# Can a smooth diffeomorphism of a Riemannian manifold have only positive Lyapunov exponents on a large set?

Let $M$ be a compact Riemannian manifold, $f: M \to M$ a diffeomorphism, and $\mu$ an ergodic measure for $M$. Suppose that the support of $\mu$ is not a finite set. Is it possible that all the Lyapunov exponents of $\mu$ will be positive?

This is a version of a previous question with one added condition, to remove a specific example.

The answer is no if $\mu$ is absolutely continuous with respect to Lebesgue measure. Then $\mu$ has a density function $d$, and there is some interval $I$, not containing $0$ or $\infty$, such that $\mu(d^{-1}(I))>0$. Hence for $x$ with $d(x)<I$, there are infinitely many $n$ such that $f^n(x)\in I$. By the change of variables formula $\left|\det \left(\frac{d f^n}{dx}\right)\right| (x) d(f^n(x))=d(x)$, there is a bounded interval that $\left|\det \left(\frac{d f^n}{dx}\right)\right|$ stays in infintely often and hence the sum of the Lyapunov exponents is $0$ so one is nonpositive.

The same thing is true if $\mu$ is merely absolutely continuous with respect to Lebesgue measure on a submanifold.

In fact, there is a much weaker condition that is sufficient to rule this out. Define $r_x(\delta)$ be the infimum of $r$ such that a ball of radius $r$ around $x$ has measure at most $\delta.$ Suppose there is a function $s(\delta)$ such that for $x$ in a set of positive measure, $r_x(\delta)$ is bounded above and below by a constant times $s(\delta)$. Then one of the Lyapunov exponents is nonpositive - we can take universal constants $C_1,C_2$ such that $c_1 s(\delta) < r_x(\delta)< C_2 s(\delta)$ on a set of positive measure. Then as long as $x$ and $f^n(x)$ are both in that set, the lowest singular value of $\frac{d f^n}{dx}(x)$ is at most $C_2/C_1$ and so one of the Lyapunov exponents is nonpositive.

So any counterexample must at the very least have quite nontrivial local structure.

The answer is no. In fact, the following result holds: if $M$ is a compact manifold, $f\in\mathrm{Diff}^{1+\alpha}(M)$ and $\mu$ is an ergodic $f$-invariant probability measure such that all its Lyapunov exponents are positive, then $\mu$ is a periodic measure, i.e. there exists a periodic source $\{p,f(p),\ldots,f^{n-1}(p)\}$ such that $\mu=1/n\sum_{i=0}^{n-1} \delta_{f^i(p)}$.