# Think of the surface of genus $k$ as a sphere with $k$ tubes sewn in. Calculate its Euler characteristic by trangulating.

Think of the surface of genus $k$ as a sphere with $k$ tubes sewn in. Calculate its Euler characteristic by trangulating.

I know that I need to make the genus covered by infinitely many triangle then use the formula

$$\chi=V-E+F$$

where $V$ is the number of vertices on the tessellated surface, $E$ is the number of edges on the tessellated surface and $F$ is number of polygonal faces.

My professor told me to remove $2k$ non-adjacent faces from $S^2$, then triangulate a closed cylinder $S^1 \times [0,1]$.

Why do I need to remove $2k$ non-adjacent faces from $S^2$? I know that $\chi(S^2)=2$, but I can't see what it has to do with this move.

I also know that the surface of genus $k$ admits a Lefschetz map homotopic to the identity with 1 source, 1 sink and $2k$ saddles. This implies that $\chi =2-2k$. So I guess this is the answer for this problem. The only thing I need to do is using triangulating to get to that result.

• He's just suggesting to delete enough so that you can attach a handle. (If there weren't an even number, the 'obvious' triangulation of the handle wouldn't actually attach to the sphere.)
– user98602
Dec 1, 2014 at 1:32
• can you explain it a little further please? My professor didn't teach this section at all, that suggestion is the only thing he gave me when I asked him. Dec 1, 2014 at 1:46

Starting from a genus $k$ surface, delete two triangles, then sow in one tube made from two triangles. What happens to the number of faces, vertices, and edges?
• For surfaces, the Euler characteristic is only $2$ for the sphere. It is a homeomorphism classifier for closed orientable surfaces (so if you tell me its Euler characteristic, I can tell you what the surface is). Dec 2, 2014 at 2:24
• Take a sphere. Cut two holes out of the sphere. Take a cylinder and attach one end to each hole. You now have a torus. We want to do all of this with triangles, so we can keep track of vertices, edges, and faces. We should be able to see that $V+F-E$ changed by exactly $2$ through this process. Then, inductively, we can get to any genus. Dec 2, 2014 at 3:36
• Just want to make sure I understand it correctly, every time you do what you said, the Euler sum will reduce by $2$, meaning for genus $k$, I have to do it $k$ time, so the Euler sum is reduce by $2k$. Start with the sphere which have $\chi =2$, and we reduce $2k$, so the final Euler sum will be $2-2k$, right? Dec 2, 2014 at 14:50
• Yes. Make sure you really understand why. You remove 2 faces. Then make a prism with a triangluar base, and glue it to each of the holes you just made. The vertices and edges of the holes are already there, so you are only adding 3 faces and 3 edges from the "sides" of the prism. The effect on $V+F-E$ is that $F \to F+1$ and $E \to E+3$, which gives a net change of $\chi \to \chi-2$. Dec 2, 2014 at 15:01