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Let $G\leq \operatorname{Homeo}^+(S_g)$ be finite, where $S_g$ is a closed, connected, orientable surface of genus at least $2$. Then I have the following questions:

(1) Can $G$ always be realized as a subgroup of $\operatorname{Diffeo}^+(S_g)$ for some smooth structure on $S_g$?

(2) Can $G$ always be realized as a subgroup of $\operatorname{Isom}^+(S_g)$ for some hyperbolic metric on $S_g$?

Note that: (a) The questions are equivalent.

(b) Nielsen-Realization theorem is not helpful in the above context as it starts with a finite subgroup of $\operatorname{Mod}(S_g)$.

(c) Originally I wanted to prove that the restriction of the natural projection map $ \operatorname{Homeo}^+(S_g)\to \operatorname{Mod}(S_g)$ to $G$ is injective. In Farb and Margalit's "A primer on mapping class groups" it is written that this follows from Theorem 6.8 (which says that the Torelli group is torsion free) but I am not able to deduce it from this theorem.

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  • $\begingroup$ For (2), can't you just average any hyperbolic metric over $G$? (This is the standard trick for unitarising a complex, linear representation of a finite group—but I'm no geometer, and maybe it breaks something about the geometry.) $\endgroup$
    – LSpice
    Commented Sep 18, 2023 at 13:20
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    $\begingroup$ mathoverflow.net/questions/225355/… $\endgroup$
    – Moishe Kohan
    Commented Sep 18, 2023 at 13:37
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    $\begingroup$ @LSpice: (1)=>(2) First, average any Riemannian metric to get a G-invariant Riemannian metric then apply the uniformization theorem to obtain a hyperbolic metric in the conformal class of the earlier metric which realizes G as isometries. (2)=>(1) is trivial. But note that, averaging a metric would require maps in G to be diffeomorphism. Here I'm starting with a group in Homeo+(S_g). $\endgroup$
    – Rajesh Dey
    Commented Sep 19, 2023 at 4:01
  • $\begingroup$ Re, ah, got it, thanks. (So it potentially breaks even the topology, not just he geometry!) $\endgroup$
    – LSpice
    Commented Sep 19, 2023 at 17:28
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    $\begingroup$ Wat you wanted to prove in (c) is true. The kernel of the projection is torsion-free and thus intersects trivially with the finite group G. To see why it is torsion-free, you can see the answers at the post here $\endgroup$
    – Lvzhou Chen
    Commented Sep 20, 2023 at 1:39

1 Answer 1

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Here is Moishe Kohan’s comment turned into an answer.

Suppose that $S$ is a closed, connected, oriented surface of genus at least two. Let $\rho \colon \mathrm{Homeo}^+(S) \to \mathrm{MCG}^+(S)$ be the homomorphism taking a homeomorphism to its isotopy class. Suppose that $G < \mathrm{Homeo}^+(S)$ is the given finite subgroup. By (c) we have that $\rho|G$ is an isomorphism. Let $G^* = \rho(G)$. By Nielsen realisation there is a metric on $S$, and a subgroup $G^{**} < \mathrm{Homeo}^+(S)$ so that $\rho|G^{**}$ is an isomorphism to $G^*$ and so that $G^{**}$ acts on $S$ via isometries. Let $\mathcal{O} = S/G^{**}$ be the resulting orbifold.

Claim: $S / G$ is homeomorphic (as an orbifold) to $\mathcal{O}$.

Proof: If $G$ acts freely then we are done. So, suppose that $x \in S$ be a point that is fixed by some non-identity element of $G$. Let $G_x$ be the stabiliser of $x$. We lift the action of $G_x$ to the universal cover of $S$ (choosing of course the lifts that fix a chosen preimage of $x$). By the result Moishe Kohan points at in the comments the lifted action is by (rational!) rotations. $\square$

Using this and "transfer of structure" the answers to (1) and (2) are "yes". I think this means that $G$ and $G^{**}$ are in fact conjugate.

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