# Thread: Quick Question

1. ## Quick Question

I originally didn't do this right, so our professor told us to try it over and see if we can figure it out. The question is:

a) Can a triangulation (using only triangles) of the 2-holed torus have exactly 6 edges coming in to each vertex? Explain. Hint: Recall that in class we showed that E = 3F/2 (after gluing) for triangulations by triangles.
b) Exactly 7 edges? Explain.
c) How many triangles will be needed in the triangulation in b)?

Another hint he gave us was: Fb (before gluing) = Fa (after gluing) = F. (In other words, the # of faces doesn't change after gluing) The hint is to compare:
F vs. Vb and F vs. Va

2. Originally Posted by Janu42
I originally didn't do this right, so our professor told us to try it over and see if we can figure it out. The question is:

a) Can a triangulation (using only triangles) of the 2-holed torus have exactly 6 edges coming in to each vertex? Explain. Hint: Recall that in class we showed that E = 3F/2 (after gluing) for triangulations by triangles.
b) Exactly 7 edges? Explain.
c) How many triangles will be needed in the triangulation in b)?

Another hint he gave us was: Fb (before gluing) = Fa (after gluing) = F. (In other words, the # of faces doesn't change after gluing) The hint is to compare:
F vs. Vb and F vs. Va
Are you allowed to use Euler's formula $F-E+V = 2(1-g)$, where g is the genus (= number of holes)? If so, then suppose that each vertex has 6 edges. Each edge joins two vertices, so the number of edges will be $E = 3V$. You also know that $F = 2E/3$. If you substitute all that information into Euler's equation, you will end with a contradiction.

On the other hand, if there are 7 edges at each vertex then the equation $E = 3V$ becomes $E = 7V/2$. This time, you don't get a contradiction. Instead, you have enough information to find V, E and F.

3. Yeah, my thought was to use Euler's, but I was confused as to how E and V were related and everything. I know E = 3F/2 after gluing but was confused on the other relationships.

So there can't be 6 edges at a vertex? But there can be 7? I'm still a little confused as to what I substitute into Euler's. I'm guessing that it doesn't come out to -2 (as it should for a 2-holed torus).

4. Originally Posted by Janu42
Yeah, my thought was to use Euler's, but I was confused as to how E and V were related and everything. I know E = 3F/2 after gluing but was confused on the other relationships.

So there can't be 6 edges at a vertex? But there can be 7? I'm still a little confused as to what I substitute into Euler's. I'm guessing that it doesn't come out to -2 (as it should for a 2-holed torus).
If there are six edges at each vertex then E = 3V. Also, E = 3F/2, so that F = 2E/3 = 2V. If you substitute E = 3V and F = 2V into Euler's equation then you get 0 = –2. That shows that it is impossible to have six edges at each vertex.

Now try to do the same with seven edges at each vertex. So you start with E = 7V/2 and follow the same procedure. This time, instead of a contradiction you come up with a value for V, from which you can then calculate E and F.

5. Thank you so much. For c), how do I know how many triangles I need? Couldn't I use different sized triangles? Isn't there endless ways to triangulate? Or is it asking how many faces I need at each vertex or something.

6. Originally Posted by Janu42
Thank you so much. For c), how do I know how many triangles I need? Couldn't I use different sized triangles? Isn't there endless ways to triangulate? Or is it asking how many faces I need at each vertex or something.
There are endless ways to triangulate, but there are not so many triangulations that have exactly seven edges at each vertex. That is a very restrictive condition, and in this problem it completely determines the number of vertices, edges and faces.