THE MULTIPLICATIVE ERGODIC THEOREM OF OSELEDETS

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1 THE MULTIPLICATIVE ERGODIC THEOREM OF OSELEDETS. STATEMENT Let (X, µ, A) be a probability space, and let T : X X be an ergodic measure-preserving transformation. Given a measurable map A : X GL(d, R), we can define a skewproduct F : X R d X R d by (x, v) (T(x), A(x) v). Write F n (x) = (T n (x), A (n) (x) v). The map F is called a linear cocycle. Fix some norm on R d, and use the same symbol to indicate the induced operator norm. Theorem (Multiplicative Ergodic Theorem of Oseledets [O]). Let T be an invertible bimeasurable ergodic transformation of the probability space (X, µ). Let A : X GL(d, R) be measurable such that log + A and log + A are integrable. Then there exist numbers () λ > > λ k and for µ-almost every x X, there exist a splitting (2) R d = E x E k x such that for every i k we have A(x) E i x = E i T(x) and (3) n ± n log A(n) (x) v i = λ i for all non-zero v i E i x. The subspaces E i x are unique µ-almost everywhere, and they depend measurably on x. The numbers () are called Lyapunov exponents, and the splitting (2) is called the Lyapunov splitting. Invariance of the bundles E i x implies that their dimensions d i are constant a.e. Measurability means the following: there exist measurable maps u,..., u d : X R d such that for each j =,..., k, the space E i x is spanned by the vectors u i (x) with d + + d i < i d + + d i. Exercise i. If T is not assumed to be ergodic then a similar result still holds. The Lyapunov exponents (and their number) become measurable T-invariant a.e. defined functions. State and prove this assuming Theorem and using ergodic decomposition.

2 2 OSELEDETS THEOREM Exercise ii. Use the last exercise, state and prove a version of Oseledets Theorem for diffemorphisms of a compact manifold. In the next section, we explain some additional conclusions that could be added to Oseledets Theorem, and also state some results that will be need later. The proofs will take the rest of the note. Some exercises are given along the way. Those that are need for the proof of Theorem are marked with the symbol. 2. ADDITIONAL INFORMATION 2.. One-sided version. There is a version of Theorem for non- (necessarily)-invertible transformations: Theorem 2. Let T be an ergodic transformation of the probability space (X, µ). Let A : X GL(d, R) be measurable such that log + A and log + A are integrable. Then there exist numbers λ > > λ k and for µ-almost every x X, there exist a flag (4) R d = V x V 2 x V k x V k+ x = {0} depending measurably on the point, and invariant by the cocycle, such that (5) n n log A(n) (x) v i = λ i for all v i Vx i Vx i+. We call (4) the Lyapunov flag of the cocycle. Exercise iii. In the case T is invertible, the flag is given in terms of the Lyapunov splitting by V i x = E x E i x Subadditivity. We will use the following result (see e.g. [L] for a proof): Theorem 3 (Kingman s Subadditive Ergodic Theorem). Assume T is ergodic. If (φ n ) n=,2,... is a sequence of functions such that φ + is integrable and φ m+n φ m + φ n T m for all m, n. Then φ n n converges a.e. to some c R { }. Moreover, c = inf n φn. n An immediate corollary of Theorem 3 in the setting of cocycles is the existence of the it n log n A(n) (x). (This is called the Furstenberg-Kesten Theorem.) It can be shown that this it is the upper Lyapunov exponent λ given by Oseledets Theorem:

3 Addendum 4. We have: λ = n + λ k = n + OSELEDETS THEOREM 3 n log A(n) (x) = n n log m(a(n) (x)) = n n log m(a(n) (x)), n log A(n) (x). Recall that the co-norm of a linear map L is m(l) = inf v = Lv ; it equals L if L is invertible Angles. The angles between the spaces of the Lyapunov splitting are sub-exponential, in the sense made precise by the following: Addendum 5. Let R d = E x E k x be the Lyapunov splitting, and assume k 2. Let J J 2 be a partition of the set of indices {,..., k} into two disjoint non-empty sets. Let F i x = j J i E j x for i =, 2. Then, for µ-a.e. x we have n ± n log sin ( F T n (x), ) F2 T n (x) = 0. We will deduce Addendum 5 directly from Oseledets Theorem. For that, we need the following result (that will also be usuful in the proof of the theorem itself): Lemma 6. Let φ : X R be a measurable function such that φ T φ is integrable in the extended sense. Then n ± (n < 0 is allowed if T is invertible.) n φ(tn x) = 0 for a.e. x X. Proof. Let ψ = φ T φ and let c = ψ [, + ]. By Birkhoff s Theorem, φ T n = φ n n + n ψ T j c a.e. n j=0 If c were non-zero then we would have φ(t n x) for a.e. x. But, by the Poincaré s Recurrence Theorem, the set of points x which satisfy the latter condition has zero measure. Therefore c = 0, as we wanted to show. If T is invertible then φ T φ is quasi-integrable as well, so the previous reasoning applies. ψ is said to be integrable in the extended sense if ψ + or ψ are integrable. Another exercise: Verify that the Birkhoff Theorem still applies.

4 4 OSELEDETS THEOREM Exercise iv. Let L : R d R d be a invertible linear map and let v, w be non-zero vectors. Then L L sin (Lv, Lw) sin (v, w) L L. Proof of Addendum 5. The function φ : X R defined by φ(x) = log sin (F x, F 2 x) is measurable. By Exercise iv, we have φ(tx) φ(x) log A(x) + log A(x), which is integrable, by the hypothesis of the Oseledets Theorem. So Lemma 6 gives φ(t n (x))/n 0 for a.e. x Conjugacy. For a fixed T : X X, we say a measurable map C : X GL(d, R) is tempered if n log n C± (T n x) = 0. We say that two cocycles A, B : X GL(d, R) are conjugate (or equivalent) if there is a tempered C : X GL(d, R) such that B(x) = C(Tx) A(x)C(x) That is, (T, B) = (C, id) (T, A) (C, id). Exercise v. If two cocycles are conjugate then they have the same Lyapunov exponents Determinants. Another addendum to Oseledets Theorem (that we will obtain along its proof) is this: Addendum 7. For a.e. x, the convergence in (3) is uniform over unit vectors in E i x. Using this, do the following: Exercise vi. Let J {,..., k} and F x = j J Ej x. Prove that n ± n log det A(n) (x) Fx = λ j dim E j. j J 3. LIMSUP OSELEDETS The proof of Theorem will take the rest of this note. It is comparatively very easy to prove Theorem 2 if its are replaced by sup s. Let λ(x, v) = sup n n log A(n) (x) v, v 0. Lemma 8. There exist numbers λ > > λ k such that for a.e. x, λ(x, ) takes only these values. For a.e. x there is a flag into linear spaces R d = V x V 2 x V k x V k+ x = {0}

5 such that OSELEDETS THEOREM 5 v V i x V i+ x λ(x, v) = λ i. Moreover, these spaces depend measurably on the point, and are invariant by the cocycle. Proof. First, the integrability of log + A ± implies that λ(x, v) is finite for a.e. x and all v 0. It is easy to see that: λ(x, av) = λ(x, v), λ(x, v + w) max ( λ(x, v), λ(x, w) ) with equality if λ(x, v) λ(x, w). It follows that for any t R, V x (t) = { v R d ; λ(x, v) t or v = 0 } is a linear subspace. Also, vectors with different λ are linearly independent, so for all x, λ(x, ) takes at most d distinct values λ (x) > > λ k(x) (x). Let Vx i = V x (λ i (x)). Exercise vii ( ). Show that everything is measurable and conclude the proof of the Lemma. 4. INTERLUDE: SOME ERGODIC THEORY OF SKEW-PRODUCTS Let P be a compact metric space. Let S : X P X P be a skewproduct map over T (this means that π S = T where π : X P X is the projection) such that S(x, ) : P P is continuous for a.e. x. Given φ : X P R, we write { ψ + ψ S + + ψ S ψ (n) n if n > 0; = ψ S ψ S 2 ψ S n if n < 0 and S is invertible. Let F be the set of measurable functions φ : X P R such that φ(x, ) is continuous for a.e. x and x X sup u P φ(x, u) belongs to L (µ). Exercise viii. Prove that the norm φ = sup u φ(x, u) dµ(x) makes F a separable Banach space. Lemma 9. Assume ψ F and c R are such that for µ-a.e. x X, sup n n ψ(n) (x, u) c for all u P.

6 6 OSELEDETS THEOREM Then the assertion remains true with the sup replaced with a inf that is uniform over P. More precisely, for µ-a.e. x X and every ε > 0 there is n 0 such that n ψ(n) (x, u) > c ε for all u P and n > n 0. Furthermore, if S is invertible then for a.e. x, inf n n ψ(n) (x, u) c uniformly in u. Corollary 0. Assume ψ F and c R are such that for µ-a.e. x X, sup n n ψ(n) (x, u) = c for all u P. Then the assertion remains true with sup replaced with uniform (or n ± if S is invertible). The proof of Lemma 9 is a Krylov Bogoliubov argument. For that, we need: Exercise ix ( ). Let M be the set of probability measures on X P that project on µ (that is, π ν = µ). There is a topology on M which makes it compact, and such that a sequence ν n converges to ν if and only if φ dν n φ dν for every φ F. Proof of Lemma 9. Let I n (x) = inf u P ψ(n) (x, u). Then I n is measurable, I is integrable, and I n+m I n T m + I m. So the Subadditive Ergodic Theorem 3 applies to I n, and thus n I n converges a.e. to a constant b. For a.e. x, let u n (x) be such that ψ (n) (x, u n (x)) = I n (x). Exercise x ( ). It is possible to choose u n measurable. Take u n measurable and define the following probability measures on X P: n ν 0 n = δ (x,un (x)) dµ(x) and ν n = S ν j 0 n. All this measures project to µ. By Exercise ix, there is a converging subsequence ν ni ν. Exercise xi ( ). ν is a S-invariant measure. 2 2 Hint: for any η F, ηd(f ν n ν n ) (2/n) η, where is the obvious norm on the Banach space F. j=0

7 OSELEDETS THEOREM 7 Moreover, ψ dν = ψ dν n = In dµ = b. By Birkhoff s n Theorem, there is a set of full ν-measure formed by points (x, u) such that the sequence n ψ(n) (x, u) converges to b. By the assumption, the sup s of those sequences are c ν-a.e. Therefore b c. Since b = n n inf u ψ (n) (x, u), we obtain the desired uniform inf. Notice that for any S-invariant probability measure ν, we have ψ dν b (by Birkhoff again). Now assume that S is invertible. Repeating the above reasoning for S, we obtain a S-invariant probability ν such that inf n u P n ψ(n) (x, u) = ψ dν. So inf n n ψ (n) (x, u) ψ dν b c uniformly with respect to u. 5. THE BUNDLE OF LEAST STRETCH Consider the exponents λ > > λ k and the flag V V k given by Lemma 8. We will focus the attention on λ k and V k. Write λ min (A) = λ k, E x = V k x We call E the bundle of least stretch for A. It has positive dimension. By definition, sup n n log A (n) v = λ min (A) for a.e. x and all non-zero v E x. Let us improve this information: Lemma. We have: n ± n log A(n) (x) v = λ min (A) for a.e. x and all v E x {0}. (n allowed if T is invertible.) Moreover the it is uniform over unit vectors. The conclusion of the lemma can be rephrased as: n ± n log A(n)(x) log m(a(n) E n ± n (x)) = λ E min(a). Proof of Lemma. Let m = dim E. By conjugating with some measurable C : X O(d, R), we can assume that E x = R m R d. Let P be the projective space of R m. Consider the induced skewproduct F : X P X P over T. Let ψ(x, v) = log A(x) v, where v v denotes the element of P corresponding to v. Then n ψ(n) (x, v) = n n j=0 log A(j+) (x) v A (j) (x) v = n log A(n) (x) v v

8 8 OSELEDETS THEOREM has a sup as n equal to λ min, for a.e. x and all v P. Thus Corollary 0 implies that the sup can be replaced with uniform. By Lemma 8, we have sup n n log A(n) (x) v λ min (A) for a.e. x and all v R d {0}. Using Lemma 9 and a similar argument as in the proof of Lemma, we conclude that this sup can by replaced by a inf that is uniform with respect to u. This can be rephrased as: On the other hand, sup n + inf n + n log m(a(n) (x)) λ min (A). n log m(a(n) (x)) sup log m(a(n) n + n (x)) λ E min(a). So we obtain n + n log m(a(n) (x)) = λ min (A). This proves one of the assertions in Addendum 4. The others are proven similarly. 6. INTERLUDE: HOW TO MAKE GOOD USE OF ASYMPTOTIC INFORMATION Let M(d) indicate the set of d d real matrices. Lemma 2. Let B : X M(d) be measurable with log + B integrable. Assume γ R is such that sup n + n B(n) (x) γ Then for any ε > 0 there exists a measurable function b ε : X R + such that a.e. B (n) (T i x) b ε (x)e (γ+ε)n+ε i for a.e. x and all n 0, i Z. (i < 0 is allowed if T is invertible.) Proof. The formula c(x) = sup e (γ+ε)n B (n) (x) n 0

9 OSELEDETS THEOREM 9 defines a measurable a.e. finite function. By definition, B (n) (x) c(x)e (γ+ε)n. Now we need to bound the expression c(t i x). We have c(tx) = sup e (γ+ε)n B (n) (Tx) n 0 B(x) sup e (γ+ε)n B (n+) (x) n 0 = e γ+ε B(x) sup e (γ+ε)n B (n) (x) n Either the rightmost sup equals c(x) or else c(x) = and then c(tx). In any case, we have and so c(tx) min ( e γ+ε B(x), ) c(x). log c T log c min ( γ + ε log + B, 0 ). is quasi-integrable. So Lemma 6 applies, giving i log c Ti 0 a.e. as i ±. We apply the sup trick a second time: Let Then b ε is the desired function. b ε (x) = sup e ε i c(t i x). i Z 7. THE QUOTIENT As before, let E be the bundle of least stretch for A. Assume the norm comes from an inner product. For each x, let E x be the orthogonal complement of E x ; this is a measurable subbundle. With respect to the splitting R d = E E, we write ( ) B 0 (6) A =. C A E So B(x) : E x E Tx gives a new cocycle. Lemma 3. λ min (B) λ k. In particular, λ min (B) > λ min (A). Proof. Notice that m(b (n) (x)) m(a (n) (x)). 3 So, by Addendum 4, λ min (B) λ min (A) = λ k. Let E E be the subbundle the bundle of least stretch for the cocicle B. The measurable bundle F = E E is A-invariant. 3 Let w be the unit vector in E x the most contracted by B (n) (x), and let v = [A (n) (x)] B (n) (x) w. By Pythagoras, v w. Hence m(a (n) (x)) A(n) (x)v B (n) (x)w w = m(b (n) (x)). v

10 0 OSELEDETS THEOREM Fix ε > 0 small. We have = B (n) 0 n E F, where C n(x) = A (n) C n A (n) E i=0 A (n i+) E (T i+ x) C(T i x) B (i) E (x). Write σ = λ min (B). Applying Lemma 2, we find measurable functions c, b, a such that for a.e. x and all n, i 0, Thus C(T i x) c(x)e εi B (n) E (x) b(x)e (ε+σ)n A (n) E (Ti x) a(x)e (ε+λ k)n+εi C n (x) n max 0 i n A(n i+) E (T i+ x) C(T i x) B (i) E (x) nd (x) max 0 i n eε(i+)+(λ k+ε)(n i ) e εi e (σ+ε)i nd 2 (x)e 4εn max 0 i n e(n i )λ k+σi nd 3 (x)e 4εn e σn. (In the last step we used that σ λ k.) So, at a.e. point, λ k sup n ( = max sup n n n log A(n) sup F n n log max ( A (n) E B(n) E C n ) log A(n), sup log B(n), sup E n n E n n log C n max(λ k, σ, σ + 4ε). So λ k σ + 4ε. As ε > 0 is arbitrary, λ k σ = λ min (B). Exercise xii. In the notation of the proof of Lemma 3, prove that n log A (n) (x) v exists and equals λ min (B) for all v F E. Also show that λ min (B) = λ k (that is, equality holds in Lemma 3) and F = V k is the penultimate space of the Lyapunov flag. Use induction to prove Theorem CONSTRUCTION OF A COMPLEMENTARY INVARIANT BUNDLE Lemma 4. There is a measurable invariant subbundle G of R d such that G E = R d. Remark. Here is the first time we need T to be invertible. Proof. Let G be any (non-necessarily invariant) measurable subbundle which is complementary to E, that is, such that G E = R d. )

11 OSELEDETS THEOREM Let L(x) : E x E x be linear maps depending measurably on x. The set of such maps is indicated by L. The graph of L, given by G x = {L(x)v + v; v E x }, is a subbundle complementary to E. The (forward) image of a subbundle G is the bundle whose fiber over x is A(T x) G T x. Let Γ : L L be the graph transform so that the image of the graph of L is the graph of ΓL. Recalling (6), we see that (7) ΓL = D + ΦL, where D L and Φ : L L are given by: D(x) = C(T x) [B(T x)], (ΦL)(x) = A E (T (x)) L(T x) [B(T x)]. To prove Lemma 4 we need to prove that Γ has a fixed point. We will prove that there is a measurable function a : X R and τ > 0 such that (8) (Φ n D)(x) a(x)e τn for a.e. x and all n 0. Then the formula L = n=0 Φ n D will define a element of L which is fixed by the graph transform (7). We have (Φ n L)(x) = (A E ) (n) (T n (x)) L(T n x) B ( n) (x), so (9) (Φ n D)(x) (A E ) (n) (T n x) D(T n x) B ( n) (x) We will bound each of the three factors using Lemma 2. Fix a small ε > 0 be small (to be specified later). For the first factor in (9) we have Secondly, (A E ) (n) (T n x) a (x)e n(λ k+2ε), for some measurable a. log + D(x) log + C(T x) + log B(T x) log + A(T x) + log A(T x). So log + D is integrable and Lemma 2 gives D(T n x) a 2 (x)e εn, for some measurable a 2. Now look to the third factor in (9). By Addendum 4 and Lemma 3, n + n log B( n) (x) = λ min (B) λ k. The function log + B log + A is integrable. Thus we can apply Lemma 2 to the cocycle B over T and conclude that B ( n) (x) a 3 (x)e ( λ k +ε)n, for some measurable a 3. Putting the three bounds together in (9) we obtain Φ n D(x) (a a 2 a 3 )(x) e ( λ k +λ k +4ε)

12 2 OSELEDETS THEOREM Taking ε small we obtain the desired (8). This proves Lemma CONCLUSION Proof of Theorem (and Addendum 7). Let V V k be given by Lemma 8. Let E k = V k. By Lemma, (3) holds for i = k (uniformly with respect to unit vectors). If k = then we are done. Otherwise, apply Lemma 4 to find an invariant subbundle G complementary to E k. Now consider the cocycle restricted to G; its Lyapunov flag is: G = V G V k G {0} with corresponding exponents λ > > λ k. Then define E k = V k G and repeat the argument. 0. COMMENTS ABOUT THE PROOF(S) There are several proofs of the Oseledets Theorem. The more conventional proofs use heavier Linear Algebra (exterior powers, singular values etc). The more conventional approach has the advantage of providing more useful information about the Lyapunov exponents and splittings: see the detailed treatment in [A], also [L]. The proof here is a combination of the proofs of [V] (which is on its turn based on [M]), and [W]. Lemmas 8 and 9 come from [W], and Lemma 4 comes from [V]. In [M, V], only the two-sided (i.e., invertible) case was considered, while [W] proves only the one-sided 4 version of the Oseledets Theorem. Unlike [V, M], we used the Subadditive Ergodic Theorem (in the proof of Lemma 9) 5. REFERENCES [A] L. Arnold. Random dynamical systems. Springer Verlag, [L] F. Ledrappier. Quelques propriétés des exposants caractéristiques. Springer Lect. Notes Math. 097 (984), [M] R. Mañé. Ergodic theory and differentiable dynamics. Springer Verlag, 987. [O] V. I. Oseledets. A multiplicative ergodic theorem: Lyapunov characteristic numbers for dynamical systems. Trans. Moscow Math. Soc. 9 (968), [V] M. Viana. A proof of Oseledets theorem. viana/out/ oseledets.pdf [W] P. Walters. A dynamical proof of the multiplicative ergodic theorem Trans. Amer. Math. Soc. 355 (993), Jairo Bochi May 27, It is not clear to me how to deduce the two-sided theorem from its one-sided version without using exterior powers. 5 Maybe this could be avoided, if desired.