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I was wondering if the following theorem still holds if one only requires $\tau$ to be a homeomorphism and not a diffeomorphism: "Let M be a smooth manifold and $\tau:M\to M$ a differentiable function such that $\tau(\tau(x))=x$ and $\tau(x)\neq x$. Then the quotient space $M/\tau$ is a smooth manifold."

I don't see where the "differentiable" part is used in the proof. The Theorem can e.g. be found in "Jänich, Vektoranalysis" and if needed I can also give you the proof.

Thanks in advance for any help!

LordOfNumbers
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  • If $\tau$ is a diffeomorphism, how do you define a differentiable map on $M/\tau$? Does this still work if $\tau$ is only an homeomorphism? – D. Thomine Apr 12 '20 at 17:49
  • When you say "$M/\tau$ is a smooth manifold" you really mean "is homeomorphic to a smooth manifold," since all what you have on $M/\tau$ is the quotient topology. In the quoted result, the right statement is "$M/\tau$ admits structure of a smooth manifold such that the quotient map $M\to M/\tau$ is a local diffeomorphism." Otherwise, this is a good question. – Moishe Kohan Apr 12 '20 at 17:57

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Such examples do not exist if $M$ has dimension $\le 3$, since in this range of dimensions all topological manifolds are (uniquely) smoothable. However, there are fixed-point free non-smoothable involutions $\tau: S^4\to S^4$ such that the fake $RP^4=S^4/\tau$ is a non-smoothable 4-manifold. See:

Daniel Ruberman, Invariant knots of free involutions of $S^4$, Topology Appl. 18, 217-224 (1984). ZBL0559.57016.

Moishe Kohan
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  • Thanks for your answer! I had a look at the paper you linked, unfortunately I'm an undergraduate student and this seems a bit too advanced for me. I will post my thought process in a separate comment, maybe you can tell me what my mistake is :-) – LordOfNumbers Apr 12 '20 at 19:35
  • With $\pi:M\to M/\tau$ ${}$ I denote the projection $\pi(x)=[x]$. Since $\tau$ is fixed-point free one has that for sufficiently small sets U, $\pi:U\to \pi(U)$ is a homeomorphism. I think this is independent of $\tau$ being smooth or continuous.

    And I also thought that for chart $\varphi:U\to \mathbb{R}^n$ of M one gets a chart $\hat{\varphi}=\varphi \circ \left.\pi\right|_{U}^{-1}:\pi(U) \to \mathbb{R}^n$ of $M/\tau$. So one has a differentiable structure. And then $\pi:U\to \pi(U)$ should be a diffeomorphism since $\hat{\varphi} \circ \left.\pi\right|_U \circ \varphi=id$.

     – LordOfNumbers Apr 12 '20 at 19:36
  • @LordOfNumbers: Yes, this is quite advanced, as any proofs in this area of topology. As for your attempt, check transition maps of your charts $\hat\phi$: The transition maps will not be smooth without assuming smoothness of $\tau$. (Just take another chart corresponding to $\tau(U)$.) – Moishe Kohan Apr 12 '20 at 19:39
  • Ah yes of course! Thank you so much. Question solved! – LordOfNumbers Apr 12 '20 at 20:04