# Very short proof of the global Gauss-Bonnet theorem

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?

• Theorem 8.5 may simply be a local Gauss-Bonnet theorem, but without knowing its notation this doesn't make much sense. What are $n_k$, $\alpha_{ki}$? Aug 5 '19 at 13:30
• @LeeMosher thanks for pointing that out. $n_{k}$ are the $n$ corners of the polygon of the local version and $\alpha_{ki}$ the angles at each such corner. Aug 5 '19 at 13:41
• What book is this from? Aug 5 '19 at 17:36
• @NateEldredge matematik.lu.se/matematiklu/personal/sigma/Gauss.pdf Aug 5 '19 at 19:18

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 $$2\pi$$.

• isnt all $n_{k}=3$, as each is a triangle? Moreover in that case suppose we have $1$ triangle, then $3=2$. Or, which is probably the case, am I misunderstanding this? Aug 6 '19 at 10:10
• According to the wording of that copied text, yes each is a triangle, and so yes we should have each $n_k=3$; I suppose this is just an editorial error in that text. I don't understand the rest of your comment, although I'll add that one triangle is not possible, in fact an odd number of triangles is not possible, because an odd number of edges cannot be identified in pairs. Aug 6 '19 at 13:06
• you are saying that the proof implies that a "triangulation" is done with an even number of triangles necessarily? and we can always archive this simple by splitting up the last one? Aug 6 '19 at 13:18
• @Lobsided: It seems that you might profit from reading about how surfaces are constructed by gluing polygons. Trying to give you a course on this topic in the comments to an answer to your question is not good practice on this site. I suggest that you look up topics on the "Classification of Surfaces"; there seem to be several good links that come up if you search on that term. Massey's book also covers this topic in detail. Aug 6 '19 at 14:42
• As a very brief answer, one you have paired up the sides of the set of triangles, for each side pair $A,B$ one chooses a homeomorphism $h : A \to B$ and one identifies each $x \in A$ to $h(x) \in B$. Aug 6 '19 at 14:43