Totally geodesic surfaces in fibered 3-manifolds

2010-09-09T15:39:08Z

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?)

Answer by HW for Totally geodesic surfaces in fibered 3-manifolds

2010-09-09T16:05:11Z

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.

Answer by Bill Thurston for Totally geodesic surfaces in fibered 3-manifolds

2010-09-09T17:32:59Z

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.

Totally geodesic surfaces in fibered 3-manifolds - MathOverflow

Totally geodesic surfaces in fibered 3-manifolds

Answer by HW for Totally geodesic surfaces in fibered 3-manifolds

Answer by Bill Thurston for Totally geodesic surfaces in fibered 3-manifolds