I asked this in stackexchange, but got no answer, so I am trying here.

Is it possible for a diffeomorphism $\phi$ (of a smooth manifold $M$) to have the following properties:

  1. All its orbits are finite.
  2. $\phi$ is not of finite order. (and hence has arbitrarily large finite orbits).

There are such examples if we allow infinitely many connected components (Take for instance countable disjoint copies of $\mathbb{S}^1$ and rotation of order $n$ in the $n$-th copy).

So let us assume $M$ is connected. (I prove below that we can reduce the case of finitely many components to one component).

I will add that this question was inspired by a previous question of mine, about global obstructions for a diffeomorphism to be an isometry (You can see the update there for the relevant connection).

Claim: existence of a self-diffeomorphism for a manifold with finitely many connected components, implies existence for a connected manifold.

Proof: In this case the components $U_i$ are clopen connected sets, hence $\phi(U_i)$ are also clopen connected sets. Now combine the following two facts:

  1. clopen sets are always a union of connected components.
  2. every connected subset of a topological space $X$ is contained in a connected component $X$.

Now it is evident that each $\phi(U_i)$ is a connected component, i.e $\phi$ permutes the components. Since there are finitely many components, if we choose some component $U_i$, we will get $\phi^n(U_i)=U_i$ for sufficiently large $n$.

Next, we observe that the if $\phi$ has the two required properties, then $\phi^n$ also has them.

So finally, $\phi^n|_{U_i}$ is a self-diffeomoprhism of a connected manifold with the required properties.


No, there is no such animal. This is due to Montgomery (1938), see my answer to this question: Nonperiodic points of homeomorphisms of a ball

| cite | improve this answer | |
  • $\begingroup$ In every example I have seen for a diffeomorphism of infinite order, the set of non-periodic points was dense. Do you know if this is necessary? (What I am saying here is that the theorem you mentioned only implies existence of one non-periodic point, and "in practice" in all cases I have looked so far, there were much more such points than one). I would like to find a case where the set of periodic points are dense. $\endgroup$ – Asaf Shachar May 3 '15 at 8:10
  • $\begingroup$ @asafshachar - it seems to me that if the set of periodic points has interior, then this interior is invariant under the map, and every connected component is obviously periodic. Hence it follows from the result cited by Igor that the map has finite order on each such component. $\endgroup$ – Lasse Rempe-Gillen May 5 '15 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.