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Let $S$ be a closed incompressible surface in a finite volume hyperbolic 3-manifold $M$ without cusps. Let $N$ be the cover of $M$ associated to $\pi_1(S) \subset \pi_1(M)$. The cover $N$ is homeomorphic to $S\times\mathbb{R}$ and has two ends. An end is convex cocompact if it has a neighborhood whose intersection with the convex core of $N$ is bounded and degenerate otherwise.

If $S$ is a fiber, then both ends of $N$ are degenerate. If the group of isometries corresponding to $\pi_1(S)$ is quasi-Fuchsian, then both ends of $N$ are convex cocompact.

My questions: Is it possible for one end of $N$ to be convex cocompact and the other to be degenerate?
What if we allow $M$ to have infinite volume? Can you point me to a description of an example or an explanation of why there are no examples?

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up vote 13 down vote accepted

It follows from Thurston's Covering Theorem that there are no such examples.

The covering theorem says that if a degenerate end is infinite-to-one under a covering map, then you are (virtually) in the fibered case. See R.D. Canary, A covering theorem for hyperbolic 3-manifolds and its applications, Topology 35:3 (1996), 751–778.

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To answer your question about allowing $M$ to have infinite volume, there exist such examples on $M=S \times \mathbb{R}$ itself. These were originally constructed by Bers, his "singly degenerate" groups.

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