I am have troubles with the following proof of the global Gauss-Bonnet which take the form;
Let $M$ be a compact regular surface in $\mathbb{R}^3$. If $K$ is the Gaussian curvature of $M$ then
$\int_{M} KdA=2 \pi \chi(M)$,
where $\chi$ is the Euler characeristic.
The proof is presented as follows
(Source: "An Introduction to Gaussian Geometry" by Sigmundur Gudmundsson, http://www.matematik.lu.se/matematiklu/personal/sigma/Gauss.pdf.)
Theorem $8.5$ is simply a local Gauss-Bonnet theorem.
Now, I can't figure out how the number of edges $E$ and vertices $V$ are obtained in the last equation, could someone help me out with this?
Best Answer
As indicated in the comments, $n_k$ is the number of corners of the $k^{\text{th}}$ polygon, and $\alpha_{ki}$ is the angle of that polygon at its $i^{\text{th}}$ corner.
Perhaps rewriting the last three equations a bit differently might help: \begin{align*} \int_M K \, dA &= \sum_{k=1}^F \bigl( (2-n_k) \pi + \sum_{i=1}^{n_k} \alpha_{ki} \bigr) \\ &= 2 \pi F - 2 \pi \cdot \frac{1}{2} \sum_{k=1}^F n_k + 2 \pi \cdot \frac{1}{2\pi} \sum_{k=1}^F \sum_{i=1}^{n_k} \alpha_{ki} \end{align*}
We have $\frac{1}{2} \sum_{k=1}^F n_k = E$, equivalently $\sum_{k=1}^F n_k = 2E$, because the set of edges is obtained from the set of sides of the polygons by identifying those sides in pairs.
We have $\frac{1}{2\pi} \sum_{k=1}^F \sum_{i=1}^{n_k} \alpha_{ki} = V$ because the sum can first be rewritten as a sum over the vertex set of the sum of all the angles meeting at each vertex, but each of the latter sums equals $2\pi$.