Timeline for Lyapunov exponent for circle diffeomorphisms
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2015 at 2:53 | vote | accept | Pengfei | ||
| Feb 20, 2015 at 2:42 | comment | added | Victor Kleptsyn | Ah yes, indeed. Nice, I did not know that it can be done that simply! | |
| Feb 20, 2015 at 2:40 | comment | added | Pengfei | Thanks for your comments. Here $m$ is the Lebesgue measure. I used $\mu_f$ for the invariant measure. | |
| Feb 20, 2015 at 2:30 | comment | added | Victor Kleptsyn | 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. | |
| Feb 20, 2015 at 1:57 | history | answered | Pengfei | CC BY-SA 3.0 |