# Can a smooth diffeomorphisms of a Riemannian manifold have only positive Lyapunov exponents?

Let $\mu$ be some ergodic measure of our compact Riemannian manifold $M$, which is preserved by $f\in Diff^{1+\beta}(M)$. Is it possible that all the Lyapunov exponents of $\mu$ will be positive? Intuitively this seems wrong, but I couldn't find any general proof without assuming that $h_\mu(f)>0$ (which I don't want to restrict myself to doing).

As Will shows, the case in which $\mu$ is absolutely continuous with respect to Lebesgue measure and has density bounded away from zero and infinity is constrained in that the Lyapunov exponents of $\mu$ must sum to zero. If $\mu$ is an arbitrary ergodic measure then the Lyapunov exponents can all be positive, for example if $\mu$ is the Dirac measure on a repelling fixed point.

• For a really explicit example, consider the doubling map on the Riemann sphere, $z\mapsto 2z$. This is as smooth as imaginable; the $\delta$-measure at the origin has all Lyapunov exponents equal to 2. Mar 5 '16 at 19:51
• What about if the support of $\mu$ contains infinitely many points? If the support is a manifold and $\mu$ is absolutely continuous on the manifold then it's possible by my argument, so the support would have to be a slightly complicated (probably Cantor set-like) shape. Mar 5 '16 at 20:29
• Will, while I can't come up with an explicit example off the top of my head I suspect that examples are possible where the measure is fully supported and not absolutely continuous. Mar 6 '16 at 1:26
• @IanMorris But by the Lebesgue decomposition theorem, we can write it as as a sum of a continuous measure and a singular measure. Since this decomposition is canonical, it should be $f$-invariant. So a measure of one of those two types should be an example. I guess it could be absolutely continuous but given by a function with poles, though. Mar 6 '16 at 14:01
• It is a classical fact that pairs of distinct ergodic measures are mutually singular, so by "where the measure is fully supported and not absolutely continuous" I essentially meant "where the measure is fully supported and singular with respect to Lebesgue measure". In many cases a dynamical system will have infinitely many singular ergodic measures. I strongly suspect that your argument can be tweaked to show that every absolutely continuous ergodic measure must have Lyapunov exponents summing to zero irrespective of the structure of the density. Mar 6 '16 at 14:44

Yes, because the system is conservative, the sum of the Lyapunov exponents is $0$, so they cannot all be positive.

Observe that the sum of the Lyapunov exponents is $\lim_{n \to \infty} \frac{1}{n} \log \left|\det \frac{ df^n}{dx}(x)\right|$. But as $\mu$ is invariant under $f^n$, $\det \frac{ df^n}{dx}$ is the ratio of the density of $\mu$ at $x$ to the density of $\mu$ at $f^n(x)$. Because the density is continuous function on a compact manifold, it is bounded, so the limit is zero.

• Why is the density function continuous? What about a delta measure on a fixed point (or something similar) ? $\mu$ is not necessarily a smooth measure.
– BOS
Mar 5 '16 at 18:57
• @BOS good point. Ian handled it. Mar 5 '16 at 19:12