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.