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Let $M$ be a closed, oriented, hyperbolic $3$-manifold which is a surface bundle over $\mathbb{S}^1$.

Is there some $\pi_1$-injective closed surface (perhaps not embedded) $S \subset M$ which is not a fiber, and which does not intersect all the fibers?

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  • $\begingroup$ For hyperbolic once-punctured torus bundles (which of course do not meet your first assumption) it is known that there are no closed incompressible surfaces except the boundary torus. This is Theorem 1a) in math.cornell.edu/~hatcher/Papers/2-bridgeKnots.pdf $\endgroup$ – ThiKu Apr 18 '18 at 5:54
  • $\begingroup$ I wouldn‘t expect a similar result in higher genus fibrations. For an approach towards incompressible surfaces in fibered 3-manifolds you may look at the proof of Theorem 3.1 in homepages.warwick.ac.uk/~masgar/Maths/2005fibre.pdf $\endgroup$ – ThiKu Apr 18 '18 at 5:56
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There is no such surface. If $S$ is a surface that misses a fiber then $S$ lies in a submanifold of $M$ that is homeomorphic to a product $F \times I$, where $F$ is the fiber. Any $\pi_1$-injective closed surface in a product is homotopic to a cover of the fiber $F$. This was shown by Waldhausen for embedded surfaces. The immersed case follows by passing to a covering space.

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  • $\begingroup$ Thank you very much for your answer. I am wondering now: is it possible to find two $\pi_1$-injective closed surfaces which are not fibers, and do not intersect? $\endgroup$ – Vanderson Lima Apr 21 '18 at 18:43
  • $\begingroup$ @VandersonLima, in this context, the existence of an immersed surface that’s not a fibre was proved by Cooper—Long—Reid. Finding two that don’t intersect doesn’t sound possible, in general. $\endgroup$ – HJRW Apr 22 '18 at 8:19
  • $\begingroup$ Why do you think is not possible in general? $\endgroup$ – Vanderson Lima Apr 23 '18 at 14:38

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