Low Dimensional Topology

December 13, 2007

Triangulations and Normal Surfaces

Filed under: 3-manifolds,Triangulations — Jesse Johnson @ 12:16 am

I’m at the AIM workshop on triangulations, Heegaard splitting and hyperbolic geometry. A lot of interesting things have been discussed, including the ongoing project by Jaco and Rubinstein to show that every atoroidal three-manifold has what’s called a 1-efficient triangulation. The idea is that the normal surfaces (those that intersect the triangulation in a locally simple way that lends itself well to algorithms and calculations) include incompressible surfaces and some Heegaard splittings, but it also contains normal surfaces that are topologically trivial, but are artifacts of the triangulation. In particular, there will be compressible normal tori, which are particularly problematic for calculations. A one vertex triangulation is 0-efficient if every normal sphere is the boundary of a regular neighborhood of the vertex. A 0-efficient triangulation is 1-efficient if every normal torus is the boundary of a regular neighborhood of either an edge of the triangulation or what’s called a layered triangulation. (I hope to discuss layered triangulations in a future post.)

Jaco and Rubinstein’s idea for constructing a 1-efficient triangulation is to take every “bad” normal torus and crush it down to a point. This creates a triangulation such that near the vertex, it looks like a cone over a torus. This is not a manifold point so we need to repair it. We can think of it as an ideal triangulation of a cusped manifold if we remove the vertex. There is a fairly simple and very clever construction for turning this ideal triangulation into a complete triangulation for a 3-manifold with boundary, into which one can glue a layered triangulation. Jaco has offered to send the section of the preprint that describes this construction to anyone who asks (it’s one of the parts that’s already written) but I want to give a rough idea of the construction here.

We start with a 3-dimensional triangulation with a single vertex whose link (the boundary of a regular neighborhood) is a torus. The regular neighborhood contains the corners of all the tetrahedra in the triangulation. If we remove the regular neighborhood, we get a sort of truncated triangulation and the truncated corners form a triangulation of the boundary. We will choose a collection of edges of this triangulation of the boundary whose union forms a graph two valence-three vertice and whose complement is a disk.

In order to shrink the truncated corners of the tetrahedra back down to vertices, we need to shrink the complementary disk down to a vertex. But first we need thicken each edge in the graph into a rectangle and thicken the two valence three vertices into triangles. Then when we shrink the disk down to a point, the two triangles will expand to give us a one vertex triangulation of the boundary torus.

Each edge in the initial triangulation of the boundary is a short arc where one of the triangles in the ideal triangulation has been truncated. To turn the edge into a rectangle, we will slice the triangle open, leaving it hinged at the far edge, then glue in an extra tetrahedron, sort of like a bellows from one side of the slice to the other. The new tetrahedron has one edge that goes from one side of the slice to the other. If we crop the new tetrahedron along this edge, then in the boundary we end up replacing the edge with a rectangle as desired.

We have to slice open all the edges of the theta graph simultaneously, then at each of the two valence three vertices we glue in a tetrahedron with three faces at the ends of the widened edges and a fourth face on the link of the vertex. We can now shrink the complementary disk down to a point. This causes each rectangle to shrink down to an edge (in the direction opposite from how we expanded it a second ago) and causes the two triangles at the vertices to expand to form a two triangle (one vertex) triangulation of the boundary torus.

Advertisements

3 Comments »

  1. tell me what ‘euler’ means

    Comment by Ashley Last — February 20, 2009 @ 2:53 pm | Reply

  2. my email is ashleylast@hotmail.com

    Comment by Ashley Last — February 20, 2009 @ 2:54 pm | Reply

  3. The wikipedia has thorough entries on both Euler characteristic and on Leonhard Euler, for whom it is named.

    Comment by Jesse Johnson — February 25, 2009 @ 9:30 am | Reply


RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: