# Hyperbolic Manifolds which fiber over the circle

If $N^2$ is a closed, orientable surface of genus at least $2$, and if $\phi$ is an (orientation-preserving) pseudo-Anosov mapping on $N$, then one can form the closed orientable 3-manifold $M^3$ by gluing the boundary components of $N^2\times [0,1]$ via $\phi$. In other words, $M^3$ is fibered over $S^1$, with fiber $N^2$. $M^3$ can also be given a hyperbolic structure (much like $N^2$ can). So my question is:

Does there exist a closed orientable hyperbolic manifold $M^k$, fibered over $S^1$ with a fiber $N^{k-1}$, when $k\neq 3$?

No such $M$ exists for $k=2$. It is also known (via Mostow rigidity) that for such an $M^k$ to exist for $k>3$, we cannot have the fiber $N^{k-1}$ hyperbolizable as well. But I don't know anything further than that.

• It is impossible for even $k$ and wide-open for odd. An example would also give an example if non hyperbolic group of finite type without Baumslag-Solitar subgroups, existence if which is unknown. – Misha Sep 19 '13 at 17:36
• @Misha: Thank you for the quick reply! Do you know a reference for the impossibility when $k$ is even? Also, do you think there is a general "leaning" among experts whether such an odd $k$ exists or not? – Steve D Sep 19 '13 at 18:01
• For even $k$ this is just the Euler characteristic obstruction: It is nonzero for hyperbolic manifolds and zero for manifolds fibering over $S^1$. For odd $k$ my gut feeling is that such manifolds do not exist, but we have no tools for proving this. – Misha Sep 19 '13 at 18:52
• The Euler characteristic of such a fibration is $0,$ so if $k$ is even, the volume of a hyperbolic such $M^k$ is also $0$ by Gauss-Bonnet... – Igor Rivin Sep 19 '13 at 18:54
• One more thing that we know is that if such a fibration exists, then the fundamental group of the fiber cannot be word-hyperbolic: This is a corollary of the Rips theory (as the fiber group has an outer automorphism of infinite order). – Misha Sep 19 '13 at 19:53

I make a remark here that is well-known to experts, that $\pi_1(M)$ is an extension of $\pi_1(N)$ by an infinite-order outer automorphism, explaining partly Misha's comment.
Consider the sequence $\pi_1(N^{k-1}) \to \pi_1(M^k) \overset{\varphi}{\to} \mathbb{Z}$. Choose $\alpha\in \pi_1(M)$ such that $\varphi(\alpha)=1\in \mathbb{Z}$. Then conjugation $g\mapsto \alpha^k g\alpha^{-k}, g\in \pi_1(N)$ gives an automorphism of $\pi_1(N)$. If this automorphism is conjugation by an element of $\pi_1(N)$ for some $k$, then there exists $h\in \pi_1(N)$ such that $\alpha^k g \alpha^{-k}=hgh^{-1}$ for all $g\in \pi_1(N)$. Let $\alpha'=h^{-1}\alpha^k$, then $\alpha' g= g \alpha'$ for all $g\in \pi_1(N)$. Consider the maximal cyclic subgroup $C<\pi_1(M)$ containing $\alpha'$ (there is a unique such group since $\pi_1(M)$ is the fundamental group of a closed hyperbolic manifold, so $C$ is the stabilizer of the axis of $\alpha'$ by the proof of Preissman's theorem). Also by Preissman's theorem, the subgroup generated by $\langle \alpha', g \rangle$ is abelian and therefore cyclic, and therefore $g\in C$. But this implies that $\pi_1(N)\leq C$, so $\pi_1(M)\leq C$, a contradiction. Thus, the conjugation by $\alpha$ maps to a infinite order element of $Out(\pi_1(N))$.
Then if $\pi_1(N)$ is hyperbolic, one can deduce that $\pi_1(N)$ splits over $\mathbb{Z}$ by Rips' theory. However, an aspherical closed $k-1$-manifold group cannot split over $\mathbb{Z}$ for $k>2$.
• Thank you for the informative post. Let me just remark that there is a (slightly) different proof about $N$ not being hyperbolic, which I suggested in my post: using Mostow rigidity we can homotope $\alpha$ to be an isometry, and then the fact that the isometry group of $N$ is finite (indeed, isomorphic to $Out(\pi_1(N))$), we get $M$ finitely covered by $N\times S^1$. – Steve D Sep 23 '13 at 16:50
• @SteveD: by $\pi_1(N)$ hyperbolic, I mean $\delta$-hyperbolic in the sense of Gromov. So, for example, this includes the case that $N$ admits a negatively curved metric, but might not be constant negative curvature (although it is more general than that). – Ian Agol Sep 23 '13 at 17:35