Euler characteristic of torus

Question

We know that the euler characteristic of the torus is $0$. Let´s say I have a torus which has a quadratic hole. As far as I understood the shape of the hole doesn't make any difference in the euler characteristic. I saw many proofs about why it is $0$, but I couldn't find anything that expresses it in terms of number of edges, vertices and faces. Is there such thing? Maybe to tile the torus in such a way that we have these vertices, edges and faces. I tried some ways but unfortunately it didn't come out $0$.

What do you mean by a "quadratic hole"? You're correct that the shape of the hole doesn't make a difference, and there is a result for the euler characteristic in terms of the number of edges/vertices/faces. You first have to fix a simplicial structure on the torus, and then you can apply the theorem you're probably familiar with: $\chi = # \text{Vertices} - # \text{Edges} + # \text{Faces}$. — HallaSurvivor, Jun 01 '21 at 16:02
@HallaSurvivor I'm inclined to guess that "quadratic" here means "square". — Andreas Blass, Jun 01 '21 at 16:07
@AndreasBlass -- aaaaah. That makes sense. Hopefully OP clarifies, but I'll start writing up an answer under that assumption — HallaSurvivor, Jun 01 '21 at 16:09

HallaSurvivor · Accepted Answer · 2021-06-01T16:43:18.383

As I mentioned in my comment, in order to talk about vertices/edges/faces/etc, we have to fix a simplicial or cellular structure on our torus. There are lots of ways to do this, but a famous one is this:

Here we have one vertex $v$, two edges $a$ and $b$, and one face $F$. These are glued together in the way indicated by the picture.

So what's our euler characteristic?

$$\chi = \# \text{Vertices} - \# \text{Edges} + \# \text{Faces} = 1 - 2 + 1 = 0.$$

So what about a torus with a square hole? Our cellular decomposition is going to be a little bit trickier (I'm restricting myself to square cells for concreteness), but we can still make it work (I've left this image big because there's more details):

First, you should convince yourself that when you make the indicated identifications you really do get at torus with a square cutout¹. It shouldn't be too hard to visualize, since we're really doing the usual identifications, and just carrying the obvious square cutout along for the ride.

Next, we can compute the euler characteristic. We have

$5$ vertices: $v,w,x,y,z$
$10$ edges: $a,b,vw,vx,vy,vz,wx,wz,xy,yz$
$4$ faces: $F_1, F_2, F_3, F_4$

So our euler characteristic is

$$\chi = 5 - 10 + 4 = -1$$

which is exactly what we would expect from any number of other calculations (for instance given a manifold of genus $g$ with $b$ boundary components and $k$ punctures, we expect $\chi = 2 - 2g - (b+k)$. Since we have genus $1$ with $1$ boundary component and $0$ punctures, our answer agrees with this formula).

In general, this is what makes the euler characteristic such a useful invariant: It's extremely easy to compute in practice, and can give a shocking amount of information about your space (for instance, it's related to curvature by the Gauss-Bonnet Theorem).

¹: It's definitely not that I tried and failed to draw a torus with a square cutout. It is actually helpful to practice visualizing these for yourself, but I would have liked to have included a picture were I able.

I hope this helps ^_^

Euler characteristic of torus

1 Answers1