In this post, topological entropy is defined. In this book, the notion of Lyapunov exponent for, let's say a self-diffeomorphism from a smooth, closed Riemannian manifold to itself, is defined (in terms of the eigenvalues/singular values of the Jacobian matrix).
How are they related? Is the topological entropy of a self-diffeo $T: M \to M$ simply the maximum Lyapunov exponent over all $x \in M$?
Thanks in advance.
UPDATE: Is this true: The topological entropy of $T = \tilde{\Phi}_{t=1}$, and therefore $\xi$, is the sup over all measures on $M$ that $T$ leaves invariant of the integral over $M$ of the sum of the positive Lyapunov exponents of $T$, $\displaystyle \sup_{\nu \in \text{ Invar}(T)} \int_{M} \sum_{i=1}^{k}\lambda_k(x,v)\ d\nu$?
Trying to understand what your notation means, in general you only have $$ h(f)\le \sup_{\nu} \int_{M} \sum_{i=1}^{k(x)}\lambda_{k(x)}(x)\ d\nu(x) $$ (really it is a bad idea to use $h$ for the map). This follows from the Margulis–Ruelle inequality.
The supremum runs over all $f$-invariant probability measures and the sum is over all positive Lyapunov exponents $\lambda_1(x)\le\cdots \le\lambda_{k(x)}(x)$ (unfortunately $k(x)$ may indeed depend on $x$).
But if $f$ is a $C^{1+\alpha}$ diffeomorphism and $\nu$ is absolutely continuous with respect to the volume, then $$ h_\nu(f)= \int_{M} \sum_{i=1}^{k(x)}\lambda_{k(x)}(x)\ d\nu(x). $$ This is Pesin's entropy formula. Therefore, under the same assumptions, $$ h(f)\ge \int_{M} \sum_{i=1}^{k(x)}\lambda_{k(x)}(x)\ d\nu(x), $$ although you cannot add a supremum in general. However, if $\nu$ is also a measure of maximal entropy you obtain $$ h(f)= \int_{M} \sum_{i=1}^{k(x)}\lambda_{k(x)}(x)\ d\nu(x) $$ and so also $$ h(f)= \sup_{\nu} \int_{M} \sum_{i=1}^{k(x)}\lambda_{k(x)}(x)\ d\nu(x). $$
Of course, all depends exactly on what you are assuming. Up to some technicalities on the $C^{1+\alpha}$ assumption this is the most general answer that you can get under general assumptions.