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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).

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up vote 5 down vote accepted

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.

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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 at 19:51
Thank you very much. This indeed settles it. – BOS Mar 5 at 20:02
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 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 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 at 14:01

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.

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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 at 18:57
@BOS good point. Ian handled it. – Will Sawin Mar 5 at 19:12

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