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The question is fairly self-explanatory. In particular, I would like to know how to triangulate the mapping cylinder arising from applying a Dehn twist to the torus. The reason is that I am thinking of this mapping cylinder as a cobordism from the torus to itself, and the TQFTs I am working with are of the state-sum variety.

Any assistance would be greatly appreciated!

Gerrit Goosen
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1 Answers1

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Let $f : M \to M$ be a diffeomorphism, and (to be explicit) define the mapping torus to be $$T_f = M \times [0,1] \,\, / \,\, (x,1) \sim (f(x),0) $$

Pick a triangulation $\tau$ of $M$.

Perturb $f$ by a small isotopy so that the triangulations $f(\tau)$ and $\tau$ are in general position with respect to each other.

It follows that there exists a triangulation $\sigma$ of $M$ containing a subcomplex $\tau'$ which is a subdivision of $\tau$ and containing another subcomplex $\tau''$ which is a subdivision of $f(\tau)$.

Triangulate $M \times [0,1]$ as follows:

  • On $M \times 0$ use $f(\tau) \times 0$.
  • On $M \times \frac{1}{2}$ use $\sigma \times \frac{1}{2}$.
  • You can now extend the triangulations on $M \times 0$ and $M \times \frac{1}{2}$ to give a triangulation of $M \times [0,\frac{1}{2}]$.
  • On $M \times 1$ use $\tau \times 1$.
  • You can now extend the triangulations on $M \times \frac{1}{2}$ and on $M \times 1$ to give a triangulation of $M \times [\frac{1}{2},1]$.

This triangulation on $M \times [0,1]$ now descends to a triangulation on $T_f$.

Lee Mosher
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  • Thanks a lot for your answer. A follow-up question/comment; you mention that "there exists a triangulation $\sigma$ of $M$ ..." It was this step I was not aware of. Can you please point me to a source where this is proved? – Gerrit Goosen Sep 17 '14 at 10:23
  • @GerritGoosen: I would say that this is a standard construction in piecewise linear topology. The classical text in that topic is Rourke and Sandersen, however I could not say if this specific result is proved there. Roughly speaking, what you do is to subdivide the skeleta of $\sigma$ by induction. When you reach dimension $n$ and it is time to subdivide an $n$-simplex $\Delta$ (whose boundary, by induction, has already been subdivided), you do so in a manner so that for each $k \le n$ the set $\Delta \cap \tau^{(k)}$ is a subcomplex of the subdivision of $\Delta$. – Lee Mosher Sep 17 '14 at 12:44