What is an example of a compact manifold which does not admit a diffeomorphism with at least one dense orbit?
Moreover, is it true to say that every isometry of $\mathbb{C}P^n$ with the Fubini-Study metric do not possess any dense orbit?
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asked Apr 25, 2018 at 20:46
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6$\begingroup$ No isometry of $\mathbb{CP}^n$ has a dense orbit, because the square of such an isometry is holomorphic and so can be regarded as an element of the compact Lie group $\mathrm{SU}(n{+}1)$ and hence is congugate to an element of the maximal torus, and no element of a maximal torus can have a dense orbit because it has to preserve a transverse collection of hyperplanes and so perserve the distances from these hyperplanes. A little more work takes care of the case of an anti-holomorphic isometry. $\endgroup$– Robert BryantCommented Apr 25, 2018 at 22:10
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4$\begingroup$ For $C^1$ diffeomorphisms preserving volume, there is a residual set which is transitive (having a dense orbit). See Theorem 1.3 of mathscinet.ams.org/mathscinet-getitem?mr=2090361 So in some sense such diffeomorphisms are "generic". See also the comments after the statement of the theorem about the homeomorphism case. $\endgroup$– Ian AgolCommented Apr 25, 2018 at 23:02
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$\begingroup$ @RobertBryant Thank you very much for your answer. I try to understand its detail for $n>1$. My motivation was $S^2$ whose all isometries are non ergodic maps. $\endgroup$– Ali TaghaviCommented Apr 26, 2018 at 14:35
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$\begingroup$ My apology, I correct: The isometries are ergodic and without dense orbit. $\endgroup$– Ali TaghaviCommented Apr 26, 2018 at 15:03
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Dolgopyat and Pesin proved (ETDS 2002) that "every compact manifold of dimension $\geq 2$ admits a Bernoulli diffeomorphism with non-zero Lyapunov exponents". That is, on any such manifold $M$ there is a $C^\infty$ diffeomorphism $f$ that preserves volume $m$ and has the property that $(M,m,f)$ is a measure-preserving transformation that is measure-theoretically isomorphic to a Bernoulli shift. In particular, by the Birkhoff ergodic theorem, $m$-a.e. point $x\in M$ has the property that its orbit equidistributes with respect to $m$, meaning that for every continuous $\phi\colon M\to \mathbb{R}$, we have $\frac 1n \sum_{k=0}^{n-1} \phi(f^kx) \to \int \phi\,dm$ as $n\to\infty$. In particular, any such $x$ has a dense orbit.
answered Apr 26, 2018 at 3:13