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An important necessary condition is that on periodic points, the determinant in the period must be one (that is not hard). I believe that what you ask has to do with Theorem 5.1.13 of Katok-Hasselbla …

To the first question the answer is negative. There are two homeomorphisms of the circle with irrational rotation number such that their composition is Morse-Smale (in fact, you can multiply two $2\ti …

Not sure if it is related, but in Lemma 1 of this paper (A.Avila, J. Bochi, A C1 generic map has no invariant absolutely continuous probability measure) it is proved that if a map has no invariant abs …

I believe that combining these two papers, you get an answer: http://arxiv.org/abs/0906.2176 and http://arxiv.org/abs/1010.3372. In the first one it is shown that partially hyperbolic sets with one …

I think a good example (which may not qualify as consise, but does as self contained) is given in the famous paper of Anosov and Katok. There they construct smooth diffeomorphisms of the disc being we …

The answer is no. A trivial example is to concentrate the measure in a "periodic orbit", this will give an invariant measure for the shift. But there are a whole lot of invariant measures (including …

For Lie groups, the only one with this property is $S^1$. To see this, consider a one parameter subgroup given by exponential of a vector in the lie algebra, and the closure of this subgroup should …

The key word you are looking for is "physical measures", sometimes known as SRB measures (because of the result.that coudy mentions). See https://link.springer.com/article/10.1023/A:1019762724717 A re …

I do not have an direct answer but since I fear the answer might not be known and you also ask for related stuff let me mention a couple of results. In the $C^1$ case, Mañe-Bochi's result asserts th …

You can just take an Anosov map on $T^2$ and multiply by identity on the circle. Then, you will have SRB measures supported on $T^2 \times pt$ which are not physical.

In $S^2$ it is possible to construct such an example by a quite different method (not as a limit of rotations) which is indeed Bernoulli with respect to Lebesgue. The idea is to quotient the cat map …

The question is why a measurable function which is invariant under the action of all $g \in G$ must be a.e. constant. But since the action of $G$ is transitive on $G/\Gamma$ it is easy to see that thi …