Lecture 16 - Local Gauss Bonnet Theorem

#math

Let $X (u, v) : R \to M$ be your chart for your surface $M$ .
Recall that:

\begin{aligned} X_{u} \cdot X_{u} & = E \\ X_{v} \cdot X_{v} & = G \end{aligned}

We want to construct an orthonormal basis for our tangent space at each point on the surface, this can easily be done by taking:

\begin{aligned} e_{1} & = \frac{X_{u}}{\sqrt{E}} \\ e_{2} & = \frac{X_{v}}{\sqrt{G}} \end{aligned}

We are assuming $X_{u} \cdot X_{v} = F = 0$ .
Now let $β : [a, b] \to U$ be a simple smooth closed curve in $R^{2}$ . The image of this curve on the surface is $α$ . That is:

X \circ β (s) = α (s)

$s$ as usual is the arc-length parameterization.

First we investigate the following integral:

I = \int_{β} e_{1} \cdot e_{2}^{'} d s

Let $δ (s)$ be the angle between the angle between $α^{'} (s)$ and $e_{1}$ at $α (s)$ , with the angle measured in the anticlockwise direction. Then we express $α^{'} (s)$ as:

α^{'} (s) = c o s (δ) e_{1} + s i n (δ) e_{2}

Now define $η = \vec{n} \times α^{'}$ then:

η = - s i n (δ) e_{1} + c o s (δ) e_{2}

Thus:

\begin{aligned} α^{″} (s) & = δ^{'} (- s i n (δ) e_{1}) + c o s (δ) e_{1}^{'} + δ^{'} (c o s (δ) e_{2}) + s i n (δ) e_{2} \\ = δ^{'} η + c o s (δ) e_{1}^{'} + s i n (δ) e_{2}^{'} \end{aligned}

The geodesic curvature:

\begin{aligned} K_{g} & = α^{″} \cdot η \\ = δ^{'} + (e_{1}^{'} c o s (δ) + e_{2}^{'} s i n (δ)) \cdot (- s i n (δ) e_{1} + c o s (δ) e_{2}) \\ = δ^{'} - e_{1} \cdot e_{2}^{'} \end{aligned}

Thus:

\begin{aligned} I & = \int_{β} e_{1} \cdot e_{2}^{'} d s \\ = \int_{β} δ^{'} d s - \int_{β} K_{g} d s \\ = 2 π - \int_{β} K_{g} d s \end{aligned}

We go back:

\begin{aligned} I & = \int_{β} e_{1} \cdot e_{2}^{'} d s \\ = \int_{β} e_{1} \cdot (u^{'} (e_{2})_{u} + v^{'} (e_{2})_{u}) d s \\ = \int_{β} (e_{1} \cdot {e_{2}}_{u}) u^{'} + (e_{1} \cdot {e_{2}}_{v}) v^{'} d s \\ = \int_{β} P u^{'} + Q v^{'} d s \\ = \int_{C} (Q_{u} - P_{v}) d u d v (By Green’s Theorem) \end{aligned}

Lemma

$Q_{u} - P_{v} = K \sqrt{E G - F^{2}}$

We are interested in this lemma since if we can prove it our integral above reduces to:

\int_{X^{- 1} (R)} (Q_{u} - P_{v}) d u d v = \int_{R} K d A

Proof

This requires computing $Q_{u} - P_{v}$ . We have:

\begin{aligned} Q_{u} - P_{v} & = {e_{1}}_{u} \cdot {e_{2}}_{v} + e_{1} \cdot {e_{2}}_{u v} - {e_{1}}_{u} \cdot {e_{2}}_{v} - e_{1} \cdot {e_{2}}_{u v} \\ = {e_{1}}_{u} \cdot {e_{2}}_{v} - {e_{1}}_{v} \cdot {e_{2}}_{u} \end{aligned}

Recall that

K \sqrt{E G - F^{2}} = \frac{L N - M^{2}}{\sqrt{E G - F^{2}}} = n \cdot (n_{u} \times n_{v})

Thus:

\begin{aligned} n \cdot (n_{u} \times n_{v}) & = (e_{1} \times e_{2}) \cdot (n_{u} \times n_{v}) \\ = (e_{1} \cdot n_{u}) (e_{2} \cdot n_{v}) - (e_{1} \cdot n_{v}) (e_{2} \cdot n_{u}) \end{aligned}

Since $e_{1} \cdot n = 0 ⟹ e_{1} \cdot n_{u} = - (e_{1})_{u} \cdot n$ (and likewise for other identities involving $e_{2}$ and $v$ ) the above equation transforms to:

= ({e_{1}}_{u} \cdot n) ({e_{2}}_{v} \cdot n) - ({e_{1}}_{v} \cdot n) ({e_{2}}_{u} \cdot n)

Notice that we have:

\begin{aligned} (e_{1})_{w} & = ((e_{1})_{w} \cdot e_{2}) e_{2} + ((e_{1})_{w} \cdot n) n \\ (e_{2})_{w} & = ((e_{2})_{w} \cdot e_{1}) e_{1} + ((e_{2})_{w} \cdot n) n \end{aligned}

Thus since $Q_{u} - P_{v} = {e_{1}}_{u} \cdot {e_{2}}_{v} - {e_{1}}_{u} \cdot {e_{2}}_{v}$ we completed our proof. $◻$

To adapt the proof to the case that $α$ has n corners (that is it is a piecewise curve). We revise our first evaluation of the integral:

I = \int δ^{'} (s) d s - \int K_{g} d s

Since $α$ has n vertices we obtain that:

\int_{β} δ^{'} (s) d s = 2 π - \sum_{i = 1}^{n} ϑ_{i}

where $ϑ_{i}$ are the external angles at the corners. Thus we have shown that:

I = \int K d A = 2 π - \int K_{g} d s - \sum_{i = 1}^{n} ϑ_{i}

We have thus proven the following theorem:

Theorem

Local Gauss Bonnet Theorem:

\sum ϑ_{i} + \int_{S} K d A + \int K_{g} d s = 2 π

Definition

Let $F = (p (x, y), q (x, y))$ be a vector field on a surface $S$ with isolated singularities $p_{i}$ (that is $F (p_{i}) = 0$ ). The index of a surface $S$ w.r.t. the vector field $F$ is defined by:

I_{F} (S) = \sum_{i = 1}^{N} I_{F} (p_{i})

I_{F} (p_{i}) = \frac{1}{2 π} \oint \frac{p d q - q d p}{p^{2} + q^{2}}

Poincare Hopf Index Theorem

The sum $I_{F} (S)$ of the indices of the vector field $F$ on a compact, connected, orientable surface $S$ is equal to the Euler characteristic of $S$ .