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Let A(x,n) be the cocycle over f, where f is an measure preserving transformation on a probability space X. Is the largest Lyapunov exponent always given by

\lim_{n\to +\infty} \log ||A(x,n)||?

Since the above limit can be bounded from above by \int_X\log ||A|| can one give an example of the cocycle where the above inequality is strict? Tnx!

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1 Answer 1

Hi. First, I suppose that the Lyapunov exponent is given by $$ \lim_{n\to\infty} \frac{1}{n} \int \log\|A(x,n)\| d\mu(x), $$ where $\mu$ is an appropriate ergodic measure. (You have some base dynamics for the cocycle, i.e. $A(x,n + m ) = A(T^n x, m) A(x,n)$ and $T$ is $\mu$ ergodic.

Then $$ \lim_{n\to\infty} \frac{1}{n} \log\|A(x,n)\|, $$ for almost every $x$, not for every. This follows from the subadditive ergodic theorem.

The inequality is very rarely strict. For the simplest example, consider a dynamic over a one-point space given by $A(x,n) = A^n$ for $$ A = B \begin{pmatrix} 2 & 0 \\\ 0 & \frac{1}{2} \end{pmatrix} B^{-1}. $$ It is easy to check that the Lyapunov exponent will be $2$, but using an appropriate choice of $B$, one can make $\|A\|$ arbitrarily large.

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Thanks for answering. In particular, the above expression or the integral form you wrote always gives the largest exponent. Am I right? –  Joachim Feb 27 '12 at 21:32
    
My integral gives the largest Lyapunov exponent. Without the integral it doesn't. Just take the base dynamics random, and choose a periodic point. –  Helge Feb 27 '12 at 23:03
    
Sorry, the term 1/n is missing. Besides this, two things should be equal when we work with ergodic measure –  Joachim Feb 27 '12 at 23:18
    
Equal almost everywhere, but not equal ;-) These are two very different notions. There are certain theorems that analyze when convergence is everywhere, the basic conclusion is that then the Lyapunov exponent vanishes (can't think of the names right now). –  Helge Feb 28 '12 at 2:01
    
Sure, equal almost everywhere. Thanks for your comments! –  Joachim Feb 28 '12 at 12:40
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