An angle between two vectors in Oseledets theorem

Question

Let $f:\Sigma \to \Sigma$ be a two side shift map, where $\Sigma=\{1,2,3,4\}^{\mathbb{Z}}$ and let $A:\Sigma \to SL(2,\mathbb{R})$ be a function such that $A((x_{n}))=A_{x_{0}}$. Assume that there are two different Lyapunov exponents $-\lambda$ and $\lambda$, so there there is a Oseledets splitting $\mathbb{R}^{2}=E_{x}^{s}+E_{x}^{u}$ such that $A(x)E_{x}^{s}=E_{f(x)}^{s}$ and $A(x)E_{x}^{u}= E_{f(x)}^{u}$.

$\textbf{Question}$: $\lim_{n\to \infty}\frac{1}{n}\log | sin(<A^{n}(x)v,E_{f^{n}}^{u}(x))|=-2\lambda(x)$ for $v \notin E^{s}.$

My attempt: Let $u(x)$ and $s(x)$ be two directions that generate $E^{s}$ and $E^{u}.$

Let $\alpha_{n}=<(A^{n}(x)v,A^{n}(x)u(x)).$

So, $A^{n}(x)v=\sin(\alpha_{n}) A^{n}(x) s(x)+\cos(\alpha_{n})A^{n}(x) u(x).$

But, I can not get somewhere.

Answer 1

Ok, let take $v$ in the bundle at $x$. We may decompose $v=v^{u}+v^{s}\in E^{u}\oplus E^{s}$. Assume without loss of generality that $\lVert v\rVert=1$.

Applying $A^{i}$, using equivariance and Osceldets' theorem we get $$A^{i}v \approx e^{i\cdot\lambda}\cdot v+e^{-i\cdot\lambda}\cdot v^{s}$$. Therefore a unit vector amounting to $A^{i}v$ would more or less amount to dividing by $e^{i\cdot\lambda}$. More correctly, we get $$ v = L(A^{i}v)^{-1}\cdot e^{i\cdot\lambda}v^{u} + L(A^{i}v)^{-1}\cdot e^{-i\cdot\lambda}v^{s},$$ where $L(A^{i}v)$ is the norm of $A^{i}v$ which equals to essentially $\sqrt{e^{2i\cdot\lambda}\lVert v^{u}\rVert^{2} + e^{-i\cdot\lambda}\lVert v^{s}\rVert^{2} + 2C(i)v^{u}\cdot v^{s}}$, where $C(i)$ is some function which growth is bounded by $e^{2\epsilon}$ (note that also the exponents $2\lambda\cdot i, -2i\cdot\lambda$, are correct only up to epsilon, and there might be also some absolute constants depending on $\epsilon$ there). The important upshot is that the norm is bounded by $\tilde{C}(i)e^{i\cdot\lambda}$ where $C$ depends on $\epsilon ,v$, but definitely it is bounded from above by a constant times $e^{i\cdot \epsilon}$.

Going back to $u$ we see that $$ u = \tilde{C}(i)^{-1}\cdot v^{u} + \tilde{C}(i)^{-1}\cdot e^{-i\cdot\lambda}\cdot e^{-i\cdot\lambda}v^{s}. $$ As we are only interested in the angle with the unstable, we can just examine the $v^s$ part.
The sine of the angle is essentially $\tilde{C}(i)^{-1}\cdot e^{-2i\cdot\lambda}$. Taking logarithm and dividing by $i$, letting $i$\to $\infty$, you see that the limit equal to $-2\lambda$.