Continuity of Lyapunov spaces

Question

The following appears in the paper "Continuity properties of entropy" by Newhouse from 1989:

Let $M$ be some smooth Riemannian compact manifold (you may assume boundary-less), and let $f\in Diff^{1+\epsilon}(M)$ for some $\epsilon>0$ (it means that $f,f^{-1}$ are differentiable, and the differential is $\epsilon$-H\"older continuous).

For all $l\in\mathbb{N}$ and $\chi>0$, define

$\Lambda_{\chi,l}:=\{x\in M:$ There is a splitting $T_xM=E^s(x)\oplus E^u(x)$, and $\forall v_u\in E^u(x),v_s\in E^s(x), n\geq0$:

$|d_xf^nv_s|\leq l e^{-\chi n}|v_s|$, $|d_xf^{-n}v_s|\geq l^{-1} e^{\chi n}|v_s|$ and

$|d_xf^{-n}v_u|\leq l e^{-\chi n}|v_u|$, $|d_xf^{n}v_u|\geq l^{-1} e^{\chi n}|v_u|$ $\}$

The following line in Newhouse's paper is that it is fairly easy to see that $E^u(x)$ change continuously for $x\in \Lambda_{\chi,l}$ on each piece with $\dim E^u(x)$ constant.

I've tried showing so myself, or finding some reference online- neither yielded results. I'm not sure I understand even what continuity in this context means (I'd rather not to go into Grassmannians or tangent bundles, and keep it analytic); therefore even a reduction to the continuity of $Jac(d_xf|_{E^u(x)})$ will be greatly appreciated. It is not clear to me if the continuity is unique to the sets $\Lambda_{\chi,l}$, or is it a property for all $\chi$-hyperbolic points, or all Lyapunov regular points, etc. As well as the regularity module- could one discuss H\"older continuity, uniform continuity etc. ?

Thanks ahead for anyone contributing... I'm sorry I couldn't present some more of my own progress on this question, I'm really quite lost on this one and would really like to understand his paper.

Answer 1

This is a standard argument in Pesin Theory, which I'll sketch below leaving details to the reader. Let's pretend $M = \mathbb T^d, d \geq 2$, so that I don't have to bother with charts (the proof is local anyway and so can be transcribed verbatim onto a single chart).

Let $x, x_n \in \Lambda_{\chi, l}, n \geq 0$ and assume (i) $x_n \to x$ and (ii) $\dim E^u(x_n) \equiv \dim E^u(x)$.

Without loss we can pass to a subsequence $x_{n}$ along which $\{E^u(x_{n})\}$ converges to a subspace $E \subset T_x M$. It suffices to show that $E = E^u(x)$. If not, then there is a unit vector $e \in E$ for which $e = e^u + e^s$ with $e^s \neq 0$ (here we split according to $E^u \oplus E^s$ at $x$).

Now, $e$ is the limit of some sequence of unit vectors $e_n \in E^u(x_n)$. Below, $k$ is fixed and $n \geq 1$:

$$ \| df^{-k}_x e\| \leq \| df^{-k}_x - df^{-k}_{x_n} \| + \| df^{-k}_{x_n}(e - e_n)\| + \| df^{-k}_{x_n} e_n\| $$ Taking $n$ to infinity, the first and second terms vanish, while the third term is $\leq l e^{- k \chi}$ for all $n$. On the other hand, $$ \| df^{-k}_x e \| \geq \| df^{-k}_x e^s\| - \| df^{-k}_x e^u\| \geq l^{-1} e^{\chi k} \| e^s\| - l e^{- \chi k} \| e^u\| , $$ which is clearly incompatible with the earlier estimate when $k$ is sufficiently large. We conclude $\|e^s\| = 0$, i.e., $E = E^u(x)$.

---

As you can see, the proof heavily depends on the uniformity of the hyperbolicity estimates across $\Lambda_{\chi, l}$, which are often called uniformity sets or Pesin sets in the literature. One expects continuity of $E^{u/s}$ to hold only along Pesin sets, and not along the whole set of Lyapunov regular points.

Generally, only Holder continuity holds-- my favorite proof is given in Brin & Stuck (Chapter 5, I think). Uniform continuity across Pesin sets can be recovered by modifying the above proof.