# A question on triangulation of compact surfaces

We consider the quotient space of the hexagon. Its edges are identified according to the string $abcab^{-1}c$. I have no idea what surface does it describe? And I also wonder does there exist some general ways to do such questions?

• There is an algorithm for reducing surfaces described by words to $S^2$, a sum of $\mathbb{T}^2$, or a sum of $\mathbb{P}^2$. See the proof of the classification theorem here math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Teo.pdf Sep 24, 2017 at 1:57
• Thank you. I will read the paper and try to do it by myself.
– JACK
Sep 24, 2017 at 2:53

## 2 Answers

Sure there is some general algorithm : first see if your surface is orientable or not, and then compute the Euler characteristic. Both can be determined from a triangulation, and such informations fully characterize the surface. We have $\chi = 2 - 2g$ if $X$ is orientable ($g$ is the number of torus appearing in the connected sum $X \cong T^2 \# \dots \# T^2$) and $\chi = 2 - k$ if $X$ is not orientable ($k$ is the number of projective planes appearing in the decomposition $X \cong \Bbb RP^2 \# \dots \# \Bbb RP^2$.)

First, take any point in $c$ and the segment connecting the two copies of these points (one for every copy of $c$) passing by the center of the hexagon : a little neighbrohood of such a segment is a Moebius band. This shows that your surface is not orientable.

If I am not mistaken I count $3$ vertices, $9$ edges and $6$ triangles i.e a zero Euler characteristic, so this should be $\Bbb RP^2 \# \Bbb RP^2$.

Edit : I knew I was a bit too tired for answer but I couldn't help it. Of course there is a much simpler way of doing this : notice that one can "glue" the segment $a$ and $c$ together, in a new segment that we call $a$ and we obtain the square $abab^{-1}$ which is the familiar presentation of $K \cong \Bbb RP^2 \# \Bbb RP^2$. Indeed $K$ is not orientable and has $\chi(K) = 0$.

• Thank you very much. I got your idea.
– JACK
Sep 24, 2017 at 2:52
• @OliverMou : you are welcome ! Sep 24, 2017 at 7:00

Augmenting Nicolas's answer: you can avoid triangulating the surface to compute Euler characteristic by thinking of it as a CW complex. Here's an algorithm that even a computer can do.

Step 1. Determine the number of vertices in the boundary of the polygon after identification: label an unlabeled vertex with a fresh name, then propagate that label to other vertices using the identification word; count the number of labels once everything is labeled.

Step 2. Let $n$ be the number of vertices after identification and $e$ the number of edge labels. Then $\chi=n-e+1$. (The $1$ is for the one polygonal face.)

Step 3. Determine whether the surface is orientable: if every letter $a$ in the word also has an $a^{-1}$ in the word, then the surface is orientable, otherwise it is unorientable.

Step 4. If orientable, $g=1-\frac{1}{2}\chi$ is the number of $T^2$'s. If unorientable, $k=2-\chi$ is the number of $\mathbb R\mathrm P^2$'s.

Example. The vertices on either end of $a$ end up being the only distinct vertices in the identification, so $n=2$, hence $\chi=2-3+1=0$. The word has $a$ twice, so it is unorientable. Hence $k=2-0=2$ is the number of $\mathbb R\mathrm P^2$'s.