Euler characteristic on torus

Question

I'm trying to compute $\chi(T^2)$:

• I know that the sectional curvature of $T^2$ is $\dfrac{\cos(t)}{2+\cos(t)}$ with the parametrization: $F(t,s)=((2+\cos(t))\cos(s),(2+\cos(t))\sin(s),\sin(t))$
• Now I want to compute $\int\limits_{T^2} K \ dx$, where $K$ is the sectional curvature of the torus.

Is that the correct approach? And how does this computation work?

Answer 1

You can triangulate the torus by splitting the identification square into nine squares with diagonals in each. Counting the number of vertices faces and edges will give you the Euler characteristic after application of the definition of the euler characteristic. I will add a picture later.

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One can also do this using a simple version of the Gauss-Bonnet Theorem. This states that for $S$ a closed surface we have $$ \int_S K dA = 2 \pi \chi(S) $$ where $K$ is the Gaussian curvature of $S$. Integrating the Gaussian curvature $$K= \frac{\cos(\theta)}{r(R+r\cos(\theta))}$$ of the torus (where I'm using notation from this question) gives $\int_T K dA = 0$ so we may conclude that $\chi(T) = 0$.

Answer 2

The Euler characteristic can be computed through a variety of ways. The simplest way is to use the fact that the Euler characteristic is the alternating sum of the number of cells in a CW decomposition of your space. Picking the simplest one gives 1-2+1=0. An alternative and equivalent approach is to take the alternating sum of the rank of the homology, again this gives 1-2+1=0.

If you would like to use Gauss-Bonnet and you know that the torus admits a locally Euclidean metric, then you can deduce it is 0 since the curvature of the plane is 0.