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:



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



Hyperbolic manifolds which fiber over the circle
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](https://en.wikipedia.org/wiki/Preissmann%27s_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 $k>2$.


