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CONVEX HAMILTONIANS WITHOUT CONJUGATE POINTS GONZALO CONTRERAS AND RENATO ITURRIAGA

Introduction.

Let M be a closed connected riemannian manifold and T M its cotangent bundle. By a convex Hamiltonian on T M we shall mean a C 3 function H : T M ! R satisfying the following conditions: H (q; p) (calculated (a) Convexity: For all q 2 M , p 2 TqM , the Hessian matrix @p@i2@p j with respect to linear coordinates on Tq M ) is positive de nite. (b) Superlinearity: H (q; p) = +1 : lim jpj!1 jpj The Hamiltonian equation for H is de ned as q 0 = Hp ; (1) p0 = ?Hq where Hq and Hp are the partial derivatives with respect to q and p. Observe that the Hamiltonian function H is constant along the solutions of (1). Its level sets = H ?1feg are called energy levels of H . Then the compacity of M and the superlinearity hypothesis imply that the energy levels are compact. Since the Hamiltonian vector eld (1) is Lipschitz, the solutions of (1) are de ned on all R . Denote by t the corresponding Hamiltonian ow on T M . Our main interest in this paper are disconjugate orbits of the Hamiltonian ow, i.e. orbits without conjugate points. Let : T M ! M be the canonical projection and de ne the vertical subspace on 2 T M by V () = ker(d). Two points 1 , 2 2 T M are said conjugate if 2 = (1 ) for some 6= 0 and d (V (1 )) \ V (2 ) 6= f0g. In this work we generalize some results of the theory of geodesics without conjugate points. A rst step is to construct the Green subspaces for disconjugate orbits: Date : December, 1996. G. Contreras was partially supported by CNPq-Brazil. R. Iturriaga was partially supported by Conacyt-Mexico, grant 0324P-E. 1

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Proposition A.

G. CONTRERAS AND R. ITURRIAGA

Suppose that the orbit of 2 T M does not contain conjugate points and H () = e is a regular value of H . Then there exist two '-invariant Lagrangian subbundles E , F T (T M ) along the orbit of given by ? E ( ) = lim d ?t V ( t ( )) ; t!+1 F ( )

= t!lim d +1

t

?

V(

?t ())

:

Moreover, E () [ F () T , E () \ V () = F () \ V () = f0g, hX ()i E () \ F () and dim E () = dim F () = dim M , where X () = (Hp; ?Hq ) is the Hamiltonian vector eld and = H ?1 feg.

These bundles where constructed for disconjugate geodesics of riemannian metrics by Green [16] and of Finsler metrics by Foulon [15]. In the Finsler case this is done by e such that the solutions of the Euler-Lagrange equation de ning a new connection r e _ ( _ ) 0. This approach leads to extensions of many important theorems in satisfy r Riemannian geometry to Finsler geometry. See Chern [6] and the references therein for a survey and an introduction to this eld. If the energy level does not have conjugate points, the Green bundles E and F can be de ned on all 2 . Nevertheless, in general they are neither continuous (cf. [4]) nor transversal in T (i.e. dim E \ F = 1). An example of the relationship between the transversality of the Green subspaces and hyperbolicity appears in the following

Proposition B.

Let ? be a periodic orbit of t without conjugate points. Then ? is hyperbolic (on its energy level) if and only if E () \ F () = hX ()i for some 2 ?, where hX ()i is the 1-dimensional subspace generated by the Hamiltonian vector eld X (). In this case E and F are its stable and unstable subspaces.

We can compare proposition A with the following theorem by G. Paternain and M. Paternain [29].

0.1. Theorem. [Paternain] [29]

Suppose that e is a regular value of H , and that t j admits a continuous invariant Lagrangian subbundle E . Then, (a) E () \ V () = f0g 8 2 : (b) () = M . (c) contains no conjugate points.

HAMILTONIANS WITHOUT CONJUGATE POINTS

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It is not dicult to see (e.g. [21]) that when the ow t j is Anosov, the stable and unstable subbundles of T are Lagrangian. Paternain's theorem then implies that Anosov Lagrangian ows do not have conjugate points. This result was proven for geodesic ows by Klingenberg [20]. An extension of Klingenberg result appears in Ma~ne [21]. De ne B () = 2 T j sup jd t() j < +1 : t2R

An important property (cf. proposition 1.11) in the proof of proposition B is that if the orbit of is disconjugate then B () E () \ F (). This in turn implies that hX ()i E () \ F () and E () [ F () T (cf. corollary 1.12). The global version of proposition B is given by

Theorem C.

Let = H ?1 feg be a regular energy level without conjugate points. Then the following statements are equivalent: (a) t j is Anosov. (b) For all 2 , E () and F () are transversal in T . (c) For all 2 , E () \ F () = hX ()i. (d) If 2 , v 2 T , v 2= hX ()i then sup jd t () vj = +1 : t2R

Where hX ()i T is the 1-dimensional subspace generated by the vector eld X () of .

This theorem was proven by Eberlein (cf. [10]) for geodesic ows. An interesting ingredient on its proof is the characterization of recurrent quasi-hyperbolic R -actions of vector bundles. Given a vector bundle : E ! B consider a continuous R action : R ! Isom(E ; E) where t : E ! E is a bundle map which is a linear isomorphism on each ber and s+t = s t. The action induces a continuous

ow t : B - such that t = t . We say that is quasi-hyperbolic if sup jt ( )j = +1 for all 2 E ; 6= 0 ; t2R

and we say that it is hyperbolic if there exists a continuous splitting E = E s E u and C > 0, > 0 such that (i) jt( )j C e?t for all t > 0, 2 E s. (ii) j?t( )j C e?t for all t > 0, 2 E u. It is easy to see that hyperbolic actions are quasi-hyperbolic. The converse is false by counterexamples in Robinson [32] and Franks & Robinson [11]. But it is true with an additional recurrency hypothesis, (that in our case is always true because the Lagrangian ow preserves the Liouville measure).

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De ne the non-wandering set () of as the set of points b 2 B such that for every neighbourhood U of b there exists T > 0 such that T (U ) \ U 6= 0. The following theorem is a slight modi cation of a theorem in Freire [12], written there for quasi-Anosov ows (i.e when t = dt and E is a continuous invariant subbundle of TB transversal to the vector eld X , see theorem 3.1). For completeness of the exposition we present a proof of the following theorem in section x3.

0.2. Theorem. Let : E ! B be a continuous vector bundle and : R ! Isom(E ; E) a continuous R -action with induced ow t: t = t . If is quasi-hyperbolic and (jB ) = B , then is hyperbolic. The reason to use this version of Freire's result is that our Hamiltonian ows are not of contact type and a priori they may not preserve any continuous bundle transversal to the vector eld X (). Indeed, in apendix I we show an example of a convex Hamiltonian without conjugate points where this phenomenon occurs. The solution is to study the behaviour of an (orbitwise) reparametrization of the ow which preserves a continuous tranversal bundle E T (T M ). This reparametrization can not? be made global but instead we use the quasi-hyperbolicity of the action t() = P t () d t () P () where P () : T T M ! E () is the projection along the direction of the vector eld X ().

Generic Lagrangians. A C 4 function L : TM ! R is called a (convex autonomous) Lagrangian if it satis es the following properties: (a) Convexity: For all x 2 M , v 2 TxM , the Hessian matrix @v@i2@vL j (x; v) (calculated with respect to linear coordinates on TxM ) is positive de nite. (b) Superlinearity: L(x; v) = +1 ; lim jvj!1 jv j uniformly on x 2 R . Observe that by the compactness of M , the Euler-Lagrange equation, that in local coordinates is: d @L (x; x_ ) = @L (x; x_ ) ; (2) dt @v @x generates a complete ow ' : TM R ! TM de ned by ? 't(x0 ; v0 ) = x(t); x_ (t) ; where x : R ! M is the maximal solution of (2) with initial condition x_ (0) = v0.


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A Lagrangian ow is conjugated to a Hamiltonian ow by the Legendre transform FL : TM ! T M de ned by @L FL(x; v) = x; @v (x; v) : The corresponding Hamiltonian is given by H (q; p) = E (FL?1(q; p)), where E : TM ! R is the energy function de ned by @L E (x; v) := @v v ? L(x; v) : (3) (x;v) Conversely, given a convex Hamiltonian H on T M , the corresponding convex Lagrangian is given by the Legendre transform of the Hamiltonian FH : T M ! T M TM which can be seen as L(x; v) = max f p(v) ? H (x; p) g = (p Hp ? H ) FH : p Observe that item (b) in theorem 0.1 implies that the energy e of an Anosov level for the Lagrangian ow satis es e > e0 := ? max L(p; 0) ; p2M because e0 is the minimal energy level e for which () = M . De ne

c(L) := ? min

Z

TM

L d

is a ' ? invariant probability :

It is proven in [25], [8], that c(L) e0. We say that a '-invariant probability is minimizing if Z

c(L) = ? L d ; and we say that it is uniquely ergodic if it is the only invariant probability measure supported inside supp(). Action minimizing curves do not contain conjugate points. This fact is known since Morse earlier works (cf. [27]). We present a simple proof of this fact in section x4 and we use proposition B to prove the following theorem, which was stated as theorem III in Ma~ne's un nished work [25].

Theorem D.

There exists a generic set O C 1 (M; R ) such that for all 2 O the Lagrangian L + has a unique minimizing measure and this measure is uniquely ergodic. When this measure is supported on a periodic orbit or a xed point, this orbit (point) ? is hyperbolic and its stable and unstable manifolds intersect transversally W s(?) t W u(?).

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On Ma~ne [25], it is conjectured that there exists a generic set O such that this unique minimizing measure is supported on a periodic orbit or an equilibrium point. The rst statement of theorem D was proved by Ma~ne in [23]. Here we prove the generic hyperbolicity of the minimizing periodic orbit and the transversality property W s(?) t W u(?). The hyperbolicity of ? was announced in Ma~ne [25] and the transversality W s(?) t W u(?) was conjectured in Ma~ne [24]. We use the Clebsch transformation [7] to derive formulas for the index form on an orbit without conjugate points. Then we use similar arguments to the Rauch comparison theorem to obtain the transversality of the Green bundles of the periodic orbit. Finally, proposition B implies the hyperbolicity of the periodic orbit.

The metric entropy. The machinery developed for the proof of theorem C can be used to obtain formulas for the metric entropy of a Hamiltonian ow without conjugate points. Oseledec's theory [28] gives a splitting T = E s() E c() E u() for a set of points 2 of total measure, such that lim 1 log jd t () j < 0 for 2 E s() ; t!1 t lim 1 log jd t () j = 0 for 2 E c() ;

t!1 t

lim 1 log jd t () j > 0 for 2 E u() :

t!1 t

Following Freire and Ma~ne [13], we see in proposition 1.14 that E u() F () E u() E c() ; E s() E () E s() E c(): Given a riemannian metric on M , de ne the horizontal subspace as the kernel of the connection map. Then there exists symmetric linear isomorphisms U : H () ! V () such that F () = graph (U()) for all 2 . It satis es the Ricatti equation (7) U_ + U Hpp U + U Hpq + Hqp U + Hqq = 0; where Hpp, Hqp, Hpq and Hqq are linear maps which coincide with the second derivatives of H in local coordinates, and U_ is the covariant derivative of U de ned as 1 U_ ( ) v = lim h U( h ( )) v ? U( ) v 2 V ( ) T M; h!0 h where h : T h () M ! T() M is the parallel transport.


For 2 de ne the unstable Jacobian by 1 log det d j u : () := T !lim T E ( ) +1 T

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We prove in proposition 1.14 that Z T 1 () = T !lim tr [Hpq + Hpp U] ( t ()) dt ; +1 T 0 where tr ( ) is trace. From Ruelle's inequality [33], for any invariant measure for t j , we obtain that the metric entropy satis es

h ( j)

Z

d

Z

tr[Hpq + Hpp U] d :

Pesin's formula states that for a smooth invariant measure of a C 1+ dieomorphism f : N -, we have that

h (f ) = In our case we obtain

Z

N

d :

Theorem E.

Let m be the Liouville measure on an energy level without conjugate points. Then Z hm ( j) = tr[Hpq + Hpp U] dm :

This theorem was proved for geodesic ows by Freire and Ma~ne [13], Finsler metrics by Foulon [15] and mechanical Lagrangians by Innami [19].

The exponential map. The quasi-hyperbolicity in the case E () t F () is obtained from proposition 1.10 which states the horizontal growth of vertical vectors, i.e.

d d t jV () = +1: lim t!+1 The bounds obtained in proposition 1.10 are not uniform, by an example in Ballman,

Brin & Burns [4]. In section x6 we give a uniform lower bound for d d t jV () , 2 , jtj 1. This bound was given by Freire and Ma~ne [13] in the case of geodesic

ows. De ne the exponential map of an energy level as expq : TqM ! M , expq (t) = ( t ()), where 2 \ TqM . This map is not dierentiable at 0 2 TqM , but its Lagrangian version is dierentiable with d expq (0) = I . An important question is

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wether the lift of this map to the universal cover, expq : TqM ! M~ , is a dieomorphism in an energy level without conjugate points and, in particular, if the universal cover M~ is homeomorphic to R n . We show in corollary 1.18 that if the energy level has no conjugate points then expq is a local dieomorphism. Let = p dq be the Liouville's 1-form (4) in T M . Given a regular energy level and q 2 (), let q := (j)?1 fqg. Since M~ is simply connected, in order to obtain a global dieomorphism expq : TqM ! M~ it is sucient to show that expq is a covering map. For this it is enough the condition (X ) > 0. Indeed, in section x6 we prove the following

Theorem F.

Let H : T M ! R be convex and = H ?1 feg a regular energy level. Suppose that q 2 () satis es (a) The positive orbit of all 2 q has no conjugate points. ? (b) inf X (#) # 2 R+ (q ) > 0. Then the (Lagrangian) exponential map expq : Tq M~ ! M~ associated to ( t ; ) is a dieomorphism.

We give an example (cf. ex. I.4) in apendix I in which expq is not surjective and where all the orbits in of t starting at ?1fqg have no conjugate points. This shows that condition (a) is not sucient in theorem F. By theorem X in Ma~ne [25] (see also [8]), this can not happen for high energy levels. Example 1 in apendix I also shows that the norm of the derivative kd (expq )k may not be bounded below on an energy level without conjugate points. We don't know if the exponential map is a dieomorphism when the whole energy level has no conjugate points. The condition (X ) = p Hp > 0 is written (X ) = v Lv > 0 in Lagrangian form. It is satis ed by Finsler Lagrangians, L(x; v) = 21 kvk2x, mechanical Lagrangians L(x; v) = 12 hv; vix ? (x) (with (X ) = 1, H (q; p) = 21 hp; piq + (q)) and all convex Hamiltonians on suciently high energy levels. It fails for magnetic Lagrangians L(x; v) = 21 hv; vix ? (x) + x(v) (with d 6= 0) on lower energy levels. If one adds a closed 1-form to a Lagrangian, the Euler-Lagrange equation and hence the Lagrangian ow does not change. This may help to make a given Lagrangian ow satisfy condition (b). The obstruction to obtain a bound on kd t() (expq )k is the angle ? X ( t ()); d t(V () \ T ) . If the orbit of is disconjugate this angle is always nonzero by proposition 1.16. On Example 3 in apendix I this angle reaches zero before the rst conjugate point appear. Moreover, there are crossings of solutions on starting at () before the rst conjugate point.


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On section x1 we prove proposition A, theorem E and we show the proof of item (c) in theorem 0.1. In section x2 we study the transversal behaviour of the Hamiltonian

ow and prove proposition B. On section x3 we show the proof of theorem 0.2 and we prove theorem C. On section x4 we give a characterization of the index form and on section x5 we prove theorem D. On section x6 we give a uniform lower bound for d d t jV () and prove theorem F. Finally, we add some examples in the apendix. We want to thank G. Paternain, M. Paternain and R.O. Ruggiero for usefull conversations and J. Sotomayor for pointing us reference [17]. We also want to thank the hospitality of CMAT, Uruguay; IMPA, Brazil; CIMAT, Mexico and PUC-Rio, Brazil.

1. The Green Bundles. For proofs of the basic results of this section see Abraham & Marsden [1]. There is a canonical symplectic structure on T M given by a closed non-degenerate 2-form ! = d, where is the Liouville's 1-form, de ned by ( ) = (d() ) ; (4) where : T M ! M is the canonical projection. The Hamiltonian vector eld X of a function H : T M ! R is de ned by !(X; ) = ?dH . The Hamiltonian

ow t preserves the function H and the 2-form !. The level sets of H are called energy levels. Given a local chart q : U M ! R n it induces a natural chart (q; p) : T U ! Rn R n of T M . In natural charts the Liouville's 1-form is written as = p dq. Hence in natural charts ! = dp ^ dq and the Hamiltonian equations are given by q_ = Hp ; p_ = ?Hq : (5) In general, a set of local coordinates (q; p) of T M is called symplectic if the canonical 2-form is written as ! = dp ^ dq. A subspace E T (T M ) is said Lagrangian if dim E = n = dim M and !(x; y) = 0 for all x; y 2 E . Fix a riemannian metric on M and the corresponding induced metric on T M . Then T T M splits as a direct sum of two Lagrangian subspaces: the vertical subspace V () = ker (d()) and the horizontal subspace H () given by the kernel of the connection map. Using the isomorphism K : T T M ! T() M T() M , 7! (d() ; r ( )), we can identify H () T() M f0g and V () f0g T() M T() M . If we choose local coordinates along t 7! t () such that t 7! @q@ i ( t ()) are parallel vector elds, then this identi cation becomes $ (dq( ); dp( )). Let E T T M be an n-dimensional subspace such that E \ V () = f0g. Then E is a graph of some linear map S : H () ! V (). It can be checked that E is Lagrangian if and only if in symplectic coordinates S is symmetric.

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Proof of Proposition A. Take 2 T M and = (h; v) 2 T T M = H () V () T() M T() M . Consider a variation

?

s(t) = qs(t); ps(t) such that for each s 2]?"; "[, s is a solution?of the Hamiltonian H such that 0 (0) = d and ds s()js=0 = . Writing dt( ) = h(t); v(t) , we obtain the Hamiltonian Jacobi equations h_ = Hpq h + Hpp v ; v_ = ?Hqq h ? Hqp v ; (6) where the covariant derivatives are evaluated along (o(t)), and Hqq , Hqp, Hpp and Hqq are linear operators on T() M , that in local coordinates coincide with the H , @ 2 H and @ 2 H . Moreover, matrices of partial derivatives @q@i2@qH j , @q@i2@p @pi @pj @pi @qj j since the Hamiltonian H is convex, then Hpp is positive de nite. We derive now the Ricatti equation. Let E be a Lagrangian subspace of T T M . Suppose that for t in some interval ] ? "; "[ we have that d t (E ) \ V ( t ()) = f0g. Then we can write d t(E ) = graph S (t), where S (t) : H ( t ) ! V ( t ) is a symmetric map. That is, if 2 E then ? d t ( ) = h(t); S (t) h(t) : Using equation (6) we have that _ + S (Hpq h + HppSh) = ?Hqq h ? HqpSh : Sh Since this holds for all h 2 H ( t ()) we obtain the Ricatti equation: S_ + SHppS + SHpq + HqpS + Hqq = 0 : (7) Take the symmetric map S (t) for which d t (V ()) = graph(S (t)), t > 0. Let Y (t) : T() M V () ! H ( t ) T t M , t 2 R , be the family of isomorphisms Y (t)w = d d t () (0; w), where (0; w) 2 H () V (). Then for t > 0, Y (t) satis es Y_ = (Hpq + Hpp S ) Y (8) Y (0) = 0 ; Y_ (0) = Hpp ; where Y_ (0) is calculated from (8) using that lim S (t); Y (t) = I; : t!0 Given u 2 T() M , de ne

h1(t) = Y (t) u ; v1(t) = S (t) Y (t) u ; t > 0 :


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Then (h1 ; v1) is a solution of equation (6) with h1 (0) = 0, v1 (0) = u. Let 2 T T M and let (h(t); v(t)) = d t ( ) 2 H V . Since d t preserves the symplectic form, we have that hh(t); v1 (t)i ? hv(t); h1(t)i = hh(0); ui Hence hY (t) S (t) h(t) ; ui ? hY (t) v(t) ; ui = hh(0) ; ui for all u 2 T() M . Therefore v(t) = S (t) h(t) ? (Y (t))?1 h(0) : (9) Since h_ = Hpq h + Hpp v we get h_ = (Hpq + HppS )h ? Hpp(Y )?1 h(0) (10) Dierentiating Y ?1 h and using equation (8), we obtain

h(t) = Y (t) Y (c)?1 h(c) + Y (t)

Z c

Y ?1 (s) Hpp (Y (s))?1 h(0) ds ; for 0 < t < c: t (11) Now consider the map d t restricted to the subspace d ?c(V ( c )), c > 0. Its projection Zc(t) to the horizontal subspace is given by the solutions of (6) with h(c) = 0. Then Zc(t) is the family of isomorphisms given by Zc(t) = Y (t)

Z c

t

Y ?1 (s) Hpp (Y (s))?1 ds ; 0 < t < c ;

(12)

because Zc(t) satis es (6) for 0 < t < c, Zc(c) = 0 and Z (t) = I : Zc(0) = tlim !0+ c

Observe that from its de nition, the linear map Zc(t) is de ned for all t 2 R and the no-conjugate points condition implies that Zc(t) is an isomorphism for all t 6= c. Moreover, since Zc(t) is a matrix solution of the Jacobi equation (6), then Z_ c(0) exists, and taking the limit in (12), we have that the linear isomorphism

Z_ d(0) ? Z_ c(0) = Hpp()

Z d

c

Y (s)?1 Hpp (Y (s))?1 ds

(13)

is symmetric and the second factor on the right is positive de nite for all 0 < c < d. In particular h

i

Zd(t) ? Zc(t) = Y (t) Hpp()?1 Z_ d (0) ? Z_ c(0) ; 0 < c < d : For " > 0 de ne Z?"(t) := Y (t) Nc + Zc(t) ; t 2 R ;

(14) (15)

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where Nc := ?Y (?")?1 Zc(?"). We have that Z?"(t) is the solution of the Jacobi equation (6) satisfying Z?"(?") = 0, Z?"(0) = I and Z_ ?"(0) ? Z_ c(0) = Hpp() Nc : (16)

1.1. Claim. Y ?1(t) Zc(t) is symmetric for all t 6= 0. In particular Nc is symmetric for all c > 0.

Proof: Using (10), we have that

Z_ c = A Zc + Hpp (Y )?1 ; Y_ = A Y + 0 ; t 6= 0 ; where A = Hpq + Hpp S . And then

Z_ ?"= A Z?" + Hpp(Y )?1 : Using (9) we have that the functions h2(t) = Z?"(t) u ; ? v2(t) = S (t) h2(t) ? Y (t) ?1 u ; are solutions of (6) with h2 (?") = 0, h2 (0) = u. Similarly, the functions h3 (t) = Zc(t) w ; ? v3 (t) = S (t) h3(t) ? Y (t) ?1 w ;

(17)

(18)

are solutions of (6) with h3 (0) = w, h3(c) = 0. Since the ow preserves the symplectic form, we have that the following expressions do not depend on t for all u, w 2 T() M : h h2(t); v3 (t) i ? h v2(t); h3(t) i h Z?" u ; (SZc ? (Y )?1 ) w i ? h (SZ?" ? (Y )?1) u ; Zc w i Zc SZ?" ? Y ?1Z?" ? Zc SZ?" + Zc (Y )?1 = Zc(Y )?1 ? Y ?1Z?" : Using formula (15), this is equal to (Y ?1 Zc) ? Y ?1Zc ? Nc = ?Nc ;

(19)

where the right hand side corresponds to its value when t = ?". Since Zc(c) = 0, evaluating the equation at t = c, we get that Nc is symmetric for all c > 0. Using (19) again, we obtain that Y ?1Zc is symmetric for all t 6= 0. 1.2. Claim. Nc is positive de nite for all c > 0.


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Proof: It is enough to show it for Nc?1 . We have that

d ? Z (t)?1 Y (t) = Z ?1 Z_ Z ?1 Y ? Z ?1 Y_ = ?H : pp c c c c t=0 t=0 dt c Therefore ?Zc(t)?1Y (t) is positive de nite for small t < 0. By claim 1.1, this matrix is symmetric for all t < 0. Since for t < 0 its determinant never vanishes, it is positive de nite for all t < 0, in particular for t = ?", Nc?1 is positive de nite.

De ne a partial order on the symmetric isomorphisms of T() M by writing A B if Hpp()?1 (A ? B ) is positive de nite. From (13) we have that the family F = fZ_ c(0) ? Z_ 1(0) j c > 1 g is monotone increasing. Moreover, by (16) and claim 1.2, we have that _Zc(0) ? Z_ 1(0) Hpp Nc + Z_ c(0) ? Z_ 1(0)

Z_ ?"(0) ? Z_ 1 (0) :

So that the family F is bounded above. It can be seen that this implies the existence of a lower upper bound which is the limit of the family. We obtain the rst part of

1.3. Claim.

(a) limc!+1 Z_ c(0) ? Z_ 1 (0) = Q exists and is symmetric. (b) limc!+1 Zc(t) = D(t) exists (uniformly for bounded t intervals). Moreover, h(t) = D(t) ; ? v(t) = S (t) D(t) + Y (t) ?1 ; t 6= 0 is a matrix solution of (6) such that D(0) = I , D_ (0) = Q + Z_ 1 (0) and det D(t) 6= 0 for all t 2 R . Part (b) is a consequence of (14) and the continuous dependence of the solutions of equation (6) on the initial data. We have that the subspace ? E (0) := u; v(0) u u 2 T() M H () V () is the limit of the subspaces ? ? Ec() := d ?c V ( c()) = h3 (); v3() u 2 T() M where h3(t) and v3 (t) are given by (18). Moreover ? ? E (t) := d t E (0) =

h(t) u; v(t) u u 2 T() M

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satis es E (t) \ V ( t ) = f0g and E (t) =

?

?

lim d V ( c()) = d!lim d ?d V ( d( t )) = E t (0) : c!+1 t?c +1 Since the vertical subspace is Lagrangian, then K (c) is Lagrangian. By the continuity of the symplectic form, the subspaces E (t) are Lagrangian. To obtain the \unstable" Green bundle F (), observe that the ow of the Hamiltonian H () = ?H () is the ow t of H with the time reversed. In this case H pp is negative de nite, but similar arguments apply to obtain the subbundle F . Applying the lemma to H , we obtain the subbundle F . Moreover, in this case the family G = f Z_ ?c(0) ? Z_ ?1(0) j c > 1 g is monotone decreasing and bounded below by Z_ "(0) ? Z_ ?1(0). This nishes the proof of the rst part of proposition A, we defer the proof of the second statement to corollary 1.12. Let

S(t) : =

v (t) h(t)?1

= S (t) + Y (t)?1 D(t)?1:

(20)

Then S(t) is a symmetric solution of the Ricatti equation (7), which is de ned on all t 2 R. The symmetry of S(t) can be checked either because E (t) = graph (S(t)) is a Lagrangian subspace or because Y (t)?1 D(t) = clim Y (t)?1 Zc(t) !1 is symmetric and this implies that (Y )?1 D?1 is symmetric. Let S(), U() be the the symmetric solutions of the Ricatti equation (7) corresponding to the green bundles E () = Graph(S()) and F () = Graph(U()). Let Kc() : H () ! V () be the symmetric linear map such that Graph(Kc()) = d ?c(V ( c())). De ne a partial order on the symmetric isomorphisms of T() M by writing A B if A ? B is positive de nite.

1.4. Proposition. For all " > 0, (a) If d > c > 0 then K?" Kd Kc. (b) If d < c < 0 then K" Kd Kc. (c) d!lim Kd = S, d!?1 lim Kd = U. +1 (d) S 4 U:

Proof:


15

Let (Zc(t); Vc(t)) be the matrix solution of the Jacobi equation (6) with Zc(0) = I , Zc(c) = 0, c > 0. From (9) we have that Vc(t) = S (t) Zc(t) ? (Y (t))?1: (21) Hence, using (14) and (13) we have that Vd(t) ? Vc(t) = S (t) [Zd(t) ? Zc(t)] (22) = S (t) Y (t)

Z d

c

Y ?1 Hpp (Y )?1 ds:

(23)

When t ! 0 we have that S (t) Y (t) ! I . Therefore

Vd(0) ? Vc(0)

Z d ? ?1 = Y H

c

pp (Y

)?1 ds:

(24)

Let Kc(t) = Vc(t) Zc(t)?1 be the corresponding solution of the Ricatti equation (7). Since Zc(0) = I , we have that Kc(0) = Vc(0). By (24), Kd(0) ? Kc(0) is positive de nite if d > c > 0, hence the sequence Kd(0), d > 0 is strictly increasing. From (17) we have that equation (22) also holds for c = ?". Using (15), we have that Vd(t) ? V?"(t) = S (t) (Zd(t) ? Z?"(t)) = S (t) (?Y (t) Nd) : Hence Vd(0) ? V?"(0) = ?Nd is negative de nite. Since Z?"(0) = I , then K?"(0) = V?"(0). Therefore K?" Kd for all d > 0. This completes the proof of item (a). Item (c) has already been proven on claim 1.3 above. Item (b) is proven similarly using the concave Hamiltonian H = ?H and item (d) is a corollary of items (a), (b) and (c).

1.5. Remark. If a bounded open segment f t() j a? < t < a+ g has no conjugate

points, the arguments above apply to obtain limit solutions limc!a Zc(t) = D(t), D(0)? = I , det D (t) 6= 0, Lagrangian subspaces E () = limt!a d ?t(V ( t )) = d a V ( a ()) and the monotonicity properties of proposition 1.4. Nevertheless, in general E () 6 T , see remark 1.17

1.6. Remark. The existence of the Green bundles, the solutions h, v, U, S and the monotonicity properties in proposition 1.4 do not need the uniform boundedness of the operators Hpp, Hpq , Hqq . But in the unbounded case we can not guarantee neither that X () 2 E () \ F () nor E () [ F () T .

16

1.7. Proposition.


(a) Any symmetric solution S (t) : H ( t()) ! V ( t ()) of the Ricatti equation (7) which is de ned on t > 0 is uniformly bounded on t 1, i.e. there exists A > 0 such that kS (t)k < A for all 2 ; t > 1: (b) Any measurable symmetric solution of the Ricatti equation de ned on the energy level is uniformly bounded. An inmediate consequence is the following

1.8. Corollary.

(a) The solution S (t) of equation (8) given by Image(Y (t) ; S (t) Y (t)) = d t V () is uniformly bounded for jtj > 1. (b) The solutions S(t), U (t) corresponding to the Green bundles are uniformly bounded on . We shall need the following lemma, whose proof is elementary: 1.9. Lemma. The function w(r) = R coth (R t ? d), R > 0, satis es (i) w_ + w2 ? R2 = 0. (ii) w_ < 0, w (?t + Rd ) = ?w (t + Rd ). (iii) limt!+1 w(t) = R, limt!?1 w(t) = ?R. (iv) limt!( d )+ w(t) = +1, limd ? w(t) = ?1. R t!( ) R

Proof of proposition 1.7:

We rst prove (a). By the spectral theorem, it is enough to prove that x S (t) x is uniformly bounded for kxk = 1 and t > 1. Since Hpq = Hqp and S = S , we have that S_ + S Hpp S + Hpq S + S Hpq + Hqq = 0: Let C := (Hpp)?1 Hpq , D := Hqq ? C HppC ? C_ and V (t) := S (t) + C ( t ()). Then V_ + V Hpp V + D = 0 : Since C () and D() are continuous, then they are uniformly bounded on the energy level. Hence it is enough to prove that x V (t) x is uniformly bounded on t > 1 and 2 . Observe that V (t) is not necessarily symmetric. Let M > 0 be such that yHpp() y M1 kyk2 for all 2 : (25) Let R > 0 be such that j yD() y j < M R2 for all 2 ; kyk = 1: (26)


17

We claim that j yV (t) y j M R coth(R) for all kyk = 1; t > 1; 2 : For, let 0 2 , kxk = 1. Write V (t) := V (t), Hpp(t) := Hpp( t ). Suppose that there exists to 2 R such that xV (t0 ) x =: M > M R coth(R): There exists d0 2 R such that R coth (R t0 ? d0 ) = . Observe that d0 > 0 and t0 > d0 > 0. Write w(t) := R coth (R t ? d ). Then w(t) is a solution of w_ + w2 ? R2 = 0 0 R d for t > R0 . In particular, M w_ + M w2 ? M R2 = 0: Let f (t) := x V (t) x ? M w(t). Then f (t0 ) = 0 and ? ? f 0(t) + x V (t)Hpp(t) V (t) x ? M w(t)2 + x D(t) x + M R2 = 0 : (27) Using Schwartz inequality and (25), we have that M w(t0)2 = M 2 = M1 (M )2 = M1 hx; V (t0 ) xi2 ? ? M1 jV (t0) xj2 V (t0 ) x Hpp(t0 ) V (t0) x

M w(t0)2 x V (t0) Hpp(t0) V (t0 ) x (28) Then (28), (26) and (27) imply that f 0(t0 ) < 0. The same argument can be applied each time that f (t) = 0. Therefore x V (t) x M w(t) for all t > t0 and x V (t) x M w(t) for all dR0 < t < t0 : Then lim x V (t) x lim M w(t) = +1 : d0 + t!( R ) t!( dR0 )+ Since dRo > 0, this contradicts the existence of V (t) for t > 0. Hence such t0 does not exists and x V (t) x M R coth(R) for all t > 0: Now suppose that there exists t1 2 R , kzk = 1 such that x V (t1) x < ?M R : We compare v(t) = x V (t) x with M w1(t), where w1(t) = R coth (R t ? c0 ) is such that M w1(t1 ) = v(t1 ). Observe that w1(t) is de ned for t < cR0 and in this case c0 > 0 and cR0 > t1 > 0. The same argument as above shows that v(t) M w1(t) for t1 t < cR0 . Since limt!( c0 )? w1(t) = ?1, this contradicts the fact that V (t) is R de ned for all t 2 R .

18


The prove of (b) is similar to that of (a). In this case we can change the MR coth(R) bound by MR. We allow t0 2 R and d0 2 R to be nonpositive if necessary. The second inequality has also the same proof, but here we allow t1 2 R and c0 2 R .

The following proposition states that limt!1 d d t ()jV () = +1 for all 2 . This limit is not uniform in 2 . A uniform lower bound for this norm is given in section x6.

1.10. Proposition. For all R > 0 there exists T > T (R; ) > 0 such that jY (t) vj > R jvj for all jtj > T and all v 2 T M n f0g. Proof: Let

M (t) :=

Z 1

t

Y ?1(s) Hpp(s) (Y (s))?1 ds :

From (12), using that D(t) = limc!1 Zc(t), we have that D(s) = Y (s) M (s) for s > 0 : (29) Consider the solutions S (t), S(t) of the Ricatti equation given on (8) and (20). From (20) and (29) we have that S(t) ? S (t) = Y (t)?1 M (t)?1 Y (t)?1 for t > 0: Since the solutions S (t), S(t) are de ned on all t > 0, by proposition 1.7, there exist t0 > 0 and k > 0 such that kS (t)k < k and kS(t)k < k for all t > t0. Then for jxj = 1 and t > t0 ,

M (t)?1 Y (t)?1 x ; Y (t)?1 x = jh S (t) x ; x ij + jh S(t) x ; x ij 2 k : Let (t) be the largest eigenvalue of M (t), t > 0. Since M (t) is positive de nite then (t) > 0 and kM (t)k = (t). Moreover,

2 k M (t)?1 Y (t)?1 x ; Y (t)?1 x (1t) Y (t)?1 x 2 ; Y (t)?1 x

p 2 k kM (t)k 12

Then if jvj = 1, we have that

jY (t) vj kY (t1)?1k p

for all

jxj = 1; t > t0 :

1 for t > t0 : 1 2 k kM (t)k 2 Since M (t) ! 0 when t ! +1, given R > 0 there exists T > t0 such that 1 (2 k kM (t)k) 2 < R1 for all t > T and hence jY (t) v j > R for all t > T and jvj = 1. Using the Hamiltonian H := ?H , we get the result for t < ?T .


For 2 TM , de ne

n

19

o

B () := 2 T T M sup jd t () j < +1 : t2R

1.11. Proposition. If the -orbit of 2 TM does not contain conjugate points, then B () E () \ F (). An example showing that in general E () \ F () 6 B () is given in Ballmann, Brin & Burns [4]. Proof: Let 2 B () and let ? 2 ET () := d ?T (V ( T ())) be such that d() T = d() , in particular ? T 2 V (). Since limT !+1 ET () = E () and E ( ) t V ( ) then there exists := limT !+1 T 2 E ( ). Moreover, d d T ( ? T ) = d d T ( ). Since d d T ( ) is bounded on T > 0, by proposition 1.10, limT !+1( ? T ) = 0. Since T ! 2 E (), then 2 E (). Similarly if T 2 FT () := d T (V ( ?T ())) is such that d() T = d() . Then limT !+1 T 2 F () and limT !+1( ? T ) = 0.

1.12. Corollary. If the orbit of has no conjugate points then (a) hX ()i E () \ F (). (b) E () [ F () T . Proof: The property hX ()i E () \ F () is a direct consequence of proposition 1.11. It remains to show that E () [ F () T . Observe that since ?dH = !(X; ), then T = f 2 T T M j !(X (); ) = 0 g : Since F () is Lagrangian and X () 2 F (), then !(X (); ) = 0 for all 2 F (). Hence F () T . Similarly E () T . 1.13. Corollary. If M is connected and is a regular energy level without conjugate points, then () = M .

Proof: For 2 , d() : F () ! T() M is an isomorphism. By corollary 1.12, F ( ) T . Hence : ! M is an open map. Since M is connected then () = M .

The metric entropy. Let be the set of total measure having the Oseledec's splitting T T M = E s() E c() E u().

20


1.14. Proposition. For all 2 (a) E s() E () E s() E c(). (b) E u() F () E u() E c(). Z 1 lim T !+1 T

(c)

T

0

tr Hpq + Hpp U

?

t ()

dt.

Proof:

We only proof

E u() F () E u() E c() ; (30) which shall be used for theorem E, the other inclusions are similar. We rst prove the second inclusion. Observe that for v = (0; v) 2 V () we have that d t () v = (Y (t) v; ). Then, by proposition 1.10, we have that V () \ E s() = f0g. For every 2 there exists a subspace R() E u() E c() and a linear map L() : R() ! E s() such that V () = graph L(). Let be an ergodic invariant measure for t . Given " > 0 let K be a compact subset such that (K ) > 1 ? " and (a) sup kL()k j 2 K < +1. (b) There exists T > 0 and 0 < < 1 < < ?1 such that for t > T and 2 K

?1

t

t jE s () < t and

d t jE u()E c () < ? Let V?s() := d s V ( ?s()) . Then ? V?s() = d s graph L( ?s ()) h?

?

= graph (d s L( ?s () d sjR(

?s ()

?1 i

?

For 2 K take a sequence sn ! +1 such that ?sn () 2 K . Since d sn R( ?sn ()) E u() E c(), by properties (a) and (b), we have that

? sn sn

d s L( ?s ( )) d ?s jd R(

sn n n n ?sn ()) sup kL( )k : Since

sn sn

! 0 when n ! +1, we have that ?

2K

E u() E c() : Therefore the second inclusion in (30) is satis ed for 2 K . Since (K ) 1 ? " F ( )

d = n!lim V?sn () = n!lim +1 +1

sn

R(

?sn ())

and " and are arbitrary, this inclusion holds for a set of total measure in . We prove now the rst inclusion. There exist (cf. [1] theorem 3.1.19) a continuous riemannian metric h ; i on T M and a continuous family of linear isomorphisms J () : T T M - such that J ()2 = ?I and the symplectic forms is written as !(x; y) = hx; J ()yi for all x; y 2 T T M and 2 T M . From now we use this metric. Let fe1 (); : : : ; en()g be an orthonormal basis of F () and de ne en+i() =


21

Jei() for i = 1; : : : ; n. Since the subspace F () is Lagrangian, the subspace J F () = span fen+1(); : : : ; e2n()g is the orthogonal complement of F () with respect to h ; i. The matrices of ! and d t () with respect to the family of basis fe1 (); : : : ; e2n()g are I A ( ) C ( ) t t J = ?I and d t () = B () t

Since d t preserves the symplectic form we have that (d t ) J (d t) = J . Hence Bt() = At()?1 and Bt() Ct () is symmetric. Since F () E u() E c() then 1 log

B ?1() w

= lim 1 log kA () wk 0 lim t t t!1 t t!1 t for all w 2 R n , 2 . This implies that 1 log

B ?1() w

0 : (31) lim t t!1 t Suppose that E u() 6 F (). Take v 2 E u() n F (). Then? v = w + z with z 2 F () and 0 6= w 2 J F (). We have that d ?t v = u?t + ; Bt?1() w with u?t 2 F ( ?t ()) E u( ?t ()) E c( ?t ()). Since Bt?1() w 2 J F ( ?t ()) and F and J F are orthogonal, we have that kd ?t(v)k Bt?1() w . Hence from (31) we obtain 1 log kd () vk 0 : lim ?t t!+1 t This contradicts the choice v 2 E u(). Finaly we prove (c). Let : F () ! H () be the restriction of d. Then and by corollary 1.8

sup ?1 < +1 : (32)

?1(v) = (v; U() v)

2

Fix 2 , let Z (t) : V () ! H ( t ()) be de ned by Z (t) v = Then d t ()jF() = ?t1() Z (t) By (30) and (32), we have that 1 log det d ()j () = T !lim t F() +1 T 1 log jdet Z (t)j : = T !lim +1 T

)

t(

d t() (v; U()v).

22


Since the isomorphisms H (t) = Z (t), V (t) = U( t ()) Z (t) satisfy the Jacobi equation (6), we have that Z_ = (Hpq + Hpp U) Z : Since det Z (t) 6= 0 for all t 2 R , then det Z (t) has constant sign and d jdet Z (t)j = d det Z (t) = tr (H + H U) jdet Z (t)j ; pq pp dt dt d log jdet Z (t)j = tr [H + H U] ; pq pp dt

Z T 1 1 tr [Hpq + Hpp U] dt : lim log det d t ()jF() = T !lim T !+1 T +1 T 0

The exponential map. The following proposition proves item (c) in theorem 0.1. We present a modern proof that shall be needed in the sequel.

1.15. Proposition. [Hartman][17] Let H : T M ! R be a convex Hamiltonian.

Then (a) A half open segment f t() j t 2 [0; a[ g, a 2]0; +1[[f+1g has no conjugate points if and only if there exists a Lagrangian subspace E T T M such that d t(E ) \ V ( t ()) = f0g for all t 2]0; a[. (b) A closed segment f t() j t 2 [0; a] g has no conjugate points if and only if there exists a Lagrangian subspace E T T M such that d t (E ) \ V ( t ()) = f0g for all t 2 [0; a]. (c) If a closed segment f t () j t 2 [0; a] g has no conjugate points, then there exists > 0 such that the segment f t () j t 2 [?; a + ] g has no cojugate points.

Proof: (a) If the half open segment is disconjugate, then the Lagrangian subspace E = V () satis es d t (V ()) \ V ( t ()) = f0g.

Conversely, suppose that such Lagrangian subspace E is given. Let E (t) := d t(E ). Then the projection d : E (t) ! H ( t ()) is an isomorphism. Let S (t) : H ( t()) ! V ( t ()) be the linear isomorphism such that E (t) = graph (S (t)) for t 2]0; a[. Since E (t) is Lagrangian, then S (t) is symmetric and it satis es the Ricatti equation (7). Let Z (t) : E (t) ! H ( t()) be given by Z (t) w = d d t () w. Then Z_ = (Hpq + Hpp S ) Z ; Z (0) = djE : (33)


23

Let w 2 E and (h1 (t); v1(t)) = d t(w) 2 E (t), then h1 (t) = Z (t) w, v1(t) = S (t) Z (t) w. Let 2 T T M and (h(t); v(t)) = d t ( ). Suppose that h(c) = 0, c 2]0; a[. Using these h1 (t), v1 (t), the same arguments as in (9), (10), (11) give that

h(t) = Z (t) Z (c)?1 h(c) + Z (t) Since h(c) = 0, then

Z (t)?1 h(t) ; Z (c)v(c) =

Z t

Z t

c

?

Z ?1(s) Hpp (Z (s))?1 Z (c)v(c) ds :

(34)

Hpp (Z (s))?1 Z (c)v(c) ; (Z (s))?1 Z (c)v(c) ds : c (35) In particular h(t) 6= 0 for all t 2]0; a[nfcg. Hence the segment ]0; a[ does not have conjugate points. It remains to prove the case c = 0. Let 0 < b < a. We prove that is not conjugate to b(). We already know that b () has no conjugate points in f t () j 0 < t < a g. For 0 < d < b let Ed = d b?d(V ( d ())) = graph (Kd), where Kd : H ( b ()) ! V ( b ()) is the corresponding solution of the Ricatti equation (7). Let " > 0 be such that b < b + " < a and let Eb+" = ?"(V ( b+"())) = graph (Kb+"). By proposition 1.4(b) and remark 1.5, the family Kd is monotone decreasing on d and has the (lower) bound Kb+". Hence there exists E := limd!0+ Ed = d b(V ()) and E \ V ( b ( )) = f0g. (c) If the closed segment [0; a] has no conjugate points, then d t (V ())\V ( t ()) = f0g for 0 < t a. Then d t(V ()) \ V ( t ()) = f0g for 0 < t a + for some > 0. By item (a), using E (t) = d t(V ()), the segment [0; a + ] has no conjugate points. The same argument using F (t) = d a+?t (V ( a+ ())) shows that there exists 1 > 0 such that the segment [?1 ; a + ] has no conjugate points. (b) Suppose that the segment [0; a] is disconjugate. By item (c) there exists > 0 such that the segment [?; a + ] is disconjugate. Now apply item (a) to the segment [? 2 ; a + [. Conversely, if there is a Lagrangian subspace E T T M with d t(E ) \ V ( t ()) = f0g for all 0 t a, then the argument of equation (35) applies for all c 2 [0; a]. It may be impossible to nd such Lagrangian subspace E of theorem 1.15 (a),(b) satisfying E T as explained in remark 1.17. Compare this with theorem 0.1. If there exists a continuous invariant Lagrangian bundle E (), one can always suppose that E () T by taking (E () \ T ) hX ()i. Then it must satisfy the transversality condition 0.1(a). 1.16. Proposition. For 2 de ne W () := (V () \ T ) h? X ()i. Suppose that ? the orbit of has no conjugate points. Then for all t 6= 0, d t W () \ V t () = f0g.

24


Proof:

Suppose that there exists b > 0 such that (0; wb) 2 d b (W ()) \ V ( b ()) 6= f0g. Let (k0; w0) := d ?b (0; wb) 2 W (). Since the segment f () j t 2 [0; +1[ g has no conjugate points, then k0 6= 0. Write Hp(t) := Hp( t ()), Hq (t) := Hq ( t ()). Since W () = (V () \ T ) hX ()i and X () = (Hp(0); ?Hq (0)), then k0 = Hp(0) for some 6= 0. Multiplying wb by 1 , we can assume that = 1. Let (h(t); v(t)) := d t (X ()) ? (k0; v0 )) 2 d t (W ()) T t () : Then h(0) = Hp(0) ? k0 = 0 ; v(0) 6= 0: h(b) = d [X ( t ()) ? (0; wb)] = Hp(b) : (36) Since (h(0); v(0)) 2 T , then 0 = dH (h(0); v(0)) = Hq h(0) + Hp v(0) = Hp(0) v(0) : (37) Let E (t) = E ( t ()) be the stable Green bundle of R(), let S(t) be from (20) and let Z (t) := h(t) : H ( t ()) ! V ( t ()) be from claim 1.3. So that Z (t) w = d d t (w; S(t) w) and Z (t) satis es (33). Using formula (34) for c = 0, (with Z (t) = h(t), Z (0) = I and h(0) = 0) we obtain that

h(t) = Z (t)

Z t 0

Z ?1(s) Hpp(s) (Z (s))?1v(0) ds :

Hence, since v(0) 6= 0, we have that

hZ (b)?1h(b);

v(0)i =

Z b 0

hHpp (Z (s)?1v(0)); (Z (s)?1v(0))i ds > 0 :

(38)

Since X () 2 E (), then (Hp(b); ?Hq (b)) = X ( b ()) = d b(X ()) = d b (Hp(0); S(0) Hp(0)) = (Z (b) Hp(0); S(b) Z (b) Hp(0)) : From the rst component we get that Z (b)?1Hp(b) = Hp(0). Using (36) we have that Z (b)?1 h(b) = Hp(0). Replacing this equation on (38), we obtain that hHp(0); v(0)i > 0 : This contradicts equation (37).

1.17. Remark.

The same proof applies for the following statement: If H : T M ! R is convex and there is a Lagrangian subspace E T such that d t (E ) \ V ( t ()) = f0g for ? ? t 2 [0; a], then d t W () \ V t () = f0g for all t 2 [0; a].


25

In particular this hypothesis holds if f t () j t 0 g or f t () j t 0 g have no conjugate points. Just take the corresponding Green bundles. The example I.2 does not satisfy the conclusion of proposition 1.16 on t 2 [0; ], nevertheless the segment f t () j t 2 [0; ] g has no conjugate points. This implies that the Lagrangian subspace given by proposition 1.15(b) may not satisfy E T .

1.18. Corollary. Let H : T M ! R be convex, = H ?1feg, q 2 () and suppose that the orbits f t () j t 0 g have no conjugate points for all 2 ?1 fqg\ . Then the exponential map expq : Tq M ! M is an inmersion. Proof: We have that expq (t) = t () for 2 q = ?1fqg \ . Consider the splitting ? ? Tt (TqM ) Tq M = T q hi = V () \ T hi : The derivative d expq (t) : Tq M ! T( )M is given by d expq (t)jV ()\T = d d t ()jV ()\T ; (39) ? ? d expq (t)() = d X ( t ()) = d d t X () : (40) t

Therefore

?

?

d expq (t) TqM = d d t V () \ T hX ()i ? = d d t W () : (41) Since dim W () = n, then dim d t(W ()) = n. By proposition 1.16 d t(W ()) \ V ( t ()) = f0g. Hence d : d t (W ()) ! T( t )M is a linear isomorphism. From (41) d expq (t) is a linear isomorphism for all t > 0, 2 q .

2. The transversal behaviour. Given 2 , consider Hp() = d X () 6= 0. Let N () T be de ned by N () := f 2 T j h d ; d X () i() = 0 g : (42) Then N ()hX ()i = T . Fix 0 2 and a smooth coordinate system (q1; : : : ; qn; t) of M R along the projection (( t (0 )); t) of the orbit of 0 such that = d X ( t (0 )). (2.a) @q@ 1 (( t 0 );t) (2.b) @q@ 2 ; : : : ; @q@ n is an orthonormal basis for d N ( t (0 )) = h @q@ 1 i? along ( t (0); t).

26


Write pi = dqijTq M , i = 1; 2; : : : ; n. From (2.a) we have that

1 = dtd q1 ((

t

0 ); t)

= Hp1 ( t (0)) 6= 0 :

Then the equation H (q1; q2; : : : ; qn; p1; p2; : : : ; pn) = h can be solved locally for p1 :

p1 = ?K (Q; P; T ; h) ; where P = (p2; : : : ; pn), Q = (q2; : : : ; qn), T = q1 . This solution can be extended to a simply connected neighbourhood W of the orbit f ( t (0 ); t) j t 2 R g T M R . From now on we omit the R -coordinate in W T M R. Let T be the reparametrization of the ow t on W such that it preserves the foliation q1 = T = constant. In particular dT (0 ) preserves the transversal bundle N (T (0 )) along the orbit of 0 . De ne (T; ) by T () = (T;) (). Then (T; 0 ) = T for all T 2 R . The following reduction appears in Arnold [2]. 2.1. Lemma. The orbits T () = (T = q1 ; Q(T ); p1(T ); P (T ); (T; )) 2 W satisfy the equations

dQ = @K dP = ? @K ; ; dT @P dT @Q where Q = (q2 ; : : : ; qn), P = (p2 ; : : : ; pn ), T = q1 and K (Q; P ; T ) is de ned by H (T; q2; : : : ; qn; ?K; p2; : : : ; pn) = h. Proof: Consider the canonical 1-form = p dq on TM and the symplectic form ! = d = dp ^ dq. On T we have that !(X (); ) = ?dH = 0. Let dTd T () = ? Y (T ()); dTd (T; ) be the vector eld of T . Then Y (T ()) is a multiple of the ? vector eld X (T ()) of t , and then ! Y (T ()); = 0 on TW . We have = p dq = P dQ ? K dT; !jT = djT =

n X i=2

[dPi ^ dQi ? KQi dQi ^ dT ? KPi dPi ^ dT ] :

The matrix of !jT is given by 2

3

0 ?I ?KQ g Q 4 I 0 ?KP 5 g P g T =q1 KQ KP 0 ?

Then the vector KP ; ?KQ; 1 = dqdT1 =: Z (T ()) satis es !(Z (T ()); ) = 0 on f (T; Q; p1; P ) j p1 = ?K (Q; P; T ) g. Since the form ! has maximal rank on , we have that Z (T ()) = Y (T ()) for 2 W . This proves the lemma.


Since for # 2 W , T (#) =

27

(T;#) (#),

then @ d dT (#) = d (T;#) + @# dt t (#) (T;#) (T;#) = d (T;#) + (T; ) X (T (#)) ;

(43)

@ where (T; ) = @# (T;#) . Let ( t (0 )) : T t(0 ) ! N ( t (0 )) be the projection along the direction of the vector eld, i.e. = + ( ) X ( t(0 )) with ( ) 2 R such that 2 N ( t (0 )) : Through the proof of lemma 2.2 all the quantities will be understood in local coordinates T = q1, Q = (q2 ; : : : ; qn), P = (p2; : : : ; pn) of W\ . In particular @ ; @ . Since TT (0) = R R n?1 R n?1 for all T 2 R , with coordinates @T@ ; @Q @P along the orbit of 0 we have that q_ = (T;_ Q_ ) = (1; 0n?1) = (Hp1 ; HP ) along T (0) = (T; 0n?1 ; 0n?1 );

then the subspace N (T (0 )) in (42) is written as N := f0g R n?1 R n?1 in coordinates. De ne P : T# = R R n?1 R n?1 ! N = f0g R n?1 R n?1 as the projection P(h1 ; H; V ) := (0; H; V ): Using = ?h1 on (43) and Hp = (1; 0), the projection is written as (h1; H; V ) = (0; H; V + h1 HQ) ; where HQ := (Hq2 ; : : : ; Hqn ). De ne the matrizes H H pq pP H (T ) = ?HQq ?HQP (2n?1)(2n?1) 2

3

01n 01(n?1) KPP 5 K (T ) = 4 KPq ?KQq ?KQP (2n?1)(2n?1)

along the orbit of 0 .

2.2. Lemma. Along the orbit T := T (0 ) of 0 , we have that (a) The matrix KPP (T ) is positive de nite. (b) For 2 N (T ), K (T ) = (T ) H (T ) . (c) The operator K (T )jN (T ) is uniformly bounded on T 2 R.

28


Proof: (b) Write T = T (0 ) = equation for T is written as

= (T; 0; 0).

T (0 )

d dT [P dT

From lemma 2.1, the Jacobi

] = K [ P dT ] ;

where dT = (h1 (T ); H (T ); V (T )) 2 R R n?1 R n?1 . Also, from (6),

R

R n?1 R n?1 . for any 2 T0 =

d dt [d t ] = H

[d t ] :

Since dT (0 ) preserves tha foliation q1 = constant, then it preserves the subspace N . Therefore P dT = dT = dT for 2 N : In particular, the rst coordinate of dT is h1(T ) 0 and also dTd h1 0. Hence if 2 N , then d d dT [P dT ] = dT [dT ] = dTd [ d T + (T ) X (T ) ] = H (d T ) + _ (T ) X (T ) + (T ) dTd X (T ) = H dT ? (T ) X (T ) + _ (T ) X (T ) + (T ) dTd X (T ) ; where (T ) = (T; ) is from (43). Since dTd X (T ) = H X (T ), then d dT [P dT ] = H

Hence K

[ dT ] + _ (T ) X (T ) :

dT = H [dT ] + _ (T ) X (T ) for 2 N :

(44)

Observe that this equation must hold in all the 2n ? 1 coordinates of T#. Since Image(K ) N , then K = K . So K (dT ) = K (dT ) = [ H dT + _ (T; ) X (T ) ] = H (dT ) for 2 N : Since dT : N (0 ) = N ! N (T ) = N is surjective, this proves item (b). (a) Let V 2 R n?1 , then

VK

PP

V=

(0; V ); 0

n?1 K

0n : V


29

Since = (0n; V ) 2 N , we can use equation (44) to obtain

VK

Hp 0 n pp V = n?1 H V + _ (T; d?T ) ?H Q ? = 0 + V HPP V + _ (0; V ) Hp = V HPP V + _ h(1; 0n?1); (0; V )i

(0; V ); 0

= V Hpp V > 0 :

We now prove (c). Since the operator H (#) is uniformly bounded on # 2 , from (b) it is enough to prove that the projection is bounded on the orbit of 0 . For, it is enough to see that the angle (X (T ); N (T )) is uniformly bounded away from 0. Given a vector

= (h; v) =

n X i=1

hi @q@ i + vi @p@ i 2 N (T ) ; P

1

written in our coordinates (2.a), (2.b), de ne the norm j j := ( ni=1 h2i + vi2) 2 . Suppose that j j = 1. Since 2 N (T ), then Hp h = 0. Moreover, Hp = @q@ 1 and hence jX (T ) j = jHp h ? Hq vj = qjHq j jvj jX (T )j (jHpj2 + jHq j2 ) 21 1 + jHq j2 x = p A < 1; p 1 0max xA 1 + x2 1 + A2 where A := supT 2R jHq (T )j < +1 with the norm in our coordinates @p@ i .

2.3. Corollary. Along the orbit T = T (0 ), we have

(a) The orbit of 0 under T has no conjugate points. (b) Existence of the Green bundles for dsjN (T ): E > (T ) = lim d?s (V (T +s ) \ N (T +s )) s!+1 F > (T ) =

lim d?s (V (T +s) \ N (T +s))

s!?1

Moreover, E > () = E () \ N () and F > () = F () \ N () for all 2 . (c) Horizontal growth of iterates of vertical vectors: For all R > 0 there exists S = S (R; T ) > 0 such that for all jsj > S (R; T ) and all 2 N (T ) \ V (T ) we have that jd(ds(T ) )j > R j j.

30


(d) De ne

B >(T ) = 2 N (T ) j sup jd d s j < +1 :

B >(T ) E > (T ) \ F > (T ).

s2R

Then (e) Let s := dsjN = d sjN , then dsjN (T ) = (T +s) d s (T ) = s ; s+t = s t :

Proof:

Let U (T ) := V (T ) \ N (T ) = V (T ) \ TT . From (43) we have that dT (0 ) = d T (0 ) + ((T; 0) ) X (0) : ? Hence d T (U (0 )) dT (W (0)), where W () := V () \ T hX ()i. From proposition 1.16, U (T ) \ dT (U (0 )) V (T ) \ d T (W (0 )) = f0g for all T 6= 0 : This implies item (a). From lemma 2.2, the rst part of item (b) and items (c) and (d) have the same proofs as propositions A, 1.10 and proposition 1.11 respectively. Since ds and d s satisfy (43) and ds preserves the bundle N (T ), we have that dsjN (T ) = (T +s) d sjN (T ) = (T +s) d s (T ) = s : Moreover, s+t = ds+tjN (T ) = dsjN (T +t) dtjN (T ) = s t : This proves item (e). Finally, we have ? E > (0 ) = lim d?T V (T ) \ N (T ) T !+1

?

d ?T V (T ) \ TT = (0) T !lim +1 (0 ) E (0 ) E (0 ) : Because X (0 ) 2 E (0 ). Thus E > (0 ) E (0 ) \ N (0 ). Since dim E > (0) = dim U = n ? 1 = dim(E (0 ) \ N (0)), we have that E > (0 ) = E (0 ) \ N (0 ). The equation F > (0 ) = F (0 ) \ N (0 ) is proven similarly. This completes the proof of item (b).

Proof of proposition B. Let 0 2 ? and let > 0 be the period of 0 . Suppose that ? is hyperbolic, then by proposition 1.14, E (0 ) E s(0 ) hX (0 )i and F (0 ) E u(0 ) hX (0 )i. Hence E (0 ) \ F (0 ) = hX (0 )i.


31

Now observe that if we construct the coordinates (q1; : : : ; qn) in (2.a), (2.b) so that they are periodic with period , then d (0 ) : N (0 ) - is the derivative of the Poincare map of t de ned on the cross-section f # j q1(#) = 0 g. Suppose that E (0 ) \ F (0 ) = hX (0)i. By corollary 2.3 (a) and (b) we have that B >(0) E > (0 ) \ F > (0 ) = hX (0)i \ N (0 ) = f0g. This implies that ? is hyperbolic.

3. Quasi-hyperbolic actions. Let B be a compact metric space and : E ! B a vector bundle provided with a continuous norm j jp on each bre ?1fpg. Let be an R -action : R ! Isom(E ), i.e. (i) There exists a continuous ow t on B such that t = t . (ii) t : E (p) ! E ( t (p)) is a linear isomorphism, where E (p) := (jE )?1fpg, p 2 B. (iii) s+t = s t. We say that the action t is quasi-hyperbolic if sup j t ( )j = +1 for all 2 E ; 6= 0: t2R

We say that is hyperbolic if there exists an invariant continuous splitting E (p) = E u(p) E s(p), p 2 B and C > 0, > 0 such that ? ? ? ? (3.a) t E u(p) = E u t (p) , t E s(p) = E s t (p) , for all x 2 B , t 2 R . (3.b) j t ( )j C e?t j j for all t > 0, 2 E u(p), p 2 B . (3.c) j ?t ( )j C e?t j j for all t > 0, 2 E u(p), p 2 B . The aim of this section is to prove the following

0.2. Theorem. If the action is quasi-hyperbolic and the non-wandering set

( jB ) = B , = , then jE is hyperbolic. Applying this theorem to the case in which B = M is a compact manifold, t is a dierentiable

ow on M , t = d t and E is a d t -invariant continuous bundle such

d that E dt t = TM , we get

3.1. Theorem. [Freire] [12]

If M is a closed manifold and is a quasi-Anosov ow on M such that ( ) = M , then is Anosov.

32


The proof of theorem 0.2 is similar to that of theorem 3.1. We include it here for completeness. Suppose that the action of is quasi-hyperbolic. For p 2 B de ne n o E s(p) := v 2 E (p) sup j t (p)(v)j < +1 ; n

E u(p) := v 2 E

(p)

t>0

o

sup j t (p)(v)j < +1 : t 0 such that for all p 2 B ,

jE s (p) < 1 ; ? jE u (p) < 1 2 2 for all p 2 M .

Proof:

Suppose that the lemma is false for E s. The proof for E u is similar. Then there exist sequences xn 2 B , vn 2 E s(xn), jvnj = 1 such that j n(vn)j 21 for all n 2 N . We claim that there exists C such that

sup t (p)jEs(p) < C < +1 : t0

for all p 2 B . Suppose the claim is true. Let wn := n(vn) and yn := n(xn). We have that 1 jw j C for all n and n 2 j t (wn)j = j t+n(vn)j C for all t ?n : Since B is compact there exists a convergent subsequence of (yn; wn). If yn ! y and n wn ! w, we have that y 2 B , w 2 E (y), jwj 21 and n

j t (w)j C

for all t 2 R :

This is a contradiction. Now we prove the claim. Suppose it is false. Then there exists xn 2 B , tn 0, vn 2 E s(xn ), jvnj = 1, such that (45) sup j tn (vn)j = +1 : n

Let sn > 0 be such that j sn (vn )j > 21 sups0 j s(vn)j 12 j tn (vn)j : By (45) we have that sn ! +1. Let yn := sn (xn) and n sn (vn) : wn := j (v )j sn n


33

Then jwnj = 1 and if t > ?sn we have 2 j t+sn (vn)j 2 : j t (wn)j = j j t+sn(v(vn)j)j sup sn n s0 j s(vn )j Since jwnj = 1 and yn 2 M , there exists a convergent subsequence (yn; wn) ! (y; w). We would have that y 2 B , w 2 E (y), jwj = 1 and j t(w)j 2 for all t 2 R : This is a contradiction.

3.3. Lemma. There exists K > 0 such that for all x 2 B , v 2 E (x) j t (v)j K ? jvj + j s(v)j for all 0 t s: Proof:

Suppose it is false. Then there exist xn 2 B , 0 6= vn 2 E (x) and 0 tn sn, such that ? j tn (vn)j n jvnj + j sn (vn)j : Then tn ! +1 and sn ? tn ! +1 when n ! +1. We can assume that j tn (vn)j = sup j t (vn)j. Let 0tsn

tn (vn) wn := j (v )j tn n

For ?tn < t < sn ? tn, we have that j t (vn)j = 1: j t (wn)j = j j t+tn(v(vn)j)j j 1(v )j 0sup tsn tn n tn n Taking a subsequence, if necessary, we can assume that xn ! x and wn ! w 2 E (x). Then j t (w)j 1 for all t 2 R , with jwj = 1. This is a contradiction.

Proof of theorem 0.2:

Standard methods (cf. Hirsch-Pugh-Shub [18]) show that the continuity of the strong stable and unstable bundles is redundant in the de nition of a hyperbolic action. So it remains to prove that for all x 2 B , we have that E s(x)E u(x) = E (x). Given x 2 B , since x 2 ( jB ), there exist xn ! x, sn ! +1 such that sn xn ! x. Let En be a subspace of E (xn) such that dim En = dim E (x) ? dim E s(x) ; E s(x) lim E = E (x) : n n

34


We claim that there exists C > 0 such that

?t s (En ) C for all 0 t sn: n Suppose the claim is true. Taking a subsequence if necessary, we can assume that the limit limn sn (En) exists. Then

?t lim s (En ) C for all t > 0: n n

Then limn sn (En) E u(x) and dim E u(x) + dim E s(x) dim En + dim E s(x) = dim E (x). Since E s(x) \ E u(x) = f0g, this completes the proof. Now we proof the claim. Suppose it is false, then there exist vn 2 sn (En), jvn j = 1 and 0 < tn < sn such that j ?tn (vn)j n. By lemma 3.3, we have that ? n j ?tn (vn)j K j ?sn (vn)j + jvnj :

Hence j ?sn (vn)j n?KK . From lemma 3.3 we also have that j t( ?sn (vn))j K + K j ?sn (vn)j j ?sn (vn)j for 0 < t < sn : ?sn (vn) . The estimates above give that Let wn := j ?sn (vn )j 2 K j t (wn)j K + n ? K for 0 < t < sn : If wn ! w, then jwj = 1 and w 2 lim En, thus w 2= E s(x). But j t(w)j K for all t 0. This is a contradiction.

Proof of theorem C. For 2 , let N () := 2 T j h d ; d X () i() = 0 ; and let () : T ! N () be the projection along the direction of X (): () = + ( ) X () such that () 2 N () : Let s() = ( s()) d s jN () . By corollary 2.3 (e), s de nes an R -action on the vector bundle : N ! . Suppose that the Green bundles satisfy E \F = hX i on . Then by corollary 2.3(b) and 2.3(d),

B >() E > () \ F > () = E () \ F () \ N () = f0g :


35

Therefore the action s is quasi-hyperbolic. Since s preserves the Liouville measure, which is positive on open sets, then ( j) = . Hence s is a hyperbolic action. This implies that t is Anosov. By the methods in Hirsch-Pugh-Shub [18], we get that t is Anosov. We sketch the proof below. For 2 let E u() N () be the subspace given by item (3.b) in the de nition of hyperbolic action. Then E u is a continuous subbundle of T . De ne F := L : E u ! R continuous L() : E u() ! R is linear 8 2 ; with kLk = sup2E unf0g jLj(j)j . To each functional in F associate a subbundle WL of T by WL() := Graph(L()) = + (L() ) X () 2 E u() : Let > 0 be such that C e? < e? < 1, where C and are from (3.b). Consider the following \graph transformation" T : F ! F , corresponding to WTL = d (WL) and de ned by ? d + L( ) X () = ( ) + T (L) ( ( )) X ( ()); 2 ; 2 E u(): We claim that T is a contraction. Indeed T (L1 ? L2 ) ( ) kX ( )k k d [(L1 ( ) ? L2 ( )) X ( )]k k k e k k

j(L1e? kL2k)( )j kX ( )k :

Hence kT (L1 ) ? T (L2)k e? kL1 ? L2 k. The xed point L of T gives the (continuous) strong unstable bundle of t . Indeed, if E u = WL , then E u is clearly d t -invariant. Moreover, if 2 E u(), then = + (L ) X () with = 2 E u(). Since L is continuous then there exists Q1 > 0 such that j j Q j j = Q j j for all 2 E u. Since N () is transversal to hX ()i and both are continuous subbundles of T , ( compact), then the angle (X (); N ()) is bounded below and hence there exists Q2 > 0 such that Q12 j j = Q12 j j < j j. Then

jd t( )j = j t( ) + (T (L) ) X ( t) j Finally,

?1 e t C et 1 Q j t ( )j Q j j C Q Q j j : 2 2 1 2

dim E u = dim Graph(L ) = dim E u = n ? 1 : The existence of a continuous strong stable subbundle of dimension n ? 1 is proven similarly.

36


4. The Index Form. Let L = p Hp ? H be the Lagrangian associated to H . Consider @L @v (x; v ) : T(x;v) (TxM ) TxM ! R as an element Lv 2 T M . The Lagrange transform FL(x; v) = (x; Lv ) identi es q = x, p = Lv . Similarly, using FL?1 = FH , we get that v = Hq under the same identi cation. Also H = v Lv ? L = p Hp ? L(q; Hp) : Hence Hq = ?Lq = ?Lx . With these identi cations the Hamiltonian equations (1) become d ! dtd q = Hp ; dt x = v d ! dtd p = ?Hq ; (46) dt Lv = Lx Hence, identifying Lx = Hq 2 T(p;q)(T M ), the Euler-Lagrange equation (46) is understood as a rst order dierential equation on T M , where dtd Lv is the tangent vector to the path t 7! Lv ( (t); _ (t)) in T M . We derive the Jacobi equation in this Lagrangian setting. Let (t) be a solution of the Euler-Lagrange equation (2). Considering a variation f (s; t) of (t) = f (0; t) made of solutions t 7! f (s; t) of the Euler-Lagrange equation (2) and taking the covariant derivative Dds , we obtain the Jacobi equation D (L k + L k0 ) = L k + L k0 ; (47) vv xx xv dt vx _ D @f where k = @f @s (0; t), k = dt @s and the derivatives of L are evaluated on (t) = = Ddt @F f (0; t). Here we have used that Dds @F @t @s for the variation map F (s; t) = ? @f D D Lv f (s; t); @s (s; t) 2 T M , where ds and dt are the covariant derivatives on the riemannian manifold T M . The linear operators Lxx, Lxv , Lvv coincide with the corresponding matrices of partial derivatives in local coordinates. The solutions of (47) satisfy ? ? D't k(0); k_ (0) = k(t); k_ (t) 2 T (t) (TM ) ; where 't is the Lagrangian ow on TM . Let T be the set of continuous piecewise C 2 vector elds along [0;T ]. De ne the index form on by Z T I (; ) = _ Lvv _ + _ Lvx + Lxv _ + Lxx dt ; (48) 0

which is the second variation of the action functional for variations f (s; t) with @f 2 . For general results on this form see Duistermaat [9]. T @s


37

The following transformation of the index form is taken from Hartman [17] and originally due to Clebsch [7]. Let 2 TM and suppose that the orbit of , 't (), 0 t T does not have conjugate points. Let E T T M be a Lagrangian subspace such that d t (E ) \ V ( t ()) = f0g for all 0 t T . Such subspace E always exists by proposition 1.15(b). Let E (t) := d t (E ) and let H (t); V (t) be a matrix solution of the Hamiltonian Jacobi equation (6) such that det H (t) 6= 0 and E (t) = Image (H (t); V (t)) T t () (T M ) is a Lagrangian subspace. In particular H (t) satis es the Lagrangian Jacobi equation (47). We have that H (t) = DF (' ()) H (t) = I 0 H (49) L t V (t) H 0(t) Lvx Lvv H 0 From (49) and (47), we have that V = Lvv H 0 + Lvx H (50) V 0 = Lxv H 0 + Lxx H Moreover, since E (t) is a Lagrangian subspace, then the Hamiltonian solution of the Ricatti equation (7) S (t) = V (t) H (t)?1 is symmetric, i.e. H (t) V (t) = V (t) H (t): (51) Let 2 T and de ne 2 T by (t) = H (t) (t). Then the integrand of I (; ) in (48) is (Lvv H 0 + V ) (H 0 + H 0) + (V 0 + Lxv H 0) H : Since (Lvv H 0) (H 0 ) = (H 0) (Lvv H 0 ) and (Lxv H 0) (H ) = (H 0) (Lvx H ), the integrand can be written as Lvv H 0 H 0 + H 0 V + V H 0 + V H 0 + V 0 H : Using (51), we have that V H 0 = H V 0 = V 0 H . hence the integrand in (48) is Lvv H 0 H 0 + (V H )0. Since and are continuous, we have that

I (; ) =

Z T ?

Lvv H 0 H 0 dt + H V T0

if = H 2 T :

(52) Let FH (q; p) = (q; Hp) be the Legendre transform of the Hamiltonian. Then FL and FH are inverse maps, in particular 0

?1 I 0 I 0 (DFH Hqp Hpp = Lvx Lvv = DFL ; and hence Lvv = (Hpp)?1. We obtain that

)?1 =

I (; ) =

Z T 0

(H 0) (Hpp)?1 (H0) dt + (H )(V ) T0 :

(53)

38


for = H 2 T , = H 2 T . This formula can also be written as

I (; ) =

Z T 0

(H 0) (Hpp)?1 (H0) dt + S T0;

(54)

where S (t) = V (t) H (t)?1 is the corresponding solution to the Ricatti equation (7).

4.1. Corollary. If 2 T M and the segment f t() j t 2 [0; T ] g has no conjugate points then the index form is positive de nite on ?T = : [0; T ] ! TM (t) 2 T t M; is piecewise C 2; (0) = 0; (T ) = 0 :

Proof:

Let 2 ?T , 6= 0. Write (t) = H (t) (t). Since det H (t) 6= 0, (0) = 0, (T ) = 0 and (t) 6 0, then 0 6 0. Now use formula (53).

We now extend formula (53) to the case in which H (t) may be singular (or 0(t), 0(t) may be discontinuous) at a nite set of points. Let ft1 ; : : : ; tN g be the points in [0; T ] such that det H (ti) = 0. Then

I (; ) =

Z T

(h 0) (H

pp

)?1 (H0) dt + (H )(V ) T

0 ?

N X

+

(H )(V ) tti?i :

(55) where = H , = H. But now , are piecewise C 2 but may be discontinuous at t1 ; : : : ; tN . Alternativelly, we can use a sum of formulas (53) or (55) using dierent Lagrangian subspaces d t (Ei) and corresponding solutions (Hi; Vi) on disjoint subintervals ]ti ; ti+1[ [0; T ]. 0

i=1

Observe that for any convex Hamiltonian on T M and any 2 M there exists " > 0 such that the segment f t () j jtj < "g has no conjugate points. Indeed, let Y (t) v = d d t() (0; v), where (0; v) 2 V (). Then from the Jacobi equation (6) we have that D Y (t) = H (): pp dt t=0 Since Y (0) = 0 and Hpp() is nonsingular, then Y (T ) is an isomorphism for jtj < ", some " > 0. In particular d t () V () \ V () = f0g for jtj < ". We say that a curve (t) 2 M , t 2 [0; T ] is minimizing if it minimizes the action functional Z T L((t); 0(t)) dt 0

over all absolutely continuous curves (t) 2 M , 0 t T , such that (0) = (0) and (T ) = (t).


39

4.2. Corollary. If 2 T M and the segment f t() j t 2 [0; S [ g is minimizing,

then it has no conjugate points.

Proof:

Suppose it is false. Let T (), T < S be the rst conjugate point in f t () j t 2 [0; S [ g. Then there exists = (0; v) 2 V () such that d T () 2 V ( T ()). Let (t) = d d t () . Using the limit of (53) on the interval [0; T [ we have that I (; )jT0 = 0. Let " > 0 be such that (cf. prop. 1.15) the segment f t () j T ? " t T + "g has no conjugate points. Let (t) = Y T +" () (t ? T ? ") w = d d t?T ?"( T ?"()) (0; w), T ? " t T + ", where w is such that (T ? ") = (T ? "). Let b = (t) if T ? " t T and b(t) = 0 if T t T + ". Then using H (t) = Y T +"() (t ? T ? ") on (53), we have that I (b; b) TT +?"" > I (; ) TT +?"": Extend and b by (t) = b(t) = (t) for 0 t T ? ". Then

I (; ) T0 +" < I (b; b) T0 +" = 0: Now consdider the variation f (s; t) = exp t() s b. We have that f (s; 0) = (), f (s; T + ") = ( T +"()), @f @s (0; t) = b(t) and if A(s) = is the action of f (s; t), then Lagrange equation and

Z T +" 0

?

L f (s; t); @f @s (s; t) dt

A0 (0) = 0

because f (0; t) = t () satis es the Euler-

A00(0) = I (b; b) < 0 : Therefore the segment f () j 0 t T + " g is not minimizing.

5. Proof of theorem D. We begin by quoting the following theorem by R. Ma~ne [23]. Given a (periodic) Lagrangian L : TM S 1 ! R and ! 2 H 1(M; R), let M! (L) be the set of minimizing measures of L + !, where ! is any 1-form in the class !. 5.1. Theorem. (Ma~ne [23]) (a) For every ! 2 H 1(M; R ) there exists a residual subset O(!) C 1(M S 1) such that 2 O(!) implies #M! (L + ) = 1. (b) There exist residual subsets O C 1(M S 1) and H H 1(M; R ) such that 2 O and ! 2 H imply #M! (L + ) = 1.

40


Now take an autonomous Lagrangian L and ! = 0 2 H 1(M; R ). Then the item (a) implies the rst part of theorem D. Let O be the residual subset given by this theorem. Let A be the subset of O of potentials for which the measure on M(L + ) is supported on a periodic orbit. Let B := O n A and let A1 be the subset of A on which the minimizing periodic orbit is hyperbolic. We prove that A1 is relatively open on A. For, let 2 A1 and M(L + ) = f g where is the invariant probability measure supported on the hyperbolic periodic orbit for the ow of L + . We claim that if n 2 A, n ! and M(L + k ) = fk g, then n ! . Indeed, since L is superlinear, the velocities in the support of the minimizing measures k := k are bounded (cf. [26], [22]), and hence there exists a subsequence k ! converging weakly* to a some invariant measure for L + . Then if 6= , lim S ( ) = SL+ ( ) > SL+ ( ) : (56) k L+k k Thus if k is the analytic continuation of the hyperbolic periodic orbit to the ow of L + k in the original energy level c(L + ), since limk SL+ k (k ) = SL+ ( ), for k large we have that, SL+k (k ) < SL+k (k ) ; which contradicts the choice of k . Therefore = . For energy levels h near to c(L + ) and potentials near to , there exist hyperbolic periodic orbits ;h which are the continuation of . Now, on a small neighbourhood of a hyperbolic orbit there exists a unique invariant measure supported on it, and it is in fact supported in the periodic orbit. Thus, since n ! , then n is hyperbolic. Hence n 2 A1 and A1 contains a neighbourhood of in A. Let U be an open subset of C 1(M; R ) such that A1 = U \ A. We shall prove below that A1 is dense in A. This implies that A1 [ B is generic. For, let V := int (C 1(M; R ) nU ), then U[V is open and dense in C 1(M; R ). Moreover, V\A = because A A1 U and V \ A A n U = . Since O = A [ B is generic and ? (U [ V ) \ (A [ B) = (U \ A) [ (U [ V ) \ B A1 [ B ; then A1 [ B is generic. Note: The perturbation to achieve hyperbolicity in the case of a periodic orbit and a singularity follow the same spirit. However to prove it in the case of a singular point is much easier because the Jacobi equation (the linear part of the ow) is autonomous. We suggest to read it rst. We have to prove that A1 is dense in A, i.e. given that a Lagrangian L has a unique minimizing measure supported on a periodic orbit , then there exists a


41

perturbation by an arbitrarily small C 1-potential , such that the new Lagrangian L + has a unique minimizing measure supported on a hyperbolic periodic orbit. Fix 2 A and let ? be the periodic orbit in M(L + ), we can assume that = 0. By the graph property (cf. [26], [22]), the projection j? : ? ! M , (x; v) = x is injective. In particular (?) M is a simple closed curve. Choose coordinates x = (x1 ; : : : ; xn) : U ! S 1 R n?1 on a otubular neighbourhood of (?) such that n x(?) = S 1 f0g and @x@ 1 ; @x@ 2 ; : : : ; @x@ n is an orthonormal frame over the points of (?). In particular @x@ 1 is paralel to (?). De ne Le = L + with (x1; : : : ; xn ) = 1 2 2 2 1 2 f (x) " (x2 + x3 + + xn ), where f (x) is a C nonnegative bump function with support in U which is 1 on a small neighbourhood of (?). Clearly can be made C 1 arbitrarily small. On (?) we have that Levv = Lvv , Levx = Lvx and Lexx = Lxx + " [ 00 0I ]. Observe that since Le L then M(L + ) = f?g. Let H and He = H ? be the Hamiltonians associated to L and Le respectively. Let ? be the corresponding periodic orbit for H and He on T M . We have to prove that the periodic orbit ? is hyperbolic for the ow of He . Since it is minimizing, by corollary 4.2 the orbit ? has no conjugate points. By proposition B it is enough to prove that the Green bundles Ee , Fe satisfy Ee () \ Fe() = hXe ()i on a point 2 ?, where Xe is the Hamiltonian vector led of He . Fix 2 ? and let ZeT (t) = d d et djd e?T (V ( eT ())) ?1 . Let (ZeT ; VeT ) and Ke T (t) = VeT (t) (ZeT (t))?1, 0 t < T , be the corresponding solutions of the Jacobi equation (6) and the Ricatti equation (7) respectively: graph(Ke T (t)) = d et (d e?T (V ( eT ))), ZeT (T ) = 0, ZeT (0) = I . De ne N () := f w 2 T() M j hw; _ i = 0 g. Then N () is the subspace of T() M generated by the vectors @x@ 2 ,: : : , @x@ n . Let vo 2 N (), jvoj = 1 and let T (t) := ZeT (t) vo. Denote by IeT and IT be the index forms on [0; T ] for Le and L respectively. Using the solution (ZeT ; VeT ) on formula (53), we obtain that

IeT ( T ; T ) = ?(ZeT (0) vo) (VeT (0) vo) = ?vo Ke T (0) vo :

(57)

Moreover, in the coordinates (x1 ; : : : ; xn; @x@ 1 ; : : : ; @x@ n ) on TU we have that

IeT ( T ; T ) = =

Z T ? T _ Levv _T 0 Z T ? T T 0

+ 2 _T Lexv T + T Lexx T dt

_ Lvv _ + 2 _T Lxv T + T Lxx T dt +

Z T 0

"

n X T 2 i=2

i

dt : (58)

42


We have that ZeT (0) = I and for all t > 0, limt!1 ZeT (t) = he (t) with he (t) given on claim 1.3 for He . Writing N ( ) = (2; 3; : : : n) then N he (0) v0 = jv0j = 1 T because v 0 2 N( ). Hence there exists > 0 and T0 > 0 such that N (t) = e N ZT (t) v0 > 12 for all 0 t and T > T0 . Therefore IeT ( T ; T ) IT ( T ; T ) + " (59) 4: ? Let (h(t); v(t)) = d t djE() ?1 be the solution of the Jacobi equation for H given by claim 1.3 and let S( t()) = v (t) h(t)?1 be the corresponding solution of the Ricatti equation, with graph[S( t()] = E ( t ()). Using formula (53), and writing T (t) = h(t) (t), we have that

IT

( T ; T ) =

Z T 0

? ? (h _) Hpp?1 (h _) dt + 0 ? h(0) (0) v(0) (0)

IT ( T ; T ) ?vo S() vo :

(60)

From (57), (59) and (60), we get that

: vo S() vo vo Ke T vo + " 4

From proposition 1.4, we have that limT !+1 Ke T (0) = eS(), where graph(eS()) = e . Therefore Ee ( ), the stable Green bundle for H vo S() vo v0 eS() vo + " (61) 4: Similarly, for the unstable Green bundles we obtain that vo U() vo + 2 vo Ue () vo for vo 2 N (); jvoj = 1: (62) for some 2 > 0 independent of vo. From proposition 1.4 we have that U() < S(). From (61) and (62) we get that e jN UjN < SjN e e ( )), we U SjN . Since Ee ( ) = graph(e S( )) and Fe ( ) = graph(U get that Ee () \ Fe() hXe ()i. Then proposition B shows that ? is a hyperbolic periodic orbit for L + . This proves that A1 is dense in A. Let A2 be the subset of A1 of potentials for which the minimizing hyperbolic periodic orbit ? has transversal intersections W s(?) t W u(?). Then A2 is relatively open on A1. Indeed, if n 2 A1 and n ! 2 A2 and M(L + n) = f n g, M(L + ) = f?g, we have seen that the n's are continuations of the hyperbolic orbit ? for nearby ows (on nearby energy levels): n = ?n;hn . Since compact subsets (fundamental domains) of the invariant manifolds W s(?;h) depend continuously in the C 1 topology on (; h),


43

then the transversality property W s(?;h) t W u(?;h) persists on a neighbourhood of ( ; h0). In particular it holds for n = ?n ;hn for (n; hn) suciently close to ( ; h0). By the same arguments as above, it is enough to prove that A2 is dense in A. For it we need the following lemma,

5.2. Lemma. Let ? be a hyperbolic periodic orbit without conjugate points of a convex Hamiltonian. Then for all 2 W s(?) there exists S = S () > 0 such that the segment f t () j t S g has no conjugate points. Proof:

Let E be the stable Green subspace of ?. Then E (#) \ V (#) = f0g for all # 2 ?. Since the weak stable manifold W s(?) is tangent to E ; then for all 2? W s(?) there s exists S = S () > 0 such that E (t) := T t ()W (?) satis es E (t) \ V t () = f0g. Moreover, dim E (t) = n and for all u 2 E (t), lims!+1 d s(u) 2 hX ()i. Since t preserves the symplectic form !, we have that !(u; v) = lims!+1? !(d s(u); d s(v)) = 0 for all u; v 2 E (t). Hence E (t) is Lagrangian and E (t) \ V t () = f0g for all t > S . Then the lemma follows from proposition 1.15.

Now we prove that A2 is dense in A. Let 2 A n A2 and make the perturbation Le = (L + ) + explained above so that + 2 A1. We can assume that + = 0 and also that + 2= A2, otherwise there is nothing? to prove. Write M(L) = f?g and let 2 W s(?) \ W u(?) be such that dim T W s(?) \ T W u(?) > 1. Observe that the and ! limit sets of are ?. In particular the orbit of has no autoacumulation points. Let S = S () > 0 be from lemma 5.2 and > 0 small. Then there exists > S () and a neighbourhood W M of ( t ()) such that f t 2 R j t () 2 W g =]; + 3[ and W \ (?) = . Let U M R be a tubular neighbourhood of f ( t();? t) j t > S ()?g. Choose coordinates y = n (y1; : : : ; yn; t) : U ! R R such that y t (); t = (t); 0; t 2 R R n?1 R , o n and @y@ 1 ; @y@ 2 ; ; @y@ n is an orthonormal frame over the points t (), t > S (). Let '(y1; : : : ; yn) = 12 f (y) (y22 +y32 +yn2 ), where f (y) is a C 1 bump function with support in W which is 1 on a small neighbourhood of t () j t 2 [ + ; + 2] . By choosing a small > 0 the function ' can be made C 1 arbitrarily small. Choose small enough such that the orbit ? remains hyperbolic. Then we still have that M(L + ) = f?g and the orbit of is the same for both ows. Write Le = L + + , Lb = Le + and IeT , IbT the corresponding index froms on f t () j t 2 [; T ] g. Since f t() j t > S () g has no conjugate points, the stable Green subspace Eb ( t ()), t > S () and the stable solution of the Ricatti equation bS( t ()) exist. The same

44


arguments as in equations (58){(61) give for T = ZbT (t) vo, vo 2 N ( ()), jvoj = 1, ZbT (t) = d d bt?T djd ?T (V ( bT + ())) ?1, that Z +2

IbT ( T ; T ) = IeT ( T ; T ) + ?

v0 S

()

for some b > 0. In particular Eb

?

()

+

?

vo vo bS ?

()

n X T 2 i

i=2

dt ;

vo + 4 ; b

\ Ee () = X ( ) :

?

(63)

By the same arguments as in proposition 1.14, we have that Eb () = ? ? ? = T () WLeu( ) . Observe that T () WLeu( ) = T () WLeu( ) didn't change. ? Write T () WLeu( ) = V W , with V Ee ( ) and W \ Ee ( ) = f0g. By diminishing if necessary we can get (63) and Eb ( ) \ W = f0g. Thus

T b () WLbs (?) \ T b () WLbu(?) = Eb ( ) \ T () W u(?) = Xb ( ) : Finally, by a nite number of these perturbations on a fundamental domain of W s(?) one can remove all tangencies of W s(?) and W u(?) and obtain a potential in A2 arbitrarily C 1-near 2 A.

The case of a singularity.

Suppose that the minimizing measure is supported on a singularity (x0 ; 0) of the Lagrangian ow. From the Euler-Lagrange equation (2) we get that Lx(x0 ; 0) = 0. Dierentiating the energy function (3) we see that (x0 ; 0) is a singularity of the energy level c(L). Moreover, the minimizing property of implies that x0 is a minimum of the function x 7! Lxx(x; 0). In particular, Lxx(x0 ; 0) is positive semide nite in linear coordinates in Tx0 M . Choose coordinates y = fy1; : : : ; yng ona neighborhood of x such that y(x0 ) = 0 0 @ @ and Lvv (x0 ; 0) is the identity on the basis @y1 x0 ; ; @yn x0 . Use the coordinates ? fy1; : : : ; yng ; @y@ 1 ; ; @y@ n on a neighbourhood of x0 . On the orbit 't (x0; 0) (x0 ; 0) the matrices Lxx, Lxv , Lvv are constant with respect to the time t. The Jacobi equation (47) at (x0 ; 0) becomes k = Lxx(x0 ; 0) k ; where Lvv (x0 ; 0) = I . Let f be a nonnegative C 1 function on R n with support on jyj < " < 1 and f 1 on a neighbourhood of 0. Let (y) = 21 f (y) (y12 + + yn2 ). Adjusting > 0 we can make arbitrarily C 1 small. Let Le := L + . Then the


45

atomic measure supported on (x0 ; 0) is still the unique minimizing measure for Le and the Jacobi equation for the orbit 'et (x0; 0) (x0; 0) is ? k = Lxx(x0 ; 0) + I k_ : (64) ? Since Lxx(x0 ; 0) is positive semide nite, then A := Lxx(x0 ; 0) + I is positive de nite. Equation (64) is linear 0 k = 0 I k (65) A 0 k0 k00 with constant coecients where k 0 is a hyperbolic singularity. To obtain the transversality property W s t W u at homoclinic orbits we can use the same arguments as above if we have a replacemnet for lemma 5.2. The argumenmts of lemma 5.2 apply if we show that the stable subspace of (x0 ; 0) is transversal to the vertical subspace. For, it is enough to prove that the eigenvectors (x; y) of the (non-singular) matrix in (65) have coordinate x 6= 0. These eigenvectors satisfy y = x, Ax = 2x, ( 2 R ). Hence if x = 0, then (x; y) = (0; 0) is not an eigenvector.

6. The Exponential map. The following lemma was proven for geodesic ows by Freire and Ma~ne [13] and has intrinsic interest (compare with proposition 1.10).

6.1. Lemma. Let H be a convex Hamiltonian. Choose a riemannian metric k k on M . If t has no conjugate points on = H ?1 feg, then there exists B > 0 such that

kY (t) vk > B kvk for all 2 , jtj > 1 and v 2 T M . Where Y (t) v = d d t () (0; v), (0; v) 2 H () V ().

We sketch the proof of theorem F. In the case in which (X ) > 0 on one can de ne the vector eld Y = j(1X )j X . This vector eld preserves the 1-form on T . Extending quadratically the Hamiltonian to T M by H (t ) = t2 , where 2 , we obtain a new Hamiltonian H , which is convex and has no conjugate points in . The Hamiltonian vector eld Y of H is called the (convex) simplecti cation of the contact vector eld Y . The transversal bundle N () = ker jT is invariant under the ow of Y. Let Eq : Tq M~ ! M~ be the exponential map for (Y; ). The derivative dEq (t) is given in formulas (39) and (40). The norm of the directional derivative in formula (40) is clearly bounded below. By lemma 6.1 the norm of the

46


derivative dEq jV ()\T in formula (39) is uniformly bounded away from zero. Since ? V () \ T N (), and the angle N (); X () is bounded below, we obtain that kdEq k is bounded away from zero. This implies that Eq : Tq M~ ! M~ is a covering map and hence a dieomorphism. In particular, the universal cover M~ of M is homeomorphic to R n . By corollary 1.18, the exponential map expq for X is a local dieomorphism. Once we know that Eq is a dieomorphism, it is easy to show that expq is bijective and hence a dieomorphism.

Proof of Lema 6.1:

Since Y (t) satis es (8) (with Hqp and Hpp uniformly bounded on ) and by corolary 1.8, S (t) is uniformly bounded for jtj > 1, then there exists D1 > 1 such that

_

Y (t) w D1 kY (t) w k for all jtj > 1; w 2 V ( ); 2 : Let

A := (1 + D1)2 sup Hpp()?1 : 2

?

?1

Let H (t) := Y ?2() (t +2) Y ?2 () (2) , then det H (t) 6= 0 for t > ?2, H (0) = I and there exists a linear homomorphism V (t) : H () ! V ( t ()) such that (H ; V ) is a matrix solution of the Jacobi equation (6) and Image(H (t); V (t)) = d t+2 (V ( ?2 ())) is a Lagrangian subspace. Let

H (t) (Hpp )?1 H (t) < +1: D2 := ?max 2t2 2

Let : R ! [0; 1] be a C 1 function such that (t) = 0 for jtj 1 and (0) = 1. Let

C := D2 Let

Z 1

?1

j0 (t)j2 dt :

0 < B < (A C ) 21 :

(66)

Suppose that kY (T ) vok < B kvok for some 2 , vo 2 T() M , kvo k = 1 and jT j > 1. Assume that T > 1, the case T < ?1 is proven similarly. Let J (t) 2 H ( t()) = T t() M be de ned by J (t) = 0 for ?1 t 0, J (t) = Y (t) vo for 0 t T and J (t) = (T + 1 ? t) t J (T ) for T t T + 1, where t : T T () M ! T t() M is the parallel transport along the path t 7! t (). Then J (t) is continuous and piecewise C 2. Let v(t) 2 V () T() M be de ned by J (t) = Y (t) v(t). Then v(t) = 0 for ?1 t 0 and v(t) = vo for 0 t T . Let fe1 ; : : : ; eng be a basis of T() M


Pn i=1 vi (t) ei ,

and write v(t) = t () of J (t) = Y (t)v(t) is

D (Y v) = D dt dt

n X

47

Yi(t) = Y (t) ei. Then the covariant derivative along !

X DYi

X dvi

dt vi + i dt Yi ; Y v0 = D (Y v) ? DY v : dt dt For T t T + 1, since J (t) = Y (t) v(t) = (T + 1 ? t) t J (T ), we have that

kY v0k Ddt (Y v) + DYdt v

Ddt J (t) + D1 kJ (t)k kJ (T )k + D1 kJ (T )k (1 + D1 ) B : Using Y (t) on formula (55) we have that i=1

vi (t) Yi(t) =

I (J; J ) =

Z T +1

T

i

(Y v0) (Hpp)?1 (Y v0) dt

(1 + D1)2 B 2 Hpp?1 = A B 2:

Let Z (t) 2 H ( t ()) = T t() M be de ned by Z (t) = (t) H (t) vo. Then Z (t) = Y (t) (t) for t 6= 0, with (t) = 0 for jtj > 1. Since d t () V () = Image (Y (t) ; S (t) Y (t)), we have that S (t) Y (t) ! I when t ! 0. Using Y (t) on formula (55), we have that the only non zero term is I (Z; J ) = ?Z (0+ ) vo + Z (0?) 0 = ? kvo k2 = ?1: Write Z (t) = H (t) (t), (t) = (t) vo. Using H (t) and V (t) on fromula (53), we have that Z 1 I (Z; Z ) = (H 0) (Hpp)?1 (H 0) dt =

?1 Z 1 ?1

D2

(0 (t))2 (H (t) vo) (Hpp?1) (H (t) vo) dt Z 1

?1

j0 (t)j2 dt = C:

Since by (48) the index form is symmetric, for 2 R we have that I (Z ? J ; Z ? J ) C + 2 + 2 A B 2 : Since by (66) 4 ? 4 A B 2 C > 0, this polynomial is has two real roots. Therefore I (Z ? J ; Z ? J ) < 0 for some value of . Hence there must be conjugate points in the orbit segment f t () j t 2 [?1; T + 1] g.

48


6.2. Proposition. Let e be a regular value of H . If (X ) > 0 on the energy level = H ?1 feg and the ow of X has no conjugate points, then there exists a Hamiltonian H on T M with vector eld Z such that (i) H is convex and H ?1 feg = . (ii) Z = (1X ) X on . (iii) The ow t of Z has no conjugate points on . (iv) t preserves the 1-form on T . (v) If N () := dsd sjs=1, then for 2 we have dt() N () = N (t ()) + t X (t()) :

6.3. Remark. This proposition holds if (X ) 6= 0 on a connected compact foward invariant set K . In this case we obtain an open connected neighbourhood U of K such that inf #2U j(X (#))j > 0 and a convex Hamiltonian H without conjugate points, de ned on a neighbourhood of U in T M . To apply this on theorem F, we ? can take K = clousure of R+ (j)?1 fqg .

Proof:

Recall that !(X; ) = ?dH . Let = p dq be the canonical 1-form on T M . Suppose that (X ) > 0 on . Let Y be the vector eld on given by Y = (1X ) X . Then Y preserves the 1-form on T because the Lie derivative LY = d iY + iY d = d[(Y )] + !(Y; ) = d(1) ? (1X ) dH = 0 on T : Let t(q; p) be the ow of Y on . De ne the simplecti cation et of t (cf. Arnold [3], [2]) on the cotangent bundle without its zero section T M n M by et(q; p) = t(q; p), (q; p) 2 . Observe that et preserves the canonical 1-form . Indeed, we have that t = et and if t(q; p) = (qt; pt ), (qt ; pt) (det) = ( pt) d (det w) = [pt (d dt w)] = [p (d w)] = (q; p) w : Let Ye be the vector eld of et. We have that LYe = d[(Ye )] + !(Ye ; ) = 0 : Therefore the function He = (Ye ) is a Hamiltonian for Ye . But Ye has conjugate points on and He is not convex.


49

Let H = 21 He 2 and let Z be the Hamiltonian vector eld of H . We have that Z = He Ye , since He j 1, Z is also an extension of Y . Let t be the ow of Z . Observe that t(q; sp) = s etHe (q;sp)(q; p) = s et 21 s2 (q; p) ; for (q; p) 2 :

Taking the derivative dsd s=1, we obtain dt() N () = N (t()) + t X (t()) for 2 : Write (t; ) = (s(t; ); ) with s(t; ) > 0, 2 . Then @s D t = D s + X ( s()) @ on T : (67) Let E T be the stable Green bundle for X on . Since X () 2 E () and E is t -invariant, then equation (67) shows that the ow t of Z preserves the Lagrangian bundle E T , which satis es E () \ V () = f0g for all 2 . By proposition 1.15, t has no conjugate points on . The following lemma completes the proof of proposition 6.2

6.4. Lemma. The Hamiltonian H is convex, i.e. H pp is positive de nite. Proof:

Let q 2 and S (q) := TqM \ . Since H is homogeneous of degree 2, then H pp is homogeneous of degree 0 and it is enough to prove that H pp (q; p) is positive de nite for p 2 S (q). Let p0 2 S (q). Let h : Tp0 S (q) R ! Tq M be h(x; ) := x + p0. Let G = H h and K = (Ye h). Then G = 21 K 2. Since D2h = 0, then D2H (q0 ) is positive de nite if D2G(0; 1) is positive de nite. Since K (0; ) = , we have that G (0; 1) = K2 + K K (0;1) = 1 : Since in the coordinates (x; ) the vectors

@ @xi (0;1) ,

i = 1; : : : ; n ? 1 are tangent at

p0 to S (q) and (Ye ) 1 on S (q), then Kx(0; 1) = 0. Hence Gxx(0; 1)(v; v) = jKx vj2 + K Kxx(v; v) = Kxx(v; v) : Observe that K is homogeneous of degree 1. Therefore Kx(0; 1) = Kx(0; 1) for all > 0. Hence Kx(0; 1) = 0, so that Gx(0; 1) (v; ) = [K (Kx v) + K (Kx v)] = 0 : And then D2G(0; 1)(v; )(2) = 2 G + 2(Gx v) + Gxx(v; v) = 2 + Kxx(0; 1) (v; v) :

50


So it is enough to see that Kxx(0; 1) is positive de nite. Let N be the unit normal vector to Tp0 S (q) such that p0 N < 0. Such N exists because H jS(q) e and p0 Hp = (X ) 6= 0. Let F : Tp0 S (q) R ! R be given 2 by F (x; y) = H (p0 + x + yN ). Since @F @y (0;0) < 0, there exists a local C function y : U Tp0 S (q) ! R such that F (x; y(x)) = e. Dierentiating this equation with respect to x, we obtain that dy = ? FFx y d2y = ? F1 D2F (I; dy)(2): y

Since the vectors are tangent to the level F e at (q; p0), we have that Fx(0; 1) = 0. Hence dy(0; 1) = 0 and d2y(0) v = ? F1 D2F (v; 0)(2) = ? F1 Hxx (v; v) > 0 : y y Now let v 2 Tp0 S (q) be a unit vector and r(s) = jp0 + s vj, z(s) = y(sv), R(s) = jp0 + sv + z(s) N j. We have that s hence 1 r(s)2 = 1 jp j2 + (p v) s + 1 s2 : r0(s) = p0 r (vs+ 0 ) 2 2 0 2 Also ) (v + z0 N ) = p0 v + s + z0 (p0 N ) + z z0 R0(s) = (p0 + sv + z(Rs)(N s) R(s) therefore, since s2 z(s) s2 for some 0 < < and jsj small. 2 1 1 1 2 1 2 2 2 R(s) = 2 jp0 j + (p0 v )s + 2 s + z (p0 N ) + 2 z = 21 r(s)2 + z(s) (p0 N ) + 21 z(s)2 21 r(s)2 ? 21 s2 ; for some > 0 and jsj small. Then for some > 0, we have that r(s) 1 + s2 R(s) 2 2 r (s) d d Kxx (v; v) = ds2 K (sv; 1) s=0 = ds2 R(s) > 0 : s=0 @ @xi (0;0)


Proof of Theorem F.

51

Trough the proof of theorem F we work with the lifted Hamiltonian to the cotangent bundle of the universal cover M~ of M . Let t be the ow of the Hamiltonian ~ and let Eq : Tq M~ ! M~ be the exponential H of proposition 6.2. Let q 2 M map associated to the ow t and the energy level . We rst prove that Eq is a dieomorphism. The derivative dEq (t) is non-singular by corollary 1.18 and is given by formulas (39) and (40). We claim that the norm kdEq (t)k is uniformly

bounded below. By lemma 6.1 the norm dEq (t )jV ()\Tq is uniformly bounded

below for t > . From equation (40) dEq (t)jhi is also uniformly bounded below. It remains to see that the angle dEq (t)(V () \ Tq ) ; d(X (t )) is uniformly bounded below. Since t preserves the 1-form , the bundle K() = ker jT is invariant under dt. Since V () \ Tq K(), then ? ? ? dEq (t) V () \ Tq d dt() K() d K(t()) : (68) For = (q; p) 2 write Hp(q; p) = p + k, with hk; pi = 0. Since (X?) = p Hp 6= 0 on there exists 0 < 1 such that j kj jHpj on . Let h 2 d K(t ) with jhj = 1, i.e. hp; hi = 0, jhj = 1. Then hh; Hpi = hp; hi + hk; hi = hk; hi : jhh; Hpij 1 jHpj : Hence cos (h; Hp) < 1. Therefore the angle d(K()); d(X ()) is uni ? ? formly bounded below on 2 . By (68) dEq (t) V () \ T ; d X (t ) > ? d K(t ) ; d(X (t )) is uniformly bounded below. This proves that kdEq (t)k is uniformly bounded below on t 2 R + , 2 . Hence the map Eq : Tq M~ ! M~ is a dierentiable covering map. Since M~ is simply connected, it must be a dieomorphism. Now let expq : Tq M~ ! M~ be the exponential map of ( t ; ). Since t is a reparametrization of t for t > 0, 2 there exists (t; ) > 0 such that t () = (t;)(). By corollary 1.18 expq is a local dieomorphism. It remains to prove that it is bijective. Since j(X )j = jjXY jj is bounded then t 7! (t; ) is a (bijective) homeomorphism of R + . This implies that the map t 7! (t; ) , with 2 TqM \ , ? is a homeomorphism of TqM . Since Eq is bijective and expq (t) = Eq (t; ) , then expq is bijective.

52


APENDIX I.

I. Examples. The following formalism for twisted Hamiltonians on surfaces was derived by G. Paternain and M. Paternain in [30]. Let M be a closed orientable riemannian surface with riemannian metric ?h ; ix. Let K : TTM ! TM be the connection d map K = rx_ v, where = dt x(t); v(t) . Let : TM ! M be the canonical projection. Let !0 be the symplectic form in TM obtained by pulling back the canonical symplectic form via the Legendre transform associated to the riemannian metric, i.e. !0(; ) = hd ; K i ? hd ; K i : Denote by the area 2-form on M . Given a smooth function F : M ! R , de ne a new symplectic form !F on TM by !F = !0 + (F ) ( ) : This is called a twisted symplectic structure on TM (cf. [30]). Let H : TM ! R be the Hamiltonian H (x; v) = 21 hv; vix :

Consider the Hamiltonian vector eld XF corresponding to (H; !F ), i.e. ? !F XF (); = dH :

(69)

De ne Y : TM ! TM as the bundle map such that

x(u; v) = hY (u); vix :

(70)

Then Y (u) = iu is the rotation of angle + 2 . From (69) and (70) we have that ? ? ? dH ( ) = !0 XF (); + F () Y d XF () ; d () : (71) Hence if = (1; 2) = (d ; K ) 2 H () V () and XF = (X1 ; X2) 2 H () V (), equation (71) becomes h; 2i = hX1; 2i ? hX2; 1i + F hY (X1 ); 1i ; ? XF () = ; F Y () : (72) ?

In particular, the orbits of the Hamiltonian ow are of the form (t); _ (t) for

: R ! M , with D _ = F ( ) Y ( _ : (73) dt

R


53

If M F = 0 this ow can be seen as a Lagrangian ow as follows. The condition M F = 0 implies that the cohomology class of F is zero. Hence there exists a 1-form on M such that d = F . Consider the Lagrangian L(x; v) = 21 hv; vix + x(v) : The corresponding Euler-Lagrange equation is D h ;_ i = d ( _ ) = hF Y ( _ ); i : x dt which gives the same dierential equation (73). This ow corresponds to a magnetic eld with Lorentz force F (x) Y (v). Observe thatR the theory developed in the previous sections also applies to the case in which M F 6= 0. Indeed, let q : U M ! R2 be a local chart de ned on a simply connected domain U M . Then the 2-form F is exact on U and there exists a 1-form jU such that d = F on U . The Hamiltonian ow on TU TM is the Lagrangian ow of L(x; v) = 21 hv; vix + x(v), x 2 U . Let (q; p) be the natural chart on T U T M corresponding to the chart q and let F ?1 be the inverse of the Legendre transform corresponding to L. Let (q; w) = F ?1 (q; p). Since F (d) = !F , then (q; w) is a symplectic chart for !F and sends the vertical bundle of Tq(U )R n to the vertical bundle of TU M . We derive the Jacobi equation. Let (c(t); c_(t)) be an orbit of the ow. Consider a variation s(t) = f (s; t), where f (0; t) = c(t) and the paths t 7! f (s; t) are solutions of (73): D _ = F Y ( _ ) : (74) s dt s Denote the variational eld of f by J (t) = @f @s (0; t). Using the formula D D @f = D D @f + R @f ; @f @f : ds dt @t dt ds @t @t @s @t and taking the covariant derivative of equation (74) in the direction J (t), we obtain J + R(_c; J ) c_ = D F Y (

_ s) : ds Since Y (v) is linear on the bers of TM and J_ = Dds @f@t = Ddt @f @s , we have D FY ( _ ) = r (F Y )( _ ) + F Y (J_) ; s J s ds = (rJ F ) Y (_c) + F Y (J_) : Hence we obtain the Jacobi equation J + R(_c; J ) c_ ? (rJ F ) Y (_c) ? F Y (J_) = 0 : (75) R

54


Consider the orthonormal basis f c_; Y (_c) = i c_ g of Tc(t) M . We have that D c_ = F Y (_c) ; dt D Y (_c) = (r Y )(_c) + Y ? D c_ c_ dt dt = ?F c_ : Write J = x c_ + y Y (_c) : Then J_ = (x_ ? F y) c_ + (y_ + F x) Y (_c) ; J = (x ? 2 F y_ ? F_ y ? F 2 x) c_ + (y + 2 F x_ + F_ x ? F 2 y) Y (_c) ; F Y (J_) = ?F (y_ + F x) c_ + F (x_ ? F y) Y (_c) : Replacing these formulas on equation (75) we obtain x ? (F y)0 = 0 ; (76) y + K y + F x_ = y (rY (c_) F ) ; (77)

?

where K = R c_; Y (_c) c_ ; Y (_c) is the sectional curvature along c(t). Fix the energy level = H ?1f 21 g = SM . If (J; J_) 2 Tc_(t) H () V (), we have that dH (J; J_) = hc_; J_i = 0. Hence x_ ? F y = 0 : (78) This implies equation (76), and from (77) and (78), we get ? y + K + F 2 ? rY (c_) F y = 0 : (79) Hence the Jacobi equations on Tc_(t) are given by (78) and (79).

Example 1. A convex Hamiltonian without conjugate points and no continuous

invariant transversal bundle.

Take K ?1, F 1. Then equations (78) and (79) become x_ = y ; y = 0 : The solutions have the form x(t) = 21 a t2 + b t + c ; y(t) = a t + b ; y_ (t) = a :

(80) (81)


The derivative of the Hamiltonian ow

t

?

55

on c(t); c_(t) is given by

? ? d t J (0); J_(0) = J (t); J_(t) 2 H ( t) V ( t ) ? = x c_ + y Y (_c) ; 0 c_ + (y_ + x) Y (_c) ? = x(t) c_ + y(t) Y (_c) ; z(t) Y (_c) ; ?

(82)

where z(t) = y_ (t) + x(t) and = c(0); c_(0) . Every orbit (c; c_) has no conjugate points because the only solution of (81) with x(0) = y(0) = 0 and y(T ) = 0 = x(T ) , T 6= 0 is x(t) = y(t) 0. The Green bundles coincide and are generated by X () = (_c; ?Y (_c)) and (Y (_c); 0). The last vector can be found by solving x(0) = 0, y(0) = 1, x(T ) = y(T ) = 0 and letting T ! 1. Now suppose that there exists an invariant continuous subbundle N () T which is transversal to the vector eld X (). Let 2 be a recurrent point . Let 2 N () n f0g. If d t ( ) = and let tn ! +1 be such that tn () ! n n ? ? x(t); y(t); z(t) = J (t); J_(t) , then 1 d ( ) = lim 1 ?x(t ); y(t ); z(t ) lim n n n n!1 jx(tn )j tn n x(tn ) = (1; 0; 1) = X () :

(83)

By the continuity of N we have 1 d ( ) 2 N () : X () = lim n x(tn ) t This contradicts the transversality hypothesis. The?same argument as in equation (83) shows that for every 2 , the angle ? d t V () \ T ; X ( t) t!?! 0. +1 In this case the trajectories of the Hamiltonian ow are the unit tangent vectors of the horospheres of the riemannian metric, i.e. t () = ?Y

?

ht(Y ()) ;

(84)

where ht is the horocycle ow. In particular, the exponential map expq : Tq M~ ! M~ is a dieomorphism for all q 2 M . This ow has no periodic orbits because the ~ 1(M ) are homeomorphic to R . horocycles W uu = T 1M=

56


~ M

example I.1

We prove (84). Let 2 Tq M~ , kk = 1. Let t : T M~ - be the geodesic ow of M~ . Consider a sphere ? ST () = q 2 M~ j dM~ q; T () = T and a parametrization by arc length T (s) of ST () with T (0) = (). Then Dds _ T (s) is orthogonal to _ T (s). Moreover, it has constant norm, because it is the image of D _ (0) by the isometry given by a rotation about the center () of S (). T T ds T Therefore T satis es D _ = Y ( _ ) ds T T T for some constant T > 0. Passing to the limit when T ! +1 we obtain that the reparamerizations by arc length of the horospheres satisfy D _ = Y ( _ ) ; (85) ds with = limT !+1 T , (in fact T ! T is strictly decreasing). We claim that = 1. Indeed, the solutions of (85) of are closed circles if is big enough. The in mum 0 of such 's corresponds to the horospheres . For > 0 the ow has conjugate points corresponding to the period (length) of these closed circles (see example 3). From the Jacobi equation (79) with F = , we see that = H ?1f 12 g has conjugate points for > 1 and the ow is Anosov for < 1. Hence 0 = 1. Indeed, for < 1 the solutionpof (78) and (79) are y = A et + B e?t, x = ? ( A et ? B e?t + C ), with = 1 ? 2. Writing = x(0); y(0); x_ (0); y_ (0) we ? have that d t ( ) = x(t); y(t); x_ (t); y_ (t) . For A 6= 0 the function t 7! kd t ( )k has exponential growth and for B 6= 0, t 7! kd ?t( )k has exponential growth. As in section x3, this implies that the ow is Anosov for < 1. For > 1, the Jacobi equations (78), (79) exhibit conjugate points, by the same arguments as for equations (87) and (88).


57

Example 2. A periodic orbit without conjugate points and no transversal invariant subspace.

We construct a convex Lagrangian which has the Jacobi equations (80) on a periodic orbit. Let 2 be such a periodic point with period > 0. Changing the coordinates (x; y; z) on T of equation (82) to (x; y; y_ ) = (x; y; z ? x), the matrix of the derivative d () is 2 3 2 3 1 t 21 t2 g x 1 1 Jordan d () = 4 1 t 5 g y 4 1 15 in form. 1 1 g y_

Let fe1; e2 ; e3g be the basis of T in which d () is written in canonical form. The transformations d and 2 3 0 1 d ? I = 4 0 15 0 have the same invariant subspaces. The vector eld corresponds to the unique eigenspace he1 i and the only invariant subspaces are he1i = hX ()i, he1 ; e2i = E () = F ( ) and he1 ; e2 ; e3 i = T . None of them is transversal to X ( ) = e1 . Now we make the construction. Consider the torus T2 = R 2 =5Z2 with the at metric and polar coordinates (r; ) centered at (0; 0) 2 R 2 =5Z2 de ned on r 2. Let F : T2 ! R Rbe a C 1 function such that F (r; ) = F (r), F (r = 1) = 1, @F 2 @r r=1 = ?1, and T2 F = 0. Let be a 1-form in T such that d = F . De ne L(x; v) = 12 hv; vix + x(v). At r = 1, the Euler-Lagrange equation (73) is D _ = F ( ) Y ( _ ) = Y ( _ ) at r = 1 : dt Hence r = 1, _ = ?1 is a periodic solution. Moreover, at r = 1 we have that F = 1, rY (c_) F = ? @F@r = +1. Hence the Jacobi equations (78), (79) are the same as (80).

Example 3. A segment without conjugate points with crossing solutions.

Let W = (x; y) 2 R 2 x2 + y2 4 and consider the Lagrangian L(x; v) = 12 hv; vi + x(v) ; where h ; i is the euclidean metric and d = ? . This Lagrangian corresponds to F jW = ?1 in the formalism above and can be embedded into a convex Lagrangian on a closed surface. The Euler-Lagrange equation (73) is v_ = ?i v : (86)

58


The solutions of (86) on the energy level jvj = 1, are all the circles contained in W with constant angular velocity _ = ?1. Fix C : R(t) = 1, (t) = ? t in polar coordinates in W with center (0; 0). Let # = (0; 1)(?1;0) be the initial vector of this solution. Then the circles which are rotations of C xing the initial point p = (?1; 0) are also solutions of (86) (see gure I.3) with the same energy E = 21 jvj = 21 . The solution C has conjugate points, but the conjugate points appear at t = 2. In 3 particular the segment t (#) 0 t 2 has no conjugate points. But the intersections with other solutions starting at p = (?1; 0) appear at t < 32 and at t = there is a vector 2 V (#) \ T# such that d (d t ) = d X ( t #) = (0; ?1). Proposition 1.16 says that these phenomena can not occur if all the positive (negative) orbit of # has no conjugate points. The claims stated above can be proved using the Jacobi equations (78) and (79): x_ + y = 0 ; y + y = 0 ; (87) whose solutions are x(t) = a cos t + b sin t + c ; y(t) = a sin t ? b cos t : (88) ? Then if x(0) = y(0) = 0 and y_ (0) 6= 0 then x(t); y(t) 6= 0 for all 0 < t < 2. Moreover, if a = ?1, b = 0, c = +1, then x(0) = y(0) = 0, y() = 0, x() = +2.

p=(-1,0)

(0,0)

example I.3

(1,0) t= π

example I.4

Example 4. A non surjective exponential map without conjugate points. Consider the

at torus T2 = R 2 =6Z = [?3; 3]2=6Z and polar coordinates (r; ) on (x; y) 2 T2 x2 + y2 < 3 centered at 0 = (0; 0). De ne the Lagrangian L(x; v) = 41 hv; vix + x(v) + (x) ; where is a C 1 function on T2 such that (r; ) = 14 r2 + cos r on r 2 and (r; ) 1 outside r 2; and (r; ) = ? 21 r2 f (r) d with f (r) a C 1 function with support on r 25 such that f (r) = 1 on r 2 and f (r) 1.


59

Observe that the path C : r(t) 1, (t) = t is static (cf. Ma~ne [25]). Indeed, the Lagrangian L(r; ; r;_ _) = 14 (r_ 2 + r2 _2 ) ? 21 r2 _ + 14 r2 + cos r on r 2 ; is minimized on r_ = 0, _ = 1, r = 1, with L = ?1; and on r 2, we have that (r) 1 and f 1, so that 1 r 2 _ 2 + (v ) r 2 1 _2 ? 1 _ ? 1 r 2 ?1 4 4 2 4 Hence L (r) 1 on r 2. By theorem VII in Ma~ne [25] (see also [8]), there exists a semistatic (hence foward minimizing) solution (t) such that (0) = (0; 0) and the !-limit of ( ; _ ) is the (hyperbolic) closed orbit C . By the rotational symmetry of the Lagrangian, for every 1-dimensional subspace L T(0;0) T2 we obtain a semistatic curve , which is tangent to L, with L(0) = (0; 0) and !-limit( L; _ L) = C . Since the semistatic solutions are minimizing, none of them Lj[0;+1[ have conjugate points. Moreover, by theorem X in Ma~ne [25] (also [8]), C and all the ( L; _ L)'s are in the same energy level E = c(L). By the graph property (theorem VII, Ma~ne [25] and also [8]), the

Lj]0;+1['s don't intersect (C ) neither intersect each other.

References

[1] R. Abraham & J. E. Marsden. Foundations of Mechanics. Benjamin: London, 1978. [2] V.I. Arnold. Mathematical Methods of Classical Mechanics. Graduate Texts in Math., 60. Springer, 1989. [3] V. I. Arnold, S.P. Novikov. (Editors) Dynamical Systems IV. Symplectic Geometry and its Applications. Encicl. of Math. Springer, 1985. [4] W. Ballman, M. Brin & K. Burns, On surfaces with no conjugate points. Jour. Dierential Geometry, 25 (1987) 249-273. [5] D. Bao & S. S. Chern. On a notable connection in Finsler geometry. Houston Jour. of Math., vol. 19, No. 1, (1993), 135-180. [6] S. S. Chern. Finsler Geometry is just Riemannian Geometry without the Quadratic Restriction. Notices of the A.M.S., 43 No. 9, sept. 1996, 959-963. die Reduktion der zweiten Variation auf ihre einfachste Form. J. Reine [7] A. Clebsch. Uber Angew. Math. 55 (1958) 254-276. [8] G. Contreras, J. Delgado, R. Iturriaga. Lagrangian Flows: The Dynamics of Globally Minimizing Orbits, part II. [9] J. J. Duistermaat. On the Morse Index in Variational Calculus. Advances in Math., 21, (1976), 173-195. [10] P. Eberlein. When is a geodesic ow of Anosov type? I. J. Diferential Geometry, 8, (1973), 437-463. [11] J. Franks & C. Robinson, A quasi-Anosov dieomorphism that is not Anosov, Trans. Amer. Math. Soc., 223, (1976), 267-278. [12] A. Freire. Fluxos geodesicos em variedades riemannianas compactas de curvatura negativa, M.Sc. Thesis, IMPA, 1981.

60


[13] A. Freire & R. Ma~ne. On the entropy of the geodesic ow in manifolds without conjugate points. Invent. Math. 69, (1982), 375-392. [14] P. Foulon, Geometrie des equations dierentielles du second order, Ann. Inst. Henri Poincare, 45 (1), (1986), 1-28. [15] P. Foulon. Estimation de l'entropie des systemes lagrangiens sans points conjugues. Ann. Inst. Henri Poincare 57, (1992), 117-146. [16] L. W. Green. A theorem of E. Hopf, Michigan Math. J. 5 (1958) 31-34. [17] P. Hartman. Ordinary Dierential Equations. John Wiley & Sons, Inc., 1964. [18] M. Hirsch, C. Pugh, M. Shub. Invariant manifolds. Lecture Notes in Math., 583. Springer, 1977. [19] N. Innami. Natural Lagrangian systems without conjugate points. Ergod. Th. & Dynam. Sys. 14, (1994), 169-180. [20] W. Klingenberg. Riemannian manifolds with geodesic ows of Anosov type, Ann. of Math., 99, (1974), 1-73. [21] R. Ma~ne. On a theorem of Klingenberg, Dynamical Systems and Bifurcation Theory, M. Camacho, M. Paci co and F. Takens eds., Pitman Research Notes in Math. 160, (1987), 319-345. [22] R. Ma~ne. On the minimizing measures of Lagrangian dynamical systems. Nonlinearity, 5, (1992), no. 3, 623-638. [23] R. Ma~ne. Generic properties and problems of minimizing measures of lagrangian systems. Nonlinearity, 9, (1996), no. 2, 273-310. [24] R. Ma~ne. Asymptotic measures of class A solutions of generic Lagrangians. Unpublished. [25] R. Ma~ne. Lagrangian Flows: The Dynamics of Globally Minimizing Orbits. To appear, Proc. Int. Congress in Dyn. Sys., Montevideo 1995, Pitman Research Notes in Math. [26] J. Mather. Action minimizing invariant measures for positive de nite Lagrangian systems. Math. Z. 207, (1991), no. 2, 169-207. [27] M. Morse. Calculus of variations in the large. Amer. Math. Soc. Colloquium Publications vol. XVIII, 1934, 1960. [28] V.I. Oseledec. A multiplicative ergodic theorem. Trans. Moscow Math. Soc. 19, (1960), 197231. [29] G. Paternain & M. Paternain. On Anosov Energy Levels of Convex Hamiltonian Systems. To appear. Math. Zeitschrift. [30] G. Paternain & M. Paternain. On the topological entropy of autonomous Lagrangian systems. To appear. [31] Ya. Pesin. Lyapounov characteristic exponents and smooth ergodic theory. Usp. Mat. Nauk, 32-4, (1977), 55-111. [32] C. Robinson. A quasi-Anosov, ow that is not Anosov. Indiana Univ. Math. J. 25, (1976), no 8, 763-767. [33] D. Ruelle. An inequality for the entropy of dierentiable maps. Bol. Soc. Bras. Mat. 9, (1978), 83-87. Depto. de Matematica. PUC-Rio, R. Marqu^es de Sa~o Vicente, 225, 22453-900 Rio de Janeiro, Brasil

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