What surface is represented by $a_1 a_2 \cdots a_n a_1^{-1} a_2^{-1} \cdots a_n^{-1}$?

Question

Question: What surface is represented by $a_1 a_2 \cdots a_n a_1^{-1} a_2^{-1} \cdots a_n^{-1}$?

Attempt: I'm not sure where to even begin except I can make a few of the following observations:

EDIT: I think these two observations are non-sensical.

(1) If $G$ is an abelian group, the relation above trivially holds.

(2) With $n=2$ we have that the relation above holds if and only if $a_1$ and $a_2$ commute.

How should I proceed from here?

Answer 1

The easiest way to determine the surface is the Euler characteristic. The surfaces are all orientable, since no the edge $a_i$ is not glued with a twist. So one can use the formula $2-2g=V-E+F$, where $g$ is the genus (number of holes) and $V,E,F$ count the number of vertices, edges and faces respectively. The number of edges and faces is easy to see from a picture $E=n$ and $F=1$. The number of vertices is trickier. There are $2n$ apparent vertices in the picture, but these are identified. If you follow around the identifications, you can see that if $n$ is odd, there are $2$ vertices (represented by alternate vertices as you go around the $2n$-gon) and if $n$ is even, there is just $1$ vertex. So, solving for $g$, we see that $g=\frac{n-1}{2}$ is $n$ is odd, and $g=\frac{n}{2}$ if $n$ is even.

Answer 2

For $n=2$, the square looks like this

. ->- .
|     |
v     v
. ->- .

If you start in the top left corner and you go around the square, first you see two arrows that go clockwise, then you see two that go counter-clockwise. When you glue them, you match the arrows. So the result is a torus.