

This induces a linear map of T fk(bk)(M) into T f1(a1)(M). Further the parallel lift of C described in the opposite direction is the parallel lift of 1 C− and hence 1 τ(C− ) τ(C) = idT (M). ◦ f1(a1) 1 and τ(C) τ(C− ) = idT (M) by the same token. ◦ fk(bk) Given a path C we can cut oﬀ the path after a point C(t) on C and then the new path gives, by the above proposition, an isomorphism of 68 Linear connections

T f1(a1)(M) onto TC(t)(M). This isomorphism will be denoted by τ(C, t), or τt. Now given a curve f and a lift g of f into T(M) we can express 69 DPg in terms of τ(C, t) and ordinary derivatives of functions. To see this let f D(I, M), t I and let x ,..., x be a ﬁxed basis of T (M). ∈ 0 ∈ { 1 d} f (t0) Then, let us denote by g1,..., gd the parallel lifts of f into T(M) with initial points x1,..., xd respectively. Then, by the previous proposition g (t),..., g (t) is a basis of T (M) and hence any lift g of f can be { 1 d } C(t) written as:

(2.7.8) g(t) = ψi(t)gi(t) X where ψ F(I). By C.D.2.-3 we have i ∈ dψ D g = i g + ψ (D g ) P dt · i i P i Xi Xi dψ = i g dt · i Xi since gi is a parallel lift and hence DPgi = 0. Now, let us deﬁne g by the equation b 1 (2.7.9) g(t) = τ(C, t)− (g(t))

Then, we have the followingb proposition

7.10 dg(t) Proposition. DPg = τ(C, t) dt ! b Proof. τ(C(t)) being linear we have

g(t) = ψi(t)gi(t) = ψi(t)gi(0) = ψi(t)xi Xi Xi Xi b b and hence dg(t) dψ = i x . dt dt · i b Xi dψ But D g = i g (t). P dt · i P 8. JACOBI FIELDS 69

7.11

Remark. If C is a loop, i.e. a path whose origin and end coincide, in 70 general, τ(C) , id (M). T f1(a1) For let X, Y C (M) be such that ∈ [X, Y] = 0 and let C be a curvilinear parallelogram in M whose sides are integral curves of X and Y. Then the horizontal lift of C is made up of the integral curves of XH and Y H. But by (5.11)

ξ([XH, Y H] ) = R(X, Y) x − x and R(X, Y) is, in general, diﬀerent from zero. Then, our contention − x follows from the geometric interpretation of the bracket of two vector ﬁelds.

8 Jacobi ﬁelds

8.1

Deﬁnition. A one-parameter family f of curves in a manifold M is an f D(I J, M) where I, J are open intervals. ∈ ×

8.2

We denote the ﬁrst coordinate of I J R2 by t and the second by × ∈ ∂ α. We denote the canonical basis of (I J) by P, Q (i.e. P = and × ∂t ∂ Q = ). The one parameter family f gives rise to a family fα α J of ∂α { } ∈ curves deﬁned by the equation 70 Linear connections

8.3

fα(t) = f (t,α), (t,α) I J, and to another family Ct t I of curves 71 ∀ ∈ × { } ∈ Ct, called transversal curves, which are deﬁned by the equation

(2.8.4) C (α) = f (t,α), (t,α) I J. t ∈ × Now setting

8.5 P = f T P and Q = f T Q, we observe that P appears as the family ◦ ◦ of tangent vectors to the curves fα′ s and Q appears as the vectors of the

variation of the fα′ s i.e. as the family of tangent vectors to the transversal curves Ct′ s. Let us note that

(2.8.6) [P, Q] = 0.

8.7

Proposition. With the above notation if C is symmetric and α J, fα ∀ ∈ is a geodesic, then DPDPQ = R(P, Q)P.

Proof. By 5.12 C.D.5.

DQDPP = DPDQP + R(Q, P)P + D[Q,P]P.

72 But since fα are geodesic by 3.5 we have DPP = 0 and by (2.8.6) we have D[P,Q]P = 0. Hence we have

(2.8.8) DPDQP + R(Q, P)P = 0.

But since is symmetric we have by (C.D.4) (see (2.3.15)) C D Q D P = [P, Q] = 0 P − Q 8. JACOBI FIELDS 71 and hence DPDPQ = DPDQP Now by (2.8.8) we have

DPDPQ = R(P, Q)P.

Now we give the following deﬁnition

8.9 Deﬁnition. A lift g of a geodesic f is called a Jacobi ﬁeld f if

DPDPg = R( f ′, g) f ′.

8.10 Example. Let f D(I, M) be a geodesic and let ψ F(I). ∈ ∈ Then ψ f ′ is, by deﬁnition, a Jacobi ﬁeld along f if and only if

DPDP(ψ f ′) = R( f ′, ψ f ′) f ′.

But by C.D.2.-3 (2.3.15)

d2ψ DPDP(ψ f ′) = f ′ dt2 · since f is a geodesic and hence DP f ′ = 0 by 3.5. Further

1 R( f ′, ψ f ′) f ′ = ψR( f ′, f ′) f ′ since R L ∈ 3 = 0 by C.T.1. 5.2.

Hence ψ f ′ is a Jacobi ﬁeld along f if and only if 73 d2ψ = 0 dt2 i.e. if and only if ψ is an aﬃne function. 72 Linear connections

8.11 Let A be an open subset of Rd with its canonical connection. Let f be a geodesic. It is a segment of a straight line by (1.20). Further R = 0 for A by (2.5.4). Hence a lift g of f is a Jacobi ﬁeld along f if and only if

DPDPg = 0. But (by (3.9)):

d2(ζ g) DPDPg = DP(ζ g)′ = (ζ g)′′ = ◦ . ◦ ◦ dt2 Hence g is a Jacobi ﬁeld along f if and only if

(2.8.12) ζ g = tx + y, x, y in R. ◦ Let us note that

(2.8.13) g(t) = tx + y.

8.14 b

Lemma. Given a geodesic f of M and x, y in T f (0)(M) there exists at most one Jacobi ﬁeld g along f such that

g(0) = x and (DPg)(0) = y.

Proof. With the notation of (7), droping t in ζt for simplicity, we have dg D g = τ( ), P dt b d dg d dg D D g = τ( )(τ( )) = τ( ( )) P P dt dt dt dt d2g b b = τ( ). b dt2 b 74 So g is a Jacobi ﬁeld along f if and only if

2 d g 1 (2.8.15) = τ− (R( f ′, g) f ′). dt2 b 8. JACOBI FIELDS 73

Now for a given t let us deﬁne a transformation

R(t) = T (M) T (M) f (0) → f (0) by setting b

1 (2.8.16) R(t)(y) = τ− (R( f ′, τ(y)) f ′), y T (M). ∀ ∈ f (0) Then g is a Jacobib ﬁeld along f if and only if

d2g (2.8.17) = R(t)g. dt2 b b b Thus, if g is a Jacobi field along f then g satisfies a second order homogeneous linear differential equation. Further, since b = = τ0 idT f (0)(M) and hence g(0) g(0) dg dg and (D g)(0) = τ ( (0))b= (0) P 0 dt dt b b the requirements g(0) = x and (DPg)(0) = y become the initial conditions dg g(0) = x and (0) = y dt b for the above differentialb equation. But under these circumstances it is known that the above differential equation can have at most one solution; the existence of the field would also follow from general theorems on differential equations, but in our case this will follow from (8.26).

8.18

Remark. If g is a Jacobi ﬁeld along f such that g(0) = 0 then 75

dg t3 dg g(t) = t (0) + R( f (0), (0)) f (0) + 0(t3), dt 6 ′ dt ′ b b b 0(t3) where 0 with t. t3 → 74 Linear connections

Proof. By (2.8.17)

d2g (0) = R(0)g(0) = 0, since g(0) = g(0) = 0; dt2 b b and b b

d3g dR dg (0) = (0)g(0) + R(0) (0) 3 dt · dt dt b b dg b b = R( f (0)b, (0)) f (0). ′ dt b Now we have only to apply Taylor’s formula with a remainder to g.

Now let us see if, given a geodesic f D(I, M) there exists a Jacobib 0 ∈ ﬁeld passing through a given point x, above the geodesic with a speed whose vertical component y is given. For simplicity, let us suppose that all intervals occurring in this article contain zero. Now let S D(J, M) be any curve in M with S (0) = x, and let U ∈ ′ (resp. y) be the parallel lift of S such that f U = = e (0) f0′(0) (resp. y(0) y). Now set f e

(2.8.19) S (α) = U (α) + αy(α) α J ∀ ∈ e f e 8.20

76 Remark. We have by (7.10) d( f (0) + αy) D S = τ ( ′ ) = τ (y) Q α dα α e and in particular e (DQS )(0) = y.

Now, since f0 is a geodesic t f ′(0)e Ω for every t in I. Let ]a, b[= I and 0 ∈ let 0 <ǫ< min a , b , {| | } 8. JACOBI FIELDS 75 and let ]a′, b′[=]a + ǫ, b ǫ[= Iǫ . − Then the compact set [a , b ]S (0) = tS (0) t [a , b ] is contained in Ω ′ ′ { | ∈ ′ ′ } and Ω is an open set. Hence there is an open neighbourhood J of α = 0 e e ǫ such that

(2.8.21) tS (α) Ω, t Iǫ and α Jǫ. ∈ ∀ ∈ ∈ Now set e

(2.8.22) f (t,α) = exp(t S (α)) for (t,α) Iǫ Jǫ. · ∈ × Then f is a one parameter familye of geodesics and

f (t, 0) is f Iǫ itself. 0| Now we claim that

(2.8.23) g(t) = Q(t, 0) t Iǫ ∀ ∈ is a Jacobi field along f Iǫ passing through x with a speed whose vertical 0| component is y, i.e. that g(0) = x and (DPg)(0) = y. By (8.7) and the 77 definition of a Jacobi field it follows that g is a Jacobi field along f0. To establish our claim we need only show that

g(0) = x and (DPg)(0) = y.

We have ∂C g(0) = Q(0, 0) = 0 (0), ∂α but C (α) = exp(0 S (α)) = S (α) and hence 0 · = = = (2.8.24) g(0) (S ′(eα))α=0 S ′(0) x. Further we have, by the deﬁnition of g,

(Dpg)(0) = (DPQ)(0, 0). 76 Linear connections

Since the connection is symmetric, using (2.3.15) C.D.4 and the fact that [P, Q] = 0 we have

(2.8.25) (DPQ)(0, 0) = DQP(0, 0).

But we know that for every α

fα : t exp(tS (α)) → is a geodesic in M whose tangent vectore at 0 is S (α) = P(0,α). Hence e

(DQP)(0, 0) = (DQS )(0) = y by the remark.

Hence we have proved the followinge proposition.

8.26 Proposition . Given a manifold M with a symmetric connection, a geodesic f0 in M, a point m on f0 and x and y in Tm(M) there exists 78 a unique Jacobi field g along f0 such that g(0) = x and (DPg)(0) = y. Furthermore, if we cut off the geodesic at both ends, then the Jacobi field can be realised as the vectors of the variations of a one parameter family f with fα for α = 0 coinciding with the corresponding restriction of f0. Now let us follow the above notation and examine two special cases.

8.27

1. Let the Jacobi ﬁeld g be such that g(0) = x and (DPg)(0) = 0. Then y 0 is a parallel lift of S with initial speed zero and so the vectors of ≡ variations of the one parameter family

exp(t U (α)) · realise g. Further the initial tangent vectorsf of the exp t U (α) : t I { · ∈ } are got simply by parallel transport of f (0) along S . 0′ f 8. JACOBI FIELDS 77

8.28

2. Let the Jacobi ﬁeld g be such that g(0) = 0. Then the trivial path f (0) can be taken for S . Then a one parameter family can be given by { } exp(t ( f (0)+αy)). Hence all the curves fα start from f (0). Furthermore · 0′ 0 we have the following result.

8.29

Corollary. Let m M, 0 , U T (M) Ω, and g be the Jacobi ﬁeld ∈ ∈ m ∩ along the geodesic f : t exp(t U ) → · = = T 1 = with g(0) 0 and (DPg)(0) y. Then expm(ξu− y) g(1).

Proof. Let us follow the notation of the previous proposition and take the one parameter family given by

exp(t S (α)) where S (α) = f ′(0) + α y. · · e e Let us consider the transversal curve 79

C = exp( f ′(0) + α y). 1 · We have

C1(α) = exp .S (α), and hence e

T g(1) = Q(1, 0) = C′ (0) = (exp S )′(0) = exp (S ′(0)) 1 ◦ e e But = 1 S ′(0) ξu− y. e 78 Linear connections

8.30 Deﬁnition. Given a non trivial geodesic f D(I, M) and t , t in I such ∈ 1 2 that t1 , t2, the points f (t1) and f (t2) are said to be conjugate points on f (or simply conjugate points when there is no possible confusion over f ) if there exists a Jacobi ﬁeld g along f such that

g(t1) = 0, g(t2) = 0 and g , 0.

8.31

Remark . Let us note that the fact that f (t1) and f (t2) are conjugate points on a geodesic f does not imply that there is a one parameter family f (t,α) of geodesics such that f (t ,α) = f (t ) and f (t ,α) = f (t ) α. 1 0 1 2 0 2 ∀ All we can say, thanks to 8.28, is that there exists one satisfying the ﬁrst condition, namely f (t ,α) = f (t ) α. 1 0 1 ∀ 8.32

80 Corollary . Given a non trivial geodesic f , and two points f (0) and T f (t) on f , then they are non-conjugate on f if and only if exp f (0) is of maximal rank at t f (0) T (M). · ′ ∈ f (0) Proof. Let g be any Jacobi field along f . Then if we replace f by f 0 = 0 1 0 f kt 1, we get f ′(0) = t− f ′(0) and g = g kt 1 is a Jacobi field along ◦ − ◦ − f 0, and g0(1) = g(t). It follows directly from definitions that f (0) and f (t) are conjugate points on f if and only if f 0(0) and f 0(1) are on f 0. So let us take t to be 1. (See (6.23)).

Let us suppose that f (0) and f (1) are conjugate. Then there is a Jacobi ﬁeld g along f such that g(0) = 0, g(1) = 0 and g , 0. 8. JACOBI FIELDS 79

Let the vertical component of the speed of g at zero be y; i.e. let (D g)(0) = y. Then y , 0 since otherwise g 0 by (8.14). Then P ≡ by the previous corollary (8.29) we have:

T 1 exp (ξ− (y)) = g(1) = 0. f (0) f ′(0)

T Hence exp f (0) would not be of maximal rank at f (0). T Now suppose that exp is not of maximal rank at f ′(0). Then there , T = exists a z in V f ′(0) such that z 0 and exp z 0, and a vector y such that 1 ξ− y = z. p′(z) Also, by the corollary, for the Jacobi ﬁeld g along f with

g(0) = 0 and (DPg)(0) = y we have g(1) = expT z = 0 and hence f (0) and f (1) are conjugate.

Chapter 3 Riemannian manifolds

1.1

Definition . Let M be a manifold. An element g L 2(M) is said to 81 ∈ define a Riemannian structure (or simply r.s.) on M, if for every m in M, g L 2(T (M)) defines a euclidean structure on T (M) (see (3.3)). m ∈ m m A manifold M with an r.s. is called a Riemannian manifold (or simply r.m.). By (M, g) we denote the r.m. M with the r.s.g.

1.2

Example. 1. As a ﬁrst example of an r.m. we endow Rd with the r.s., denoted by ǫ; deﬁned by the equation

ǫ(X, Y) = (ζ X) (ζ Y), X, Y C (Rd) ◦ · ◦ ∀ ∈ where denotes the usual scalar product on Rd.

1.3

Let (M, g) be any r.m., and let U be an open subset of M. Then the restriction of g to U deﬁnes an r.s. on U, denoted sometimes by g U. | Whenever we consider an open subset of an r.m. as an r.m. it is the structure above we have in mind.

81 82 Riemannian manifolds

1.4

Remarks. Given an r.s. g on M, we have, for every m of M, a positive deﬁnite symmetric bilinear form gm on Tm(M) which depends diﬀeren- tiably on m, i.e. the map g : m g → m belongs to D(M, L2(T(M)). The converse of this statement follows from (0.2.3).

The above examples make it possible to deﬁne an r.s. on every man- 82 ifold M. To see this, let (Ui, ri) be a family of charts of M such that the U cover M and let ϕ be a partition of unity (see (1)) subordinate i { i} to the covering U . By (1.3) on each U we have an r.s. namely r (ǫ ) { i} i i∗ i where ǫ = ǫ r (U ) is the restriction of ǫ to r (U ). Now let us extend i | i i i i the form ϕiri∗(ǫi) to the whole of M by deﬁning it to be zero outside Ui. Then 2 ϕ r∗(ǫ ) D(M, L (T(M)), i · i i ∈ and set = h ϕiri∗(ǫi). Xi Then we have for every X, Y in C (M), and m in M,

h(X, Y) = ϕ r∗(ǫ )(X, Y) = i · i i Xi = ϕiri∗(ǫi)(Y, X) since ǫ is symmetric Xi = h(Y, X),

and

= = h(X, X)(m) ϕi(m)ri∗(ǫi)(X, X) Xi = T T ϕi(m)ǫ(ri (X), ri (X))(m). Xi Riemannian manifolds 83

= , Since ϕi(m) 1 there is an i0 such that ϕi0 (m) 0 and since ϕi 0 ff ≥ and ri Pare di eomorphisms we have T T T T ϕi(m) ǫ(r X(m), r X(m)) ϕi (m) ǫ(r X(m), r X(m)) > 0, · i i ≥ 0 · i0 i0 Xi if X(m) , 0. Hence h is symmetric and positive definite and clearly 83 bilinear. Hence every manifold admits an r.s. Given two vectors x and y such that p(x) = p(y) then g(x, y) is defined and we call g(x, y) the scalar product of x and y. Associated to this scalar product there is a norm, , on T(M); namely || || 1 (3.1.5) x = (g(x, x)) 2 . || || 1.6 We denote g(x, x) = x 2 by E(x) and note that E : T(M) R is || || → differentiable and is continuous (E stands for energy). || || 1.7

On the tangent space Tm(M) of M at m the topology induced by T(M) and that induced by the norm are the same. || || 1.8 Lemma. x T(M) and z V we have ∀ ∈ ∈ x z(E) = 2g(x,ξ(z)). Proof. Let i : T (M) T(M) p(x) → be the canonical injection. Then T 1 z(E) = (i )− (z)(E i) ◦ where E i is the quadratic form on T (M) associated to g . Then ◦ p(x) p(x) (see [36] and (0.4.11)): T 1 z(E) = 2g(x, ζ((i )− )(z)) = 2g(x,ξ(z)). 84 Riemannian manifolds

1.9

84 Deﬁnition. We say that an r.m. (M g) is isometric to an r.m. (N, h) if · there exists an f D(M, N) such that ∈ i) f is a diﬀeomorphism,

ii) f ∗h = g. Then f is called an isometry between (M, g) and (N, h). We say that (M, g) is locally isometric to (N, h) if, m M, there exists an open ∀ ∈ neighbourhood U of m in M and a map f D(U, N) such that: ∈ i) f (U) is open in N, ii) f is an isometry between (U, g U) and ( f (U), h f (U)). | | 1.10 Orthogonal vector ﬁelds. ∂ ∂ Let us consider (Rd, ǫ). Then the canonical basis ,..., has the ∂u1 ∂ud following two properties: ∂ ∂ i) orthogonality: ǫ , = 0 if i , j, ∂ui ∂u j ! ∂ ∂ ii) commutativity: , = 0. "∂ui ∂u j #

1.11 In the case of a manifold (M, g), locally, we can select an orthonormal basis for C (M). To see this let (U, r) be a chart of M and let X1,..., Xd be a basis of C (U), say for example the one got by pulling back the

canonical basis on r(U). The deﬁne Yi′ and Yi inductively by setting = = X1 85 Y1′ Y1 , division by X1 is possible since X1 is a basis element X1 || || and hence never|| || zero,

i 1 − Y′ = X g(X , Y )Y i i − i k k Xk=1 2. EXAMPLES 85

Yi′ Y = , (Y′ is never zero since X is a basis). i Y i { i} || i′|| Then Y1,..., Yd is an orthonormal basis of C (U). By pulling back the canonical basis on C (r(U)) we get a commutative basis of C (U). But, in general, there is not any local orthonormal basis of C (M) which is also commutative. For, relative to (U, r) a chart of M, let Y1,..., Yd be an orthonormal basis of C (U) such that

[Yi, Y j] = 0.

Then by Frobenius’s theorem there exists a local system of coordinates y1,..., yd such that ∂ Y = i = 1,..., d. i ∂yi Then the map f : m (y1(m),..., yd(m)) → gives a local isometry between U and Rd. For clearly f is a diﬀeomor- phism, and

T T ( f ∗ǫ)(Yi, Y j) = ǫ( f (Yi), f (Y j)) ∂ ∂ = ǫ , = δi j = g(Yi, Y j), ∂yi ∂y j !

d and hence g = f ∗ǫ. But in general r.m.’s are not locally isometric to R 86 (see (5.2) and (5.4)).

2 Examples

2.1 Suppose that (M, g) is an r.m. and N any manifold such that there exists a map f D(N, M): ∈ f : N M → such that f T is injective for every n N. Then f g L 2(N) and ( f g) n ∈ ∗ ∈ ∗ n is symmetric on T (N). Moreover for x T (N) if ( f g)(x, x) = 0 n ∈ n ∗ 86 Riemannian manifolds

= = T T then 0 ( f ∗g)(x, x) g( fn (x), fn (x)) and since g is positive definite T = T = we have fn (x) 0. But fn is injective and hence x 0. Hence f ∗g is positive definite. Therefore f ∗g defines an r.s. on N. We apply this remark to the following cases. A) Let N be an sub-manifold of (M, g). Then the injection i : N M → has the properties stated above for f and hence we have an r.s., called the induced r.s., on N. We denote it by (g N). | The study, initiated by Gauss, of surfaces S in R3, i.e. of two dimensional sub-manifolds S with the induced structure was the starting point for Riemann’s original investigations on this subject. On the other hand, the theorem of Nash ([34]) states that if (M, g) is a connected r.m. then there exists an integer d′ and a diffeomorphism f between M and a sub-manifold N of Rd′ such that f is an isometry between (M, g) and (N, ǫ N). | 87 B) Let us consider the sphere Sd: Sd = x Rd+1 x x = 1 (Rd+1, ǫ). { ∈ | · }⊂ Then ǫ Sd is called the canonical r.s. on Sd. Let us note that any | + element of the orthogonal group of Rd 1 induces an isometry on (Sd, ǫ Sd). | C) Suppose that (N, h) is an r.m., M is any manifold and f D(M, N) ∈ is a covering map i.e. a map such that to each point n of N there is a neighbourhood U such that f restricted to each connected com- 1 ponent of f − (U) is a homeomorphism between that component and U. Then f ∗h defines an r.s. on M and this situation is described by saying that f is a Riemannian covering. Note that, in this case, (M, f ∗g) and (N, h) are locally isometric.

2.2 ) Now let us suppose that (M, g) is an r.m. and that G is a discrete group of isometries of (M, g) without ﬁxed points. If the quotient M/G is a 3. SYMMETRIC PAIRS 87 manifold, then the canonical map p:

p : M M/G → is a covering map, and there is a unique r.s. on M/G, denoted by g/G, for which p : M M/G is a Riemannian covering. → To see this, let n M/G and let m p 1(n). Then since p is a ∈ ∈ − covering map it is a local diﬀeomorphism, and we naturally deﬁne

1 hn = (p− )∗gm and hn defines a euclidean structure on Tm(M). Since G is a group of 88 isometries it follows that hn is independent of the m chosen. Thus we have a map h : n h → n of N into L2(T(M/G)). Furthermore since p is a local diffeomorphism, and since the differentiability is a local property it follows that h is differentiable. Hence there is an r.s. with the above properties.

D) Let us note two particular cases.

1. For (M, g) take (Sd, ǫ Sd) and for G take the group generated | by the antipodal map of Sd, i.e. the one induced on Sd by d+1 d the map idRd+1 of R . Then S /G is the real projective − space (Pd(R)), can). 2. For (M, g) take (Rd, ǫ) and for G take a discrete subgroup of the group of all translations of Rd such that the rank of G is d. Then the torus Rd/G considered as an r.m. with the r.s. Rd/G is called a ﬂat torus and denoted by (Rd/G, ǫ/G).

3 Symmetric pairs

3.1 Let G be a Lie group and let λ(s) and ρ(s) denote the left and the right multiplications by s G. Further let G be the Lie algebra of G and ∈ 88 Riemannian manifolds

exp : G G the associated exponential map. Let X G; then the → ∈ vector ﬁeld deﬁned by the one parameter group

λ(exp(tX)), t R ∈ is the right invariant vector ﬁeld whose value at e (the neutral element 89 of G) is X. To see this we have only to compute the speed of the curve t (exp(tX))g, at t = 0. → But (exp(tX))g = ρ(g)(exp(tX)), so that T (ρ(g) (exp(tX)))′(0) = (ρ(g)) (exp(tX))′(0) ◦ and by the deﬁnition of the exp map

(exp(tX))′(0) = X.

Hence the result.

3.2 Deﬁnition. A symmetric pair (G, H,σ) consists of

1) a connected Lie group G

2) a compact subgroup H of G and

3) an involutive automorphism σ of G such that

H ⊂ ⊂ X0 X

where is the subgroup of elements in G ﬁxed by σ, and 0 is the connectedP component of the identity, e of . P P When no confusion is possible, we shall speak of the symmetric pair (G, H) without any reference to σ. 3. SYMMETRIC PAIRS 89

Let M be the homogeneous space of left cosets of H in G, p be the projection map from G to G/H. Set m0 = p(e), and let τ denote the left action of G on M, i.e.

τ(a)(xH) = ((ax).H), a G. ∈ Let us denote the Lie algebra of H by H and identify it with the corresponding subalgebra of G. Let us denote by M the set of elements X in G such that

(3.3.3) σT (X) = X. − Then we have 90

(3.3.4) G = H + M where the right hand side denotes the direct sum of vector spaces. We have further

(3.3.5) pT M :M T (M) is an isomorphism. | → m0 3.6 Let X M. Then for the two homomorphisms ∈ t σ exp(tX) → ◦ t exp(t( X)) → − from R into G the diﬀerential maps are the same and hence

σ exp(tX) = exp(t( X)). ◦ − = T For every element a of G let us denote by Ad(a) (Int(a))e the automorphism of the Lie algebra G corresponding to the automorphism

1 Int(a) : u aua− → of G. Since Int(h) and σ commute h H we have ∀ ∈ (3.3.7) (Ad H)(M) M. ⊂ 90 Riemannian manifolds

3.8 Deﬁnition. The map σ from M to M is deﬁned by the equation

σ pb= p σ i.e. σ(aH) = σ(a)H. ◦ ◦ Now one sees thatb b

(3.3.9) pT Ad h = (τ(h))T pT , h H, ◦ ◦ ∀ ∈ T (3.3.10) σ = idT (M), m0 − m0 (3.3.11) σ τ(a) = τ(σ(a)) σ, a G. ◦ b ◦ ∀ ∈ 91 b b

3.12 Proposition. With the above notation, there exists an r.s. γ on M such that the transformations σ and τ(a) for every a in G are isometries of (M, γ); moreover γ is unique (upto a positive constant) if Ad H acts irreducibly on M. b

Proof. a) Existence. Since H is compact there is a euclidean structure γe on M which is invariant under the action of Ad H. Now deﬁne b γ L 2(T (M)) m0 ∈ m0 by the equation

(3.3.13) p∗(γ ) M = γ . m0 | e Then by (3.3.9) we have b

(3.3.14) (τ(h))∗(γ ) = γ , h H. m0 m0 ∀ ∈ Now, for m M, choose an a G such that ∈ ∈

τ(a)(m0) = m 3. SYMMETRIC PAIRS 91

and deﬁne γm by the equation

= (3.3.15) (τ(a))γm γm0 .

Then if a′ is another element of G such that

τ(a′)(m0) = m

we have a = ah for some h H, and (3.3.14) shows that γ is ′ ∈ m independent of the a chosen. Now we can see that the map

γ : m γ → m defines an r.s. on M. By the very definition of γ it follows that 92 τ(a) is an isometry of (M, γ) for every a of G. Now let us consider σ. Since σ is a diffeomorphism, to show that σ is an isometry it is enough to show that γ is invariant under σT . To see this let bm M. Thenb since M is a homogeneous spaceb there exists an a ∈ in G such that b m = τ(a) m . · 0 Now let x and y be in Tm(M), and then define x0 and y0 by the equations T T x = τ(a) (x0), y = τ(a) (y0). Then

T T T (σ ) (x, y) = γ 1 (σ (x), σ (y)) m σ− (m) T T T T = γb 1 (σ (τ(a)) (x0), σ (τ(a)) (y0)) b σ− (m) b ◦ b ◦ T T T T = γb 1 (τ(σ(a)) σ (x0), τ(σ(a)) σ (y0)) σ− (m) b ◦ b ◦ by (3.3.11) b b b T T = γ 1 (τ(σ(a)) ( x0), τ(σ(a)) ( y0)) σ− (m) − − bby (3.3.10) = γm0 (x0, y0) by the deﬁnition of γ.

Hence the existence. 92 Riemannian manifolds

b) Uniqueness. Let γ1 be any r.s. for which τ(a), for every a in G, 93 is an isometry. Then, in particular τ(h) is an isometry of (M, γ1) for every h of H and now by (3.3.9) it follows that p (γ ) M is ∗ 1 | a euclidean structure on M which is invariant under the action of Ad h for every h in H. Now the uniqueness is a consequence of the well known lemma of Schur.

Remark . The transformation σ deﬁned above is called the symmetry around m0. By (3.3.10) it follows that it is an involution having m0 as isolated ﬁxed point. Since G actsb transitively on M, given any point m of M, there exists an a of G such that

τ(a)(m0) = m.

94 Then we see that the transformation

1 (3.3.16) τ(a) σ τ(a)− ◦ ◦ is an involution having m as isolatedb ﬁxed point. We call this involution the symmetry around m and denote it by

(3.3.17) σm.

Concerning these symmetriesb we have the following: 3. SYMMETRIC PAIRS 93

3.18 Proposition. If m,m M are suﬃciently near then n M such that ′ ∈ ∃ ∈

σn(m) = m′.

Proof. Since G acts transitivelyb on M we can suppose that m = m0. Now since the map

(exp M)T pT pT (exp M)T :M | T (G) T (N), ◦ | −−−−−−−→ e −−→ m0 is an isomorphism, the inverse function theorem shows, since m′ is close enough to m ; that there exists an X M such that 0 ∈

m′ = exp(X).

X Now p(exp( )) can be taken for n. 2

3.19 The homogeneous Riemannian manifold (M, γ) is called the symmetric space associated to the symmetric pair (G, H).

3.20 Remark. 1. Let f : t exp(tX). → Then from the above proof it is clear that 95

σ ( f (t)) = f (2t t). f (t0) 0 − b 3.21

Remark. 2. In fact, the condition that m and m′ be suﬃciently near is superﬂuous: (see 4.3). 94 Riemannian manifolds

4 The S.C.-manifolds

By K we denote one of the following ﬁelds: R (the real numbers), C (the complex numbers), H (the quaternions); the conjugation in K n is z z and we set k = dimR K(k = 1, 2, 4). We, consider K (for an → integer n) as a left vector space over K, denote by e1,..., en its canonical basis and use freely the two identiﬁcations:

n+1 n K = K K : (z1,..., zn, zn+1) = (z = (z,..., zn), zn+1) = (z, zn+1). × + (3.4.1) Kn Kn 1 : (z ,..., z ) = (z ,..., z , 0). ⊂ 1 n 1 n On Kn we consider the canonical hermitian structure <,>:

< (z1,..., zn), (w1,..., wn) >= ziwi. Xi

By U(Kn) we denote the set of all K-endomorphisms of Kn which leave <,> invariant; for K = R, C we denote by SU(Kn) the subgroup of U(Kn) of elements of determinant equal to one. We also set:

(3.4.2) S 0(n) = SU(Rn), 0(n) = U(Rn). (3.4.2) SU(n) = SU(Cn), U(n) = U(Cn), n 1 ≥ S p(n) = (SU(Hn) =)U(Hn).

96 All these are compact Lie groups, and, with the exception of 0(n), all are connected. The component of the identity of 0(n) is S 0(n).

4.3

Proposition. SU(Kn) acts transitively on the K-directions and also on the R-directions of Kn; also S 0(n) acts transitively on the planes of Rn; if K = C, then n > 1.

In particular SU(Kn) acts transitively on the sphere

+ (3.4.5) S(Kn) = x Kn < x, x >= 1 ; Sn = S(Rn 1). { ∈ | } 4. THE S.C.-MANIFOLDS 95

On Kn+1 0 the equivalence relation R : z z if and only if λ K with − ∼ ′ ∃ ∈ z = λz′ yields the projective space

Kn+1 0 ar[d] −{ } p + (3.4.6) Pn(K) = (Kn 1 0 )/R, −{ } Pn(K) on which SU(Kn+1) acts transitively. So Pn(K) is a homogeneous space; and is a connected compact manifold, of dimension k.n. We write Pn(K) = M = G/H where G = SU(Kn+1) and H is the subgroup of G leaving the point m0 = p(en+1) ﬁxed. We set

(3.4.7) H = S(U(Kn) U(K)) = SU(Kn+1) (U(Kn) U(K)) × ∩ × where U(Kn) U(K) is embedded in U(Kn+1) by (3.4.1). For elements × of H we use the notation ( f, λ) where f U(Kn) and λ U(K) stands ∈ ∈ for the map µ µλ of K into itself which is associated to the element 97 → b b λ K (with <λ,λ>= 1), i.e. ∈

(3.4.8) ( f, λ)(z, zn+1) = ( f (z), zn+1λ).

From (3.4.7) and (3.4.7)b we get the ﬁbrations:

Sn S2n+1 S4n+3

(3.4.10) Z2 S1 S3 Pn(R) Pn(C) Pn(H)

We show now that (G, H) is canonically a symmetric pair; deﬁne an endomorphism s of Kn+1 and an automorphism σ of G by:

s(ei) = ei for i = 1,..., n and s(en+1) = en+1 (3.4.11) − σ : g σ(g) = s g s. → ◦ ◦ As is easily veriﬁed, the subgroup of elements ﬁxed under σ is nothing but H = S(U(Kn) U(K)). × 96 Riemannian manifolds

We study now the complement M in G = H + M as deﬁned in (3), by introducing the map r : Kn G deﬁned by: → (3.4.12) r(z)(e + ) = z, r(z)(z) = e + , r(z)(u) = 0 n 1 − n 1 u < u, z > = 0 ∀ | when < z, z >= 1 and extension by linearity. Hence the one-parameter subgroup of G associated to r(z) : t exp(t r(z)) is nothing but: → · e + (cos t)e + + (sin t)z n 1 → n 1 (3.4.13) exp(t r(z)) : z (sin t)en+1 + (cos t)z ·  →− u u u with < u, z >= 0.  → ∀  Remark that dim(Kn) = k.n = dim G/H = dim M and that r(Kn) M ⊂ for a direct computation yields:

(exp(t.r(z)) = exp( t.r(z)). − 98 Hence:

(3.4.14) r(Kn) = M, pT r : Kn T (M) ◦ → m0 is an isomorphism. To ﬁnd the action of Ad H on M we use (3.3.9) and (3.4.14). To z Kn we associate the curve l : t p(e + + tz) in Pn(K). ∈ → n 1 For the image τ(h) l under h = ( f, λ) H, by (3.4.8) and deﬁnition of ◦ ∈ Pn(K) we have: b t τ(h)(p(e + + tz)) = p(h(e + + tz)) = p(λ e + + t f (z)) = → n 1 n 1 · n 1 · 1 = p(e + + t λ− f (z)) n 1 · · hence by (5.7) and (3.3.9):

T 1 T (τ(h) l)′(0) = (p r)(λ− f (z)) = (τ(h)) (l′(0)) = ◦ ◦ · = ((τ(h))T pT r)(z) = (pT Ad(h))(r(z)) ◦ ◦ ◦ which yields:

1 1 (3.4.15) Ad(( f, λ)) = r (λ− f ) r− . ◦ ◦ b 4. THE S.C.-MANIFOLDS 97

4.16

Remarks . In the case K = R, C the homomorphism H = S(U(Kn) n 1 × U(K))) U(K ) deﬁned by ( f, λ) λ− f in onto because the con- → + → · dition ( f, λ) SU(Kn 1) is equivalent to the condition λ det f = 1, ∈ b · so that λ 1 f = (det f ) f (and this is injective when n > 1.). This −b · · explains why one sometimes writes: Pn(R) = S 0(n + 1)/0(n); Pn(C) = SU(n + 1)/U(n). One also writes:

Pn(H) = Sp(n + 1)/ Sp(n) Sp(1) × which is correct because SU(Hn) = U(Hn). For the sphere S 0(n + 1)/S 0(n) the computation as above (except that there is no projection p) shows that Ad H is isomorphic to the action of S 0(n) in Rn (under 99 (3.4.1)). From these remarks and from (4.3) we get:

4.17

Proposition. For Pn(K) = G/H the action of Ad H is transitive on the directions of M; for Pn(R) or Sn the action of Ad H is transitive on the planes of M.

In particular Ad H acts irreducibly on M so by (3.12) there exists (upto a positive constant) a canonical r.s. on G/H; in fact there exists a canonical r.s. on G/H, for we see from (3.4.14) and (3.4.15) that the canonical euclidean structure on Kn (deduced from its canonical hermitian form) gives a euclidean structure on M invariant by Ad H. Sowe require by deﬁnition, that:

4.18 r : Kn M be a euclidean isomorphism. We write (Pn(K), can) for this → symmetric space and leave the reader to check in the case of Sd = S 0(d+ 1)/S 0(d) that this procedure yields nothing but (Sd, can) as deﬁned in 2.1. 98 Riemannian manifolds

4.19 Remarks. we leave to the reader the proofs of the following facts: A) the imbedding Kn Kn+1 yields an imbedding of pn 1(K) as a → − sub manifold of Pn(K). Prove the canonical r.s. on the Pn(K)’s n 1 are hereditary in the sense that the induced r.s. on P − (K) by the canonical r.s. of Pn(K) is its canonical r.s.;

B) the ﬁrst ﬁbration in (3.4.10) is a riemannian covering. The two last ones are excellent in the sense that for

100 ﬁrst the r.s. induced on the ﬁbres is the canonical one; second: if one writes at any m E the orthogonal decomposition ∈ 1 Tm(E) = Tm(p− (p(m))) + N then pT N : N T (B) is a euclidean isomorphism. → p(m) To those symmetric spaces we add the symmetric space P2(Γ) = F4/ Spin(9) the Cayley projective plane (see: [37]). We consider on it the canonical r.s. (P2(Γ), can) such that all its geodesics are closed and of length π (see [14], p.356).

4.20 Deﬁnition. Any of the (Pn(K), can), (Sd, can), (P2(Γ), can) is called as S.C.-manifold. We will use also a non compact symmetric space deﬁned as follows. Let S 00(d, 1) be the identity component of the linear group of Rd+1 which leaves invariant the quadratic form:

2 2 2 (x ,..., x , x + ) x + + x x + 1 d d 1 → 1 ··· d − d 1 4. THE S.C.-MANIFOLDS 99

For the same s and σ as in (3.4.2) we get the symmetric pair (G, H) = d (S 00(d, 1), S 0(d)). Again S 0(d) acts by Ad on M as S 0(d) on R ; in particular:

4.21

Ad H acts transitively on the planes of M. In particular by 3.12 we get (upto a positive constant) a canonical r.s. on M = G/H. In fact M is homeomorphic to Rd; for,

d+1 A = z = (x ,..., x x + ) R x + 0,< z, z >= 1 and { 1 d d 1 ∈ | d 1 ≥ } deﬁne the injection 101

d i : A P (R) by i(x ,..., x , x + ) = p(x ,..., x , 1). → 1 d d 1 1 d

Claim . S 00(d, 1) acts transitively on i(A); in fact, for any z with < z, z >= 1, we have in S 00(d, 1) the one parameter subgroup

e + (ch t) e + + (sh t) z n 1 → · n 1 · t g(t) z (sh t) en+1 + (ch t) z →  → · · u u u with < u, z >= 0,  → ∀  so that

g(t))(p(e + )) = p(g(t)(e + ) = p(ch t) e + + (sh t) z) = n 1 n 1 · n 1 · = p(e + + (th t) z). n 1 · Having an r.s. on i(A), we use a diﬀeomorphism between A and Rd to get a canonical r.s. on Rd which we denote by (Rd, hyp).

4.22

(Rd, hyp) is called the hyperbolic space of dimension d. 100 Riemannian manifolds

5 Volumes

Let (M, g) be an r.m. Then for every m M, T (M) has a euclidean ∈ m structure by g i.e. the one given by the symmetric positive deﬁnite bilinear form gm. Hence it has a canonical volume (see 3.6) tm, and further if (M, g) is oriented then Tm(M) admits a canonical volume form S m induced by the oriented form σ′, i.e. the orientation given by σm′ .

5.1

102 Proposition. With the above notation the map

θ : m t → m is a volume element on M (called canonical volume element on M) and if M is oriented the map σ : m S → m is a volume form (called the canonical volume form).

Proof. Let us consider the second case first. We know that for X1,..., Xd belonging to C (M), σ(X1,..., Xd) is F(M)-multilinear. Hence to show that σ is a volume form we need only show that σ(X1,..., Xd) is differentiable. To see this, since differentiability is a local property, let (U, r) be a chart of M, and let Y1,..., Yd be a basis of C (U), orthonormal relative to the restriction of g to U. We can suppose that Yi in that order are positive relative to σ′. Then we have σ(Y ,..., Y )(m) = 1, m U, 1 d ∀ ∈ and hence, since σ is F(M)-multilinear, it is differentiable on U.

Now let us take up the ﬁrst part. We know that tm is a volume element on Tm(M). So to show that θ is a volume element, it is enough to show that θ is, locally the modulus of a volume form. To see this let us

take an orientation σ1′ on U and then, clearly, θ = σ , | 1|

where σ1 is related to σ1′ as σ is to σ′. 5. VOLUMES 101

5.2 103 Let us note that if X ,..., X is any local basis of C (M) the formula { 1 d} (3.6) gives 1 θ(X1,..., Xd) = (det(g(Xi, X j))) 2 . If (M, g) is a compact manifold then we know that

θ Z M exists. We call this number the volume of (M, g) and denote it by Vol (M, g). Generally, if no confusion is possible, we omit the reference to g and say the volume of M and write Vol(M). If the dimension of the manifold is one, volume is called length and if it is two volume is called area. They are denoted by lg(M) and ar(M) respectively. Let us note that in the oriented case, with the notation of (5.1),

θ = σ Z Z M M If N is a compact sub manifold of M then there is an induced r.s. on N (see (2.1)) A). Whenever we talk of volume of N without any reference to the r.s. on N this induced structure is the one that is considered. And further, we set: Vol(N, g N) = Vol(N). | 5.3 In particular, suppose the dimension of N is equal to one; then if N admits the parametric representation N = f (I) where I = [0, 1] and f is a curve f : I M, we have: → 1

lg( f (I)) = λ = f ∗λ = ( f ∗λ)(P) Z Z Z f (I) I 0 where λ is the length form of (N, g N). But ( f λ)(P) = λ( f T P) = 104 | ∗ ◦ 102 Riemannian manifolds

λ( f ) = f so that ′ || ′|| 1

lg(N) = f ′ dt = lg( f (I)). Z || || 0

This leads to the following deﬁnition for paths (and a fortiori for curves):

Deﬁnition . Let C = f D(I , M), [a , b ] I be a path in an r.m. { i ∈ i i i ⊂ i} (M, g). Then the length of C, denoted by lg(C), is by deﬁnition:

lg(C) = f ′ dt = lg( f [a , b ]). || i || i i i Xi Z Xi ai

Now let us examine the volume element more closely.

5.4

Let (M, g) be an r.m. and let ϕ be a diﬀerentiable function on M which is everywhere positive. Then ϕg deﬁned by the equation

ϕ g(X, Y) = ϕ(p (X)) g(X, Y), X, Y C (M) · M · ∈ deﬁnes an r.s. on M. If we denote the volume element of (M,ϕ, g) by θϕ and that of M by θ then, from (5.2) it follows that they are related by the equation d/2 θϕ = ϕ θ ·

5.5

B. Let (M, g) and (N, h) be two compact r.m.’s which are isometric. They by the deﬁnitions of isometry, volume and by (0.3.9) we have Vol(M, g) = Vol(N, h). 5. VOLUMES 103

5.6 105 C. Let (M, g) and (N, h) be two r.m.’s and let

p : (M, g) (N, h) → be a riemannian covering. Further let us suppose that N is compact and the covering has a ﬁnite number, k, of sheets. Then Vol(M, g) exists and further Vol(M, g) = k, Vol(N, h).

To see this let us take an open covering (Ui, ri) by coordinate neigh- { 1} bourhoods of N, Ui taken so small as to make p− (Ui) the union of k dis-

joint components Ui1 ,..., Uik (this is possible because p is a k-sheeted covering map.) Then U , r = r p U is a complete atlas for (M, g). { i j i j i ◦ | i j} Now let us take a partition of unity ϕ on N subordinate to the covering { i} Ui. Then ψi j deﬁned by

ϕ p on U ψ = i ◦ i j i j  0 outside Ui  j  is a partition of unity on M subordinate to the covering U . Now since { i j } p is a riemannian covering map, denoting the volume element of (M, g) and (N, h) by θM and θN respectively, we have

ψi θM = (ϕi p)(p∗θN) = p∗(ϕiθ) Z j Z ◦ Z Ui j Ui j Ui j

= ϕiθ by (3.5.5). Z Ui

Hence 106

= = Vol(M, g) ( ψi j θM) ( θ) Xj Xi Z Xj Z Ui j N = k. Vol(N, h). 104 Riemannian manifolds

Now let us note a particular case. If we take the real projective space (Pd(R), can) for (N, h) and (Sd, can) for (M, g) and the natural map of Sd onto Pd(R) for p then we have k = 2. Hence 1 Vol(Pd(R), can) = Vol(Sd, can). 2

5.7 D. Let us consider the case of a ﬂat torus Rd/G (see (2.2) D.2). Let (τ1,...,τd) be a set of generators for G. Then τ1(0),...,τd(0) is a basis d of R . Let (v1,..., vd) be the corresponding coordinate system. Now let

W = x Rd 0 < vi(x) < 1, for i = 1,..., d . { ∈ | } Then W is open in Rd, W is compact and the restriction of

p : Rd Rd/G → to W is a diﬀeomorphism of W with p(W). Also p(W) = Rd/G and W W is of measure zero. Hence − Vol(Rd/G, ǫ/G) = Vol(p(W), ǫ/G) = (chap0:0.3.2) = Vol(W, ǫ) = det(τ (0),...,τ (0)) by (). | 1 d | 6 The canonical forms α and dα

Let (M, g) be an r.m. Then there is a map

♯ g : C (M) C ∗(M) → 107 given by the equation

g♯(X)(Y) = g(X, Y), Y C (M). ∀ ∈ Clearly it is an F(M)-linear map and since g is non-degenerate it follows that g♯ is an isomorphism of F(M)-modules. Let us note that

g♯(X) = i(X)g. 6. THE CANONICAL FORMS α AND Dα 105

6.1

♭ ♯ 1 ♯ We set g = (g )− . Now evaluating g at each point of T(M) we get a vector bundle isomorphism of T(M) with T ∗(M). We denote this map also by the symbol g♯ and it will be clear from the context which we mean by g♯.

g♯ T(M) T ∗(M) < ♭ << g ÑÑ << ÑÑ (3.6.2) << ÑÑ p << ÑÑ << ÑÑ p∗ << ÑÑ < ÐÑÑ M

6.3 Remark. Let us note that g♯ and g♭ give rise, in a natural, way, to isomorphisms of L r i L r s ±i with s ∓ between tensor spaces over (M, g) (these isomorphisms describe the op- eration usually called “raising” or “lowering” subscripts).

Now we utilise the maps g♭ and g♯ to associate an element of C (M) 108 with every ϕ and to pull µ from T ∗(M) to T(M) where µ is deﬁned in (0.4.24).

6.4 Deﬁnition. For every ϕ F(M) we set ∈ grad ϕ = g♭(dϕ).

If U is an open subset of (Rd, ǫ) and ϕ F(U) then we have ∈ ∂ϕ ∂ grad ϕ = . ∂ui ∂ui Xi 106 Riemannian manifolds

6.5

♯ Deﬁnition. α = (g )∗(µ). If we have to consider the α’s associated to several manifolds (M, g), (N, h),... at the same time, we write Mα,Nα for the corresponding α’s.

6.6 For every z in T(T(M)) we have

T α(z) = g(p′(z), p (z)).

Proof. For

♯ ♯T α(z) = (g )∗(µ(z)) = µ(g (z)) = ♯T T ♯T = (p′′(g (z)) = (p∗ (g (z)).

T(T(M)) / T(T ∗(M)) (g♯)T

p′ p′′

g♯ T(M) / T (M) J ∗ JJ tt JJ tt JJ tt p J tt JJ tt p∗ JJ tt J$ tt M y

109 But p T g♯T = (p g♯)T = pT and p g♯T = g♯ p . Hence ∗ ◦ ∗ ◦ M ′′ ◦ ◦ M ♯ T T (z) = (g p′ (z))(p (z)) = g(p′(z), p (z)). ◦ M

6.7 If f : (M, g) (N, h) is an isometry, then → M T N M = T N and d( α) = ( f ) (d( α)). α ( f )∗( (α)) ∗ 6. THE CANONICAL FORMS α AND Dα 107

Proof. In view of the formula (2.4) IV. We need only prove the ﬁrst equality. We have, by (6.6),

M = T and α(z) g(p′M (z),pM (z)) T N = N T = T T T T T ( f )∗( α)(z) α( f (z)) h(p′N ( f ) (z)), pM ( f ) (z)

( f T )T T(T(M)) / T(T(N))

T p T pM p′M ′N pN

f T T(M) / T(N)

pM pN

f M / N But we have

T T T T T T T T p′ ( f ) = f p′ and p ( f ) = f p . N ◦ ◦ M N ◦ ◦ M Hence 110

T ∗ N = T T T = f ( α) h( f (p′M(z)), f (pM(z)) = T = ( f ∗h)(p′M(z), pM (z)) = T g(p′M(z), pM (z)) since f is an isometry.

Hence the result.

♯ ♯ Since g is an isomorphism so is (g )∗. Now by lemma (4.27) we get the following result.

6.8 Proposition. dα is non-degenerate everywhere on T(M). 108 Riemannian manifolds

6.9 α In particular it follows that dα is a volume form for T(M) (see also ∧ (2.5)).

6.10 Lemma. For every x T(M), z V and z T (T(M)) we have ∈ ∈ x ′ ∈ x T (dα)(z, z′) = g(ξ(z), p (z′)).

Proof. Let us choose Z and Z′ in C (T(M)) such that i) pT (Z) = 0 and Z(x) = z (for example let us take any Y C (M) = = 1 ∈ such that Y(p(x)) ξ(z) and set Z(x) ξU− (Y(p(U)) for every U in T(M)); and

T ii) there exists an X in C (M) for which p (Z′) = X; (for example let T T us take any Y in C (M) such that Y(p(x) = p (z′) and set Z′ = Y (see (4.19)).

Then

dα(z, z′) = (dα)(Z, Z′)x

= Z(α(Z′)) Z′(α(Z)) α([Z, Z′]) . x − x − x 111 But by (6.6) we have

T (Z) = g(p′ Z′, p Z) = g(p′ Z, 0) = 0; ◦ ◦ ◦

T α([Z, Z]) = g(p′ [Z, Z′], p [Z, Z′]) ◦ ◦ T T = g(p′ [Z, Z′], [p (Z), p (Z′)]) ◦ = g(p [Z, Z′], 0) = 0, ◦ and

T T α(Z′) = g(p′ Z′, p Z′) = g(id)T(M), p Z′), ◦ ◦ ◦ 7. THE UNIT BUNDLE 109

T i.e. α(Z′) = g(y, (p Z′)(p(y)). Hence y ◦ dα(z, z′) = Z(α(Z′))x = z(α(Z′)).

Now let us consider the map of T(M) into R deﬁned by

T(M) y α(Z′) = g(y, X(p(y)). → → y Clearly it is a linear function on T(M). Hence, by (4.16) we have

z(α(Z′)) = g(ξ(z), X(p(ξ(Z)) T = g(ξ(z), p (z′)).

Note that this gives another proof of the non-degeneracy of dα, already proved in (6.8).

7 The unit bundle

7.1 Deﬁnition. The subset U(M) = E 1(1) = x T(M) x = 1 of T(M) − { ∈ || || } is called the unit bundle of M. Let us set W = E 1( ] , 1[ ) = x T(M) x < 1 and observe − −∞ { ∈ || || } that U(M) is the boundary of W in T(M).

7.2

Lemma. W is a nice domain. 112 Proof. We will be through if we show that x U(M) : (dE) , 0. But ∀ ∈ x by (1.8) we have 1 = (dEx)(ξx− x) 2(g(x, x)) , 0. Now by (3) we have the following result.

7.3 Proposition. U(M) is a sub manifold of T(M) of dimension 2d 1. − 110 Riemannian manifolds

7.4

Let us note that U(M) is compact if M is. For, the ﬁbre Um at m is

U = x T (M) g(x, x) = 1 m { ∈ m | }

and hence it is a sphere in the euclidean space (Tm(M), gm). We denote the pullbacks of α and dα to the sub manifold U(M) by α and dα themselves.

8 Expressions in local coordinates

∂ Let (M, g) be an r.m. and let (U, r) be a chart of (M, g). Let Xi = (∂xi ) be the basis of C (U) dual to the basis dxi = d(ui r) of C (U). Then ◦ ∗ (dxi dx j) (for this multiplication see (0.2.2)) is a basis of L 2(U). { · } Hence there exists local functions g such that { i j} 8.1

i j g = gi jdy dx , with g ji = gi j since g is symmetric. Now let us take i, j · P i ♯ ♯ i 113 any X = p Xi and compute g (X). Setting g (X) = pidx , we have, i by deﬁnition,P P

♯ j j (3.8.2) pi = g (X)(Xi) = g( p X j, Xi) = gi j p . Xj Xj

Since g is positive deﬁnite it follows that det(gi j) is never zero and hence i p can be calculated in terms of p j. So let

i i j (3.8.3) p = g p j. Xj

Then we have il g gl j = δi j. X1 8. EXPRESSIONS IN LOCAL COORDINATES 111

8.4 Considering the set x1,..., xd, p ,..., pd of coordinates on T(U), and { 1 } by (6.6).

(3.8.5) α U = g pidx j | i j Xi, j and hence

∂gi j i k j i j (3.8.6) dα U = p dx dx + g , dp dx . | ∂xk ∧ i j ∧ Xi, j,k Xi, j

Chapter 4 Geodesics

1 The first variation 114 In this article, we study the following question; given two points m, n in an r.m. (M, g), look for a curve f with end points m, n which has min- imal length among all curves in M with end points m, n. Then the first variation has to be zero for all one-parameter families of curves with fixed ends m, n. This will lead us to a sufficient condition involving the acceleration vector f ′′ of f and from there to a spray on (M, g) canonically, so to geodesics in (M, g). The necessity of this condition could be proved directly but we will not do it, for it will be a consequence of Chapter VII. We shall, in the course of these considerations, obtain a useful formula for a geodesic in (M, g), the first variation formula. Let us consider an r.m. (M.g) and two points m, n M; let f be ∈ a one-parameter family of curves, for which we use the notation introduced in (8). We suppose that f : (I =] ǫ, 1 + ǫ[ ) J M with 0 J − × → ∈ and ǫ > 0; moreover we suppose that f (t) 1 t. || 0′ || − ∀ For the lengths of the fs’s we define the first variation l′(0) by: d (4.1.1) l′(0) = (lg( f [0, 1])) = . ds s| s 0

To compute l′(0), if α is the canonical form on T(M) (deﬁnition in (6.5), we set

(4.1.2) β = P∗(α).

113 114 Geodesics

115 We have:

(4.1.3) E P = α(PT P) = β(P); ◦ ◦ for, α(PT P) = g(p PT P, PT PT P). But p P P = P and ◦ ′ ◦ ◦ ◦ ◦ ′ ◦ ◦ pT P P = P. ◦ ◦ ∂ Since f (t) = 1 we have, by the very deﬁnition of Q = , the || 0′ || ∂s following commutative diagram:

T(T(M)) oo7 T oo P ooo oo T oo p′ p ooo ooo oo f T T(I J) / T(M) 7 O ×O ooo 7 Q oo ooo oooooo oo oo p P Q oooooo oooooo P oooooo oooo I J / M × f Further,

1 1 1 lg( fs [0, 1]) = f ′(t) dt = (E P) 2 (t, s) dt = | Z || s || Z ◦ 0 0 1 1 = β(P) 2 (t, s) dt, Z 0 and 1 1 l′(0) = Q( (β(P)) 2 (t, s) dt) Z 0 1 1/2 = (1/2) Q(β(P))(t, 0) (β(P)(t, 0))− dt = Z · 0 1. THE FIRST VARIATION 115

1 = (1/2) Q(β(P))(t, 0) dt . Z 0

But 116

dβ(P, Q) = Pβ(Q)) Q(β(P)) β([P, Q]) and [P, Q] = 0 − − so we get

1 1

l′(0) = (1/2) P(β(Q))(t, 0) dt (1/2) dβ(P, Q)(t, 0) dt . Z − Z 0 0 ∂ From the deﬁnition P = : ∂t

1 P(β(Q))(t, 0) dt = [β(Q)]1,0 Z (0,0) 0 but

T T T β(Q) = α(P Q) = (see (6.6)) = g(p′ P Q, p P Q) = ◦ ◦ ◦ ◦ ◦ = g(P, Q) to the eﬀect:

1 (1,0) (4.1.4) l′(0) = (1/2)[g(P, Q)] (1/2) dβ(P, Q)(t, 0) dt . (0,0) − Z 0 We can also write it as:

1 (1,0) T (4.1.5) l′(0) = (1/2)[g(P, Q)] (1/2) dα( f ′′(t), (P Q)(t, 0)) dt (0,0) − Z 0 ◦ 0 for the acceleration f0′′(t) of f0 (see (5.8)). Clearly: 116 Geodesics

1.6 If f (0, s) = m and f (l, s) = n s J, then ∀ ∈ (1,0) g(P, Q) = 0. (0,0) = But this does not imply that f0′′(t) should be such that i( f0′′(t))(dα) 0 for PT Q(t, 0) is not an arbitrary element in T(T(M)). For, adding ◦ (5.9), (6.10) and (1.8) we get:

1.7 117 if z vertical then (dα)( f (t), z) = (1/2)z(E). 0′′ − But (PT Q)E = Q(β(P)) so we have ◦ T (i( f ′′(t))(dα) + (1/2)dE)(P Q)(t, 0) 0 ◦ = (P(β(Q)) Q(β(P)) + (1/2)Q(β(P)))(t, 0) = − = (P(β(Q)) (1/2)Q(β(P))(t, 0) − and ﬁnally: (4.1.8) 1 (1,0) T l′(0) = [g(P, Q)] ((i( f ′′(t)(dα) + (1/2)dE))(P Q)(t, 0) dt (0,0) − Z 0 ◦ 0 from which we get:

+ = ﬃ 1.9 i( f0′′(t)(dα) (1/2)dE 0 is a su cient condition for f0 to be of critical length among all curves with ends m, n. And in particular

1.10 + = if the curve f0 is such that i( f0′′)(dα) (1/2)dE 0 then we have the = (1,0) ﬁrst variation formula: l′(0) [g(P, Q)](0,0). We are so lead to the following: 2. THE CANONICAL SPRAY ON AN R.M. (M, G) 117

2 The canonical spray on an r.m. (M, g)

By Proposition (6.8) it follows that there exists one and only one G ∈ C (T(M)) such that

1 (4.2.1) i(G)(dα) = dE. −2

Regarding the G we prove the following proposition.

2.2

Proposition. G is a spray on M.

Proof. First let us prove that pT G = id . For any x T(M) and ◦ T(M) ∈ z V we have, on the one hand, by (1.8) ∈ x 1 (dα)(G(x), z) = z(E) = g(x,ξ(z)) −2 − and, on the other by (6.10) 118

(dα)(G(x), z) = g(T (G(x)),ξ(z)) − and hence g(pT (G(x)) x,ξ(z)) = 0. −

Since ξ is an isomorphism between Vx and Tx(M) and g is non - degenerate we have pT (G(x)) = x, x T(M). ∀ ∈ Now let us prove the homogeneity.

First let us note that

α hT = θ α, ◦ θ · 118 Geodesics

In fact, z T(T(M)) we have: ∀ ∈ hT T(T(M)) θ / T(T(M))

T T p′ p p′ p

h T(M) θ / T(M)

p p

M / M idM

T T T T α(h (z)) = g(p (h (z)), p′ h (z)) = θ θ ◦ θ T T = g((p hθ) (z), hθ p′(z)) = g(p (z),θ p′(z)) = ◦ ◦ · T = θg(p (z), p′(z)) = θ α(z). · 119 Hence, we have, for every z, z′ in T(T(M)),

T T dα(h (z), h (z′)) = θ dα(z, z′). θ θ · Now for any Z C (T(M)) we have ∈ dα(θ(hT G), hT Z) = θ dα(hT G, hT Z)) θ ◦ θ ◦ · θ ◦ θ ◦ 1 = θ2 dα(G, Z) = θ2 Z(E), and also · −2 ·

T 1 T dα(G hθ, h Z) = h Z(E) ◦ θ ◦ −2 θ · 1 1 2 1 2 = Z(E hθ) = Z(θ E) = θ Z(E). −2 ◦ −2 · −2 · Again, as above, since dα is non-degenerate we have

T G hθ = θ(h G). ◦ θ ◦ Now we give the following deﬁnition. 2. THE CANONICAL SPRAY ON AN R.M. (M, G) 119

2.3 Deﬁnition. The spray G of the above proposition is called the canonical spray of the r.m. (M, g). Whenever we talk of a spray on an r.m. we always mean the canonical spray. In view of this we can talk of geodesics on an r.m. By the deﬁnition of isometry E is invariant under isometry and by (6.7), dα is also invariant under isometry and we have the following:

2.4

Proposition. If 120 λ : (M, g) (N, h) → is an isometry, and MG (respectively NG) is the canonical spray of (M, g) (respectively on (N, h)) then N T T M 1 T G = (λ ) G (λ− ) . ◦ ◦ Proof. Let any Z C (T(N)); since λ is an isometry and hence in ′ ∈ particular, a diﬀeomorphism, there exists a Z C (T(M)) such that ∈ Z = (λT )T Z (λ 1)T . Set moreover: ′ ◦ ◦ − T T M 1 T G′ = (λ ) G (λ− ) . Then, by ((6.7)), (0.2.6), (2.4), ◦ ◦ we have: N 1 T (i(G′) d( α))(Z′) = d(((α− ) )∗(Mα))(G′, Z′) = · 1 T M = ((λ− ) )∗(d( α))(G′, Z′) = M 1 T T T 1 T T T = d( α)(((λ− ) ) G′ λ , ((λ− ) ) Z′ λ ) ◦ ◦ ◦ ◦ M M M = d( α)(G′, Z′) = (i( G) d( α))(Z). ·

Now, by the deﬁnition of MG and because λ is an isometry implies M M E = N E λT , we have: ◦ M M 1 M 1 N T T 1 T (i( G) d( α))(Z) = d( E)(Z) = d( E)((λ ) Z (λ− ) ) = · −2 −2 ◦ ◦ 1 N = d( E)(Z′). −2 120 Geodesics

2.5 Corollary . If λ : (M, g) (N, h) is an isometry and f a geodesic in → (M, g), then λ f is a geodesic in (N, h). ◦ 121 Proof. Let (I, f ) be an integral curve of MG in T(M). Then (I, λT f ) ◦ is an integral curve of NG = (λT )T MG (λ 1)T in T(N). ◦ ◦ − For:

T T T T T T T M (λ f )′ = (λ f ) P = (λ ) f ′ = (λ ) G f = ◦ ◦ ◦ ◦ ◦ ◦ N T N T = ( G λ ) f = G (λ f )′. ◦ ◦ ◦ ◦ 2.6 Corollary . If λ : (M, g) (N, h) is an isometry, then λT (MΩ) = NΩ → and the following diagram is commutative:

λT MΩ / NΩ

Mexp Nexp

λ M / N Proof. Apply (4.5).

We slightly generalize the above results in the:

2.7 Proposition. Let (M, g) and (N, h) be two r.m.’s of the same dimension and a map λ D(M, N) be such that λ h = g. Then λ is a local isometry; ∈ ∗ moreover, if (I, f ) is a geodesic in (N, h) and f : I M such that → λ f = f , then f is a geodesic in (M, g). ◦ b Proof.b We noteb ﬁrst the injectivity of λT m M, which comes from m∀ ∈ λ∗h = g and g positive deﬁnite; the equality of dimensions then implies λT is an isomorphism m M; hence the inverse function theorem and m ∀ ∈ λ∗h = g imply the local isometry. 3. FIRST CONSEQUENCES OF THE DEFINITION 121

Remark. 2.7 can be applied, in particular, to a riemannian covering. 122

2.8

Example. Let us determine the canonical spray of (Rd, ǫ).

First of all from the deﬁnition of the scalar product of g we can identify α and µ, and dα and dµ. Now by (0.4.26) we have, for every z, z′ of T(M),

T T T T dα(z, z′) = (ζ (z)) (p (z′) (ζ (z)) (p (z)), · − · and hence x T(M) we have ∀ ∈ dα(G(x), z) = (ζT (G(x))) (pT (z)) (ζT (z)) (pT (G(x))) = · − · = (ζT (G(x))) (pT (z)) ζT (z) ζ(x). · − · On the other hand E(x) = (ζ(x)) (ζ(x)), · and hence for z T (T(M)), by (0.4.5) ∈ x z(E) = 2(ζ(x)) (ζT (z)). · Hence (ζT G(x)) pT (z) = 0, z T (T(M)). ◦ · ∀ ∈ x T But since g is non-degenerate and p is onto Tx(M) we have

(4.2.9) ζT (G(x)) = 0.

Consequently, by ((2.1.22)), the geodesics are segments of straight lines. A more geometrical proof would be to use the symmetry around a line in Rd (which is an isometry) and apply the local uniqueness of geodesics and the lemma (2.6). We shall do this in a more general situation in 4. § 122 Geodesics

3 First consequences of the deﬁnition 123 3.1

Proposition. G(E) = 0.

Proof. In fact, by deﬁnition, we have

G(E) = (dE)(G) = 2dα(G, G) = 0. −

3.2

Corollary. E is constant along a geodesic f in particular f is con- || ′|| stant.

Proof. In fact, we have, with usual notation,

T P(E f ′) = ( f ′ P)(E) = f ′′(E) ◦ ◦ = (G f ′)(E) = G(E) f ′ = 0. ◦ ◦ This corollary shows that any geodesic is parametrised by either the arc length or by a constant times the arc length. Hence for a geodesic f we have

(4.3.3) lg( f [t , t ]) = (t t ) f ′(t ) . | 1 2 2 − 1 || 1 ||

Hence follows the

3.4

Corollary. For every x in Ω, exp(x) is the end point of the geodesic f with f (0) = x and of length equal to x . ′ || || 3. FIRST CONSEQUENCES OF THE DEFINITION 123

3.5 Now, for every positive number r and every point m of (M, g) we set

B(m, r) = x T (M) x < r { ∈ m || || } and call it the ball of radius r in Tm(M).

3.6 By (4.2) and (4.7) it follows that, to every point m of (M, g) there exists an r(m) > 0 such that B(m, r) Ω ⊂ and 124 exp B(m, r) exp(B(m, r)) m | → is a diﬀeomorphism. We set:

B(m, r) = exp(B(m, r)) for such an image set and describe the situation when we have such ﬀ a di eomorphism as expm is r-O.K. Hence to every point m of (M, g) there is an r(m) > 0 such that expm is r(m)-O.K. In fact, something more is true. For, let us take, in the proof of (6.8), instead of an arbitrary euclidean structure on Tm(M) the one that is induced by g. Then the following result is obtained:

3.7 For every m of (M, g) there exists an r(m) > 0 such that r [0, r] one ∀ ′ ∈ has: B(m, r′) is convex.

3.8 Definition. The flow of the vector field G on T(M) is called the geodesic flow of the r.m. (M, g). 124 Geodesics

3.9 Remark . Combining (5.7) and (3.1) we conclude that ﬂow leaves the unit bundle U(M) of M invariant. Furthermore we have the following result.

3.10 Proposition . The geodesic ﬂow leaves dα on T(M) and α on U(M) invariant. Proof. By (0.6.9) we have only to show that on T(M)

θ(G)(dα) = 0,

125 and on U(M): θ(G) α = 0. · Clearly, by 0.6.12, we have 1 θ(G)(dα) = (i(G)d + d i(G))dα = d(i(G)dα) = d( dE) = 0. ◦ −2 Also 1 θ(G)α = d(i(G)α) + i(G)dα = d(α(G)) dE. − 2 By (6.6)

T α(G) = g(p G, p′ G) = g(id , id ) = E. ◦ ◦ T(M) T(M) Hence 1 θ(G) α = dE. · 2 But E 1 on U(M). Hence the result. ≡ 4 Geodesics in a symmetric pair

A symmetric pair is an example of a manifold where one can write down all the geodesics explicitly. With the notations of (3) we prove the following proposition. 4. GEODESICS IN A SYMMETRIC PAIR 125

4.1 Proposition. The geodesics of (M, γ) are the curves

f : t (τ(g) p)(exp tX) → ◦ where g runs through G and X through M.

Proof. Since, for every element g of G, τ(g) is an isometry and G acts transitively on the manifold we need only consider the situation at m0, i.e. for the curves t p(exp(t X)). → · Let us ﬁx X M; and choose a positive r for which exp m is r-O.K. ∈ 0 and moreover such that

(i) B(m0, r) = V is convex, and 126

(ii) the only point in V ﬁxed under σ is m0.

(This is possible: see the remark followingb the proof of (3.12)). Let us note that by our construction

σ(V) = V and denote the map b t p(exp(t X)) → · by f . Let t0 be such that

f [ t , t ] V. | − 0 0 ⊂

Then, since V is convex, for any t1 such that 0 < t1 < t0 there is a geodesic g from f ( t ) to f (t ). By reparametrising g if necessary, we − 1 1 can assume that

g( t ) = f ( t ) and g(t ) = f (t ). − 1 − 1 1 1 Further from (6.5) it follows that g is unique. Clearly by (2.5), (3.12), (3.20), σ g is a geodesic in V from f (t ) to f ( t ) and hence σ ◦ 1 − 1 ◦ b b 126 Geodesics

g k 1 (where k 1 = idR) is a geodesic from f ( t1) to f (t1). By the ◦ − − − − uniqueness of g we have

σ g k 1 = g. ◦ ◦ − In particular, b σ(g(0)) = g(0),

and since m0 is the only pointb in V ﬁxed by σ we have

g(0) = m0 = f (0)b.

127 That is to say: the unique geodesic from f ( t1) to f (t1) has to go through − t1 f (0). Because the τ’s are isometries and τ(exp( 2 X)) sends f ( t1) into t1 t1 · − f ( 2 ) and f (0) into f ( 2 ), we see, by the above argument (applied for t1 − t1 2 instead of t1) that g should also go through f ( 2 ). By the same token p p it follows that f and g coincide at all points t1 q where q is a fraction in [ 1, 1] and q is a power of 2. Since such p are dense in [ 1, 1] we see, − q − using the continuity of f and g, that f and g coincide everywhere.

4.2 By (4.3) it follows that

p exp :M M ◦ m0 → is onto. Hence given any point m of M there is an X in M such that

m = p exp X ◦ m0 for (4.1) implies in particular: Ω T (M). ⊃ m0 4.3 X Now clearly the symmetry around p(expm0 ( 2 )):

σp(exp ( X )) m0 2

takes m0 onto m and it follows that the assumption that m, m′ be suﬃ- ciently close in (3.19) can be dropped. 5. GEODESICS IN S.C.-MANIFOLDS 127

5 Geodesics in S.C.-manifolds

5.1 Deﬁnition. A curve f D(R, M) in an r.m. (M, g) is said to be a closed ∈ geodesic if

(i) it is a geodesic,

(ii) it is periodic i.e. t > 0 f (t + t ) = f (t) t R. ∃ 0 | 0 ∀ ∈

5.2

Deﬁnition . A closed geodesic is said to be simply closed if moreover 128 there exists a period t1 of f such that f is injective on [0, t1[.

In this article, we consider the S.C.-manifolds (4.19) but the state- ments of (5.4) and (5.7) relative to P2(Γ) will not be proved; one can ﬁnd some of them in [14] p.355.358 and the remaining ones in [38], Nos. 107 and 151.

5.4 Proposition. In an S.C. manifold all geodesics are simply closed and of the same length. This length is 2π for (Sd, can) and π for the others. Moreover, the geodesics for (Pd(R), can) are projective lines and for (Sd, can) they are great circles.

Proof. Because of the existence of a transitive family of isometries it is enough to examine the situation at one point, say m0. By (4.1) (4.18) and (3.4.13) the geodesics through m are t cos t.e + + sin t.z with 0 → d 1 < z, z >= 1 and z Rd for (Sd, can), and t p(cos t e + + sin t z) ∈ → · n 1 · with z Kn, < z, z >= 1 for the Pn(K) s and in both cases they are ∈ ′ parametrized by arc length. Hence the result for (Sd, can) is proved as also for Pd(R). For a Pn(K) if

p(cos t.en+1 + sin t.z) = p(cos t′.en+1 + sin t′.z) 128 Geodesics

then λ K with ∃ ∈

cos t.en+1 + sin t.z. = λ(cos t′.en+1 + sin t′.z)

which implies that cos t = λ cos t , sin t = λ sin t and hence λ R · ′ · ′ ∈ 129 and λ2 = 1; so λ = 1 and all geodesics in Pn(K) have π as the smallest ± period.

5.5 This result explains the terminology S.C.-manifolds (S.C. stands for “symmetric circled”). 5. GEODESICS IN S.C.-MANIFOLDS 129

5.6

Remark. In (Sd, can) two geodesics through m Sd meet again at the ∈ first time at the antipodal point m of Sd. For the other S.C.-manifolds − the situation will be described in the proposition below; in its statement 2 in the case of P (Γ) one should replace “z, z′ are or not K-dependent” by the following: “z, z′ belong or not to the same fibre of the canonical 130 fibration S15 S8”. →

5.7

Proposition . Let (M, g) be an S.C.-manifold other than an (Sd, can); T T then two geodesics through m0 with tangent vectors p (r(z)), p (r(z′)) at m (for z, z Kn, < z, z >=< z , z >= 1 and z , z) 0 ′ ∈ ′ ′ ′ ±

i) never meet except at m0 if z, z′ are K-independent π ii) meet exactly at m and at their mid-point at distance if z, z are 0 2 ′ K-dependent.

Proof. By the argument in the proof of (5.4) our two geodesics meet at distances t, t from m if and only if λ K such that ′ 0 ∃ ∈

cos t.e + + sin t.z = λ(cos t′ e + + sin t′ z′) n 1 · n 1 · hence

cos t = λ cos t′ and sin t.z = λ sin t′ z′. · · · We can suppose that t, t ]0, π[. Since λ = 1 either cos t = cos t = 0 or ′ ∈ π | | ′ λ R. In the ﬁrst case t = t = and z = λ z so z, z are K-dependent. ∈ ′ 2 · ′ ′ In the second λ R, < z, z >=< z , z >= 1 and sin t z = λ sin t z ∈ ′ ′ · · ′ · would imply z = z , a contradiction. ± ′ Remark. For Pd(R) in fact z , z , implies z, z are never R-dependent ± ′ ′ so two distinct geodesics in Pd(R) meet at most at one point as they should since they are projective lines. 130 Geodesics

5.12 Proposition. For every point m of an S.C.-manifold (M, g) we have Sd 131 1) expm is π-O.K. if (M, g) is an ( , can) π 2) exp is -O.K. in the other cases. m 2 = Sd Proof. By (5.6) we see that for M , expm is one-one on. B(m0,π) 0 π and if M , Sd by (5.7) that exp is one-one on B(m , ). Thus it m0 0 2 ﬃ su ces to show that expm is of maximal rank on B(m0,π) in the case π 0 of Sd and on B(m , ) in the other cases. We show, by example for the 0 2 Pn(K)’s, that the expression we had for geodesics proves in particular that the map h = exp pT r : Kn Pn(K) m0 ◦ ◦ → is nothing but

(4.5.13) h : tz p(cos t.e + + sin t.z) with < z, z >= 1, t R. → n 1 ∈ T 1 T We know that htz(ζtz− z) , 0 by (6.32), and hence we study htz for z′ with n < z′, z′ >= 1 and orthogonal to z in the euclidean structure of K . The curve α t(cos α z + sin α z′) (whose speed for s = 0 is 1 → · · ζtz− (tz′)) has for image under h the curve

α h(t(cos α z + sin α z′) = p(cos t e + + sin t(cos α z + sin α, z′)) → · · · n 1 · π whose speed for z = 0 is pT (ζ 1(sin t z )). Since t ]0, [ and < tz− · ′ ∈ 2 z , z >= 1 one has sin t z . , 0. From (pT ) 1(0) Kn = 0 one gets ′ ′ · ′ − ∩ { } the proof. Remark. (5.12) would also follows from (7.1).

6 Results of Samelson and Bott

In the previous article we have seen that in an S.C.-manifold all geodesics are simply closed and of the same length. Now let us examine the converse. Given a connected riemannian manifold (M, g) such that for each point m of (M, g) 6. RESULTS OF SAMELSON AND BOTT 131

(i) each geodesic through m is simply closed; 132 (ii) the length of each geodesic through m is a number 1 independent of the geodesic; and (iii) 1 is independent of m; does it follow that (M, g) is isometric with an S.C.-manifold? (see article 8). More generally we give the:

6.1 Deﬁnition. Given an r.m. (M, g) and some m M we say that (M, g) is a ∈ Cm-manifold if it is connected and if all geodesics through m are simply closed and of same length.

Though there are no isometry relations between an arbitrary Cm- manifold and an S.C.-manifold we shall see that the cohomology ring of a Cm-manifold resembles that of an S.C.-manifold. Let us recall some deﬁnitions and results from algebraic topology. Given a compact topological manifold M we denote its graded cohomology ring over a ring A by H∗(M, A).

6.2

Deﬁnition. The manifold M is said to be a TR(A)-manifold if H∗(M, A) is a truncated polynomial ring.

This means H∗(M, A) is isomorphic to A[X]/A where A is an ideal generated by a positive power of X. The image under that isomorphism of X is a homogeneous element of H∗(M, A); call its degree α. If d = dim M, then deﬁne λ by d = α λ. The following results are standard. · Sd is a TR(Z) manifold with α = d and λ = 1 Pn(C) is a TR(Z) manifold with α = 2 and λ = n (4.6.3) Pn(H) is a TR(Z) manifold with α = 4 and λ = n P2(Γ) is a TR(Z) manifold with α = 8 and λ = 2 d P (R) is a TR(Z2) manifold with α = 1 and λ = d. 132 Geodesics

133 We shall prove a result due to Samelson to the eﬀect that a Cm-manifold is a TR(Z2)-manifold.

6.4

Remark . In the deﬁnition of a Cm-manifold we have assumed that all the geodesics through m are of the same length. This property is not a consequence of the former namely that each geodesic is simply closed. To see this one can take either a suitable lens space or more precisely the following example due to I.M. Singer. A compact Lie group G is always a symmetric space, when associated to the symmetric pair (G G, G) × for the involutive automorphism σ(g, h) = (h, g). For r.s. as in (3) the geodesics through e are the one-parameter subgroups (as follows from (4.1)). Take now G = S 0(3) and such an r.s. on it; if H is the subgroup generated by an element of order two then the quotient G/H is endowed with an r.s. by (3), which makes G G/H a riemannian covering. Then → the geodesics through p(e) are all of same length (as in S 0(3)) with one exception (the one which corresponds to that in S 0(3) which contains H) which is of length half that of the others.

134 However it is not known whether the assumption of equal length is, or is not, superﬂuous in the case of simply connected manifolds.

6.5

Proposition . Let (M, g) be a Cm-manifold for which the (common) length of the geodesics through m equals 1. Then, 1) Ω T (M) and M = exp(B(m, 1/2)) in (particular M is compact) ⊃ m 2) if dim M 2, then:either M is simply connected, or the uni- ≥ versal riemannian covering (M, g) (M, g) is two-sheeted and → (M, g) is aC -manifold m p 1(m), with common length for m ∀ ∈e − the geodesics through m equal toe 21. e e e e Proof. Because each geodesice is of length 1 we have 1 B(m, ) Ω. 2 ⊂ 6. RESULTS OF SAMELSON AND BOTT 133

Since every geodesic, and hence expm, is periodic with period 1 we have

T (M) Ω. m ⊂ Since M is connected the Hopf-Rinow theorem (4.1) implies that M = exp(Tm(M)). Now the periodicity of expm implies that = = M expm(Tm(M)) expm(B(m, 1/2)). ff Since B(m, 1/2) is a closed ball and hence compact and expm is de erentiable and hence continuous, it follows that M is compact (for M is Hausdorff). This proves the first part.

The second part of 2) follows from the remark after (2.7). To prove 135 the ﬁrst part of 2) we remark ﬁrst that all the closed geodesics of period 1 in a Cm-manifold are homotopic as loops based at m (hereafter we use notions and notations of (6)) if dim M 2; in fact if γ , γ are two such ≥ 1 2 geodesics and λ : [0, 1] U (M) a path connecting γ (0) to γ (0) in → m 1′ 2′ the unit tangent sphere Um(M) at m then the required homotopy is

(t,α) exp(t λ(α)) → · for exp(0, λ(α)) = exp(1 λ(α)) = m α [0, 1]. · ∀ ∈

In particular the homotopy class in π1(M, m) of a closed geodesic γ of 1 period 1 through m is independent of γ, say a. moreover γ− is again 1 1 2 such a geodesic and belongs to a− hence a = a− or a = 1. Now we claim: any b π (M, m)(b , 0) is a power of a (which yields the ∈ 1 proposition). In fact let b π (M, m), m p 1(m) and m , m. By ∈ 1 ∈ − ′ (2.7) the covering (M, g) is complete hence by (4.10) there exists γ ∈ S (m, m ). The projection γ = p γ is a geodesice from m toem ine (M, g) ′ ◦ so by (5.2) has to be a multiple of a closed geodesic of period 1 throughe m;e hencee b is a power of a. e

6.6

Theorem H. Samelson.. ACm-manifold is a TR(Z2) manifold. 134 Geodesics

6.7

Lemma. There exists a positive number r such that

1 1 B(m, ) exp− (B(m, r)) = B(m, r). 2 ∩ m

136 Proof. Let r′ be a positive number such that expm is r′-O.K. Then let us set

1 W = exp B m, B(m, r′) . m  2! −      1 Note that B(m, ) B(m, r ) is a closed subset of the compact ball 2 − ′ 1 B(m, ) and hence compact. Therefore W is compact. Since the geo- 2 desics through m are simply closed it follows that m < W. Hence there is a positive number r such that

i) 0 < r < r′ and

ii) B(m, r) W = . ∩ ∅

1 Now let n B(m, r) and let n = exp (x) where x < . Since B(m, r) ∈ m || || 2 ∩ 1 W = we see that x < B(m, ) B(m, r ) and hence x B(m, r ). But ∅ 2 − ′ ∈ ′ since expm is r′-O.K. and r < r′ we have expm is one-one on B(m, r) and since n B(m, r) we have x B(m, r). ∈ ∈ 6. RESULTS OF SAMELSON AND BOTT 135

Proof of the proposition. By (5.4) we can multiply the r.s. on (Pd(R), can) by a constant so as to make the common length of its geodesics equal to 1, the length of each closed geodesic of (M, g) through m. Let 137 us ﬁx a point n on Pd(R) and a euclidean isomorphism

u : T (Pd(R)) T (M). n → m Now we deﬁne a map f D(Pd(R), M) by requiring the commutativity ∈ of the diagram: 1 u B(n, ) / Tm(M) 2

exp exp

Pd(R) / M f 1 By (5.12) exp is -O.K. and hence the map n 2

1 1 exp u (exp B(n, ))− m ◦ ◦ n | 2 136 Geodesics

1 1 is well deﬁned on B(n, ). On B(n, ) we deﬁne f to be this map. Now 2 2 1 1 1 let n exp(B(n, ) B(n, )). Then if x B(n, ) and exp x = n , ′ ∈ 2 − 2 ∈ − 2 ′ since the length of every geodesic is equal to 1 we have exp( x) = n . − ′ By (5.7) no two geodesics through n meet outside n and hence it follows x are the only inverse images of n in ± ′ 1 1 B(n, ) B(n, ). 2 − 2 138 Since 1 u(x) = u( x) = || || || − || 2 and since the common length of geodesics through m is 1, we conclude that exp(u(x)) = exp(u( x)). − 1 Now, we extend f to B(n, ) by setting 2 f (x) = exp(u(x)).

By (4.5.13) f is a diﬀerentiable map. (We need only the fact that f is continuous). Now we claim that the topological degree mod 2 of f is one. First let us recall the process of getting the topological degree of a map. First we can deﬁne the fundamental class mod 2 of the compact manifold M as follows. We pick up a neighbourhood U of an arbitrary point m of M such that U is homeomorphic to an open ball. Then we consider the exact sequence

i 0 Hd(U, Z ) Hd(M, Z ) Hd(X U, Z ) → c 2 −→ 2 → − 2 →··· d Z ([12]: p. 190. th. 4.10.1 and the following lines), in which Hc (U, 2) consists of exactly two elements 0 and γU . Then ϕM, the fundamental class mod 2 of M, is i(γU ). Then, , the degree mod 2 of a map

f : N M → 6. RESULTS OF SAMELSON AND BOTT 137 from a compact manifold N into M is by deﬁnition such that

f ∗(ϕ ) = δ ϕ . M · N In our case we define ϕM with the help of the ball U = B(m, r) given by 139 d the lemma and for N = P (R) and ϕN with U′ = B(n, r). Let us note that the lemma together with the definition of f implies that 1 f − (U) = U′ and that f U is a diffeomorphism. Then we have ([12]: 4.16., p. 199) | ′ the following commutative diagram

i 0 / Hd(U) / Hd(M) / c c ···

fc∗ f ∗

i 0 / Hd(U ) ′ / Hd(Pd(R)) / c ′ ··· Since f : U U is a homeomorphism we have ′ → = f ∗(γU ) γU′ . = = Since ϕM i(γU ) and ϕN i′(γU′ ) the commutativity of the diagram gives

(i′ f ∗)(γ ) = i′( f ∗(γ )) = i′(γ ) = ϕ = ( f ∗ i)(γ ) = ◦ c U c U U′ N ◦ U = f ∗(ϕ ) = δ ϕ . M · M Hence δ = 1. Now we assert that

d f ∗ : H∗(M, Z ) H(P (R), Z ) 2 → 2 is injective. For Poincar´eduality asserts that given any non-zero e 140 ∈ H (M, Z ), e H (M, Z ) such that ∗ 2 ∃ ′ ∈ ∗ 2 e e′ = ϕ . ∪ M 138 Geodesics

Hence we have

f ∗(e) f ∗(e′) = f ∗(e e′) = f ∗(ϕ ) = ϕ ∪ ∪ M N and hence f ∗(e) , 0 e i.e. f ∗ is injective. ∀ Hence H∗(M, Z2) is isomorphic to a homogeneous sub-ring A of the truncated polynomial ring

d d+1 H∗(P (R), Z2) = Z2[X]/(X ) (by (4.6.3)).

d θ Let X be the image of X in H∗(P (R), Z2) and X be the least positive θ power of X that is in A. Then e′ H∗(M, Z2) such that X = f ∗(e′) ∃ ∈ Z = and by Poincar´eduality e1′ H∗(M, 2) such that e′ e1′ ϕM. Hence d θ ∃ ∈ ∪ kθ = − θ f (e1′ ) X A. By the same token, for k with d > k we have X A d kθ ∈ ∈ and X − A. So the choice of θ implies k d = k θ and H(M, Z2) is ∈ ∃ | · k+1 isomorphic to the truncated polynomial ring Z2[X]/(X ).

6.9 Remark. In fact Samelson’s result is sharper. But by using Morse theory Bott proved the following theorem:

6.10

Theorem [6]: p.375. A simply connected Cm-manifold is a TR(Z)-manifold. The universal covering of an r.m. which is not simply connected is a homotopy sphere.

141 We will not go into the proof of this theorem but be content with the remark that recent theorems in algebraic topology and (4.6.3) imply the following:

1) For a simply connected TR(Z)-manifold the cohomology ring H∗ (M, Z) is isomorphic to that of an S.C. manifold. 7. EXPRESSIONS FOR G IN LOCAL COORDINATES 139

2) a simply connected Cm-manifold of dimension d odd and d > 5 is homeomorphic to Sd.

3) a Cm-manifold, of dimension greater than or equal to 5 and which is not simply connected is homeomorphic to Pd(R).

7 Expressions for G in local coordinates

Given (M, g) let us express G, locally, in terms of g. Let us follow the conventions of (4). On T(U) let us set

(4.7.1) G • = (g♯)T G g♭ and compute G •. ◦ ◦ Now for 1 d ω = (x ,..., x ; p ,..., p ) T ∗(U) 1 d ∈ let

• 1 d 1 d (4.7.2) G (ω) = (x ,..., x ; p1,..., pd; v ,..., v ; w1,..., wd).

1 d Let Y1,..., Yd; P ,..., P be the basis of C (T ∗(U)) dual to the basis 1 { d } dx ,..., dx ; dp1,..., dpd of C ∗(T ∗(U)). Now set 1 (4.7.3) H = E g♭. 2 ◦ Then by the deﬁnition of G we have

i(G •)(dµ) = dH. − Hence 142

• ∂H (4.7.4) dµ(G , Y ) = i − ∂xi ∂H (4.7.5) dµ(G, Pi) = −∂pi 140 Geodesics

and hence by (0.4.26) we have ∂H (4.7.6) vi = ∂pi ∂H (4.7.7) w = i ∂xi Now we compute G on U. By the deﬁnition of g♯ we have

♯ 1 d 1 d 1 d (4.7.8) g (x ,..., x ; p ,..., p ) = (x ,..., x ; p1,..., pd) where j pi = gi j p (see (8)). Xj Now let

(4.7.9) (g♯)T (x1,..., xd; p1,..., pd; v1,..., vd; w1,..., wd) = 1 d 1 d = (x ,..., x ; p1,..., pd; v ,..., v ; w1,..., wd).

Then a straightforward calculation of the Jacobian Dg♯ gives that

∂gi j (4.7.10) w = vk p j + g w j. 1 ∂xk · · i j Xi,k Xj Now let

G(x1,..., xd; p1,..., pd) = (x1,..., xd; p1,..., pd; p1,..., pd; G1,..., Gd).

With the notation of ((1.3.2)) we have

(4.7.11) Gi = Γi p j pk. − jk Xj,k Γi 143 Hence we have to compute jk. We have

1 d 1 d 1 d G(x ,..., x ; p1,..., pd) = (x ,..., x ; p1,..., pd; p ,..., p ; G1,..., Gd). From (3.8.3) it follows that 1 (4.7.12) H = g jk p p 2 j k Xj,k 7. EXPRESSIONS FOR G IN LOCAL COORDINATES 141 and hence ∂H 1 ∂g jk (4.7.13) = p p . ∂xi 2 ∂xi j k Xj,k By (3.8.3) we have il g gl j = δi j Xl and hence for every i

∂g jk ∂g jk (4.7.14) p p = p j pk. ∂xi j k − ∂xi Xj,k Xj,k Hence by (4.7.7), (4.7.10) and (4.7.14) we have

∂gi j 1 ∂g jk (4.7.15) pk p j + g G j = p j pk, ∂xk i j 2 ∂xi Xj,k Xj Xj,k or

i 1 ∂gi j ∂gik j k (4.7.16) gi jG + 2 p p = 0. 2 ∂xk − ∂xi ! Xj Xj,k

Now if we set 144

1 ∂gi j ∂gik ∂g jk (4.7.17) Γ jik = + 2 ∂xk ∂x j − ∂xi ! then we have

j j k (4.7.18) gi jG + Γ jik p p = 0. Xj Xj,k

Hence by the deﬁnition of gi j we have Γi = ilΓ (4.7.19) jk g jlk Xl where the Γ jlk are given by (4.7.17). Remark. (4.7.6) and (4.7.7) are nothing but Hamilton’s equations. 142 Geodesics

7.1 Lagrange’s equation

In local coordinates (4.2.1) also yields Lagrange’s equations. On T(U) we introduce the local vector ﬁelds

∂ ∂ Xi = , Pi = and have : ( ∂xi ) ( ∂pi ) T T p ([Xi, G]) = 0,α ([Xi, G]) = 0, p (Xi) = ξ(pi)

So by (1.8) 1 ∂E α(X ) = . i 2 ∂pi And

1 ∂E dα(G, Xi) = G(α(Xi)) Xi(α(G)) α([G, Xi]) = G Xi(E) = − − 2 ∂pi ! − 1 = dE(X ) −2 i

145 so: ∂E ∂E G = . ∂pi ! ∂xi d Working along a geodesic parametrized by t implies G = hence the dt Lagrange equations: d ∂E ∂E = . dt ∂pi ! ∂xi

8 Zoll’s surface

In this article we answer negatively the question which was posed at the beginning of article 6 by giving a counter example due to Zoll of a r.s. on S2 for which S2 is a C -manifold m S 2 but not isometric to m ∀ ∈ − (S2, can). 8. ZOLL’S SURFACE 143

To construct the surface in the (x, y, z)-space we take in the (z, r)-plane + two curves r f (r) (i = 1, 2) with f , f D([0, 1[, R ) and f (1) = 0, 146 → i 1 2 ∈ i f ′(0) = 0, f ′ 0 and f ′(r) (i = 1, 2). i i ≤ i → ∞1 We generate a surface of revolution M by setting:

2 2 2 2 M1 = (x, y, z) z = f1( x + y ) , M2 = (x, y, z) z = f2( x + y ) ( | q ) ( | − q ) hence M = M M is a C1-manifold. We have two charts (U , s )(i = 1 ∪ 2 i i 1, 2) on U = M (z 1(0) (x 1(0) y 1(0))) deﬁned as follows: i i − − ∪ − ∩ − if p(x, y, z) = (x, y) we set s (m) = ( p(m) , angle (e , p(m))) R2(i = 1, 2) i || || 1 ∈

(polar coordinates). On Ui the local coordinates will be denoted by x1 = r = u1 s , x2 = ϕ = u2 s (i = 1, 2) so that ◦ i s ◦ i 1 1 2 1 2 1 2 1 (4.8.1) s− (u , u ) = (u cos u , u sin u , f (u )). 1 · · i 144 Geodesics

We drop the i in si and in fi. We have the vector ﬁelds

X , X dual basis of dr, dϕ on C (U ), { 1 2} { } i Y , Y dual basis of du1, du2 on C (R2). { 1 2} { } We endow M with the r.s. induced by (R3, can) as in (2.1). For g = i j gi jdx dx we have by deﬁnition of ǫ (see (1.2)): i, j P

1 T • 1 T 1 1 g = g(X , X ) = (ζ (s− ) Y ) (ζ (s− ) Y ) = (Ds− Y ) (Ds− Y ) i j i j ◦ ◦ i · ◦ ◦ j ◦ i · ◦ j 1 1 where Ds− is the (Jacobian)− of s, i.e. by (4.8.1):

cos u2 u1 sin u2 1 2 − 1 · 2 Ds− = sin u u cos u   1 ·   f ′(u ) 0      d f 147 hence, setting f = : ′ dr

2 2 2 2 2 2 (4.8.2) g = (1 + f ′ )dr + r dϕ , g = 1 + f ′ , g = 0, g = r · 12 12 22 hence by (3.8.3)

1 1 g11 = , g12 = 0, g22 = ; + 2 2 1 f ′ r thus by (4.7.17) and (4.7.19):

1 Γ = r, Γ =Γ = 0, Γ2 = , Γ2 =Γ2 = 0. 122 121 222 12 r 11 22 Hence if ψ : t (r(t),ϕ(t)) → is a geodesic, then by (1.3.5) we have

d2ϕ 2 dr dϕ + = 0 dt2 r · dt · dt 8. ZOLL’S SURFACE 145 i.e. d dϕ r2 = 0. dt dt ! Thus, along a given geodesic, there exists an a such that dϕ (4.8.3) r2 = a. dt We suppose moreover that ψ is parametrized by arc length, so that by (4.3.3) and (4.8.2): dr 2 dϕ 2 1 + f 2 + r2 = 1 ′ ds ds ! ! and hence 148 dϕ 1 . ds ≤ r

Now by (4.8.3) we have

1 a r2(s) = r(s), ≤ r(s) · and hence the geodesic never crosses the curve given by the section of M by the planes z = fi(a) (i = 1, 2). Moreover: 146 Geodesics

1/2 dϕ a 1 + f 2 (4.8.4) = ′ . dr r r2 a2 ! −

8.5

149 Let us note that (by continuity arguments, for our computations are not valid outside U U ) that the geodesics corresponding to a = 0 are 1 ∪ 2 the meridians and that to a = 1 corresponds the equator. So to ensure that M is a C -manifold m M we have only to take care of the m ∀ ∈ case 0 < a < 1. Let then m = (x , y , f (a)) and m = x , y , f (a)) 0 0 0 1 1 1 1 − 2 be extreme points of ψ such that the part of ψ between them does not meet again the planes z = f (a), z = f (a). Then - thanks to the 1 − 2 symmetry of M-showing that ψ is simply closed amounts to showing that the variation Φ(a) of the angle ϕ from m0 to m1 is equal exactly to π. Since the integral

1 1/2 a 1 + f 2 ′ dr Z r r2 a2 ! · a −

exists we have by (4.8.4) applied for the part of ψ in U1 and for the part in U2:

1 1/2 2 1/2 a (1 + f ′) + (1 + f ′ ) (4.8.6) Φ(a) = 1 2 dr. Z r · (r2 a2)1/2 · a −

Then we have shown:

8.7

Theorem Darboux. The surface (M, ǫ M) is aC -manifold m M if | m ∀ ∈ and only if Φ(a) = π a ]0, 1[, with Φ(a) as in (4.8.6). ∀ ∈ 8. ZOLL’S SURFACE 147

8.8 Theorem Zoll: [30]). There exists f , f D([0, 1[, R+) such that: 1 2 ∈ 150 i) M is a real-analytic sub manifold of R3 ii) MisaC -manifold m M m ∀ ∈ ii) M is not isometric to (S2, can).

Proof. A. To ﬁnd f1, f2 we derive relations from Φ(a) = π a. Set 1 1 · g = (1 + f 2)1/2 + (1 + f 2)1/2, = x + 1, = α + 1, x = α u, ′1 ′2 r a2 · r g(r) = 2 θ(x); then we must have · · α θ(x) dx = π α Z (α x)1/2 · ∀ 0 − and thus 1 θ(α u) · √α du = π α. Z (1 u)1/2 · ∀ 0 − Diﬀerentiating the above equation with respect to α we have

(4.8.9) θ(αu) + 2α uθ′(αu) = 0. · But the function k θ(x) = √x satisﬁes the above equation, and since it is known that

1 dx = π Z (x(1 x))1/2 0 − 1 it follows that the function can be taken for θ(x). Hence we should √x try to ﬁnd f1 and f2 such that:

+ 2 1/2 + + 2 = 2 (1 f ′1) (1 f ′ ) . (1 r2) − p 148 Geodesics

(Note that the choice f (r) = f (r) = (1 r2)1/2 makes the associated 1 2 − 151 surface a sphere). We set:

2 1/2 2 1/2 (1 + f ′ ) = (1 r ) + λ(r) 1 − and 2 1/2 2 1/2 (1 + f ′ ) = (1 r ) λ(r) 2 − − where in order that the equations have meaning we should have 1 λ(r) 1. 1 r2 ± ≥ − There exist such functions λ, for example the binomial expansion of (1 r2)1/2 gives that the function − 1 k r2 for 0 < k < · 2 is such a function. Hence we set:

2 1/2 2 1/2 2 (1 + f ′ ) = (1 r )− + kr 1 − 2 1/2 2 1/2 2 (1 + f ′ ) = (1 r )− kr , 2 − − and we know by (8.7) that M is a C -manifold m M. m ∀ ∈ B. We check (i) now. Analyticity has to be checked only at r = 1 and we wish to express r as a function of z (instead of z = f (r), z = f (r)). 1 − 2 We have (since f1′ < 0, f2′ < 0) and for z > 0:

dz = = r + 2 1/2 + 2 2 2 f1′ − (1 2k(1 r ) k r (1 r )) dr (1 r2) − − − p so if we set s = (1 r2)1/2: − ds 1 = = b(s) dz 1 + 2ks + k2 s2(1 s2) − ds 152 with b(s) analytic in s. By Cauchy’s theorem = b(s) has a unique dz analytic solution with s(0) = 0. 8. ZOLL’S SURFACE 149

For z < 0 and t = (1 r2)1/2: − − dz r = = 2 + 2 2 2 f2′ (1 2k (1 r ) k r (1 r )) −dr − (1 r2) − − − − p p hence dt 1 = = b(t) dz 1 + 2kt + k2t2(1 t2) − with the same b. So the unique analytic solution is the same for z > 0 and for z < 0. C. We shall prove that (M, ǫ M) is not isometric to (S2, can). One | can either compute the curvature of (M, ǫ M) at a simple point or argue | as follows: in (S2, can) the meridians are the geodesics. The geodesics through two antipodal points m, m′ have a closed geodesic as orthogonal trajectory, and this geodesic is the locus equidistant from m, m′. The same should hold in a surface isometric to (S2, can). But for Zoll’s surface and the two antipodal points m = (0, 0, f (0)), m = (0, 0, f (0)) 1 ′ − 2 the only orthogonal trajectory is the equator, which is not equidistant from m, m′ since the lengths of the meridians from m to the equator and from the equator to m′ are respectively:

1 1 1 2 1/2 2 1/2 2 k (1 + f ′ ) dr = (1 r )− dr + kr = 2 + Z 1 Z − Z 3 0 0 0 1 1 1 2 1/2 2 1/2 2 k (1 + f ′ ) dr = (1 r )− dr kr = 2 . Z 2 Z − − Z − 3 0 0 0

153

8.10 3 Remarks. A. Check with the choice λ(r) = r4 that the meridians have 2 inﬂexions 150 Geodesics

So there exist surfaces (M, ǫ M) which are C -manifolds m M | m ∀ ∈ and which contain points with strictly negative curvature (Zoll: [30]). B. We will see ((9)) the striking fact that if (M, g) is a Cm-manifold m M for M homeomorphic to P2(R) then (M, g) is isometric to ∀ ∈ (P2(R), can). C. Note (4.8.3) is classical for motions with central acceleration (as it is the case for geodesics of a surface of revolution since the normal meets the z-axis). Note also the condition

α θ(x) dx = π α Z (α x)1/2 · ∀ 0 −

154 is equivalent to the search of tautochronous motions on a vertical curve and yields the cycloidal pendulum. Chapter 5 Canonical connection

1.1

Deﬁnition . By (2.2) an r.m. (M, g) has a canonical spray G. BY (2.3) 155 there is a unique symmetric connection associated to G. Whenever we consider (M, g) we consider it with the above connection and refer to C as the connection or canonical connection on (M, g). Hence given an r.m. (M, g) we can speak canonically of the concepts associated with a connection, namely, the derivation law, the curvature tensor, the parallel transport and Jacobi ﬁelds in (M, g).

1.2 Example. Now let us examine the connection on (Rd, ǫ). By (2.1.21) it follows that the spray G′ associated to the canonical connection satisﬁes the equation

T (5.1.3) ζ G′ = 0. ◦ By (4.2.9) the canonical spray G on (Rd, ǫ) satisﬁes the equation

(5.1.4) ζT G = 0. ◦ = T 1 + T 1 We have the direct sum Tz(T(M)) (ζ )z− (0) (p )z− (0) and since T T p (G(z)) = p (G′(z)) = p′(z) we have G = G′. Hence by the uniqueness of the associated symmetric connection we see the canonical connection on Rd is the same as the connection on (Rd, ǫ). Now we shall prove that

151 152 Canonical connection

the connection is carried into the connection by isometries. We know by 156 (2.4) that the canonical spray is carried into the canonical spray and we have only to use ((2.2.5)).

1.5 Proposition. If (M, g) and (N, h) are r.m.’s and

f : (M, g) (N, h) → is an isometry, then

( f T )T (MC(x, y)) = NC( f T (x), f T (y)) (x, y) T(M) T(M). ∀ ∈ ×M Proof. By (4.17) it is enough to show that both sides of the equality have the same eﬀect on dϕ F(T(N)) for every ϕ F(N). We have ∈ ∈ T f T (MC(x, y))(dϕ) = MC(x, y)(dϕ f T ) = MC(x, y)(d(ϕ f )) = ◦ ◦ 1 = (MG(x + y)(d(ϕ f )) MG(x)(d(ϕ f )) MG(y)(d(ϕ f ))) = 2 ◦ − ◦ − ◦ by (2.2.5) 1 = (MG(x + y)(dϕ f T ) MG(x)(dϕ f T ) MG(dϕ f T )) = 2 ◦ − ◦ − ◦ 1 = ( f T )T (MG(x + y))(dϕ) ( f T )T (MG(x))(dϕ) 2 − ( f T )T (MG(y))(dϕ)) = − 1 = (NG( f T (x + y))(dϕ) NG( f T (x))(dϕ) NG( f T (y))(dϕ)) 2 − − = NC( f T (x), f T (y))(dϕ) by ((2.2.5)) and since f T (x + y) = f T (x) + f T (y).

1.6

157 Corollary. If f : (M, g) (N, h) is an isometry, then → Canonical connection 153

i) Nv ( f T )T = f T Mv ◦ ◦ ii) if S is a manifold, X C (S ) and ϕ D(S, T(M)), then N ( f T ∈ ∈ DX ◦ ϕ) = f T D ◦ X iii) For every X ,Y C (M) 1 1 ∈ N T 1 T M 1 1 = D f T X1 f ( f Y1 f − ) f ( DX1 Y1) f − . ◦ ◦ − ◦ ◦ ◦ ◦ Proof. (i) For every Z in C (T(M)) we have M M T v Z = ξ(Z C(p′ Z, p Z)). ◦ − ◦ ◦ But from the deﬁnition of ξ it follows that the following diagram is commutative:

Tx(Tp(x)(M)) P iT rr PP ζ x rr PPP x rr PPP rrr PPP xrr ' Vx / Tp(x)(M) Mξ

( f T )T

( f T )T f T N f T (x)Tp( f T (x))(N) iT r OO ζ f T (x) rr OO f T (x) rr OOO rrr OOO yrr N O' ξ T V f T (x) / Tp( f (x))(N)

Hence 158 f T Mζ = Nξ ( f T )T . ◦ ◦ Hence T M N T T T f v Z = ζ ( f ) (Z C(p′ Z, p Z)) = ◦ ◦ ◦ ◦ − ◦ ◦ N T T T T T = ζ(( f ) (Z) ( f ) (C(p′ Z, p Z))) = − ◦ ◦ N T T T T T = ζ (( f ) Z C( f p′ Z, f p Z)) by (1.5) = ◦ ◦ − ◦ ◦ ◦ ◦ N T T T T T T T = ζ(( f ) Z C(p′ ( f ) Z, p ( f ) Z)) = ◦ − ◦ ◦ ◦ ◦ =N v ( f T )T Z. ◦ ◦ 154 Canonical connection

(ii) We have N D ( f T ϕ) = Nv ( f T ϕ)T X by deﬁnition, = X ◦ ◦ ◦ ◦ = Nv ( f T )T ϕT X = ◦ ◦ ◦ = f T Mv ϕT X by (i) = ◦ ◦ ◦ f T D ϕ by the deﬁnition of D . ◦ X X (iii) We have only to apply (ii) and (3.14)

1.7 An application. Let f : (M, g) (N, h) be an isometry and let ϕ be a curve in (M, g) → M and ψ a parallel lift of ϕ. Then DPψ = 0. Hence by (ii) it follows that f T ψ is a parallel lift of f ϕ in (N, h). ◦ ◦ 1.8 Example. Let (G, H) be a symmetric pair and (M, γ) be an associated 159 riemannian homogeneous manifold, as in (3). Let X M and let ∈ ϕ : t p(exp(t.X)) → be the associated geodesic and ψ a parallel lift of ϕ. Then we claim : ψ(t) = (τ(exp(t.X)))T (ψ(0)) i.e. the parallel transport is given by the tangent maps of the one parameter family of diﬀeomorphisms induced by X. Proof. We know that for every t , τ(exp(t X)) takes the image of ϕ 0 0 · into itself and is an isometry. Hence (τ(exp(t X)))T takes ψ again into 0 · a parallel lift thanks to (1.7). But we do not know if it coincides with

ψ. On the other hand we know that the map σϕ(t0) is an isometry and hence preserves parallel lifts, maps the image of ϕ into itself and further T b (σϕ(t0)) is an isometry and acts as the negative of identity on T(t0)(M). Hence ψ(t) goes into ψ( t). Therefore we try to represent τ(exp(tX)) − − asb the composition of an even number of symmetries. 2. RIEMANNIAN STRUCTURES ON T(M) AND U(M) 155

T Clearly (σϕ ) ψ is a parallel lift of σϕ ϕ. But by (3.20) we (t0) ◦ (t0) ◦ have b b (σϕ f )(t) = f (2t t), (t0) ◦ 0 − and hence with the notationb of (5) = σϕ(t0) f f τ2t0 k 1. ◦ ◦ ◦ − Now consider b 160 = T ψt0 σϕ(t0) ψ τ2t0 k 1. − ◦ ◦ ◦ − 2 = Since (τ2t0 k 1) idR, it followsb that ψt0 is a parallel lift of ψ and at ◦ − t0 is equal to σϕ ψ(t ) = ψ(t ). − (t0) ◦ 0 0 Hence by the uniqueness ofb parallel lifts (7.4) we have = ψ ψt0 .

Now if we set t0 = 0 we obtain

T ψ = σ ψ k 1 − ◦ ◦ − and hence b

= T ψ σϕ(t0) ( σ ψ k 1) τ2t0 k 1. − ◦ − ◦ ◦ − ◦ ◦ − But b b

σϕ σ = τ(exp(2t )X)) and hence (t0) ◦ 0 = ψ τ(exp 2t0X) ψ k 1 τ2t0 k 1 b b ◦ ◦ − ◦ ◦ − = τ(exp 2t0X) ψ τ 2t0 since ◦ ◦ − 2 = (τ2t0 k 1) idR . ◦ − Hence in particular we have

ψ(2t ) = (exp(2t X)))T ψ(0). 0 0 · 156 Canonical connection

2 Riemannian structures on T(M) and U(M) 161 Let (M, g) be an r.m. For every element x of T(M) the tangent space at x is the direct sum of Hx and Vx: Tx(T(M)) = Hx + Vx.

The map ξ deﬁnes an isomorphism between Vx and Tp(x)(M) and so does T the restriction of p to Hx between Hx and Tp(x)(M):

ξ : V T (M) x → p(x) pT H : H T (M). | x x → p(x) By means of these isomorphisms and the euclidean structure on Tp(x)(M) we deﬁne a euclidean structure g on Tx(T(M)). Precisely:

2.1 Deﬁnition. The canonical r.s. on T(M), g, is deﬁned by the equation

T T g(z, z′) = g(v(z), v(z′)) + g(p (z), p (z′)) z, z′ T (T(M)) ∀ ∈ x x T(M). ∀ ∈

2.2 Lemma. With respect to this structure g for the function E on T(M) we have (see 6.4): grad(E) = Ξ C (T(M)) ∈ where 1 Ξ(x) = 2ξ− x, x T(M). x ∈ 162 Proof. Let Z be any vector ﬁeld on T(M). Then

(5.2.3) Z(E) = (dE)(Z) = g♯(grad(E))(Z) = g(grad(E), Z). 2. RIEMANNIAN STRUCTURES ON T(M) AND U(M) 157

Let z be a horizontal vector. Then we have for y = pT (z):

z = C(x, pT (z)) where C is the canonical connection on (M, g). But by ((2.2.5)) and (3.1) we have 1 z(E) = (G(x + y)(E) G(x)(E) G(y)(E)). 2 − − Hence for every horizontal vector ﬁeld Z we have

g(grad(E), Z) = 0.

Hence grad(E) is a vertical vector ﬁeld. Let x T(M) and let z V . ∈ ∈ x Then, by (1.8) z(E) = 2g(x,ξ(z)), and by (2.1), since z and grad E are vertical, we have:

g(grad(E)x, z) = g(ξ(grad(E)),ξ(z)).

Hence by (5.2.3) we have

g(2x ξ(grad(E) ),ξ(z)) = 0 − x for every vertical vector z. Since ξ is an isomorphism between Vx and Tp(x)(M) and g is non-degenerate we have

2x ξ(grad(E) ) = 0. − x Hence: = 1 = 1 grad(E)x ξx− (2x) 2ξx− (x).

Since U(M) is a sub manifold of T(M) there is an induced r.s. on U(M). 163

2.4

Deﬁnition. The canonical r.s. gonU(M) is the structure g U(M) = g. | 158 Canonical connection

2.5 Proposition. The volume elements θ and θ of T(M) and U(M) are respectively given by the equations

θ = σ , θ = σ | | | | where d(d 1) 1 d σ = ( 1) 2− (dα) − · d! · ∧ and d(d 1) 1 d 1 σ = ( 1) 2− α ( − (dα)). − · (d 1)! · ∧ ∧ − 2.6 Let us note, incidentally, that this implies that T(M) and U(M) are oriented canonically.

Proof. In view of (5.1) and (3.4) it is enough to show that θ and θ take value one on some orthonormal basis of Tx(T(M)) and Tx(U(M)) for x in T(M) and U(M) respectively.

a) Let us consider θ. Let x T(M) and let x1,..., xd be an or- ∈ = 1 { = } = thonormal basis of Tp(x)(M). Set pi ξx− xi and zi C(x, xi) (i 1,..., d). Then, by the deﬁnition of g, it follows that p ,..., p , { 1 d z ,..., z is an orthonormal basis of T (T(M)). Further by (6.10) 1 d} x we have

(dα)(pi, p j) = 0 (5.2.7) (dα)(p , z ) = δ i, j i j i j ∀ 164 Hence by the deﬁnition of an exterior product through shuﬄes we have

d d(d 1) d − ( 1) 2 (dα)(p ,..., p , z ,..., z ) = (dα)(pσ , zσ ) − ∧ 1 d 1 d (i) (i) σXPd Yi=1 ∈ 2. RIEMANNIAN STRUCTURES ON T(M) AND U(M) 159

where σ runs through all possible permutations of 1,..., d . { } Hence

d(d 1) d ( 1) 2− (dα)(p ,..., p , z ,..., z ) = d!. − ∧ 1 d 1 d

b) Now let us consider U(M). We want to look at the tangent vectors to U(M) among those to T(M). Let us consider the injection i : U(M) T(M). If x U(M) and z Tx(T(M)) is a vector in the → ∈ T ∈ image of Tx(U(M)) by i then there exists z1 in Tx(U(M)) such T that i (z1) = z. But then since E is constant on U(M) we have by (5.2.3) 0 = z (E i) = iT (z )(E) = z(E) = g(grad(E), z). Hence 1 ◦ 1 every such vector z is orthogonal to grad(E). But the subspace in Tx(T(M)) of vectors orthogonal to grad(E)x is of dimension 2d 1 T − which is equal to the dimension of U(M). Since ix is injective we have

(5.2.8) iT (T (U(M)) = z g(grad(E) , z) = 0 . x x { | x }

2.9

Now let x U(M) and let x1,..., xd be an orthonormal basis of ∈ = { }= 1 = Tp(x)(M) with x1 x. Set, as above, pi ξx− xi and zi C(x, xi) for = = 1 i 1,..., d. Then, since grad(E)x 2ξx− by (2.2), we see, by (5.2.7), that p ,..., p , z ,..., z { 2 d 1 d} is an orthonormal basis of Tx(U(M)) and the relations (5.2.7) are valid. 165 Moreover (see 6.6): α(p ) = 0 i and α(z ) = g(x , x ) = 1. Hence i ∀ 1 1 1 d(d 1) d 1 ( 1) − α ( − (dα))(p ,..., p , z ,..., z ) − 2 ∧ ∧ 2 d 1 d d 1 − = α(z1) (dα)(pσ(i), zσ(i)) σXPd 1 Yi=2 ∈ − 160 Canonical connection

where σ runs through all permutations of 2,..., d . Hence { } (d 1) α ( − dα)(p ,..., p , z ,..., z ) = (d 1). ∧ ∧ 2 d 1 d − Hence the result.

2.10 Now let us compute vol(U(M), g) when M is compact. First let us suppose that M is oriented by a form σ ξd(M), and 1 ∈ denote (see (5)) the canonical orienting form corresponding to σ1 by σ. Let p ,..., p ; z ,..., z be as in the proof (a) of (2.5). Then we deﬁne { 1 d 1 d} τ ξd(T(M)) by setting ∈ τ(p ,..., p ) = , 1 d ± according as ξp ,...,ξp is a positive basis of T (M) or not, and zero { 1 d} m for any other combination from p ,..., p ; z ,..., z , and extending, { 1 d 1 d} multilinearly. First let us note that the restriction of τ to Tm(M) is the canonical 166 volume form of the euclidean space (Tm(M), gm). Then we have

d(d 1) − (5.2.11) σ = ( 1) 2 τ p∗ σ − ∧ M where p : T(M) M. For M → (τ p∗ σ)(p ,..., p , z ,..., z ) ∧ M 1 d 1 d = τ(p ,..., p ) σ(pT (z ),..., pT (z )) = 1 d · 1 d = 1 by the deﬁnition of τ.

Now let p ,..., p , z ,..., z be as in (2.9). Then deﬁne ω by set- { 1 d 1 d} ting ω(p ,..., p ) = 1 according as ξp ,ξp ,...,ξp is positive or 2 d ± { 1 2 d} negative with respect to σ, and zero for any other combination from p ,..., p , z ,..., z and extending linearly. We note that the restric- { 1 d 1 d} tion of ω to Um(M) is the canonical volume form on the sphere (Um(M), g U (M)) and check (as in the case of τ p σ) that m| m ∧ ∗M

(5.2.12) σ = ω p∗ σ. ∧ M 2. RIEMANNIAN STRUCTURES ON T(M) AND U(M) 161

Now we prove that ¨ (5.2.13) Vol(U(M), g) = d 1 . vol(M, g), − © where we use the notation:

(5.2.14) d = Vol(Sd, ǫ Sd). | Proof. First let us assume that M is oriented. Then ω p σ is a volume ∧ ∗M form on U(M) and we have

Vol(U(M), g) = ω p∗ σ = Z ∧ M U(M)

=  ω Um(M) σ, by (3.17) Z  Z |  x M U (M)  ∈  m    But 167

d 1 d 1 ω Um(M) = Vol(S − (Tm(M), gm S − (Tm(M))) = (by (3.15)) Z | | Um(M) and because all euclidean structures on vector spaces of the same ﬁnite dimension are isometric, this

d 1 d 1 ¨ = Vol(S − , ǫ S − ) = d 1 . | − © Hence

¨ ¨ Vol(U(M), g) = d 1 σ = d 1 Vol(M, g). Z − © − ©· x M ∈ b) In the general case we take a partition of unity and proceed. Let W ,ϕ be a partition of unity on M, the W s being so small that they { i i} i′ can be oriented. Then we have

(5.2.15) (ϕi p) = 1 on U(M) X ◦ 162 Canonical connection

where p : U(M) M. Now let σ be an orienting form on W such that → i i

θ U(W ) = σ | i | i| where

(5.2.16) σ = ω (p∗σ ). i ∧ i Then we have

(ϕi p)θ = (ϕi p)σ = Z ◦ Z ◦ U(Wi) U(Wi)

= (ϕi p)ω p∗σi. Z ◦ ∧ U(Wi)

¨ (5.2.17) =  ω Um(M) ϕi σi = d 1 ϕi σi . Z  Z |  | | − ©Z | | m Wi U (M)  Wi ∈  m    168 Now by (5.2.17) we have

Vol(U(M), g) = θ = (ϕ p)θ = i ◦ Z Xi Z U(M) U(Wi) ¨ ¨ = d 1 ϕi σi = d 1 Vol(M, g). − © | | − © Xi Z Wi

3 Dg = 0

Let us denote the canonical derivation law of the r.m. (M, g) by D. Since g L 2(M) and by 4.7, we can introduce the covariant derivative Dg ∈ ∈ L 3(M). We shall prove that the 3-form Dg is zero. 3. Dg = 0 163

3.1 First let us note that the form Dg is symmetric in the last two variables (because g is symmetric):

(5.3.2) Dg(X, Y, Z) = Dg(X, Z, Y) X, Y, Z C (M). ∀ ∈ To start with we shall prove a lemma.

3.3 Lemma.

Dg(y, x, x) = 2Dg(x, x, y) (x, y) T(M) T(M). ∀ ∈ ×M Proof. With the notation of (8) let us take a one-parameter family f such that P(0, 0) = x, Q(0, 0) = y, and f0 is a geodesic.

Now let us follow the notations of (1). Then since f0 is a geodesic by (4.2.1), and (1.7) we have;

(5.3.4) P(β(Q))(0, 0) = 1/2Q(β(P))(0, 0).

Further since the connection is symmetric and [P, Q] = 0 we have 169

(5.3.5) DPQ = DQP and since f0 is a geodesic

(5.3.6) (DPP)(t, 0) = 0.

Further, by the deﬁnition of D, we have, at the point (t, 0):

P(β(Q)) = P(g(P, Q))

= Dg(P, P, Q) + g(DpP, Q) + g(P, DpQ) 164 Canonical connection

= Dg(P, P, Q) + g(P, DpQ) by (5.3.6) and Q(β(P)) = Q(g(P, P))

= Dg(Q, P, P) + 2g(DQP, P)

= Dg(Q, P, P) + 2g(DPQ, P) by (5.3.5).

Hence by (5.3.4) we have, at the point (t, 0) 2 Dg(P, P, Q) = Dg(Q, P, P),

and hence by the choice of the one parameter family 2Dg(x, x, y) = Dg(y, x, x). Proof of Dg = 0. Clearly, by the lemma (5.3.7) Dg(x, x, x) = 0 x T(M). ∀ ∈ Further, we have, by the lemma, Dg(y, x + y, x + y) = 2Dg(x + y, x + y, y), i.e. Dg(y, x, x) + Dg(y, y, y) + 2 Dg(y, x, y) = 2Dg(x, x, y) + 2Dg(x, y, y) + 2 Dg(y, x, y) + 2 Dg(y, y, y). Again by the lemma and (5.3.7) we have (5.3.8) Dg(x, y, y) = 0.

170 Hence Dg(x, y + z, y + z) = 0 i.e. Dg(x, y, y) + Dg(x, y, z) + Dg(x, z, y) + Dg(x, z, z) = 0. By (3.1) and (5.3.8) we have 2 Dg(x, y, z) = 0. Hence Dg = 0. Hence we have the following 4. CONSEQUENCES OF Dg = 0 165

3.9 Corollary.

(D.L.5) X(g(Y, Z)) = g(D Y, Z) + g(Y, D Z) X, Y, Z in C (M). x x ∀ 4 Consequences of Dg = 0

Now we shall complete the lemma (6.10) in the:

4.1 Proposition. For every x of T(M) we have

T T (dα)(z, z′) = g(v(z)), p (z′)) g(v(z′), p (z)) z, z′ T (T(M)). − ∀ ∈ x

Proof. We know that the result holds good if at least one of z and z′ is a vertical vector (see (6.10)). In view of the bilinearity of both the sides and the direct sum decomposition

Tx(T(M)) = Hx + Vx it is enough to prove the result z, z H . Now let Z, Z C (M) be ∀ ′ ∈ x ′ ∈ such that

T T (5.4.2) Z(p(x)) = p (z) and Z′(p(x)) = p (z′).

Then we contend that

H H dα(Z , Z′ ) = 0 (see (5.8)).

We have by (0.2.10) 171

H H H H H H H H (5.4.3) dα(Z , Z′ ) = Z (α(Z′ ) Z′ (α(Z ) α([Z , Z′ ]). − − But for every X C (M) we have ∈ H H T H α(Z ) X = g(p′ Z X, p Z X) ◦ ◦ ◦ ◦ ◦ = g(X, Z) by (1.5) C1., 166 Canonical connection

and hence (5.4.4) α(ZH) = g♯(Z). Now by (5.4.4) and (5.4.3) we have H H H ♯ H ♯ H H dα(Z , Z′ ) = Z (g (Z′)) Z′ (g (Z)) α([Z , Z′ ]), − − ♯ ♯ H H = D (g (Z′)) D (g (Z)) α([Z , Z′ ]) by (5.9). Z − Z′ − Hence for every X C (M) we have ∈ (5.4.5) H H ♯ ♯ H H dα(Z , Z′ ) X = D (g (Z′))(X) D (g (Z))(X) α([Z , Z′ ] X) ◦ Z − Z′ − ◦ = Z(g(Z′, X)) g(Z′, D X) − Z

H H Z′(g(Z, X)) g(Z, D X) α([Z , Z′ ] X) by (2.4.8) − − Z′ − ◦ H H = g(D Z′, X) g(D , Z, X) α([Z , Z′ ] X) by (3.9). Z − Z − ◦ But H H H H T H H (5.4.6) α([Z , Z′ ] X) = g(p′ [Z , Z′ ] X, p [Z , Z′ ] X) ◦ ◦ ◦ ◦ ◦ = g(X, [Z, Z′]) (see the proof of (5.10)). Hence we have by (5.4.5) and (5.4.6) H H (dα)(Z , Z ) X = g(D Z′ D Z, X) g(X, [Z, Z′]) ◦ Z − Z′ − = g([Z, Z′], X) g(X, [Z, Z′]) by (4.1) D.L.4. − = 0 since g is symmetric.

172 This being so for every X we have H H (dα)(Z , Z′ ) = 0. Now we shall extend the result (3.9). Let N be any manifold and let h, h D(N, T(M)) be such that p h = p h . Then we can deﬁne a ′ ∈ ◦ ◦ ′ function g(h, h′) on N by setting

(5.4.7) g(h, h′)(n) = g(h(n), h′(n)), n N. ∈ Since h(n), h (n) T (M), we have the following result. ′ ∈ p(h(n)) 5. CURVATURE 167

4.8 Corollary. With the above notation we have

(C.D.7) X(g(h, h′)) = g(D h, h′) + g(h, D h′) X C (N). X X ∀ ∈ Proof. This follows directly from (4) and Dg = 0.

4.9 Corollary. Parallel transport preserves g. Proof. Clearly it is enough to prove the result for a curve instead of for a path. Let f D(I, M) be a curve and let h and h be parallel lifts of f ∈ ′ such that h(0) = x and h′(0) = y. Then we have d (g(h(t), h (t)) = P(g(h, h )) = dt ′ ′ g(Dph, h′) + g(h, Dph′) by (5.4.7)

= 0 since Dph = Dph′ = 0.

Hence g(h, h′) is constant on I. Hence 173

g(h(t), h′(t)) = g(h(0), h′(0)) t I. ∀ ∈

5 Curvature

From the deﬁnition of R (see (5.1)) and from (1.6) iii) it follows, directly, that R is invariant under isometries, i.e.

5.1 Proposition. If f : (M, g) (N, h) → is an isometry then X,Y,Z C (M) we have ∀ ∈ T M 1 N T 1 T 1 T 1 f R(X, Y)Z f − = R( f X f − , f Y f − )( f Z f − ). ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ 168 Canonical connection

5.2 Remark. Let us note that this result in particular implies that an r.m. in general, is not locally isometric to Rd. For, for Rd, R = 0 by (5.4). Now let us consider the map

(5.5.3) (X, Y, Z, T) g(R(X, Y)Z, T); C 4(M) R → → and prove

(5.5.4) C.T.3 g(R(X, Y)Z, T) = g(R(X, Y)T, Z) − (5.5.5) C.T.4 g(R(X, Y)Z, T) = g(R(Z, T)X, Y)

Proof. C.T.3. We have by the deﬁnition of R

(5.5.6) g(R(X, Y)Z, Z) = g(D D Z, Z) g(D D Z, Z) g(D , Z, Z). X Y − Y X − [X Y] By by (3.9) D.L.5. we have 1 g(D Z, Z) = [X, Y](g(Z, Z)) [X,Y] 2 g(D D Z, Z) = X(g(D Z, Z)) g(D Z, D Z) X Y Y − Y X and g(D D Z, Z) = Y(g(D Z, Z)) g(D Z, D Z) Y X X − X Y 174 and hence by (5.5.6) (5.5.7) 1 g(R(X, Y)Z, Z) = X(g(D Z, Z)) Y(g(D Z, Z)) [X, Y](g(Z, Z)); Y − X − 2 and again by (3.9) 1 1 1 = X( Yg(Z, Z)) Y( Xg(Z, Z)) [X, Y]g(Z, Z) = 0. 2 − 2 − 2 Now if we replace Z by Z + T in (5.5.7) and carry out the cancellation we get C.T.3.

Note . We sketch here the geometrical reason for which (C.T.3) holds: because parallel transport leaves g invariant, and thanks to (7.11), the endomorphism R(X, Y) belongs to the tangent space of the orthogonal group (i.e. its Lie algebra); this fact is nothing but (C.T.3). 6. JACOBI FIELDS IN AN R.M. 169

C.T.4 We use C.T.1., C.T.2. and C.T.3. and prove the result

In the octahedron, the sum of the terms at the three vertices of every 175 shaded triangle is zero by (C.T.1), (C.T.2) and (C.T.3). Now adding the terms corresponding to the triangles above and subtracting the similar sum for the lower ones from it we get

2(g(R(X, Y)Z, T)) 2g(R(Z, T)X, Y)) = 0. −

6 Jacobi ﬁelds in an r.m.

Suppose that f D(I, M) is a geodesic in an r.m. (M, g) and that h is a ∈ Jacobi ﬁeld along f . Then we have the following proposition.

6.1 Proposition. If for some t I, 0 ∈

g( f ′(t0), h(t0)) = g( f ′(t0), (DPh)(t0)) = 0 170 Canonical connection

then g( f ′, h) = 0 on I.

Proof. We have d (5.6.2) (g( f , h)) = P(g( f , h)) dt ′ ′ = g(DP f ′, h) + g( f ′, DPh) by (4.8) (C.D.7) =

= g( f ′, DPh) by (7.3)

and again by (4.8) and (7.3).

d2 d d (5.6.3) (g( f ′, h)) = (g( f ′, DPh)) = (g( f ′, DPh)) dt2 dt dt = g( f ′, R( f ′, h) f ′) by the deﬁnition of a of a Jacobi ﬁeld (see (8.9)) = 0 by ((5.5.5) C.T.4) C.T.3.

Hence g( f ′, DPh) is constant on I.

176 But, by hypothesis g( f ′, DPh)(0) = 0, and hence g( f ′, h)) is constant on I. But g( f ′, h)(0) = 0. Hence g( f, h) = 0.

6.4 Deﬁnition. An r.m. of dimension two is called a surface. As an application of the above proposition we prove the existence of nice coordinate neighbourhoods for surfaces.

6.5 Application. Given a surface (M, g) and a point m of M there exists a chart (U, r) such that m is in U and if we denote the local coordinates with respect to (U, r) by x and y then

g U = dx2 + K(x, y)dy2. | 6. JACOBI FIELDS IN AN R.M. 171

Proof. Let x , y be an orthonormal base of T (M) and let S { 0 0} m ∈ D(J′′, M) be any curve in M such that

(5.6.6) S ′(0) = x and S ′(α) = 1. 0 || ||

6.7

Then choose the lift S (α) of S through y0 such that at each point α, S ′(α) and S (α) form an orthonormal basis for T (M) (see (1.2)). Now by e S (α) taking a suitable sub interval J of J and an interval I we can consider e ′ ′′ the one parameter family

(5.6.8) f : I J′ M × → f (t,α) = exp(t S (α)). · As in the proof of (8.26) we get e

T P(0, 0) = ( f P)(0, 0) = x0 (5.6.9) ◦ Q(0, 0) = ( f T Q)(0, 0) = y ◦ 0 and hence f T (0, 0) is an isomorphism. Therefore by choosing I and J suﬃciently small we see, by the inverse function theorem, that

(5.6.10) f : I J f (I J) = U × → × is a diﬀeomorphism. Now set 177

1 (5.6.11) f − U = r | and

(5.6.12) x = t r, y = α r. ◦ ◦ Then we have

(5.6.13) g U = g(P, P)dx2 + 2g(P, Q)dx dy + g(Q, Q)dy2. | 172 Canonical connection

We have

g(P, P)(0,α) = S (α) = 1 by the deﬁnition of S (α) || || e e and g(P, P)(t,α) = g(P, P)(0,α)

by (3.2) since for each α, fα(t) is a geodesic, and hence

(5.6.14) g(P, P) = 1.

Further from the proof of (8.26) it follows that for each α

(5.6.15) t Q(t,α) = hα(t) →

is a Jacobi ﬁeld along the geodesic fα(t). Further

(5.6.16) g( f ′(0), hα(0) = g(P, Q)(0,α)

= g(S (α), S ′(α)) = 0

by our construction; and also e

= (5.6.17) g( fα′(0), DPhα(0)) g(P, DPQ)(0,α)

= g(P, DQP)(0,α) by 4.1 D.L.4.,

= g(S , DQS )(0,α) 1 = Qe(g(S e, S )) by (4.8) 2 = 0 by thee e construction of S . e 178 Now (6.1) gives that

(5.6.18) g(P, Q)(t,α) = 0

Now the result follows if we set K(x, y) = g(Q, Q). 6. JACOBI FIELDS IN AN R.M. 173

6.19 Proposition. Let (M, g) be an r.m. let f D(I, M) be a geodesic and let ∈ h be a Jacobi ﬁeld along f . Then d(E h) i) ◦ = 2g(D h, h) dt P d2(E h) ii ) ◦ = E DPh + g(R( f ′, h) f ′, h) dt2 ◦ iii) if h(0) = 0 and (D h)(0) = y with y = 1 then P || || 3 t 3 h(t) = t + g(R( f ′(0), y) f ′(0), y) + 0(t ), || || 6 where 0(t3) 0 with t. t3 → Proof. By (4.8) we have

d (5.6.20) (E h) = P(g(h, h)) = 2g(D h, h) dt ◦ P and again by (4.8), we have

d2 (E h) = 2P(g(DPh, h)) dt2 ◦ = 2g(DPDPh, h) + 2g(DPh, DPh)

= 2g(R( f ′, h) f ′, h) + 2E D h by the deﬁnition (8.9). ◦ P Now with the notation of (2.7.9) by (8.18) we have 179

t3 (5.6.21) h(t) = ty + (R( f (0), y) f (0) + 0(t3) 6 ′ ′ b and hence

4 2 2 t 4 h(t) = t g(y, y) + g(R( f ′(0), y) f ′(0), y) + 0(t ) || || 3 b 174 Canonical connection

t2 = t2 1 + g(R( f (0), y) f (0), y) + 0(t2) 3 ′ ′ Since h(t) is the parallel transport of h(t) along f we have by (4.9)

h(t) =bh(t) || || || || Now the rest is the expansion formulab for

√1 + δ.

Sometimes it is convenient to re parametrise the arcs by their arc lengths and then deal with them. In the case of geodesics which are non trivial this re parametrisation is always possible. Let f be a geodesic. Then we know that f ′ is constant, by (3.2), say θ. Then θ , 0 and for f k θ || || ◦ − we have ( f k θ)′ = 1 (see (5.11)). || ◦ − ||

6.22 Now for any vector x in U(M) let us denote the curve

t exp(tx) →

by γx(t). By applying a change of parameter, from (8.29) we get the following result.

6.23

180 Proposition. For every non zero x in Ω and y T (M), we have ∈ p(x) T 1 exp (ζ− y) = h( x ) m x || ||

where h is the Jacobi ﬁeld along γx/ x such that h(0) = 0 and (D h)(0) = y/ x . || || P || || 6. JACOBI FIELDS IN AN R.M. 175

Proof. By (8.29) we have

T 1 = ◦ expm(ζx− y) h(1) where h◦ is the Jacobi ﬁeld along

◦f : t exp(t x) → · such that ◦ ◦ h(0) = 0 and (DPh)(0) = y.

x ◦ ◦ ◦ But γ = f k x 1 and if we set f k x 1 = f and h k x 1 = h x/ x ! ◦ || || − ◦ || || − ◦ || || − || || then h is a Jacobi ﬁeld along f such that h(0) = h◦(0) = 0 and h( x ) = || || ◦ = T 1 h(1) expm ζx− y . ◦ 1 ◦ But DPh = DP(h k x 1) = x − DPh, so that ◦ || || − || ||

1 ◦ D h(0) = x − D h = y/ x . P || || P || ||

6.24

Gauss Lemma. For every x and y in Tm(M) such that

(5.6.25) x , 0, x Ω and g(x, y) = 0 ∈ we have T 1 T 1 = g(expm(ζx− x), expm(ζx− y)) 0.

Proof. Let h be the Jacobi ﬁeld along γx/ x such that h(0) = 0 and 181 y || x|| (D h)(0) = . Then we have γ (0) = and hence P x ′x/ x x || || || || || || x g(γ′ , h)(0) = g , 0 = 0, x/ x x ! || || || || 176 Canonical connection

x y g(γ′ , DPh)(0) = g , = 0 by (5.6.25). x/ x x x ! || || || || || ||

Hence by (6.1) we have = (5.6.26) g(γ′x/ x , h) 0, || || and hence in particular = g(γ′x/ x , h)(x) 0. || || But by (6.23) we have

T 1 (5.6.27) h( x ) = exp (ζ− y). || || m x Further x γ x (t) = exp t = (exp S )(t) x m || || · x ! ◦ || || 182 where 6. JACOBI FIELDS IN AN R.M. 177

x S (t) = t . · x || || Hence = T γ′x/ x expm S ′. || || ◦ But

1 x (5.6.28) S ′(t) = ζ−tx x x ! || || || || and hence 1 (5.6.29) γ = expT (ζ 1 x). ′x/ x x m x− || || || || Now our result follows from (5.6.26), (5.6.27) and (5.6.29). Now for an m in M and a non zero x in Tm(M) set

1 (5.6.30) L = kζ− x k R x x | ∈ = n 1 o= (5.6.31) Nx ζx− y g(x, y) 0 . n | o Then we have the following corollary.

6.32 Corollary. With the above notation, we have

i) expT (u) = u , if u L || m || || || ∈ x T T = ii) g(expm(Lx), expm(Nx)) 0.

Proof. The proof of the ﬁrst part follows from the fact that Lx is one dimensional and by (5.6.29), (7.3) and (4.9). The second part follows from Gauss’ lemma.

Chapter 6 Sectional Curvature

Let (M, g) be an r.m. of dimension d greater than one, and let x and y 183 be two linearly independent tangent vectors at a point m of M. Let us denote the subspace generated by x and y by P(x, y). Let e1 and e2 be 2 an orthonormal basis of P(x, y). We recall that the norm on T (M) is ∧ m given by x y 2 = x 2 y 2 (g(x, y))2. || ∧ || || || · || || − Now we shall prove the following proposition.

0.33 Proposition . For (x, y) T(M) T(M), such that x and y are linearly ∈ ×M independent the quantity g(R(x, y)x, y) (6.1.2) A(x, y) = , − x y 2 || ∧ || depends only on P(x, y).

Proof. Let x′ and y′ be two vectors such that

P(x, y) = P(x′, y′).

Then we have numbers a, b, a′, b′ such that

x′ = ax + by and y′ = a′ x + b′y.

179 180 Sectional Curvature

Then

R(x′, y′) = R(ax + by, a′ x + b′y) = (ab′ ba′)R(x, y), by (5.2) (C.R.1). − Hence

g(R(x′, y′)x′, y′) = (ab′ ba′)g(R(x, y)x′, y′) − = (ab′ ba′)g(R(x′, y′)x, y) by ((5.5.5) C.T.4)(C.T.4) − 2 = (ab′ ba′) g(R(x, y)x, y) −

184 and further

x′ y′ = (ab′ a′b)(x y) ∧ − ∧ and hence 2 2 2 x′ y′ = (ab′ ba′) x y . || ∧ || − · || ∧ ||

0.3

For any two dimensional subspace P of the tangent space Tm(M) we denote the above quantity by

A(P) = A(x, y)

and call the “function” A the sectional curvature of (M, g).

0.4

We set

P(M) = P T (M) m M and dim P = 2 { ⊂ m | ∈ } A(M) = A(P) P P(M) . { | ∈ } 2. EXAMPLES 181

0.5 Proposition. Let λ : (M, g) (N, h) → be an isometry, then

M A(P) = N A(λT (P)) P P(M). ∀ ∈ Proof. By (5.1) the curvature is invariant under isometry and by the deﬁnition of isometry the r.s. is invariant under it. Hence the result.

2 Examples 185 If (M, g) is a surface then the tangent space at each point m of M is two dimensional and hence A assumes only one value on the set of two dimensional subspaces of Tm(M). Hence A can be considered as a function on M itself. Now let us compute A and verify that it is nothing but the so-called Gaussian (or total) curvature of our surface (M, g). Let us take a coordinate system x, y which has been constructed in { } (6.5) and follow the notations of that proof. We use propositions (8.7) and (7.10) and compute C. Let

(6.2.1) ω : t ωα(t) → be the parallel lift of fα such that

(6.2.2) ωα(0) = Q(0,α).

Then by (4.9) we have

g(ωα(t),ωα(t)) = 1.

Since fα′(t) is a parallel life of fα by (4.9) we have = = = g(ωα(t), fα′(t)) g(ωα(0), fα′ (0)) g(Q, P)(0,α) 0.

Therefore Q(t,α) = θ(t,α)ωα(t) 182 Sectional Curvature

and hence

(6.2.3) Q(t,α) = θ(t,α)ωα(0). (see (2.7.9)). b 186 But g(Q(t,α), Q(t,α)) = g(Q(t,α), Q(t,α)) = k(t,α)

and hence by (6.2.3)b b

(6.2.4) θ(t,α) = (k(t,α))1/2.

Now by (8.7)

∂2θ(t,α) (6.2.5) R(P, Q)P = D D Q = ωα(t). P P ∂t2 · We have ∂2θ(t,α) ∂2θ(t,α) g(R(P, Q)P, Q) = g ωα(t),θ(t,α)ωα(t) = θ(t,α) ∂t2 ! ∂t2

Further P Q 2 = θ(t,α)2. || ∧ || Hence by the deﬁnition of A we have

t ∂2θ(t, ) (6.2.6) A(t,α) = . −θ(t,α) ∂t2

2.7 This is nothing but the value of the Gaussian curvature when computed in a local coordinate system of the type of (6.5) (see, for example [31]: formula (3, 2) on p.196).

2.8 The concept of sectional curvature is very powerful. Very simple restrictions on C can have deep implications. For example: 2. EXAMPLES 183

2.9 Proposition . If A(M) ] , 0] then for every m of M, exp is of ⊂ − ∞ m maximal rank on T (M) Ω. m ∩ Proof. To prove this we use (8.32). Let x T (M) Ω and let f : t 187 ∈ m ∩ → exp(t x) be a geodesic, and let h be a Jacobi ﬁeld along f such that · (6.2.10) h(0) = 0 and h . 0.

Then we have by (6.19) ii)

d2(E h) (6.2.11) ◦ = 2E DPh + 2g(R( f ′, h) f ′, h) ... dt2 ◦ ≥ 2 ... 2g(R( f ′, h) f ′, h) = A( f ′, h) f ′ h 0. ≥ − || ∧ || ≥ Hence the function E h is a non-negative convex function which is not ◦ identically zero and hence it can vanish at most at one point. Hence h cannot vanish outside 0 and we are through by (8.32).

2.12 By 5.4 R = 0 for (Rd, ǫ). Hence A(Rd) = 0 . { } 2.13 Now let us consider the sectional curvature of the three symmetric pairs (Sd, can), (Pd(R), can), (Rd, hyp) (see (4)). We write any of them M = G/H; since G operates transitively and by isometries on M, (0.5) yields

(6.2.14) A(M) = A(T (M)) m M. m0 ∀ 0 ∈ By (4.17) and (4.21) we have in fact: A(M) consists of a single real number, say k.

2.15 Deﬁnition. An r.m. (M, g) for which k R with A(M) = k is called ∃ ∈ { } an r.m. of constant sectional curvature. 184 Sectional Curvature

Hence the three symmetric pairs are of constant sectional curvature and so is (Rd, can). The corresponding constant is 0 for (Rd, can), 1 for (Sd, can) or (Pd(R), can) by (6.3.12) and (4.19) B. We shall ********** • 188 move that the constant is < 0 for (Rd, hyp). We do this by global arguments and use the results of Chapters VII and VIII. First (Rd, hyp) is complete in the sense of (4) because geodesics through m0 are deﬁned on the whole Rd (see 4.21 to 4.22). Now this remark together with the fact that the sectional curvature is constant and is equal to k and (3.7) en- ables us to exclude the case k > 0; and we are left to throw out the case k = 0. If k = 0 then (Rd, hyp) would be locally isometric to (Rd, can) by (6.4.35). By (1.8) the τ(exp X) for X M in the symmetric space ∈ (Rd, hyp) are products of symmetries around points of M; those symmetries are determined by the r.s. of (M, g) hence those symmetries are the same (locally) as in (Rd, can). In conclusion the τ(exp X) are isomorphic to those in (Rd, can); but the latter are the translations and hence commute. That would imply [M, M] = 0; it is easy to check that for (S 00(d, 1), S 0(d)) this is not the case. By using (6.3.12) we normalise k in (Rd, hyp) so as to have A(Rd, hyp) = 1 . {− } 2.16 Proposition. For any k R there exists a simply connected r.m. (M, g) ∈ such that A(M, g) = k . We may take { } d 1/2 (S , k− can) for k > 0 · (Rd, can) for k = 0 d 1/2 (R , ( k)− hyp) for k < 0. − ·

3 Geometric interpretation 189 3.1 Deﬁnition. For P P(M) set ∈ S(P, r) = x P x = r ∈ || || n o

3. GEOMETRIC INTERPRETATION 185 and when S(P, r) Ω set: ⊂ S (P, r) = exp(S(P, r)).

We call S (P, r) the circle of radius r in P.

3.2

Note. Let us note that if P is a subspace of Tm(M) and expm is r′-O.K. where r < r′, then the image of the circle S(P, r) is a compact subman- ifold of M of dimension one. So for r suﬃciently small by 5.3 we can consider the length (lg(S (P, r))). To compute lg(S (P, r)) let us take an orthonormal basis x, y of P and the parametric representation { } (6.3.3) v : [0, 2π] α r.e.(α) S(P, r) ∋ → ∈ where e(α) = cos α x + sin α y. · · Now let us set

(6.3.4) S = exp v. ◦ Then we have

2π

(6.3.5) lg(S (P, r)) = S ′ dα. Z || || 0 We have T S ′ = exp v′ ◦ and by the deﬁnition of v π ζ(v′(α)) = r.e. α + 2 and hence

= T 1 + π (6.3.6) S ′ exp ζr.e(− α) r.e. α . 2 186 Sectional Curvature

3.7 190

Now if we denote by hα the Jacobi ﬁeld along γe(α) such that

π hα(0) = 0 and (DPhα)(0) = e α + , 2 then by (6.23) and by (6.3.6) we have

S ′(α) = hα(r).

Then by (6.19) iii) we have

3 r π π 3 (6.3.8) S ′(α) = r + g(R(e(α), e(α + ))e(α), e(α + )) + 0(r ) = || || 6 2 2 r3 = r A(P) + 0(r3). − 6

Hence by (6.3.5) we have

πr3 (6.3.9) lg(S (P, r)) = 2πr A(P) + 0(r3). − 3

Therefore we get

3 (6.3.10) A(P) = lim (2πr lg(S (P, r))). r 0 πr3 − →

3.11

Application . As an application of the above formula we note the following: Let (M, g) be an r.m. and let us denote the A of the manifold (M, kg) where k > dbyA′. Then we have by (5.4) and (6.3.10).

1 (6.3.12) A′(P) = A(P). k2 4. A CRITERION FOR LOCAL ISOMETRY 187

3.12 Example. Now let us calculate A for (Sd, can). We have

exp(S (α)) = cos r e + + sin r.e(α) · d 1 by the proof of (5.12) and hence

S ′ = sin r. || || Therefore 191

lg(S (P, r)) = 2π sin r r3 = 2π r + 0(r3) − 6 ! and hence A(P) = 1 P. Instead of proceeding with circles we can pro- ∀ ceed, in a similar manner, with discs B(m, r) P (for some P P(M)) ∩ ∈ and thus get 12 (6.3.15) A(P) = lim (πr2 ar(exp B(m, r) P)). r 0 πr4 − ∩ → The proof is left to the reader as an exercise.

4 A criterion for local isometry

In this article we give a result of Elie Cartan which gives a criterion for local isometry. To start with we shall prove a result which is purely algebraic and asserts, essentially, that the sectional curvature determines the curvature tensor.

4.1

Proposition. Let V and V′ be vector spaces with euclidean structures g and g′, and let v : V V′ → 188 Sectional Curvature

be a euclidean isomorphism. Let R and R′ be bilinear maps

R : V V End(V) × → R′ : V′ V′ End(V′) × → such that properties C.T.1, C.T.2, C.T.3 (and hence C.T.4) hold for R and R′. Let A and A′ be the maps deﬁned by means of R and R′ as in (6.1.2). 192 Then

i) if an element x in V is such that for every y which is not linearly dependent on x we have

(6.4.2) A(x, y) = A′(v(x), v(y))

then R′(v(x), v(y))v(x) = v(R(x, y)x) y V. ∀ ∈ and

ii) if A′(v(x), v(y)) = A(x, y) for every x and y in V which are linearly independent then

R′(v(x), v(y))v(x) = v(R(x, y)z) x, y, z V. ∀ ∈

Proof. i) Let y and z be elements in V such that the pairs x, y , { } x, y and x, y + z are linearly independent. Then by hypothesis { } { } we have

(6.4.2) A(x, y + z) = A′(v(x), v(y + z)) = A′(v(x), v(y) + v(z)). − Hence using trilinearity of R and C.T.4 we have by deﬁnition of A and A′ A(x, y) + A(x, z) 2g(R(x, y)x, z) (6.4.3) − = ... x (y + z) 2 || ∧ || A (v(x), v(y)) + A (v(x), v(z)) 2g(R(v(x), v(y))v(x), v(x)). ... = ′ ′ − v(x) (v(y) + v(z)) 2 || ∧ || 4. A CRITERION FOR LOCAL ISOMETRY 189

By (6.1.2) and the fact that V is a euclidean isomorphism and (6.4.3) we have

(6.4.4) g(R(x, y)x, z) = g′(R′(v(x), v(y))v(x), v(z)).

Again since v is a euclidean isomorphism we have

1 g(R(x, y)x, z) = g(v− (R(v(x), v(y))v(x)), z)

and, since g is non-degenerate, varying z, we have 193

(6.4.5) v(R(x, y)x) = R′(v, (x), v(y))v(x).

The proof is trivial in case x and y are linearly dependent (C.T.1).

ii) Substituting x + z for x in (6.4.5) and using (6.4.5) we have

v(R(x, y)z) + v(R(z, y)x) = R′(v(x), v(y))v(z) + R′(v(z), v(y))v(x).

i.e. (6.4.6) v(R(x, y)z) R′(v(x), v(y))v(z) = v(R(z, y)x) R′(v(z), v(y))v(x). − − − This means that if we denote the map

(x, y, z) v(R(x, y)z) R′(v(x), v(y))v(z) → − by θ we have

(6.4.7) θ(x, y, z) = θ(z, y, x). − But by C.T.1 we also have

(6.4.8) θ(x, y, z) = θ(y, x, z). − Now by (6.4.7) and (6.4.8) we have

(6.4.9) θ(x, y, z) = θ(z, y, x) = θ(y, z, x) − i.e. a cyclic permutation of x, y, z does not alter θ. Hence we have

3θ(x, y, z) = θ(x, y, z) + θ(y, z, x) + θ(z, x, y). 190 Sectional Curvature

But by using C.T.2. we have

θ(x, y, z) + θ(y, z, x) + θ(z, x, y) = 0.

Hence θ(x, y, z) = 0.

194 Example. As an example let us note that if A(M) = 0 then R = 0. { } 4.10 In what follows, if x U(M), we denote the parallel transport along the ∈ geodesic γ : t exp tx x → from γx(0) to γx(t) by τ(x, t). Then the main theorem can be stated as follows:

4.11 Theorem. Let (M, g) and (N, h) be two r.m.’s. Let m and n be points in M and N such that there exists

i) a euclidean isomorphism u between (Tm(M), gm) and (Tn(N), gn) and

ii) an r > 0 such that exp is r-O.K. and B(n, r) NΩ and further m ⊂ that for every x, y in Um(M) which are linearly in dependent and every t in ]0, r[ we have (6.4.12) NC(τ(u(x), t)u(x), τ(u(x), t)u(y)) = MC(τ(x, t)x, τ(x, t)y).

Then there exists λ D(B(m, r), B(n, r)) such that ∈

λ∗(h) B(n, r) = g (B(m, r)). | | Proof. First let us prove a lemma. 4. A CRITERION FOR LOCAL ISOMETRY 191

4.13 Lemma. Under the assumptions of the above theorem, if x U (M) ∈ m and h and h1 are Jacobi ﬁelds along γx and γu(x) respectively such that

(6.4.14) h1(0) = u(h(0)) and (DPh1)(0) = u((DPh)(0)) then 1 h (t) = (τ(u(x), t) u τ(x, t)− )h(t). 1 ◦ ◦ Proof. With the notations (2.7.9) and (2.8.16), by the deﬁnition of τ(x, t) 195 we have d2h (6.4.15) = MR(t)h(t) = τ(x, t) 1(MR(γ (t), h(t)γ (t)). 2 − ′x ′x dtb b b b Since γ′x(t) is a parallel lift of γx(t) and τ(x, t) is the parallel transport along γx we have = = (6.4.16) γ′x(t) τ(x, t)γ′x(0) τ(x, t)x. Hence, by (6.4.15) we have

d2h (6.4.17) = τ(x, t) 1(MR(τ(x, t)x, τ(x, t)h(t))(τ(x, t)x)). 2 − dtb Let us note that the map b

1 (6.4.18) v = τ(u(x), t) u τ(x, t)− t ◦ ◦ From Texp tx(M) to Texp t u(x)(N) is a euclidean isomorphism since parallel transport preserves the· euclidean structure (see (4.9)) and since u is a euclidean isomorphism. Now applying the ﬁrst part of (4.1) to v we have

NR(τ(u(x), t)u(x), τ(u(x), t)u(h(t)))(τ(u(x), t))u(x)) = ... (6.4.19) M = vt R(τ(x, t)x, τ(x, tb)h(t))(τ(x, t)x)) . h i Hence since u is linear we have b d2(u h) d2h ◦ = u = ... 2 ◦ 2 dt b dt 192 Sectional Curvature

(6.4.20) 1N ... = τ(u(x), t)− R(τ(u(x), t)u(x), τ(u(x), t)u(h(t))τ(u(x), t)u(x).

196 Now let us note, as above (6.4.16), that b = γu′ (x)(t) τ(u(x), t)u(x) and hence by (6.4.20)

d2(u h) (6.4.21) ◦ = NR(t)((u h)(t)). 2 ◦ dt b b b But by (2.8.4) we have

d2h (6.4.22) 1 = N R(t)h (t). 2 1 dtb b b Hence u h and h satisfy the same second order diﬀerential equation; ◦ 1 and further the assumptions (6.4.14) are the same as b b dh dh d(u h) h (0) = (u h)(0) and t (0) = u (0) = ◦ (0) 1 ◦ dt  dt  dt b  b  b b b   since u is linear and this means that h and u h satisfy the same diﬀer- 1 ◦ ential equation with the same initial conditions. Hence b b (6.4.23) h = u h 1 ◦ and hence b b

(6.4.24) h (t) = v h(t) t. 1 t ◦ ∀

Proof of the theorem. By (2.6) we have to deﬁne λ by:

N M 1 (6.4.25) λ = exp u ( exp B(m, r))− ; n ◦ ◦ m | we check that λ is indeed a local isometry. To show this we have only 197 to prove that the map 4. A CRITERION FOR LOCAL ISOMETRY 193

T λ : T (M) Tλ (N) a a → (a) is a euclidean isomorphism a B(m, r). We ﬁrst note, since u is linear, ∀ ∈ that

T 1 1 (6.4.26) u ζ− = ζ− u, x T (M). ◦ x u(x) ◦ ∈ m Now, for a = m, using (4.6) we have

T T 1 1 λ = ζ0 u ζ− = ζ0 ζ− u = u m n ◦ ◦ 0m n ◦ 0n ◦ and hence we are through in case a = m. In the other case t ]0, r[ ∃ ∈ and x U (M) such that ∈ m (6.4.27) a = M exp (t x). m · Now let z be any element of T (M). Since M exp is r-O.K. and a a m ∈ B(m, r) there exists an ω in Tt x(Tm(M)) such that · = M T (6.4.28) z expm(ω).

Now set

(6.4.29) y = ζ(ω).

By (6.23) we have

(6.4.30) z = h( t x ) = h(t) || · || where h is the Jacobi ﬁeld along γx such that y y (6.4.31) h(0) = 0 and D h(0) = = . P t x t || · || Further by the deﬁnition of λ and y we have (6.4.32) T N T T 1 N T 1 λ (z) = ( exp ) u (ζ− (y)) = ( exp ) ζ− (u(y)) by (6.4.25) n ◦ ◦ tx n ◦ tx and hence, again by (6.23), we have 198 194 Sectional Curvature

(6.4.33) λT (z) = h ( t u(x) ) = h (t) 1 || · || 1

(since u is a euclidean isomorphism) where h1 is the Jacobi ﬁeld along λu(x) with

u(y) u(y) y (6.4.34) h (0) = 0 and D h (0) = = = u . 1 P 1 t u(x) t t || · || Now considering vt, h, h1 and applying (because of (6.4.31) and (6.4.34)) the lemma we have

(6.4.35) h (t) = v h(t) t 1 t ◦ ∀

where vt is given by (6.4.18). But vt is a euclidean isomorphism and hence the result follows by (6.4.33), (6.4.33) and (4.36).

4.35 Corollary. Any two r.m.’s of constant sectional curvature k are locally isometric. Hence, by (6.3.12) any r.m. of constant sectional curvature k is locally isometric to

d 1/2 (S , k− , can) if k > 0 (Rd, ǫ) if k = 0 d 1/2 (R , k− , hyp) if k < 0.

4.36 Remarks. 1) For a global result on manifolds of constant sectional curvature see (7)

2) For a global result in general see [1]

3) In general the knowledge of sectional curvature (which is the same by (4.1) as that of curvature tensor) does not determine, even 199 locally, the r.s. In fact there exist r.m.’s (M, g) and (N, h) such that there exists a diﬀeomorphism λ such that 5. JACOBI FIELDS IN SYMMETRIC PAIRS 195

i) M A(P) = N A(λT (P)) P P(M) and ∀ ∈ ii) λ is not an isometry.

The theorem (4.11) says that if for all geodesics through a point, λ preserves the sectional curvature of planes corresponding under parallel transport along geodesics, then λ is a local isometry at this point.

5 Jacobi ﬁelds in symmetric pairs

In this article we assume that (M, g) is an r.m. associated to a symmetric pair (G, H) and follow the notation of (3) and of (4). Let f be a geodesic in M. Then by (1.8) the parallel transport τ( f ′, t) along f is given by the tangent maps of isometries. Hence by (1.6) we have the:

5.1 Proposition. Let m be a point of (M, g). Then

i) τ(x, t)(R(x, y)x) = R(τ(x, t)x, τ(x, t)y)τ(x, t)x x U (M), y ∀ ∈ m ∀ ∈ Tm(M) and ii) A(x, y) = A(τ(x, t)x, τ(x, t)y) if x, y are linearly independent.

5.2 Remark. The converse of (5.1) ii) is true: an r.m. where parallel transport preserves sectional curvature is locally isometric to a symmetric space: see Chapter IV of [14]. The relation (5.1) i) gives a complete in- 200 formation about Jacobi ﬁelds of M. For now the formula (6.4.15) gives

d2h (6.5.3) = R(x, h(t))x. 2 dtb b Now let us introduce the endomorphism R(x) of Tm(M) by setting:

(6.5.4) R(x)y = R(x, y)x y T (M). ∀ ∈ m 196 Sectional Curvature

Then by ((5.5.5) C.T.4) we have

g(R(x)y, z) = g(R(x, y)x, z) = g(R(x, z)x, y) = g(R(x)z, y)

and hence R(x) is symmetric. Now by the reduction process of quadratic forms to diagonal form we see that there exist real numbers λ1,...,λd and an orthonormal basis x ,..., x of T (M) such that { 1 d} m

(6.5.5) R(x)xi = λi xi, i = 1,..., d.

But since R(x)x = 0, and x is in Um(M) we can assume that x1 = x and λ1 = 0. For any lift h of f into T(M) there correspond functions ψ1,...,ψd such that h(t) = ψi(t)xi Xi b and by (7.10) we have

d2h(t) d2ψ (t) = i x . 2 2 · i dtb Xi dt Further if h is a Jacobi ﬁeld then by (6.5.3) we have

2 d ψi (6.5.6) = λiψi. dt2

201 In particular we deﬁne for i = 1,..., d a Jacobi ﬁeld hi by

hi(0) = 0 and (DPhi)(0) = xi.

Then we have

( λ ) 1/2 sin( √ λ t) x if λ < 0 − i − · − i · · i i (6.5.7) h (t) = t x if λ = 0 i  i i  · 1/2 (λi)− sh( √λi t) xi if λi > 0.  · · ·  Let us note that by ((5.5.5) C.T.4)

(6.5.8) λ = g(R(x)x , x ) = g(R(x, x )x, x ) = A(x, x ). i i i i i − i 6. SECTIONAL CURVATURE OF S.C.-MANIFOLDS 197

6 Sectional curvature of S.C.-manifolds

In this article we assume that (M, g) stands for an S.C.-manifold and follow the notations of (4). We already know (by (2)) that A(M) = 1 { } for (Sd, can) and (Pd(R), can). In the following proposition we think of the remark (5.6) for the case of P2(Γ): then “y is orthogonal to K x” · stands for “y is orthogonal to the ﬁbre through x”. Note that the proof below, for the case of P2(Γ), uses only (5.7).

6.1 Proposition. Let (M, g) be an S.C.-manifold, which is (Pn(K), can) with n > 1,K , R or P2(Γ). Then A(M) = 1, 4 ; { } more precisely, if x, y are R-independent: 202 i) A(x, y) = 4 if x, y are K-dependent ii) A(x, y) = 1 if y is orthogonal to K x. · Proof. ***** (6.6.2). By (5.4) all geodesics through m0 are closed and are of length π. By (8.26) every Jacobi ﬁeld along a geodesic γx can be realised as the variations of vectors along a family of geodesic curves. Hence it follows that any Jacobi ﬁeld along a geodesic through m0 which vanishes at m for t = 0 vanishes again for t = π. Now let x U (M) 0 ∈ m0 and the corresponding objects x = x , x ,..., x , λ = 0, λ ,...,λ , h , h ,..., h { 1 2 d} { 1 2 d} { 1 2 d} be as in (5). Hence hi(π) = 0 implies, necessarily by (6.5.7) that λi < 0 and we have 1/2 (6.6.3) h (t) = ( λ )− sin( λ t) x . i − i · − i · · i p Again hi(π) = 0 implies λi 1, 4, 9,... . Now by (8.32) and (5.12) ∈ {− − π − } we know h (t) cannot vanish on ]0, [, hence λ = 1 or 4. i 2 i − − This implies already that A(M) 1, 4 for the eigen values of R(x) ⊂{ } on the orthogonal complement of x are 1 or 4 (by (6.5.8)). 198 Sectional Curvature

(i) Now let xi be K-dependent on x and let h be the Jacobi ﬁeld along γx given by the variations of vectors along the one parameter family of geodesics

γ(cos α) x+(sin α) xi (t). · · 203 Then, since each (cos α) x + (sin α) xi is K-dependent on x by · · π (5.7) (ii), it follows that all of them meet at t = and hence 2 π hi = 0, 2 hence λ = 4. i − (iii) Suppose conversely that λ = 4 and set i − π e(α) = (cos α) x + (sin α) xi and S (α) = exp e(α) . · · 2 · Then (see (6.3.6)) we have

T 1 π π α = ζ π α + . S ′( ) expm0 − e(α) e 2 · 2 · 2 But by (6.23) we have π (6.6.6) S ′(α) = h 2 where h is the Jacobi ﬁeld along γe(α) satisfying the equations π h(0) = 0 and DPh(0) = e α + . 2 π Since the subspaces generated by the pairs e(α), e( + α) and { 2 } x, x are the same we have { i} π A e(α), e + α = A(x, xi) = 4. 2 Since 4 is the maximum value of the quadratic form A(e(α), ), it π ∗ follows that e(α + ) is an eigen vector of R(e(α)). Hence in view 2 of (6.6.3) we have π h = 0 2 7. VOLUMES OF S.C. MANIFOLDS 199

and hence by (6.6.6) we have

S ′(α) = 0 α. ∀ π Hence S (α) = S (0) = exp( x) for every α, and hence in partic- 204 2 · ular π π exp xi = S (1) = S (0) = exp x . 2 · 2 · Then by (5.7) we see that x, x are K-dependent. We conclude { i} that λ = 4 if and only if x is K-dependent on x. Since x i − i { i} is an orthogonal basis, we see that λ = 1 if and only if x is i − i orthogonal to K x. Hence R(x) is 4 times the identity on the · − orthogonal complement of x in Kx and 1 times the identity on − the orthogonal complement of Kx(in T ). Thus, if K , R, i m0 ∃ such that λ = 4; if n > 1, then i such that λ = 1. i − ∃ i −

6.8 Remark. Let (M, g) be any S.C.-manifold and x U(M). Then (upto a ∈ permutation of the i’s) we have for the associated set λ (i 2) { i} ≥ Pd(R), Sd(R) : λ = 1 i i − ∀ Pn(C) : λ 4, λ = 1 i 3 2 − i − ≥ Pn(H) : λ = λ = 4, λ = 1 i 5 2 3 − i − ≥ P2(Γ) : λ = 4 : i = 2,..., 8, λ = 1 : i = 9,..., 16. i − i − 7 Volumes of S.C. manifolds

Now we give an example of calculation. We show how the exponential map and Jacobi ﬁelds can be used to estimate the volume of an r.m. In sharper recent results this has been systematically done to get bounds for Vol(M, g) in terms of those of A(M) (see [33], Ch. 11)). Here we shall be content with the treatment of a case where the computation can be explicitly carried out and, at the same time, is simple enough to exhibit 205 200 Sectional Curvature

clearly the method involved. The same method works in more general situations but we do not have space here to handle them. We compute the volume of an S.C. manifold. Since we know (see (5.6)) that Vol(Sd, can) = 2 Vol(Pd(R), can), we may assume that (M, g) is diﬀerent from Sd. First let us note that the fact that any point m can be joined to m0 by a geodesic (see proof of (5.4)) together with the fact that each geodesic is of length π implies that , π/ = . expm0 (B(m0 2)) M But exp is a C -map and the set B(m , π/2) B(m , π/2) has measure m0 ∞ 0 − 0 zero. Hence we have

, = , π/ Vol(M g) Vol(expm0 (B(m0 2)) = , π/ Vol(expm0 (B(m0 2)) = Vol(B(m0, π/2)). π/ By (5.12) expm0 is 2-O.K., and if we ﬁx an orientation on Tm0 (M) and pass to the canonical volume form σ on B(m0, π/2), (see (5))we have

Vol(B(m0, π/2)) = σ = (exp )∗σ. Z Z m0 π B(m0,π/2) B(m0, 2 )

206 To compute the latter we look upon B(m, π/2) 0 as diﬀeomorphic to −{ } the product manifold

W =]0, π/2[ U (M), × m0 under the map

n : W (t, x) t x B(m , π/2) 0 . ∋ → · ∈ 0 −{ } Here n clearly is a diﬀeomorphism and since the point m0 has zero measure we have

vol(M, g) = (exp )∗σ Z m0 B(m0,π/2) 7. VOLUMES OF S.C. MANIFOLDS 201

= (exp n)∗σ. Z m0 ◦ W Now let us introduce two projection maps on W deﬁned by the equations

p(t, x) = t and q(t, x) = x.

Denoting the canonical volume form on Um0 (M) by θ we get a volume form ω = p∗(dt) q∗(θ) ∧ on W. But then there exists a function ϕ such that

(exp n)∗σ = ϕω. m0 ◦ and we will be through if we know ϕ.

7.1 Lemma. With the above notation α sin 2t d α 1 ϕ(t, x) = (sin t) − − 2 !

d 1 for every x in S − and t in ]0, π/2[, where

α = 0 for Pd(R); α = 1 for Pd(C); α = 3 for Pd(H); α = 7 for P2(Γ).

207

Proof. Let us follow the notation of article 6. The tangent space T(t,x) (W) is in a natural way isomorphic to T ( ]0, π/2[ ) T (U (M)). But t × x m0 (see 2.9) the vectors ζ 1 x ,...,ζ 1 x form a basis for T (U (M)) and { x− 2 x− d} x m0 hence 1 1 (P, 0), (0, ζ− x ),..., (0, ζ− x ) { x 2 x d } is a basis for T(t,x)(M) such that 1 1 = ω((P, 0), (0, ζx− x2),..., (0, ζx− xd)) 1. 202 Sectional Curvature

Hence we have 1 1 ϕ(t, x) = ((exp n)∗)((P, 0), (0, ζ− x ),..., (0, ζ− x )) = m0 ◦ x 2 x d = T T , , T T , ζ 1 ,..., (expm0 (n (P 0) expm0 (n (0 x− x2) ) ..., T T , ζ 1 . expm0 (n (0 x− xd))) Since the map n : (t, x) t x → · is bilinear, we have T 1 = 1 T = 1 n (0, ζx− xi) t ζt−x xi and n (P, 0) ζtx− x. · · Further by (6.32) the images T ζ 1 , T ζ 1 ,... expm0 ( tx− x1) expm0 ( tx− x2) are mutually orthogonal and T 1 1 exp (ζ− x) = ζ− x = 1. || m tx || || tx || 208 Hence we have T 1 ϕ(t, x) = exp (ζ− t x ) . || m0 tx · i || Yi>2 But by (6.23) we have T ζ 1 = = expm0 ( t−x (t xi)) hi( tx ) hi(t) · · || || where hi is the Jacobi ﬁeld along γx such that t x h (0) = 0 and (D h )(0) = · i = x . i P i t x i || · i|| But then by (6.5.7) we have 1/2 h (t) = ( λ )− sin( λ t) x i − i · − i · · i p and hence 1/2 (6.7.2) ϕ(t, x) = ( λ )− sin( λ t) − i · − i · Yi>2 p and we are through by (6.8). 7. VOLUMES OF S.C. MANIFOLDS 203

Now let us note that since all euclidean structures of the same dimension are isomorphic we have

d 1 Vol(U (M), g ) = Vol(S − , can) m M. m m ∀ ∈ We set

(6.7.3) Vol(Sd, can) = d .

Hence by (3.17):

π/2

ϕ(t, x) p∗(dt) q∗(θ) =  ϕ(t, x) dt θ = · ∧ · · Z Z Z  W Um0 (M)  0    π/2   =  θ ϕ(t, x) dt  Z  · Z · Um (M)  0  0    since ϕ is independent of x. 209 Hence

π/2 α ¨ sin 2t d α 1 (6.7.4) Vol(M, g) = d 1 sin − − t dt . − ©· Z 2 ! · · 0

First (6.7.4) gives for the d ’s the recurrence formula:

π/2 d ¨ d 1 d = 2 Vol(P (R), can) = 2 d 1 sin − t dt . · − ©Z · 0

π/2 d 1 The value of sin − t dt is well known. Using this, we get 0 · R ¨ (2π)n ¨ πn (6.7.5) 2n = 2 , 2n 1 = 2 © · (2n 1) (2n 3) ... 3.1 − © · (n 1)! − · −n − n 2n (2π) 2n 1 π Vol(P (R), can) = , Vol(P − (R), can) = (2n 1) (2n 3) ... 3.1 (n 1)! − · − − 204 Sectional Curvature

Note, in particular, that (6.7.6) 2 = Vol(S2, can) = 4π, Vol(P2(R), can) = 2π. For Pn(C) we compute π/2 π/2 sin 2t 2n 2 2n 1 1 sin − t dt = sin − t cos t dt = ; Z 2 · · Z · · 2n 0 0 with (6.7.5) this gives πn (6.7.7) Vol(Pn(C), can) = n! 210 By the same kind of straightforward computations, π2n (6.7.8) Vol(Pn(H), can) = , (2n + 1)! π8 (6.7.9) Vol(P2(Γ), can) = . 11.10.9.8.7.6.5.4.

7.10 Remark. Using (3.4.10) and (4.19), and the formula (3.17) one would get ¨ Vol(Pn(C), can) = 2n + 1 / 1 , © ¨ Vol(Pn(H), can) = 4n + 3 / 3 , © which agrees with (6.7.5), (6.7.6), (6.7.7). Note also that P2(Γ) = 23 / 7 . But here it is impossible to get that formula through a ﬁbration because one knows from algebraic topology (non existence of elements of Hopf invariant one) that there does not exist a ﬁbration S23

S7 P2(Γ) 7. VOLUMES OF S.C. MANIFOLDS 205

Next, we give a formula which will be used later on:

2 ¨ (6.7.11) I = x x0 σ = d 1 , Z | · |· d − © x Sd ∈

d where σ is the canonical volume form of (S , can) and x0 a ﬁxed point in Sd.

Proof. If x is at the distance t from x on Sd then x x = cos t. We 211 0 · 0 make the computations on Pd(R) instead of Sd using the remark that x x x is invariant under the antipodal map of Sd. → | · 0|

Thus I = 2 p( x x ) σ where σ denotes the volume element on pd(R) | · 0| · pd(R). We followR the computation of Vol(Pd(R), can) as above, taking p(x0) = m0. This gives

π/2 I = 2  ϕ(t, x) cos t dt = ... · | |· Z Z  Um0 (M)  0    π/2  ¨ d 1 2 ¨ = 2 d 1 sin − t cos t dt = d 1 . − ©Z · · d − © 0 206 Sectional Curvature

8 Ricci and Scalar curvature

In this article we assume that the dimension d of the manifold (M, g) is greater than one. For every point m of M and positive number r we get

S (m, r) = x T (M) x = r { ∈ m || || } and when S (m, r) Ω we set ⊂ S (m, r) = exp(S (m, r)).

If expm is r-O.K. then for every r′ in ]0, r[S (m, r′) is a sub manifold of 212 M of dimension d 1. Now we calculate the volume −

σ(m, r′) = Vol(S (m, r′), g S (m, r′)). |

First let us relate Um(M) and S (m, r) by deﬁning the map

r′ : U (M) x r′ x S (m, r′) m ∋ → · ∈

so that if expm is r-O.K. then taking some orientation on Tm(M) we get 1 by means of exp B(m, r) and r′ a volume form σ on S (m, r ). Then m−0 | ′ we have

σ(m, r′) = σ = (exp r′)∗σ. Z Z m ◦ S (m,r′) Um(M)

But if we denote the canonical volume form on Um(M) by θ then there exists a ϕ in F(U(M)) such that

(6.8.1) (exp r′)∗σ = ϕ θ, m ◦ · and we will be through if we can compute ϕ. Now let x = x , x ,..., x { 1 2 d} be an orthonormal basis of Tm(M). Then by the deﬁnition of θ we have 1 1 = θ(ζx− x2, ζx− x3,...) 1.

Now let us note that

T 1 1 r′ (ζ− x ) = ζ− (r′ x ) for i = 2,..., d x i r′ x i 8. RICCI AND SCALAR CURVATURE 207

and that Gauss’ Lemma (see (6.24)) gives that the vectors

T 1 = T 1 (expm r′) (ζx− xi) expm(ζ− r′ xi), i 2 ◦ r′ x ≥ T 1 213 are all orthogonal to the vector exp (ζ− x) and hence are tangent to m r′ x S (m, r′). Hence we have

1 1 ϕ(x) = ((exp r′)∗σ)(ζ− x ,...,ζ− x ) m ◦ x 2 x d T 1 T 1 = σ(exp (ζ− rx ),..., exp (ζ− rx ). m r′ x 2 m r′ x d But by (6.23) T 1 exp (ζ− r′ x ) = h (r′) m r′ x i i

where hi is the Jacobi ﬁeld along γx satisfying the conditions r x h (0) = 0 and (D h )(0) = ′ i = x . i P i r x i || ′ i|| Hence

ϕ(x) = σ(h (r′),..., h (r′)) = h (r′) ... h (r′) = ... (by (4.9)) 2 d || 2 ∧ ∧ d || 1 1 = τ(x, r′)− h (r′) ... τ(x, r′)− h (r′) = ... || 2 ∧ ∧ d || (6.8.2)

= h (r′) ... h (r′) . || 2 ∧ ∧ d || But by (8.18)b we have b

r 3 h (r ) = r x + ′ R(x, x )x + 0(r 3). i ′ ′ i 6 i ′ Hence we have (for R as introduced in (5)):

d+1 d 1 (r′) ϕ(x) = (r′) − (x ... x ) + || 2 ∧ ∧ d 6 d d+1 (6.8.3) x2 ... xi 1 R(x)xi xi+1 ... xd + 0(r′ ) . ∧ ∧ − ∧ ∧ ∧ || Xi=2

But 214 208 Sectional Curvature

d x2 ... xi 1 R(x)xi xi+1 ... xd = ... ∧ ∧ − ∧ ∧ ∧ ∧ Xi=2 d = g(R(x)x , x )(x x ... x ) i i 2 ∧ 3 ∧ ∧ d Xi=2 = (Trace (R(x))) (x ... x ). · 2 ∧ ∧ d 8.4 Deﬁnition. We set for x T(M): ∈ Ric(x) = Trace(R(x)) − and call it the Ricci curvature of (M, g) at x. Then we have 2 d 1 (r′) 2 (6.8.5) ϕ(x) = (r′) − 1 Ric(x) + 0((r′) ) − 6 !

8.6 Note. If x U(M), we have, since x and x are orthonormal, ∈ i

g(R(x)xi, xi) = g(R(x, xi)x, xi) = A(x, x ). − i Hence x = x , x ,..., x are orthonormal and we have { 1 2 d} d Ric(x) = A(x, xi). Xi=2

8.7 Remarks. i) Ric is a quadratic form on C (M). To see this let us deﬁne for X, Y C (M) a map R(X, Y) : C (M) C(M) by the ∈ → equation R(X, Y)(Z) = R(X, Z)Y. 8. RICCI AND SCALAR CURVATURE 209

215 Then the trace of the map R(X, Y) is a symmetric bilinear map on C (M) C (M): for, if x ,..., x is an orthonormal base of × { 1 d} Tm(M), then we have

Trace(R(x, y)) = g(R(x, xi)y, xi) = Xi

g(R(y, xi)x, xi by C.T.4 = Trace(R(y, x)). Xi Hence we have

Ric(X) = Trace(R(X)) = Trace(R(X, X)).

ii) Since Ric can be considered as a quadratic form on Tm(M) the trace of Ric gives a function on M.

8.8 Deﬁnition. We set Γ= Trace(Ric) and call it the scalar curvature of M. Then if x is an orthonormal basis of T (M) we have { i} m

(6.8.9) Γ(m) = Ric(xi) = A(xi, x j). Xi Xi, j

In particular if the dimension is 2 we have Γ= 2A where A, the sectional curvature, is considered as a function on the manifold.

Now let us take up the calculation of σ(m, r). By (6.8.1) we have

σ(m, r) = ϕ(x)θ = Z Um(M) 2 d 1 r′ 2 (6.8.10) = (r′) − 1 Ric(x) + 0(r′) = Z − 6 ! Um(M) 210 Sectional Curvature

d+1 d 1 ¨ (r′) d+1 = (r′) − d 1 Ric(x) θ + 0((r′) ). · − ©− 6 Z · Um(M)

Since Ric is a quadratic form on Tm(M), there exist constants λ1,...,λd 216 1 and a system of orthonormal vectors x1,..., xd such that, if x = x x1 + d + x xd, then ··· i 2 Ric(x) = λi(x ) . Then X i 2 Ric(x)θ = λi (x ) θ. Z Xi Z Um(M) Um(M) But since by symmetry

(xi)2θ = (x j)2θ i, j Z Z ∀ Um(M) Um(M) we have

d (x1)2θ = ((x1)2 + (x2)2 + + (xd)2)θ = ... · Z Z ··· Um(M) Um(M) = θ since x = 1. Z || || Um(M) Hence we have ¨ ¨ d 1 d 1 (6.8.11) Ric θ = λi − ©= − ©Trace(Ric) = ... Z X d d · Um(M) ¨ d 1 =Γ(m) − ©. · d Hence by (6.8.10) we have 2 ¨ d 1 (r′) 2 (6.8.12) σ(m, r′) = d 1 (r′) − 1 Γ(m) + 0(r′) . − © − 6 ! Now by (6.8.12) we have

Γ = 6d d 1 ¨ (6.8.13) (m) lim + (r − d 1 σ(m, r)). r 0 rd 1 · − ©− → 8. RICCI AND SCALAR CURVATURE 211

8.14

Remark. When the dimension of the manifold is two the Gaussian cur- 217 vature has deep implications on the topology of the manifold (see Gauss- Bonnet formula (6.1)). But for dimension greater than two the situation is quite diﬀerent. For example on S3 we can deﬁne a homogeneous r.s. for which Γ is constant and is of given sign. Further let M be a compact manifold which is oriented by a form ω. Then there exists an r.s. g on M such that, if we denote the canonical orientation of (M, g) by ω1, then

Γω1 Z M has a preassigned sign, (see [3]). However, if Γ is everywhere positive then there are certain topological restrictions on M (see [20]: theorem 2, p.9).

Chapter 7 The metric structure

1.1 218 In this chapter we assume that all the manifolds we consider are connected. 1. Identity of balls

1.2 Deﬁnition. For every pair of points m and n in (M, g) we set

P(m, n) = C C is a path from m to n . { | } Since we have assumed that M is connected it follows that M is arc wise connected and hence that

P(m, n) , m, n M. ∅ ∀ ∈ Now set

(7.1.3) d(m, n) = inf lg(C). C P(m,n) ∈ 1.4 Then, since the length of a path is non-negative we have

d(m, n) 0; ≥ 213 214 The metric structure

since every path from m to n gives rise to a path from n to m with the same length we have d(m, n) = d(n, m); and since a path from m to p and another from p to n give rise to one from m to n we have

d(m, n) d(m, p) + d(p, n). ≤ Hence d has all the properties of a metric structure except perhaps the one which asserts that d(m, n) = 0 only if m = n. Now we shall proceed 219 to show that this property is also valid; in fact, we shall prove much more, namely, that locally geodesics realise the distance d.

1.5 Now throughout this article, let us ﬁx a point m and a positive number r such tht expm is r-O.K., and set:

1 λ = (exp B(m, r))− . m | 1.6

Lemma. Let n and n′ be points in B(m, r) and let the map

C : [0, t ] M 0 → be a path such that

i) the image of C does not contain m,

ii) the image of C lies wholly in B(m, r).

Then we have lg(C) λ(n′) λ(n) , ≥ || || − || ||

and equality holds if and only if

i) the three points 0m, λ(n) and λ(n′) are in a straight line and The metric structure 215

ii) λ C lies in the segment [λ(n), λ(n )] and is injective. ◦ ′ Proof. Since C is made up of ﬁnite number of curves and for any three real numbers a, b and c we have

a c a b + b c | − | ≤ | − | | − | we need only prove that results for curves. So let us assume that C is a curve: C D([0, t ], M), ∈ 0 and set 220

(7.1.8) b = λ C. ◦ b Since by i) 0 is not in the image of C the map makes sense and we m b write x for it: || || b (7.1.9) x = b || || Then we have

1 (7.1.10) b′ = b ′ ζ− x + b x′. || || · x || || By the deﬁnition of x we have x = 1 and hence we have || || 1 = g(ζx− x, x′) 0 and hence by Gauss’ lemma (see (6.24)) we have

T 1 T (7.1.12) g(exp ( b ζ− x), exp ( b x′)) = 0. m || || · x m || || · Now by (7.1.8) and (7.1.10) we have

2 T 2 C′ = exp b′ = || || || m ◦ || (7.1.13) T 1 2 T 1 T = exp ( b ′ ζ− x) + 2g(exp ( b ′ζ− x), exp ( b x′)) + || m || || · x || m || || x m || || · T 2 + exp ( b x′) = || || || · || 216 The metric structure

T 1 2 T 2 = exp ( b ′ ζ− x) + exp ( b x′) , by (7.1.12). || m || || · x || || m || || || But by (6.32) i) we have

T 1 exp (ζ− x) = x = 1 || m x || || || and hence

2 2 T 2 2 (7.1.14) C′ = ( b ′) + exp ( b x′) , ( b ′) || || || || || m || || || ≥ || || 221 equality holding if and only if expT x = 0. || m ′|| 1.15

Since expm is r-O.K., this is equivalent to saying that

C′ b ′, || || ≥ || ||

and that equality holds if and only if x′ = 0. Now we have

t0 t0

(7.1.16) lg(C) = C′ dt b ′ dt λ(n′) λ(n) . Z || || ≥ Z || || ≥ || || − || || 0 0

1.17 Further the last inequality becomes equality if and only if b is of the || ||′ same sign.

1.18 Hence by (1.15), (7.1.16) and (1.17) we have

lg(C) λ(n′) λ(n) ≥ || || − || ||

where equality holds if and only if x = 0 and b is of the same sign. ′ || ||′ This means that equality holds if and only if x is constant, and b is injective. This means that b should lie on the line joining 0m and λ(n) and b is injective. The metric structure 217

1.19 Note . In the case λ C lies on a line through 0 and is injective we ◦ m say, for simplicity, that C is a monotonic image by expm of the segment [(λ C)(0), (λ C)(t )]. ◦ ◦ 0 1.20

Lemma. If 0 < r′ < r, then 222 = exp(B(m, r′)) B(m, r′) and the latter is a compact set. ﬀ Proof. Since B(m, r′) is compact, exp continuous and M Hausdor , exp(B(m, r′)) is compact. In particular exp(B(m, r′)) is closed. But since exp(B(m, r )) exp(B(m, r )) it follows that ′ ⊂ ′ exp(B(m, r )) exp(B(m, r )). ′ ⊂ ′ In our case the other inclusion follows from the fact that exp is a diﬀeo- morphism on B(m, r).

1.21 Lemma. Let

C : [0, 1] M → 218 The metric structure

be a path in M such that

C(0) = m and C[(0, 1]) 1 B(m, r).

Then r ]0, r[, t ]0, 1[ such that ∀ ′ ∈ ∃ 0 ∈

λ(C(t )) = r′ and C([0, t [) B(m, r′). || 0 || 0 ⊂ Proof. Let J = s ]0, 1[ C(s) < B(m, r ) . By (1.20) and the conti- { ∈ | ′ } nuity of C, J is closed so that t0 = inf J > 0. We claim that t0 has = the property required above. By continuity: C(t0) limt t0 C(t); and −→< C(t) B(m, r ) t < t since, J being closed, t J. ⊂ ′ ∀ 0 0 ∈ Hence by (1.20): C(t ) B(m, r ) = exp(B(m, r )), i.e. λ(C(t )) 0 ∈ ′ ′ || 0 || ≤ 223 r . But t J implies that λ(C(t )) r . Further, C([0, t [) B(m, r ) ′ 0 ∈ || 0 || ≥ ′ 0 ⊂ ′ by deﬁnition of t0.

1.22 Lemma. Let n B(m, r) and let ∈ C : [0, t ] M 0 →

be a path, such that C(0) = n and C(t0) = m and whose image lies in B(m, r). Then

i) lg(C) λ(n) and ≥ || || ii) in i) the equality holds if and only if C is a monotonic image by expm of the segment [λ(n), 0m]. = = Proof. Let t1 inft [0,t0] t C(t) m. ∈ | Then for every suﬃciently small positive number ǫ we have

(7.1.23) lg(C) lg(C [0, t ]) lg(C [0, t ǫ]). ≥ | 1 ≥ | 1 − Since m < C([0, t ǫ]) by (1.6) we have 1 − (7.1.24) lg(C [0, t ǫ]) λ(n) (λ C)(t ǫ) . | 1 − ≥ || || − || ◦ 1 − ||

The metric structure 219

Since lg(C [0, t]) is a continuous function of t, taking limit as ǫ 0 we | → have

lg(C [0, t ]) lim λ(n) (λ C)(t ǫ) = | 1 ≥ || || − || ◦ 1 − || (7.1.25) = λ(n) λ(m) = λ(n) . || || − || || || ||

1.26 To prove ii) let us note that if equality holds above then (i) lg(C [t , t ]) = 0 and hence C([t , t ]) = m and | 1 0 1 0 { } (ii) in the following sequence of inequalities equality should hold everywhere:

lg(C [0, t ]) = lg(C [0, t ǫ]) + lg(C [t ǫ, t ]) | 1 | 1 − | 1 − 1 λ(n) (λ C)(t ǫ) + (λ C)(t ǫ) (by i) λ(n) . ≥ || || − || ◦ 1 − || || ◦ 1 − || ≥ || ||

Hence it follows that the above inequalities are equalities. Now 224 from (1.6) it follows from the ﬁrst equality that ǫ: ∀ (λ C)(t ǫ) [0 , λ(n)] and that C [0, t ǫ] ◦ 1 − ∈ m | 1 − is a monotone image by exp of the segment [λ(n), (λ C)(t ǫ)]. m ◦ 1 − 1.28 Lemma. Let n be a point in M and let

C : [0, t ] M 0 → be a path from m to n such that

C([0, t0]) 1 B(m, r).

Then lg(C) r. ≥ 220 The metric structure

Proof. Let 0 < r < r. Then by (1.21) t such that ′ ∃ 1

λ(C(t )) = r′ and C([0, t [) B(m, r′). || 1 || 1 ⊂ Hence, by (1.22) we have

lg(C) lg(C [0, t ]) λ(C(t )) = r′. ≥ | 1 ≥ || 1 ||

This being true for every r′ < r we are through.

1.29 Theorem. Let n B(m, r). Then ∈ i) d(m, n) = λ(n) , and the geodesic || || γ λ(n)/ λ(n) [0,λ(n)] || ||

realises the distance d(m, n), and

ii) it is the only one, i.e. if C P(m, n) is such that lg(C) = d(m, n), ∈ then C is a monotone image by expm of the segment [0m, λ(n)].

225 Proof. The geodesic deﬁned above has length λ(n) : see (4.3.3). || || Hence d(m, n) λ(n) . ≤ || || But if C P(m, n) then by (1.22) i) and (1.28) we have ∈ lg(C) λ(n) . ≥ || || Hence d(m, n) λ(n) and the ﬁrst part follows. ≥ || || To prove ii) let C P(m, n) and let lg(C) = λ(n) . Then by (1.28) ∈ || || the image of C lies in B(m, r) and the part ii) states the same thing as (1.22) ii).

1.30 Notation. We set

D(n, s) = n′ M d(n, n′) < s . { ∈ | } 2. THE METRIC STRUCTURE 221

1.31

Corollary. If expn is s-O.K. then B(n, s) = D(n, s).

Proof. We have only to prove that D(n, s) B(n, s); so let n D(n, s). ⊂ ′ ∈ Then s′ = d(n, n′) < s and hence by the deﬁnition of d there exists a C P(n, n ) such that s < lg(C) < s. Then by (1.28) the image of C ∈ ′ ′ has to lie in B(n, s). Hence

n′ B(n, s). ∈

2 The metric structure

2.1 Proposition. d is a metric structure.

Proof. By (1.4) we have only to prove that d(m, n) = 0 implies m = n. = Now let d(m, n) 0. Then there exists an r > 0 such that expm is r-O.K. Hence by (1.31) it follows that n B(m, r). Then by (1.29) the result ∈ follows.

2.2 226 Whenever we consider the manifold (M, g) as a metric space, it is this metric we deal with. Hence we speak of distance without any mention of the metric structure involved. In particular, we may speak of bounded sets, of completeness. . . , in a r.m. (M, g).

2.3 Remark. Throughout we have assumed that M is Hausdorﬀ. This is a necessary condition for a topology to come from a metric; Hausdorﬀ- ness was used only in proofs of (1.20) and (1.21). 222 The metric structure

2.4 Proposition . For a riemannian manifold (M, g), the underlying topology of M considered as a diﬀerentiable manifold and that induced by the metric structure are the same.

Proof. Let M be a point of M and let expm be r-O.K. Then by (1.31) we have D(m, r′) = B(m, r′) if 0 < r′ < r. But every neighbourhood of m in the topology induced by the metric d contains D(m, r′) for sufficiently small r′, and every neighbourhood of m in the original topology contains B(m, r′) for sufficiently small r′ because λ is a diffeomorphism. Hence the neighbourhood system of m in either topology is the same as that in the other. Hence the result.

The following is a direct consequence of the above proposition.

2.5 Corollary. The function d : M M R is continuous. × → 2.6

227 Corollary. We have

D(m, r) = n M d(m, n) r r > 0. { ∈ | ≤ }∀ Proof. In view of (1.31) and (2.5) we have only to prove that

n M d(m, n) = r D(m, r). { ∈ | }⊂ Suppose n be such that d(m, n) = r; ǫ > 0 (small enough) C ∀ ∃ ∈ P(m, n) such that lg(C) r + ǫ; by continuity we can ﬁnd nǫ on the ≤ image of C with d(m, nǫ ) = r ǫ and so: nǫ D(m, r); moreover: − ∈

d(n, nǫ ) lg(C n to nǫ ) = lg(C) lg(C m to nǫ ) r + ǫ d(m, nǫ ) 2ǫ ≤ | − | ≤ − ≤ ǫ 0 in particular nǫ → n, hence n D(m, r). −−−→ ∈ 3. NICE BALLS 223

2.7 Example . Let m be a point of (Sd, can) and σ(m) its antipodal point. Then since σ(m) < B(m,π) and expm is π-O.K. (see (5)) we have d(m,σ(m)) π. ≥ But since there exists a great circular arc (geodesic) of length π joining m and σ(m) we have d(m,σ(m)) π ≤ Hence d(m,σ(m)) = π.

3 Nice balls = If expm is r-O.K. then we know that D(m, r) B(m, r) and further that for any point n in B(m, r) the distance d(m, n) is realised by a geodesic and that such a realisation is essentially unique. Further (see (6.8) and 228 (3.7) we know that for suﬃciently small r, B(m, r) is convex and hence any two points n and n′ of it can be joined by a geodesic fn,n′ which is completely contained in B(m, r) and that such a curve is unique. So a natural question is whether fn,n′ realises the distance between n and n′ and if so if it is the unique path with that property. Before trying to answer this question we give a name to the balls for which this (and more) is true.

3.1 Deﬁnition. A ball B(m, r) is called a nice ball if i) B(m, r) is convex ii) exp is r-O.K. n B(m, r) and n ∀ ∈

iii) n, n′ B(m, r) the geodesic fn,n′ realises d(n, n′) uniquely, i.e., ∀ ∈ = that d(n, n′) lg( fn,n′ ) and upto a re parametrisation by a func-

tion fn,n′ is the only path that realises d(n, n′). Now we shall prove that there are nice balls with assigned centres. 224 The metric structure

3.2 Lemma. Suppose that x U (M) and that t > 0 such that ∈ m ∃ 0 i) [0, t [ x Ω and 0 · ⊂ ii) there exists a sequence t of numbers in [0, t [ and a point p in { n} 0 M such that t t and γ (t ) p as n . n → 0 x n → −∞ Then t x Ω. 0 ∈ 229 Proof. a) First let us prove that

γ (t) P as t t . x → → 0

t t0 Since d is a metric it is enough to prove that d(γ (t), p) → 0. x −−−→ We have

d(γ (t), p) d(γ (t), γ (t )) + d(γ (t ), p). x ≤ x x n x n

Since d is continuous by (2.5) and since γx(tn) tends to p as n tends to inﬁnity by hypothesis the second term on the right hand side tends to zero as n tends to inﬁnity. Further, by (4.3.3),

lg(γ [t , t]) t t x| n ≤ | − n| since x U (N). Hence ∈ m lg(γ [t , t]) t t + t t , x| n ≤ | − 0| | 0 − n| and hence

d(γ (t), p) d(γ (t ), p) + t t + t t . x ≤ x n | n − 0| | − 0| So for t in [t ǫ, t ] we have 0 − 0 d(γ (t), p) d(γ (t ), p) + t t + ǫ, x ≤ x n | n − 0| and hence the result. 3. NICE BALLS 225

b) Let W be a convex ball with centre at p, and let

s : W W W Ω × →

t t0 be the associated map (see (6)). Since γ (t) → t then t 230 x −−−→ 0 ∃ 1 ∈ [0, t0[ such that γ ([t , t ) W. x 1 0 ⊂ = Set u γ′x(t1); by (6.6) we have s(γ (t ), γ (t)) = (t t ) u for t < t < t . x 1 x − 1 · 1 0 Now s being continuous we have

s(γx(t1), p) = lim s(γx(t1), γx(t)) = lim(t t1) u = (t0 t1) u. t t0 t t0 → → − · − · Hence by the deﬁnition of s it follows that (t t ) u W Ω. Since W Ω 0 − 1 · ∈ is an open set there exists a positive number δ such that

[t , t + δ] u W Ω 1 0 · ⊂ and since W Ω Ω we have ⊂ [t , t + δ] u Ω. 1 0 · ⊂

But this means simply that the geodesics γx can be extended to the point exp(t + δ) x and hence t x Ω. 0 · 0 · ∈

Lemma 7.3.2 bis. Let m be a point of (M, g) such that expm is r-O.K. Then Ω B(n, r d(m, n)) nB(m, r). ⊃ − · 226 The metric structure

Proof. Let n B(m, r), x U (M). ∈ ∈ n

We have to show that (see (6.4))

s = t+(x) r d(m, n). G ≥ − We suppose the contrary and deduce a contradiction. 231 Let s < r d(m, n). − Then setting r′ = d(m, n) + s we have r′ < r. Further t [0, s], we have ∀ ∈ d(m, exp(tx)) d(m, n) + d(n, exp(tx)) ≤ d(m, n) + s = r′ < r. ≤ Now we take a sequence t of numbers in [0, s[ tending to s as n . { n} →∞ Then the sequence of points exp tn x lies in B(m, r′) and since B(m, r′) { }= is compact (see (1.20)) and B(m, r′) exp(B(m, r′)), a sub sequence of exp t x converges to a point p B(m, r). Hence the above lemma gives { n } ∈ that s x Ω · ∈ = + and this contradicts the deﬁnition of s tG(x).

3.3 Theorem. For every m of M there exists an r > 0 such that B(m, r) is a nice ball. 3. NICE BALLS 227

Proof. Let r1 be a positive number such that expm is r1-O.K. and B(m, r1) is convex. r We claim that we have only to set r = 1 . 3 i) By (3.7) B(m, r) is convex.

ii) Let n be in B(m, r). Then, by (1.31) we get d(m, n) < r. Hence by (28): r1 B n, r1 = B(n, 2r) Ω. − 3 ⊂ Further by (6.5) expn is 2r-O.K. and in particular r-O.K.

iii) Let n, n B(m, r). Then d(n, n ) d(n, m) + d(m, n ) < 2r so we ′ ∈ ′ ≤ ′ are through because of (1.29).

3.4

Corollary . Let K be a compact subset of M. Then there exists δ > 0 232 such that exp is δ O.K. m K. m − ∀ ∈ Proof. By the compactness of K we can cover it with a ﬁnite number of nice balls B(mi, ri). Then we are through if we set δ = infi ri.

3.5 Corollary. Let n, n M and C P(n, n ) be such that ′ ∈ ∈ ′

lg(C) = d(n, n′).

Then (upto injective re parametrisation) C is a geodesic.

Proof. Let C = [a , b ] I , f D(I , M) . { i i ⊂ i i ∈ i } Since being a geodesic is a local property, it is enough to show that for each point n = fi(ti) there is a neighbourhood of ti in which C is a geodesic. Now let B(n, r′) be a nice ball. 228 The metric structure

Then since C is continuous there exists a δ> 0 such that

C([t δ, t + δ]) B(n, r′). i − i ⊂ Now let f be the geodesic in B(n, r ) from C(t δ) to C(t + δ). ′ i − i Then by the deﬁnition of a nice ball, if C restricted to (t δ, t + δ) i − i were not a geodesic, we would have lg(C) > lg( f ), and hence replacing that part of C by f we would get

d(n, n′) < lg(C [a , t δ]) + lg(C [t + δ, b ]) + lg f < lg(C), | 1 i − | i k a contradiction.

233 As an application of the above corollary we prove the following result called “the corner condition” or “the strict triangle inequality”.

3.6

Application. Let f be a geodesic from n to n′ and g be a geodesic from n′ to n′′ both being parametrised by the arc length. Then if f ′(n′) , g′(n′) then d(n, n′′) < d(n, n′) + d(n′, n′′).

Proof. For if d(n, n′′) = d(n, n′) + d(n′, n′′) the above corollary gives that the path from n to n′′ given by f from n to n′ together with g from n′ to n′′ is a geodesic. Then since both are parametrised by arc length we should have f ′(n′) = g′(n′). 3. NICE BALLS 229

3.7 We insert here a result we will need later on (see (10)). Let B(m, r) be a nice ball, n, n B(m, r). We associate to n, n and t [0, 1] the point ′ ∈ ′ ∈ (n, n′, t) deﬁned as follows: (n, n′, t) is on the geodesic from n to n′ in B(m, r) and such that d(n, (n, n , t)) = t d(n, n ). In particular (n, n , 1 ) ′ · ′ ′ 2 can be called the mid-point of n, n′.

3.8 Proposition . Let (M, g) be an r.m. such that A(M) ] , 0] and let ⊂ −∞ B(m, r) be a nice ball. Then

d((m, n, t), (m, n′, t)) t d(n, n′) t [0, 1], n, n′ B(m, r). ≤ · ∀ ∈ ∀ ∈

Proof. Set 1 = d(n, n′) and deﬁne a one parameter family of curves f : [0, 1] [0, 1] M by: × → α f (t,α) = m, n, n′, , t . 1

Then with the notations of (8): each fα is a geodesic, so t Q(t,α) is a → 2 1 d Jacobi ﬁeld along fα. By A(M) ] , 0] and (6.19) we have (E 234 ⊂ −∞ 2 dt2 ◦ Q) E D Q. We normalize Q in H, so that Q = ϕ H with E H = 1. ≥ ◦ P · ◦ dϕ 2 dϕ 2 Then: g(H, DPH) = 0 so that E DPQ = + E DPH as ◦ dt ! ◦ ≥ dt ! 230 The metric structure

d2ϕ E Q = ϕ2, we get > 0. So ϕ is a positive function, vanishing at ◦ dt2 t = 0, in a particular ϕ(t) t ϕ(1), which reads Q(t,α) t Q(1,α) . ≤ · || || ≤ || || But Q(t,α) is the speed of the transverse curve ct which connects (m, n, t) 1 and (m, n , t) so: d((m, n, t), (m, n , t)) lg(c ) = Q(t,α) dα ′ ′ ≤ t 0 || || R 1

t Q(1,α) dα = t lg(c1) = t d(n, n′). ≤ Z || || · · 0

4 Hopf-Rinow theorem

4.1

Deﬁnition. We set: T (n, n ) = C P(n, n ) lg(C) = d(n, n ) . Then ′ { ∈ ′ | ′ } each path is a geodesic by (3.5) and we will always normalize it requiring it to be parametrized by its arc length. We put only such paths in T (n, n′).

Let us note that, in general, T (n, n′) is neither empty nor consists of a single element, as can be seen from the manifold (Sd p, can) obtained − from the sphere Sd by deleting one point p: T (m, m ) = and T (n, n ) ′ ∅ ′ has continuous elements. 4. HOPF-RINOW THEOREM 231

4.2 235 Suppose that T (n, n′) is non-empty. Then there exists an

f : [0, 1] M → such that f is a geodesic parametrised by the arc length, f (0) = n and f (1) = n′. Let x be the initial speed of f . Then the curve

t γ (t) = exp(tx) → x is a geodesic through n with the initial speed x. Hence it follows that f is the restriction of γx to [0, 1] and hence that

n′ = exp(1 x)

i.e. n′ exp(B(n, 1)) and 1 x Ω, ∈ · ∈ since f is parametrised by the arc length and hence x is a unit vector.

4.3 Theorem 4.1 (Hopf-Rinow). The following four properties are equivalent.

i) Every closed bounded set in M is compact.

ii) M is complete.

iii) Ω= T(M).

iv) There exists a point m in M such that T (M) Ω. m ⊂ Proof. We shall prove that

(i) (ii) (iii) (iv) (i). ⇒ ⇒ ⇒ ⇒ Since every Cauchy sequence is bounded the ﬁrst implication is clear. 232 The metric structure

To prove that Ω = T(M) it is enough to show that x U(M) one + ∀ ∈ has t (x) =+ . ∞+ Suppose t (x) < ; then there exists a Cauchy sequence ∞ + t [0, t (x)[ { n}⊂ 236 such that + t t (x). n → n →∞ Then we have: d(exp(t x), exp(t x)) t t x = t t and hence i · j · ≤ | i − j| · || || | i − j| exp ti x is a Cauchy sequence. So if M is complete this sequence has a limit point and an application of (3.2) gives the second implication The third implication is clear. So we are left with proving the fourth implication. It will be a consequence of the following:

4.4 Proposition. If for some point m in M

T (M) Ω m ⊂ then T (m, n) , n M ∅ ∀ ∈ (In fact if K is closed and bounded, r K D(m, r) but D(m, r) ∃ | ⊂ ⊂ exp(B(m, r)) which is compact. So K is closed in a compact, hence compact) Proof of the proposition. Set:

D = D(m, t), F = n D T (m, n) , , I = t 0 D = F . t t { ∈ t| ∅} { ≥ | t t} We remark that F is closed: this follows easily from the fact that Ω t ⊃ T (M). Moreover r I [0, r], for (1.29) implies that I [0, r] as soon m ∃ | ⊃ ⊃ as expm is r-O.K. We are going to prove that I is both closed and open. 4. HOPF-RINOW THEOREM 233

4.5

Lemma. Let r I, n < D ; then p D such that d(m, p) = r and 237 ∈ r ∃ ∈ r d(m, n) = d(m, p) + d(p, n).

Proof of the lemma. D = F being compact, p D such that r r ∃ ∈ r d(p, n) = inf d(n, q) q D . { | ∈ r} We claim: p is the right one. For ǫ > 0, C P(m, n) such that ∀ ∃ ∈ lg(C) < d(m, n) + ǫ. Because t d(m, C(t)) is continuous b such that → ∃ d(C(b), m) = r.

But:

r + d(p, n) r + d(C(b), n) ≤ ≤ lg(C m to C(b)) + lg(C C(b) to n) = lg(C) ≤ | | ≤ ǫ + d(m, n) ≤ so : r + d(p, n) ǫ + d(m, n). ≤ Letting ǫ 0, we obtain → r + d(p, n) d(m, n) d(m, p) + d(p, n) r + d(p, n) ≤ ≤ ≤ hence d(m, p) + d(p, n) = d(m, n). 234 The metric structure

4.6

I is closed. Let ri I be such that lim ri = r, and let n Dr. If 238 { } ⊂ i ∈ d(m, n) < r, then i d(m, n) < r so T (m→∞, n) , . If d(m, n) = r and if ∃ | i ∅ n < D i then by (4.5) p D with d(m, n) = r = r + d(p , n). When ri ∀ ∃ i ∈ ri i i i : d(p , n) 0, so p n. But i : p F F and F is → ∞ i → i → ∀ i ∈ ri ⊂ r r closed; so n F . In any case then n F i.e. D = F . ∈ r ∈ r r r

4.7

I is open. Suppose I = [0, r0] and get a contradiction as follows. From = the compactness of Fr0 Dr0 we can by (3.4) ﬁnd r > 0 such that expp is r-O.K. p D . ∀ ∈ r0 Claim: F = D r r < r < r + r. In fact if n M with r < r′ r′ ∀ ′| 0 ′ 0 ∈ 0 d(m, n) < r < r + r then n < D . By (4.5) p with d(m, p) = r ′ 0 r0 ∃ 0 and d(m, p) + d(p, n) = d(m, n) hence d(p, n) < r + r r = r so by 0 − (1.29) g T (p, n) and f T (m, n) and lg( f g) = d(m, n) so ∃ ∈ ∃ ∈ ∪ f g T (m, n). Let us note the results proved in the course of the ∪ ∈ proof of the Hopf-Rinow theorem as corollaries.

4.8 Corollary. If (M, g) satisﬁes any one of the four equivalent conditions of (4.3) then T (n, n′) , n, n′ M. ·∅ ∀ ∈

4.9 Corollary. If (M, g) satisﬁes any one of the four equivalent conditions of (6.4.3) then for any point n of M and every s > 0 we have

D(n, s) = exp(B, (n, s)).

Corollary. If (M, g) is complete and n, n′ and n′′ are points of M such that d(n, n ) = d(n, n ) + d(n , n ) then f T (n, n ) such that the ′′ ′ ′ ′′ ∃ ∈ ′ image set of f contains n′. 5. A COVERING CRITERION 235

239 For one can take f T (n, n ) and f T (n , n ) and joining them 1 ∈ ′ 2 ∈ ′ ′′ get the element f f of T (n, n ) by (3.5). 1 ∪ 2 ′′ 4.10 Remarks. 1) The converse of (4.8) is false. For example in a euclidean ball for any two points n and n we have T (n, n ) , . ′ ′ ∅ But it is not complete.

2) The completeness, in general, depends on g. For example the manifold S1 R with the product r.s. is homeomorphic to the × manifold R2 0 with the r.s. induced from R2. But the former −{ } is complete and the later is not. But if M is compact then for any r.s. g(M, g) is complete by (2.4) and hence complete.

3) The Hopf-Rinow theorem and corollary (4.7) are the starting points for obtaining global results in riemannian geometry. We shall give typical examples in the next articles and in Chapter VIII.

4.12 Symmetric pairs give r.m.’s. M = G/K which are always complete: in fact use (4.1) to check (iv) in (4.3) for m = m0 = p(e).

5 A covering criterion

5.1 Proposition. Let two r.m.’s (M, g) and (N, h) of the same dimension and a map p D(M, N) be such that ∈ i) p is onto,

ii) g = p∗h,

iii) (M, g) is complete. 236 The metric structure

Then p is a covering map.

Proof. By the deﬁnition of a covering map we have to show that given any point n of N there exists a neighbourhood V of n such that each 1 connected component of p− (V) is homeomorphic to V through p. We 240 shall show that if expn is s-O.K., then B(n, s) is a neighbourhood with the above property.

1 T a) Let m p− (n); then from ii) it follows that pm is injective, and ∈ T since the manifolds are of the same dimension that pm is bijec- T tive. Hence by (ii) it follows that pm is a euclidean isomorphism between Tm(M) and Tn(N). Let us set

1 λ = (exp B(n, s))− . n | Then since (M, g) is complete and hence T(M) = MΩ we can deﬁne the map

M T 1 q = exp (p )− λ : B(n, s) M. ◦ m ◦ → T Let us note, since pm is a euclidean isomorphism, that the image of q is B(m, s). Further since g = p∗h by (2.6) we have

p M exp = N exp pT ◦ ◦ m and hence

T 1 M T 1 N T (p )− λ p exp = (p )− λ exp p m ◦ ◦ ◦ m ◦ ◦ ◦ m T 1 T = (p )− id , p = id , m ◦ B(n s) ◦ m B(m s) and hence (pT ) 1 λ p is the inverse of M exp on B(m, s). In par- m − ◦ ◦ ticular M exp is s-O.K., and p and q are diﬀeomorphisms which are inverses of each other.

b) Now suppose that m is any point in p 1(n). Let u B(m, s) ′ − ∈ ∩ B(m′, s). Since by a) expm is s-O.K, by (1.29) there is a unique geodesic γ1 parametrised by the arc length from m to u realising 241 the distance between m and u, and a similar one γ2 from m′ to 5. A COVERING CRITERION 237

u. Then since p is a local isometry by ii) it follows that p γ is ◦ 1 a geodesic (parametrised by the arc length) realising the distance between n and p(u), and p γ is also one such. Hence p γ = ◦ 2 ◦ 1 p γ . Therefore γ and γ are lifts of the same curve having a ◦ 2 1 2 common point u. Since p is local diﬀeomorphism γ1 = γ2 and hence m = γ (0) = γ (0) = m . So we have B(m, s) B(m, s ) = 1 2 ′ ∩ ′ ∅ as soon as m , m′. c) We now prove that

1 p− (B(n, s)) = B(m, s). m [p 1(n) ∈ −

Let u p 1(B(n, s)) and let ∈ − γ : [0, 1] N → be an element of T (p(U), n) and γ be the geodesic in (M, g) such that = T b1 γ′(0) (pu )− (γ′(0)). Since (M, g) is completeb the geodesic γ is deﬁned for every value of the parameter. Since p γ has the initial speed γ (0) it follows ◦ ′ that b bp γ = γ ◦ b 238 The metric structure

and hence (p γ)(1) = γ(1) = n ◦ 1 and hence γ(1) is in p− (n).b Hence u is in B(m, s). m p 1(n) ∈S− b

5.2 Remarks. 1) We have assumed that M and N are of the same dimension. T 242 Actually this can be deduced from ii) Since we have seen that pm is injective for every m of M it follows that given any point m there exists a neighbourhood U of m which is mapped by a C∞-map into p(U) by means of p. Hence the dimension of N is greater than or equal to that of M. If it is greater, then U will be mapped into a set of measure zero in N, and since M is paracompact and connected it is the union of countable number of the neighbourhoods of the type U; it follows that p(M) is of measure zero. But by i) p(M) = N and hence this is a contradiction. 2) Completeness of (M, g) is essential as this picture shows:

5.3 Proposition . Let (M, g) be a complete r.m, and let m be a point of M such that expm is of maximal rank everywhere. Then the map expm is a 6. CLOSED GEODESICS 239 covering map. = Proof. First let us note that expm is onto by (4.8). By (2.1) g (expm)∗g is a r.s. on the manifold Tm(M). Hence if we prove that (Tm(M), g) is complete then by (5.1) we shall be through. Let x T (Mb), x , 0 and 243 ∈ m v : R T (M) be deﬁned by v(t) = t.x. Then exp v is a geodesic in → m m ◦ (M, g) (because (M, g) is complete). Hence by (2.7), v is a geodesic in (T (M), g). This being so T (M) we have fulﬁlled condition (iv) m ∀× ∈ m of (4.3) for 0m Tm(M). b ∈ 5.4

Remark. If we suppose, in addition to the maximality of rank for expm and completeness of (M, g), that M is simply connected then the above ﬀ result gives that expm is actually a di eomorphism.

5.5 Corollary . If (M, g) is complete and simply connected and further A(M) ] , 0] then M is diﬀeomorphic to Rd. ⊂ −∞

Proof. By (2.9) it follows that expm is of maximal rank everywhere for every m. Now we have only to apply the above remark.

5.6 Remark. 1) The completeness assumption on (M, g) cannot be re- moved as the example ((R3 0 ), can) shows. −{ } 2) There is a converse to the above corollary see (7.2).

6 Closed geodesics

In this article we prove that if (M, g) is a compact r.m, then in every non-zero free homotopy class there exists a closed geodesic (see (5)) the length of which is the least among the lengths of all paths in that class. Given a non-zero homotopy class, (M, g) being compact we construct a 240 The metric structure

curve in that class which is of the smallest length and then show that it is a closed geodesic. Before proceeding further we recall a few deﬁnitions and results.

6.1 244 1) For any points m1 and m2 of M let us denote the class of all continuous maps σ of I = [0, 1] into M such that

σ(0) = m1,σ(1) = m2

by Lm1,m2 . If m1, m2 and m3 are points of M and, σ and ζ are

elements of Lm1,m2 and Lm2,m3 respectively then there is an element ζ σ L , which corresponds to the pair (σ, ζ) in a natural ◦ ∈ m1 m3 way: σ(2t) if t 1 (ζ σ)(t) = ◦≤ ≤ 2 ◦ ζ(2t 1) if 1 t 1  − 2 ≤ ≤  If m1 and m2 are points ofM then there is a homotopy relation in

the class Lm1,m2 : σ and ζ in Lm1,m2 are homotopic if there exists a map f : I I M × → such that f (t, 0) = σ(t), f (t, 1) = ζ(t) t I ∀ ∈ and f (0,α) = m , f (1,α) = m α I. We denote this by σ ζ. 1 2 ∀ ∈ ∼

The particular case when m1 and m2 coincide is of special impor- = tance for us. In this case, we set Lm1 Lm1,m1 and call elements of Lm1 loops based at m ; in L there is a law of composition and the relation 1 m ◦ . This relation is an equivalence relation in L and the law of com- ∼ m position passes down to the quotient set which is denoted by π1(M, m). Further this law of composition in the quotient set deﬁnes the structure 245 of a group on this quotient set. π1(M, m) is called the fundamental group of M at m. 6. CLOSED GEODESICS 241

6.2 Deﬁnition . Let m , m M, σ L , ζ L . Then they are free 1 2 ∈ ∈ m1 ∈ m2 homotopic if there exists a map

f : I I M × → such that

f (t, 0) = σ(t), f (t, 1) = ζ(t), f (0,α) = f (1,α) t I, α I. ∀ ∈ ∀ ∈ If this is so then the map F:

f (0, 3α t) if 0 t 1 · ≤ ≤ 3 I I (t,α) F(t,α) =  f (3t 1,α) if 1 t 2 × ∋ →  − 3 ≤ ≤ 3  2  f (0, 3(1 t) α) if 3 t 1  − · ≤ ≤  is a homotopy between σ and f (0,α) 1 ζ f (0,α), where f (0,α) − ◦ ◦ ∈ Lm1,m2 . The quotient set is denoted by F(M); it does not carry a natural group structure. But it has a zero element, the class of loops reduced to a point. Conversely if γ is a path from m1 to m2 and Γ is a homotopy between σ and γ 1 ζ γ then the map Φ: − ◦ ◦ γ(α 3t α) if 0 t 1 − · ≤ ≤ 3 I I (t,α) Φ(t,α) = Γ(3t 1,α) if 1 t 2 × ∋ →  − 3 ≤ ≤ 3  2 γ((3t 2) α) if 3 t 1  − · ≤ ≤  is a free homotopy between σ andζ. Hence, we obtain (6.3). An element

σ of Lm1 is free homotopic to an element ζ of Lm2 if and only if there exists a path γ from m to m such that the loops σ and γ 1 ζ γ are 1 2 − ◦ ◦ homotopic. Thus there is a map

f : π (M, m) F(M) 1 → m[M ∈ from π1(M, m) into the free homotopy group of M. Further from 246 m M S∈ 242 The metric structure

6.3

if follows that the image f (u) of an element u in π1(M, m) is not zero in F(M) if u is not. Further from (6.3) we conclude that for every a in F(M), if we write a for π (M, m) f 1(a), then m 1 ∩ −

1 (7.6.4) va v− = a v π (M, m). m m ∀ ∈ 1

6.5

3. Let p : M M → e be the universal covering of M. Then let σ be an element of π1(M, m), m p 1(m) and m be the end point of the lift σ of σ through m. Then ∈ − ′ m depends only on the class of σ in π (M, m), not on σ itself. This ′ f 1 ise related to the deck transformations of the coveringb p : M e M as f → follows: a deck transformation is a map f : M M such that p f = p; → e ◦ they operate transitively on a given fibre p 1(m); and there exists a map − e e from π (M, m) into the set of all deck transformations, denoted by u u 1 → such that, given u π (M, m), then u(m) is the common end point m of ∈ 1 ′ lifts through m of all loops σ u. b ∈ b f We endow M with the r.s. g = p g (see (2.1) C) so that (M, g) e ∗ → (M, g) is a riemannian covering; and denote by d the metric structure e e of (M, g). Note g = p g andep f = p imply f g = g, i.e.e deck ∗ ◦ e ∗ transformations are isometries. e e e e e We fix from now on an a F(M) such that a , 0 and define a ∈ function ϕ on M by setting: ϕ(m) = inf d(m, u(m)), m p 1(m), u a { ∈ − ∈ } 247 the geometrical meaning of which is: ϕ(m) is the infimum of lengths of e e b e e all loops in Lm whose free homotopy class is in a; we are going to show that ϕ is continuous on M and that its minimum on the compact set M is realized by a closed geodesic. We remark that, for a lift σ in (M, g) of a curve σ in (M, g), we have: lg(σ) = lg(σ). e e e e 6. CLOSED GEODESICS 243

6.6 Lemma. Let m be a point of M. Then

1 ϕ(m) = inf d(m, u(m) u a m p− (m). { | ∈ m}∀ ∈ Proof. Let m p 1(m); sincee e b decke transformationse act transitively and ′ ∈ − each deck transformation is induced by an element of π (M, m), then f 1 v π (M, m) such that m = v(m). ∃ ∈ 1 ′ Then we have f b e d(m , u(m )) = d(v(m), u(v(m))) = d(m, (v 1 u v)(m)) ′ ′ − ◦ ◦ since eache f deckb f transformationeb e b isb ane isometry.e e But, byd (6.3), ev 1 u v − ◦ ◦ ∈ a for u a . Hence the result. m ∈ m 6.7 Lemma. If p : (N, h) (M, g) is an r. covering and (M, g) is complete → then (N, h) is complete. Proof. We check iii) of (4.3); let y U(N) and x = pT (y); the geodesic ∈ t M exp(tx) is deﬁned for all values of t since (M, g) is complete; its → lift in (N, h) is by (2.7) nothing but the geodesic t N exp(ty), which is → therefore deﬁned for all values of t.

6.8 Lemma. If (M, g) is complete, then m M, m p 1(m), u a ∀ ∈ ∀ ∈ − ∃ ∈ m such that ϕ(m) = d(m, h(m)). e Proof. By (6.6) e e b e 248 ϕ(m) = inf d(m, d(m)) u a . { | ∈ m} Looking for an infimum we cane worke b insidee a given ball D(m, r) for a suitable r, which is compact by (5.3); then p 1(m) D(m, r) is a discrete − ∩ set in the compact set D(m, r), so is finite; a fortiori the points u(m) in D(m, r) are finite in number. b 244 The metric structure

6.9

Lemma. ϕ is continuous on M.

Proof. Let m be a point of M and let r be a positive real number such 1 that B(m, r) is a nice ball. Let n B(m, r), m p− (m). By the proof of 1 ∈ ∈ (5.1) n with B(m, r) p− (n) = n , and further d(m, n) = d(m, n). But ∃ ∩ { } e u π1(M, m): ∀ ∈ e e e e e e d(m, u(m)) d(m, n) + d(n, u(n)) + d(u(m), u(n)) ≤ e e e e and since u ise anb isometrye wee e have e b e b e be

b d(u(n), u(m)) = d(n, m) = d(n, m). e e Hence we have b e b e e e

d(m, u(m)) d(n, u(n)) + 2d(m, n) ≤ e e Hence by the deﬁnitione b ofeϕ we havee be

ϕ(m) d(n, u(n)) + 2d(m, n) ≤ ee be and hence taking the inﬁmum on the right as u varies through am,

ϕ(m) ϕ(n) + 2d(m, n). ≤ Since B(m, r) is a nice ball we can interchange the roles of m and n and get ϕ(n) ϕ(m) + 2d(m, n). ≤ Hence ϕ(m) ϕ(n) 2d(m, n) < 2r. | − |≤ 249 This holds for any r small enough, so we are through. 6. CLOSED GEODESICS 245

6.10 Theorem . Let (M, g) be a compact r.m., and let a be a non zero free homotopy class on M. Then γ a such that: ∃ ∈ i) γ is a closed geodesic,

ii) lg(ξ) lg(γ) ζ a. ≥ ∀ ∈ Proof. Since M is compact and ϕ is continuous m M such that ∃ ∈ ϕ(m) ϕ(n) n M. ≤ ∀ ∈ Let m p 1(m); then by (6.8) u a such that ∈ − ∃ ∈ m e ϕ(m) = d(m, u(m)).

By (6.7) and by (4.8), T (m, h(m))e,e .b Lete ∅ γ : [0,e1]b eM, γ T (m, u(m)); → ∈ and set γ = p γ. Wee check ii) ﬁrst:e e we havee b lg(eγ) = ϕ(m) by construc- ◦ tion. Then let τ a with τ u; and let n p 1(n) and τ be the lift of τ ∈ ∈ ∈ − through n. Thene by (6.6) e e e lg(τ) = lg(τ) d(n, u(n)) ϕ(n) ϕ(m) = lg(γ). ≥ ≥ ≥ We prove i) now: wee havee toe provebe that γ has no corner. Suppose, by contradiction, that γ did have a corner at m; choose a nice ball B(m, r). Suppose, to simplify the notation, that m = γ(0) = γ(1); denote by fǫ the unique geodesic in B(m, r) from γ(ǫ) to γ(1 ǫ) (see (3.1)); because − γ′(0) , γ′(1), we have, by (3.6),

lg(γ [0, ǫ]) + lg(γ [1 ǫ, 1]) < lg( fǫ). | | −

The family of loops fǫ (γ [ǫ, 1 ǫ]) for ǫ [0, ǫ ], with ǫ < r, is a free { ∪ | − } ∈ 0 0 homotopy, so fǫ (γ [ǫ , 1 ǫ ]) a and lg( fǫ (γ [ǫ , 1 ǫ ]) < lg(γ), 250 0 ∪ | 0 − 0 ∈ 0 ∪ | 0 − 0 contradicting ii), 246 The metric structure

6.11 Remarks. 1) Let us note that the compactness condition on M is, in general, necessary. For if we take the curve

2 x (x, y) R y = e− { ∈ | } and take the surface of revolution it generates in R3 by rotation around the x-axis theP theorem does not hold for . For the cross sections of the surface by the planes orthogonalP to the x-axis are all in the same non zero free-homotopy class but the inﬁnimum of their lengths iz zero. 2) From (6.10) one sees that if (M, g) is any non-simply-connected compact m., it has at least one closed geodesic. This remains true when this (M, g) is simply connected compact but the proof is far more diﬃcult, for it uses Morse theory: the oldest proof is in [11]

7 Manifolds with constant sectional curvature 251 In (4) we have seen that every r.m. of constant sectional curvature is upto a scalar, locally isometric to one of the three standard r.m. of constant sectional curvature:

(Sd, can), (Rd, ǫ), (Rd, Hyp). 7. MANIFOLDS WITH CONSTANT SECTIONAL CURVATURE247

These standard r.m.s are complete. Use (4.12) and the fact that the geodesics in (Rd, ǫ) are well known (see (4.2.9)). Further they are simply connected. Now we prove that every simply connected r.m. with constant sectional curvature is essentially one of the manifolds listed above.

7.1 Proposition . Let (M, g) be an r.m. of constant sectional curvature k which is complete and simply connected. Then (M, g) is isometric to

d 1 d d 1 (R , ( k)− 2 Hyp)), (R , ǫ), (S , k− 2 can) − · · (k < 0, k = 0, k > 0). Proof. Let us denote each of the simply connected r.m.s of constant sectional curvature by (N, h). a) Suppose ﬁrst that k 0. Then by (2.9) exp has maximal rank ≤ everywhere, and hence, by (5.4), we conclude that for any point n of N, exp is -O.K. Let m be any point in M. Then, since (M, g) n ∞ is complete, expm is deﬁned on the whole of Tm(M). Now if u is any euclidean isomorphism from Tn(N) to Tm(M), then by the proof of (4.11) (see (6.4.25)) it follows that the map

1 λ = exp u exp− m ◦ ◦ n is a local isometry between N and M. Then by (5.1) it follows 252 that λ is a covering map. But since (M, g) is simply connected λ is one-one and hence λ is an isomorphism.

b) Now suppose that k > 0. Let n be a point of N and let n′ , n and diﬀerent from the antipodal point of n. Then by (5.12) and (4.18) it follows that exp and exp are (π/ √k)-O.K. Now let m be any n n′ point of (M, g) and let u : T (N) T (M) be any euclidean n → m isomorphism. Then since (M, g) is complete expm is deﬁned on Tm(M) and hence by the proof of Cartan’s theorem (4.11) the map

M N 1 λ = exp u ( exp B(n, π/ √k))− m ◦ ◦ n | 248 The metric structure

is a local isometry; in particular λT is an euclidean isomorphism, n′ so again

M T N 1 λ′ = exp λ ( exp B(n′, π/ √k))− λ(n′) ◦ n′ ◦ n′ | is a local isometry.

We claim that λ, λ′ coincide on their common domain of deﬁnition, T T i.e. B(n, π/ √k) B(n′, π/ √k). For we have λ = (λ′) hence the asser- ∩ n′ n′ tion follows from the diagram (2.6). Now since B(n, π/ √k) together with B(n′, π/ √k) covers N and λ and λ coincide on the intersection B(n, π/ √k) B(n , π/ √k) we have a ′ ∩ ′ map λ′′ from N to M. Since λ and λ′ are local isometries it follows that λ′′ is. Now if we can show that λ′′(N) = M then by (5.1) it follows that 253 λ′′ is a covering map, but since M is simply connected we get that λ′′ is an isometry. Since N is compact λ′′(N) is compact (M is Hausdorﬀ) and hence λ′′(N) is closed in M. But since λ and λ′ are local isometries and hence in particular open maps we have

λ′′(N) = λ(B(n, π/ √k)) λ′(B(n′, π/ √k)) ∪ is open in M. But M is connected and hence:

λ′′(N) = M. Chapter 8 Some formulas and applications

In this chapter we assume that all the manifolds we consider are of di- 254 mension greater than or equal to two.

1 The second variation formula

To investigate whether a geodesic γ with γ(0) = m, γ(1) = n belongs to T (m, n) or not, we are led to proceed as follows (where we use the notations of (8) and (1).

1.1 Let f be a one parameter family

f :] ǫ, 1 + ǫ[ J M(ǫ > 0) − × → with ﬁxed end points, i.e.

f (0,α) = γ(0) and f (1,α) = γ(1) α J, ∀ ∈ and such that f0 = γ. Then the ﬁrst variation formula (see (4.1.2), (2.1)) gives that ∂ l′ (0) = (lg( fα [0, 1]) = 0. f ∂α | 249 250 Some formulas and applications

Then if γ is in T (γ(0), γ(1)), we have

lg( fα [0, 1]) lg(γ [0, 1]) = lg( f [0, 1]); | ≥ | 0|

therefore the so-called second variation l′′f (0) veriﬁes: ∂2 l′′(0) = (lg( fα [0, 1])) 0. f ∂α2 | ≥ < T Later on we will use the fact that l′′f (0) < 0 implies that γ (m, n) (see (1.12). So we wish to compute l′′f (0).

1.2 255 Now let us assume that f is such that

i) f0 is a geodesic parametrised by the arc length, = ii) g(P, Q)(t, 0) 0 and then compute l′′f (0) (we do not require that the end points be ﬁxed). We have ∂2 l′′(0) = (lg( fα [0, 1])) f ∂α2 | 1 = ...... = Q(Q g(P, P)1/2 dt)) Z 0 1 (8.1.3) = Q(Q(g(P, P)1/2)) dt; Z 0

1/2 1 Q(g(P, P) ) = 1/2 Q(g(P, P))g(P, P)− 2

1 (8.1.4) = g(DQP, P)g(P, P)− 2 by (4.8) (C.D.7).

Hence, using the fact that

1/2 1/2 Q(Q(g(P, P) )) = Q(g(DQP, P))g(P, P) , 1. THE SECOND VARIATION FORMULA 251 we obtain

3/2 (8.1.5) (1/2)g(D P, P) Q(g(P, P)) g(P, P)− = ... − Q · · 1 2 3/2 ... = g(D D P, P) + g(D P, D P)g(P, P) 2 g(D P, P) g(P, P)− . Q Q Q Q − Q Now we use (1.2). We have

g(DQP, P) = g(DPQ, P) since [P, Q] = 0 and D is symmetric = P(g(Q, P)) g(Q, D , P) by (4.8). − P

Since f0 is a geodesic (DPP)(t, 0) = 0 and hence by (1.2) ii) we have

(8.1.6) g(DQP, P)(t, 0) = 0.

Again since D is symmetric and [P, Q] = 0 we have 256

g(DQDQP, P) = g(DQDPQ, P) = g(DPDQQ, P) + g(R(Q, P)Q, P) by (5.12) (C.D.5) = P(g(D Q, P)) g(D Q, D P) + g(R(P, Q)P, Q). Q − Q P

Now since f0 is a geodesic we have (DPP)(t, 0) = 0 and hence we have

(8.1.7) g(DQDQP, P)(t, 0) = P(g(DQQ, P))(t, 0)+g(R(P, Q)P, Q)(t, 0).

Combining (8.1.5), (8.1.6) and (8.1.7) with the fact that f0 is parametrised by the arc length and hence g(P, P)(t, 0) = 1 we have the following proposition. Proposition 1.8. If f is a one parameter family of curves satisfying the conditions (1.2) (i) and (ii) then

1 (1,0) = + 2 + l′′f (0) g(DQQ, P) ( DPQ g(R(P, Q)P, Q))(t, 0) dt (0,0) Z || || 0

Noticing the fact that the integral in (1.8) depends only on the curve f0 and on the lift Q(t, 0) of that curve we give the following deﬁnition. 252 Some formulas and applications

Deﬁnition 1.9. Let γ : [0, 1] M → be a curve and let h be a lift of γ into T(M). Then we deﬁne

1 2 l′′(0) = ( DPh + g(R(γ′, h)γ′, h))(t) dt . h Z || || 0

257 In general it is easier to construct a lift h along a curve γ and com-

pute lh′′(0) then to construct a family f of curves and compute l′′f (0).A relation between the two processes is given by the:

Proposition 1.10. Let

γ : [0, 1] I (M, g) ⊂ → be a geodesic parametrised by the arc length and let h be a lift of γ into T(M) such that g(γ′, h) = 0. Then, there exists an interval J containing zero and a one parameter family f :] ǫ, 1 + ǫ[ J M(ǫ > 0) − × → of curves such that

a) f0 = γ and (1.2) i) ii) hold for f = b) l′′f (0) lh′′(0). Proof. We wish to set:

f (t,α) = exp(α h(t)). · But the right hand side makes sense only if α h(t) Ω. So we adjust J · ∈ so that this happens. Let 0 < ǫ′ < ǫ; the map

[ ǫ′, 1 + ǫ′] J (t,α) α h(t) T(M) − × ∋ → · ∈ 1. THE SECOND VARIATION FORMULA 253 is continuous. Under this map the compact set [ ǫ , 1 + ǫ ] 0 goes − ′ ′ ×{ } into the open set Ω since Ω 0 p M. Hence there exists r > 0 such ∋ P∀ ∈ that under the above map [ ǫ , 1 + ǫ ] [ r, r] goes into Ω. Hence we − ′ ′ × − can set

f (t,α) = exp(α h(t)) (t,α)ǫ[ ǫ′, 1 + ǫ′] [ r, r]. · ∀ − × − Then 258

f (t) = exp(0 h(t)) = exp(0 ) = exp(0γ ) = γ(t) 0 · p(h(t)) (t) and hence f0 is a geodesic parametrised by the arc length; and since Q(t, 0) = h(t) we have

g(P, Q)(t, 0) = g(γ′, h)(t) = 0.

1.11 Hence f satisﬁes a) i). For ii) we see that h(t) = Q(0, t) (as in (8)), and hence:

1 (1,0) = + 2 + l′′f (0) g(DQQ, P) ( DPQ g(R(PQ)P, Q))(t, 0) dt . (0,0) Z || || 0 But since the map α exp(α h(t)) → · is a geodesic we have (DQQ) = 0.

Hence we are through by (1.11) and the deﬁnition of lh′′(0). Corollary 1.12. Let m and n be points of M and let γ P(m, n) be a ∈ geodesic in (M, g); suppose that there exists a lift h(t) of γ such that

1) g(h, γ′) = 0, 254 Some formulas and applications

2) h(0) = h(1) = 0,

3) lh′′(0) < 0. Then γ < T (m, n). Proof. Let us take the one parameter family deﬁned in the above proposition. Then we have ∂ (lg( fα [0, 1])) = 0 ∂α | 259 and ∂2 (lg( fα [0, 1])) = l′′(0) = l′′(0) < 0. ∂α2 | r h Hence using Taylor’s formula with a remainder we see that there exist α’s such that

(8.1.13) g( fα [0, 1]) < g( f [0, 1]). | 0| But we have

f (0,α) = exp(α h(0)) = exp(α 0γ ) = γ(0) = m · · (0) and f (1,α) = exp(α h(1)) = exp(α 0γ ) = γ(1) = n · · (1) and hence by (8.1.13) there exist curves from m to n with length less than that of γ. Hence γ is not in T (m, n).

2 Second variation versus Jacobi ﬁelds

2.1 Notation. For x U(M) let us set (see (8.30)): ∈ j(x) = inf t > 0 γ (0) and γ (t) are conjugate on γ . { | x x x}

If expm is r-O.K. then by (8.32) it follows that j(x) r x U (M). ≥ ∀ ∈ m Hence j(x) is a strictly positive function on U(M). 2. SECOND VARIATION VERSUS JACOBI FIELDS 255

2.2 Proposition. Let x U(M), l [0, j(x)] and let ∈ ∈ h : [0, 1] T(M) → be a lift of γ such that h(0) = h(1) = 0. Then l (0) 0 and equality x h′′ ≥ holds if and only if h is a Jacobi ﬁeld.

Proof. Let x = x , x ,..., x be a basis of T = (M) and let h be the { 1 2 d} m p(x) i Jacobi ﬁeld along γx such that hi(0) = 0 and (DPhi)(0) = xi (see (8.26)). Then t [0, 1] : h (t),..., h (t) form a basis of Tγ (M): for if for ∀ ∈ { 1 d } x(t) some t0 > 0 we have: aihi(t0) = 0. Xi then the Jacobi ﬁeld k : t aih(t) vanishes at t = 0 and at t = t0 so 260 → i by (8.30): k = 0. P Hence

DPk = aiDPhi = 0; Xi in particular

(DPk)(0) = ai xi = 0. Xi Since x ,..., x is a basis of T (M) we have { 1 d} m

a1 = ... = ad = 0. Q.E.D.

Hence corresponding to every lift h of γx into T(M) we have functions f ,..., f on ]0, 1[ such that { 1 d}

(8.2.3) h(t) = fi(t)hi(t). Xi

Then we have = + DPh fi′hi fi(DPhi) Xi Xi 256 Some formulas and applications

and setting

= = (8.2.4) u fi′hi and v fi(DPhi) Xi Xi we have DPh = u + v. Hence

(8.2.5) D h 2 = u 2 + 2g(u, v) + v 2 || P || || || || || and

= = (8.2.6) g(R(γ′x, h)γ′x, h) g(R(γ′x, fihi)γ′x, h) X = = fig(R(γ′x, hi)γ′x, h) fig(DPDPhi, h), Xi Xi

(since hi are Jacobi ﬁelds).

2.6 261

= g( f D D h , h) = g(D v f ′D h , h) i P P i P − i P i Xi Xi

= g(D v, h) f ′g(D h , h). P − i P i Xi

= P(g(v, h)) g(v, D h) f ′ f g(D h , h = − P − i j P i j Xi, j by (4.8) (C.D.7)

= P(g(v, h)) g(v, u + v) f ′ f g(D h , h ). − − i j P i j Xi, j

But if we set

ξ = g(D h , h ) g(D h , h ) (a classical Sturmain argument) P i j − P j i 2. SECOND VARIATION VERSUS JACOBI FIELDS 257 then

ξ′ = P(ξ) = g(DPDPhi, h j) + g(DPhi, DPh j) g(D D h , h ) g(D h , D h ) − P P j i − P j P i = g(D D h , h ) g(D D h , h ) = P P i j − P P j i = g(R(γ′ , h )γ′ , h ) g(R(γ′ , h )γ′ , h ) = 0 x i x j − x j x i by ((5.5.5) C.T.4).

Hence ξ is constant, but ξ(0) = 0 for h (0) = 0 i. So ξ = 0. Hence we i ∀ have

= fi′ f jg(DPhi, h j) fi′ f jg(DPh j, hi) Xi, j Xi, j = = g( f jDPh j, fi′hi) g(u, v). Xj Xi

Hence by (8.2.5) and (8.2.6) we have

2 2 D h + g(R(γ′ , h)γ′ , h) = P(g(v, h)) + u . || P || x x || ||

Hence 262 1 1 2 l′′(0) = g(v, h) + u dt, h 0 Z || || 0 and since h(0) = h(1) = 0 we have

1 2 (8.2.7) l′′(0) = u dt . h Z || || 0

Hence l (0) 0 and equality holds if and only if u is identically zero. h′′ ≥ But since h (t),..., h (t) from a basis for every t, u is identically zero { 1 d } if and only if each fi′ is zero, i.e. if and only if each fi is a constant. Hence h = fihi is a Jacobi ﬁeld. i P 258 Some formulas and applications

2.9 Proposition. Let x U(M) and l > j(x); then ∈ γ [0, 1] T (γ (0), γ (1)). x| ∈ x x Proof. First we prove two lemmas.

2.10

Lemma. Let h be a Jacobi ﬁeld along γx. Then

1 l′′(0) = g(D h, h) . h p 0 h i Proof. We have by deﬁnition

l′′(0) = g(Dph, DPh) + g(R(γ′ , h)γ′ , h) dt = h Z x x 0 1

= g(DPh, DPh) + g(DPDPh, h) dt (since h is a Jacobi ﬁeld) Z 0 1 1 = P(g(DPh, h)) dt by (4.8) (C.D.7) = g(DPh, h) . Z 0 0

2.11

Lemma. Let h be a Jacobi ﬁeld along γx such that

h(0) = 0 and (DPh)(0) , 0.

Then given any positive number η we can choose a positive number ǫ 263 less than η and a Jacobi ﬁeld k along γ [ ǫ, ǫ] such that x| − i) k( ǫ) = h( ǫ) − − 2. SECOND VARIATION VERSUS JACOBI FIELDS 259

ii) k(ǫ) = 0 iii) g(D h, h)( ǫ) < g(D k, k)( ǫ). P − P − Proof. Let r be a positive number such that B(m, r) is convex. Then t ]0, r[ there exists a Jacobi ﬁeld k along γ [ t, t] such that ∀ ∈ t | − (8.2.12) k ( t) = h(t) and k (t) = 0. t − t

In fact we set fα = the geodesic from exp(α h( t)) to γ(t) parametri- · − sed by the arc length, then the family fα of the curves fα so obtained is { } a one-parameter family of geodesics and the k(s) = Q(s, 0) satisfy our requirements. Now let us set

u = (DPh)(0) and kt(t′) = at + t′bt + 0(t′). Then b h(t′) = t′ u + 0(t′), h′(t′) = u + 0(1) · and b b = + kt′(t′) bt 0(1). Then by (8.2.12) we have b a tb + 0(t) = k ( t) = h( t) = tu + 0(t) t − t t − − − and b b kt(t) = at + tbt + 0(t) = 0. Hence 264 b u b = + 0(t). t 2 260 Some formulas and applications

2.13 Hence 2 g(D h, h)( t) = g(h′( t), h( t)) = g(u+0(1), tu+0(t)) = t u +0(t), P − − − − − || || and b b

g(D k , k )( t) = g(k′( t), k ( t)) P t t − t − t − u u t = g + 0(1), h( t) = g + 0(1)b, tu +b0(t) = u 2 + 0(t). 2 − 2 − 2|| || Since u is non zerob by our hypothesis we have u > 0 and hence by || || (2.13) we can choose t so small that

g(D h, h)( t′) < g(D k, k)( t′) P − P − for 0 < t′ < t. Now let us take up the proof of (2.6). Since 1 > j(x) there exists a t ]0, 1[ and a Jacobi field h along γ such that 0 ∈ x h , 0, h(0) = h(1) = 0. Set η = t . Then applying the lemma (2.11) for the Jacobi field h − 0 around the point t we get ǫ ]0, η[ and a Jacobi field kǫ along γ [t 0 ∈ x| 0 − ǫ, t0 + ǫ]. Now let f be the lift of γx such that: h = f [0, t ǫ] = h [0, t ǫ] 1 | 0 − | 0 − h = f [t ǫ, t + ǫ] = kǫ 2 | 0 − 0 h = f [t + ǫ, 1] = 0. 3 | 0 265 Then we have l (0) = l (0) + l (0) + l (0). ′′f h′′1 h′′2 h′′3 But by (2.10) we have 1 ǫ l′′ (0) = g(DPh1, h1) − = g(DPh1, h1)(1 ǫ) since h1(0) = h(0) = 0; h1 0 − and = 1+ǫ = + = lh′′ (0) g(DPh2, h2) 1 ǫ g(DPh2, h2)(1 ǫ) since h2(1 ǫ) 0 2 − − − and l (0) = 0 since h = 0. h′′3 3

Hence l′′f (0) < 0 so we are done with (1.12). 3. THE THEOREMS OF SYNGE AND MYERS 261

2.14 Remark . The two propositions (2.2) and (2.6) are essential parts of Morse’s index theorem.

3 The theorems of Synge and Myers

First let us recall that Ricci curvature (Ric) is a function from T(M) to R.

3.1 Proposition. Let (M, g) be an r.m. of dimension d such that

Ric(U(M)) [k, [ ⊂ ∞ for some k > 0. Let γ be a geodesic in P(m, n) where m and n are points of M. Now if d 1 1/2 lg(γ) >π − k ! then γ < T (m, n). 266

Proof. Let us suppose that γ is parametrised by the arc length and then take an orthonormal basis x = γ (0), x ,..., x of Tγ (M). Let h be { 1 ′ 2 d} (0) i the parallel lift along γ with hi(0) = xi, and set π si(t) = sin t hi(t). 1 ·

Then since hi is a parallel lift we have DPhi = 0 and hence by (4.1) D.L.3 we have π πt (DP si)(t) = cos hi(t), 1 1 · and since parallel transport preserves g we have

2 2 π 2 πt (8.3.2) (DP si)(t) = cos . || || 12 1 262 Some formulas and applications

Again since parallel transport preserves g by the deﬁnition of A (see (6.1.2)) and since γ′(0) and hi(0) are orthogonal we have

2 2 πt g(R(γ′, h)γ′, h) = A(γ′, s ) s = A(γ′, h ) sin . − i || i|| − i 1 Hence by the deﬁnition (1.9) we have

1 2 = π 2 π 2 πt (8.3.3) l′′s (0) cos t sin A(γ′, hi) dt . i Z 12 1 − 1 0

By (8.6) we have d A(γ′, h ) = Ric(γ′) k i ≥ Xi=2 and hence (8.3.4) d π2 πt πt 1 π2 l (0) (d 1) cos2 k sin2 dt = (d 1) k . ′′si 2 2 = ≤ Z l · − 1 − 1 2 1 − − ! Xi 2 0

1 d 1 2 d 267 Hence if l >π − then l′′si (0) < 0 and hence k ! i=2 P

i l′′(0) < 0; ∃ | si hence, by (1.12), γ < T (m, n).

3.5

Deﬁnition. The real number d(M, g) deﬁned as

d(M, g) = sup d(m, n) m, n M { | ∈ } is called the diameter of the r.m. (M, g). 3. THE THEOREMS OF SYNGE AND MYERS 263

3.6 Remarks. 1) If d(M, g) < and the manifold (M, g) is complete ∞ then by Hopf-Rinow theorem it follows that M is compact.

2) Let us note that (8.6) implies that if

A(M) [k, [ then Ric(U(M)) [(d 1)k, [ ⊂ ∞ ⊂ − ∞

3.7 Corollary Myers. Suppose that (M, g) is a complete r.m. such that there exists a k > 0 satisfying

Ric(U(M)) [k, [. ⊂ ∞ Then 1 d 1 2 d(M, g) π − . ≤ k ! In particular M is compact and the fundamental group of M is ﬁnite.

Proof. Suppose that there exist m and n in M such that

1 d 1 2 d(m, n) >π − . k !

Then by (8.4.8) there exists a geodesic γ joining m and n such that

1 d 1 2 lg(γ) >π − . k !

1 d 1 2 But this contradicts (3.1); hence d(M, g) π − . 268 ≤ k ! By (3.6) it follows that M is compact. Now let

p : (M, g) (M, g) → e e 264 Some formulas and applications be the universal riemannian covering of (M, g). Then since p is a local isometry we obtain

Ric(U(M)) = Ric(U(M)) [k, [. ⊂ ∞ Hence the first part of thee corollary gives that M is compact. Hence the covering p : M M has finite number of sheets. → e e 3.8 Remarks. 1) This result was first proved by Bonnet for two dimensional manifolds.

2) In the above inequality (3.7) the equality occurs for

1 2 Sd d 1 Rd+1  , − can .  k ! ·  ⊂     The proof here would not yield easily the result that equality in d d 1 1 (3.7) is attained only if (M, g) is isometric to (S , ( − ) 2 can). k · However the validity of this result can be deduced easily from [33]: Corollary 4, p.256-257.

3.9 Theorem. (Synge) Let (M, g) be an r.m. such that

i) M is compact

ii) M is orientable

iii) the dimension of M is even

iv) A(M) ]0, [. ⊂ ∞ Then M is simply connected. 3. THE THEOREMS OF SYNGE AND MYERS 265

269 Proof. By contradiction: suppose that M is not simply connected; then (see (6.3)) there exists in M at least one non-zero free homotopy class. Let γ denote the closed geodesic which fulﬁlls the conditions i) and ii) of theorem (6.10). Let l = lg(γ) and τ(t) denote the parallel transport along γ from 0 to t. Since γ(0) = γ(1) then τ(1) is an endomorphism of Tγ(0)(M); by (4.9) we see that τ(1) belongs to the orthogonal group of Tγ0(M). In fact τ(1) is a rotation. For: suppose that σ is the orienting form and that γ (0) = x ,..., x is a positive orthonormal base of { ′ 1 d} Tγ(0)(M). Then the function

σ(τ(t)(x1),...... ,τ(t)(xd)) never vanishes since σ is never zero and τ(t) is an isomorphism; hence

σ(τ(1)(x1),...,τ(1)(xd)) = 1. Set H = y Tγ(0)(M) g(γ′(0), y) = 0 ; n ∈ | o then since parallel transport preserves g and τ(1)(γ′(0)) = γ′(1) = γ′(0) by (3.5) we get τ(1)(H) = H. But H is odd dimensional and hence y H, y , 0, with τ(1)(y) = y (because τ(1) is a rotation). ∃ ∈ Let h be the parallel lift of γ such that h(0) = y. since τ(1)(y) = y we have h(0) = h(1) = y. Now let f be the parameter family of curves associated to h in the man- 270 ner of (1.10). Then by the deﬁnition of f we see that fα is a closed curve α and (see (1.9)): ∀ 1 2 l′′(0) = l′′(0) = A(γ′, h) h dt < 0 f h − Z || || 0 since h(t) is non zero and A(M) ]0, [. Hence α such that ⊂ ∞ ∃ 0

lg( fα0 ) < lg(γ).

Moreover the restriction f [0, 1] [0,α0] deﬁnes a free homotopy be- | × tween γ, and fα0 . This contradicts our choice of γ. 266 Some formulas and applications

3.12

Remark . The manifold (P3(R), can) shows that this result cannot be improved.

4 A formula

We prove an integral formula (4.7) which will be used in articles 7 and 9 (see [13]: theorem 5.1). In this article we assume that the r.m. (M, g) is complete. In particular (4.3) the geodesic flow is defined on the whole R M. We denote the × flow by Gt. We recall the fact that Ric is the Ricci curvature of (M, g) (see (8)).

4.1

Deﬁnition. We say that an r.m. (M, g) is NCB(k) if k is a positive number and (see (2.1)) j(U(M)) [k, [. ⊂ ∞ 271 Now for any positive number k and for any x U(M) we set: ∈ k/2 2 π 2 πt 2 πt (8.4.2) fk(x) = (d 1) sin cos Ric(Gt(x)) dt . Z − k2 k − k · ! k/2 −

Note that by the deﬁnition of a ﬂow and because γ′x is an integral curve of G we have: = Gt(x) γ′x(t).

4.3

Proposition. If (M, g) is NCB(k) then

f (x) 0 x U(M), k ≥ ∀ ∈ 4. A FORMULA 267

k k equality occurring for a given x if and only if t , and y which ∀ ∈ "−2 2# ∀ is linearly independent of Gt(x) we have

2 2 A(Gt(x), y) = π /k .

Proof. As in the proof of (3.1) for a given x let x = x , x ,..., x be { 1 2 d} an orthonormal basis of Tp(x)(M), let hi be the parallel lift of γx through xi and let si be the lift deﬁned by the equation πt k k (8.4.4) si(t) = cos hi(t) for t . k · − 2 ≤ ≤ 2 Then (1.9):

k/2 2 = π 2 πt 2 πt (8.4.5) l′′s (0) sin A(Gt(x), hi(t)) cos dt . i Z k2 k − k ! k/2 − Hence by (8.6) and (8.4.2) we have

d = (8.4.6) fk(x) l′′si (0). Xi=2 k Now let us note that, by the definition, s vanishes at and the fact 272 i ±2 k that (M, g) is an NCB(k) implies that j(γ ( )) k. Hence by (2.1) and ′ −2 ≥ (2.2) we have l′′(0) 0 i = 2,..., d si ≥ ∀ and equality occurs if and only if si is a Jacobi field. Now suppose that f (x) = 0. Then s is a Jacobi field i. Hence k i ∀

DPDP si = R(Gt(x), si)(Gt(x)).

But π2 πt DPDPS i = cos hi(t) − k2 k · 268 Some formulas and applications

and πt R(Gt(x), si)Gt(x) = cos R(Gt(x), hi(t))Gt(x). k · πt k k Further cos( ) does not vanish in ] , [ and hence we have k − 2 2 π2 R(G (x), h (t))G (x) = h (t) i. t i t − k2 · i ∀

and since h (t), h (t),..., h (t) form a basis of Tγ (M) (see the proof { 1 2 d } x(t) of (2.2)) we have (see (6.8.2)) π2 R(G (x))y = y t − k2 whenever y is orthogonal to G (x). In particular if G (x), y are orthonor- t { t } mal we have π2 π2 A(G (x), y) = g(R(G (x), y)G (x), y) = g(y, y) = . t − t t k2 k2 Before stating the next proposition let us recall that Γ denotes the scalar curvature of (M, g) (see (8))

4.7

273 Proposition. Let (M, g) be an oriented compact manifold. Then if (M, g) is NCB(k) we have k2/π2 Vol(M, g) Γ σ ≥ d(d 1) · Z · − M (where σ is the canonical volume form of (M, g)) and the equality occurs if and only if A(M) = π2/k2 . { } Further, if k = , Γ σ = 0, then A(M, g) = 0 . ∞ · { } R Proof. Let σ = ω (p σ) (see (6.2.5)) be the volume form of U(M) so ∧ ∗ that for every m of M, ω U (M) is the volume form of U (M). Now set | m m 2 π 2 πt 2 πt ϕk(x, t) = (d 1) sin cos Ric(Gt(x)) x U(M), t R − k2 k − k · ∀ ∈ ∈ 4. A FORMULA 269 and deﬁne I by the equation

k 2 = =   I fk(x) σ  ϕk(x, t) dt σ. Z · Z Z ·  U(M) U(M)  k  − 2    Since M is compact we may interchange the order of integration and obtain

k/2

(8.4.8) I =  ϕk(x, t)σ dt = A B, Z  Z  − k/2 U(M)  −     where  

k/2 π2 πt (8.4.9) A = (d 1) sin2 σ dt  2  Z  Z − k k  k/2 U(M)  −     and  

k/2 2 πt (8.4.10) B =  cos Ric(Gt(x))σ dt . Z  Z k  k/2 U(M)  −     But 274

k/2 2 πt (8.4.11) A = (d 1) sin2 σ dt . 2   − k · Z k  Z  k/2 U(M)  −    k/2 π2 ¨ πt = (d 1) d 1 Vol(M, g) sin2 dt = − k2 − © Z k · k/2 − 2 ¨ k = (d 1) d 1 Vol(M, g) by (5.2.13). − k2 − © · 2 By (3.10) and (2.5) we have:

Gt(σ) = σ. 270 Some formulas and applications

Hence by (0.3.9) we have

Ric(Gt) σ = (Ric Gt)Gt(σ) = Gt(Ric σ) = (by (0.3.9)) Z · Z ◦ Z ◦ U(M) U(M) U(M)

= Ric σ = Ric (ω p∗σ) = Z ◦ Z ◦ ∧ U(M) U(M)

=  Ric Um(M) ω))σ by (3.17). Z  Z |  · m M U (M)  ∈  m    But by (6.8.11) we get ¨ ¨ d 1 d 1 Ric σ = − ©Γ(m) σ = − © Γ σ Z Z d · d Z · U(M) M M

and hence

k/2 ¨ ¨ πt d 1 d 1 k (8.4.12) B = cos2 − © Γ σ dt = − © Γ σ. Z k ! d Z · d 2 Z · k/2 M M − But k/2 k/2 πt πt k sin2 dt = cos2 dt = . Z k Z k 2 k/2 k/2 − − 275 Hence we have ¨ d 1 k (d 1) π2 1 I = − © Vol(M, g) Γ σ .  − 2 ·  2  k − d Z ·   M     ¨  But by (4.3) I 0 and since k d 1 > 0 we have ≥ · − © k2π 2 (8.4.13) Vol(M, g) − Γ σ ≥ (d 1)d Z · − M 4. A FORMULA 271

where the equality occurs if and only if I = 0 i.e. if and only if fk(x) = 0 for every x. But by (4.3), this happens if and only if whenever

k k x U(M), t ] , [ and y is linearly independent of G (x), ∈ ∈ − 2 2 t we have π2 A(G (x), y) = . t k2 This is clearly equivalent to the fact that

A(M) = π2/k2 . n o Now suppose that (M, g) is NCB( ) and Γ σ = 0. First by (4.3) ∞ · MR letting k tend to inﬁnity we have f (x) 0 x U(M) and then by ∞ ≥ ∀ ∈ (8.4.8), (8.4.10) and (8.4.12) we have ¨ k d 1 I = f (x) σ = − © Γ σ = 0. Z ∞ · − 2d Z · U(M) M

Hence

(8.4.14) f (x) = 0 x U(M). ∞ ∀ ∈ Now by (8.2.7) we have

k/2 2 = π 2 πt 2 πt 0 l′′s (k) sin A(Gt(x), hi(t)) cos dt ≤ i Z k2 k − k k/2 − and because of (8.4.6) and (8.4.14) 276

d l′′ (0) 0 as k . si(k) → →∞ Xi=2 272 Some formulas and applications

Since each of the terms is non-negative we have

k/2 2 π 2 πt 2 πt lim sin A(Gt(x), hi(t) cos dt = 0. k Z k2 k − k ! →∞ k/2 − But by (8.2.7) the integrand is non-negative, and hence

2 2 πt π 2 πt cos A(Gt(x), hi(t)) sin . k ≤ k2 k Fixing t and letting k to inﬁnity we have

A(G (x), h (t) 0. t i ≤

Hence the sequence fn(t) of functions deﬁned by the equation

2 πt n n A(Gt(x), hi(t)) cos n in ] 2 , 2 [ fn(t) = − − 0 outside   is a non decreasing sequence of functions. Hence

0 = lim fn(t) dt = lim fn(t) dt = A(Gt(x), hi(t) dt . n n →∞ Z Z →∞ Z

Since A(Gt(x), hi(t)) is of the same sign we have

A(G (x), h (t)) = 0 t. t i ∀ Putting t = 0 we get A(x, xi) = 0

whenever x and xi are orthogonal. But the sectional curvature is a function of the two dimensional subspaces and hence the result.

5 Index of a vector ﬁeld 277 In the case of a compact oriented r.m. (M, g), we have seen that Γ σ · MR plays a role. In the two dimensional cases we have Γ = 2A, and the 5. INDEX OF A VECTOR FIELD 273

Gauss-Bonnet formula (see next article) says that

A σ = 2π χ(M), Z · · M where χ(M) is the Euler-Poincarécharacteristic of M. In the next section we give that part of the proof which involves riemannian geometry. Now we collect some facts from algebraic topology. 1. Let us take Rd with its canonical r.s. ǫ and denote by ω E d(U(Rd)) ∈ the form called ω in (5.2.11). Let m Rd and B an open set of Rd ∈ b containing m and X a vector field on Rd which does not vanish in B except at m. We define X on B m by the equation −{ } X = X/ X . || || Then X D(B m , U(Rd)). ∈ −{ } Let r > 0 be sufficiently small; we consider the number 1 (8.5.1) ¨ (X)∗(ω) d 1 Z − ©S (m,r) b ¨ (for d 1 see (6.7.3) and for S (m, r) see (8). It is known that the above − © number is an integer depending only on X and m but not on r. This number is called the index of the vector field X at m and is denoted by i(X, m). In fact this index can be viewed as the degree (in the sense of 278 algebraic topology; see the proof of (6.6) or [35]: theorem 4.2 p. 127) of the map:

1 d 1 d 1 (8.5.1.bis) ζ X ζ− : S − S − . ◦ ◦ m → 2. Let M be an oriented manifold and let X be an element of C (M) having an isolated zero at the point m of M. Now let (A, U) be a positive chart around m, such that X does not vanish on A m . Then it is known −{ } that for a suﬃciently small positive number r

depends only on X and m but not on (A, U). This number is called the index of X at m and is denoted by i(X, m). If we deﬁne ω0 by: ω = ω U (Rd) b 0 | 0 then b b

= 1 T 1 (8.5.3) i(X, m) ¨ (U X U− )∗(ζ∗(ω0)). d 1 Z ◦ ◦ ©S (U(m),r) − b 3. It is known that on a compact oriented r.m. (M, g) there exist vector fields which vanish only at a finite number of points, and that for any such vector field X, if we denote the set of points where it vanishes by ms, we have

(8.5.4) i(X, ms) = χ(M) Xx

279 where χ(M) is the Euler-Poincar´echaracteristic of M. A proof of this fact is usually got either by using the Lefschetz ﬁxed point theorem or by using Morse theory.

6 Gauss-Bonnet formula

In this section we assume that (M, g) is a compact oriented surface, that σ denotes its canonical volume form and that A is its sectional curvature. We follow the notation of the preceding article.

6.1

Theorem. (Gauss-Bonnet formula)

A σ = 2π χ(M). Z · · M 6. GAUSS-BONNET FORMULA 275

6.2 Lemma. For the form ω E 1(U(M)) (deﬁned in (5.2.11)) we have ∈

dω = p∗(A σ). − · Proof of the lemma. For every element x of U(M) let x be the unique element of T (M) such that x, x forms a positive orthonormal basis p(x) { } of Tp(x)(M). By the deﬁnition of ω.

1 (8.6.3) ω(z) = 0 z H ,ω(ζ− x) = 1. ∀ ∈ x x

First let us note that a basis of Tx(U(M)) is given by

1 a = ζ− x, a = C(x, x), a = C(x, x) { 1 x 2 3 } and then that to prove the lemma we have only to check the equality on pairs of these vectors. To compute dω let us extend a1, a2, a3 to vector ﬁelds on U(M).

6.4 280 We take arbitrary vector ﬁelds X, X on M such that

X(p(x)) = x, X(p(x)) = x and [X, X] = 0 and set 1 Z(y) = ξ− y y U(M). y ∀ ∈ H H Then clearly Z, X , X are extensions of a1, a2 and a3 respectively. By construction we have

XH, Z = XH, Z = 0, h i h i for pT and v vanish on both since they can be taken inside the brackets. Hence by (0.2.10) we have

(8.6.5) dω(XH, Z) = 0 and dω(XH, Z) = 0, 276 Some formulas and applications

and further

H T T p∗(A σ)(X , Z) = A σ(p X, p Z) = 0 · · ◦ ◦ since Z is vertical, and similarly

H p∗(A σ)(X , Z) = 0. · Now on the one hand we have dω(a , a ) = dω(XH, XH) = ω(R(x, x)x) by (5.10) 2 3 x − = g(R(x, x)x, x) = A(x, x) − since x and x are orthonormal. On the other hand

T T p∗(A σ)(a , a ) = A(p(x))σ(p (a ), p (a )) · 2 3 2 3 = A(p(x))σ(x, x) = A(p(x)) since x, x is a positive orthonormal basis of T (M). Hence the { } p(x) lemma. 281 Proof of the theorem. a) Let X be a vector ﬁeld on M which vanishes only on the ﬁnite set X m . Let M = M m and on M set X = . We choose a { s} −{ s} X positive number ǫ such that || || i) exp is ǫ-O.K. s ms ∀ ii) B(m , ǫ) B(m , ǫ) = φ if s , t. s ∩ t In particular, ms is the only zero of X in B(ms, ǫ), hence M B(m , ǫ) = B(m , ǫ) m s. ∩ s s −{ s}∀

b) For every r in ]0, ǫ[, the set Mr = M B(ms, r) is a nice domain − s ǫ , S of M, for expms is -O.K. and B(ms r) is a nice domain in Tms (M). The boundary of Mr is the union S (m, r). Note that the set s S (ms, r) as the boundary of B(ms, r)S has the orientation opposite to that on it as a part of the boundary of B(m , ǫ) B(m , r). s − s 6. GAUSS-BONNET FORMULA 277

c) By (6.2) noting that p X = id we have ◦ M (X)∗dω = X(p∗(C σ)) = A σ; − · − · and further by (2.4) iv,

(X)∗(dω) = d((X)∗ω).

Hence

A σ = d((X)∗ω) = i∗(X)∗ω Z · − Z − Z • • • Mr Mr b(Mr) by Stokes formula ((3.13)) and hence

(8.6.6) A σ = i∗((X)∗ω). Z · ZS (ms,r) • Xs Mr

Let λ = (exp B(m , ǫ)) 1. Then λ is a diﬀeomorphism of 282 s ms | s − s S (ms, r) onto S (ms, r). Hence by (0.3.9) we have

T 1 T 1 (8.6.7) i∗((X)∗ω) = (λ X λ )(((λ )− )∗(ω). Z Z ◦ ◦ − S (ms,r) S (ms,r)

By (8.5.3) we have (since 1 = 2π)

1 T 1 (8.6.8) i(X, ms) = λ X λ− ∗ λ∗(ω0) , 2π Z ◦ ◦ S (m ,r) s b T 1 d) Now let us look at ((λ )− ∗ )(ω) and ζ∗(ω0). We have λ 1 = exp and hence i− mi b T 1 T ((λ )− )∗(ω) = (exp )∗(ω);

from (4.6) it follows that

(expT ) ω = ζ (ω) = ζ (ω ). ∗ ∗ 0m ∗ 0 0mi i h i b 278 Some formulas and applications

Hence

T 1 ((λ )− )∗(ω) ζ∗ (ω ) 0 as n m . − ms 0 n → → s Now m M and z T (Tb(M)) we have (4.6), ∀ ∈ ∀ ∈ 0m m 1 T 1 λ− (z) = exp (z) = ζ− (z) = z || || || || || 0m || || || and hence

T 1 T 1 (ζ X λ ) (λ X λ− ) 0 as n m . ◦ ◦ − n − ◦ ◦ n → → s Hence by (8.6.7) and (8.6.8) it follows that 1 ((X∗)(ω) i(X, ms))ω 0 as r 0. 2π Z − → → S (ms,r)

283 Hence we have

lim A σ = 2π i(X, ms) = 2π χ(M) by (8.5.4). r 0 Z ·   · → • Xs   Mr   But since m has measure zero we have { s} lim A σ = A σ. r 0 Z · Z · → • M Mr

6.9 Corollary. If M is homeomorphic to S2 then

A σ = 4π, Z · M and if M is homeomorphic to S1 S1 then × A σ = 0. Z · M 6. GAUSS-BONNET FORMULA 279

6.10 Remarks. a) In the case of dimensions greater than three we have a formula: χ(M) = ϕ σ Z · M where ϕ is an element of F(M) depending on the curvature tensor R of (M, g). The relation between ϕ and R is complicated. The proof of the above formula proceeds analogously; one proves that there exists a form ψ such that

dψ = p∗(ϕ σ) · on U(M); here ψ is not the form ω of (5.2.11) but one far more complicated, which involves R: for details see [9]: pp. 38-41.

b) The formula A σ = 2π( i(X, ms)) proves that the sum · s MR P i(X, ms) does not depend on the vector ﬁeld X on M (put any s Pr.s. on M) so we know that this sum is an invariant. Then (8.5.4) gives us its value.

c) The formula A σ = 2π( i(X, ms)) proves that A σ does 284 · s · MR MR not depend on the r.s. chosenP on M but only on the diﬀerentiable manifold M.

d) In order to be self-contained, we prove (6.9) without appeal to (8.5.4); we will, in fact, use (6.1) in the next articles only for the sphere and for the torus. When M is homeomorphic to the torus we use the fact there exists on M a nowhere zero vector ﬁeld. Then there is a map X : M U(M) and, by Stokes formula, →

A σ = d((X)∗ω) = 0. Z · − Z M M

When M is homeomorphic to the sphere, we compute i(X, ms) s for a particularly simple vector ﬁeld on S2 and apply theP remark 280 Some formulas and applications

c). Let p denote the projection (x, y, z) (x, y) in R3 and X the → vector ﬁeld on S2 such that

2 T 1 m S : p (X(m)) = ζ− (p(m)) ∀ ∈ p(m) π where x x denotes rotation by in R2. Because the restric- → 2 tions of p to the northern and southern hemispheres are charts, the formula (8.5.2) (8.5.1.bis) give the index of X at the north and the south poles P, P′:

i(X, P) = i(X, P′) = 1. π In fact the maps S1 S1 so obtained are the rotations by or by 2 π → , and the degree of either is 1. −2 7 E. Hops’s theorem 285 Using (2.6) we prove the:

7.1 Lemma. If (M, g) is a complete r.m., and if m M and r > 0 are such ∈ that exp B(m, r) is injective, then exp is r-O.K. m | m Proof. If the result were false, x U (M) with 1 = j(x) < r. Let ǫ be ∃ ∈ m so that 1 < 1 + ǫ < r and let n = γx(1 + ǫ). Then, by (2.6), the restriction γ [0, 1+ǫ] < T (m, n). Hence 1+ǫ > d(m, n). By (8.4.8) we see that y x| ∃ such that y = d(m, n) and exp(y) = n. Hence exp(y) = exp((1 + ǫ)x) || || and y , (1 + ǫ)x, which is the required contradiction.

7.2 Proposition. Let m be a point of a complete r.m. (M, g). Then the following conditions are equivalent:

1) exp : T (M) M is injective. m m → 7. E. HOPS’S THEOREM 281

T 2) M is simply connected and expm is everywhere of maximal rank.

Proof. The fact that (2) implies (1) is (5.3). Now by (7.1) from (1) it fol- ﬀ lows that expm is of maximal rank, and hence expm is a di eomorphism between Tm(M) and M. Hence (1) implies (2). Now suppose that (M, g) is complete and that for every point m of M, exp is injective on T . Then it follows that m, n M the set m m ∀ ∈ T (m, n) consists of exactly one element. Then the geodesics have the set theoretic properties of straight lines in euclidean or hyperbolic geometry. It is of interest to see whether there exist compact r.m.s (M, g) whose universal riemannian covering (M, g) have geodesics sharing that property. In the case of surfaces any surface with A(M) ] , 0] will 286 e ⊂ −∞ do. But if we require that M be homeomorphice to a torus we have the following theorem due to E. Hopf ([15]).

7.3

Theorem. If (M, g) is an r.m. such that

1) M is NCB( ) ∞ 2) M is homeomorphic to S1 S1 × then A(M, g) = 0 . { } Proof. By the inequality in the ﬁrst part of (4.7), letting k tend to inﬁnity, we get A σ = 0, Z · M and then by the last part of (4.7) and (6.9) we get:

A(M) = 0 . { } 282 Some formulas and applications

7.4 Remarks. 1) For the dimensions greater than two the question is open.

2) For further considerations see article 10.

8 Another formula

In this article, we prove an inequality of integral geometry, which is slightly weaker than a formula of Santalo: see the remark at the end of the article. Let us consider (M, g) a complete, oriented r.m. of dimension d and A M a sub manifold of M of dimension d 1. For points n A we ⊂ − ∈ identify Tn(A) with its image in Tn(M) and deﬁne

U(A) = (U(M) T (A)) ∩ n [n A ∈ U(M) A = T (M). | n [n A ∈

287 Clearly the orthogonal of Tn(A) is one-dimensional and we assume from now on that we can choose unit vectors in this one dimensional subspace such that they can be extended into a vector ﬁeld N of M; then, (see (2.9)) we have (8)

♯ 1 U(A) = (U(M) A) (g (N)− (0)) | ∩ 8.1 Lemma. The subset

W = U(M) A U(A) { | }− of U(M) is a sub manifold of dimension 2d 2. − Proof. Suppose that locally A is given by ϕ 1(0) where ϕ F(M). Then − ∈ U(M) A is given by (ϕ p) 1(0) (for p : U(M) M). Further U(M) | ◦ − → 8. ANOTHER FORMULA 283 is a (2d 1) dimensional. Hence it follows that U(M) A is (2d 2) − | − dimensional. But U(A) is closed in U(M) A since, by (1.10), U(A) = | (g♯(X)) 1(0). Hence the given set is an open sub manifold of U(M) A. − | Hence the result. In this section we compute the volume of the set

(8.8.2) E = G (W) 0 < t < 1 , 1 { t | } where Gt is the geodesic ﬂow. To do this we introduce the map

f : R W U(M) × → deﬁned by the equation

(8.8.3) f (t, x) = Gt(x)

((Gt(x) makes sense, since M is complete and hence Gt is deﬁned for all t.). We deﬁne an orientation on A in a natural way by means of N: a basis x ,..., x of T (A) is said to be a positive basis if N(n), x ,..., x { 2 d} n { 2 d} is a positive basis for Tn(M) relative to the orientation on M. Let α be 288 the volume form of this oriented r.m. (A, g A). We use the symbol ω | exclusively for the restriction of ω (see (5.2.12)) to U(M) A. Then the | canonical volume form α of (U(M) A)) is given by the equation, | (8.8.4) α = ω (p∗(α)). ∧ On the other hand a volume form β on R W is given by × (8.8.5) β = p∗(dt) p∗(α) 1 ∧ 2 where p and p denote the projections from R W to R and W respec- 1 2 × tively. Hence ϕ F(R W) such that ∃ ∈ × (8.8.6) f ∗(σ) = ϕ β · where σ is the canonical volume form of (U(M), g). Clearly the computation of Vol(W, g) amounts to that of ϕ. 284 Some formulas and applications

8.7 Lemma. x W, t R we have ∀ ∈ ∈ ϕ(t, x) = ϕ(0, x).

Proof. This merely asserts the invariance of σ under the geodesic ﬂow. To be more precise let τ denote the transformation of R W into itself t × taking (s, x) into (s + t, x). Then since G + = G G we have the t s t ◦ s following commutative diagram:

f R W / V(M) ×

τt Gt

R W / V(M) × f

Hence we have

ϕ β = f ∗(σ) = f ∗(G∗(σ)) = (G f )∗(σ) [by (3.10) and (5.2.7)] · t t ◦ = ( f τ )∗(σ) = τ∗( f ∗(σ)) = τ∗(ϕ β) = τ∗(ϕ)τ∗(β). ◦ t t t ◦ t t

289 But by construction β is invariant under τt and hence we have

ϕ β = τ∗(ϕ) β. · t · Since β is never zero we have

ϕ(0, x) = ϕ(t, x).

8.8

Lemma. x W : ϕ(0, x) = g(x, N(p(x))). ∀ ∈ 8. ANOTHER FORMULA 285

Proof. Let p(x) = n and let y ,..., y and x ,..., x be orthonormal { 1 d} { 1 d} bases of Tn(M) with y1 = N(n) and x1 = x. Then if we set = = 1 = ui C(x, yi) and vi ζx− xi for i 1,..., d, we see, by deﬁnition of the r.s. on T(M), that u , u ,..., u ; v ,..., v { 2 3 d 2 d} is an orthonormal basis of Tx(W). Hence by the deﬁnition of β we have

β((P, 0), (0, u2), (0, ud); (0, v2),..., (0, vd)) = 1. Hence

ϕ(0, x) = f ∗(σ)(P, 0); (0, u2),..., (0, ud); (0, v2),..., (0, vd)) = ... T T T ... = ( f (P, 0), f (0, u2),..., f (0, vd)).

Since G0 = idU(M) we have f T (0, z) = z z T (W); (0,x) ∀ ∈ x and f T (P, 0) is the initial speed of the curve t G (x). But this is (0,x) → t C(x, x) by the deﬁnition of Gt (see (0.33)) and hence we have

ϕ(0, x) = σ(C(x, x), u2,..., ud, v2,..., vd).

Now let us write 290 x = g(x, N(n)) N(n) + k with g(k, N(n)) = 0. · Then C(x, x) = g(x, N(n)) C(x, N(n)) + C(x, k) = g(x, N(x)) u + C(x, k). · · 1

Since k is orthogonal to N(n) it is a linear combination of y2,..., yd and hence C(x, k) is that of u2,..., ud. Hence we have

σ(C(x, k), u2,..., ud, v2,..., vd) = 0. Hence we have ϕ(0, x) = g(x, N(n)) σ(u ,..., u , v ,..., v ) = g(x, N(n)). · 2 d 2 d 286 Some formulas and applications

8.9 Remarks. This lemma together with (8.7) in particular shows that f is of maximal rank. Hence E is open in U(M).

Now we come to our computation.

8.10 Proposition. If Vol(A, g A) is ﬁnite, then Vol(E , g) is ﬁnite and | 1 ¨ 21 d 2 Vol(E , g) · − ©Vol(A, g A). 1 d 1 · | − Proof. We know (8.9) that f T is everywhere of maximal rank and that f is onto. Hence we have by (3.10)

Vol(E1, g) f ∗(σ) ≤ Z | | [0,1] W × But the deﬁnition of β and ϕ and (3.17) gives

f ∗(σ) = ϕ (p∗(dt) p∗(α)) = ϕ p∗(α) dt Z | | Z | | 1 ∧ 2 Z Z | | 2 [0,1[ W ]0,1[ W 0 t W × × { }×

291 But neither ϕ nor p2∗(α) depends on t by (8.7), and hence we have

ϕ p∗(α) = ϕ p∗(α) = ϕ α Z | | 2 Z | | 2 Z | | t W 0 W W { }× { }× where we have identiﬁed 0 W with W (since G = id ). { }× 0 U(M) Hence we have

Vol(E1, g) 1 ϕ α. ≤ · Z | | W 8. ANOTHER FORMULA 287

2 ¨ g(x, N(n)) (ω Un(M)) = d 2 n; Z | | | d 1 · − ©∀ Un(M) − hence ¨ 2 ¨ 2 d 2 ϕ α = d 2 α = · − ©Vol(A, g A). Z | | d 1 · − ©Z d 1 · | U(M) − A −

8.11 Remark . If we assume that A, in addition to satisfying the conditions of the above sections, is convex (i.e. everywhere locally A is given by 1 ϕ− (0) with ϕ having a non-negative Hessian form D dϕ (see (4.11)) and if we set

+ W = x U (M) g(x, N(p(x))) > 0 , { ∈ n | } then, one can show that, for suﬃciently small l, actually f is a diﬀeo- 292 morphism between ]0, l[ W+ and f (]0, l[ W+), so that one has the × × equality ¨ 1 d 2 Vol( f (]0, l[ W+)) = · − ©Vol(A, g A) × d 1 · | − (see [27]: formula (21) p. 488). 288 Some formulas and applications

9 L.W. Green’s theorem

In (8) we have built up an r.m. (M, g) which is C -manifold m M, m ∀ ∈ which is homeomorphism to S2 but is not isometric to (S2, k. can) for any real number k. It is somewhat striking that if we replace “homeomorphic to S2” by “homeomorphic to P2(R)”, then there exists a positive real number k such that (M, g) is isometric to (P2(R), k. can). This theorem (9.5) is due to L.W. Green: [13] For the rest of this section let us assume that (N, h) is a Cm-manifold m N and that N is homeomorphic to P2(R). ∀ ∈ Under the above assumption it follows that the common length of the geodesics through a point n is independent of n. For if n′ is any other point, then by the Hopf-Rinow theorem and (4.8) it follows that there exists a geodesic γ through n and n′, and all geodesics through n′ have the same length as γ. Hence, by multiplying, if necessary, the r.s. on (N, h) by a positive constant we can assume that the length of each geodesic is π. Now let (M, g) be the universal riemannian covering of (N, h). Then by (6.5) it follows that

293 i) M is homeomorphic to S2

ii) (M, g) is a Cm-manifold for every m of M, and every geodesic starting from m and of length π ends at σ(m), where σ is the non trivial deck transformation of (M, g).

9.1

Lemma. m, m M, ∀ ′ ∈ d(m, m′) π, ≤

and equality occurs if and only if m′ = σ(m).

Proof. follows from ii). Now d(m,σ(m)) < π is absurd for take γ ≤ ∈ T (m,σ(m)). Then p γ would be a geodesic from p(m) to p(m) and of ◦ length <π contradicting the deﬁnition of a Cm-manifold. 9. L.W. GREEN’S THEOREM 289

9.2

Lemma. For every m of M expm is π-O.K.

Proof. Because of (7.1) we need only show that exp B(m,π) is injec- m | tive. Now let γ1 and γ2 be two geodesics meeting at a point m′ diﬀerent from both m and σ(m). Then we can suppose that the length of the curve γ1 between m and σ(m) is less than or equal to that of γ2. Then since γ1 , and γ2 are distinct geodesics we have γ1′ (m′) γ2′ (m′). Hence by (3.8) we have

d(m,σ(m)) < lg γ from m to m′ + lg γ from m′ to σ(m) 1| 2| lg γ from m to σ(m) = π. ≤ 2|

But this contradicts (9.1). 294

9.3

Corollary. One has: ar(M, g) 4π and equality occurs if and only if ≥ A(M, g) = 1 . { } Proof. By (9.2), (M, g) is NCB(π). Then the results follows from (4.7) and (6.9). 290 Some formulas and applications

9.4

Lemma . Any closed geodesic γ1 in (M, g) meets any geodesic γ of length π.

Proof. Let us suppose that they do not meet and that γ is a geodesic from m to σ(m). Then since M is homeomorphic to the sphere and γ1 is a simple closed curve (by Jordan curve theorem) it divides the rest of M into two connected components. Since γ is connected it is in one connected component. Let m′ be a point in the other. Then let γ2 be a geodesic curve from m to σ(m) passing through m′. Then m′ and σ(m) are in two diﬀerent connected components of M γ and γ from m − 1 2 ′ to σ(m) joins them. Hence γ2 from m′ to σ(m) meets γ1 and similarly γ2 from m to m′ does. Hence γ2 meets γ1 in at least two points m1 and m2. Then we have a part of the geodesic curve γ2, of length less than π, joining m1 and m2; and a part of the geodesic curve γ1 with the same property. But this contradicts the fact that exp B(m ,π) is injective. m1 | 1 Hence the result.

9.5 Theorem. (L.W. Green) Let (N, h) be an r.m. such that

i) N is homeomorphic to P2(R),

ii) (N, h) is a C -manifold n N (the common length being π). n ∀ ∈ Then (N, h) is isometric to (P2(R), can .).

295 Proof. (a) Let A be a closed geodesic of length 2π chosen once for all. It is a sub manifold of (M, g). Using the notation of (8) and (8.10) we have

2 (8.9.6) Vol(Eπ, g) 8π . ≤

(b) Now we claim that

Eπ = U(M) U(M) A. − | 9. L.W. GREEN’S THEOREM 291

Since, any two geodesics through a point m meet only at σ(m) by the deﬁnition of Eπ it follows that

Eπ U(M) U(M) A. ⊂ − | Now let x U(M) U(M) A, and let γ be the curve: ∈ − | t exp t.x, t [ π, 0]. → ∈ − But (9.4) it follows that t [ π, 0] such that exp(t x) A and ∃ 0 ∈ − 0 ∈ hence, if we set y = γ (t ) we have y U(M) A. Clearly since A ′ 0 ∈ | is a geodesic we have

Gt(U(A)) = U(A) and hence y < U(A). Further t , π. Hence y W and 0 − ∈ = G t0 (y) x Eπ. − ∈ (c) Now since U(M) A is of measure zero in U(M) we have | by (5.2.13)

Vol(Eπ, g) = Vol(U(M), g) = 2π ar(M, g). · But this together with (8.9.6) gives

ar(M, g) 4π. ≤ But then by (9.3) we have ﬁrst

ar(M, g) = 4π, and then A(M, g) = 1 . { } Now by (7.1) it follows that (M, g) is isometric to (S2, can) and, 296 since the deck transformation σ is such that

d(m,σ(m)) = π,

σ is nothing but the antipodal map of S2 hence (N, h) is isometric to (P2(R), can). 292 Some formulas and applications

9.7 Remarks. 1) The theorem was proved by Green under a slightly dif- ferent assumption, namely that (M, g) is a complete two dimensional r.m. such that j(x) = π for every x in U(M). In this form, the theorem was conjectured by Blaschke a long time ago.

2) In the proof of (9.5) one can replace the use of (8.10) by that of (12.9).

3) The question is open for the Pd(R)’s, d 3. ≥ 10 Concerning G-spaces

In order to bring out the results that are special to riemannian geometry and do not belong to distance geometry in general, H.Busemann was led to introduce metric spaces having the following properties.

i) Every closed bounded set is compact.

ii) Any two points can be joined by a geodesic.

iii) Geodesics can be prolonged locally.

iv) These prolongations are unique.

The results that follow are in [7]; the precise deﬁnition of the spaces is as follows.

10.1 297 A G- space M is a metric space M whose distance map d veriﬁes the following conditions:

i) every closed bounded set is compact;

ii) m, n M, p such that ∀ ∈ ∃ p , m, p , n and d(m, n) = d(m, p) + d(p, n); 10. CONCERNING G-SPACES 293

iii) a M r > 0 such that m, n M with ∀ ∈ ∃ a ∀ ∈ d(a, m) < r , d(a, n) < r , p with a a ∃ p , m, p , n and d(m, p) = d(m, n) + d(n, p);

iv) if m, n M and p , p are such that ∈ 1 2 d(m, pi) = d(m, n) + d(n, pi)(i = 1, 2), d(n, p1) = d(n, p2)

then p1 = p2. Under the above conditions i) and ii) one can define a set analogous to T (m, n) and the notion of geodesics. Then iii) gives local prolongability and iv) uniqueness of such prolongations. Whether a G-space is necessarily a topological manifold is an open question in dimensions greater than two. Now any complete r.m. is indeed a G-space; for i) comes from (4.3); ii) follows from (4.8), iii) comes from the existence of nice balls and iv) from (3.6). The essential result of Busemann is that the theorem (5.5) pertains to G-spaces theory, whereas the theorems (7.3) and (9.5) pertain to riemannian geometry. Concerning (5.5) the first thing to do is to find a metric definition equivalent to that of non positive curvature for a r.m.; such a definition is suggested by (3.8); in a G-space M and inside a ball D(a, r ) = m 298 a { ∈ M d(a, m) < r one proves the uniqueness of a geodesic between two | a} points, in particular of a midpoint (n, n , 1 ) n, n D(a, r ), defined by ′ 2 ∀ ′ ∈ a 1 1 d(n, (n, n , )) + d((n, n , ), n ) = d(n, n ). ′ 2 ′ 2 ′ ′ Then

10.2 Deﬁnition . A G-space is said to have non positive curvature if a ∀ ∈ M < s < r such that ∃ ◦ a a 1 1 1 d((m, n, ), (m, n′, )) d(n, n′) n, n′ D(a, s ). 2 2 ≤ 2 ∀ ∈ a Then Busemann proved the following theorems [7] (38.2) and (39.1): 294 Some formulas and applications

10.3 Theorem. A simply connected G-space M with non positive curvature is such that one can take r = in (10.1) iii) a M. In particular any a ∞ ∀ ∈ two geodesics in M meet in at most one point and M is homeomorphic to some Rd. On the other hand the theorems of L.W. Green and E. Hopf are not true for G-spaces, as shown by the following

10.4 Theorem. ([7], theorem (33.3)) There exists on S S a G-space struc- ′ × ture whose universal covering is not isomorphic to the canonical G- space R2, such that any two geodesics meet at most at one point. Theorem (33.3) of [7] gives far more: it characterizes completely the systems of curves in R2 which are the set of geodesics of the universal covering of a G-space structure on S1 S1; then it is easy to pick out × such a system which is not arguesian, although the canonical G-space structure on R2 is arguesian.

10.5

299 Theorem . (Skornyakov: [28], see also [8]) Let be any system of 2 2 closed Jordan curves in P (R) such that any two distinctP points in P (R) lie exactly on one curve of . Then there exists a G-space structure on 2 P (R) whose set of geodesicsP is . P 11 Conformal representation

If (M, g) is an oriented surface we can deﬁne a map

J : T(M) T(M) → by extending linearly the map

x x → 11. CONFORMAL REPRESENTATION 295 appearing in the beginning of the proof of (6.2). It follows directly from the deﬁnition that the map J has the following properties:

(i) J2 = id − T(M) (ii) g J = g ◦ i.e. g(J X, J Y) = g(x, Y), X, Y C (M) ◦ ◦ ∀ ∈ (iii) If f : M M is an isometry, then → f T J = J f T if f preserves orientation, ◦ ◦ f T J = J f T if f reverses orientation. ◦ − ◦ (iv) If h is any other r.s. on M such that h J = h then there exists a ◦ ϕ in F(M) such that h = ϕ g. Now suppose that on the tangent · bundle T(M) of an even dimensional manifold M a map J

J : T(M) T(M) → satisfying the equation 300

J2 = id − T(M) is given. Then in order that there exists on M a complex analytic structure for which J coincides with the multiplication by i it is necessary and suﬃcient that the map deﬁned by

[X, Y] J [J X, J Y] + J [J X, Y] + [J X, J Y] = 0 − ◦ ◦ ◦ ◦ ◦ ◦ ◦ X, Y C (M) ∀ ∈ be the zero map: see [23].

In the case of our (M, g), because of the bilinearity of the expression above, we can assume that J X = Y in the expression above; then it is ◦ zero. Hence it follows that every two dimensional oriented r.m. admits a complex analytic structure the associated J of which satisﬁes ii). This construction, applied to (S2, can) leads to the Riemann sphere. 296 Some formulas and applications

11.1 Remark. In out two dimensional case for the existence, we do not actually need the deep theorem of [23]. What we need is “the existence of isothermal coordinates” which is easier: see [10]. Now for the rest 2 of this article, let h0 denote the canonical r.s. on S , and h0 the canonical r.s. on a ﬂat torus (see (2.2)) or that on P2(R). It is clear that a e ﬂat torus admits a canonical complex structure, coming down from that of R2, and we denote the associated multiplication by √ 1 on its tan- − gent bundle by J0; we denote also by J0 the multiplication by √ 1 on 2 − the tangent bundle of the Riemann sphere (S , h0). Then the conformal 301 representation theorem for Riemann surface gives, in particular the e

11.2 Theorem. Let M be a one dimensional complex manifold. Then

i) if M is homeomorphic to S1 S1 there exists a flat torus (N, h ) × 0 and a diffeomorphism λ : N M → such that λT J = J λT ◦ 0 ◦ ii) if M is homeomorphic S2 there exists a diffeomorphism λ : S2 → M such that λT J = J λT ◦ 0 ◦ (the condition λT J = J λT simply means indeed that λ is ◦ 0 ◦ holomorphic).

Now suppose that we take the complex analytic structure associated 2 to the r.s. g on (M, g). Then on the ﬂat torus (N, h0) or on (S , can)

T T T λ∗(g) J = g λ J = g J λ = g λ = λ∗(g) ◦ 0 ◦ ◦ 0 ◦ ◦ ◦ which together with (iv) gives the 11. CONFORMAL REPRESENTATION 297

11.3 Proposition. Let (N, g) be an r.m. Then (1) if M is homeomorphic to S1 S1 then there exists a flat torus × (N, h ) and a diffeomorphism λ : N M and a ϕ F(N) such 0 → ∈ that λ (g) = ϕ h . ∗ · 0 (2) if M is homeomorphic to S2, then there exists a diffeomorphism λ : S2 M and a ϕ F(S2) such that λ (g) = ϕ h . → ∈ ∗ · 0 e 11.4

Corollary. Let (M, g) be an r.m. that is homeomorphic to P2(R). Then 302 there exists a diﬀeomorphism λ : P2(R) M and a ϕ F(P2(R)) such that → ∈ λ (g) = ϕ h . ∗ · 0 Proof. Let (M, g) be the universal riemannian covering of (M, g)

e e (M, g) e pe (p∗(g) = g)

(M, g)

Then M is homeomorphic to S2 and for the J associated to g and by (8.2.3), there exists a diﬀeomorphism µ : S2 M such that e → µT J = J µT ◦ 0 ◦ Let us denote the non trivial deck transformation of M by σ and set

1 σ = µ− σ µ e e ◦ ◦ The deck transformation σbis an isometrye (see in (6.5)) reversing the orientation and hence by (iii) we have e σT J = J σT ◦ − ◦ e e 298 Some formulas and applications

which gives σT J = J σT ◦ 0 − ◦ S2 Hence σ is an automorphismb of whichb reverses the sign of J, i.e. an antiholomorphic map, and further σ is an involution without fixed points (sinceb σ is). All antiholomorphic maps of the Riemann sphere being of the type b e az + b z → cz + d 303 one deduces that there exists a holomorphic map θ : S2 S2 such that → σ = θ σ θ 1 where σ = id R3 S2 is the antipodal map on S2. Now ◦ ◦ − − | | if we set λ = µ θ then λ is a diffeomorphism between S2 and M and ◦ bwe have e e 1 λ− σ λ = σ ◦ ◦ and hence we have a map λedefinede bye the following commutative diagram

λ 2 (M, g) o (S , h0) e ep e ep λ 2 (M, g) o (P (R), h0) Since λ is holomorphic by (iv) there exists a ψ in F(S2) such that e (λ)∗(g) = ψh0.

We have e e e (λ)∗(g) = λ∗(p∗(g)) = p∗(λ∗)(g)

and hence e e e p∗((λ∗(g)) = ψp∗(h0) which implies, since pT is of maximal rank, that there exists a ϕ in F(P2(R)) such that

ψ = ϕ p and λ∗(g) = ϕ h · · 0 Hence the result. 12. THE THEOREMS OF LOEWNER AND PU 299

12 The theorems of Loewner and Pu

In this article we consider a given manifold M of dimension two and various r.s. on it. We set a(g) = ar(M, g) and in case F(M) is non trivial (see (6)) 304

c(g) = inf lg(τ)) τ F(M) 0 { | ∈ −{ }} i.e. c(g) is the inﬁmum of the lengths of all closed curves in (M, g) which are not homotopic to zero. The theorem of Loewner gives a lower bound a(g) for for any r.s. on S1 S1 and that of P in the case of P2(R). (c(g))2 × u Before giving them we compute a(g)/(c(g))2 in the standard cases.

2 A) Let us take P (R) and denote its canonical r.s. by h0. By (6.7.6) a(h0) = 2π. To compute c(h0) we look at the universal riemannian covering: 2 2 (S , can) (P (R), h0). For every τ in F(P2(R)) 0 the lift τ of τ through m ends in 2 −{ } σ(m)(σ = idR3 S ) so that lg(τ) π and equality occurs if τ is − | ≥ a geodesic. Hence we have e e a(h ) 2 (8.12.1) 0 = . 2 2 c (h0)) π

2 B) Now let us take a ﬂat torus (T, h0) = (R /G, ǫ/G). Let us denote the metric on (Rd, ǫ) by ρ. Now let s, t be a system { } of generators of G such that

ρ(0, s(0)) = inf ρ(0, k(0)) k , 0, k G . { | ∈ }

We set a = s(0) and b = t(0). To compute c(h0) we look at the universal riemannian covering:

(R2, ǫ) (T, h ). → 0 300 Some formulas and applications

By our selection of s it follows that the lift of any closed curve τ is such that lg(τ) = lg(τ) ρ(0, a), ≥ 305 and equality is attained. Furthere we have

ρ(0, a) ρ(0, b), and ρ(0, a) ρ(a, b) ≤ ≤ because s t G, we have − ∈ 1 c(h ) (perimeter of 0ab). 0 ≤ 3

On the other hand we have by (5.7)

a(h0) = det(0a, 0b) = 2 (area of 0ab).

But the well known isoperimetric inequality for triangles gives that

√3 (8.12.2) (area of 0ab) (perimeter of 0ab)2 ≤ 36

where equality occurs if and only if 0ab is equilateral.

12.3

Deﬁnition . A ﬂat torus in R2 is said to be equilateral if there exists a similitude of R2 transforming it into the torus got by the group G generated by

1 3 s : 0 s = (0, 1) and t : 0 t = , . → → 2 2!

Then by (8.12.2) we have 12. THE THEOREMS OF LOEWNER AND PU 301

12.4 Lemma. For a ﬂat torus √3 a(h )/c2(h ) 0 0 ≥ 2 where equality occurs if and only if the torus is equilateral. C) The underlying idea of the proofs of the theorems of Loewner and Pu is as follows: ﬁrst using 11 we get an isometry between (M, g) and an r.s. of the type ϕ h on the standard manifold. Then we · 0 use the transitive group of isometries these standard manifolds possess to average the function ϕ in such a way a(h) decreases and c(h) increases. First we prove two lemmas, relative to the 306 following situation: (N, ǫ) is a compact surface and C a compact Lie group operating transitively, by isometries, on (N, ǫ); denote by τ any volume form on C such that

Vol(C ) = τ = 1. Z C We suppose given on N an everywhere strictly positive function and we consider the average ϕ, by C , of its square root, i.e. we set 2 b 1 ϕ =  (ϕ t) 2 τ . Z ◦ ·  t C  b ∈  In this context we have the following two lemmas.

12.5 Lemma . a(ϕ ǫ) a(ϕ ǫ) and the equality holds if and only if the · ≤ · function ϕ is constant on N. b Proof. Let σϕǫ be the volume element of the r.m. (N,ϕ ǫ); by (5.4), · σϕǫ = ϕσǫ so that

1 2 a(ϕ ǫ) = σϕǫ = ϕσǫ = (ϕ t) 2 τ) σǫ; · Z Z Z Z ◦ · N N N C b b b 302 Some formulas and applications

by Schwarz’ inequality

2 1  (ϕ t) 2 τ (ϕ t) τ τ = (ϕ t) τ Z ◦ ·  ≤ Z ◦ · Z Z ◦ · C  C C C     and

a(ϕ ǫ) (ϕ t)(τ σǫ) = (ϕ t) σǫ) τ. · ≤ Z ◦ × Z Z ◦ · · N C t C N b × ∈ But the fact that t is an isometry implies that t∗σǫ = σǫ and hence

(ϕ t) σǫ = t∗ϕ t∗σǫ = t∗(ϕ σǫ) = ϕσǫ (by (3.15)) Z ◦ · Z · Z · Z N N N N = a(ϕ ǫ); · 307 Consequently

a(ϕ ǫ) a(ϕ ǫ) τ = a(ϕ ǫ) Vol(C ) = a(ϕ ǫ). · ≤ Z · · · · · t C b ∈ Equality occurs if and only if it occurs in Schwarz’ inequality, ie. if and only if the function t ϕ t on C is constant, which implies that ϕ → ◦ itself is constant because C acts transitively on N.

12.6 Lemma. c(ϕ ǫ) c(ϕ ǫ). · ≥ · Proof. Let bγ be any curve in N; we suppose it given in the form γ : [0, 1] N. Then → 1 1 1 1 1 lg(γ, ϕ ǫ) = ((ϕ ǫ)(γ′, γ′)) 2 dt = ϕ 2 (ǫ(γ′ γ′) 2 dt = · Z · · Z · 0 0 b 1b b 1 1 = 2 2 =  (ϕ t) τ (ǫ(γ′, γ′)) dt Z Z ◦ ·  · 0 t C  ∈    12. THE THEOREMS OF LOEWNER AND PU 303

1 1 = (ϕ t) 2 (ǫ(γ′, γ′)) 2 τ dt = Z ◦ · × [0,1] C × 1 1 =  (((ϕ t) ǫ)(γ′, γ′)) 2 dt τ Z Z ◦ · ·  C  0    But since t is an isometry 

((ϕ t)ǫ)(γ′, γ′) = (t∗ϕ t∗ǫ)(γ′, γ′) = t∗(ϕ ǫ)(γ′, γ′) = ... ◦ · · T T ... = (ϕ ǫ)(t γ′, t γ′) = (ϕ ǫ)(t γ)′, (t γ)′). · ◦ ◦ · ◦ ◦ Now 1 1 ((ϕ ǫ)(t γ)′, (t γ)) 2 dt = lg(t γ,ϕ ǫ) Z · ◦ ◦ · ◦ · 0 so we get lg(γ, ϕ ǫ) = lg(t γ,ϕ ǫ) τ. · Z ◦ · · t C b ∈ Suppose now that τ is closed and non homotopic to zero in N; then the 308 same holds for the curves t γ, because the t’s are diffeomorphisms; ◦ hence by the definition of c(ϕ ǫ), · lg(γ, ϕ ǫ) c(ϕ ǫ) τ = c(ϕ ǫ) Vol(C ) = c(ϕ ǫ). · ≥ Z · · · · · C b Now this holds for any such γ, so that, by the definition of c(ϕ ǫ), we · get c(ϕ ǫ) c(ϕ ǫ). b · ≥ · b 12.7 Proposition. With the hypothesis of the lemmas, we have a(ϕ ǫ) a(ǫ) · (c(ϕ ǫ))2 ≥ (c(ǫ))2 · and equality holds if and only if ϕ is constant. 304 Some formulas and applications

Proof. We have only to remark that the transitivity of C implies the average ϕ is constant on N; let k be that constant value, then a(ϕ ǫ) = · k2 a(ǫ) and c(ϕ ǫ) = k c(ǫ); hence the two lemmas yield · b · · b b a(ϕ ǫ) k2 a(ǫ) a(ǫ) · · = (c(ϕ ǫ))2 ≥ k2 (c(ǫ))2 (c(ǫ))2 · ·

12.8 Corollary. (Loewner, unpublished) Let (M, g) be such that M is homeomorphic to S1 S1. Then × a(g) √3 (c(g))2 ≥ 2 where equality holds if and only if (M, g) is isometric to an equilateral ﬂat torus.

Proof. By (8.2.3) there exists a ﬂat torus (N, h), a diﬀeomorphism f from N to M and ϕ F(N) such that ∈

f ∗(g) = ϕ h. · 309 So that (M, g) and (N,ϕ, h) are isometric under f and in particular

a(g) = a(ϕ h), c(g) = c(ϕ h). · · Now the corollary follows from (12.7) combined with (12.4).

12.9 Corollary . (Pu: [25]) Let (M, g) be such that M is homeomorphic to P2(R). Then a(g) 2 (c(g))2 ≥ π where equality holds if and only if there is k > 0 such that (M, g) is isometric to (P2(R), k. can). 12. THE THEOREMS OF LOEWNER AND PU 305

Proof. This time we get the transitive group C of isometries of (P2(R) · can) by writing P2(R) as the homogeneous space

P2(R) = S 0(3)/0(2).

12.10 Remarks. A. The result (12.8) has been generalized by Blatter in [5], to compact orientable manifolds of genus γ 2; precisely: ≥ for any integer γ 2 there exists a real number nγ such that ≥ a(g) nγ (c(g))2 ≥

for any surface (M, g) where M is compact, orientable, of genus γ. The proof is definitely deeper than the one above. But here the equality is never attained. We sketch here the proof (based on an idea of N. Kuiper): suppose (M, g) is such that the equality is attained and call m-curve in (M, g) a curve of length equal to c(g) (they are closed geodesics); then first: given any point m M ∈ and any neighbourhood V of m there exists an m-curve which 310 meets V (in fact, if not, change slightly the r.s. inside V in order to have a smaller a(g); this change would not affect c(g) and so it is impossible). By continuity one shows now that m M there ∀ ∈ exists a one-parameter family f of m-curves such that f (0, 0) = m. Looking at the corresponding Jacobi fields, such a Jacobi field either never vanishes or vanishes at least at two points (thanks to the fact that M is oriented); but it cannot vanish at two points for (2.6) would imply that the geodesic is not an m-curve. Finally following such a family by continuity one would get a vector field on M, which is never zero, contradicting the fact that the genus of M is greater than one.

B. The theorems of Loewner and Pu are answers to particular cases of the following general question: given a compact diﬀerentiable 306 Some formulas and applications

manifold M and some homology or homotopy class or set of classes δ, and an r.s. g on M, set

a(g) = Vol(M, g) and c(g) = inf Vol(N, g N) N δ . { | | ∈ } Then: does one have an inequality a(g) k (c(g))n, valid for any ≥ · g; if so, what are the cases of equality? (the integer n is deﬁned a(g) so that the quotient be homogeneous of degrees zero). (c(g))n The good candidates for generalizing the Pu theorem are the S.C.- manifolds (excepting the spheres), when one takes for δ the class of a 311 projective subspace. Except in the case of P2(R) the question is completely open. However, the answer cannot be so simple; in fact, for the complex projective space either k has not the value of the canonical r.s. or the equality can be attained for an r.s. non isometric to the canonical one. The proof of this fact consists simply in deforming the canonical k¨ahlerian r.s. by adding to it √ 1 d d f (where f is a real function); − · ′ ′′ Stokes formula (3.13) and an inequality of Wirtinger show easily that both a(g) and c(g) are unchanged. Bibliography

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