## Examples of acylindrical 3-manifolds

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Let $C$ be the compact cylinder $S^1\times [0,1]$. A 3-manifold $M$ with incompressible boundary is called acylindrical if every map $(C,\partial C)\to (M,\partial M)$ that sends the components of $\partial C$ to essential curves in $\partial M$ is homotopic rel $\partial C$ into $\partial M$.

I'm looking, for each $g\geq 2$, for examples of compact, orientable, acylindrical, hyperbolic 3-manifolds $M_g$ with non-empty, incompressible boundary such that each component of $\partial M_g$ is homeomorphic to the surface of genus $g$.

I'm sure such things should be well known to the experts.

Here's a little motivation. Such examples would be useful because, given an arbitrary hyperbolic 3-manifold $N$ with incompressible boundary, you can glue copies of the $M_g$ to the non-toroidal boundary components of $N$ and the result, by Geometrization (for Haken 3-manifolds, so you only need Thurston, not Perelman), is a hyperbolic 3-manifold of finite volume.

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Sorry, the last paragraph is a little garbled.  You may need to use several copies of $N$, too. – HW Jun 5 at 2:07
Thanks for all the great answers!  I accepted Richard's only because it included very convenient references that are easy to cite. – HW Jun 9 at 18:44

## 4 Answers

The exterior of Suzuki's Brunnian graph on $n$-edges, here pictured with $n=7$, is irreducible, atoroidal, boundary incompressible, and acylindrical. See

Luisa Paoluzzi and Bruno Zimmermann. On a class of hyperbolic 3-manifolds and groups with one defining relation. Geom. Dedicata, 60(2):113–123, 1996

or

Akira Ushijima. The canonical decompositions of some family of compact orientable hyperbolic 3-manifolds with totally geodesic boundary. Geom. Dedicata, 78(1):21–47, 1999.

(I think these manifolds may be contained in Bruno's list also.)

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You are looking for

compact 3-manifolds that admit a hyperbolic metric with geodesic boundary

equivalently

compact 3-manifolds that do not contain any essential surface with $\chi \geq 0$

with the additional requirement that every boundary component has the same genus $g$.

To construct such manifolds you may draw pictures of suffficiently knotted graphs in $S^3$ consisting of some copies of genus-$g$ graphs, and take their complements. Then you can use orb to check whether the complement has a hyperbolic structure with geodesic boundary.

An alternative construction uses ideal triangulations, extending Thurson's original "knotted y" example from his notes. Pick a bunch of tetrahedra and pair their faces so that every edge in the resulting triangulation has valence $> 6$. Then remove an open star at each vertex. Geometrization guarantees that the resulting manifold admits a hyperbolic metric with geodesic boundary (because you can put an angle structure à la Casson which excludes any normal surface with $\chi \geq 0$).

For example, you can take $g\geqslant 2$ tetrahedra and pair the faces in such a way that the resulting triangulation consists of one vertex and one edge only (which has thus valence $6g$). The resulting manifold is a hyperbolic 3-manifold with connected genus-$g$ geodesic boundary. Its hyperbolic structure is simply obtained by giving each tetrahedron the structure of a truncated regular hyperbolic tetrahedron with all dihedral angles of angle $\pi/(3g)$. Thurston's knotted y is obtained in this way for $g=2$.

The manifolds constructed in this way are "the simplest ones" among those having a connected genus-$g$ boundary, from different viewpoints: they have smallest volume (as a consequence of a result of Myiamoto) and smallest Matveev complexity: we have investigated these manifolds here. There are many such manifolds because there are many triangulations with one vertex and one edge: their number grow more than exponentially in $g$.

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Miyamoto points out that one can construct such manifolds as cyclic branched covers over the single edge in the Gieseking manifold. The even order covers will be orientable. The Tripus manifold is the double branched cover. – Agol Jun 5 at 17:54
Thanks for this answer, Bruno. Could you explain quickly why an acylindrical manifold has totally geodesic boundary, or give a reference? – HW Jun 9 at 18:42

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@Henry: If $M$ is acylindrical, the double $DM$ is hyperbolic with an orientation reversing involution fixing $\partial M$. By Mostow rigidity, this involution may be taken to be an isometry with fixed point set a geodesic surface homotopic to $\partial M$. If you like, there's a more detailed sketch in Leininger's "Small curvature surfaces in 3-manifolds," J. Knot Theory Ramifications 15 (2006), 379--411. There's a much different proof in McMullen's "Iteration on Teichmueller Space," Inventiones mathematicae, 99(2), 425--445. – Richard Kent Jun 9 at 20:29

I believe Bob Brooks constructs really cool examples in this paper:

MR0860677 (88b:32050) Brooks, Robert(1-UCLA) Circle packings and co-compact extensions of Kleinian groups. Invent. Math. 86 (1986), no. 3, 461–469.

The idea is that given a circle-packed hyperbolic surface (such are dense in teichmuller space, by an earlier theorem of Brooks) one can manufacture a hyperbolic manifold whose boundary consists of four copies of the surface.

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See the proof of theorem 19.8 in my book. I explain two constructions, one via orbifold trick as Igor explained and the other using Meyers' theorem. Meyers' idea is: Take genus $g$ handlebody $H$ and take a knot $K\subset H$ which busts all essential annuli and disks in $H$. Then do a Dehn surgery on $H$ along $K$.

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