Proving the Gauss-Bonnet theorem for embedded surfaces using triangulations

Question

I heard this really neat elementary proof of the "Gauss-Bonnet Theorem" :

Let $S$ be a surface embedded in $\mathbb{R}^3$. Now take a triangulation of that surface, and approximate the surface with a triangular mesh ("flatten" the triangles).

Call $\delta(v)$ the angle defect at a vertex $v$, that is $2\pi$ minus the sum of the angles around the vertex $v$. Now, the Euler characteristic is defined at $\chi(S) = V - E + F$. Also, if we assume that the surface is without boundary then every face has $3$ bounding edges and every edge bounds $2$ faces, which means $2E=3F$, so $F=2(E-F)$.

Let's compute the sum of all the angle defects at all the vertices of the mesh in two different ways. First, sum the angle defect at each vertex, that is $\sum_{v} \delta(v)$. Second, for each triangle the sum of the angles is exactly $\pi$ since they are euclidean triangles in $\mathbb{R}^3$, so we can sum of the angles of all the faces and get $F\pi$, and subtract that from what the total angle sum would be if the surface were flat to get $2 \pi V - F \pi$. Using the formula for the Euler characteristic, we get

$\sum_{v} \delta(v) = 2 \pi V - \pi F$

$\sum_{v} \delta(v) = 2\pi (V - E + F) = 2\pi\chi(S)$

Now, to make that into the "real" Gauss-Bonnet Theorem I heard the argument "take an infinitesimal triangulation, then the angle defect becomes the curvature". Is there any way to make this rigorous with a reasonable amount of not too advanced tools? If so, can I get a reference?

Thanks!

Answer 1

Yes, there's close relationship between angle defect and curvature. It gets called the Bertrand–Diquet–Puiseux theorem:

$$\kappa(p) = \lim_{r\to 0^+} 3\frac{2\pi r-C_p(r)}{\pi r^3}$$

$$\kappa(p) = \lim_{r\to 0^+}12\frac{\pi r^2-A_p(r)}{\pi r^4 } $$

where $C_p(r)$ is the circumference of the circle of radius $r$ centred at $p$ and $A_p(r)$ the area.

These formulas are for smooth surfaces but you can think of the first formula as describing $\kappa$ as an angle defect, and the 2nd formula as describing how $\kappa$ gets concentrated at the vertices for a triangular mesh.

Answer 2

You only need to take a geodesic traingulation, i.e., a triangulation such that the edges are geodesic arcs. Then you can use a local version of the Gauss-Bonnet theorem that says that the integral of the curvature over such a triangle equals the angular defect, defined as

$$ 2\pi - \mbox{sum of the exterior angles at the vertices}. $$

Then you proceed following your outline. For more details, see my favorite source on curves and surfaces

Dirk J. Struik, Lectures on Classical Differential Geometry, 2nd Edition, Dover, 1988

The proof that I have just outlined can be found in Sec. 4.8 in the above reference. On page 157 of the same book you will also find a description of the curvature as an "angular defect per unit of area". This description follows by considering a sequence of geodesic triangles that concentrates around a point.