Gauss Bonnet Theorem

The Gauss Bonnet Theorem.

These notes are set so that you get to prove the main results by solving smaller problems that when put together give the big result. The answers to the problems are in the videos. You will get the most out of these notes if you do (or try) the problems before looking at the videos.

Polygon

He we use the Gauss-Bennett Theorem to for a simply connected region to generalize Euler's formula

V - E + F = 2

which we proved for the sphere to more general surfaces. Recall that we have have stated so far is that if

D

is a simply connected region on an surface,

M

, with boundary

\partial D

and with some corners as shown in the figure here then we have the basic Gauss-Bonnet Theorem

\iint_{D} K d A + \int_{\partial D} κ d s + \sum_{j} β_{j} = 2 π .

Here

\begin{aligned} K & = Gauss curvature of the surface \\ d A & = Area measure on surface \\ κ & = geodesic curvature of \partial D \\ d s & = arclength measure \partial D \\ β_{i} & = Exterior angle at the i -th corner of D \end{aligned}

We will also what to use the interior angle

α_{i}

of the

i

-th corner. Then we have

α_{i} + β_{i} = π .

In the case where the surface is the usual plane

R^{2}

(so that

K \equiv 0

) and

D

is a triangle with straight sides, this just says that the sum of the interior angles is

π

. If

D

is a convex region in

R^{2}

with no corners, then this

\int_{\partial D} κ d s = 2 π

reflecting that

\int_{\partial D} κ d s

is the total angle that the sweeps through while moving around the boundary of

D

.

We now want to take a surface with no boundary and ``triangulate'' as in the pictures here. The reason for the quotes on triangulate is that the domains in the decomposition do not have to be triangles, what is important is that they are simply connected. Genus 3 surface.

So in the picture on the left, it is a strict triangulation in the sense that all the faces in the decomposition are triangles, in the torus to the right all the faces are rectangles. And the faces do not all have to have the same number of sides. So it is ok of there are triangles, rectangles, pentagons, etc. all in the same ``triangulation''.

Let $M$ be a compact oriented surface and let fix a triangulation of $M$ . Let $\begin{aligned} V & = The set of vertices of the triangulation, \\ E & = The set of edges of the triangulation, \\ F & = The set of faces of the triangulation, \end{aligned}$ and a triangulation of $M$ . Let $\begin{aligned} V & = # (V) = The number of vertices in the triangulation, \\ E & = # (E) = The number of edges in the triangulation, \\ F & = # (F) = The number of faces in the triangulation, \end{aligned}$
Anther numerical quantity related to the triangulation is the degree of a vertex. This is the number of edges that are on it. It also tells us how many faces come together at a vertex.

Proposition 1 . Let $p$ be a vertex. Then the number of faces that are have is $p$ as a vertex is $\deg (p)$ . That is $# {D \in V : p is a corner of D} = \deg (p) .$ Degrees

Problem 1. Prove this.
Solution:

Proposition 2. If $\deg (p)$ is the degree of the vertex $p$ then $\sum_{p \in V} \deg (p) = 2 E .$ That is the sum of the degrees of the vertices is twice the number of edges.

Problem 2: Prove this.
Solution:

For each face $D \in F$ (so $D$ is a simply connected domain on the surface $M$ let $V_{D} = The vertices, p \in V that are on D .$ and for each $p \in V_{D}$ let $\begin{aligned} β (p, D) & = exterior angle of D at p \\ α (p, D) & = interior angle of D at p \end{aligned}$ Then by the Basic Gauss-Bonnet Formula for any $D \in F$ we have $\iint_{D} K d A + \int_{\partial D} κ d s + \sum_{p \in V_{D}} β (p, D) = 2 π .$
The idea now is to sum this over all $D \in F$ . As the faces cover the surface $M$ we have $\sum_{D \in F} \iint_{D} K d A = \iint_{D} K d A .$
Lemma 1. $\sum_{D \in F} \int_{\partial D} κ d s = 0.$
Problem 3. Prove this.
Solution:

Lemma 2 $\sum_{D \in F} (\sum_{p \in V_{D}} β (p, D)) = 2 π (E - V) .$
Problem 4. Prove this.
Solution:

Recalling a formula above we have for all $D \in F$ that by the Basic Gauss Bonnet Theorem the formula $\iint_{D} K d A + \int_{\partial D} κ d s + \sum_{p \in V_{D}} β (p, D) = 2 π$ holds.
Gauss-Bonnet Theorem. For a triangulated oriented surface $M$ $\iint_{M} K d A = 2 π (V - E + F) .$
Problem 5. Prove this by summing the basic Gauss-Bonnet formula over the collection of faces $D$ .
Solution:

Let us look a what $\iint_{M} K d A = 2 π (V - E + F)$ tells us. First it tells us that for all triangulations of $M$ $V - E + F$ always has the same value ( $= \frac{1}{2 π} \iint_{M} K d A$ ). So on a sphere for any triangulation $V - E + F = 2$ . Likewise on a torus for any triangulation we have $V - E + F = 0$ . To prove this you just need to find one triangulation where $V - E + F = 0$ holds (I leave this to you) and then the Gauss-Bonnet Theorem tell us it will be the same for all triangulations. And one a $g$ -holed torus $V - E + F = 2 - 2 g$ holds for all triangulations.
The formula information when read in the other direction. Fix a triangulation. Then for all metric, or if you like all embeddings of $M$ into $R^{3}$ the integral $\iint_{D} K d A = 2 π (V - E + F)$ . Therefore if $M$ is a sphere, we always have $\iint_{D} K d A = 4 π$ . For a torus it is always the case that $\iint_{D} K d A = 0$ .