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It is known that if a finite group $G$ admits a faithful topological action on the 3-sphere $S^3$, then $G$ admits a faithful action on $S^3$ by isometries. (Pardon proved that a topological action implies a smooth action, and Dinkelbach & Leeb proved that a smooth action implies an isometric one.) I wonder if this extends to infinite groups acting on $R^3$:

Question: Let $G$ be a finitely generated group that admits a faithful, co-compact, topological action on $R^3$, such that no orbit has an accumulation point. Must $G$ admit an action by isometries on one of Thurston’s geometries, preserving the above properties (i.e. faithful, co-compact, accumulation-free)?

Update: The comments below suggest that the answer is negative in this generality (an "official" answer with references and explanation would be welcome). What if G is assumed to be Gromov-hyperbolic? I'm most interested in the 1-ended case, anticipating an isometric action on $\mathbb{H}^3$. (1-endedness excludes $\mathbb{S}^2 \times \mathbb{R}$.) This is partly motivated by Cannon's conjecture.

By topological action I mean an action by homeomorphisms.

Update: instead of just assuming that no orbit has an accumulation point, I'm happy with stronger discreteness conditions such as proper discontinuity

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  • $\begingroup$ Could you define "cocompact"? By "the orbit has no accumulation point" do you mean that it has no accumulation in the orbit, or in the ambient space? $\endgroup$
    – YCor
    Commented Mar 11, 2022 at 17:19
  • $\begingroup$ @YCor: By cocompact I mean that there is a compact subset K of R^3 such that the image of K under the action of G covers R^3. The other condition says that the orbit of any point has no accumulation point in R^3. $\endgroup$
    – Agelos
    Commented Mar 11, 2022 at 17:25
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    $\begingroup$ @Agelos the group is the fundamental group of the resulting closed aspherical manifold acting on the universal cover by deck transformations. it can not act isometrically discretely and cocompactly on any of the Thurston geometries because it's neither virtually solvable nor is it Gromov Hyperbolic. $\endgroup$
    – Vitali Kapovitch
    Commented Mar 11, 2022 at 19:07
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    $\begingroup$ @VitaliKapovitch: you’re right that graph manifolds provide counterexamples, but it’s not true that the fundamental group of a geometric 3-manifold is either virtually solvable or Gromov hyperbolic. The fundamental groups of Seifert fibres manifolds with hyperbolic base are neither of these. $\endgroup$
    – HJRW
    Commented Mar 12, 2022 at 23:31
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    $\begingroup$ In the present form, assuming tameness of the action, the answer is positive. It is an easy corollary of the Geometrization Theorem. $\endgroup$
    – Moishe Kohan
    Commented Mar 17, 2022 at 20:39

2 Answers 2

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I believe (though have not checked carefully) that the argument in my paper proves:

If $\Gamma$ (discrete) acts continuously and properly discontinuously on a smooth three-manifold $M$, then that action can be uniformly approximated by a smooth action.

The point is simply that each step in the argument is local on the quotient space $M/\Gamma$ (which is a reasonable topological space given proper discontinuity).

Here is a (sketched) better argument, which proves the indented statement above as a consequence of my paper. Fix $x\in M$, and consider the stabilizer $\Gamma_x\leq\Gamma$, which is finite. Choose coset representatives $g_i\in\Gamma/\Gamma_x$, so $\Gamma x=\{g_ix\}_i$. Fix a $\Gamma_x$-invariant open neighborhood $U$ of $x$ whose translates $g_iU$ are all disjoint (should exist by proper discontinuity). Now smooth the action of $\Gamma_x$ on $U$ using my paper, and smooth the homeomorphisms $g_i:U\to g_iU$ using Bing--Moise. This determines a smoothing of the action of $\Gamma$ on $\Gamma U\subseteq M$. By making the approximations sufficiently $C^0$-close, we ensure that this smoothed action of $\Gamma$ on $\Gamma U\subseteq M$ splices together with the original action of $\Gamma$ on $M\setminus\Gamma U$ to define a new action of $\Gamma$ on $M$, which is now smooth over $\Gamma U$. Now iterate a (locally) finite number of times to cover all of $M$.

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  • $\begingroup$ @HJRW : does this combine with your answer to give an action by isometries? $\endgroup$
    – Agelos
    Commented Apr 19, 2022 at 8:04
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My earlier attempt at an answer was a bit of a mess -- let me have another go.

The hypotheses of the question introduce several technical difficulties, but I'm unsure which are crucial and which can be relaxed. Certainly, if we're willing to relax them slightly then we can get a positive answer, so I'll give an answer under certain hypotheses that seem reasonable to me.

The usual discreteness hypothesis in this context is not an absence of accumulation points, but proper discontinuity, and it seems to me that this is the natural way to generalise Pardon's theorem. With the hypothesis of absence of accumulation points (which @YCor rightly points out is strictly weaker) it's not clear to me what happens even for smooth actions. Apologies if this strictly weaker properness hypothesis is the point of the question (but I don't see a connection with Cannon's conjecture).

So let's suppose that $\Gamma$ is a hyperbolic group acting properly discontinuously and cocompactly by homeomorphisms on $\mathbb{R}^3$. To keep things simple, let's also assume that $\Gamma$ has a (wlog normal) torsion-free subgroup $\Gamma_0$ of finite index.

Since the action is properly discontinuous and $\Gamma_0$ is torsion-free, the action of $\Gamma_0$ is free and so the quotient $M_0=\Gamma_0\backslash\mathbb{R}^3$ is a closed topological 3-manifold.

By Moise's theorem $M_0$ has a smooth structure, and now $M_0$ is an aspherical 3-manifold whose fundamental group has no $\mathbb{Z}^2$ subgroups, so $M_0$ admits a hyperbolic metric by the geometrisation theorem. This metric pulls back to realise the action of $\Gamma_0$ as an action by isometries on $\mathbb{H}^3$.

Pardon's theorem shows that the action of the finite deck group $\Gamma\backslash\Gamma_0$ on $M_0$ can be approximated by smooth actions, and a theorem of Gabai implies that this action is isotopic to an action by isometries. As a result, the action of the whole group $\Gamma$ on $\mathbb{H}^3$ is also by isometries, as desired.

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    $\begingroup$ I think Tukia's 1986 "quasiconformal" paper (DOI link) directly implies that every discrete group quasi-isometric to $X=H^n(\mathbf{R})$ (for $n\ge 3$) has a geometric action on $X$, without passing to any finite index subgroup. (This is also true for $n=2$ but was proved later by other authors.) $\endgroup$
    – YCor
    Commented Mar 17, 2022 at 22:23
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    $\begingroup$ @Agelos Say, this one: for $n=2$, is every discrete group acting properly cocompactly continuously on $\mathbf{R}^n$ virtually torsion-free? What if assumed in addition to be hyperbolic? $\endgroup$
    – YCor
    Commented Mar 21, 2022 at 9:48
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    $\begingroup$ @YCor: For n=2 this is true, using the classification of compact, 2-dimensional orbifolds and the fact that all orbifold groups are "good in the sense of Serre". (This latter implies that finite extensions of orbifold groups are residually finite, hence also virtually torsion free.) For n=3 and a smooth action it's probably also true by a suitable orbifold version of geometrisation and the fact that 3-manifold groups are also good. The case of a non-smooth action seems interesting, and one should probably ask Pardon. For $n\geq 4$ I suspect it's false, even in the smooth case. $\endgroup$
    – HJRW
    Commented Mar 21, 2022 at 14:01
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    $\begingroup$ @Agelos: If your proof produces an equivariant embedding of the Cayley graph into the plane then the action on the plane can easily be made piecewise linear, so all the difficulties arising from only assuming the action to be by homeomorphisms go away. The quotient is then visibly an orbifold, and the result follows from the fact that orbifolds admit geometric structures. Virtual torsion-freeness is then a byproduct (and one would still probably use Selberg's lemma to handle the case of triangle orbifolds). $\endgroup$
    – HJRW
    Commented Mar 22, 2022 at 12:25
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    $\begingroup$ @Agelos: That's right. Since $G$ is one-ended with no $\mathbb{Z}^2$ subgroups, the orbifold and geometrisation theorems imply that the orbifold is hyperbolic, so $G$ is a finite extension of a Kleinian group. One consequence of Agol's Virtual Haken theorem is that Kleinian groups are also "good in the sense of Serre", from which it follows that $G$ is residually finite and hence virtually torsion-free. $\endgroup$
    – HJRW
    Commented Mar 22, 2022 at 16:46

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