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I have been trying to understand the proof of the following result, which is considered well-known.

Theorem: Fix a compact metric space $X$, a homeomorphism $T:X \to X$, and a continuous map $ A : X \to \mathrm{SL}_2(\mathbb{R}) $. Define the skew-product $$ (T,A): X \times \mathbb{R}^2 \to X \times \mathbb{R}^2, \quad (x,v) \mapsto (Tx, A(x)v), $$ and set $ A_n(x) = A(T^{n-1}x) \cdots A(x) $ for $x \in X, n > 0$ and similarly for $n \leq 0 $ so that $ (T,A)^n = (T^n,A_n) $. If there are uniform constants $ C > 0, \lambda > 1 $ such that $$ \| A_n(x) \| \geq C \lambda^{|n|} $$ for all $n,x$, then the cocycle $(T,A)$ is uniformly hyperbolic. More precisely, there are continuous maps $ \Lambda^s,\Lambda^u: X \to \mathbb{RP}^1 $ and constants $ c >0 , L>1$ such that $$ A(x) \Lambda^{\bullet}(x) = \Lambda^{\bullet}(Tx), \quad \bullet \in \{ s,u \} $$ and $$ \| A_n(x) v_s \| \leq cL^{-n} \|v_s\|, \quad \| A_{-n}(x) v_u \| \leq cL^{-n} \| v_u \| $$ for all $n \geq 0, x \in X, v_s \in \Lambda^s(x), v_u \in \Lambda^u(x)$.

The consruction of $\Lambda^s$ goes like this: Given $ x \in X, n >0 $, let $ \Lambda_n^s(x) $ be the most contracted subspace of $ A_n(x) $ (i.e. the eigenspace of $ A_n(x)^* A_n(x) $ corresponding to the eigenvalue $ \|A_n(x) \|^{-2} $). Of course, one needs $ \| A_n(x) \| > 1 $ for this subspace to be one-dimensional, but the growth condition assures us that this happens for sufficiently large $n$.

One then proves readily that there are constants $C_0,C_1$ independent of $x$ and $n$ such that the angles between these singular subspaces obey $$ \angle \left( \Lambda_n^s(x), \Lambda_{n+1}^s(x) \right) \leq C_0 \| A_n(x) \|^{-2} \leq C_1 \lambda^{-2n}. $$ In particular, $ \Lambda_n^s(\cdot) $ converges (uniformly) to a limiting map $ \Lambda^s(\cdot) $. Continuity of $\Lambda^s$ is immediate, and the $A$-invariance condition follows from a straightforward calculation.

Now, here is the part of the proof with which I am having difficulties - the exponential decay estimates. Pick a unit vector $v_s \in \Lambda^s(x)$, and let $ \theta_n = \theta_n(x) $ denote the (smallest nonnegative) angle between $\Lambda_n^s(x)$ and $\Lambda^s(x)$. One can check that $$ \| A_n(x) v_s \|^2 = \| A_n(x) \|^{-2} \cos^2(\theta_n) + \| A_n(x) \|^{2} \sin^2(\theta_n) $$ (simply decompose $v_s$ in an orthonormal basis consisting of a unit vector from $\Lambda_n^s(x)$ and a unit vector from $\Lambda_n^u(x)$). We want to see that this decays exponentially. That the first term on the RHS decays exponentially is obvious, but the second term is bothersome. The factor $ \| A_n(x) \|^2 $ grows exponentially, and the factor $ \sin^2(\theta_n) $ decays exponentially, but it is not clear to me why the exponential decay of $\sin^2(\theta_n)$ should necessarily ``win'' and produce a net result of exponential decay.

I have a nice reference for this result, namely Yoccoz' article ``Some questions and remarks about $ \mathrm{SL}_2(\mathbb{R}) $ cocycles.'' Unfortunately for me, the decay estimate I want to understand is referred to as something easily checked, so I am likely missing something very obvious. I would be very grateful for any helpful remarks.

EDIT: Due to lack of interest, I have cross-posted this question on math stack exchange: https://math.stackexchange.com/questions/467478/uniform-hyperbolicity-decay-estimate

EDIT (Inspired by A. Blumenthal's comment on Math Stack Exchange): The bound on the angle between $ \Lambda_n $ and $\Lambda_{n+1}$ implies that $$ \theta_n(x) \leq C_0 \sum_{m=n}^\infty \| A_m(x) \|^{-2}, $$ so it would be enough to prove that $$ \| A_n(x) \|^2 \left( \sum_{m=n}^\infty \| A_m(x) \|^{-2} \right)^2, $$ decays exponentially. If we define $ a_n(x) = \| A_n(x) \| $ and $ B = \sup_{x\in X} \|A(x) \| $, then we have a sequence of functions $ a_n:X \to \mathbb{R}_{\geq 0} $ with the following properties:

$\bullet$ $ C \lambda^n \leq a_n(x) \leq B^n $

$\bullet$ $ B^{-1} \cdot a_n(x) \leq a_{n+1}(x) \leq B\cdot a_n(x) $

$\bullet$ $ B^{-2} \cdot a_n(x) \leq a_n(Tx) \leq B^2 \cdot a_n(x) $

for all $x \in X$, $n \geq 0$.

We would then like to show that $$ a_n(x) \sum_{m=n}^\infty a_m(x)^{-2} $$ decays exponentially (uniformly in $x$). This looks more promising, but the desired estimate remains elusive.

In particular, if the sequence $a_n(x)^{-2}$ decreases monotonically, or if $ B<\lambda^2 $, the desired estimate is obvious. However, if $B $ is much larger than $\lambda^2$ and the sequence is wildly non-monotonic, then the waters remain murky to me.

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    $\begingroup$ It is an interesting question. Your argument actually covers most interesting cases. $\endgroup$ – Pengfei Aug 15 '13 at 3:09
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I nearly asked this exact question earlier this year, after coming to the same series and being confounded by the problems which arise when the decay of $\|A_m(x)\|^{-2}$ is too irregular. I discussed this result this summer with an expert in the field and we agreed that behind Yoccoz's "easily checked" lurks perhaps half of the entire proof.

You can find a complete proof of this result in Bochi and Gourmelon's article "Some characterizations of domination" where it arises as a special case of a more general theorem: if a cocycle taking values in $GL_d(\mathbb{R})$ has the property that the $k^{\mathrm{th}}$ singular value of $A_m(x)$ grows faster than the $(k+1)^{\mathrm{st}}$ singular value at an exponential rate which is uniform in $x$ and $m$, then the cocycle has a continuous splitting into a $k$-dimensional invariant bundle and a $(d-k)$-dimensional invariant bundle, and the growth rate in the top bundle always dominates the growth in the bottom bundle along the same orbit by a uniform exponential factor. To overcome the problem of the series they use ergodic theory. Once an ergodic measure is chosen, the series is well-behaved because there is a well-defined exponential growth rate at almost every $x$, and we obtain the desired growth rate almost everywhere. This can be done for every ergodic measure, and we obtain an upper bound on the asymptotic exponential growth rate of $\|A_m(x)s(x)\|$ which is uniform across all measures. By appealing to abstract results such as Schreiber's theorem ("On growth rates of subadditive functions for semiflows", Journal of Differential Equations 1998) one can obtain an exponential growth estimate which is uniform across all $x$.

I believe that it is possible to prove the result without ergodic theory, but this gets quite involved.

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    $\begingroup$ This is a very interesting approach. I had thought about using ergodic theory, but shied away from such a tactic exactly because I didn't think that one could do anything with "bad x's" that didn't live in the full-measure sets guaranteed by the multiplicative ergodic theorem. I do have a follow-up question: In Bochi-Gourmelon's proof of the result, they make reference to a "Krylov-Bogoliubov" argument. Do you know to what they are referring? The only KB argument with which I am familiar is the standard one that people use to prove that invariant measures exist. $\endgroup$ – Jake Fillman Aug 24 '13 at 1:14
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    $\begingroup$ At this step they use the following lemma: if $T \colon X \to X$ is a continuous transformation of a compact metric space, $f \colon X \to \mathbb{R}$ is continuous, and $\int f\,d\mu \leq \lambda$ for all invariant measures $\lambda$, then $\lim_{n \to \infty} \sup_{x \in X}(1/n)\sum_{k=0}^{n-1}f(T^kx) \leq \lambda$. This lemma is proved in the same way as uniform convergence in the ergodic theorem is proved for uniquely ergodic transformations, see e.g. Walters: that is, if the conclusion is false then we can find long orbit stretches with large average and use them to build a limit measure. $\endgroup$ – Ian Morris Aug 24 '13 at 20:18
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    $\begingroup$ To be more concrete: if the conclusion is false then there exist $\varepsilon>0$, a sequence $(x_i)$ in $X$, and a subsequence $(n_i)\to \infty$ such that $\sum_{k=0}^{n_i-1} f(T^kx_i) \geq n_i(\lambda+\varepsilon)$ for all $i$. Let $\mu$ be a weak-* accumulation point of the sequence of measures $m_i:=(1/n_i)\sum_{k=0}^{n_i-1}\delta_{T^kx_i}$, then $\mu$ is invariant and $\int f\,d\mu\geq \lambda+\varepsilon$, a contradiction. Bochi and Gourmelon mean that this argument resembles the typical textbook proof of Krylov-Bogolioubov. For a powerful general version of this lemma see Schreiber. $\endgroup$ – Ian Morris Aug 24 '13 at 20:22

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