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Every non-zero element in $H_2(M,\mathbb Z)$ corresponds to an incompressible surface. So these surfaces are non-separating. But I'm interested in knowing about separating incompressible surfaces. A result of Peter Shalen guarantees that a compact, connected, orientable, irreducible 3-manifold $M$ [such a manifold is known as a Haken manifold] will have a non-trivial separating incompressible surface provided that $H_1(M;\mathbb Q)$ is carried by the boundary of M and that some boundary component of M has genus $> 1$.

Question 1: Does there exist any closed Haken 3-manifold with all its incompressible surfaces are non-separating.

My intuition is NO. If we consider $S^1\times S^1\times S^1$, then all its incompressible surfaces has to be torus since this is the only surface with abelian fundamental group. And then any incompressible torus transverse to the fiber (thinking it as a torus bundle over circle) has to intersect it with a non-trivial simple closed curve. And by some more argument we can prove that it is non separating since it can't separate any fiber.

Question 2: What if the fundamental group is non-abelian? Moreover what if we consider a hyperbolic Haken manifold, then can anyone give me an explicit example or reference of such a manifold whose all incompressible surfaces are non-separating?

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    $\begingroup$ The three-torus is indeed an answer to your first question. It is closed, Haken, and all of its incompressible surfaces are two-tori, and are non-separating. Your second question is more difficult. $\endgroup$ – Sam Nead Oct 31 '17 at 17:40
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    $\begingroup$ For the first part of Q2, plenty of Bieberbach manifolds should work, no? Say the orientable Bieberbach (euclidean) manifold that is a circle bundle over the Klein bottle. $\endgroup$ – Ryan Budney Oct 31 '17 at 22:21
  • $\begingroup$ Here is a revised version of my previous comment. The specific example I gave did not work out, as there's a separating incompressible torus (boundary of a neighbourhood of the incompressible klein bottle). But if you take the torus bundle over the circle where the torus is "square" and the monodromy is order 4 then this manifold does appear to work as I believe the only incompressible surfaces are horizontal, i.e, the fiber. $\endgroup$ – Ryan Budney Nov 1 '17 at 0:52
  • $\begingroup$ @RyanBudney whats it's fundamental group? Is it non-abelian? $\endgroup$ – Anubhav Mukherjee Nov 2 '17 at 14:25
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    $\begingroup$ The first two sentences of your question are not quite correct. If a non-zero element of $H_2(M;\mathbb{Z})$ which is not primitive is represented by an embedded incompressible surface, then the surface must be disconnected, consisting of homologically parallel surfaces, hence separating. So of course in your question you want to restrict to connected incompressible surfaces. $\endgroup$ – Ian Agol Nov 8 '17 at 4:10
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There exist closed orientable hyperbolic 3-manifolds that are surface bundles such that the fiber is the only incompressible surface in the manifold (up to isotopy). Such manifolds can be obtained by Dehn surgery on certain 2-bridge knots, using the classification of incompressible surfaces in 2-bridge knot complements in a paper of Bill Thurston and myself in Inventiones math. 79 (1985), 225-246. Here are a few more details:

It is a well-known general fact that if one does Dehn filling of slope $p/q$ on the exterior $X$ of a knot $K\subset S^3$ to produce a closed manifold $M=M_{p/q}(K)$, then each incompressible surface in $M$ comes from an incompressible surface in $X$ that is either closed (and not the peripheral torus) or has boundary consisting of curves of slope $p/q$ in $\partial X$.

Suppose $K$ is a fibered knot, so its exterior $X$ is fibered with Seifert surface fibers. Then if one does Dehn filling with slope $p/q=0/1$ the resulting manifold $M$ will also be fibered with fibers obtained by capping off the Seifert surface fibers with disks. We would like to choose a fibered knot $K$ whose fiber is the only incompressible surface (up to isotopy) in $X$ of boundary slope $0$, and with no closed incompressible surfaces in $X$. All 2-bridge knots have the latter property, as shown in the paper cited above where it is also shown that all Dehn surgeries on nontrivial 2-bridge knots produce irreducible manifolds. The paper gives an algorithm for computing all the incompressible surfaces in $X$ and their boundary slopes. The incompressible surfaces correspond to certain continued fraction expansions of the rational number $\beta/\alpha$ classifying the 2-bridge knot, where the terms in the continued fractions are allowed to be negative as well as positive. Incompressibility corresponds to no terms being $\pm 1$. There are finitely many incompressible surfaces for each continued fraction. Seifert surfaces correspond to the unique continued fraction for $\beta/\alpha$ with all terms even, and the ones that are fibers of surface bundle structures on $X$ correspond to the case that all the terms in the continued fraction are $\pm 2$. In these cases the fiber surface is the unique surface associated to the continued fraction. There is a simple formula for the boundary slope of each surface in terms of the continued fraction.

A few examples are worked out on page 231 of the paper, including the knots $6_2$ and $6_3$ which correspond to $\beta/\alpha = 4/11$ and $5/13$, respectively. These are fibered knots, and although they have other incompressible surfaces, only the fiber surface has boundary slope $0$. This means that when we do the slope $0$ Dehn surgery we obtain a fibered closed manifold $M$ with a unique incompressible surface, the fiber. For these two knots the fiber has genus $2$ (its Euler characteristic is easy to compute from the continued fraction) so $M$ has a good chance of being hyperbolic. (It is irreducible with $H_1={\mathbb Z}$ and contains no incompressible tori.) Using SnapPea for example one can check that $M$ is actually hyperbolic in the cases of $6_2$ and $6_3$, but perhaps there's a general reason for this. The monodromy of the bundle structure should be computable and maybe it can be checked that this has infinite order in the mapping class group of the fiber, which is all that is necessary since the monodromy would have finite order if $M$ was a Seifert manifold.

It seems likely that many more choices for $\beta/\alpha$ should also work, with fiber surfaces of arbitrarily large genus.

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Ian Agol, in the last sentence of this paper

https://arxiv.org/pdf/math/0205091.pdf

suggests the possibility of a hyperbolic fibered three-manifold where the fiber is the only incompressible surface. Such an example would be a very pleasing answer to your second question.

This would be an excellent project for a graduate student!

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