4
$\begingroup$

Let $f:S^1\to S^1$ be an orientation-preserving circle diffeomorphism with irrational rotation number (see here). Then the system $(S^1,f)$ admits a unique invariant measure, say $\mu_f$.

Let $\displaystyle \lambda(f,x)=\lim_{n\to\infty}\frac{1}{n}\log D_xf^n$ be the Lyapunov exponent at $x$. Then $\lambda(f,x)$ is independent of $x$, and is denoted by $\lambda(f)$.

Question. Is $\lambda(f)$ always zero?

Note that $\lambda(f)$ equals to the integral $\int \log D_x f\; d\mu_f(x)$. Take the Denjoy map for example (a non-transitive diffeomorphism with irrational rotation number).

Remark. If $\log Df$ has bounded variation, then Denjoy proved that $e^{-V}\le Df^{q_n}(x)\le e^V$ for all $x$ (a much stronger property), where $q_n, n\ge 1$ are the denominators of rational approximates. In particular this implies $\lambda(f,x)=0$ for all $x$.

Edit again. I thought the integral might be easier to compute. Then I realized that the estimation of $\lambda(f)$ is quite straight forward, and answered my own question.

Improve this question
$\endgroup$
2
  • $\begingroup$ Herman improved Denjoy's result, and proved that if $f$ is $C^3$, then $Df^{q_n}\to 1$ uniformly on $S^1$. $\endgroup$
    – Pengfei
    Commented Feb 20, 2015 at 22:55
  • $\begingroup$ It seems difficult for critical circle maps. A paper with title 'The Lyapunov exponent of a critical circle map' is listed at ime.usp.br/~edson/publications.html $\endgroup$
    – Pengfei
    Commented Feb 21, 2015 at 23:11

1 Answer 1

Reset to default
3
$\begingroup$

Note that $\int \log D_xf^n \, dm(x)\le \log \int D_xf^n \, dm(x)=\log 1=0$ for all $n\ge 1$. Therefore (by uniform convergence for unique ergodic system)

$\displaystyle \lambda(f)=\int \lambda(f,x) dm(x)=\int\lim_{n\to\infty}\frac{1}{n}\log D_x f^n dm(x)=\lim_{n\to\infty}\frac{1}{n}\int\log D_x f^n dm(x)\le 0$.

Then applying to the inverse $f^{-1}$, one gets $\lambda(f^{-1},x)\le 0$, too. Therefore, $\lambda(f,x)=0$ for all $x$. Or equally, $\mu_f(\log Df)=0$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 0) The conclusion is correct: indeed, at least in C^1-regularity the Lyapunov exponent vanishes. 1) But you should be more careful: $D_x f^n$ is the derivative of $f^n$ in the sense of the Lebesgue measure, and you are integrating it w.r.t. the invariant one, dm(x). I do not see an immediate way to say that the integral is then equal to 1. $\endgroup$
    – Victor Kleptsyn
    Commented Feb 20, 2015 at 2:30
  • 1
    $\begingroup$ Thanks for your comments. Here $m$ is the Lebesgue measure. I used $\mu_f$ for the invariant measure. $\endgroup$
    – Pengfei
    Commented Feb 20, 2015 at 2:40
  • $\begingroup$ Ah yes, indeed. Nice, I did not know that it can be done that simply! $\endgroup$
    – Victor Kleptsyn
    Commented Feb 20, 2015 at 2:42

You must log in to answer this question.

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