Topological conjugacy between homeomorphisms and diffeomorphisms - MathOverflow

Topological conjugacy between homeomorphisms and diffeomorphisms

2010-05-26T17:29:19Z

Consider a compact differentiable manifold $M$. We say that $f:M\to M$ and $g: M \to M$ are topologically conjugated if there exists $h:M\to M$ a homeomorphism such that $f\circ h= h \circ g$. The conjugacy class of a homeomorphism $f$ is the set of all $g$ such that $g$ is topologically conjugated to $f$.

If a homeomorphism $f: M\to M$ has infinite topological entropy (which is an invariant under topological conjugacy), then the conjugacy class of $f$ has no diffeomorphisms.

Is there any other known obstruction for a homeomorphism not to have diffeomorphisms in its conjugacy class? I would guess that yes, but I could not find one.

Is there a restriction on the dimension of the manifold?

(Remark: In dimension one, that is, in the circle, every homeomorphism has a diffeomorphism-of class $C^1$ in its conjugacy class. However, Denjoy counterexamples have no $C^2$ diffeomorphisms in theirs).

Answer by coudy for Topological conjugacy between homeomorphisms and diffeomorphisms

2010-05-26T18:40:12Z

My two cents. There is a famous open problem in ergodic theory, which is somehow related to your question.

Given a measure preserving transformation T of a Lebesgue space, does there exist an isomorphism (that is, a measure preserving measurable conjugacy) between T and a volume preserving diffeomorphism defined on a compact riemannian manifold ?

Finite entropy is a necessary condition, but no other invariant is known at the moment. It is expected that there are indeed other invariants, but nobody found them yet. I would be surprised if the problem was simpler in the topological setting than in the measure preserving setting.

Answer by Sergei Ivanov for Topological conjugacy between homeomorphisms and diffeomorphisms

2010-05-26T19:31:50Z

Here is an example. Consider a map $f:\mathbb R^2\to\mathbb R^2$ given by $(r,\varphi)\mapsto (r,\varphi+\sin(1/r))$ in polar coordinates $(r,\varphi)$, $0<r\le 1/\pi$ . For $r>1/\pi$, let $f$ be identity, then close up the plane to make a compact manifold.

This map has zero topological entropy but has no conjugate $C^1$ diffeomorphism. Indeed, we may assume that 0 is mapped to itself by the conjugation. Then the circles centered at the origin are mapped to Jordan curves winding around the origin. There are arbitrarily small circles made of fixed points, hence the derivative of the diffeomorphism at the origin is the identity. On the other hand, there are arbitrary small circles on which our map is periodic with rotation number, say, 1/10. It is easy to see that this cannot happen on a Jordan curve winding around the origin, unless some point changes is angular coordinate by $2\pi/10$. This contradicts the fact that the derivative at 0 is the identity.