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Suppose we have a plane model for torus. We want to decompose the plane model of torus into cell complex which satisfies

1)every face of the cell complex has exactly 6 boundary edges 2)at each vertex, there are 3 faces meet at it.

How to draw such cell complex? I have try to squeeze a set of hexagons which intersect each other along side, but I can't manage to put this set into plane model of torus to satisfy the conditions above.

Can anyone enlighten me?

Paulo Henrique
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Idonknow
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3 Answers3

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I'm not sure what you mean by 'decompose a plane model for a torus' so I'll just outline how to give a CW-complex construction of a torus which satisfies your conditions.

A hexagon with side edge pairings $[abca^{-1}b^{-1}c^{-1}]$ is homeomorphic to the torus (seen by cutting along the line from the center of one $a$ edge to the center of the other $a$ edge and then gluing along the $c$ edges). This doesn't yet satisfy your first condition because the face only has three boundary edges. We can consider a union of fundamental regions though which does satisfy the conditions.

I drew the region in paint shown below, but the side pairing sequence is given as (the letters used refer to the colours in the image e.g $r=\color{red}{red}$, $m=\color{purple}{mauve}$, $p=\color{pink}{pink}$, ...) $$[gbpmyrb^{-1}g^{-1}m^{-1}p^{-1}r^{-1}y^{-1}]$$ which uses three of the above hexagons glued together to give a different fundamental region of a free/transitive $\mathbb{Z}^2$-action on the plane. This better satisfies the conditions you ask for because each vertex shares three different faces, and we have that each face properly has three boundary edges.

enter image description here

Dan Rust
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Here's a picture of such a cellulation of the torus. Identify opposite edges of the hexagon as in Daniel Rust's answer.

enter image description here

Cheerful Parsnip
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  • But midpoints do not meet at three faces. They only meet at two faces. – Idonknow Nov 08 '13 at 02:49
  • @Idonknow: No they meet at three. Look at the bottom midpoint. It is the same as the top one. It meets all three visible hexagons. – Cheerful Parsnip Nov 08 '13 at 02:53
  • are you using hexagon as plane model for torus? If so, then how you define the orientation of sides of hexagon? Because I can't seem to define an orientation such that it will become a square plane model of torus. – Idonknow Nov 08 '13 at 06:13
  • Yes, I am using the hexagon plane model. As you go around the hexagon you have identifications $abca^{-1}b^{-1}c^{-1}$. This is different than the square plane model which has identifications $aba^{-1}b^{-1}$. You can prove that the hexagon is a model for the torus by first gluing top an bottom. Then you have a cylinder with two edges on each boundary circle. These are then glued together to give a torus. – Cheerful Parsnip Nov 08 '13 at 12:02
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Draw (or find on the net) a honeycomb pattern of hexagons. Find four hexagons whose centres lie on the corners of a parallelogram. Draw in the edges of the parallelogram, Warp the parallelogram so it becomes a square.

Empy2
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