CONVEX HAMILTONIANSWITHOUT CONJUGATE POINTSGONZALO CONTRERAS AND RENATO ITURRIAGAIntroduction.LetM be a closed connected riemannian manifold and T �M its cotangent bundle.By a convex Hamiltonian on T �M we shall mean a C3 function H : T �M ! Rsatisfying the following conditions:(a) Convexity: For all q 2M , p 2 T �qM , the Hessian matrix @2H@pi @pj (q; p) (calculatedwith respect to linear coordinates on T �qM) is positive de�nite.(b) Superlinearity: limjpj!1 H(q; p)jpj = +1 :The Hamiltonian equation for H is de�ned asq0 = Hp ;p0 = �Hq (1)where Hq and Hp are the partial derivatives with respect to q and p. Observethat the Hamiltonian function H is constant along the solutions of (1). Its levelsets � = H�1feg are called energy levels of H. Then the compacity of M andthe superlinearity hypothesis imply that the energy levels are compact. Since theHamiltonian 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 projectionand 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 conju-gate 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

2 G. CONTRERAS AND R. ITURRIAGAProposition A.Suppose that the orbit of � 2 T �M does not contain conjugate points and H(�) = eis a regular value of H. Then there exist two '-invariant Lagrangian subbundles E ,F � T (T �M) along the orbit of � given byE (�) = limt!+1 d �t�V ( t(�))� ;F(�) = limt!+1 d t�V ( �t(�))� :Moreover, E (�)[F (�) � T��, E (�)\V (�) = F(�)\V (�) = f0g, hX(�)i � E (�)\F (�)and dim E (�) = dim F(�) = dimM , where X(�) = (Hp;�Hq) is the Hamiltonianvector�eld and � = H�1feg.These bundles where constructed for disconjugate geodesics of riemannian metricsby Green [16] and of Finsler metrics by Foulon [15]. In the Finsler case this is done byde�ning a new connection er such that the solutions of the Euler-Lagrange equationsatisfy er _ ( _ ) � 0. This approach leads to extensions of many important theorems inRiemannian geometry to Finsler geometry. See Chern [6] and the references thereinfor a survey and an introduction to this �eld.If the energy level � does not have conjugate points, the Green bundles E and Fcan 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 subspacesand hyperbolicity appears in the followingProposition B.Let � be a periodic orbit of t without conjugate points. Then � is hyperbolic (onits energy level) if and only if E (�)\F(�) = hX(�)i for some � 2 �, where hX(�)i isthe 1-dimensional subspace generated by the Hamiltonian vector�eld X(�). In thiscase E and F are its stable and unstable subspaces.We can compare proposition A with the following theorem by G. Paternain andM. Paternain [29].0.1. Theorem. [Paternain] [29]Suppose that e is a regular value of H, and that tj� admits a continuous invariantLagrangian subbundle E. Then,(a) E(�) \ V (�) = f0g 8� 2 �:(b) �(�) =M .(c) � contains no conjugate points.

HAMILTONIANS WITHOUT CONJUGATE POINTS 3It is not di�cult to see (e.g. [21]) that when the ow tj� is Anosov, the stableand unstable subbundles of T� are Lagrangian. Paternain's theorem then impliesthat Anosov Lagrangian ows do not have conjugate points. This result was provenfor geodesic ows by Klingenberg [20]. An extension of Klingenberg result appearsin Ma~n�e [21].De�ne B(�) = �� 2 T�� j supt2R jd t(�) �j < +1 :An important property (cf. proposition 1.11) in the proof of proposition B is thatif the orbit of � is disconjugate then B(�) � E (�) \ F(�). This in turn implies thathX(�)i � E (�)\F(�) and E (�)[F(�) � T�� (cf. corollary 1.12). The global versionof proposition B is given byTheorem C.Let � = H�1feg be a regular energy level without conjugate points.Then the following statements are equivalent:(a) tj� 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 supt2R jd t(�) � vj = +1 :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 interestingingredient on its proof is the characterization of recurrent quasi-hyperbolic R-actionsof 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 linearisomorphism 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 ifsupt2R j�t(�)j = +1 for all � 2 E; � 6= 0 ;and we say that it is hyperbolic if there exists a continuous splitting E = Es � Euand C > 0, � > 0 such that(i) j�t(�)j � C e��t for all t > 0, � 2 Es.(ii) j��t(�)j � C e��t for all t > 0, � 2 Eu.It is easy to see that hyperbolic actions are quasi-hyperbolic. The converse is falseby counterexamples in Robinson [32] and Franks & Robinson [11]. But it is truewith an additional recurrency hypothesis, (that in our case is always true becausethe Lagrangian ow preserves the Liouville measure).

4 G. CONTRERAS AND R. ITURRIAGADe�ne the non-wandering set (�) of � as the set of points b 2 B such that forevery neighbourhood U of b there exists T > 0 such that �T (U) \ U 6= 0. Thefollowing theorem is a slight modi�cation of a theorem in Freire [12], written therefor quasi-Anosov ows (i.e when �t = d�t and E is a continuous invariant subbundleof TB transversal to the vector�eld X, see theorem 3.1). For completeness of theexposition 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 � isquasi-hyperbolic and (�jB) = B, then � is hyperbolic.The reason to use this version of Freire's result is that our Hamiltonian owsare not of contact type and a priori they may not preserve any continuous bundletransversal to the vector�eld X(�). Indeed, in apendix I we show an example of aconvex Hamiltonian without conjugate points where this phenomenon occurs. Thesolution is to study the behaviour of an (orbitwise) reparametrization of the owwhich preserves a continuous tranversal bundle E � T (T �M). This reparametriza-tion 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 projectionalong the direction of the vector�eld X(�).Generic Lagrangians.A C4 function L : TM ! R is called a (convex autonomous) Lagrangian if itsatis�es the following properties:(a) Convexity: For all x 2M , v 2 TxM , the Hessian matrix @2L@vi @vj (x; v) (calculatedwith respect to linear coordinates on TxM) is positive de�nite.(b) Superlinearity: limjvj!1 L(x; v)jvj = +1 ;uniformly on x 2 R.Observe that by the compactness ofM , the Euler-Lagrange equation, that in localcoordinates is: ddt @L@v (x; _x) = @L@x (x; _x) ; (2)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.

HAMILTONIANS WITHOUT CONJUGATE POINTS 5A Lagrangian ow is conjugated to a Hamiltonian ow by the Legendre transformFL : TM ! T �M de�ned byFL(x; v) = �x; @L@v (x; v)� :The corresponding Hamiltonian is given by H(q; p) = E(F�1L (q; p)), where E :TM ! R is the energy function de�ned byE(x; v) := @L@v ����(x;v) � v � L(x; v) : (3)Conversely, given a convex Hamiltonian H on T �M , the corresponding convex La-grangian is given by the Legendre transform of the Hamiltonian FH : T �M !T ��M � TM which can be seen asL(x; v) = maxp f p(v)�H(x; p) g = (pHp �H) � FH :Observe that item (b) in theorem 0.1 implies that the energy e of an Anosov levelfor the Lagrangian ow satis�ese > e0 := �maxp2M L(p; 0) ;because e0 is the minimal energy level e for which �(�) =M . De�nec(L) := �min�ZTM Ld� ��� � is a '� invariant probability� :It is proven in [25], [8], that c(L) � e0.We say that a '-invariant probability � is minimizing ifc(L) = � Z L d� ;and we say that it is uniquely ergodic if it is the only invariant probability measuresupported inside supp(�).Action minimizing curves do not contain conjugate points. This fact is knownsince Morse earlier works (cf. [27]). We present a simple proof of this fact insection x4 and we use proposition B to prove the following theorem, which wasstated as theorem III in Ma~n�e's un�nished work [25].Theorem D.There exists a generic set O � C1(M;R) such that for all � 2 O the LagrangianL+� has a unique minimizing measure and this measure is uniquely ergodic. Whenthis 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(�) tW u(�).

6 G. CONTRERAS AND R. ITURRIAGAOn Ma~n�e [25], it is conjectured that there exists a generic set O such that thisunique minimizing measure is supported on a periodic orbit or an equilibrium point.The �rst statement of theorem D was proved by Ma~n�e in [23]. Here we provethe generic hyperbolicity of the minimizing periodic orbit and the transversalityproperty W s(�) t W u(�). The hyperbolicity of � was announced in Ma~n�e [25]and the transversality W s(�) t W u(�) was conjectured in Ma~n�e [24]. We use theClebsch transformation [7] to derive formulas for the index form on an orbit withoutconjugate points. Then we use similar arguments to the Rauch comparison theoremto 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 formu-las for the metric entropy of a Hamiltonian ow without conjugate points. Oseledec'stheory [28] gives a splittingT�� = Es(�)� Ec(�)� Eu(�)for a set of points � 2 � � � of total measure, such thatlimt!�1 1t log jd t(�) �j < 0 for � 2 Es(�) ;limt!�1 1t log jd t(�) �j = 0 for � 2 Ec(�) ;limt!�1 1t log jd t(�) �j > 0 for � 2 Eu(�) :Following Freire and Ma~n�e [13], we see in proposition 1.14 thatEu(�) � F(�) � Eu(�)� Ec(�) ;Es(�) � E (�) � Es(�) � Ec(�):Given a riemannian metric on M , de�ne the horizontal subspace as the kernel ofthe 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 + UHpp U + UHpq +Hqp U +Hqq = 0;where Hpp, Hqp, Hpq and Hqq are linear maps which coincide with the second deriva-tives of H in local coordinates, and _U is the covariant derivative of U de�ned as_U(�) v = limh!0 1h��h U( h(�)) v � U(�) v� 2 V (�) � T ��M;where �h : T �� h(�)M ! T ��(�)M is the parallel transport.

HAMILTONIANS WITHOUT CONJUGATE POINTS 7For � 2 � de�ne the unstable Jacobian by�(�) := limT!+1 1T log ��det d T jEu(�)�� :We prove in proposition 1.14 that�(�) = limT!+1 1T Z T0 tr [Hpq +Hpp U] ( t(�)) dt ;where tr ( ) is trace. From Ruelle's inequality [33], for any invariant measure � for tj�, we obtain that the metric entropy satis�esh�( j�) � Z� � d� � Z� tr[Hpq +Hpp U] d� :Pesin's formula states that for a smooth invariant measure � of a C1+� di�eomor-phism f : N -, we have that h�(f) = ZN � d� :In our case we obtainTheorem E.Let m be the Liouville measure on an energy level � without conjugate points.Then hm( j�) = Z� tr[Hpq +HppU] dm :This theorem was proved for geodesic ows by Freire and Ma~n�e [13], Finslermetrics 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 1.10which states the horizontal growth of vertical vectors, i.e.limt!+1 d� � d tjV (�) = +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 tjV (�) ,� 2 �, jtj � 1. This bound was given by Freire and Ma~n�e [13] in the case of geodesic ows.De�ne the exponential map of an energy level � as expq : T �qM !M , expq(t�) =�( t(�)), where � 2 � \ T �qM . This map is not di�erentiable at 0 2 T �qM , but itsLagrangian version is di�erentiable with d expq(0) = I. An important question is

8 G. CONTRERAS AND R. ITURRIAGAwether the lift of this map to the universal cover, expq : T �qM ! ~M , is a di�eomor-phism in an energy level without conjugate points and, in particular, if the universalcover ~M is homeomorphic to Rn . We show in corollary 1.18 that if the energy levelhas no conjugate points then expq is a local di�eomorphism.Let � = p dq be the Liouville's 1-form (4) in T �M . Given a regular energy level� and q 2 �(�), let �q := (�j�)�1fqg. Since ~M is simply connected, in order toobtain a global di�eomorphism expq : T �qM ! ~M it is su�cient to show that expq isa covering map. For this it is enough the condition �(X) > 0. Indeed, in section x6we prove the followingTheorem F.Let H : T �M ! R be convex and � = H�1feg a regular energy level. Supposethat 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 : T �q ~M ! ~M associated to ( t;�) is adi�eomorphism.We give an example (cf. ex. I.4) in apendix I in which expq is not surjective andwhere all the orbits in � of t starting at ��1fqg have no conjugate points. Thisshows that condition (a) is not su�cient in theorem F. By theorem X in Ma~n�e [25](see also [8]), this can not happen for high energy levels. Example 1 in apendix Ialso shows that the norm of the derivative kd�(expq)k may not be bounded belowon an energy level without conjugate points. We don't know if the exponential mapis a di�eomorphism when the whole energy level has no conjugate points.The condition �(X) = pHp > 0 is written �(X) = v Lv > 0 in Lagrangian form.It is satis�ed by Finsler Lagrangians, L(x; v) = 12 kvk2x, mechanical LagrangiansL(x; v) = 12hv; vix� �(x) (with �(X) = 1, H(q; p) = 12hp; piq + �(q)) and all convexHamiltonians on su�ciently high energy levels. It fails for magnetic LagrangiansL(x; v) = 12hv; vix � �(x) + �x(v) (with d� 6= 0) on lower energy levels. If oneadds a closed 1-form to a Lagrangian, the Euler-Lagrange equation and hence theLagrangian ow does not change. This may help to make a given Lagrangian owsatisfy 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 alwaysnonzero by proposition 1.16. On Example 3 in apendix I this angle reaches zerobefore the �rst conjugate point appear. Moreover, there are crossings of solutionson � starting at �(�) before the �rst conjugate point.

HAMILTONIANS WITHOUT CONJUGATE POINTS 9On 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 andwe prove theorem C. On section x4 we give a characterization of the index form andon section x5 we prove theorem D. On section x6 we give a uniform lower bound ford� � d tjV (�) 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 usefullconversations and J. Sotomayor for pointing us reference [17]. We also want to thankthe 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]. Thereis 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�eldX 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 calledenergy levels. Given a local chart q : U � M ! Rn it induces a natural chart(q; p) : T �U ! Rn � Rn of T �M . In natural charts the Liouville's 1-form is writtenas � = p dq. Hence in natural charts ! = dp ^ dq and the Hamiltonian equationsare given by _q = Hp ; _p = �Hq : (5)In general, a set of local coordinates (q; p) of T �M is called symplectic if the canonical2-form is written as ! = dp ^ dq. A subspace E � T (T �M) is said Lagrangian ifdimE = n = dimM 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 sub-space V (�) = ker (d�(�)) and the horizontal subspace H(�) given by the kernel ofthe connection map. Using the isomorphism K : T�T �M ! T�(�)M � T ��(�)M , � 7!(d�(�) � ; r�(� �)), we can identifyH(�) � T�(�)M�f0g and V (�) � f0g�T ��(�)M �T�(�)M . If we choose local coordinates along t 7! � t(�) such that t 7! @@qi (� t(�))are parallel vector�elds, then this identi�cation becomes � $ (dq(�); dp(�)). LetE � T�T �M be an n-dimensional subspace such that E \ V (�) = f0g. Then E is agraph of some linear map S : H(�)! V (�). It can be checked that E is Lagrangianif and only if in symplectic coordinates S is symmetric.

10 G. CONTRERAS AND R. ITURRIAGAProof 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 HamiltonianH such that �0(0) =� and dds�s(�)js=0 = �. Writing d�t(�) = �h(t); v(t)�, we obtain the HamiltonianJacobi equations _h= Hpq h+Hpp v ;_v = �Hqq h�Hqp v ; (6)where the covariant derivatives are evaluated along �(�o(t)), and Hqq, Hqp, Hppand Hqq are linear operators on T�(�)M , that in local coordinates coincide with thematrices of partial derivatives � @2H@qi@qj�, � @2H@qi @pj�, � @2H@pi @pj� and � @2H@pi @qj�. Moreover,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 asymmetric map. That is, if � 2 E thend t(�) = �h(t); S(t) h(t)� :Using equation (6) we have that_Sh + S(Hpqh+HppSh) = �Hqqh�HqpSh :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. LetY (t) : T�(�)M � V (�) ! H( t�) � T� t �M , t 2 R, be the family of isomorphismsY (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 thatlimt!0 S(t); Y (t) = I; :Given u 2 T�(�)M , de�neh1(t) = Y (t) u ;v1(t) = S(t)Y (t) u ; t > 0 :

HAMILTONIANS WITHOUT CONJUGATE POINTS 11Then (h1; v1) is a solution of equation (6) with h1(0) = 0, v1(0) = u. Let � 2 T�T �Mand let (h(t); v(t)) = d t(�) 2 H � V . Since d t preserves the symplectic form, wehave that hh(t); v1(t)i � hv(t); h1(t)i = hh(0); uiHence hY �(t)S(t) h(t) ; ui � hY �(t) v(t) ; ui = hh(0) ; uifor all u 2 T�(�)M . Thereforev(t) = S(t) h(t)� (Y �(t))�1 h(0) : (9)Since _h = Hpq h+Hpp vwe get _h = (Hpq +HppS)h�Hpp(Y �)�1h(0) (10)Di�erentiating Y �1 h and using equation (8), we obtainh(t) = Y (t)Y (c)�1 h(c) + Y (t) Z ct Y �1(s)Hpp (Y �(s))�1 h(0) ds ; for 0 < t < c:(11)Now consider the map d t restricted to the subspace d �c(V ( c �)), c > 0. Itsprojection Zc(t) to the horizontal subspace is given by the solutions of (6) withh(c) = 0. Then Zc(t) is the family of isomorphisms given byZc(t) = Y (t) Z ct 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 andZc(0) = limt!0+Zc(t) = I :Observe that from its de�nition, the linear map Zc(t) is de�ned for all t 2 R andthe 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 _Zc(0)exists, and taking the limit in (12), we have that the linear isomorphism_Zd(0)� _Zc(0) = Hpp(�) Z dc Y (s)�1Hpp (Y �(s))�1 ds (13)is symmetric and the second factor on the right is positive de�nite for all 0 < c < d.In particularZd(t)� Zc(t) = Y (t)Hpp(�)�1 h _Zd(0)� _Zc(0)i ; 0 < c < d : (14)For " > 0 de�ne Z�"(t) := Y (t)Nc + Zc(t) ; t 2 R ; (15)

12 G. CONTRERAS AND R. ITURRIAGAwhere Nc := �Y (�")�1 Zc(�"). We have that Z�"(t) is the solution of the Jacobiequation (6) satisfying Z�"(�") = 0, Z�"(0) = I and_Z�"(0)� _Zc(0) = Hpp(�) Nc : (16)1.1. Claim. Y �1(t)Zc(t) is symmetric for all t 6= 0. In particular Nc is symmetricfor all c > 0.Proof: Using (10), we have that_Zc = AZc+Hpp (Y �)�1 ;_Y = AY + 0 ; t 6= 0 ;where A = Hpq +Hpp S. And then_Z�"= AZ�"+ Hpp(Y �)�1 :Using (9) we have that the functionsh2(t) = Z�"(t) u ;v2(t) = S(t) h2(t)� �Y �(t)��1 u ; (17)are solutions of (6) with h2(�") = 0, h2(0) = u. Similarly, the functionsh3(t) = Zc(t)w ;v3(t) = S(t) h3(t)� �Y �(t)��1 w ; (18)are solutions of (6) with h3(0) = w, h3(c) = 0.Since the ow preserves the symplectic form, we have that the following expres-sions do not depend on t for all u, w 2 T�(�)M :h h2(t); v3(t) i � h v2(t); h3(t) ihZ�" u ; (SZc � (Y �)�1)w i � h (SZ�" � (Y �)�1) u ; Zcw iZ�cSZ�" � Y �1Z�" � Z�cSZ�" + Z�c (Y �)�1 = Z�c (Y �)�1 � Y �1Z�" :Using formula (15), this is equal to(Y �1Zc)� � Y �1Zc �Nc = �N�c ; (19)where the right hand side corresponds to its value when t = �". Since Zc(c) = 0,evaluating the equation at t = c, we get thatNc 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.

HAMILTONIANS WITHOUT CONJUGATE POINTS 13Proof: It is enough to show it for N�1c . We have thatddt � Zc(t)�1 Y (t)���t=0 = Z�1c _ZcZ�1c Y � Z�1c _Y ��t=0 = �Hpp :Therefore �Zc(t)�1Y (t) is positive de�nite for small t < 0. By claim 1.1, thismatrix is symmetric for all t < 0. Since for t < 0 its determinant never vanishes, itis positive de�nite for all t < 0, in particular for t = �", N�1c is positive de�nite.De�ne a partial order on the symmetric isomorphisms of T�(�)M by writingA � B if Hpp(�)�1 (A � B) is positive de�nite. From (13) we have that thefamily F = f _Zc(0)� _Z1(0) j c > 1 g is monotone increasing. Moreover, by (16) andclaim 1.2, we have that_Zc(0)� _Z1(0)�HppNc + � _Zc(0)� _Z1(0)�� _Z�"(0)� _Z1(0) :So that the family F is bounded above. It can be seen that this implies the existenceof a lower upper bound which is the limit of the family. We obtain the �rst part of1.3. Claim.(a) limc!+1 _Zc(0)� _Z1(0) = Q exists and is symmetric.(b) limc!+1Zc(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= 0is a matrix solution of (6) such that D(0) = I, _D(0) = Q+ _Z1(0) and detD(t) 6= 0for all t 2 R.Part (b) is a consequence of (14) and the continuous dependence of the solutionsof equation (6) on the initial data.We have that the subspaceE � (0) := � �u;v(0) u� �� u 2 T�(�)M � H(�)� V (�)is the limit of the subspacesEc(�) := d �c�V ( c(�))� = � �h3(�); v3(�)� �� u 2 T�(�)M where h3(t) and v3(t) are given by (18). MoreoverE � (t) := d t�E � (0)� = � �h(t) u;v(t) u� ��u 2 T�(�)M

14 G. CONTRERAS AND R. ITURRIAGAsatis�es E � (t) \ V ( t�) = f0g andE � (t) = limc!+1 d t�c�V ( c(�))� = limd!+1 d �d�V ( d( t�))� = E t �(0) :Since the vertical subspace is Lagrangian, then K�(c) is Lagrangian. By the conti-nuity of the symplectic form, the subspaces E � (t) are Lagrangian.To obtain the \unstable" Green bundle F(�), observe that the ow of the Hamil-tonian H(�) = �H(�) is the ow t of H with the time reversed. In this caseHpp 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 thefamily G = f _Z�c(0) � _Z�1(0) j c > 1 g is monotone decreasing and bounded belowby _Z"(0)� _Z�1(0). This �nishes the proof of the �rst part of proposition A, we deferthe proof of the second statement to corollary 1.12.Let S(t) : = v(t) h(t)�1= S(t) + Y �(t)�1D(t)�1: (20)Then S(t) is a symmetric solution of the Ricatti equation (7), which is de�ned onall t 2 R. The symmetry of S(t) can be checked either because E (t) = graph (S(t))is a Lagrangian subspace or becauseY (t)�1D(t) = limc!1Y (t)�1 Zc(t)is symmetric and this implies that (Y �)�1D�1 is symmetric.Let S(�), U(�) be the the symmetric solutions of the Ricatti equation (7) cor-responding 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�(�)Mby 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) limd!+1Kd = S, limd!�1Kd = U.(d) S4 U:

HAMILTONIANS WITHOUT CONJUGATE POINTS 15Proof: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 thatVc(t) = S(t)Zc(t)� (Y �(t))�1: (21)Hence, using (14) and (13) we have thatVd(t)� Vc(t) = S(t) [Zd(t)� Zc(t)] (22)= S(t)Y (t) Z dc Y �1Hpp (Y �)�1 ds: (23)When t! 0 we have that S(t)Y (t)! I. ThereforeVd(0)� Vc(0) = Z dc �Y �1Hpp (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 positivede�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 havethat 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 1.3above. Item (b) is proven similarly using the concave Hamiltonian H = �H anditem (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 conjugatepoints, the arguments above apply to obtain limit solutions limc!a� Zc(t) = D�(t),D�(0) = I, detD(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 1.171.6. Remark. The existence of the Green bundles, the solutions h, v, U, S andthe monotonicity properties in proposition 1.4 do not need the uniform boundednessof the operators Hpp, Hpq, Hqq. But in the unbounded case we can not guaranteeneither that X(�) 2 E (�) \ F(�) nor E (�) [ F(�) � T��.

16 G. CONTRERAS AND R. ITURRIAGA1.7. 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 > 0such that kS�(t)k < A for all � 2 �; t > 1:(b) Any measurable symmetric solution of the Ricatti equation de�ned on the energylevel � is uniformly bounded.An inmediate consequence is the following1.8. 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 uniformlybounded 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 + dR) = �w (t + dR).(iii) limt!+1w(t) = R, limt!�1w(t) = �R.(iv) limt!( dR)+ w(t) = +1, limt!( dR)�w(t) = �1.Proof of proposition 1.7:We �rst prove (a). By the spectral theorem, it is enough to prove that x� S�(t) xis uniformly bounded for kxk = 1 and t > 1. Since H�pq = Hqp and S� = S, we havethat _S + S�Hpp S +H�pq S + S�Hpq +Hqq = 0:Let C := (Hpp)�1Hpq, 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 theenergy level. Hence it is enough to prove that x�V�(t) x is uniformly bounded ont > 1 and � 2 �. Observe that V�(t) is not necessarily symmetric. Let M > 0 besuch that y�Hpp(�) y � 1M kyk2 for all � 2 �: (25)Let R > 0 be such thatj y�D(�) y j < M R2 for all � 2 �; kyk = 1: (26)

HAMILTONIANS WITHOUT CONJUGATE POINTS 17We claim thatj y�V�(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 thatthere exists to 2 R such thatx�V (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 >d0R > 0. Write w(t) := R coth (R t� d0). Then w(t) is a solution of _w+w2�R2 = 0for t > d0R . In particular, M _w +M w2 �M R2 = 0:Let f(t) := x� V (t) x�M w(t). Then f(t0) = 0 andf 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 thatM w(t0)2 =M �2 = 1M (M �)2 = 1M hx; V (t0) xi2� 1M 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 appliedeach time that f(t) = 0. Therefore x� V (t) x �M w(t) for all t > t0 andx� V (t) x �M w(t) for all d0R < t < t0:Then limt!( d0R )+ x� V (t) x � limt!( d0R )+M w(t) = +1 :Since doR > 0, this contradicts the existence of V (t) for t > 0. Hence such t0 doesnot 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 thatx� 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 suchthat M w1(t1) = v(t1). Observe that w1(t) is de�ned for t < c0R and in this casec0 > 0 and c0R > t1 > 0. The same argument as above shows that v(t) � M w1(t)for t1 � t < c0R . Since limt!( c0R )� w1(t) = �1, this contradicts the fact that V (t) isde�ned for all t 2 R.

18 G. CONTRERAS AND R. ITURRIAGAThe prove of (b) is similar to that of (a). In this case we can change theMR coth(R) bound by MR. We allow t0 2 R and d0 2 R to be nonpositive ifnecessary. The second inequality has also the same proof, but here we allow t1 2 Rand 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 isgiven 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 1t Y �1(s)Hpp(s) (Y �(s))�1 ds :From (12), using that D(t) = limc!1Zc(t), we have thatD(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 thatS(t)� S(t) = Y �(t)�1M(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 existt0 > 0 and k > 0 such that kS(t)k < k and kS(t)k < k for all t > t0. Then forjxj = 1 and t > t0,��M(t)�1 Y (t)�1 x ; Y (t)�1 x ��� = jhS(t) x ; x ij+ jhS(t) x ; x ij � 2 k :Let �(t) be the largest eigenvalue of M(t), t > 0. Since M(t) is positive de�nitethen �(t) > 0 and kM(t)k = �(t). Moreover,2 k � ��M(t)�1 Y (t)�1 x ; Y (t)�1 x ��� � 1�(t) ��Y (t)�1 x ��2 ;��Y (t)�1 x �� � p2 k kM(t)k 12 for all jxj = 1; t > t0 :Then if jvj = 1, we have thatjY (t) vj � 1kY (t)�1k � 1p2 k kM(t)k 12 for t > t0 :Since M(t) ! 0 when t ! +1, given R > 0 there exists T > t0 such that(2 k kM(t)k) 12 < 1R 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 .

HAMILTONIANS WITHOUT CONJUGATE POINTS 19For � 2 TM , de�neB(�) := n� 2 T�T �M ��� supt2R jd t(�) � �j < +1o: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!+1ET (�) = E (�) andE (�) 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. Itremains to show that E (�) [ F(�) � T��. Observe that since �dH = !(X; �), thenT�� = 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 conju-gate 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 = Es(�)�Ec(�)� Eu(�).

20 G. CONTRERAS AND R. ITURRIAGA1.14. Proposition. For all � 2 �(a) Es(�) � E (�) � Es(�)� Ec(�).(b) Eu(�) � F(�) � Eu(�)� Ec(�).(c) limT!+1 1T Z T0 tr �Hpq +Hpp U�� t(�)� dt.Proof:We only proof Eu(�) � F(�) � Eu(�)� Ec(�) ; (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 havethat d t(�) v = (Y�(t) v; �). Then, by proposition 1.10, we have that V (�)\Es(�) =f0g. For every � 2 � there exists a subspace R(�) � Eu(�) � Ec(�) and a linearmap L(�) : R(�)! Es(�) such that V (�) = graphL(�).Let � be an ergodic invariant measure for t. Given " > 0 let K � � be a compactsubset 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 tjEs(�) < �t and d �1t jEu(�)�Ec(�) < �tLet V�s(�) := d s�V ( �s(�))�. ThenV�s(�) = d s�graphL( �s(�))�= graph h�(d s � L( �s(�)� � �d sjR( �s(�)��1iFor � 2 K take a sequence sn ! +1 such that �sn(�) 2 K. Since d sn�R( �sn(�))� �Eu(�)� Ec(�), by properties (a) and (b), we have that d sn � L( �sn(�)) � �d �snjd snR( �sn (�))� � �sn �sn sup�2K kL(�)k :Since �sn�sn ! 0 when n! +1, we have thatF(�) = limn!+1V�sn(�) = limn!+1d sn�R( �sn(�))� � Eu(�)� Ec(�) :Therefore the second inclusion in (30) is satis�ed for � 2 K. Since �(K) � 1 � "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 continuousriemannian metric h ; i on T �M and a continuous family of linear isomorphismsJ(�) : 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 thismetric. Let fe1(�); : : : ; en(�)g be an orthonormal basis of F(�) and de�ne en+i(�) =

HAMILTONIANS WITHOUT CONJUGATE POINTS 21Jei(�) for i = 1; : : : ; n. Since the subspace F(�) is Lagrangian, the subspace JF(�) =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(�)gare J = � I�I � and d t(�) = �At(�) Ct(�)Bt(�)�Since d t preserves the symplectic form we have that (d t)� J (d t) = J . HenceB�t (�) = At(�)�1 and B�t (�)Ct(�) is symmetric.Since F(�) � Eu(�)� Ec(�) thenlimt!�1 1t log B�t �1(�)w = limt!�1 1t log kAt(�)wk � 0for all w 2 Rn , � 2 �. This implies thatlimt!�1 1t log B�1t (�)w � 0 : (31)Suppose that Eu(�) 6� F(�). Take v 2 Eu(�) n F(�). Then v = w + z withz 2 F(�) and 0 6= w 2 JF(�). We have that d �t v = u�t + ��; B�1t (�)w� withu�t 2 F( �t(�)) � Eu( �t(�)) � Ec( �t(�)). Since B�1t (�)w 2 JF( �t(�)) and Fand JF are orthogonal, we have that kd �t(v)k � B�1t (�)w . Hence from (31) weobtain limt!+1 1t log kd �t(�) vk � 0 :This contradicts the choice v 2 Eu(�).Finaly we prove (c). Let �� : F(�) ! H(�) be the restriction of d�. Then��1� (v) = (v; U(�) v) and by corollary 1.8sup�2� ��1� < +1 : (32)Fix � 2 �, let Z�(t) : V (�)! H( t(�)) be de�ned by Z�(t) v = � t(�) d t(�) (v;U(�)v).Then d t(�)jF(�) = ��1 t(�) � Z�(t) � ��By (30) and (32), we have that�(�) = limT!+1 1T log ��det d t(�)jF(�)��= limT!+1 1T log jdetZ�(t)j :

22 G. CONTRERAS AND R. ITURRIAGASince the isomorphisms H(t) = Z�(t), V (t) = U( t(�))Z�(t) satisfy the Jacobiequation (6), we have that _Z� = (Hpq +HppU)Z� :Since detZ�(t) 6= 0 for all t 2 R, then detZ�(t) has constant sign andddt jdetZ�(t)j = ddt detZ�(t) = tr (Hpq +Hpp U) jdetZ�(t)j ;ddt log jdetZ�(t)j = tr [Hpq +HppU] ;limT!+1 1T log ��det d t(�)jF(�)�� = limT!+1 1T Z T0 tr [Hpq +HppU] dt :The exponential map.The following proposition proves item (c) in theorem 0.1. We present a modernproof 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 conjugatepoints if and only if there exists a Lagrangian subspace E � T�T �M such thatd 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 thereexists a Lagrangian subspace E � T�T �M such that d t(E) \ V ( t(�)) = f0gfor 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)) fort 2]0; a[. Since E(t) is Lagrangian, then S(t) is symmetric and it satis�es the Ricattiequation (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) = d�jE: (33)

HAMILTONIANS WITHOUT CONJUGATE POINTS 23Let 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 thath(t) = Z(t)Z(c)�1 h(c) + Z(t) Z tc Z�1(s)Hpp (Z�(s))�1�Z(c)�v(c)� ds : (34)Since h(c) = 0, thenZ(t)�1 h(t) ; Z(c)�v(c)� = Z tc Hpp (Z(s)�)�1 Z(c)�v(c) ; (Z(s)�)�1 Z(c)�v(c)� ds :(35)In particular h(t) 6= 0 for all t 2]0; a[nfcg. Hence the segment ]0; a[ does not haveconjugate points.It remains to prove the case c = 0. Let 0 < b < a. We prove that � is not conjugateto 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 besuch that b < b + " < a and let Eb+" = �"(V ( b+"(�))) = graph (Kb+"). Byproposition 1.4(b) and remark 1.5, the family Kd is monotone decreasing on d andhas the (lower) bound Kb+". Hence there exists E := limd!0+ Ed = d b(V (�)) andE \ 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 conjugatepoints. The same argument using F (t) = d a+��t(V ( a+�(�))) shows that thereexists �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 thesegment [� �2 ; a+ �[. Conversely, if there is a Lagrangian subspace E � T�T �M withd 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 sup-pose that E(�) � T�� by taking (E(�) \ T��) � hX(�)i. Then it must satisfy thetransversality condition 0.1(a).1.16. Proposition. For � 2 � de�neW (�) := (V (�)\T��)�hX(�)i. Suppose thatthe orbit of � has no conjugate points. Then for all t 6= 0, d t�W (�)�\ V � t(�)� =f0g.

24 G. CONTRERAS AND R. ITURRIAGAProof: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 noconjugate points, then k0 6= 0. Write Hp(t) := Hp( t(�)), Hq(t) := Hq( t(�)). SinceW (�) = (V (�) \ T��)� hX(�)i and X(�) = (Hp(0);�Hq(0)), then k0 = �Hp(0) forsome � 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��, then0 = 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 thath(t) = Z(t) Z t0 Z�1(s)Hpp(s) (Z�(s))�1v(0) ds :Hence, since v(0) 6= 0, we have thathZ(b)�1h(b); v(0)i = Z b0 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 havethat Z(b)�1h(b) = Hp(0). Replacing this equation on (38), we obtain thathHp(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 convexand there is a Lagrangian subspace E � T�� such that d t(E)\V ( t(�)) = f0g fort 2 [0; a], then d t�W (�)� \ V � t(�)� = f0g for all t 2 [0; a].

HAMILTONIANS WITHOUT CONJUGATE POINTS 25In particular this hypothesis holds if f t(�) j t � 0 g or f t(�) j t � 0 g have noconjugate 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 impliesthat 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 supposethat the orbits f t(�) j t � 0 g have no conjugate points for all � 2 ��1fqg\�. Thenthe exponential map expq : T �qM !M is an inmersion.Proof:We have that expq(t�) = �� t(�) for � 2 �q = ��1fqg\�. Consider the splittingTt�(T �qM) � T �qM = T���q�� h�i = �V (�) \ T���� h�i :The derivative d expq(t�) : T �qM ! T�( t �)M is given byd expq(t�)jV (�)\T�� = d� � d t(�)jV (�)\T�� ; (39)d expq(t�)(�) = d��X( t(�))� = d� � d t�X(�)� : (40)Therefore d expq(t�)�T �qM� = d� � d t�V (�) \ T��� hX(�)i�= d� � d t�W (�)� : (41)Since dimW (�) = n, then dimd 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 byN(�) := f � 2 T�� j h d� �; d� X(�) i�(�) = 0 g : (42)ThenN(�)�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(2.a) @@q1 ���(�( t�0);t) = d�X( t(�0)).(2.b) @@q2 ; : : : ; @@qn is an orthonormal basis for d�N( t(�0)) = h @@q1 i? along �( t(�0); t).

26 G. CONTRERAS AND R. ITURRIAGAWrite pi = dqijTqM , i = 1; 2; : : : ; n. From (2.a) we have that1 = ddt 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 toa 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 thereparametrization of the ow t on W such that it preserves the foliation q1 =T = constant. In particular d�T (�0) preserves the transversal bundle N(�T (�0))along the orbit of �0. De�ne �(T; �) by �T (�) = �(T;�)(�). Then �(T; �0) = T forall 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 satisfythe equations dQdT = @K@P ; dPdT = �@K@Q ;where Q = (q2; : : : ; qn), P = (p2; : : : ; pn), T = q1 and K(Q;P ;T ) is de�ned byH(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 ddT �T (�) =�Y (�T (�)); ddT �(T; �)� be the vector�eld of �T . Then Y (�T (�)) is a multiple of thevector�eld X(�T (�)) of t, and then !�Y (�T (�)); � � = 0 on TW . We have� = p dq = P dQ�K dT;!jT� = d�jT� = nXi=2 [dPi ^ dQi �KQi dQi ^ dT �KPi dPi ^ dT ] :The matrix of !jT� is given by24 0 �I �K�QI 0 �K�PKQ KP 0 35 g Qg Pg T=q1Then the vector �KP ;�KQ; 1 = dq1dT � =: 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.

HAMILTONIANS WITHOUT CONJUGATE POINTS 27Since for # 2 W , �T (#) = �(T;#)(#), thend�T (#) � � = d �(T;#) � � + � @�@# ����(T;#) � �� ddt 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 thevector�eld, i.e.� � = � + �(�) X( t(�0)) with �(�) 2 R such that � � 2N( t(�0)) :Through the proof of lemma 2.2 all the quantities will be understood in localcoordinates T = q1, Q = (q2; : : : ; qn), P = (p2; : : : ; pn) of W \ �. In particularT�T (�0)� = R � Rn�1 � Rn�1 for all T 2 R, with coordinates � @@T ; @@Q ; @@P �. Sincealong 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 � Rn�1 � Rn�1 incoordinates. De�ne P : T#� = R � Rn�1 � Rn�1 ! N = f0g � Rn�1 � Rn�1 as theprojection 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 matrizesH (T ) = � Hpq HpP�HQq �HQP�(2n�1)�(2n�1)K (T ) = 24 01�n 01�(n�1)KPq KPP�KQq �KQP 35(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 � 2N(�T ), K (�T ) � = �(�T ) H (�T ) �.(c) The operator K (�T )jN(�T ) is uniformly bounded on T 2 R.

28 G. CONTRERAS AND R. ITURRIAGAProof:(b) Write �T = �T (�0) = T (�0) = (T; 0; 0). From lemma 2.1, the Jacobiequation for �T is written as ddT [P d�T �] = K [P d�T �] ;where d�T � = (h1(T ); H(T ); V (T )) 2 R � Rn�1 � Rn�1 . for any � 2 T�0� =R � Rn�1 � Rn�1 . Also, from (6),ddt [d t �] = H [d t �] :Since d�T (�0) preserves tha foliation q1 = constant, then it preserves the subspaceN . Therefore P d�T � = d�T � = � d�T � for � 2 N :In particular, the �rst coordinate of d�T � is h1(T ) � 0 and also ddT h1 � 0. Hence if� 2 N , thenddT [P d�T �] = ddT [d�T �]= ddT [ d T � + �(T )X(�T ) ]= H (d T �) + _�(T )X(�T ) + �(T ) ddT X(�T )= H �d�T � � �(T )X(�T )�+ _�(T )X(�T ) + �(T ) ddT X(�T ) ;where �(T ) = �(T; �) is from (43).Since ddTX(�T ) = H X(�T ), thenddT [P d�T �] = H [d�T �] + _�(T )X(�T ) :Hence K d�T � = H [d�T �] + _�(T )X(�T ) for � 2 N : (44)Observe that this equation must hold in all the 2n � 1 coordinates of T#�. SinceImage(K ) � N , then � K = K . SoK (d�T �) = � K (d�T �) = � [ H d�T � + _�(T; �)X(�T ) ]= � H (d�T �) for � 2 N :Since d�T :N(�0) = N !N(�T ) = N is surjective, this proves item (b).(a) Let V 2 Rn�1 , thenV � K PP V = �(0; V �); 0n�1�K �0nV � :

HAMILTONIANS WITHOUT CONJUGATE POINTS 29Since � = (0n; V ) 2 N , we can use equation (44) to obtainV �Kpp V = �(0; V �); 0n�1� �H �0nV �+ _�(T; d��T �) � H�p�H�Q��= 0 + V �HPP V + _� �(0; V �) �H�p�= V �HPP V + _� h(1; 0n�1); (0; V �)i= 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, itis enough to see that the angle �(X(�T );N(�T )) is uniformly bounded away from0. Given a vector � = (h; v) = nXi=1 �hi @@qi + vi @@pi� 2 N(�T ) ;written in our coordinates (2.a), (2.b), de�ne the norm j�j := (Pni=1 h2i + v2i ) 12 .Suppose that j�j = 1. Since � 2 N(�T ), then Hp � h = 0. Moreover, Hp = @@q1 andhence jX(�T ) � �jjX(�T )j = jHp � h�Hq � vj(jHpj2 + jHqj2) 12 = jHqj jvjq1 + jHqj2� 1 � max0�x�A xp1 + x2 = Ap1 + A2 < 1 ;where A := supT2R jHq(�T )j < +1 with the norm in our coordinates @@pi .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 d�sjN(�T ):E>(�T ) = lims!+1d��s (V (�T+s) \N(�T+s))F>(�T ) = lims!�1d��s (V (�T+s) \N(�T+s))Moreover, E>(�) = E (�) \N(�) and F>(�) = F(�) \N(�) for all � 2 �.(c) Horizontal growth of iterates of vertical vectors: For all R > 0 there existsS = S(R; �T ) > 0 such that for all jsj > S(R; �T ) and all � 2 N(�T ) \ V (�T )we have that jd�(d�s(�T ) � �)j > R j�j.

30 G. CONTRERAS AND R. ITURRIAGA(d) De�ne B>(�T ) = � � 2N(�T ) j sups2R jd� � d s � �j < +1 :Then B>(�T ) � E>(�T ) \ F>(�T ).(e) Let s := d�sjN = � � d sjN , thend�sjN(�T ) = �(�T+s) � ds � �(�T ) = s ;s+t = s �t :Proof:Let U(�T ) := V (�T ) \N(�T ) = V (�T ) \ T�T�. From (43) we have thatd�T (�0) � � = d T (�0) � �� + (�(T; �0) � �)X(�0)� :Hence d T (U(�0)) � d�T (W (�0)), where W (�) := �V (�) \ T��� � hX(�)i. Fromproposition 1.16,U(�T ) \ d�T (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 sameproofs as propositions A, 1.10 and proposition 1.11 respectively.Since d�s and d s satisfy (43) and d�s preserves the bundle N(�T ), we have thatd�sjN(�T ) = �(�T+s) � d sjN(�T ) = �(�T+s) � d s � �(�T ) = s :Moreover, s+t = d�s+tjN(�T ) = d�sjN(�T+t) � d�tjN(�T ) = s �t :This proves item (e).Finally, we have E>(�0) = limT!+1 �d��T�V (�T ) \N(�T )��= �(�0) limT!+1 �d �T �V (�T ) \ T�T���� �(�0) � E (�0) � E (�0) :Because X(�0) 2 E (�0). Thus E>(�0) � E (�0)\N(�0). Since dim E>(�0) = dimU =n � 1 = dim(E (�0) \N (�0)), we have that E>(�0) = E (�0) \N(�0). The equationF>(�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, thenby proposition 1.14, E (�0) � Es(�0)�hX(�0)i and F(�0) � Eu(�0)�hX(�0)i. HenceE (�0) \ F(�0) = hX(�0)i.

HAMILTONIANS WITHOUT CONJUGATE POINTS 31Now observe that if we construct the coordinates (q1; : : : ; qn) in (2.a), (2.b) sothat they are periodic with period � , then d�� (�0) : N(�0) - is the derivative ofthe Poincar�e 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 havethat B>(�0) � E>(�0) \ F>(�0) = hX(�0)i \N(�0) = f0g. This implies that � ishyperbolic. 3. Quasi-hyperbolic actions.Let B be a compact metric space and � : E ! B a vector bundle provided with acontinuous 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 ifsupt2R jt(�)j = +1 for all � 2 E; � 6= 0:We say that is hyperbolic if there exists an invariant continuous splitting E(p) =Eu(p)� Es(p), p 2 B and C > 0, � > 0 such that(3.a) t�Eu(p)� = Eu� t(p)�, t�Es(p)� = Es� t(p)�, for all x 2 B, t 2 R.(3.b) jt(�)j � C e��t j�j for all t > 0, � 2 Eu(p), p 2 B.(3.c) j�t(�)j � C e��t j�j for all t > 0, � 2 Eu(p), p 2 B.The aim of this section is to prove the following0.2. 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 adi�erentiable ow on M , t = d t and E is a d t-invariant continuous bundle suchthat E � ddt t� = TM , we get3.1. Theorem. [Freire] [12]IfM is a closed manifold and is a quasi-Anosov ow onM such that ( ) =M ,then is Anosov.

32 G. CONTRERAS AND R. ITURRIAGAThe proof of theorem 0.2 is similar to that of theorem 3.1. We include it here forcompleteness.Suppose that the action of is quasi-hyperbolic. For p 2 B de�neEs(p) := n v 2 E(p) ��� supt>0 jt(p)(v)j < +1o ;Eu(p) := n v 2 E(p) ��� supt<0 jt(p)(v)j < +1o :Observe that by the quasi-hyperbolicity we have thatEs(p) \ Eu(p) = f0g for all p 2 B:3.2. Lemma. There exists � > 0 such that for all p 2 B, � jEs(p) < 12 ; �� jEu(p) < 12for all p 2M .Proof:Suppose that the lemma is false for Es. The proof for Eu is similar. Then thereexist sequences xn 2 B, vn 2 Es(xn), jvnj = 1 such that jn(vn)j � 12 for all n 2 N .We claim that there exists C such thatsupt�0 t(p)jEs(p) < C < +1 :for all p 2 B.Suppose the claim is true. Let wn := n(vn) and yn := n(xn). We have that12 � jwnj � C for all n andjt(wn)j = jt+n(vn)j � C for all t � �n :Since B is compact there exists a convergent subsequence of (yn; wn). If yn !n y andwn !n w, we have that y 2 B, w 2 E(y), jwj � 12 andjt(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 Es(xn), jvnj = 1, such thatsupn jtn(vn)j = +1 : (45)Let sn > 0 be such thatjsn(vn)j > 12 sups�0 js(vn)j � 12 jtn(vn)j :By (45) we have that sn !n +1. Let yn := sn(xn) andwn := sn(vn)jsn(vn)j :

HAMILTONIANS WITHOUT CONJUGATE POINTS 33Then jwnj = 1 and if t > �sn we havejt(wn)j = jt+sn(vn)jjsn(vn)j � 2 jt+sn(vn)jsups�0 js(vn)j � 2 :Since jwnj = 1 and yn 2M , there exists a convergent subsequence (yn; wn)! (y; w).We would have that y 2 B, w 2 E(y), jwj = 1 andjt(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)jt(v)j � K � jvj+ js(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 jtn(vn)j � n� jvnj+ jsn(vn)j � :Then tn ! +1 and sn � tn ! +1 when n ! +1. We can assume thatjtn(vn)j = sup0�t�sn jt(vn)j. Let wn := tn(vn)jtn(vn)j �For �tn < t < sn � tn, we have thatjt(wn)j = jt+tn(vn)jjtn(vn)j � 1jtn(vn)j sup0�t�sn jt(vn)j = 1:Taking a subsequence, if necessary, we can assume that xn ! x and wn ! w 2 E(x).Then jt(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 thestrong stable and unstable bundles is redundant in the de�nition of a hyperbolicaction. So it remains to prove that for all x 2 B, we have that Es(x)�Eu(x) = E(x).Given x 2 B, since x 2 ( jB), there exist xn ! x, sn ! +1 such that snxn ! x.Let En be a subspace of E(xn) such thatdimEn = dimE(x)� dimEs(x) ;Es(x)� limn En = E(x) :

34 G. CONTRERAS AND R. ITURRIAGAWe claim that there exists C > 0 such that �t��sn(En) � C for all 0 � t � sn:Suppose the claim is true. Taking a subsequence if necessary, we can assume thatthe limit limnsn(En) exists. Then �t�� limn sn(En) � C for all t > 0:Then limnsn(En) � Eu(x) and dimEu(x) + dimEs(x) � dimEn + dimEs(x) =dimE(x). Since Es(x) \ Eu(x) = f0g, this completes the proof.Now we proof the claim. Suppose it is false, then there exist vn 2 sn(En),jvnj = 1 and 0 < tn < sn such that j�tn(vn)j � n. By lemma 3.3, we have thatn � j�tn(vn)j � K �j�sn(vn)j+ jvnj� :Hence j�sn(vn)j � n�KK . From lemma 3.3 we also have thatjt(�sn(vn))jj�sn(vn)j � K + Kj�sn(vn)j for 0 < t < sn :Let wn := �sn(vn)j�sn(vn)j . The estimates above give thatjt(wn)j � K + K2n�K for 0 < t < sn :If wn ! w, then jwj = 1 and w 2 limEn, thus w =2 Es(x). But jt(w)j � K for allt � 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 �(�) � � 2N(�) :Let s(�) = �( s(�)) � d sjN(�). By corollary 2.3 (e), s de�nes an R-action on thevector bundle � :N ! �.Suppose that the Green bundles satisfy E\F = hXi on �. Then by corollary 2.3(b)and 2.3(d), B>(�) � E>(�) \ F>(�) = E (�) \ F(�) \N(�) = f0g :

HAMILTONIANS WITHOUT CONJUGATE POINTS 35Therefore the action s is quasi-hyperbolic. Since s preserves the Liouville mea-sure, which is positive on open sets, then ( j�) = �. Hence s is a hyperbolicaction.This implies that �t is Anosov. By the methods in Hirsch-Pugh-Shub [18], we getthat t is Anosov. We sketch the proof below.For � 2 � let Eu(�) �N(�) be the subspace given by item (3.b) in the de�nitionof hyperbolic action. Then Eu is a continuous subbundle of T�. De�neF := �L : Eu ! R continuous �� L(�) : Eu(�)! R is linear 8� 2 �;with kLk = sup�2Eunf0g jL(�)jj�j .To each functional in F associate a subbundle WL of T� byWL(�) := Graph(L(�)) = � � + (L(�) � �)X(�) �� � 2 Eu(�):Let � > 0 be such that C e��� < e�� < 1, where C and � are from (3.b). Considerthe following \graph transformation" T : F ! F , corresponding toWTL = d � (WL)and de�ned byd ��� + L(�)X(�)� = � (�) + �T (L) (� (�))�X( � (�)); � 2 �; � 2 Eu(�):We claim that T is a contraction. Indeed����T (L1 � L2) (��)k� �k ���� kX( ��)k � k d � [(L1(�)� L2(�))X(�)]ke� k� k� j(L1 � L2)(�)je� k�k kX( ��)k :Hence kT (L1)� T (L2)k � e�� kL1 � L2k.The �xed point L� of T gives the (continuous) strong unstable bundle of t.Indeed, if Eu = WL� , then Eu is clearly d t-invariant. Moreover, if � 2 Eu(�),then � = � + (L��)X(�) with � = �� 2 Eu(�). Since L� is continuous then thereexists Q1 > 0 such that j�j � Q j�j = Q j��j for all � 2 Eu. Since N(�) istransversal 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 suchthat 1Q2 j��j = 1Q2 j�j < j�j. Thenjd t(�)j = jt(�) + (T (L�) � �)X( t�) j� 1Q2 jt(�)j � C�1 e� tQ2 j�j � e�tC Q1Q2 j�j :Finally, dimEu = dimGraph(L�) = dimEu = n� 1 :The existence of a continuous strong stable subbundle of dimension n� 1 is provensimilarly.

36 G. CONTRERAS AND R. ITURRIAGA4. The Index Form.Let L = pHp � 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 transformFL(x; v) = (x; Lv) identi�es q = x, p = Lv. Similarly, using F�1L = FH, we get thatv = Hq under the same identi�cation. AlsoH = v Lv � L = pHp � L(q;Hp) :Hence Hq = �Lq = �Lx. With these identi�cations the Hamiltonian equations (1)become ddt x = v ! ddt q = Hp ;ddt Lv = Lx ! ddt p = �Hq ; (46)Hence, identifying Lx = Hq 2 T(p;q)(T �M), the Euler-Lagrange equation (46) isunderstood as a �rst order di�erential equation on T �M , where ddtLv is the tangentvector to the path t 7! Lv( (t); _ (t)) in T �M .We derive the Jacobi equation in this Lagrangian setting. Let (t) be a solutionof 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 thecovariant derivative Dds , we obtain the Jacobi equationDdt (Lvx k + Lvv k0) = Lxx k + Lxv k0 ; (47)where k = @f@s (0; t), _k = Ddt @f@s and the derivatives of L are evaluated on (t) =f(0; t). Here we have used that Dds @F@t = Ddt @F@s for the variation map F (s; t) =Lv�f(s; t); @f@s (s; t)� 2 T �M , where Dds and Ddt are the covariant derivatives on theriemannian manifold T �M . The linear operators Lxx, Lxv, Lvv coincide with thecorresponding 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 C2 vector�elds � along [0;T ]. De�nethe index form on byI(�; �) = Z T0 � _� Lvv _� + _� Lvx � + � Lxv _� + � Lxx �� dt ; (48)which is the second variation of the action functional for variations f(s; t) with@f@s 2 T . For general results on this form see Duistermaat [9].

HAMILTONIANS WITHOUT CONJUGATE POINTS 37The following transformation of the index form is taken from Hartman [17] andoriginally 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 Lagrangiansubspace such that d t(E) \ V ( t(�)) = f0g for all 0 � t � T . Such subspace Ealways exists by proposition 1.15(b). Let E(t) := d t(E) and let H(t); V (t) be amatrix solution of the Hamiltonian Jacobi equation (6) such that detH(t) 6= 0 andE(t) = Image (H(t); V (t)) � T t(�)(T �M) is a Lagrangian subspace. In particularH(t) satis�es the Lagrangian Jacobi equation (47). We have that�H(t)V (t)� = DFL ('t(�)) � �H(t)H 0(t)� = � I 0Lvx Lvv� �HH 0� (49)From (49) and (47), we have thatV = Lvv H 0 + LvxHV 0 = LxvH 0 + LxxH (50)Moreover, since E(t) is a Lagrangian subspace, then the Hamiltonian solution of theRicatti 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 (LvvH � 0 + V �) (H 0 � +H � 0) + (V 0 � + LxvH � 0) H� :Since (LvvH� 0) � (H 0�) = (H� 0) � (Lvv H 0�) and (LxvH� 0) � (H�) = (H� 0) � (LvxH�),the integrand can be written asLvvH� 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 LvvH� 0 �H� 0 + (V � �H�)0. Since � and � are continuous, we have thatI(�; �) = Z T0 �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 FLand FH are inverse maps, in particular(DFH)�1 = � I 0Hqp Hpp��1 = � I 0Lvx Lvv� = DFL ;and hence Lvv = (Hpp)�1. We obtain thatI(�; �) = Z T0 (H� 0)� (Hpp)�1 (H�0) dt+ (H�)�(V �)��T0 : (53)

38 G. CONTRERAS AND R. ITURRIAGAfor � = H� 2 T , � = H� 2 T . This formula can also be written asI(�; �) = Z T0 (H� 0)� (Hpp)�1 (H�0) 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 conjugatepoints then the index form is positive de�nite on�T = � � : [0; T ]! TM �� �(t) 2 T� t�M; � is piecewise C2; �(0) = 0; �(T ) = 0 :Proof:Let � 2 �T , � 6= 0. Write �(t) = H(t)�(t). Since detH(t) 6= 0, �(0) = 0, �(T ) = 0and �(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; : : : ; tNg be the pointsin [0; T ] such that detH(ti) = 0. ThenI(�; �) = Z T0 (h� 0)� (Hpp)�1 (H�0) dt+ (H�)�(V �)��T0 � NXi=1 (H�)�(V �)��t+it�i : (55)where � = H�, � = H�. But now �, � are piecewise C2 but may be discontinuousat t1; : : : ; tN . Alternativelly, we can use a sum of formulas (53) or (55) using di�er-ent Lagrangian subspaces d t(Ei) and corresponding solutions (Hi; Vi) on disjointsubintervals ]ti; ti+1[� [0; T ].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, letY�(t) v = d� d t(�) (0; v), where (0; v) 2 V (�). Then from the Jacobi equation (6)we have that Ddt Y�(t)����t=0 = Hpp(�):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 actionfunctional Z T0 L(�(t); �0(t)) dtover all absolutely continuous curves �(t) 2 M , 0 � t � T , such that �(0) = (0)and �(T ) = (t).

HAMILTONIANS WITHOUT CONJUGATE POINTS 394.2. 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 thatI(�; �)jT0 = 0.Let " > 0 be such that (cf. prop. 1.15) the segment f t(�) jT�" � t � T +"g hasno 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) ifT � " � 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�)��T+"T�" > I(�; �)��T+"T�":Extend � and b� by �(t) = b�(t) = �(t) for 0 � t � T � ". ThenI(�; �)��T+"0 < I(b�; b�)��T+"0 = 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 ifA(s) = Z T+"0 L �f(s; t); @f@s (s; t)� dtis the action of f(s; t), then A0(0) = 0 because f(0; t) = � t(�) satis�es the Euler-Lagrange equation and 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~n�e [23]. Given a (peri-odic) Lagrangian L : TM � S1 ! R and ! 2 H1(M;R), let M!(L) be the set ofminimizing measures of L + !, where ! is any 1-form in the class !.5.1. Theorem. (Ma~n�e [23])(a) For every ! 2 H1(M;R) there exists a residual subset O(!) � C1(M � S1)such that 2 O(!) implies #M!(L+ ) = 1.(b) There exist residual subsets O � C1(M � S1) and H � H1(M;R) such that 2 O and ! 2 H imply #M!(L+ ) = 1.

40 G. CONTRERAS AND R. ITURRIAGANow take an autonomous Lagrangian L and ! = 0 2 H1(M;R). Then theitem (a) implies the �rst part of theorem D. Let O be the residual subset given bythis theorem. Let A be the subset of O of potentials for which the measure onM(L + ) is supported on a periodic orbit. Let B := O n A and let A1 be thesubset of A on which the minimizing periodic orbit is hyperbolic. We prove that A1is relatively open on A. For, let 2 A1 andM(L+ ) = f� gwhere � is the invariant probability measure supported on the hyperbolic periodicorbit for the ow of L + . We claim that if �n 2 A, �n ! andM(L+ �k) =f��kg, then �n ! . Indeed, since L is superlinear, the velocities in the supportof the minimizing measures �k := ��k are bounded (cf. [26], [22]), and hence thereexists a subsequence �k ! � converging weakly* to a some invariant measure � forL+ . Then if � 6= � ,limk SL+�k(�k) = SL+ (�) > SL+ (� ) : (56)Thus if �k is the analytic continuation of the hyperbolic periodic orbit to the owof 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 toc(L+ ) and potentials � near to , there exist hyperbolic periodic orbits �;h whichare the continuation of . Now, on a small neighbourhood of a hyperbolic orbit thereexists a unique invariant measure supported on it, and it is in fact supported in theperiodic orbit. Thus, since �n ! , then �n is hyperbolic. Hence �n 2 A1 and A1contains a neighbourhood of � in A.Let U be an open subset of C1(M;R) such that A1 = U \ A. We shall provebelow that A1 is dense in A. This implies that A1 [ B is generic. For, let V :=int (C1(M;R)nU), then U[V is open and dense in C1(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 anda singularity follow the same spirit. However to prove it in the case of a singularpoint is much easier because the Jacobi equation (the linear part of the ow) isautonomous. We suggest to read it �rst.We have to prove that A1 is dense in A, i.e. given that a Lagrangian L hasa unique minimizing measure supported on a periodic orbit , then there exists a

HAMILTONIANS WITHOUT CONJUGATE POINTS 41perturbation by an arbitrarily small C1-potential �, such that the new LagrangianL+ � has a unique minimizing measure supported on a hyperbolic periodic orbit.Fix � 2 A and let � be the periodic orbit inM(L+�), we can assume that � = 0.By the graph property (cf. [26], [22]), the projection �j� : � ! M , �(x; v) = xis injective. In particular �(�) � M is a simple closed curve. Choose coordinatesx = (x1; : : : ; xn) : U ! S1 � Rn�1 on a tubular neighbourhood of �(�) such thatx(�) = S1 � f0g and n @@x1 ; @@x2 ; : : : ; @@xn o is an orthonormal frame over the pointsof �(�). In particular @@x1 is paralel to �(�). De�ne eL = L+� with �(x1; : : : ; xn) =12 f(x) " (x22 + x23 + � � �+ x2n), where f(x) is a C1 nonnegative bump function withsupport in U which is 1 on a small neighbourhood of �(�). Clearly � can bemade C1 arbitrarily small. On �(�) we have that eLvv = Lvv, eLvx = Lvx andeLxx = Lxx + " [ 0 00 I ]. Observe that since eL � L thenM(L+ �) = f��g.Let H and eH = H � � be the Hamiltonians associated to L and eL respectively.Let � be the corresponding periodic orbit for H and eH on T �M . We have to provethat the periodic orbit � is hyperbolic for the ow of eH. Since it is minimizing, bycorollary 4.2 the orbit � has no conjugate points. By proposition B it is enough toprove that the Green bundles eE , eF satisfy eE (�) \ eF (�) = h eX(�)i on a point � 2 �,where eX is the Hamiltonian vector�led of eH.Fix � 2 � and let eZT (t) = d��de t��d�jd e �T (V ( e T (�)))��1. Let ( eZT ; eVT ) and eKT (t) =eVT (t) ( eZT (t))�1, 0 � t < T , be the corresponding solutions of the Jacobi equation(6) and the Ricatti equation (7) respectively: graph( eKT (t)) = de t(de �T (V ( e T �))),eZT (T ) = 0, eZT (0) = I.De�ne N(�) := fw 2 T�(�)M j hw; _ i = 0 g. Then N(�) is the subspace ofT�(�)M generated by the vectors @@x2 ,: : : , @@xn . Let vo 2 N(�), jvoj = 1 and let�T (t) := eZT (t) vo. Denote by eIT and IT be the index forms on [0; T ] for eL and Lrespectively. Using the solution ( eZT ; eVT ) on formula (53), we obtain thateIT (�T ; �T ) = �( eZT (0) vo)� (eVT (0) vo) = �vo� eKT (0) vo : (57)Moreover, in the coordinates (x1; : : : ; xn; @@x1 ; : : : ; @@xn ) on TU we have thateIT (�T ; �T ) = Z T0 � _�T eLvv _�T + 2 _�T eLxv �T + �T eLxx �T� dt= Z T0 � _�T Lvv _�T + 2 _�T Lxv �T + �T Lxx �T� dt+ Z T0 " nXi=2 ���Ti ��2 dt :(58)

42 G. CONTRERAS AND R. ITURRIAGAWe have that eZT (0) = I and for all t > 0, limt!1 eZT (t) = eh(t) with eh(t) givenon claim 1.3 for eH. Writing �N(�) = (�2; �3; : : : �n) then ����N eh(0) v0��� = jv0j = 1because v0 2 N(�). Hence there exists � > 0 and T0 > 0 such that ���N�T (t)�� =����N eZT (t) v0��� > 12 for all 0 � t � � and T > T0. ThereforeeIT (�T ; �T ) � IT (�T ; �T ) + "�4 : (59)Let (h(t); v(t)) = d t � �d�jE(�)��1 be the solution of the Jacobi equation forH given by claim 1.3 and let S( t(�)) = v(t)h(t)�1 be the corresponding solutionof the Ricatti equation, with graph[S( t(�)] = E ( t(�)). Using formula (53), andwriting �T (t) = h(t) �(t), we have thatIT (�T ; �T ) = Z T0 (h _�)�H�1pp (h _�) dt+ 0� �h(0) �(0)���v(0) �(0)�IT (�T ; �T ) � �v�o S(�) vo : (60)From (57), (59) and (60), we get thatv�o S(�) vo � v�o eKT vo + "�4 :From proposition 1.4, we have that limT!+1 eKT (0) = eS(�), where graph(eS(�)) =eE (�), the stable Green bundle for eH. Thereforev�o S(�) vo � v�0 eS(�) vo + "�4 : (61)Similarly, for the unstable Green bundles we obtain thatv�o U(�) vo + �2 � v�o eU(�) vo for vo 2N(�); 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 thateUjN � UjN < SjN � eSjN . Since eE (�) = graph(eS(�)) and eF (�) = graph(eU(�)), weget that eE (�) \ eF (�) � h eX(�)i. Then proposition B shows that � is a hyperbolicperiodic 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 intersectionsW 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� ng, M(L + ) = f��g, we have seen thatthe n's are continuations of the hyperbolic orbit � for nearby ows (on nearbyenergy levels): n = ��n;hn. Since compact subsets (fundamental domains) of theinvariant manifolds W s(��;h) depend continuously in the C1 topology on (�; h),

HAMILTONIANS WITHOUT CONJUGATE POINTS 43then the transversality property W s(��;h) t W u(��;h) persists on a neighbourhoodof ( ; h0). In particular it holds for n = ��n;hn for (�n; hn) su�ciently close to( ; h0).By the same arguments as above, it is enough to prove that A2 is dense in A. Forit we need the following lemma,5.2. Lemma. Let � be a hyperbolic periodic orbit without conjugate points of aconvex Hamiltonian. Then for all � 2 W s(�) there exists S = S(�) > 0 such thatthe 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(�) thereexists S = S(�) > 0 such that E(t) := T t(�)W s(�) satis�es E(t) \ V � t(�)� = f0g.Moreover, dimE(t) = n and for all u 2 E(t), lims!+1 d s(u) 2 hX(�)i. Since tpreserves 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 allt > 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 perturbationeL = (L+ �)+� explained above so that �+� 2 A1. We can assume that �+� = 0and 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 � hasno autoacumulation points. Let S = S(�) > 0 be from lemma 5.2 and � > 0small. Then there exists � > S(�) and a neighbourhood W � M of �( t(�)) suchthat f t 2 R j � t(�) 2 W g =]�; � + 3�[ and W \ �(�) = �. Let U � M � Rbe a tubular neighbourhood of f (� t(�); t) j t > S(�) g. Choose coordinates y =(y1; : : : ; yn; t) : U ! Rn � R such that y�� t(�); t� = ��(t); 0; t� 2 R � Rn�1 � R,and n @@y1 ; @@y2 ; � � � ; @@yno is an orthonormal frame over the points � t(�), t > S(�).Let '(y1; : : : ; yn) = 12 f(y) � (y22+y23 � � �+y2n), where f(y) is a C1 bump function withsupport 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 C1 arbitrarily small. Choose� small enough such that the orbit � remains hyperbolic. Then we still have thatM(L+�) = f��g and the orbit of � is the same for both ows. Write eL = L+�+�,bL = eL + and eIT , bIT the corresponding index froms on f t(�) j t 2 [�; T ] g. Sincef t(�) j t > S(�) g has no conjugate points, the stable Green subspace bE ( t(�)),t > S(�) and the stable solution of the Ricatti equation bS( t(�)) exist. The same

44 G. CONTRERAS AND R. ITURRIAGAarguments as in equations (58){(61) give for �T = bZT (t) vo, vo 2N( � (�)), jvoj = 1,bZT (t) = d� � db t�T � �d�jd �T (V ( b T+� (�)))��1, thatbIT (�T ; �T ) = eIT (�T ; �T ) + Z �+2��+� � nXi=2 ���Ti ��2 dt ;v�0 S� � (�)� vo � vo bS� � (�)� vo + � b�4 ;for some b� > 0. In particularbE � � (�)� \ eE � � (�)� = X( ��)� : (63)By the same arguments as in proposition 1.14, we have that bE � � (�)� == T � (�)�W ueL( ��)�. Observe that T � (�)�W ueL( ��)� = T � (�)�W ueL( ��)� didn't change.Write T � (�)�W ueL( ��)� = V � W , with V � eE ( ��) and W \ eE ( ��) = f0g. Bydiminishing � if necessary we can get (63) and bE ( ��) \ W = f0g. ThusT b � (�)W sbL(�) \ T b � (�)W ubL(�) = bE ( ��) \ T � (�)W u(�) = bX( ��)� :Finally, by a �nite number of these perturbations on a fundamental domain ofW s(�) one can remove all tangencies of W s(�) and W u(�) and obtain a potentialin A2 arbitrarily C1-near � 2 A.The case of a singularity.Suppose that the minimizing measure � is supported on a singularity (x0; 0) of theLagrangian ow. From the Euler-Lagrange equation (2) we get thatLx(x0; 0) = 0. Di�erentiating the energy function (3) we see that (x0; 0) is a singu-larity of the energy level c(L). Moreover, the minimizing property of � implies thatx0 is a minimum of the function x 7! Lxx(x; 0). In particular, Lxx(x0; 0) is positivesemide�nite in linear coordinates in Tx0M .Choose coordinates y = fy1; : : : ; yng on a neighborhood of x0 such that y(x0) = 0and Lvv(x0; 0) is the identity on the basis � @@y1 ��x0; � � � ; @@yn ��x0 . Use the coordinates� fy1; : : : ; yng ; � @@y1 ; � � � ; @@yn � 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 Jacobiequation (47) at (x0; 0) becomes �k = Lxx(x0; 0) k ;where Lvv(x0; 0) = I. Let f be a nonnegative C1 function on Rn with support onjyj < " < 1 and f � 1 on a neighbourhood of 0. Let �(y) = 12 � f(y) (y21 + � � �+ y2n).Adjusting � > 0 we can make � arbitrarily C1 small. Let eL := L + �. Then the

HAMILTONIANS WITHOUT CONJUGATE POINTS 45atomic measure � supported on (x0; 0) is still the unique minimizing measure for eLand the Jacobi equation for the orbit e't(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 positivede�nite. Equation (64) is linear�k0k00� = �0 IA 0� �kk0� (65)with constant coe�cients where k � 0 is a hyperbolic singularity.To obtain the transversality property W s t W u at homoclinic orbits we canuse the same arguments as above if we have a replacemnet for lemma 5.2. Theargumenmts of lemma 5.2 apply if we show that the stable subspace of (x0; 0) istransversal 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 eigenvectorssatisfy y = �x, Ax = �2x, (� 2 R). Hence if x = 0, then (x; y) = (0; 0) is not aneigenvector. 6. The Exponential map.The following lemma was proven for geodesic ows by Freire and Ma~n�e [13] andhas intrinsic interest (compare with proposition 1.10).6.1. Lemma. Let H be a convex Hamiltonian. Choose a riemannian metric k kon M . If t has no conjugate points on � = H�1feg, then there exists B > 0 suchthat kY�(t) � vk > B kvkfor 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 cande�ne the vector�eld Y = 1j�(X)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 ofthe contact vector�eld Y . The transversal bundle N(�) = ker �jT�� is invariantunder the ow of Y. Let Eq : Tq ~M ! ~M be the exponential map for (Y;�). Thederivative dEq(t�) is given in formulas (39) and (40). The norm of the directionalderivative in formula (40) is clearly bounded below. By lemma 6.1 the norm of the

46 G. CONTRERAS AND R. ITURRIAGAderivative dEqjV (�)\T�� in formula (39) is uniformly bounded away from zero. SinceV (�) \ T�� � N(�), and the angle ��N(�); X(�)� is bounded below, we obtainthat kdEqk is bounded away from zero. This implies that Eq : Tq ~M ! ~M is acovering map and hence a di�eomorphism. In particular, the universal cover ~M ofM is homeomorphic to Rn . By corollary 1.18, the exponential map expq for X is alocal di�eomorphism. Once we know that Eq is a di�eomorphism, it is easy to showthat expq is bijective and hence a di�eomorphism.Proof of Lema 6.1:Since Y�(t) satis�es (8) (with Hqp and Hpp uniformly bounded on �) and bycorolary 1.8, S�(t) is uniformly bounded for jtj > 1, then there exists D1 > 1 suchthat _Y�(t) � w � D1 kY�(t) � wk for all jtj > 1; w 2 V (�); � 2 �:Let A := (1 +D1)2 sup�2� Hpp(�)�1 :Let H�(t) := Y �2(�)(t+2)��Y �2(�)(2)��1, then detH�(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. LetD2 := max�2�t�2�2� H�(t)� (Hpp)�1H�(t) < +1:Let � : R ! [0; 1] be a C1 function such that �(t) = 0 for jtj � 1 and �(0) = 1.Let C := D2 Z 1�1 j�0(t)j2 dt :Let 0 < B < (AC) 12 : (66)Suppose that kY�(T ) vok < B kvok for some � 2 �, vo 2 T�(�)M , kvok = 1and jT j > 1. Assume that T > 1, the case T < �1 is proven similarly. LetJ(t) 2 H( t(�)) = T� t(�)M be de�ned by J(t) = 0 for �1 � t � 0, J(t) = Y�(t) vofor 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(�). ThenJ(t) is continuous and piecewise C2.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

HAMILTONIANS WITHOUT CONJUGATE POINTS 47and write v(t) = Pni=1 vi(t) ei, Yi(t) = Y (t) ei. Then the covariant derivative along� t(�) of J(t) = Y (t)v(t) isDdt (Y v) = Ddt nXi=1 vi(t)Yi(t)! =Xi �DYidt � vi +Xi �dvidt �Yi ;Y v0 = Ddt (Y v)� DYdt v :For T � t � T + 1, since J(t) = Y�(t) v(t) = (T + 1� t) �t J(T ), we have thatkY v0k � Ddt (Y v) + DYdt v � DdtJ(t) +D1 kJ(t)k� kJ(T )k+D1 kJ(T )k� (1 +D1)B :Using Y�(t) on formula (55) we have thatI(J; J) = Z T+1T (Y v0)� (Hpp)�1 (Y v0) dt� (1 +D1)2B2 H�1pp = AB2:Let Z(t) 2 H( t(�)) = T� t(�)M be de�ned by Z(t) = �(t)H�(t) vo. ThenZ(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 isI(Z; J) = �Z(0+) � vo + Z(0�) � 0 = �kvok2 = �1:Write Z(t) = H(t) �(t), �(t) = �(t) vo. Using H(t) and V (t) on fromula (53), wehave that I(Z;Z) = Z 1�1(H� 0)� (Hpp)�1 (H� 0) dt= Z 1�1(�0(t))2 (H(t) vo)� (H�1pp ) (H(t) vo) dt� D2 Z 1�1 j�0(t)j2 dt = C:Since by (48) the index form is symmetric, for � 2 R we have thatI(Z � � J ; Z � � J) � C + 2�+ �2AB2:Since by (66) 4 � 4AB2C > 0, this polynomial is � has two real roots. ThereforeI(Z � �J ; Z � �J) < 0 for some value of �. Hence there must be conjugate pointsin the orbit segment f t(�) j t 2 [�1; T + 1] g.

48 G. CONTRERAS AND R. ITURRIAGA6.2. Proposition. Let e be a regular value of H. If �(X) > 0 on the energylevel � = H�1feg and the ow of X has no conjugate points, then there exists aHamiltonian H on T �M with vector�eld Z such that(i) H is convex and H �1feg = �.(ii) Z = 1�(X) �X on �.(iii) The ow �t of Z has no conjugate points on �.(iv) �t preserves the 1-form � on T�.(v) If N(�) := ddss�js=1, then for � 2 � we haved�t(�) �N(�) = N(�t(�)) + tX(�t(�)) :6.3. Remark. This proposition holds if �(X) 6= 0 on a connected compact fowardinvariant set K � �. In this case we obtain an open connected neighbourhood U ofK such that inf#2U j�(X(#))j > 0 and a convex Hamiltonian H without conjugatepoints, de�ned on a neighbourhood of U in T �M . To apply this on theorem F, wecan take K = clousure of R+�(�j�)�1fqg� � �.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 = 1�(X) �X.Then Y preserves the 1-form � on T� because the Lie derivativeLY� = d iY �+ iY d� = d[�(Y )] + !(Y; �)= d(1)� 1�(X) dH = 0 on T�:Let �t(q; p) be the ow of Y on �. De�ne the simplecti�cation e�t of �t (cf. Arnold[3], [2]) on the cotangent bundle without its zero section T �M nM by e�t(q; �p) =��t(q; p), (q; p) 2 �. Observe that e�t preserves the canonical 1-form �. Indeed, wehave that � � �t = � � e�t and if �t(q; p) = (qt; pt),�(qt; � pt) � (de�t) = (� pt) � d� (de�tw)= � [pt � (d� d�tw)]= � [p � (d� w)] = �(q; p) � w :Let eY be the vector�eld of e�t. We have thatLeY� = d[�(eY )] + !(eY ; �) = 0 :Therefore the function eH = �(eY ) is a Hamiltonian for eY . But eY has conjugatepoints on � and eH is not convex.

HAMILTONIANS WITHOUT CONJUGATE POINTS 49Let H = 12 eH2 and let Z be the Hamiltonian vector�eld of H . We have thatZ = eH � eY , since eHj� � 1, Z is also an extension of Y . Let �t be the ow of Z.Observe that�t(q; sp) = s e�t eH(q;sp)(q; p) = s e�t 12 s2(q; p) ; for (q; p) 2 � :Taking the derivative dds ��s=1, we obtaind�t(�) �N(�) = N(�t(�)) + tX(�t(�)) for � 2 �:Write �(t; �) = (s(t; �); �) with s(t; �) > 0, � 2 �. ThenD �t = D s +X( s(�)) @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 Lagrangianbundle 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 6.26.4. Lemma. The Hamiltonian H is convex, i.e. H pp is positive de�nite.Proof:Let q 2 � and S(q) := T �qM \�. Since H is homogeneous of degree 2, then H pp ishomogeneous of degree 0 and it is enough to prove that H pp(q; p) is positive de�nitefor p 2 S(q). Let p0 2 S(q). Let h : Tp0S(q)�R ! T �qM be h(x; �) := x+� p0. LetG = H � h and K = �(eY � h). Then G = 12K2. Since D2h = 0, then D2H (q0) ispositive de�nite if D2G(0; 1) is positive de�nite. Since K(0; �) = �, we have thatG��(0; 1) = �K2� +K �K��� ��(0;1) = 1 :Since in the coordinates (x; �) the vectors @@xi ���(0;1), i = 1; : : : ; n� 1 are tangent atp0 to S(q) and �(eY ) � 1 on S(q), then Kx(0; 1) = 0. HenceGxx(0; 1)(v; v) = jKx � vj2 +KKxx(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 thatGx�(0; 1) � (v; �) = � [K� (Kx � v) +K (Kx� � v)] = 0 :And then D2G(0; 1)(v; �)(2) = �2G�� + 2�(G�x � v) +Gxx(v; v)= �2 +Kxx(0; 1) � (v; v) :

50 G. CONTRERAS AND R. ITURRIAGASo it is enough to see that Kxx(0; 1) is positive de�nite.Let N be the unit normal vector to Tp0S(q) such that p0 �N < 0. Such N existsbecause HjS(q) � e and p0 � Hp = �(X) 6= 0. Let F : Tp0S(q) � R ! R be givenby F (x; y) = H(p0 + x + yN). Since @F@y ���(0;0) < 0, there exists a local C2 functiony : U � Tp0S(q) ! R such that F (x; y(x)) = e. Di�erentiating this equation withrespect to x, we obtain that dy = �FxFyd2y = � 1Fy D2F � (I; dy)(2):Since the vectors @@xi ���(0;0) are tangent to the level F � e at (q; p0), we have thatFx(0; 1) = 0. Hence dy(0; 1) = 0 andd2y(0) � v = � 1Fy D2F � (v; 0)(2) = � 1Fy Hxx � (v; v) > 0 :Now let v 2 Tp0S(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 thatr0(s) = p0 � v + sr(s) hence 12 r(s)2 = 12 jp0j2 + (p0 � v) s+ 12 s2 :Also R0(s) = (p0 + sv + z(s)N) � (v + z0N)R(s) = p0 � v + s+ z0 (p0 �N) + z z0R(s)therefore, since � s2 � z(s) � �s2 for some 0 < � < � and jsj small.12R(s)2 = 12 jp0j2 + (p0 � v)s+ 12 s2 + z (p0 �N) + 12 z2= 12 r(s)2 + z(s) (p0 �N) + 12 z(s)2� 12 r(s)2 � 12 s2 ;for some > 0 and jsj small. Then for some � > 0, we have thatr(s)R(s) � 1 + � s2Kxx � (v; v) = d2ds2 K(sv; 1)��s=0 = d2ds2 � r(s)R(s)�����s=0 � � > 0 :

HAMILTONIANS WITHOUT CONJUGATE POINTS 51Proof of Theorem F.Trough the proof of theorem F we work with the lifted Hamiltonian to the cotan-gent bundle of the universal cover ~M of M . Let �t be the ow of the HamiltonianH of proposition 6.2. Let q 2 ~M and let Eq : Tq ~M ! ~M be the exponentialmap associated to the ow �t and the energy level �. We �rst prove that Eq isa di�eomorphism. The derivative dEq(t�) is non-singular by corollary 1.18 and isgiven by formulas (39) and (40). We claim that the norm kdEq(t�)k is uniformlybounded below. By lemma 6.1 the norm dEq(t�)jV (�)\Tq� is uniformly boundedbelow for t > �. From equation (40) dEq(t�)jh�i is also uniformly bounded below.It remains to see that the angle ��dEq(t�)(V (�) \ Tq�) ; d�(X(�t �))� is uniformlybounded below.Since �t preserves the 1-form �, the bundle K(�) = ker �jT�� is invariant underd�t. Since V (�) \ Tq� � K(�), thendEq(t�) �V (�) \ Tq�� � d� d�t(�) �K(�)� � d��K(�t(�))� : (68)For � = (q; p) 2 � write Hp(q; p) = � p+ k, with hk; pi = 0. Since �(X) = p �Hp 6= 0on � there exists 0 � � < 1 such that j kj � � jHpj on �. Let h 2 d��K(�t �)� withjhj = 1, i.e. hp; hi = 0, jhj = 1. Thenhh;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�)kis uniformly bounded below on t 2 R+ , � 2 �. Hence the map Eq : Tq ~M ! ~M is adi�erentiable covering map. Since ~M is simply connected, it must be a di�eomor-phism.Now let expq : Tq ~M ! ~M be the exponential map of ( t;�). Since �t is areparametrization of t for t > 0, � 2 � there exists �(t; �) > 0 such that t(�) =��(t;�)(�). By corollary 1.18 expq is a local di�eomorphism. It remains to provethat it is bijective. Since j�(X)j = jXjjY j is bounded then t 7! �(t; �) is a (bijective)homeomorphism of R+ . This implies that the map t � 7! �(t; �) �, with � 2 T �qM\�,is a homeomorphism of T �qM . Since Eq is bijective and expq(t�) = Eq��(t; �) ��, thenexpq is bijective.

52 G. CONTRERAS AND R. ITURRIAGAAPENDIX I.I. Examples.The following formalism for twisted Hamiltonians on surfaces was derived byG. Paternain and M. Paternain in [30]. Let M be a closed orientable riemanniansurface with riemannian metric h ; ix. Let K : TTM ! TM be the connectionmap K� = r _xv, where � = ddt�x(t); v(t)�. Let � : TM ! M be the canonicalprojection. Let !0 be the symplectic form in TM obtained by pulling back thecanonical symplectic form via the Legendre transform associated to the riemannianmetric, 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 anew symplectic form !F on TM by!F = !0 + (F ��) (��) :This is called a twisted symplectic structure on TM (cf. [30]). Let H : TM ! R bethe Hamiltonian H(x; v) = 12hv; 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 thatx(u; v) = hY (u); vix : (70)Then Y (u) = iu is the rotation of angle +�2 . From (69) and (70) we have thatdH�(�) = !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) becomesh�; �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 Ddt _ = F ( ) Y ( _ � : (73)

HAMILTONIANS WITHOUT CONJUGATE POINTS 53If RM F = 0 this ow can be seen as a Lagrangian ow as follows. The conditionRM F = 0 implies that the cohomology class of F is zero. Hence there exists a1-form � on M such that d� = F . Consider the LagrangianL(x; v) = 12hv; vix + �x(v) :The corresponding Euler-Lagrange equation isDdt h _ ; � i = d�x( _ ) = hF Y ( _ ); � i :which gives the same di�erential equation (73). This ow corresponds to a magnetic�eld with Lorentz force F (x)Y (v).Observe that the theory developed in the previous sections also applies to thecase in which RM F 6= 0. Indeed, let q : U � M ! R2 be a local chart de�nedon a simply connected domain U � M . Then the 2-form F is exact on U andthere exists a 1-form �jU such that d� = F on U . The Hamiltonian ow onTU � TM is the Lagrangian ow of L(x; v) = 12hv; 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 bethe 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 verticalbundle of Tq(U)Rn to the vertical bundle of TUM .We derive the Jacobi equation. Let (c(t); _c(t)) be an orbit of the ow. Consider avariation s(t) = f(s; t), where f(0; t) = c(t) and the paths t 7! f(s; t) are solutionsof (73): Ddt _ s = F Y ( _ s) : (74)Denote the variational �eld of f by J(t) = @f@s (0; t). Using the formulaDds Ddt @f@t = Ddt Dds @f@t +R�@f@t ; @f@s� @f@t :and taking the covariant derivative of equation (74) in the direction J(t), we obtain�J +R( _c; J) _c = Dds�F Y ( _ s)� :Since Y (v) is linear on the �bers of TM and _J = Dds @f@t = Ddt @f@s , we haveDds FY ( _ s) = �rJ(F Y )�( _ s) + F Y ( _J) ;= (rJF )Y ( _c) + F Y ( _J) :Hence we obtain the Jacobi equation�J +R( _c; J) _c� (rJF )Y ( _c)� F Y ( _J) = 0 : (75)

54 G. CONTRERAS AND R. ITURRIAGAConsider the orthonormal basis f _c; Y ( _c) = i _c g of Tc(t)M . We have thatDdt _c = F Y ( _c) ;Ddt Y ( _c) = (r _cY )( _c) + Y �Ddt _c�= �F _c :Write J = x _c+ y Y ( _c) :Then _J = ( _x� F y) _c+ ( _y + F x)Y ( _c) ;�J = (�x� 2F _y � _F y � F 2 x) _c+ (�y + 2F _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 energylevel � = H�1f12g = SM . If (J; _J) 2 T _c(t)� � H(�)�V (�), we have that dH(J; _J) =h _c; _Ji = 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 T _c(t)� are given by (78) and (79).Example 1. A convex Hamiltonian without conjugate points and no continuousinvariant transversal bundle.Take K � �1, F � 1. Then equations (78) and (79) become_x = y ;�y = 0 : (80)The solutions have the form x(t) = 12 a t2 + b t+ c ;y(t) = a t+ b ; (81)_y(t) = a :

HAMILTONIANS WITHOUT CONJUGATE POINTS 55The derivative of the Hamiltonian ow t on �c(t); _c(t)� is given byd 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) withx(0) = y(0) = 0 and y(T ) = 0 = x(T ) , T 6= 0 is x(t) = y(t) � 0. The Greenbundles coincide and are generated by X(�) = ( _c;�Y ( _c)) and (Y ( _c); 0). The lastvector can be found by solving x(0) = 0, y(0) = 1, x(T ) = y(T ) = 0 and lettingT ! �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 pointand let tn !n +1 be such that tn(�) !n �. Let � 2 N(�) n f0g. If d t(�) =�x(t); y(t); z(t)� = �J(t); _J(t)�, thenlimn!�1 1jx(tn)j d tn(�) = limn 1x(tn) �x(tn); y(tn); z(tn)�= (1; 0; 1) = X(�) : (83)By the continuity of N we haveX(�) = limn 1x(tn)d t(�) 2N(�) :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!+1 0.In this case the trajectories of the Hamiltonian ow are the unit tangent vectorsof 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 ! ~Mis a di�eomorphism for all q 2 M . This ow has no periodic orbits because thehorocycles W uu � � = T 1 ~M=�1(M) are homeomorphic to R.

56 G. CONTRERAS AND R. ITURRIAGAM~example I.1We prove (84). Let � 2 Tq ~M , k�k = 1. Let �t : T ~M - be the geodesic ow of ~M .Consider a sphere ST (�) = � q 2 ~M j d ~M�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 ofDds _ T (0) by the isometry given by a rotation about the center ��T (�) of ST (�).Therefore T satis�es Dds _ T = �T Y ( _ T )for some constant �T > 0. Passing to the limit when T ! +1 we obtain that thereparamerizations by arc length of the horospheres satisfyDds _ = � Y ( _ ) ; (85)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 conjugatepoints corresponding to the period (length) of these closed circles (see example 3).From the Jacobi equation (79) with F = �, we see that � = H�1f12g has conjugatepoints for � > 1 and the ow is Anosov for � < 1. Hence �0 = 1.Indeed, for � < 1 the solution of (78) and (79) are y = Ae�t + B e��t, x =�(A� e�t � B� e��t + C), with � = p1� �2. Writing � = �x(0); y(0); _x(0); _y(0)� wehave that d t(�) = �x(t); y(t); _x(t); _y(t)�. For A 6= 0 the function t 7! kd t(�)khas 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, theJacobi equations (78), (79) exhibit conjugate points, by the same arguments as forequations (87) and (88).

HAMILTONIANS WITHOUT CONJUGATE POINTS 57Example 2. A periodic orbit without conjugate points and no transversal invariantsubspace.We construct a convex Lagrangian which has the Jacobi equations (80) on aperiodic orbit. Let � 2 � be such a periodic point with period � > 0. Changing thecoordinates (x; y; z) on T�� of equation (82) to (x; y; _y) = (x; y; z � x), the matrixof the derivative d � (�) isd � (�) = 241 t 12 t21 t1 35 g xg yg _y � 241 11 1135 in Jordanform.Let fe1; e2; e3g be the basis of T�� in which d � (�) is written in canonical form. Thetransformations d � and d � � I = 240 10 1035have the same invariant subspaces. The vector�eld corresponds to the uniqueeigenspace he1i and the only invariant subspaces are he1i = hX(�)i, he1; e2i = E (�) =F(�) and he1; e2; e3i = T��. None of them is transversal to X(�) = e1.Now we make the construction. Consider the torus T2 = R2=5Z2 with the atmetric and polar coordinates (r; �) centered at (0; 0) 2 R2=5Z2 de�ned on r � 2.Let F : T2 ! R be a C1 function such that F (r; �) = F (r), F (r = 1) = 1,@F@r ��r=1 = �1, and RT2 F = 0. Let � be a 1-form in T2 such that d� = F . De�neL(x; v) = 12 hv; vix + �x(v). At r = 1, the Euler-Lagrange equation (73) isDdt _ = F ( ) Y ( _ ) = Y ( _ ) at r = 1 :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 R2 ��x2 + y2 � 4 and consider the LagrangianL(x; v) = 12 hv; vi+ �x(v) ;where h ; i is the euclidean metric and d� = �. This Lagrangian corresponds toF jW = �1 in the formalism above and can be embedded into a convex Lagrangianon a closed surface. The Euler-Lagrange equation (73) is_v = �i v : (86)

58 G. CONTRERAS AND R. ITURRIAGAThe solutions of (86) on the energy level jvj = 1, are all the circles contained inW with constant angular velocity _� = �1. Fix C: R(t) = 1, �(t) = � � t in polarcoordinates in W with center (0; 0). Let # = (0; 1)(�1;0) be the initial vector of thissolution. 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 = 12 jvj = 12 . Thesolution C has conjugate points, but the conjugate points appear at t = 2�. Inparticular the segment � t(#) �� 0 � t � 3�2 has no conjugate points. But theintersections with other solutions starting at p = (�1; 0) appear at t < 32� andat 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 provedusing 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)t =

(1,0)

πexample I.3 example I.4Example 4. A non surjective exponential map without conjugate points.Consider the at torus T2 = R2=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 LagrangianL(x; v) = 14 hv; vix + �x(v) + �(x) ;where � is a C1 function on T2 such that �(r; �) = 14r2 + cos �r on r � 2 and�(r; �) � 1 outside r � 2; and �(r; �) = �12 r2 f(r) d� with f(r) a C1 function withsupport on r � 52 such that f(r) = 1 on r � 2 and f(r) � 1.

HAMILTONIANS WITHOUT CONJUGATE POINTS 59Observe that the path C : r(t) � 1, �(t) = t is static (cf. Ma~n�e [25]). Indeed, theLagrangianL(r; �; _r; _�) = 14 ( _r2 + r2 _�2)� 12 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 that14 r2 _�2 + �(v) � r2 �14 _�2 � 12 _�� � �14 r2 � �1Hence L � �(r) � 1 on r � 2.By theorem VII in Ma~n�e [25] (see also [8]), there exists a semistatic (hence fowardminimizing) solution (t) such that (0) = (0; 0) and the !-limit of ( ; _ ) is the(hyperbolic) closed orbit C. By the rotational symmetry of the Lagrangian, forevery 1-dimensional subspace L � T(0;0)T2 we obtain a semistatic curve , whichis tangent to L, with L(0) = (0; 0) and !-limit( L; _ L) = C. Since the semistaticsolutions are minimizing, none of them Lj[0;+1[ have conjugate points. Moreover,by theorem X in Ma~n�e [25] (also [8]), C and all the ( L; _ L)'s are in the same energylevel E = c(L). By the graph property (theorem VII, Ma~n�e [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 itsApplications. Encicl. of Math. Springer, 1985.[4] W. Ballman, M. Brin & K. Burns, On surfaces with no conjugate points. Jour. Di�erentialGeometry, 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.[7] A. Clebsch. �Uber die Reduktion der zweiten Variation auf ihre einfachste Form. J. ReineAngew. Math. 55 (1958) 254-276.[8] G. Contreras, J. Delgado, R. Iturriaga. Lagrangian Flows: The Dynamics of Globally Mini-mizing 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. Di�ferential Geometry, 8, (1973),437-463.[11] J. Franks & C. Robinson, A quasi-Anosov di�eomorphism that is not Anosov, Trans. Amer.Math. Soc., 223, (1976), 267-278.[12] A. Freire. Fluxos geod�esicos em variedades riemannianas compactas de curvatura negativa,M.Sc. Thesis, IMPA, 1981.

60 G. CONTRERAS AND R. ITURRIAGA[13] A. Freire & R. Ma~n�e. On the entropy of the geodesic ow in manifolds without conjugatepoints. Invent. Math. 69, (1982), 375-392.[14] P. Foulon, G�eom�etrie des �equations di�er�entielles du second order, Ann. Inst. Henri Poincar�e,45 (1), (1986), 1-28.[15] P. Foulon. Estimation de l'entropie des syst�emes lagrangiens sans points conjugu�es. Ann. Inst.Henri Poincar�e 57, (1992), 117-146.[16] L. W. Green. A theorem of E. Hopf, Michigan Math. J. 5 (1958) 31-34.[17] P. Hartman. Ordinary Di�erential 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~n�e. On a theorem of Klingenberg, Dynamical Systems and Bifurcation Theory, M. Ca-macho, M. Paci�co and F. Takens eds., Pitman Research Notes in Math. 160, (1987), 319-345.[22] R. Ma~n�e. On the minimizing measures of Lagrangian dynamical systems. Nonlinearity, 5,(1992), no. 3, 623-638.[23] R. Ma~n�e. Generic properties and problems of minimizing measures of lagrangian systems.Nonlinearity, 9, (1996), no. 2, 273-310.[24] R. Ma~n�e. Asymptotic measures of class A solutions of generic Lagrangians. Unpublished.[25] R. Ma~n�e. 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 Publicationsvol. XVIII, 1934, 1960.[28] V.I. Oseledec. A multiplicative ergodic theorem. Trans. Moscow Math. Soc. 19, (1960), 197-231.[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 di�erentiable maps. Bol. Soc. Bras. Mat. 9, (1978),83-87.Depto. de Matem�atica. PUC-Rio, R. Marques de S~ao Vicente, 225, 22453-900 Riode Janeiro, BrasilE-mail address : gonzalo@mat.puc-rio.brCIMAT, A.P. 402, 3600, Guanajuato. Gto., M�exico.E-mail address : renato@fractal.cimat.mx