3
$\begingroup$

There is a step in the proof of Theorem 4.11 of this set of notes that I don't quite see.

The set up is that $f$ is a $C^2$-diffeomorphism on some Riemannian manifold $M$, and that $E \oplus F = T M$ is a dominated splitting for $f$, and it is concluded that the 'stable' bundle $E$ is $\theta$-Hölder for some $\theta \in (0,1)$.

In the course of the proof it is shown that for every $k$ there is an $\epsilon_k > 0$ so that if $d(x, y) < \epsilon_k$ then $$ d(E_x, E_y) \le c(d(x,y)^\theta + \lambda^k), $$ where $c, \lambda$ and $\theta$ are uniform in $x,y$ and $k$, and $\lambda < 1$. In the next line it is claimed that the Hölder continuity of $E$ follows by taking $k$ large enough, presumably so that $\lambda^k \approx d(x,y)^\theta$, but then the value of $k$ depends on $x$ and $y$, which are in turn constrained by $k$ (via $\epsilon_k$).

To me this seems like a mistake, but I know that these kind of results are well-known, so perhaps I am missing something very simple. I would appreciate anybody explaining this step of the proof, or pointing me to a more detailed reference.

$\endgroup$

1 Answer 1

1
$\begingroup$

thanks for the comments. The notes were at several points written quite in a hurry, we plan to work on them some day (but don't know when).

For this specific proof, notice that $\epsilon_k$ can be chosen to be of the order $\epsilon_1 \|Df^k|_{F}\|^{-1}$ and so the fact that $\lambda \|Df|_F\|^\theta <1$ allows to be able to make the choices without circularity (i.e. as you say, so that $\lambda^k$ is roughly $d(x,y)^\theta$). (This is indeed the place where the proof is different with the case where $F$ is actually expanding, where one can consider a fixed value of $\epsilon$, so clearly the part that we should have explained in more detail.)

I have not been able to find a precise reference for this elsewhere, but it is definitely well known. Another (sketch of) the proof in a different but very similar context can be found in appendix A.7.2 of this paper. Also, I imagine that it is also proved in this paper (or at least it contains similar computations).

$\endgroup$
1
  • $\begingroup$ Thank you for the explanation (and, also, for writing the notes in the first place)! $\endgroup$ Commented Oct 2, 2019 at 23:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .