# 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. – Anthony Quas 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. – Will Sawin 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. – Ian Morris 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. – Will Sawin 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. – Ian Morris 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. – Will Sawin Mar 5 '16 at 19:12