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Does there exist a closed smooth manifold $M$ and a diffeomorphism $f\colon M \to M$ such that:

  1. $f$ is isotopic to the identity,
  2. $f$ is of finite order, $f^n=ID$, and
  3. $f$ is not contained in the image of any circle action, $S^1\to \operatorname{Diff}(M)$?

If the answer is yes, what is an example?

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@Igor this doesn't give you a subgroup because you don't know that you can choose $\gamma_0$ to be a part of a one-parameter subgroup of diffeos. (the exponential map for $Diff_0(M)$ is not onto). – Vitali Kapovitch May 6 '12 at 19:18
What happens if you restrict the $S^1$ action on $S^7$ to $Z/n$ and connect a homotopy sphere to each point in a $Z/n$ orbit? If $n$ divides 28, then the result is a $Z/n$ action on $S^7$ s.t. $f$ is isotopic to the identity, but it doesnt seem likely that the action embeds in an $S^1$ action. – Paul May 6 '12 at 21:35
Another possible example is the following: Start with, say $E_8\oplus E_8$ topological manifold $N^4$, let $M^5=M^4\times S^1$. Then $M^5$ has a smooth structure but no smooth product structure. Now, let $L^5$ be the $2$-fold cover of $M^5$. Let $f: L^5\to L_5$ be the deck-transformation (of order $2$). Then $L^5$ will not have a smooth $S^1$-action containing $f$. The map $f$ is homotopic to the identity. Unfortunately, I do not know enough higher-dimensional topology to conclude if $f$ is also smoothly isotopic to $id$, if it is, then one would get the desired example. – Misha May 7 '12 at 3:54
cf. Corollary 2 of this paper:… I'm not sure though whether the generator of the cyclic group is isotopic to the identity. – Ian Agol May 7 '12 at 5:22
Note that in the topological category, the example exists already for $M=S^3$: Bing constructed finite order homeomorphisms whose fixed-point sets are wild knots, while Raymond proved that the fixed-point set of every nontrivial $S^1$-action on $S^3$ is topologically conjugate to an orthogonal action. – Misha May 7 '12 at 22:02
up vote 18 down vote accepted

Such examples exist in dimension 5, they are contained in the paper by Cameron Gordon "On the higher-dimensional Smith conjecture", Proc. London Math. Soc. (3) 29 (1974), 98–110. Namely, Gordon proves in Theorem 5 of this paper that (for every $n\ge 5$) there are infinitely many smooth knots $K=S^{n-2}\subset S^n$ so that $K$ is the fixed-point set of a ${\mathbb Z}_p$-action $\alpha_p$ for every prime $p$. He also proves (Theorem 4) that, given $K$, if every action $\alpha_p$ extends to a circle action on $S^n$ then $\pi_1(S^n\setminus K)\cong {\mathbb Z}$. He then notes (a theorem by Levine) that, for $n\ge 5$, if $\pi_1(S^n\setminus K)\cong {\mathbb Z}$ then $K$ is smoothly unknotted in $S^n$. Since Gordon's theorem 5 yields infinitely many smooth isotopy classes of knots $K$, it then follows for every such (nontrivial) knot, at least for some prime $p$, one of Gordon's actions $\alpha_p$ does not extend to a smooth circle action.

Finally, every diffeomorphism of $S^5$ is PL isotopic to the identity (by the Alexander's trick). Since in dimensions $<7$, PL=DIFF, we conclude that the generator of $\alpha_p({\mathbb Z}_p)$ is smoothly isotopic to the identity.

Lastly, note that in the topological category, the examples exists already for $M=S^3$: Bing ("Inequivalent Families of Periodic Homeomorphisms of $E^3$", Ann. of Math. 80 (1964) 78-93.) constructed finite order homeomorphisms whose fixed-point sets are wild knots, while F.Raymond (Classification of the actions of the circle on $3$-manifolds. Trans. Amer. Math. Soc. 131 (1968) 51-78.) proved that the every $S^1$-action on $S^3$ is topologically conjugate to an orthogonal action.

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I think you mean $K=S^{n-2} \subset S^n$. – Ian Agol May 8 '12 at 12:48
Of cours, thanks, Ian! – Misha May 8 '12 at 12:59
Dear Misha and all who contributed, Thank you very much for the comments and the answer. This is my first experience with mathoverflow and I have to say it is very positive. THANKS! – Jarek Kedra May 10 '12 at 15:04

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