Let $M^3$ be a smooth three-manifold, and let $\gamma:G\to\operatorname{Homeo}(M)$ be a finite group action on $M$ by homeomorphisms.

Can $\gamma$ can be $C^0$-approximated by smooth group actions $\tilde\gamma:G\to\operatorname{Diff}(M)$?

Note that Bing and Moise proved (independently) that any homeomorphism $h:M\to M$ can be $C^0$-approximated by diffeomorphisms $\tilde h:M\to M$, however this does not a priori imply a positive answer to the question of approximation of group actions.

(I presume the answer to the analogous question in two dimensions is positive, but if this is not the case, that would be interesting as well.)

EDIT: Bing in this paper defined a continuous involution $\sigma:S^3\to S^3$ whose fixed set is a wildly embedded $S^2\hookrightarrow S^3$ (so, in particular, $\sigma$ is not topologically conjugate to a smooth involution). Bing also showed (in the very same paper!) that $\sigma$ is a $C^0$-limit of smooth involutions. Indeed, Bing considers a smooth involution $r:S^3\to S^3$ fixing a smooth $S^2\subseteq S^3$ and a small unknot $K\subseteq S^3$ stabilized by $r$ and intersecting the fixed locus transversally in two points. He then considers a sequence of diffeomorphisms $\varphi_n:S^3\to S^3$ which shrink the $n$th iterated Bing doubles $B^n(K)$ of $K$, namely every component of $\varphi_n(B^n(K))$ has diameter at most $\varepsilon_n>0$, where $\varepsilon_n\to 0$ as $n\to\infty$. Bing shows that (for judiciously chosen $\varphi_n$), the limit $\sigma:=\varphi_n\circ r\circ\varphi_n^{-1}$ exists and is the desired wild involution of $S^3$ (although, of course, the conjugating diffeomorphisms $\varphi_n$ do not converge to a homeomorphism, otherwise $\sigma$ would be topologically conjugate to $r$).

  • $\begingroup$ A similar shrinking/decomposition space argument probably works for the involutions of Montgomery-Zippen (see also Alford), but I haven't thought it through carefully. jstor.org/stable/2031959 jstor.org/stable/2035086 $\endgroup$
    – Ian Agol
    Oct 3, 2017 at 23:26

2 Answers 2


See Corollary 3.1 of the following paper by Robert Francis Craggs for the case of any involution of $S^3$ whose fixed set is homeomorphic to $S^2$:




I am a little confused. It is a theorem of Bing that there are periodic homeomorphisms of $S^3$ which are not conjugate to orthogonal actions (one of these has the Alexander horned sphere as the fixed point set). This being the case, how could this group be approximated by smooth maps (since any smooth finite order map is conjugate to an orthogonal map by the Smith conjecture). Am I missing something?

Bing, R.H., A homeomorphism between the 3-sphere and the sum of two solid horned spheres, Ann. Math. (2) 56, 354-362 (1952). ZBL0049.40401.

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    $\begingroup$ I don't insist $\gamma$ and $\tilde\gamma$ be topologically conjugate, just $C^0$ close. $\endgroup$ Sep 30, 2017 at 0:13
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    $\begingroup$ A $C^0$-limit of conjugates of orthogonal maps need not be conjugate to an orthogonal map. It's tempting to think that by passing to subsequences you can ensure that the corresponding sequence of orthogonal maps converge and that the conjugating elements converge (this was also my first impulse when I read the problem), but if you try to write this out carefully it doesn't work. $\endgroup$ Sep 30, 2017 at 0:39
  • $\begingroup$ @AndyPutman Yes, I thought that it was something like this, but, as you say, it is not obvious either way. $\endgroup$
    – Igor Rivin
    Sep 30, 2017 at 22:47
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    $\begingroup$ @IgorRivin: in fact Bing himself showed in the very same paper that his exotic involution is a $C^0$-limit of smooth involutions (see my edit to the question for a summary of his construction). $\endgroup$ Oct 3, 2017 at 15:16

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