6
$\begingroup$

I am trying to prove an analytic result for gesodesic flows on negatively curved manifolds and I encountered the following dynamical-system porblem.

Let $B^n$ be $n$-dimensional balls and $h:B^{n-k}\times B^{k} \to U$ a $k$-dimensional foliation chart with $C^\infty$ leaves. i.e. For $x\in B^{n-k}$, $W_x := h(\{x\}\times B^k)\subset U$ are $C^\infty$ submanifolds. Furthermore let $y\in B^{n-k}$ such that $L=h(\{y\}\times B^{n-k})$ is a smooth transversal. Then according to Brin-Stuck "Introduction to Dynamical Systems" the foliation is absolutely continuous if there is a measurable family of positive measurable functions $\delta_x: W_x\to\mathbb R$ such that for all measurable sets $A\subset U$ $$\int_U \mathbf 1_A(x,y) dm_{euc} = \int_L \int_{W_x} \mathbf 1_A(x,y) \delta_x(y) dm_{W_x}(y)dm_L(x).$$ Here $dm_{euc}$ is the Eucildean measure on $U$ and $dm_{W_x}$ and $dm_L$ the measures obtained by restricting $dm_{euc}$ on the smooth submanifolds.

If the foliation ist the stable foliation of an Anosov differomorphism then I found that it is absolutely continuous. It think it is clear that one cannot hope that $\delta_x(y)$ depends $C^\infty$ on $x$ unless the foliation is smooth.

My question is: Under which conditions on the foliation can one obtain, that for any $x$ one has $\delta_x \in C^\infty(W_x)$(ideally I would also like to have, that $\delta_x$ depends continuousely on $x$ w.r.t. to the $C^\infty$-topology, but maybe I can also handle the case if it is only measureable)? And more precisely: Is such a statement always true for the strong stable foliation of a $C^\infty$ Anosov flow. If it doesn't hold for general Anosov Flows: What are the weakest known assumption that it holds? (If I understand it correctly then $C^1$ foliations implies this statement via Fubinis theorem).

$\endgroup$

2 Answers 2

2
$\begingroup$

In general, for the geodesic flow of a compact negatively curved manifold for example, the strong stable foliation is Holder continuous; in particular, even if each leaf is (locally) a smooth manifold, transversally, the foliation (and therefore the map $x\to\delta_x$) is no more than Hölder continuous.

In the case of geodesic flows in constant negative curvature, it is smooth, and if I remember well, it is $C^1$ in the case of surfaces.

I guess that for general Anosov flows, it is similar: you cannot expect better regularity than Hölder, except in very particular cases.

$\endgroup$
1
  • $\begingroup$ Dear Barbara, thanks for your reply. I understand, that foliations are in general only Hölder and that I cannot expect $\delta_x(y)$ to be smooth in $x$. My question is rather if for given $x$ the conditional measure is smooth along the sheat $W_x$ $\endgroup$
    – twch
    Commented May 19, 2015 at 21:05
0
$\begingroup$

Due to helpfull discussions with several dynamical systems experts I am now able to give an answer to the question, that I asked about one year ago and provide some references:

The short answer is, that for the smoothness of the conditional measures it is sufficiant to know the smoothness of the holonomy maps with respect to a smooth transversal foliation. A detailed explanation of this argument can for example be found in Appendix A of arxiv-version but the techniques are rather standard, see e.g. Brin-Stuck "Introduction to dynamical systems" Section 6.2.

Concerning the stable and unstable foliations of an $C^\infty$-Anosov flow, a statement on the smoothness of the holonomy maps can for example found in Giulietti-Liverani-Pollicott Ann.Math (2013) (Appendix E). Thus one obtains the smoothness of the conditional densities for $C^\infty$-Anosov flows. A precise formulation of this statement is now given in Theorem 7 arxiv-version.

If one is interested in an analogouse statement for Anosov map, there is a remark in Chernov "Invariant measures for hyperbolic dynamical systems" (Handbook of Dynamical systems 1A), stating the smoothness of the conditional densities and giving a rough idea how to prove it.

$\endgroup$

You must log in to answer this question.

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