3
$\begingroup$

Let $S$ be a closed surface endowed with a Riemannian metric of negative curvature and let $US$ be the unit tangent bundle. Let $\mu$ be the Liouville measure on $US$.

Let $f: US\rightarrow\mathbb{R}$ be a smooth function and $M=\int_{US}f(v)d\mu(v)/\int_{US}d\mu$ be its mean value.

For each $v\in US$, let $(v(t))_{t\in\mathbb{R}}\subset US$ denote the geodesic with initial velocity $v$. Ergodicity of geodesic flow and Birkhoff theorem imply that for almost all $v$ in $US$ we have $$ (\star)\quad\quad\quad\lim_{T\rightarrow+\infty}\frac{1}{T}\int_0^Tf(v(t))\, dt=M. $$

In general, one can not expect this to hold for every $v\in US$. However, for some reason I want an answer to the following questions.

Questions

Does there exist $x\in\Sigma$ such that $(\star)$ holds for every $v\in U_xS$ ?

If yes, is there a $x$ such that the convergence in $(\star)$ is uniform on $U_xS$?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ I think the answer is no: The stable manifold of a periodic geodesic should pass through every $x$ in its vicinity. $\endgroup$
    – user25199
    Commented Jun 20, 2014 at 12:56
  • 2
    $\begingroup$ By the Bernoullicity theorem (Ornstein-Weiss,Adler-Flato, and if I recall correctly Ratner have also done some work about it in her PhD thesis), this dynamical system is isomorphic to a Bernoulli shift (by constructing suitable Markov partitions), and your question is absolutely false in the Bernoulli settings, due to appearance of Cantor sets. $\endgroup$
    – Asaf
    Commented Jun 20, 2014 at 16:08

1 Answer 1

Reset to default
8
$\begingroup$

The answer is NO. There is no chance for such a thing.

The reason is the following. The asymptotic behaviour of the ergodic average depends on the asymptotic geometric behaviour of the ... geodesic you are looking at! And the hyperbolicity of the geodesic flow means that starting from everywhere (for example from $x$) you can get every possible ergodic behaviour.

In your problem, consider a nonconstant continuous map $f$. Assume for example that there exists another probability measure $\nu$ invariant under the geodesic flow, such that $\int_{US} f\,d\nu\neq \int_{US} f\,d\mathcal{L}$, where $\mathcal{L}$ is the normalized Liouvillle measure. Without loss of generality, you can assume $\nu$ to be ergodic.

Choose a generic vector $v\in US$ for $\nu$. Then the ergodic average of $f$ along the geodesic $(g^t v)_{t\ge 0}$ defined by $v$ converges to $\int f d\nu\neq M$.

But now, by hyperbolicity, given any $x\in US$ there exists (a lot of) $w\in U_xS$ which is in the (weak) stable manifold of $v$. (If you are not convinced, draw a picture in the universal cover, lifting $x, v$ and find $w$...) In particular, the geodesics of $v$ and $w$ are asymptotic, and the ergodic averages converge to the same limit.

It is the (negative) answer to your question.

Bests

Barbara

Improve this answer
$\endgroup$
2
  • $\begingroup$ Hi Barbara, thanks for the detailed answer. The next natural question is, what can be said about the existence of time average along $g^tv$? For example, it there a $x\in S$ such that the time average of $f$ along $g^tv$ exists for every $v\in U_xS$? B.T.W. for my purpose we can assume $f$ to be the lift of a function $S\rightarrow\mathbb{R}$. $\endgroup$
    – Xin Nie
    Commented Jun 20, 2014 at 17:25
  • 1
    $\begingroup$ No chance either... By the same argument, sorry. Consider two measures $\nu_1$ and $\nu_2$ such that $\int f\,d\nu_1\neq\int f\,d\nu_2$. Now, consider two generic vectors $v_1$ and $v_2$ for $\nu_1$ and $\nu_2$ respectively. The hyperbolicity of the geodesic flow allows you to build an orbit $(g^t w)$ accumulating along the two orbits of $v_1$ and $v_2$. In particular, the ergodic averages along $(g^t w)$ have (as probability measures) at least two limit points $\nu_1$ and $\nu_2$. Now, the same argument as above allows you to find $u\in U_xM$ in the stable manifold of $w$... $\endgroup$
    – Barbara Schapira
    Commented Jun 21, 2014 at 20:39

You must log in to answer this question.

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