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



Fixed point set of an isometric group action on an hyperbolic manifold
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](http://www.math.ucdavis.edu/~kapovich/EPR/gt.pdf). Namely, I explain why manifolds in the 2nd construction are not homotopy-equivalent to hyperbolic ones. 


