# Manifold acted on by every finite group

Is there a (second-countable, connected) manifold that admits a faithful continuous group action by every finite group?

• Thank you @josé-carlos-santos. Do you have any particular suggestions on how to improve this question? I'm having a hard time improving it, since the definitions are standard, I could not find the question in the literature, but I expect that it has a fairly straightforward, reference-style answer. – Z. A. K. Feb 18 '19 at 14:20
• @Z.A.K. In my opinon, it lacks context. – José Carlos Santos Feb 18 '19 at 16:11
• @José: I think it is a perfectly natural question. Consider the analogue for Lie groups: is there a Lie group $G$ into which every finite group embeds? It turns out that the answer is no. Here is one argument: every finite subgroup of a Lie group is contained in a maximal compact, so we can assume WLOG that $G$ is compact. Then it embeds into some $U(n)$ by Peter-Weyl, so we can assume WLOG that $G$ is $U(n)$. Now it remains to exhibit a finite group whose smallest faithful representation has dimension greater than $n$, which can be done several ways. This question is for diffeomorphism... – Qiaochu Yuan Feb 18 '19 at 22:16
• ...or homeomorphism groups of manifolds rather than Lie groups. I think it would be quite interesting to see what facts about manifolds get used in either a positive or a negative answer. For example, here is at least the beginning of a reduction similar to the reduction above: in the smooth case, by averaging a Riemannian metric we can reduce from "diffeomorphism group of a smooth manifold" to "isometry group of a Riemannian manifold." – Qiaochu Yuan Feb 18 '19 at 22:17
• Elementary action of $p$-groups on manifolds by Mann and Su have that the answer is no for the closed case, I might post(or update) an answer to include this information, but I want to understand some of the ideas before posting. I guess it basically uses Smith theory – user29123 Feb 18 '19 at 23:06

Yes. There is a universal finitely presented group $$G$$ that contains an isomorphic copy of every finitely presented group; this follows from Higman's embedding theorem. Choose a finite generating set of this group, and construct the Cayley graph. Choose a nice embedding of this graph in Euclidean 3-space and take a regular neighborhood $$N$$; let $$M$$ be the boundary of $$N$$. Then $$G$$ acts faithfully on a manifold homeomorphic to $$M$$. You can construct such a homeomorphic copy by replacing the vertices of the Cayley graph by spheres with twice as many holes as the number of generators. Then attach a cylinder for each edge to the spheres corresponding to the end vertices. $$M$$ is a surface of infinite genus, but it is connected and 2nd countable. Each finite group acts by choosing an embedding into $$G$$. Since no vertex sphere is fixed, except by the identity, $$F$$ acts faithfully.
I will argue that the infinite genus surface, $$\Sigma_C$$, with the space of ends being a Cantor set has this property(think an regular infinite tree except genus).
Let $$\Sigma_g$$ be the surface of genus $$g$$ and $$\mathrm{Sym}_n$$ symmetric group on $$n$$ elements. There is a standard construction to show that for any $$\mathrm{Sym}_n$$ there is a $$g$$ large enough so that $$\mathrm{Homeo} (\Sigma_g$$) contains that symmetric groups: consider a Cayley graph for $$\mathrm{Sym}_n$$, now "thicken" it up to a surface by "thickening" each vertex to a torus, than the edges will correspond to connect sum of two tori(connecting with a tube/annulus). Now an action on the graph can be turned into an action on the surface that has been constructed.
On the surface we have constructed, choose a small disk on one of the tori/vertices, and look at the orbit of the disk, cut out the orbit and attach a thickened (under the same process as above) infinite rooted binary tree. Your group still acts on this space, where the group takes the trees you planted to trees, and preserves the thickened Cayley graph. For each $$n$$ these surfaces are homeomorphic by the "classification of surfaces" above.