# Fixed point set of an isometric group action on an hyperbolic manifold

Good morning,

I'm trying to understand the following fact, that is stated in Gromov and Thurston's paper "Pinching constants for hyperbolic manifolds" :

Let $M$ be a (at least) 3-dimensional compact oriented hyperbolic manifold, with an action of $\mathbb{Z}_i$

by diffeomorphism. Suppose the set of fixed points of the action is a compact oriented submanifold of codimendion $2$, name it $V$, such that its homology class in $H_{n-2}(M, \mathbb{R})$ is $0$.

Mostow's Rigidity theorem ensures that $\mathbb{Z}_i$ acts by isometry.

The claim that I don't understand is : "The fixed point set of this new action is a codimension $2$ submanifold of $M$ diffeomorphic to $V$"

It is not clear in the article whether those hypothesis are sufficient or not, since the submanifold $V$ is the set of ramification of a certain covering map over an hyperbolic manifold in which the image of $V$ is totally geodesic, an the action of $\mathbb{Z}_i$ is an action by deck transformation.

Thanks for the help

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What is $\mathbb{Z}_i$? – Lee Mosher Apr 22 '13 at 12:11
Sorry I should have precised, it's the notation used in the article for $\mathbb{Z} / i\mathbb{Z}$ – Selim G Apr 22 '13 at 12:28

Here is the setup with which you are dealing: You have a smooth closed manifold $M$ and a finite cyclic branched covering $p: M\to M'$ ramified over a codimension 2 totally geodesic submanifold $V'\subset M'$, where $M'$ is hyperbolic. Let $V=p^{-1}(V')$. Furthermore, you know (from the construction) that $p: V\to V'$ is a diffeomorphism. In particular, $V$ is diffeomorphic to a closed hyperbolic manifold $V'$. Let $\beta$ denote the generator of the deck group of $p$. Now, you are trying to prove that the topological manifold $M$ does not admit a hyperbolic metric $g$. Suppose it does (smooth structure might change at this point). Then, by Mostow Rigidity theorem, $\beta: M\to M$ is homotopic to an isometry $\alpha: (M,g)\to (M,g)$. In particular, the fixed-point set $F$ of $\alpha$ is a closed hyperbolic submanifold. Note that $V$ is typically not connected, so you have to argue for each component $V_j$ of $V$. First, you note that each inclusion map $V_j\to M$ is $\pi_1$-injective (since this is the case for $V'$). Next, let's lift everything to the universal cover of $M$, which is supposed to be the hyperbolic $n$-space $H^n$, $M=H^n/\Gamma$. Then the lifts $\tilde\beta, \tilde\alpha$ can be chosen so that they agree on the sphere at infinity. The subgroup $\pi_1(V_j)\subset \Gamma$ consists of elements which commute with $\tilde\beta$. The same applies to $\pi_1(F_j)$ and to $\tilde\alpha$. Therefore, we have a natural isomorphism $\pi_1(V_j)\cong \pi_1(F_j)$. By construction, $V$ is diffeomorphic to $V'$, so $V_j$ is diffeomorphic to a hyperbolic manifold. Now, it follows from Mostow rigidity (if $dim(V)\ge 3$) or classification of surfaces (if $dim(V)=2$) that $V_j$ is diffeomorphic to $F_j$, since they are both diffeomorphic to hyperbolic manifolds with isomorphic fundamental groups.

With a bit more care, the same argument goes through if you merely assume that $M$ is homotopy-equivalent to a hyperbolic manifold.

Now, one can ask if something stronger holds: Let $M$ be a closed hyperbolic manifold and $\beta: M\to M$ be a finite order diffeomorphism. Let $\alpha: M\to M$ be an isometry homotopic to $\beta$. Is it true that fixed-point sets of $\alpha, \beta$ are diffeomorphic? Note that both orbifold $M/(\alpha), M/(\beta)$ are aspherical and have isomorphic fundamental groups. In dimension 3 it suffices to conclude that they are diffeomorphic. In dimensions $\ge 5$, it probably follows from some equivariant form of Farrel and Jones but I am not sure (I have not looked at their papers for a long time). Situations in smooth category and dimension 4 seem totally unclear.

Lastly: The paper you are reading has two constructions of negatively curved manifolds not homotopy-equivalent to locally symmetric ones. The second construction is less known since people usually get tired from reading the first one. Some details for the second construction could be found here. Namely, I explain why manifolds in the 2nd construction are not homotopy-equivalent to hyperbolic ones.

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Thank you very much for your clear and detailed answer – Selim G Apr 24 '13 at 9:18
I have been looking at the detail of your answer and two details remain not clear to me, maybe because I don't know enough hyperbolic geometry : _ You claim that $\Pi_1(V_j)$ consists of elements commuting with $\tilde{\beta}$, but not of THE elements commuting with $\tilde{\beta}$ right ? In that case, where does the isomorphism $\Pi_1(V_j) \simeq \Pi_1(F_j)$ comes from ? _ Why do $F_j$ and $V_j$ have the same dimension ? Or maybe you don't use that fact, in that case if $dim(V_j) = 2$ and $dim(F_j) = 3$ how do you conclude ? – Selim G Apr 25 '13 at 12:21
Yes, $\pi_1(V_j)$ is exactly the set of elements commuting with $\tilde\beta$. One inclusion is clear. This shows that $\tilde\beta$ fixes pointwise a codimension 2 round sphere $S$ in the sphere at infinity. If there are extra elements which commute with $\tilde\beta$ there are two possibilities: (1) Their fixed points are not in $S$. Then $\tilde\beta$ is either identity or reverses orientation. This is not the case. (2) Their fixed points are in $S$. Then you take such element $g$ and consider a group $H$ it generates together with $F=\pi_1(V_j)$. Then $H$ is a finite extension of $F$... – Misha Apr 25 '13 at 12:36
But then $\pi_1(V_j)\subset \pi_1(M')$ also admits a finite index proper extension. This contradicts the fact that $V_j'$ is embedded in $M'$. The same arguments work for $\tilde\alpha$. – Misha Apr 25 '13 at 12:39
ok those last points are clear. Can I bother you with one last question ? Why is the fixed point set of $\tilde\beta$ a round sphere ? It is the accumulation set of $\tilde{V_j}$ in $\tilde{\mathbb{H}^n}$ but why can't it be degenerate ? – Selim G Apr 25 '13 at 14:44