barreira_pspm

Proceedings of Symposia in Pure Mathematics

Lectures on Lyapunov Exponents and

Smooth Ergodic Theory

L. Barreira and Ya. Pesin

Contents

Introduction

1. Lyapunov Exponents for Differential Equations

2. Abstract Theory of Lyapunov Exponents

3. Regularity of Lyapunov Exponents Associated with Differential Equations

4. Lyapunov Stability Theory

5. The Oseledets Decomposition

6. Dynamical Systems with Nonzero Lyapunov Exponents. Multiplicative Ergodic

Theorem

7. Nonuniform Hyperbolicity. Regular Sets

8. Examples of Nonuniformly Hyperbolic Systems

9. Existence of Local Stable Manifolds

10. Basic Properties of Local Stable and Unstable Manifolds

11. Absolute Continuity. Holonomy Map

12. Absolute Continuity and Smooth Invariant Measures

13. Ergodicity of Nonuniformly Hyperbolic Systems Preserving Smooth Measures

14. Local Ergodicity

15. The Entropy Formula

16. Ergodic Properties of Geodesic Flows on Compact Surfaces of Nonpositive Curvature

Appendix A. Holder Continuity of Invariant Distributions, by M. Brin

Appendix B. An Example of a Smooth Hyperbolic Measure with Countably Many Ergodic

Components, by D. Dolgopyat, H. Hu and Ya. Pesin

2000 Mathematics Subject Classification. Primary: 37D25, 37C40.Key words and phrases. Lyapunov exponents, nonuniformly hyperbolic dynamical

systems, smooth ergodic theory.L. Barreira was partially supported by FCTs Funding Program and the NATO grant

CRG 970161. Ya. Pesin was partially supported by the National Science Foundation grant#DMS-9704564 and the NATO grant CRG 970161.

c2000 American Mathematical Society

1

2 L. BARREIRA AND YA. PESIN

Introduction

This manuscript is based on lectures given by Ya. Pesin at the AMSSummer Research Institute (Seattle, Washington, 1999). It presents the coreof the nonuniform hyperbolicity theory of smooth dynamical systems. Thistheory was originated in [26, 27, 28, 29] and has since become a mathemat-ical foundation for the paradigm which is widely-known as deterministicchaos the appearance of irregular chaotic motions in pure determinis-tic dynamical systems. We follow the original approach by Ya. Pesin makingsome improvements and necessary modifications.

The nonuniform hyperbolicity theory is based on the theory of Lyapunovexponents which was originated in the works of Lyapunov [19] and Perron[25] and was developed further in [7]. We provide an extended excursioninto this theory. This includes the abstract theory of Lyapunov exponents that allows one to introduce and study the crucial concept of Lyapunov-Perron regularity (see Section 2) as well as the advanced stability theoryof differential equations (see Sections 1, 3, and 4).

Using the language of the theory of Lyapunov exponents one can viewnonuniformly hyperbolic dynamical systems as those where the set of pointswhose Lyapunov exponents are all nonzero is large, for example, has fullmeasure with respect to an invariant Borel measure (see Sections 6 and 7).In this case the fundamental Multiplicative Ergodic theorem of Oseledets[24] implies that almost every point is LyapunovPerron regular. Thus, thepowerful theory of Lyapunov exponents applies and allows one to carry outa thorough analysis of the local stability of trajectories.

The crucial difference between the classical uniform hyperbolicity andits weakened version of nonuniform hyperbolicity is that the hyperbolicityconditions can get worse when one moves along the trajectory of a nonuni-formly hyperbolic point. However, if this point is LyapunovPerron regularthen the worsening occurs with subexponential rate and the contraction andexpansion along stable and unstable directions prevail.

One of the crucial manifestations of this fact is the fundamental StableManifold theorem that was established in [27] and is a generalization of theclassical HadamardPerron theorem. In Section 9 we present the proof ofthe Stable Manifold theorem following the original approach in [27] whichis essentially an elaboration of the Perron method. In Section 10 we sketchthe proof of a slightly more general version of the Stable Manifold theorem(known as the Graph Transform Property) which is due to Hadamard. Wealso describe several main properties of local stable manifolds of which oneof the most important is that their sizes may decrease along trajectories onlywith subexponential rate (and thus the contraction prevails).

There are several methods for establishing nonuniform hyperbolicity.One of them, which we consider in these lectures, is to show that the Lya-punov exponents of the system are nonzero. One of the first examples was

LECTURES ON LYAPUNOV EXPONENTS AND SMOOTH ERGODIC THEORY 3

constructed in [26]. It is a three dimensional flow on a compact smooth Rie-mannian manifold which is a reconstruction of an Anosov flow. It revealssome mechanisms for the appearance of zero Lyapunov exponents (see thediscussion in Section 8).

Another example are geodesic flows on compact smooth Riemannianmanifolds of nonpositive curvature. Let us stress that geodesic flows havealways been a good source of examples and have provided inspiration fordeveloping the hyperbolicity theory. In these lectures we consider only ge-odesic flows on surfaces of nonpositive curvature and show that they arenonuniformly hyperbolic on an open and dense set (see Section 8).

One of the main goals of the nonuniform hyperbolicity theory is to de-scribe the ergodic properties of a smooth dynamical system preserving asmooth invariant measure. This is done in Sections 11, 12, 13, and 15 wherewe show that such a system has ergodic components of positive measure andalso establish the entropy formula that expresses the entropy of the systemvia its positive Lyapunov exponents. Finally, in Section 16 we apply theseresults to the geodesic flows on compact surfaces of nonpositive curvature.

Acknowledgment. We want to express our deep gratitude to M. Brin,D. Dolgopyat, C. Pugh, and J. Schmeling for their valuable comments anddiscussions.

The lectures were used as a handout for a graduate course on DynamicalSystems with Nonzero Lyapunov Exponents that Ya. Pesin taught during theFall 2000 semester at The Pennsylvania State University. Ya. Pesin wouldlike to thank students for their patience during the course and numerousfruitful remarks which helped improve the text. The authors are speciallygrateful to students C. Carter, T. Fisher, R. Gunesch, I. Ugarcovici, andA. Windsor for many valuable comments and corrections.

1. Lyapunov Exponents for Differential Equations

Consider a linear differential equation

v = A(t)v, (1.1)

where v(t) Cn and A(t) is a nn matrix with complex entries dependingcontinuously on t R. We assume that the matrix function A(t) is bounded,i.e.,

sup{A(t) : t R}


In order to characterize the stability of the trivial solution in the gen-eral case we introduce the Lyapunov exponent + : Cn R {} ofEquation (1.1) by the formula

+(v) = lim supt+

1

tlogv(t), (1.3)

for each v Cn, where v(t) is the unique solution of (1.1) satisfying the ini-tial condition v(0) = v. It follows immediately from (1.3) that the Lyapunovexponent + satisfies:

1. +(v) = +(v) for each v Cn and 6= 0;2. +(v + w) max{+(v), +(w)} for each v, w Cn;3. +(0) = .

The function + can take on only finitely many distinct values +1 < 0 there exists a constant C > 0 such that for every solution v(t) ofEquation (1.1) and any t 0 we have

v(t) Ce(+s +)tv(0). (1.4)

It follows from (1.4) that if

+s < 0 (1.5)

then for any sufficiently small > 0, every solution v(t) 0 as t +with an exponential rate. In other words the trivial solution v(t) = 0 isasymptotically (and indeed, exponentially) stable.

We now consider a nonlinear differential equation

u = A(t)u+ f(t, u), (1.6)

which is a perturbation of (1.1). We assume that f(t, 0) = 0 and hence,u(t) = 0 is a solution of (1.6). We also assume that there exists a neigh-borhood H of 0 in Cn such that f is continuous on [0,)H and that forevery u1, u2 H and t 0, we have

f(t, u1) f(t, u2) Ku1 u2q (1.7)for some constants K > 0 and q > 1. This means that the perturbationf(t, u) is small in H. The number q is called the order of the perturbation.

One of the main problems in the Lyapunov stability theory is whetherCondition (1.5) implies that the solution u(t) = 0 of the perturbed equa-tion (1.6) is asymptotically (and exponentially) stable.

Perron showed that in general the answer is negative (see [25]).

Example 1.1. Consider the following nonlinear system of differentialequations in R2:

u1 = [ a(sin log t+ cos log t)]u1,u2 = [ + a(sin log t+ cos log t)]u2 + |u1|+1,

(1.8)


for some positive constants , a, and . It is a perturbation of the followinglinear system of differential equations

v1 = [ a(sin log t+ cos log t)]v1,v2 = [ + a(sin log t+ cos log t)]v2. (1.9)

We assume that

a < < (2e + 1)a and 0 < 1.

The general solution of (1.8) is given by

u1(t) = c1etat sin log t,

u2(t) = c2et+at sin log t + c1

2et+at sin log t tt0

e(2+)a sin log d,

(1.11)

while the general solution of (1.9) is given by

v1(t) = d1etat sin log t, v2(t) = d2e

t+at sin log t.

Here c1, c2, d1, d2, and t0 are arbitrary numbers.It is easy to check that the values of the Lyapunov exponent associated

with Equation (1.9) are +1 = +2 = + a < 0.

Let u(t) = (u1(t), u2(t)) be a solution of the nonlinear system of differ-ential equations (1.8). In view of (1.11) it is also a solution of the linearsystem of differential equations

u1 = [ a(sin log t+ cos log t)]u1,u2 = [ + a(sin log t+ cos log t)]u2 + (t)u1, (1.12)

where(t) = sgn c1|c1|etat sin log t.

Note that|(t)| |c1|e(+a)t

and thus by (1.10), Condition (1.2) holds for Equation (1.12). Fix 0 <

tktk

e(2+)a sin log + d.

For every [tk, tk] we have2k

2 log 2k

2,

(2 + )a cos (2 + )a sin log .


This implies that tktk

e(2+)a sin log + d tktk

e(2+)a cos d.

Set r = (2+ )a cos . It follows that if k N is sufficiently large then tkt0

e(2+)a sin log + d >

tktk

er d > certk ,

where c = (1 e)/r. Settk = tke

= e2k+12.

We obtain

eatksin log t

k

tk

t0

e(2+)a sin log + d > eatk

tkt0

e(2+)a sin log + d

> ceatk+rtk = ce(a+re

)tk .

It follows from (1.10) that if c1 6= 0 and is sufficiently small, then theLyapunov exponent of any solution u(t) of (1.8) (which is also a solution of(1.12)) satisfies

+(u) + a+ re = + a+ [(2 + )a cos ]e > 0.Therefore, the solution u(t) is not asymptotically stable. This completes theconstruction of the example.

Lyapunov introduced regularity conditions which guarantee asymptotic(and indeed, exponential) stability of the solution u(t) = 0 of the perturbedequation (1.6). Although there are many different ways to state the regu-larity conditions (which we discuss below), for a given differential equationthe regularity is often difficult to verify.

We now state the Lyapunov Stability Theorem (see [7]). It claims thatunder an additional assumption known as forward regularity, Condition (1.5)indeed implies the stability of the trivial solution u(t) = 0 of (1.6).

Theorem 1.2. Assume that the Lyapunov exponent + of Equation(1.1), with the matrix function A(t) satisfying (1.2), is forward regular andsatisfies Condition (1.5). Then the solution u(t) = 0 of the perturbed equa-tion (1.6) is asymptotically and exponentially stable.

The notion of forward regularity will be introduced and discussed inSections 2 and 3.

2. Abstract Theory of Lyapunov Exponents

Before we proceed with the proof of the Lyapunov Stability Theorem 1.2we discuss the notions of Lyapunov exponent and regularity. It is moreconvenient to do this in a formal axiomatic setting using the basic properties1, 2, and 3 of the Lyapunov exponent + described in Section 1. This allows


one to study Lyapunov exponents in other situations such as sequences ofmatrices, cocycles, dynamical systems with continuous and discrete time,etc.

Let V be an n-dimensional real vector space. A function : V R {} is called a Lyapunov characteristic exponent or simply a Lyapunovexponent on V if:

1. (v) = (v) for each v V and R \ {0};2. (v + w) max{(v), (w)} for each v, w V ;3. (0) = (normalization property).We describe some basic properties of Lyapunov exponents.

Theorem 2.1. If is a Lyapunov exponent, then the following proper-ties hold:

1. if v, w V are such that (v) 6= (w) then(v + w) = max{(v), (w)};

2. if v1, . . ., vm V and 1, . . ., m R \ {0} then(1v1 + + mvm) max{(vi) : 1 i m};

if, in addition, there exists i such that (vi) > (vj) for all j 6= ithen

(1v1 + + mvm) = (vi);3. if for some v1, . . ., vm V \ {0} the numbers (v1), . . ., (vm) are

distinct, then the vectors v1, , vm are linearly independent; if, inaddition, m = n, then the vectors v1, . . ., vn form a basis of V ;

4. the function can take no more than n distinct finite values.

Proof. Suppose that (v) < (w). We have

(v + w) (w) = (v + w v) max{(v + w), (v)}.It follows that if (v + w) < (v) then (w) (v) which contradicts ourassumption. Hence, (v + w) (v), and thus, (v + w) = (w). State-ment 1 follows. Statement 2 is an immediate consequence of Statement 1and Properties 1 and 2 in the definition of Lyapunov exponent.

In order to prove Statement 3 assume on the contrary that the vectorsv1, . . ., vm are linearly dependent, i.e., 1v1 + + mvm = 0 with notall constants i equal to zero, while (v1), . . ., (vm) are distinct. ByStatement 2 and Property 3 in the definition of Lyapunov exponent, weobtain

= (1v1 + + mvm) = max{(vi) : 1 i m and i 6= 0} 6= .This contradiction implies Statement 3. Statement 4 follows from State-ment 3.


By Theorem 2.1, the Lyapunov exponent can take on only finitelymany distinct values on V \ {0}. We denote them by

1 < < sfor some s n. In general, 1 may be . For each 1 i s, define

Vi = {v V : (v) i}. (2.1)Put V0 = {0}. It follows from Theorem 2.1 that Vi is a linear subspace of Vfor each i, and

{0} = V0 $ V1 $ $ Vs = V. (2.2)We call a collection V = {Vi : i = 0, . . ., s} of linear subspaces of V satisfy-ing (2.2) a linear filtration or simply a filtration of V .

The following result gives an equivalent characterization of Lyapunovexponents in terms of filtrations.

Theorem 2.2. A function : V R {} is a Lyapunov exponentif and only if there exist numbers 1 < < s for some 1 s n, and afiltration V = {Vi : i = 0, . . ., s} of V such that:

1. (v) i for every v Vi;2. (v) = i for every v Vi \ Vi1 and 1 i s;3. (0) = .Proof. If is a Lyapunov exponent then the filtration

V = {Vi : i = 0, . . ., s}defined by (2.1) satisfies Conditions 1 and 3 of Theorem 2.2. Moreover, forany v Vi \ Vi1 we have i1 < (v) i. Since takes no value strictlybetween i1 and i, we obtain (v) = i and Condition 2 follows.

Now suppose that a function and a filtration V satisfy the conditionsof the theorem. Observe that v Vi\Vi1 if and only if v Vi\Vi1 for any R \ {0}. Therefore, by Condition 2, (v) = (v). Choose now vectorsv1, v2 V . Let (vj) = ij for j = 1, 2. It follows from Conditions 1 and 2that (2.1) holds. Therefore, vj Vij for j = 1, 2. Without loss of generalitywe may assume that i1 < i2. This implies that v1 + v2 Vi1 Vi2 = Vi2 .Hence, by Condition 1, we have

(v1 + v2) i2 = max{(v1), (v2)},and thus, is a Lyapunov exponent.

We refer to the filtration V = {Vi : i = 0, . . ., s} of V , defined by (2.1),as the filtration of V associated to and denote it by V. We call thenumber

ki = dimVi dimVi1the multiplicity of the value i, and the collection of pairs

Sp = {(i, ki) : 1 i s}the Lyapunov spectrum of .


Given a filtration V = {Vi : i = 0, . . ., s} of V and numbers 1 < (i), then there exists k < i such that i (k). Otherwise, we wouldhave (1), . . ., (i 1) i 1 and hence, (i) i. It follows that

min{i + (i) : 1 i n} k + (k) i + i.The desired result now follows.

We proceed with the proof of the theorem. Consider dual bases v andw. Without loss of generality, we may assume that (v1) (vn).Let be a permutation of {1, . . ., n} such that the numbers (i) = (wi)satisfy 1 n. We have (vi) i and i i. By Lemma 2.7 weobtain

max{(vi) + (wi) : 1 i n} max{(vi) + i : 1 i n} max{i + i : 1 i n}= (, ).

Therefore, (, ) (, ) and Statement 1 is proven.We assume now that . It is not difficult to show that one can

choose bases v and w, which are dual and both normal; furthermore, we


assume that v is ordered. It follows that (vi) = i and i =

i for each i.

Thus,

(, ) max{(vi) + (wi) : 1 i n}

ni=1

((vi) + (wi)) =

ni=1

(i + i)

nmax{i + i : 1 i n} = n(, ).Finally, since we have (, ) 0. This implies that (, ) 0,and Statement 2 follows.

We now introduce the crucial concept of regularity of a pair of Lyapunovexponents and in dual vector spaces V and W . Roughly speaking reg-ularity means that the filtrations V and V are well-adapted to each other(essentially they are orthogonal; see Theorem 2.8 below), and thus impliessome special properties which determine its role in the stability theory. Ona first glance the regularity requirements seem quite strong and even a bitartificial. However, they hold in typical situations.

The pair of Lyapunov exponents (, ) is called regular if and(, ) = 0. By Theorem 2.6, this holds if and only if (, ) = 0, and alsoif and only if i = i.

Theorem 2.8. If the pair (, ) is regular, then the filtrations V = {Vi :i = 0, . . ., s} and V = {Wi : i = 0, . . ., r} are orthogonal, that is, s = r,dimVi + dimWsi = n, and v, w = 0 for every v Vi and w Wsi.

Proof. Set mi = n dimWsi + 1. ThenWsi = {w W : (w) mi}

and mi = mi in view of Theorem 2.6. Let v be a normal ordered basisof V and w the basis of W dual to v. Since we obtain

Wsi = {w W : mi + (w) < 0}= {w W : (vi) + (w) < 0 if and only if i < mi}= span{wmi , . . ., wn}.

This implies that w is normal and ordered. Now for each i and any normalordered basis v = (v1, . . ., vni , vni+1, . . ., vn) of V and the dual basis wof W the last n ni components of w have to coincide with those of w.This implies that r = s, and

Wsi = span{wni+1, . . ., wn} = V i .The desired result now follows.

With slight changes one can also consider Lyapunov exponents on com-plex vector spaces.


3. Regularity of Lyapunov Exponents Associated with

Differential Equations

We now discuss the regularity properties of the Lyapunov exponent +

defined by (1.3) for Equation (1.1) provided the matrix function A(t) satisfies(1.2). Consider the linear differential equation which is dual to (1.1),

w = A(t)w, (3.1)where w(t) Cn and A(t) denotes the complex-conjugated transpose ofA(t). Let w(t) be a unique solution of this equation such that w(0) = w.The function + : Cn R {} given by

+(w) = lim supt+

1

tlogw(t)

defines a Lyapunov exponent associated with Equation (3.1). We note thatthe exponents + and + are dual. To see that, let v(t) be a solution of theequation (1.1) and w(t) a solution of the dual equation (3.1). Observe thatfor every t R,

d

dtv(t), w(t) = A(t)v(t), w(t)+ v(t),A(t)w(t)

= A(t)v(t), w(t) A(t)v(t), w(t) = 0,where , denote the standard inner product in Cn. Hence

v(t), w(t) = v(0), w(0)for any t R. Choose now dual bases (v1, . . ., vn) and (w1, . . ., wn) of Cn.Let vi(t) be the unique solution of (1.1) such that vi(0) = vi, and wi(t) theunique solution of (3.1) such that wi(0) = wi, for each i. We obtain

vi(t) wi(t) 1for every t R, and hence, +(vi) + +(wi) 0 for every i. It follows thatthe exponents + and + are dual.

We will study the regularity of the pair of exponents (+, +). Letv = (v1, . . ., vn) be a basis of Cn. We denote by m(t) = vm(t) the m-volume of the parallelepiped defined by the vectors vi(t), for i = 1, . . ., m,that are solutions of (1.1) satisfying the initial conditions vi(0) = vi. LetVm(t) be themmmatrix whose entries are the scalar products vi(t), vj(t)for i, j = 1, . . ., m. Then

m(t) = v

m(t) = |detVm(t)|1/2.In particular, 1(t) = |v1(t)| and n(t) = n(0)|detA(t)|, where

A(t) = exp

( t0A() d

).

Note that

detA(t) = exp

( t0trA() d

). (3.2)


The following theorem provides some crucial criteria for the pair (+, +)to be regular.

Theorem 3.1. Assume that the matrix function A(t) satisfies (1.2). Thefollowing statements are equivalent:

1. the pair (+, +) is regular;2.

limt+

1

tlog|detA(t)| =

si=1

ki+i . (3.3)

3. for any normal ordered basis v of Cn and any 1 m n thefollowing limit exists:

limt+

1

tlog vm(t).

In addition, if the pair (+, +) is regular then for any normal ordered basisv of Cn and any 1 m n we have

limt+

1

tlog vm(t) =

mi=1

+(vi). (3.4)

Proof. We adopt the following notations in the proof. Given a functionf : (0,) R we set

(f) = lim supt+

1

tlog |f(t)| and (f) = lim inf

t+

1

tlog |f(t)|.

If, in addition, f is integrable we shall also write

f = lim supt+

1

t

t0f() d and f = lim inf

t+

1

t

t0f() d.

We first show that Statement 1 implies Statement 2. We start with anauxiliary result.

Lemma 3.2. The following statements hold:

1. (detA) = Re trA and (detA) = Re trA;2. if (v1, . . ., vn) is a basis of Cn, then

ni=1

+(vi) (detA) (detA) ni=1

+(vi).

Proof of the lemma. It follows from (3.2) that

(detA) = lim inft+

1

tRe

t0trA() d = Re trA

and

(detA) = lim supt+

1

tRe

t0trA() d = Re trA.

This establishes the first statement.


Since n(0)|detA(t)| gives the volume of the parallelepiped determinedby the vectors v1(t), . . ., vn(t), we have |detA(t)|

ni=1vi(t), and hence,

(detA) ni=1

+(vi).

In a similar way,

(detA) = Re trA = Re tr(A) ni=1

+(vi).

The lemma follows.

Let i and i be the values of the Lyapunov exponents

+ and +,counted with their multiplicities. Choosing a normal basis (v1, . . ., vn) ofCn, it follows from Lemma 3.2 that

ni=1

i (detA) (detA) ni=1

i.

Therefore,

(detA) (detA) ni=1

(i + i) n(+, +),

This shows that if the pair (+, +) is regular then (3.3) holds.We now show that Statement 3 implies Statement 1. We split the proof

into two steps.Step 1. For every t 0 consider a linear coordinate change in Cn given

by a matrix U(t). We assume that the matrix function U(t) is differentiable.Setting z(t) = U(t)1v(t) we obtain

v(t) = U(t)z(t) + U(t)z(t) = A(t)v(t) = A(t)U(t)z(t).

It follows that z = B(t)z, where the matrix B(t) = (bij(t)) is defined by

B(t) = U(t)1A(t)U(t) U(t)1U(t). (3.5)We need the following lemma of Perron. Its main manifestation is to showhow to reduce Equation (1.1) with a general matrix function A(t) to a lineardifferential equation with a triangular matrix function.

Lemma 3.3. There exists a differentiable matrix function U(t) such that:

1. U(t) is unitary for each t 0;2. the matrix B(t) is upper triangular for each t;3. sup{|bij(t)| : t 0, i 6= j}


Proof of Lemma 3.3. Given a basis v = (v1, . . ., vn) we constructthe desired matrix function U(t) by applying the GramSchmidt orthogo-nalization procedure to the basis vi(t), with i = 1, . . ., n, where vi(t) is thesolution of (1.1) satisfying the initial condition vi(0) = vi. Thus we obtain acollection of functions u1(t), . . ., un(t) such that ui(t), uj(t) = ij where ijis the Kronecker symbol. Let V (t) and U(t) be the matrices with columnsv1(t), . . ., vn(t) and u1(t), . . ., un(t), respectively. The matrix U(t) is unitaryfor each t. Moreover, the GramSchmidt procedure can be effected in sucha way that each function uk(t) is a linear combination of functions v1(t), . . .,vk(t). It follows that the matrix Z(t) = U(t)

1V (t) is upper triangular foreach t.

The columns z1(t) = U(t)1v1(t), . . ., zn(t) = U(t)

1vn(t) of the matrixZ(t) form a basis of the space of solutions of the linear differential equation

z = B(t)z. Furthermore, B(t) = Z(t)Z(t)1, and as Z(t) is upper triangularso is the matrix B(t).

Since U(t) is unitary, using (3.5) we obtain

B(t) +B(t) = U(t)(A(t) +A(t))U(t) (U(t)U(t) + U(t)U(t))

= U(t)(A(t) +A(t))U(t) ddt(U(t)U(t))

= U(t)(A(t) +A(t))U(t).

Since B(t) is triangular we conclude that |bij(t)| 2A(t)


In the general case (when the entries of Z(t) are not necessarily real) we canwrite vk(t)/

v

k1(t) = |zkk(t)| andd

dtlog |zkk(t)| = 1

2

d

dtlog(zkk(t)zkk(t))

=1

2

(zkk(t)

zkk(t)+zkk(t)

zkk(t)

)

=1

2(bkk(t) + bkk(t)) = Re bkk(t).

This completes the proof of the lemma.

We define the nn matrix function Z(t) = (zij(t)) as follows: zij(t) = 0if j < i,

zij(t) = e t0bii() d

if j = i, and

zij(t) =

taij

jk=i+1

bik(s)zkj(s)e tsbii() dds

if j > i.

Lemma 3.4. For any constants aij, with 1 i < j n, the columns ofthe matrix Z(t) form a basis of solutions of the equation z = B(t)z.

Proof of Lemma 3.4. For each i we have zii(t) = bii(t)zii(t), and

zij(t) =

jk=i+1

bik(t)zkj(t) + bii(t)zij(t) =

jk=i

bik(t)zkj(t)

for each j > i. This shows that Z(t) = B(t)Z(t) and hence the columnsof Z(t) (i.e., the vectors zi(t) = (z1i(t), . . ., zni(t))) are solutions of theequation z = B(t)z. Since Z(t) is upper triangular, we have

detZ(t) = exp

(ni=1

t0bii() d

)6= 0,

and hence the vectors zi(t) form a basis.

Step 2. Assume that (vm) = (v

m) for any normal ordered basis

v and any 1 m n. We show that the pair (+, +) is regular. ByLemma 3.3, it is sufficient to consider the equation z = B(t)z, where B(t)is a n n upper triangular matrix for every t.

Lemma 3.5. If Re bii = Re biidef= Bi for each i = 1, . . ., n then:

1. the pair of Lyapunov exponents corresponding to the equations z =B(t)z and w = B(t)w is regular;

2. the numbers B1, . . ., Bn are the values of the Lyapunov exponent +;

3. the numbers B1, . . ., Bn are the values of the Lyapunov expo-nent +.


Proof of Lemma 3.5. We consider the solutions of the equation z =B(t)z described in Lemma 3.4 and show that each column

zi(t) = (z1i(t), . . ., zni(t))

of Z(t) satisfies

+(zi) = lim supt+

1

tlogzi(t) = Bi

for some choice of the constants aij . Since Re bii = Bi, we clearly have+(zii) = Bi. Assume now that

+(zkj) Bj for each i + 1 k j. Weshow that +(zij) Bj . Observe that for each > 0, we have

+(zij) lim supt+

1

t

(loge t0 bii() d

+ log

taij

jk=i+1

bik(s)zkj(s)e s0bii() dds

)

Bi + lim supt+

1

tlog

taij

Kne(BjBi+)s ds

.We exploit here the fact that, by Lemma 3.3, |bij(t)| K for some con-stant K independent of i, j, and t. For each j > i set aij = 0 if Bj Bi 0and aij = + if Bj Bi < 0. Then for every sufficiently small > 0, wehave

+(zij) Bi + lim supt+

1

tlog

Kn(e(BjBi+)t 1)Bj Bi +

if Bj Bi 0 and

+(zij) Bi + lim supt+

1

tlog

Kne(BjBi+)t

Bj Bi + if Bj Bi < 0. Therefore,

+(zij) Bi +Bj Bi + = Bj + .Since is arbitrary we obtain +(zij) Bj . This shows that +(zi) = Bifor each 1 i n.

In a similar way, one can show that there exists a lower triangular matrixW (t) such that W (t) = B(t)W (t). The entries of the matrix W (t) aredefined by wij(t) = 0 if j > i,

wij(t) = e t0bjj() d

if j = i, and

wij(t) = taji

i1k=j

bki(s)wkj(s)e tsbii() d ds


if j < i where the constants aji are chosen as above. Since Re bii = Bi, thecolumns w1(t), . . ., wn(t) of W (t) satisfy

+(wi) = lim supt+

1

tlogwi(t) = Bi = +(zi).

Note that +(zi) + +(wi) = 0 for each i. In order to prove that the pair

(+, +) is regular, it remains to show that the bases z and w are dual.Clearly, zi(0), wj(0) = 0 for every i < j. Moreover, zi(0), wj(0) = 1 foreach 1 i n. Fix i > j and t > 0. We have that

zi(t), wj(t) =i

k=j

zki(t)wkj(t). (3.6)

Since +(zij) Bj and +(wij) Bj for every i, j, and > 0, we obtain(zi(0), wj(0)) max

j+1ki1(zkiwkj)

maxj+1ki1

lim supt+

1

t

(log

taki

Kne(BiBk+)s ds

+ log

tajk

Kne(Bj+Bk+)s ds

)

maxj+1ki1

(Bi Bk Bj +Bk + 2) = Bi Bj + 2.

Since is arbitrary, if Bi Bj < 0 we obtain (zi(0), wj(0)) < 0 andzi(0), wj(0) = lim

t+zi(t), wj(t) = 0.

If Bi Bj 0, then aji = 0. Moreover, for each k we have Bi Bk 0 orBkBj 0, and hence aki = 0 or ajk = 0. Letting t 0 in (3.6) we obtain

zi(0), wj(0) = zji(0)wjj(0) + zii(0)wij(0) +i1

k=j+1

zki(0)wkj(0).

Since i > j and aji = 0, we obtain zji(0) = wij(0) = 0. Moreover, for eachk such that j + 1 k i 1 we have aki = 0 or ajk = 0, and hence,zki(0) = 0 or wkj(0) = 0. Therefore each term in the above sum is zero.Thus zi(0), wj(0) = 0, and the lemma follows.

By Lemma 3.3 and Statement 3, we have

1

t

t0Re bii() d =

1

t

t0

d

dlog

vi (t)

vi1(t)d

=1

tlog

vi (t)/v

i1(t)

vi (0)/v

i1(0) (vi ) (vi1)

as t +. By Lemma 3.5 we conclude that the Lyapunov exponentcorresponding to the equation z = B(t)z is regular.


We now show that Statement 2 implies Statement 3. By Lemma 3.3 wehave

vm(t) = z

m(t) =mk=1

zkk(t). (3.7)

By Lyapunovs construction of normal bases (see the description after The-orem 2.3) there exists a normal basis (v1(0), . . ., vn(0)) of Cn such thatzk(0) = ek + fk for some fk span{ek+1, . . . , en}, where (e1, . . ., en) isthe canonical basis of Cn. Since the matrix solution Z(t) is upper tri-angular for each t, the vectors ek and Z(t)fk are orthogonal. Therefore,Z(t)fk |zkk(t)|, and

+(zk) lim supt

1

tlog|zkk(t)| def= k. (3.8)

Without loss of generality we are considering the norm in Cn given by(w1, . . . , wn) = |w1|+ + |wn|. By (3.7) we obtain

limt

1

tlog vn(t) =

si=1

ki+i

nk=1

+(zk) n

k=1

k. (3.9)

Furthermore, by (3.2) and Lemma 3.4 we have

zn(t)/z

n(0) = exp

( t0trB() d

)=

nk=1

zkk(t),

and hence, by (3.7),

limt

1

tlog vn(t)

nk=1

k. (3.10)

It follows from (3.8), (3.9), and (3.10) that

+(vk) = +(zk) = lim

t+

1

tlogzk(t) = k

for each k = 1, . . ., n. Thus, again by (3.7), for each 1 m n we concludethat

(vm) = (v

m) =

mk=1

+(vk).

This completes the proof of the theorem.

The following example illustrates that Statement 2 of Theorem 3.1 can-not be weakened. Consider the system of differential equations

v1 = p(t)v1, v2 = p(t)v2for t > 0, where p(t) = cos log t sin log t 1. The general solution of thesystem can be written in the form

v1(t) = C1q(t)1, v2(t) = C2q(t),


for some constants C1 and C2, where

q(t) = exp

( t1p() d

)= exp(t(cos log t 1)).

We have

lim inft+

1

t

t1p() d = 2 and lim sup

t+

1

t

t1p() d = 0. (3.11)

Observe that detA(t) = 1 for every t, and hence, the limit in (3.3) exists.On the other hand, it follows from (3.11) that the the limit of 1-volumesmay not exist, and hence the pair of Lyapunov exponents (+, +) is notregular. In this case the Lyapunov exponent is equal to +(v) = 2 for everyv = (v1, v2) 6= 0 with C1 6= 0, and +(v) = 0 otherwise. Hence,

0 = limt+

1

tlog|detA(t)| < +1 + +2 = 2.

We also illustrate that Statement 2 of Theorem 3.1 cannot be weakenedby replacing the limit in (3.3) by the upper limit. Consider a system ofdifferential equations

v1 = v2, v2 = p(t)v2for t > 0. Observe that the general solution of the system can be written inthe form

v1(t) = C1 + C2

t1q() d, v2(t) = C2q(t),

for some constants C1 and C2. One can easily show that for every vectorv = (v1, v2) 6= 0, we have

+(v) = limt+

1

tlogv(t) = 0.

On the other hand, it follows from (3.11) that the limit in (3.3) does notexist, and hence, the pair of Lyapunov exponents (+, +) is not regular.

We say that the Lyapunov exponent + is forward regular if the pairof Lyapunov exponents (+, +) is regular. We also say that the Lyapunovexponent + is exact if (vm) = (

v

m) for any 1 m n and any basisv of Cn. We emphasize that exactness does not require the identity in(3.4), but only the existence of the limit in (3.4). By Theorem 3.1, if thefunction t 7 A(t) is bounded (i.e., if (1.2) holds) then forward regularity isequivalent to exactness. In the case of unbounded matrix functions exactnessmay turn out to be weaker then forward regularity as the following exampleillustrates.

Consider the system of differential equations

v1 = av1 + ebtv2, v2 = av2

for t R, where a and b are positive constants such that 2a < b. The generalsolution of the system can be written in the form

v1(t) = C1eat +

C2b 2ae

(ba)t, v2(t) = C2eat,


for some constants C1 and C2. One can easily show that the Lyapunovexponent is exact. In particular, for every vector v = (v1, v2) 6= 0 we have+(v) = a whenever C1 6= 0, and +(v) = b a otherwise. By (3.2) weobtain detA(t) = 1 and

0 = limt+

1

tlog|detA(t)| < +1 + +2 = b.

Therefore, the identity in (3.4) does not hold, and the Lyapunov exponentis not forward regular.

One can also show that forward regularity is equivalent to exactnessprovided the function t 7 A(t) is integrable.

4. Lyapunov Stability Theory

We now proceed with the proof of the Lyapunov Stability Theorem 1.2.Consider the perturbed differential equation (1.6) where the function f(t, u)satisfies (1.7). Let v1(t), . . ., vn(t) be n linearly independent solutions of(1.1) and V (t) the corresponding monodromy matrix. We assume thatV (0) = Id. Denote by V(t, s) the Cauchy matrix of (1.1) defined by V(t, s) =V (t)V (s)1.

Theorem 4.1. Assume that there are two continuous functions R andr on [0,) such that for every t 0, t

0R() d D1 < +, (4.1)

0e(q1)

s0(R+r)() d ds = D2 < +, (4.2)

and for every 0 s t,V(t, s) D3e

tsR() d+(q1)

s0r() d . (4.3)

Then there exist D > 0 and > 0 such that for every u0 Cn with u0 < ,there is a unique solution u of Equation (1.6) which satisfies:

1. u is well-defined on the interval [0,), and u(0) = u0;2. for every t,

u(t) u0De t0R() d . (4.4)

Proof. Equation (1.6) is equivalent to the integral equation

u(t) = V(t, 0)u0 +

t0

V(t, s)f(s, u(s)) ds. (4.5)

Moreover, by (1.7) for every u0 Cn with u0 H there exists a uniquesolution u(t) satisfying the initial condition u(0) = u0.

We consider the linear space BC of Cn valued continuous functions x onthe interval [0,) such that

x(t) Ce t0R() d


for every t 0, where C > 0 is a constant. We endow this space with thenorm

xR = sup{x(t)e t0R() d : t 0}.

It follows from (4.1) that x(t) xReD1 . Therefore, if xR is sufficientlysmall, then x(t) H for every t 0. This means that the operator

(Jx)(t) =

t0

V(t, s)f(s, x(s)) ds

is well-defined on BC . By (1.7), (4.2), and (4.3) we have

(Jx1)(t)(Jx2)(t)

t0D3e

tsR() d+(q1)

s0r() dK(x1 x2R)qeq

s0R() d ds

=D3K(x1 x2R)qe t0R() d

t0e(q1)

s0(R+r)() d ds

D3KD2(x1 x2R)qe t0R() d .

It follows that

Jx1 Jx2R D3KD2(x1 x2R)q= D3KD2(x1 x2R)q1x1 x2R.

Choose > 0 such that

def= D3KD2(2)

q1 < 1. (4.6)

Whenever x1, x2 B we haveJx1 Jx2R x1 x2R. (4.7)

Choose a point u0 Cn and set (t) = V(t, 0)u0. It follows from (4.3) that(t) u0D3e

t0R() d

and hence, R u0D3. Consider the operator J defined by(Jx)(t) = (t) + (Jx)(t)

on the space B. Note that B is a complete metric space with the distanced(x, y) = x yR. We have

JxR D3u0+ xR < D3 + < provided that is sufficiently small. Therefore J(B) B. By (4.6) and(4.7), the operator J is a contraction, and hence, there exists a unique func-tion u B which is a solution of (4.5). Furthermore, u can be obtained by

u(t) = limn

(Jn0)(t) =n=0

(Jn)(t)


and hence, by (4.7),

uR n=0

JnR n=0

nR = R1

D3u01 .

Therefore, the function u(t) satisfies (4.4).

As an immediate consequence of this theorem we obtain the followingstatement due to Malkin (see [20]).

Theorem 4.2. Assume that the Cauchy matrix of (1.1) admits the es-timate

V(t, s) De(ts)+s, (4.8)for any 0 s t, and some constants , , and D, such that

(q 1)+ < 0. (4.9)Then the trivial solution of Equation (1.6) is asymptotically and exponen-tially stable.

Proof. Set

R(t) = and r(t) =

q 1 .

It follows from (4.8) and (4.9) that Conditions (4.2) and (4.3) hold. In orderto verify Condition (4.1) note that by setting t = s in (4.3) we obtain

1 = V(t, t) D3e(q1) t0r() d .

It follows that for every t,

t

q 1 = t0r() d C, (4.10)

where C 0 is a constant. Therefore, by (4.9),

t =

t0R() d C +

t0(R() + r()) d C (4.11)

for all t. This implies Condition (4.1). Therefore, by (4.4), any solution uof (1.6) satisfies

u(t) u0Det u0DeC . (4.12)This shows that the trivial solution of (1.6) is asymptotically stable.

If every solution of (1.6) is defined for all t then by (4.10) and (4.11), 0 and 0. In view of (4.9), < 0, and in view of (4.12), the solutionu(t) = 0 is exponentially stable. The desired result follows.

Another important consequence of Theorem 4.1 is the following state-ment due to Lyapunov (see [19]).


Theorem 4.3. Assume that the maximal value max of the Lyapunovexponent of (1.1) is strictly negative (see (1.5)), and that

(q 1)max + < 0,where is the regularity coefficient. Then the trivial solution of Equation(1.6) is asymptotically and exponentially stable.

Proof. Consider a normal basis (v1(t), . . ., vn(t)) for the space of so-lutions of (1.1), such that the numbers

1 n = maxare the values of the Lyapunov exponent + for the vectors v1(t), . . ., vn(t).We may assume that the matrix V (t) whose columns are v1(t), . . ., vn(t) issuch that the columns w1(t), . . ., wn(t) of the matrixW (t) = [V (t)

]1 forma normal basis for the space of solutions of the dual equation w = A(t)w.

Let 1, . . ., n be the values of the Lyapunov exponent + for the vectors

w1(t), . . ., wn(t). For every > 0 there exists a constant D > 0 such that

vj(t) De(j+)t and wj(t) De(j+)t

for every t. We also have

= max{j + j : j = 1, . . ., n}.It follows that j + j for every j = 1, . . ., n. Consider the Cauchymatrix

V(t, s) = V (t)V (s)1 = V (t)W (s).

Its entries are

vik(t, s) =nj=1

vij(t)wkj(s),

where vij(t) is the i-th coordinate of the vector vj(t) and wkj(s) is the k-thcoordinate of the vector wj(s). It follows that

|vik(t, s)| nj=1

|vij(t)| |wkj(s)| nj=1

vj(t) wj(s)

nD2e(j+)t+(j+)s = nD

2e(j+)(ts)+(

j+j+2)s.

Therefore, there exists a constant D > 0 such that

V(t, s) De(max+)(ts)+(+2)s. (4.13)Since can be chosen arbitrarily small the desired result follows from The-orem 4.2 if we set = max + and = + 2.

We conclude by observing that the Lyapunov Stability Theorem 1.2 isan immediate corollary of Theorem 4.3. Moreover, if + is forward regular,setting = 0 in (4.13) we obtain that for every solution u(t) of (1.6) andevery s R,u(t) De(max+)(ts)+2su(s) = De2se(max+)(ts)u(s). (4.14)


In other words the contraction constant may get worse along the orbit of asolution of (1.6). This implies that the size of the neighborhood at times where the asymptotic (and exponential) stability of the trivial solutionis guaranteed may decay with subexponential rate. In Section 6 we willestablish an analog of (4.14) for dynamical systems with discrete time (seeStatement (L7) of Theorem 6.3) and illustrate the crucial role that it playsin the nonuniform hyperbolicity theory.

We now describe a generalization of the Lyapunov Stability Theoremin the case when not all but only some Lyapunov exponents are negative.Namely, let +1 < < +s be the distinct values of the Lyapunov exponent+ for Equation (1.1). Assume that

+k < 0, (4.15)

where 1 k < s and the number +k+1 can be negative, positive, or equalto 0 (compare to (1.5)).

Theorem 4.4 (see [7]). Assume that the Lyapunov exponent + forEquation (1.1) with the matrix function A(t) satisfying (1.2), is forwardregular and satisfies Condition (4.15). Then there exists a local smoothsubmanifold V s Cn which passes through 0, is tangent at 0 to the linearsubspace Vk (defined by (2.1)), and is such that the trivial solution of (1.6)is asymptotically and exponentially stable along V s.

This statement is a particular case of a more general and strong resultthat we discuss in Section 9.

5. The Oseledets Decomposition

Using results of Section 2, we conclude that the Lyapunov exponent +

associated with Equation (1.1), as defined by (1.3), takes on only finitelymany values on Cn \ {0}. To stress that we deal with positive time we shallnow denote its values by

+1 < < +s+ ,where s+ n. We also denote by V+ the filtration of Cn associated to +.We have

{0} = V +0 $ V +1 $ $ V +s+ = Cn,where V +i = {v Cn : +(v) +i }. Finally we denote by k+i = dimV +i dimV +i1 the multiplicity of the value

+i . Note that

s+i=1 k

+i = n.

An important advantage of Theorem 3.1 is that one can verify the regu-larity property of the pair (+, +) dealing only with the Lyapunov exponent+. In this case we say that the Lyapunov exponent + is forward regularto stress that we allow only positive time.

Reversing the time we introduce the Lyapunov exponent : Cn R {} by

(v) = lim supt

1

|t| log v(t),


where v(t) is the solution of (1.1) satisfying the initial condition v(0) = v.The function takes on only finitely many values on Cn \ {0}, which wedenote by

1 > > s ,where s n. We denote by V the filtration of Cn associated to . Wehave

Cn = V 1 % V1 % % V s % V s+1 = {0},

where V i = {v Cn : (v) i }. We also denote by ki = dimV i dimV i+1 the multiplicity of the value

i , and we have

si=1 k

i = n.

We also consider the Lyapunov exponent : Cn R{} given by

(w) = lim supt

1

|t| log w(t),

where w(t) is the solution of (3.1) satisfying the initial condition w(0) = w.We say that the Lyapunov exponent is backward regular if the pair

(, ) is regular. We say that the filtrations V+ and V comply if thefollowing properties hold:

1. s+ = sdef= s;

2. there exists a decomposition

Cn =s

i=1

Ei (5.1)

into subspaces Ei such that if i = 1, . . ., s then

V +i =i

j=1

Ej and Vi =

sj=i

Ej ;

3. +i = idef= i for i = 1, . . ., s;

4. if i = 1, . . ., s and v Ei \ {0} then

limt

1

tlog v(t) = i

with uniform convergence on {v Ei : v = 1} (recall that v(t) isthe solution of Equation (1.1) with initial condition v(0) = v).

The decomposition (5.1) is called the Oseledets decomposition associatedwith the Lyapunov exponent + (or with the pair of Lyapunov exponents(+, )).

We say that the Lyapunov exponent + (or the pair of Lyapunov expo-nents (+, )) is Lyapunov regular if the exponent + is forward regular,the exponent is backward regular, and the filtrations V+ and V comply.

Remark 5.1. We stress that simultaneous forward and backward regu-larity of the Lyapunov exponents does not imply the Lyapunov regularity.Roughly speaking, whether the Lyapunov exponent is forward (respectively,


backward) regular does not depend only on the backward (respectively, for-ward) behavior of the system. On the other hand Lyapunov regularity re-quires some compatibility between the forward and backward behavior whichis expressed in terms of the filtrations V+ and V.

The regularity property is quite strong and requires a quite rigid behav-ior of the system.

Theorem 5.2. If the Lyapunov exponent + is Lyapunov regular thenthe following statements hold:

1. the exponents + and are exact;

2. dimEi = k+i = k

i

def= ki;

3. if v = (v1, . . ., vki) is a basis of Ei, then

limt

1

tlog vki(t) = iki;

4. if vi(t) and vj(t) are solutions of Equation (1.1) such that vi(0) Ei \ {0} and vj(0) Ej \ {0} with i 6= j then

limt

1

tlog(vi(t), vj(t)) = 0.

Proof. The first three statements follow from Theorem 3.1. For the laststatement, let (t) denote the area of the rectangle formed by the vectorsvi(t) and vj(t). We have

(t) = vi(t) vj(t) sin(vi(t), vj(t)).Since sin(vi(t), vj(t)) 1, and the exponent + is exact we obtain

+i + +j = limt+

1

tlog (t)

+(vi) + +(vj) + lim inft+

1

tlog sin(vi(t), vj(t))

+i + +j + lim supt+

1

tlog sin(vi(t), vj(t)) +i + +j .

A similar argument applies to the exponent and we obtain the desiredstatement.

One can construct examples of nonstationary linear differential equa-tions whose Lyapunov exponent is forward and backward regular but is notLyapunov regular. In dimension 1, an example is given by the equationv = a(t)v, where a : R R is a bounded continuous function such thata(t) a+ as t + and a(t) a as t , for some constantsa+ 6= a. The general solution of this equation is

v(t) = exp

( t0a(s) ds

)v(0),


and hence the forward and backward exponents are respectively equal to a+and a. In dimension 2, an example is given by

v1 = a(t)v1, v2 = a(t)v2,with a as above. In this case the values of the exponents + and co-incide (up to the change of sign required for Lyapunov regularity), but thefiltrations V+ and V do not comply.

6. Dynamical Systems with Nonzero Lyapunov Exponents.

Multiplicative Ergodic Theorem

Let t be a smooth flow on a smooth compact Riemannian p-manifoldM . It is generated by the vector field X on M given by

X(x) =dt(x)

dt|t=0.

For every x0 M the trajectory {x(x0, t) = t(x0) : t R} represents asolution of the nonlinear differential equation

v = X(v) (6.1)

on the manifoldM . This solution is uniquely defined by the initial conditionx(x0, 0) = x0.

Given a point x M and the trajectory {t(x) : t R} passing throughx we introduce the variational differential equation

w(t) = A(x, t)w(t), (6.2)

where

A(x, t) = dt(x)X.

This is a nonstationary linear differential equation along the trajectory{t(x) : t R} known also as the equation of the first linear approximation.

With the flow t one can associate a certain collection of single lineardifferential equations (6.1) along each trajectory of the flow. The stabilityof a given trajectory can be described by studying small perturbations ofthe variational differential equation (6.2). The perturbation term is of type(1.6) and satisfies (1.7). The results of the previous section can be applied tostudy the stability of trajectories via the Lyapunov exponents. Although it isstill a very difficult problem to verify whether a given trajectory is Lyapunovregular it turns out that most trajectories (in the sense of measure theory)have this property.

A similar approach is used to establish the stability of trajectories ofdynamical systems with discrete time. Since it is technically simpler wedescribe such an approach here.

Let f : M M be a diffeomorphism of a compact smooth Riemannianp-manifoldM . Given x M , consider the trajectory {fmx}mZ. The familyof maps

{dfmxf}mZ


can be viewed as an analog of the variational differential equation (6.2) inthe continuous time case. Given x M and v TxM the formula

(x, v) = lim supm+

1

mlog dxfmv

defines the Lyapunov exponent specified by the diffeomorphism f at thepoint x. The function (x, ) takes on only finitely many values on TxM\{0},which we denote by

1(x) < < s(x)(x),where s(x) p. We denote by Vx the filtration of TxM associated to (x, ).We have

{0} = V0(x) $ V1(x) $ $ Vs(x)(x) = TxM,where Vi(x) = {v TxM : (x, v) i(x)}. We also introduce ki(x) =dimVi(x) dimVi1(x), which is the multiplicity of the value i(x). Wehave

s(x)i=1 ki(x) = p. Finally, the collection of pairs

Sp(x) = {(i(x), ki(x)) : 1 i s(x)}is called the Lyapunov spectrum of the exponent (x, ).

We note that the functions i(x), s(x), and ki(x) are invariant under f ,and (Borel) measurable (but not necessarily continuous).

We say that a point x M is forward, or backward, or Lyapunov regularif so is the Lyapunov exponent (x, ). Note that if x is regular then so isthe point f(x) and thus one can speak on the whole trajectory {fm(x)} asbeing forward, backward, or Lyapunov regular.

Summarizing our discussion in the previous section we come to the fol-lowing result.

Theorem 6.1. If a point x M is Lyapunov regular, then there existsthe Oseledets decomposition

TxM =

s(x)i=1

Ei(x)

into subspaces Ei(x) such that:

1. the subspaces Ei(x) are invariant under the differential dxf , i.e.,dxfEi(x) = Ei(f(x)), and depend (Borel) measurably on x;

2. if v Ei(x) \ {0} then

limm

1

mlog dxfmv = i(x)

with uniform convergence on {v Ei(x) : v = 1};3. if v = (v1, . . ., vki(x)) is a basis of Ei(x), then

limn

1

nlog vki(x)(n) = i(x)ki(x);

in particular, the Lyapunov exponent (x, ) is exact;


4. there exists a decomposition of the co-tangent bundle

T xM =

s(x)i=1

Ei (x)

into subspaces Ei (x) associated with the Lyapunov exponent +; the

subspaces Ei (x) are invariant under the co-differential dxf , that is

dxfEi (x) = E

i (f(x)), and depend (Borel) measurably on x; more-

over, if {vi(x) : i = 1, . . ., p} is a normal basis with vi(x) Ej(x)for nj1(x) < i nj(x), and {vi (x) : i = 1, . . ., p} is a dual basis,then vi (x) Ej (x) for nj1(x) < i nj(x).

The following theorem due to Oseledets (see [24]) is a key result instudying the regularity of trajectories of dynamical systems. It shows thatregularity is typical from the measure-theoretical point of view.

Theorem 6.2 (Multiplicative Ergodic Theorem). If f is a C1 diffeomor-phism of a compact smooth Riemannian manifold, then the set of Lyapunovregular points has full measure with respect to any f-invariant Borel proba-bility measure on M .

We stress that there may exist trajectories which are both forward andbackward regular but not Lyapunov regular. However, such trajectoriesform a negligible set with respect to any f -invariant Borel probability mea-sure.

We also emphasize that the notion of Lyapunov regularity does not re-quire any invariant measure. Consider the set L M of points that areLyapunov regular. Due to Theorem 6.2 this set is nonempty and indeedhas full measure with respect to any f -invariant Borel probability measureon M .

We note that only in some exceptional cases every point in M is Lya-punov regular. To illustrate this consider a Smale horseshoe, i.e., a diffeo-morphism f : R2 R2 which maps the unit square S onto a horseshoe-likeset (see for example [14]). The map f has the locally maximal invariant

set Adef=nZ f

nS S, and f |A is topologically conjugate to the fullshift on two symbols. Moreover, the map f is uniformly hyperbolic on Aand hence, TxR2 = E1(x) E2(x) at every point x A, where E1(x)and E2(x) are the one-dimensional stable and unstable subspaces at x. Set(x) = logdxf |E1(x) < 0 and +(x) = logdxf |E2(x) > 0. For everyx A and v E1(x) we have

(x, v) = lim supn

1

n

n1k=0

(fkx),

and for every x A and v E2(x) we have

(x, v) = lim supn

1

n

n1k=0

+(fkx).


For a Smale horseshoe the stable and unstable subspaces are uniformlytransverse, and hence a point x is regular if and only if the Lyapunovexponent at x is exact (see Statement 3 of Theorem 6.1). Furthermore,Theorem 6.2 is a consequence of the Birkhoff ergodic theorem applied to theHolder continuous functions and +. Therefore, in the case of a linearhorseshoe (when the functions and + are constant on A), or more gen-erally when the functions and + are cohomologous to constants (i.e., = a + f and + = a+ + + f + for some constantsa, a+ R and some continuous functions and + on A), every pointx A is regular. On the other hand, when at least one of the functions and + is noncohomologous to a constant (and thus for a generic C1 surfacediffeomorphism with a Smale horseshoe), it follows from work of Barreiraand Schmeling [3] that the set of nonregular points is nonempty and hasHausdorff dimension equal to the Hausdorff dimension of the set A.

Recall that an f -invariant measure is ergodic if for each f -invariantmeasurable set A M either A or M \A has zero measure. One can showthat is ergodic if and only if every f -invariant (mod 0) measurable function(i.e., a measurable function such that f = almost everywhere) isconstant almost everywhere.

Let be a Borel f -invariant ergodic measure. There exist numbersi =

i and ki = k

i for i = 1, . . ., s (where s = s

) such that

i(x) = i, ki(x) = ki, s(x) = s (6.3)

for -almost every x. The collection of pairs

Sp() = {(i, ki) : 1 i s}is called the Lyapunov spectrum of the measure .

We now consider dynamical systems whose spectrum of the Lyapunovexponent does not contain zero on some subset of M . More precisely, let

= {x L : there exists 1 k(x) < s(x)with k(x)(x) < 0 and k(x)+1(x) > 0}. (6.4)

Note that the set is f -invariant. We say that f is a dynamical systemwith nonzero Lyapunov exponents if there exists a Borel ergodic f -invariantprobability measure on M such that () = 1. The measure is called ahyperbolic measure for f . Note that if is hyperbolic then there is a number1 k = k < s such that k < 0 and k+1 > 0.

Consider the set = of those points in which are Lyapunov regular

and satisfy (6.3). By the Multiplicative Ergodic Theorem we have () = 1.

For every x , we set

Es(x) =ki=1

Ei(x) and Eu(x) =

si=k+1

Ei(x).


Theorem 6.3 (see [27]). Let be an ergodic hyperbolic measure . The

subspaces Es(x) and Eu(x), x have the following properties:L1. they depend measurably on x;L2. they form a splitting of the tangent space, i.e., TxM = E

s(x) Eu(x);

L3. they are invariant:

dxfEs(x) = Es(f(x)) and dxfE

u(x) = Eu(f(x));

Furthermore, there exist 0 > 0 and measurable functions C(x, ) > 0 and

K(x, ) > 0, x and 0 < 0 such that:L4. the subspace Es(x) is stable: if v Es(x) and n > 0, then

dxfnv C(x, )e(k+)nv;L5. the subspace Eu(x) is unstable: if v Eu(x) and n < 0, then

dxfnv C(x, )e(k+1)nv;L6.

(Es(x), Eu(x)) K(x, );L7. for every m Z,C(fm(x), ) C(x, )e|m| and K(fm(x), ) K(x, )e|m|.

Remark 6.4. Condition (L7) is crucial and is a manifestation of theregularity property (it is an analog in the discrete time case of Condition(4.14)). Roughly speaking it means that the estimates (L4), (L5), and (L6)may get worse as |m| but only with subexponential rate. We stressthat the rate of the contraction along stable subspaces and the rate of theexpansion along unstable subspaces are exponential and hence, are substan-tially stronger.

Proof of Theorem 6.3. We begin with the following general state-ment.

Lemma 6.5. Let X M be a Borel f-invariant set and A(x, ) a Borelpositive function on X [0, 0), 0 < 0 < 1 such that for every 0 > 0,x X, and m Z,

M1(x, )e|m| A(fm(x), ) M2(x, )e|m|,

where M1(x, ) and M2(x, ) are Borel functions. Then one can find Borelpositive functions B1(x, ) and B2(x, ) such that

B1(x, ) A(x, ) B2(x, ), (6.5)and for m Z,

B1(x, )e2|m| B1(fm(x), ), B2(x, )e2|m| B2(fm(x), ). (6.6)


Proof of the lemma. It follows from the conditions of the lemmathat there exists m(x, ) > 0 such that if m Z is such that |m| > m(x, )then

2 1|m| logA(fm(x), ) 2.

Set

B1(x, ) = minm(x,)im(x,)

{1, A(f i(x), )e2|i|

},

B2(x, ) = maxm(x,)im(x,)

{1, A(f i(x), )e2|i|

}.

The functions B1(x, ) and B2(x, ) are Borel. Moreover, if n Z thenB1(x, )e

2|n| A(fn(x), ) B2(x, )e2|n|. (6.7)Furthermore, if b1 1 b2 are such that

b1e2|n| A(fn(x), ) (6.8)

and

b2e2|n| A(fn(x), ) (6.9)

for all n Z then b1 B1(x, ) and b2 B1(x, ). In other words,B1(x, ) = sup{b 1 : inequality (6.8) holds for all n Z},B2(x, ) = inf{b 1 : inequality (6.9) holds for all n Z}. (6.10)

Inequalities (6.5) follow from (6.7) (with n = 0). We also have

A(fn+m(x), ) B2(x, )e2|n+m| B2(x, )e2|n|+2|m|,A(fn+m(x), ) B1(x, )e2|n+m| B1(x, )e2|n|2|m|,

Comparing these inequalities with (6.7) written at the point fm(x) andtaking (6.10) into account we obtain (6.6). The proof of the lemma is com-plete.

We now apply Lemma 6.5 to construct the functionK(x, ). LetK : R be a Borel function. It is called tempered at the point x if

limm

1

mlogK(fm(x)) = 0.

In other words the Lyapunov exponent of the function K(x) is exact andis 0. It follows that the function A(x, ) = K(x) satisfies the conditions ofLemma 6.5 for all > 0.

Fix > 0 and consider the function (x) = (Es(x), Eu(x)) for x .Since this function is tempered (see Theorem 6.1) applying Lemma 6.5 weconclude that the function K(x, ) = B1(x,

12) satisfies Conditions (L6) and

(L7) of the theorem.We will now show how to construct the function C(x, ). The proof

is an elaboration for the discrete time case of arguments in the proof ofTheorem 4.3 (see (4.14)).


Lemma 6.6. There exists a Borel positive function D(x, ) (x and > 0 is sufficiently small), such that if m Z and 1 i s, then

D(fm(x), ) D(x, )2e2|m| (6.11)and

dfnix D(x, )e(i+)n, dfnix D(x, )1e(i+)nfor any n 0, where dfnix = dxfn|Ei(x).

Proof of the lemma. Let x . By Theorem 6.1 (we use the no-tation of that theorem) there exists a number n(x, ) N such that ifn n(x, ), then

i 1nlog dfnix i + , i

1

nlog dfnix i + ,

and

i 1nlog dfnix i + , i

1

nlog dfnix i + ,

where dfnix = dxf

n|Ei (x) (recall that Ei (x) T xM is the dual space toEi(x) and d

f is the co-differential). Set

D+1 (x, ) = min1is

min0jn(x,)

{1, df jixe(i+)j , df jixe(i+)j

},

D1 (x, ) = min1is

minn(x,)j0

{1, df jixe(i)j , df jixe(i)j

},

D+2 (x, ) = max1is

max0jn(x,)

{1, df jixe(i)j , df jixe(i)j

},

D2 (x, ) = max1is

maxn(x,)j0

{1, df jixe(i+)j , df jixe(i+)j

},

and

D1(x, ) = min{D+1 (x, ), D1 (x, )}, D2(x, ) = max{D+2 (x, ), D2 (x, )},D(x, ) = max{D1(x, )1, D2(x, )}.

The function D(x, ) is measurable, and if n 0 and 1 i s thenD(x, )1e(i)n dfnix D(x, )e(i+)n,

D(x, )1e(i)n dfnix D(x, )e(i+)n,D(x, )1e(i)n dfnix D(x, )e(i+)n,D(x, )1e(i)n dfnix D(x, )e(i+)n.

(6.12)

Moreover, if d 1 is a number for which Inequalities (6.12) hold for alln 0 and 1 i s with D(x, ) replaced by d then d D(x, ). Therefore,

D(x, ) = inf{d 1 : the inequalities (6.12) hold for all n 0and 1 i s with D(x, ) replaced by d}. (6.13)


We wish to compare the values of the function D(x, ) at the points x andfmx for m Z. First, let us notice that for every x M , v TxM , and T xM with (v) = 1 we have

(dxf)(dxfv) = ((dxf)

1dxfv) = (v) = 1. (6.14)

Second, using the Riemannian metric on the manifold M we introduce theidentification map x : T

xM TxM such that x(), v = (v) where

v TxM and T xM .Let {vnk : k = 1, . . ., p} be a basis of Ei(fn(x)) and {wnk : k = 1, . . ., p}

be the dual basis of Ei (fn(x)). We have fn(x)(w

nk ) = v

nk . Let A

in,m and

Bin,m be matrices corresponding to the linear maps dfn+1ifm(x) and d

fn+1ifm(x)with respect to those bases. It follows from (6.14) that

Ai0,m(Bi0,m)

= Id

and hence, for every n > 0 the matrix corresponding to the map dfnifm(x) is

Ain,m = Ai0,m+n(A

i0,m)

1 = Ai0,m+n(Bi0,m)

.

It follows from here and (6.12) that:

1. if n > 0 then

dfnifm(x) D(x, )2e(i+)(n+m)+(i+)m = D(x, )2e2me(i+)n,dfnifm(x) D(x, )2e(i)(n+m)+(i)m = D(x, )2e2me(i)n,

2. if n > 0 and m n 0 thendfnifm(x) D(x, )2e(i+)(mn)+(i+)m = D(x, )2e2me(i+)n,dfnifm(x) D(x, )2e(i)(mn)+(i)m = D(x, )2e2me(i)n,

3. if n > 0 and nm 0 thendfnifm(x) D(x, )2e(i+)(nm)+(i+)m = D(x, )2e2me(i+)n,dfnifm(x) D(x, )2e(i)(nm)+(i)m = D(x, )2e2me(i)n,

Similar inequalities hold for the maps dfnifm(x) for each n, m Z. Compar-ing this with the inequalities (6.12) applied to the point fm(x) and using(6.13) we conclude that if m 0, then

D(fm(x), ) D(x, )2e2m. (6.15)Similar arguments show that if m 0, then

D(fm(x), ) D(x, )2e2m. (6.16)It follows from (6.15) and (6.16) that if m Z, then

D(fm(x), ) D(x, )2e2|m|.This completes the proof of the lemma.


We now proceed with the proof of the theorem. Replacing in (6.11) mby m and x by fm(x) we obtain

D(fm(x), ) D(x, )e|m|. (6.17)

Consider two disjoint subsets 1, 2 [1, s] N and setL1(x) =

i1

Ei(x), L2(x) =i2

Ei(x)

and 12(x) = (L1(x), L2(x)). By Theorem 6.1 the function 12 istempered and hence, in view of Lemma 6.5 one can find a function K12(x)satisfying Condition (L7) such that

12(x) K12(x).Set

T (x, ) = minK12(x),

where the minimum is taken over all pairs of disjoint subsets 1, 2 [1, s]N. The function T (x, ) satisfies Condition (L7).

Let v Es(x). Write v =ki=1 vi where vi Ei(x). We havev

ki=1

vi LT1(x, )v,

where L > 1 is a constant. Let us set C (x, ) = LD(x, )T (x, )1. It fol-lows from (6.11) and (6.17) that the function C (x, ) satisfies the conditionof Lemma 6.5 with

M1(x, ) =2

LD(x, ) and M2(x, ) = LD(x, )

2T (x, )1.

Therefore, there exists a function C1(x, ) C (x, ) for which the state-ments of Lemma 6.5 hold.

Applying the above arguments to the inverse map f1 and the subspaceEu(x) one can construct a function C2(x, ) for which the statements ofLemma 6.5 hold. The desired function C(x, ) can now be defined by

C(x, ) = max{C1(x, /2), C2(x, /2)}.This completes the proof of the theorem.

7. Nonuniform Hyperbolicity. Regular Sets

Let f : M M be a diffeomorphism of a compact smooth RiemannianmanifoldM . A measurable f -invariant subset R M is called nonuniformlyhyperbolic if there exist: (a) numbers and such that 0 < < 1 < ; (b)a number and real functions C, K : R (0,); (c) subspaces Es(x) andEu(x) for each x R, which satisfy the following conditions:

H1. the subspaces Es(x) and Eu(x) depend measurably on x and forman invariant splitting of the tangent space, i.e.,

TxM = Es(x) Eu(x), dxfEs(x) = Es(f(x)), dxfEu(x) = Eu(f(x));


H2. the subspace Es(x) is stable: if v Es(x), m Z, and n > 0, thendfm(x)fnv C(fm(x))nenv;

H3. the subspace Eu(x) is unstable: if v Eu(x), m Z, and n < 0,then

dfm(x)fnv C(fm(x))ne|n|v;H4. for n Z,

(Es(fn(x)), Eu(fn(x))) K(fn(x));H5. for m, n Z,C(fm+n(x)) C(fm(x))e|n|, K(fm+n(x)) K(fm(x))e|n|.

We can summarize the discussion in the previous section by saying that forany hyperbolic measure the set of Lyapunov regular points with nonzeroLyapunov exponents contains a nonuniformly hyperbolic set of full -mea-sure with

= ek , = e

k+1 , C(x) = C(x, ), K(x) = K(x, )

for any fixed 0 < 0 (see Conditions (L1)-(L7) in Section 6). In fact,finding trajectories with nonzero Lyapunov exponents seems to be a univer-sal approach in establishing nonuniform hyperbolicity.

We emphasize that the set of points (trajectories) with nonzero Lya-punov exponents whose regularity coefficient is sufficiently small (but maynot necessarily be zero) is nonuniformly hyperbolic for some > 0.

We now provide a more detailed description of a nonuniformly hyperbolicset R. Fix > 0. Given > 0, we introduce the regular set (of level ) by

R =

{x R : C(x, ) , K(x, ) 1

}. (7.1)

Regular sets can be viewed as noninvariant uniformly hyperbolic sets for thediffeomorphism f . They have the following basic properties:

R1. R R+1;R2. if m Z, then fm(R) R , where = exp(|m|);R3. the subspaces Es(x) and Eu(x) depend continuously on x R;

moreover, they depend Holder continuously on x R (see Appen-dix A by Brin). This means that

d(Es(x), Es(y)) C(x, y) and d(Eu(x), Eu(y)) C(x, y),where C > 0 and (0, 1] are constants, and d is the distancein the Grassmannian bundle of TM generated by the Riemannianmetric.

We consider the sets Q = R that are the closures of the sets R.Set Q =

1 Q

. It is easy to see that Q Q+1 and that the set Q isf -invariant.

Given a point x Q choose a sequence of points xn R which con-verges to x. Passing to a subsequence, we may assume that the sequences


of subspaces Es(xn) and Eu(xn) converge to some subspaces at x which we

denote by Es(x) and Eu(x) respectively. It is easy to see that they satisfythe following properties:

R4. TxM = Es(x) Eu(x);

R5. if v Es(x) and n > 0, thendxfnv nenv;

R6. if v Eu(x) and n < 0, thendxfnv ne|n|v;

R7. (Es(x), Eu(x)) 1 .This implies that the subspaces Es(x) and Eu(x) are uniquely defined (inparticular, they do not depend on the choice of the sequence xn x). Theydepend continuously on x Q (and indeed, Holder continuously). Further-more, the subspaces Es(x) and Eu(x), for x Q, determine a nonuniformlyhyperbolic structure on the set Q with C(x, ) = and K(x, ) = 1 if

x Q \ Q1.We now consider a smooth flow t on a compact smooth Riemannian

manifold M which is generated by a vector field X(x). A measurable t-invariant subset R M is called nonuniformly hyperbolic if there exist:(a) numbers and such that 0 < < 1 < ; (b) real functions C,K : R (0, 1) (0,); (c) subspaces Es(x) and Eu(x) for each x R,which satisfy Conditions (H2)-(H5) and the following condition:

H1. the subspaces Es(x) and Eu(x) depend measurably on x and to-gether with the subspace E0(x) = {X(x) : R} form an invari-ant splitting of the tangent space, i.e.,

TxM = Es(x) Eu(x) E0(x),

with

dxfEs(x) = Es(f(x)) and dxfE

u(x) = Eu(f(x)).

One can extend the notion of regular set to flows on nonuniformly hyperbolicsubsets.

We say that a dynamical system (with discrete or continuous time) isnonuniformly hyperbolic if it possesses an invariant nonuniformly hyperbolicsubset.

Remark 7.1. One can generalize the notion of nonuniformly hyper-bolicity from complete to partial. More precisely, we say that a set R isnonuniformly partially hyperbolic if there exist: (a) numbers and suchthat 0 < < and min{, 1} < 1; (b) real functions C, K : R (0, 1)(0,); (c) subspaces Es(x) and Eu(x) for each x R, which satisfy Condi-tions (H1)(H5). Let us emphasize that in the case of partial hyperbolicitythe vectors in the unstable subspaces Eu(x) may indeed contract and havenegative Lyapunov exponents since the number is not necessarily greater


than one. In these lectures we consider only the case of complete hyper-bolicity although many results can be generalized (sometimes literally) tononuniformly partially hyperbolic sets.

One can further generalize nonuniform (complete or partial) hyperbolic-ity by requiring that and be f -invariant measurable functions : R Rand : R R such that 0 < (x) < 1 < (x) for each x R (in the case ofpartial hyperbolicity one should assume that instead 0 < (x) < (x)) andmin{(x), (x)1} < 1 for each x R). In other words, the rates (x) and(x) may vary from trajectory to trajectory. In fact, one can easily reducethis more general case to the previous one by considering the (invariant)sets of points x where (x) and (x) for fixed constants and .

Remark 7.2. Let be a hyperbolic measure for a diffeomorphism f andR the set of Lyapunov regular points with nonzero Lyapunov exponents.The regular sets R consist of Lyapunov regular points which satisfy (7.1)while the sets Q may contain some nonregular points with sufficiently smallregularity coefficient (compare to (4.13) and (4.14)).

8. Examples of Nonuniformly Hyperbolic Systems

We present some examples of dynamical systems with continuous timewhich are nonuniformly hyperbolic. The first such example was constructedin [26] by a surgery of an Anosov flow.

8.1. Let t be an Anosov flow on a compact 3-dimensional manifoldMwhich is defined by a vector field X and preserves a smooth ergodic measure. Fix a point p0 M . One can introduce a coordinate system x, y, z ina ball Bd(p0) (for some d > 0) such that p0 is the origin (i.e., p0 = 0) andX = /z.

For each > 0, let T = S1 D Bd(0) be the solid torus obtained byrotating the disk

D = {(x, y, z) Bd(0) : x = 0 and (y d/2)2 + z2 (d)2}around the z-axis. Every point on the solid torus can be represented as apair (, y, z) with S1 and (y, z) D.

For every 0 2, we consider the cross-section of the solid torus = {(, y, z) : = }.

Let X be a smooth vector field on M \ T and t the flow generated byX. One can construct a vector field X such that:

1. X|(M \ T2) = X|(M \ T2);2. for any 0 , 2, the vector field X| is the image of the

vector field X| under the rotation around the z-axis that moves onto ;

3. for every 0 2, the unique two fixed points of the flow t|are those in the intersection of with the hyperplanes z = d;


Figure 1. A cross-section of the solid torus and the flow t

4. for every 0 2 and (y, z) D2 \ intD, the trajectory ofthe flow | passing through the point (y, z) is invariant under thesymmetry (, y, z) 7 (, y,z);

5. the flow t| preserves the measure that is the conditional mea-sure generated by the measure on the set (see [5] for details).

See Figure 1. One can see that the orbits of the flows t and t coincideon M \ T2, that the flow t preserves the measure , and that the onlyfixed points of this flow are those on the circles {(, y, z) : z = d} and{(, y, z) : z = d}.

On the set T2 \ intT we introduce natural coordinates 1, 2, r with0 1, 2 < 2 and d r 2d such that the set of fixed points of t iscomposed of those for which r = d, and 1 = 0 or 1 = .

Consider the flow on T2 \ intT defined by(1, 2, r, t) 7 (1, 2 + [2 r/(d)]4t cos 1, r),

and let X be the corresponding vector field. Consider now the flow t onM \ intT generated by the vector field Y on M \ intT defined by

Y (x) =

{X(x), x M \ intT2X(x) + X(x), x intT2 \ intT

.

Proposition 8.1. The following properties hold:

1. The flow t preserves the measure and is ergodic.2. The flow t has no fixed points.


3. The flow t is nonuniformly hyperbolic (but not uniformly hyper-bolic). Moreover, for -almost every x M \ T2,

(x, v) < 0 if v Es(x), and (x, v) > 0 if v Eu(x),where Eu(x) and Es(x) are respectively stable and unstable subspacesof the flow t at the point x.

Proof. By the construction of t, it preserves . Since the vector fields

X and X commute the flow is ergodic. Therefore, the first statement holds.The second statement follows from the construction of the flow t. In orderto prove the third statement consider the function

T (x, t) =

t0IT2(x) d,

where IT2 denotes the indicator (the characteristic function) of the set T2.By the Birkhoff Ergodic Theorem,

limt+

T (x, t)

t= (T2)

for -almost every x M .Fix a point x M \ T2. Consider a moment of time t1 at which the

trajectory tx enters the set T2 and the next moment of time t2 at whichthis trajectory exits the set T2. Given a vector v Eu(x) denote by vi theorthogonal projection of the vector dxtiv onto the (x, y) plane for i = 1, 2.It follows from the construction of the flows t and t (see Property 4) thatv1 v2. Since the unstable subspaces Eu(x) depend continuously on xthere exists K 1 (independent of x, t1, and t2) such that

dxtv KdxtT (x,t)v.It follows that for -almost every x M \ T2 and v Eu(x),

(x, v) = lim supt+

1

tlog dxtv (1 (T2)) lim sup

t+

1

tlog dxtv > 0.

provided is sufficiently small. Repeating the above argument with respectto the inverse flow t one can show that (x, v) < 0 for -almost everyx M \ T2 and v Es(x).

Set M1 =M \T and consider a copy (M1, t) of the flow (M1, t). Onecan glue the manifolds M1 and M1 along their boundaries T and obtain athree-dimensional smooth Riemannian manifold D without boundary. Wedefine a flow Ft on D by

Ftx =

{tx, x M1tx, x M1

.

It is clear that the flow Ft is nonuniformly hyperbolic and preserves themeasure .


8.2. Our next example is geodesic flows on compact smooth Riemann-ian manifolds of nonpositive curvature. Let M be a compact smooth p-dimensional Riemannian manifold. We assume that for any x M and anytwo vectors v1, v2 TxM the sectional curvature Kx(v1, v2) satisfies

Kx(v1, v2) 0. (8.1)We then say that M has nonpositive curvature.

The geodesic flow gt acts on the tangent bundle TM by the formula

gtv = v(t),

where v(t) is the unit tangent vector along the geodesic v(t) defined bythe vector v (i.e., such that v(0) = v; this geodesic is uniquely defined).The geodesic flow generates a vector field V on TM given by

V (v) =d(gtv)

dt

t=0

.

Since M is compact the flow gt is well-defined for all t R.We endow the second tangent space T (TM) with a special Riemannian

metric. Let : TM M be the natural projection (i.e., (x, v) = x for eachx M and each v TxM) and K : T (TM) TM the linear (connection)operator defined by K = (Z)(t)|t=0, where Z(t) is any curve in TM suchthat Z(0) = d, ddtZ(t)|t=0 = and is the covariant derivative. Thecanonical metric on T (TM) is given by

, v = dv, dvv + K,Kv.The set SM TM of the unit vectors is invariant with respect to the

geodesic flow, and is a compact manifold of dimension 2p1. In what followswe consider the geodesic flow restricted to the subset SM .

The study of hyperbolic properties of the geodesic flow is based uponthe description of solutions of the variational equation (6.2) for the flow.One can show that this equation along a given trajectory gtv of the flow isthe Jacobi equation along the geodesic v(t):

Y (t) +RXYX(t) = 0. (8.2)

Here Y (t) is a vector field along v(t), X(t) = (t), and RXY is the curvaturetensor. More precisely, the relation between the variational equations (6.2)and the Jacobi equation (8.2) can be described as follows. Fix a vectorv SM and an element TvSM . Let Y(t) be the unique solution ofEquation (8.2) satisfying the initial conditions Y(0) = dv and Y

(0) =

K. One can show that the map 7 Y(t) is an isomorphism for whichdgtvdvgt = Y(t) and Kdvgt = Y

(t) (see [9]). This map establishes

the identification between solutions of the variational equation (6.2) andsolutions of the Jacobi equation (8.2).

Recall that the Fermi coordinates {ei(t)}, for i = 1, . . ., p along thegeodesic v(t) are obtained by the time t parallel translation along v(t) of


an orthonormal basis {ei(0)} in Tv(0)M where e1(t) = (t). Using thesecoordinates we can rewrite Equation (8.2) in the matrix form

d2

dt2A(t) +K(t)A(t) = 0, (8.3)

where A(t) = (aij(t)) and K(t) = (kij(t)) are matrix functions, with entrieskij(t) = Kv(t)(ei(t), ej(t)). The boundary value problem for Equation (8.3)has a unique solution, i.e., for any numbers s1, s2 and any matrices A1, A2there exists a unique solution A(t) of Equation (8.3) satisfying A(s1) = A1and A(s2) = A2.

Proposition 8.2 (see [9, 29]). Given s R, let As(t) be the uniquesolution of Equation (8.3) satisfying the boundary conditions: As(0) = Id(where Id is the identity matrix) and As(s) = 0. Then there exists a limit

limt

d

dtAs(t)

t=0

= A+.

We define the positive limit solution A+(t) of Equation (8.3) as thesolution that satisfies the initial conditions:

A+(0) = Id andd

dtA+(t)

t=0

= A+.

One can show (see [9, 29]) that this solution is nondegenerate (i.e., thatdetA+(t) 6= 0 for every t R) and that A+(t) = lims+As(t). Moreover,it can be written in the form

A+(t) = C(t)

t

C(s)1(C(s)1)ds,

where C(t) is the solution of Equation (8.3) satisfying the initial conditions

C(0) = 0 andd

dtC(t)

t=0

= Id .

Similarly, letting s we can define the negative limit solution A(t)of Equation (8.3).

For every v SM let us setE+(v) = { TvSM : , V (v) = 0 and Y(t) = A+(t)dv},E(v) = { TvSM : , V (v) = 0 and Y(t) = A(t)dv},

where V is the vector field generated by the geodesic flow.

Proposition 8.3 (see [9]). The sets E(v) and E+(v) are subspaces ofTvSM and satisfy the following properties:

1. dimE(v) = dimE+(v) = p 1;2. dvE

(v) = dvE+(v) = {w TvM : w is orthogonal to v};

3. the subspaces E(v) and E+(v) are invariant under the differentialdvgt, i.e., dvgtE

(v) = E(gtv) and dvgtE+(v) = E+(gtv);

4. if : SM SM is the involution defined by v = v, thenE+(v) = dvE(v) and E(v) = dvE+(v);


5. if Kx(v1, v2) a2 for some a > 0 and all x M , then K adv for every E+(v) and every E(v);

6. if E+(v) or E(v), then Y(t) 6= 0 for every t R;7. E+(v) (respectively, E(v)) if and only if

, V (v) = 0 and dgtvdvgt cfor every t > 0 (respectively, t < 0), for some constant c > 0;

8. if E+(v) (respectively, E(v)) then the function t 7 Y(t)is nonincreasing (respectively, nondecreasing).

In view of Properties 5 and 7 we have E+(v) (respectively, E(v)) if and only if , V (v) = 0 and dvgt c for t > 0 (respectively,t < 0), for some constant c > 0. This observation and Property 3 justify tocall E+(v) and E(v) the stable and unstable subspaces.

In general, the subspaces E(v) and E+(v) do not span the whole secondtangent space TvSM . If they do span TvSM for every v SM , then thegeodesic flow is Anosov (see [9]). This is the case when the curvature isstrictly negative. However, for a general manifold of nonpositive curvatureone can only expect that the geodesic flow is nonuniformly hyperbolic. Tosee this consider the set

=

{v SM : lim sup

t

1

t

t0Kgsv(v, w) ds < 0

for every w SM orthogonal to v}. (8.4)

It is easy to see that the set is measurable and invariant under the flow gt.

Theorem 8.4 (see [28, 29]). If the Riemannian manifold M has non-positive curvature then for every v , we have (v, ) < 0 if E+(v)and (v, ) > 0 if E(v).

Proof. Let : R+ R be a continuous function. Set

= lim supt

1

t

t0(s) ds, = lim inf

t

1

t

t0(s) ds,

= lim inft

1

t

t0(s)2 ds.

We need the following lemma.

Lemma 8.5. Assume that c = supt0 |(t)| 0 then < 0;2. if (t) 0 for all t 0 and > 0 then > 0.Proof of the lemma. Assume that (t) 0. Then 0. On the

other hand, if c > 0 then

c=

c (c

)=

c2> 0.


This implies that < 0 and completes the proof of the first statement. Theproof of the second statement is similar.

We proceed with the proof of the theorem. Fix v , E+(v), andconsider the function (t) = 12Y(t)2. Using (8.3) we obtain

d2

dt2(t) =

1

2

d2

dt2Y, Y = K(t)(t) + Y (t)2,

where K(t) = Kv(t)(Y(t), w(t)) and w(t) = v(t). It follows from Proposi-

tion 8.3 that (t) 6= 0, and ddt(t) 0 for all t 0. Set

z(t) = ((t))1d

dt(t).

It is easy to check that the function z(t) satisfies the Ricatti equation

d

dtz(t) + z(t)2 ((t))1Y (t)2 +K(t) = 0. (8.5)

By Proposition 8.3, ddt(t) = 12

ddtY, Y = |Y, Y |

= |dgtvdvgt,Kdvgt| adgtvdvgt2 = 2a(t).It follows that supt0 |z(t)| 2a. Integrating the Ricatti equation (8.5) onthe interval [0, t] we obtain that

z(t) z(0) + t0z(s)2 ds =

t0((s))1Y (s)2 ds

t0K(s) ds.

It follows that for v (see (8.4)) we have

lim inft

1

t

t0z(s)2 ds lim inf

t

1

t

t0((s))1Y (s)2 ds

lim supt

1

t

t0K(s) ds > 0.

Therefore, in view of Lemma 8.5 we conclude that

lim supt

1

t

t0z(s) ds < 0.

On the other hand, using Proposition 8.3 we find that

(v, ) = lim supt

1

tlog dvgt = lim sup

t

1

tlog dgtvdvgt

= lim supt

1

tlog Y(t) = 1

2lim supt

1

t

t0z(s) ds.

This completes the proof of the first statement of the theorem. The secondstatement can be proved in a similar way.


The geodesic flow preserves the Liouville measure on the tangentbundle which is induced by the Riemannian metric. We denote by theLebesgue measure on M . It follows from Theorem 8.4 that if the set has positive Liouville measure then the geodesic flow gt| is nonuniformlyhyperbolic. It is, therefore, crucial to find conditions which would guaranteethat has positive Liouville measure.

We first consider the two-dimensional case.

Theorem 8.6. Let M be a smooth compact surface of nonpositive cur-vature K(x) and genus greater than 1. Then () > 0.

Proof. By the GaussBonnet formula the Euler characteristic of M is12

M K(x) d(x). It follows from the condition of the theorem that

MK(x) d(x) < 0. (8.6)

Choose two orthogonal vectors v, w SM . It is easy to see that

lim supt

1

t

t0K(gsv)(v, w) ds = lim sup

t

1

t

t0K((gsv)) ds.

By the Birkhoff ergodic theorem we obtain that for -almost every v SMthere exists the limit

limt

1

t

t0K((gsv)) ds = (v)

and that SM

(v) d(v) =

MK(x) d(x).

The desired result follows from (8.6).

In the multidimensional case one can establish the following criterion forpositivity of the Liouville measure of the set .

Theorem 8.7. Let M be a smooth compact Riemannian manifold ofnonpositive curvature. Assume that there exist x M and a vector v SxMsuch that

Kx(v, w) < 0 (8.7)

for any vector w SxM which is orthogonal to v. Then () > 0.Proof. Condition (8.7) holds in a small open neighborhood of (x, v).

The desired result now follows from the Birkhoff ergodic theorem.

We remark that Condition (8.7) is a multidimensional version of (8.6).It is easy to see that the set is everywhere dense. One can also show thatit is open (see Theorem 16.4 below).


9. Existence of Local Stable Manifolds

Consider a nonuniformly (partially) hyperbolic f -invariant set R, andthe associated sets R defined by (7.1). According to the results in Section 7we may replace everywhere in the sequel the sets R by the compact sets

R.The hyperbolicity conditions allow one to describe the asymptotic be-

havior of trajectories which start in a small neighborhood of a hyperbolictrajectory. More precisely, applying Theorem 4.4 one can construct at everypoint x R local stable and unstable manifolds V s(x) and V u(x) such thatevery trajectory fn(y) with y V s(x) approaches fn(x) with an exponen-tial rate, and every trajectory fn(y) with y V u(x) approaches fn(x)with an exponential rate. Roughly speaking, the behavior of trajectoriesthat start in a neighborhood of x resembles that of the trajectories in aneighborhood of a fixed (or periodic) hyperbolic point. However, the rateof exponential convergence may vary from orbit to orbit and may also getworse along the orbit but with subexponential rate (due to Condition (H5)in Section 7).

A more precise description is given by the following theorem which isone of the key results in the hyperbolic theory.

Theorem 9.1 (Stable Manifold Theorem). Let R be a nonuniformlyhyperbolic set for a C1+ diffeomorphism f . Then for every x R thereexists a local stable manifold V s(x) such that x V s(x), TxV s(x) = Es(x),and if y V s(x) and n 0 then

(fn(x), fn(y)) T (x)nen(x, y), (9.1)where is the distance induced in M by the Riemannian metric and T : R (0,) is a Borel function such that if m Z then

T (fm(x)) T (x)e10|m|. (9.2)Inequality (9.2) should be compared to Condition (H5) in Section 7.The Stable Manifold Theorem was first established by Pesin in [27]. His

approach was an elaboration of the classical work of Perron. This approachwas extended by Katok and Strelcyn in [15] to smooth maps with singular-ities (they include billiard systems and other physical models). Ruel

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