# Spaces that are finitely covered by manifolds

Suppose $X \to Y$ is a finite-sheeted covering of CW-complexes. Moreover, assume that the total space is homotopy equivalent to a (closed, connected, smooth) manifold $M$. I am interested in conditions to impose on the map $X \to Y$ so that we can find a manifold $N \simeq Y$ and a covering $M \to N$ making the diagram $$\begin{array}{ccc} X & \simeq & M \\ \downarrow && \downarrow \\ Y & \simeq & N \end{array}$$ commute. In general, this is not possible, for example we could take any finite group $G$ and $$X = M \times EG \to M \times BG = Y.$$ Are there counterexamples for $Y$ a finite CW-complex? On the other side, are there conditions to impose so that we actually can deduce the existence of $N$?

• If I remember correctly, there is a 3-dimensional compact (3D) Poincare duality complex with finite fundamental group not homotopy equivalent to a 3-dimensional manifold. The universal cover of this complex is homotopy-equivalent to $S^3$. This will provide a counter example. Check C. Thomas, 3-Manifolds and PD(3)-groups, an example might be there. – Misha Jan 30 '17 at 1:16
• @Misha: I looked it up, it's in the appendix to a paper of Swan. jstor.org/stable/… ams.org/mathscinet-getitem?mr=124895 – Ian Agol Jan 30 '17 at 5:27
• @IanAgol: Good, so there is an example with $\pi_1(Y)\cong S_3$. – Misha Jan 30 '17 at 16:29

Conditions where $N$ exists are for example when $M$ is a non-positively curved locally symmetric manifold, by a theorem of Mostow (for hyperbolic manifolds) and Margulis in general. Mostow's theorem is generalized to manifolds with hyperbolic fundamental group whose Gromov boundary is a sphere of dimension at least 5. So if one started with such a manifold, any group containing the fundamental group with finite index and torsion-free would have to be a manifold group.
More generally, it is conjectured that finitely presented groups which are $PD(n)$ groups are the fundamental groups of closed $n$-manifolds (Conjecture 3.4). If true, this would show that $N$ exists when $X$ is aspherical. This conjecture would follow for $n\geq 5$ from the Novikov conjecture. Many cases of this conjecture are known (for groups satisfying certain restrictions). See also Sections 5 and 8 of this paper.
• I am aware of Mostow's theorem which says that every homotopy equivalence of a hyperbolic manifold is homotopic to an isometry. But how exactly does that imply what I want? The analogue of Mostow's result for $S^1$ is obvious, but nevertheless I do not see how this implies that every finite complex that is finitely covered by some $X \simeq S^1$ has to be itself homotopy equivalent to $S^1$. – Jens Reinhold Jan 31 '17 at 5:45
• @JensReinhold: The point is that any homotopy equivalence of a hyperbolic manifold is homotopy equivalent to an isometry. If $M$ is hyperbolic, the finite-sheeted cover $X\to Y$ might not be regular, but we can pass to a finite cover so that $\tilde{X}\to Y$ is regular, and still homotopy equivalent to a hyperbolic manifold $\tilde{M}$. Then the covering translations of $\tilde{X}\to Y$ give homotopy equivalences of $\tilde{M}$, which in turn is realized by a finite group of isometries. The quotient is the desired manifold $N$. – Ian Agol Jan 31 '17 at 5:51
• @JensReinhold: The point is that since $Y$ (and X) are finite complexes, their fundamental groups are torsion-free. So the automorphisms of $\tilde{M}$ I described cannot have fixed points, hence $N$ is a manifold, not an orbifold. There are some details here, but I think that's where finiteness of Y is used. – Ian Agol Jan 31 '17 at 6:01
Indeed, C. Thomas, 3-Manifolds and PD(3)-groups, in "Novikov conjectures", Vol. 2, gives a reference to Swan's example. Namely, there is a finite complex $X$ which is a homotopy 3-sphere, a free $S_3$-action $S_3\times X\to X$, which cannot be realized (up to homotopy) as a free $S_3$-action on $S^3$, since $S_3$ cannot act freely on $S^3$ (the 3-sphere).