most recent 30 from http://mathoverflow.net 2013-06-18T03:42:20Z http://mathoverflow.net/feeds/question/89696 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89696/largest-lyapunov-exponent Joachim 2012-02-27T19:26:04Z 2012-02-28T01:59:03Z

<p>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 </p> <p>\lim_{n\to +\infty} \log ||A(x,n)||?</p> <p>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!</p>

http://mathoverflow.net/questions/89696/largest-lyapunov-exponent/89701#89701 Helge 2012-02-27T21:15:50Z 2012-02-28T01:59:03Z

<p>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.</p> <p>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.</p> <p>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 $$ A = B \begin{pmatrix} 2 &amp; 0 \\ 0 &amp; \frac{1}{2} \end{pmatrix} B^{-1}. $$ 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.</p>