Ken Baker has posted some pictures of the one tetrahedron triangulation of the solid torus to his blog. This is a surprisingly hard construction to visualize, given how simple it is. (Ken showed me the 3D models at the Georgia topology conference and after ten minutes of looking at them from different angles, I still couldn’t quite see it.) Start with a tetrahedron and choose two of the sides (it doesn’t matter which two.) There are three orientation reversing ways to glue the two sides together (an orientation preserving gluing would produce a non-orientable manifold) – You can just fold them over the common edge, or you can glue them by a 2pi/3 rotation clockwise or counter-clockwise. The first gluing gives you a 3-ball. The other two produce a solid torus, and one of these is shown in Ken’s pictures. (The blue triangles in the second picture show a Mobius band formed by the two glued triangles.)

This triangulation has a single vertex, which is in the boundary of the solid torus. The boundary is triangulated with two triangles (the two that aren’t glued to anything). One can produce other triangulations by taking another tetrahedron and gluing two of its faces to the two faces in the boundary. This produces a new solid torus with a different triangulation of its boundary. The process can be repeated to produce any desired triangulation of the boundary. The resulting triangulation of the solid torus is called a *layered triangulation*. Gluing the last two faces to each other produces a lens space, and the induced triangulation, also called a layered triangulation, is studied in a paper by Jaco, Rubinstein and Tillman that just hit the arXiv a few days ago [1]. They show that for a certain family of lens spaces, the minimal layered triangulation is in fact minimal among all triangulations of the lens space. They conjecture that this is true for all lens spaces, so I’m adding that to the open problem list as a question.

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Rather that seeing the 1-tetrahedron layered solid torus as a tetrahedron with gluings, I find it easier to think about it as a Moebius band with a tetrahedron attached. I believe I learned this from Ben Burton.

The idea is to start with a Moebius band with its 1-triangle triangulation. If a triangle has sides coherently oriented and labelled a, b, c, you get the triangulation of the Moebius band by gluing a to b, preserving the orientations of a and b. The boundary of the Moebius band is given by c, whose endpoints are glued together because of the previous operation on a and b.

Now think of the orientable 3-manifold which is the total space of an I-bundle over the Moebius band. The fibrewise boundary of the intervals (the I in I-bundle) is a cylinder, and since it covers the Moebius band in a 2:1 way, it inherits a 2-triangle triangulation.

Now we’re going to attach the tetrahedron. The mental picture of the tetrahedron you need to have is of *two* of its faces sitting flat on a plane. So there’s a 2-triangle triangulated square in the plane, the remaining edge of the tetrahedron we’ll call the non-planar edge NPE. Pick this tetrahedron up, and glue its two flat triangles to our 2-triangle triangulated annulus, so that the diagonal of the square gets sent to one of the two internal edges of the triangulated annulus (pick one). Once you compose with the projection from the annulus to the Moebius band, that’s the attaching map.

One of the reasons this 1-tetrahedron triangulation of the solid torus is non-intuitive is that all the curves a, b, c from the triangle that we started with are on the boundary of the solid torus. So the triangulated solid torus has 1 zero-cell (from the moebius band), three 1-cells (two from the Moebius band, plus the NPE), three 2-cells (one from the Moebius band, plus two on the boundary from the parts of the tetrahedron not glued to the Moebius band), and a single tetrahedron. So the boundary torus has *all* the 1-cells.

Comment by Ryan Budney — July 27, 2011 @ 10:05 pm |

That’s a good point. Not only is this much easier to “see”, but it’s also easier to generalize to higher genus handlebodies: Start with any surface with a single boundary component and a one-vertex triangulation, then layer on tetrahedra until the link of the vertex becomes a disk.

Comment by Jesse Johnson — July 29, 2011 @ 4:01 pm |

I find this picture easier to understand than Ken Baker’s:

http://postimage.org/image/2r06y5dd0/

In the picture you see a central circle with various dots on it, and arrows from the dots to various discs that have arcs drawn in them, red and blue. This is meant to be a planar depiction of the space S^1 x D^2, the central circle is the S^1 factor, and the 6 discs represent the {6 roots of unity in S^1}xD^2. The blue arcs are the cross-sections of a Moebius band sitting in S^1 x D^2, but this is *not* in the above triangulation. The red arcs also describe a Moebius band sitting in S^1 x D^2 and this *is* the initial Moebius band we use to build our triangulation. In the topmost disc, the red and blue arcs are meant to coincide, but I could only draw one. So as you see, other than the topmost case, the red arcs cut the disc into 3 regions. And if you “track” one of these regions, you’ll notice that it exists for three orbits around the S^1 factor in S^1 x D^2, and only for the middle “orbit” does the cross-section look like a triangle, otherwise it’s a bigon. The Moebius band carries all the 1-cells from the cell structure other than the NPE. The NPE is always transverse to the {z}xD^2 factors, and intersects the discs on their boundaries, one point between any two consecutive parts of the red arcs. The 2-cells consist of the triangle in the Moebius band plus two boundary triangles, i.e. they intersect {z}xD^2 in {z}xS^1, generically as three arcs.

Comment by Ryan Budney — July 27, 2011 @ 11:57 pm |

Can you see the other 2-cells in a picture like this, or does it get too messy?

Comment by Jesse Johnson — July 29, 2011 @ 4:09 pm |

The other 2-cells are on the boundary of the solid torus, so yes. It’s just that the 1-cell that separates those two triangles is transverse to the {z}xD^2 slices.

I’m going to put all these observations into an animation that slowly builds-up and strips away the cells for this triangulation of the solid torus. Here are a couple preliminary shots (I still have to adjust the lighting and tweak a few things here and there). In particular, my parametrization of the Moebius band has a singularity near the 0-cell (running all the way across the band). I’ll have to fix that before the finished product, but here they are:

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I used geodesics in the Poincare disc to construct the two arcs in the {z}xD^2 slices… maybe I should have used straight lines in a Euclidean disc instead? Perhaps making the 1-skeleton thicker would help…

Okay, so a summary: the first five shots have the 1-skeleton and the Moebius band (the internal triangle).

The second five shots have the 1-skeleton, the Moebius band and *one* of the triangles that sits on the boundary of S^1 x D^2. These pictures look red and blue.

The last five shots have the 1-skeleton, the Moebius band and the *other* of the boundary triangles. These pictures look red and green.

Comment by Ryan Budney — July 31, 2011 @ 1:38 pm |

Very cool. You should post this as a regular blog entry because I don’t know how many people are looking at these old comment sections.

Comment by Jesse Johnson — August 4, 2011 @ 4:36 pm |

Yes, I’ve been working on an animation that I hope demonstrates most everything fairly clearly. But I write the code in c++, which when compiled generates PoVRay scripts, which then render…. so it takes time. The images above are early experiments. I hope the animation will be done near the end of September.

Comment by Ryan Budney — August 16, 2011 @ 3:37 pm |

Could you do a picture with just the 1-skeleton?

Comment by Jessica Banks — August 22, 2011 @ 3:49 pm |

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Just the 1-skeleton is a little disorienting. But that’s it. The 0-cell is white. Then there’s the three 1-cells. The orangy one is slope zero. The yellow one is slope 1/2, and the light brown one is slope 1/3. So by “slope 0” I mean the slope of the longitude of the torus — parallel to a planar curve. Slope 1/0 would be the meridian of the torus.

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Up to an isomorphism of the torus (S^1xS^1) this is just your standard triangulation with two triangles — i.e. cut the “square” CW-decomposition in half. There’s two triangles, three edges, one vertex, but it does not sit on S^1xD^2 in the standard way in that the meridian is not part of this triangulation.

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The above image is the 1-skeleton but with the two triangles on the boundary torus coloured, one blue, the other red. They’re transparent so if you’re looking through both the colour filters.

Comment by Ryan Budney — August 24, 2011 @ 1:52 pm |

Why are you all attempting to visualize these triangulations, rather than their dual standard spines? The dual is vastly easier to see…

Comment by Dylan Thurston — December 6, 2012 @ 12:20 pm |

Hi Dylan, I came across these triangulations quite naturally. In my paper with Burton and Hillman we triangulated the complement of a knotted S^2 in S^4, a Cappell-Shaneson knot. The triangulation has two 4-dimensional simplices. The induced triangulation of the “cusp” S^1xS^2 has the above as a subcomplex. I needed to visualize this triangulation to see how the whole S^1xS^2 was built — actually, in order to prove the cusp was really an S^1xS^2.

Comment by Ryan Budney — December 10, 2012 @ 11:46 pm |

These triangulations are also the basic building block in the layered triangulations studied by Jaco-Rubinstein-Tillman, and are used in the inflation/filling theory that Jaco-Rubinstein use to construct triangulations of filled knot complements. There’s always a trade-off in clarity between the two perspectives – triangulations or special spines – and while the spines may be easier to visualize in this case, there are benefits to understanding the triangulation later on in these constructions.

Comment by Jesse Johnson — December 11, 2012 @ 11:51 am |