$n$-gons gluing diagram/proof?

Question

I am confused about this how I would visualize/prove this.

Given that we have a regular $n$-gon that preserves orientation, what would the resulting surface be?

Answer 1

Suppose that one glues the opposite sides of a $2n$-gon in way that preserves orientation, to produce a closed oriented surface $S$.

If $n=2k$ is even then $S$ is a closed oriented surface of genus $k$.

If $n=2k+1$ is odd then $S$ is also obtains a closed oriented surface of genus $k$.

One proof is simply to apply the classification of surfaces: by using the gluing diagram, you count the numbers

• $V = \#\text{vertices of $S$}$
• $E = \#\text{edges of $S$}$
• $F = \#\text{faces of $S$}$

You then compute $$\chi(S)=V-E+F $$ and you use the theorem that a closed, oriented surface $S$ has genus $g$ if and only if $\chi(S)=2-2g$.

So, let's start counting. Two of the terms are easy. First, $E = n$ because $2n$ edges of the polygon are glued in pairs to give $n$ edges of $S$. Also, $F=1$ because the one polygon itself gives $1$ face of $S$.

To count $V$, you must count the "vertex cycles" of the gluing polygon (which I'm sure you learned to do, but if not then I can add an explanation). If you do this you'll discover that there are two outcomes:

• If $n=2k$ is even then all of the $2n=4k$ vertices of the polygon are in a single vertex cycle, hence $V=1$. We get $$\chi(S)=1-n+1=2-n=2-2k $$ and so $S$ has genus $k$.
• If $n=2k+1$ is even then the $2n=4k+2$ vertices of the polygon fall into exactly $2$ vertex cycles, which alternate around the boundary of the polygon. Thus $V=2$. We get $$\chi(S) = 2 - n + 1 = 3 - n = 3 - (2k+1) = 2-2k $$ and so $S$ has genus $k$.

Here's a picture for gluing the hexagon ($n=3$) to produce a torus, which I found by googling "hexagon gluing to give a torus".

https://www.researchgate.net/figure/Gluing-the-edges-of-a-hexagon-into-a-torus_fig5_324889547