Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know the Birkhoff Ergodic theorem; and I know what is a Riemannian manifold and what a geodesic is.

I also read the definition of geodesic flow on the tangent bundle of one such. But I do not yet know the meaning of the statement "the geodesic flow is ergodic", of course with some added conditions, such as constant negative curvature.

Could someone please give me a reference for me for an introduction to this topic, and how does this ergodicity relate to Birkhoff's ergodic theorem?

share|improve this question
    
I am far from this area, but my guess would be that the geodesic flow is a continuous-time dynamical system, which ergodicity is similar to the one discussed in Birkhoff's Theorem for discrete-time systems. –  Ilya Jan 17 '13 at 8:56

1 Answer 1

this is my first ever comment on stackexchange, just saying.

So the geodesic flow $\phi: SM\rightarrow SM$ is a flow on the unit tangent bundle $SM$ that preserves a measure, the Liouville measure $\mu$. So if $M$ is compact it makes a fine measure preserving dynamical system of which we can naturally ask whether it is ergodic.

The geodesic flow is ergodic if \begin{equation} A=\phi A \qquad\Rightarrow \qquad \mu (A)^2=\mu(A) \end{equation} for $A\subset TM$.

There is an alternative definition where a flow is ergodic if any $\phi$-invariant measurable function is constant apart from a set of zero measure.

This has to be shown to hold true for the geodesic flow.

A legible proof can be found in "Lectures on Spaces of Nonpositive Curvature" by Werner Ballmann (in the appendix) "Nonuniform Hyperbolicity" by Pesin and Barreira (in the geodesic flow chapter)

good luck.

share|improve this answer
    
Birkhoffs theorem applies to ergodic transformations. So if we can show that a geodesic flow is ergodic, then Birkhoffs theorem holds true. –  Leo Apr 24 '13 at 17:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.