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Is there an easy example of a (closed) hyperbolic 3-manifold that fibers over the circle but contains some totally geodesic surface?

(Of course such manifolds exist if the 'Virtually Fibered Conjecture' were correct, since a geodesic surface lifts to the fibered cover. But is there something more eplicit?)

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Actually, Jorgensen's orginal example of a closed 3-manifold fibering over the circle has a geodesic immersed surface (although this wasn't proven by him). – Ian Agol Sep 9 '10 at 21:03
up vote 23 down vote accepted

There are many specific known examples. Here is one construction:

Start with the 3-torus $T^3$, parametrize in the standard way as $R^3/Z^3$. It fibers over the circle in many ways. Let $a$, $b$ and $c$ be three disjoint circles, coming form lines parallel to the x, y and z axes. For most fibrations, these three circles are transverse to the fibers. Form a branched cover of the torus with two-fold branching over all preimages of these 3 circles. The resulting manifold has a hyperbolic structure that can be constructed from right-angled hyperbolic dodecahedra, and is commensurable with the 4-fold branched cover of $S^3$ over the Borromean rings. You can think of it this way: you can take a unit cube as fundamental domain for the torus, and arrange that a, b and c lie on faces of the cube, each bisecting a pari of (glued together) opposite facce. This induces a subdividision of the boundary of the cube into what look like rectangles, but are really pentagons.

The map (x,y,z) -> x+y+z gives a fibration over the torus, also works for any branched cover as described. The preimage of any face of the cube is an extended face plane of a dodecahedron, and is always a totally geodesic immersed surface, but it splits into two embedded surfaces for suitable branched covers of $T^3$ (perhaps the one you first come up with.)

The tiling of hyperbolic space by right-angled dodecahedra has a cameo appearance in the video "Not Knot" we made at the Geometry Center, available together with "Outside In" on DVD from AKPeters. In the 1984 Scientific American Article The Mathematics of three-dimensional manifolds that Jeff Weeks and I wrote, a manifold in this family (constructed from right-angled hyperbolic dodecahedra and having the properties you asked for) was described as the configuration space of a mechanical linkage. I don't think these particular properties were pointed out in Scientific American.

This and other examples that are counterintuitive at first were a good part of my motivation when I raised the question whether all hyperbolic 3-manifolds virtually fiber over the circle, which at the time was a radical idea.

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And the right-angled pentagons glue together to give the totally geodesic surface, yes? A question: How much harder is it to describe a cover where some totally geodesic surface is embedded? – Sam Nead Sep 9 '10 at 20:03
Yes on gluing the pentagons. It's not hard to make a cover, I was just trying to keep it shorter. To get an easy-to-describe cover, first pass to the 8-fold cover of the torus that unwraps the torus mod 2. Now the axes each come in quadruplicate, so their sum is trivial in $Z/2$ homology. Choose a surface that they bound (mod 2). It doesn't need to embedded, so just use a union of annuli pairing up parallel axes. Take two copies of this manifold, and cross-connect at the surfaces: think of them as magic portals, so that when you pass through it transports you to an alternate universe. – Bill Thurston Sep 10 '10 at 18:31
I think I see it. I could equip the double cover of the $8$--fold cover with a singular Euclidean metric with $4\pi$ cone locus about the axes. Since "parallel to the $zx$--plane" makes sense, the annuli have to unwrap to give embedded surfaces. Drawing a picture and staring at it convinces me that they are genus two surfaces. – Sam Nead Sep 11 '10 at 10:52
Right. There are choices of which annuli to use in the 8-fold cover which don't change the topology of the branched cover, but do change descriptions of surfaces. Each annulus is intersected by two axes, so there is at least one other annulus cutting through it. You can choose the rest of the annuli to stay away from that one. To form the 2-fold branched cover, the annulus has a single arc in the interior as branch cut. Open the slit, giving a pair of pants; in the branched cover, you have two pairs of pants joined to mke a double-torus. Can you also see the fibers of (x,y,z) -> (x+y+z)? – Bill Thurston Sep 11 '10 at 13:28
I had to draw another picture. The fibre $F$ of the map from $T^3$ to $S^1$ is the hex torus you get by gluing a hexagon and two triangles (these are obtained by connecting the midpoints of the edges of the cube $[0,2]^3$ in the right way). Any cone locus (parallel to an axis) meets $F$ in just one point. So the fibre of the actual map is obtained by taking the double branched cover of $F$ at twelve points. So, a surface of genus seven. I need to think about the monodromy some more. – Sam Nead Sep 12 '10 at 7:49

This example is neither particularly easy nor explicit, but it is at least a definite family of examples.

It follows from a paper of Bergeron--Haglund--Wise and work of Agol that any 'standard' arithmetic hyperbolic 3-manifold virtually fibres. These examples contain lots of totally geodesic surfaces which, as you say, will lift to totally geodesic surfaces in the fibred cover.

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