Lecture 13 - Curvature

#math

Gauss (theorema egregium)

The Gaussian curvature of the surface is invariant under isometries

By definition isometries preserve the 1st fundamental form, it suffices to prove that $K$ can be expressed in local coordinates purely in terms of the coefficients $(E, F, G)$ of the 1st fundamental form and its derivatives.

Proof

Let $X : U \to S$ be an isometry with basis vectors $(X_{u}, X_{v}, n)$
Let

\begin{aligned} X_{u u} & = Γ^{1}_{11} X_{u} + Γ^{2}_{11} X_{v} + L n \\ X_{u v} & = Γ^{1}_{12} X_{u} + Γ^{2}_{12} X_{v} + M n \\ X_{v u} & = Γ^{1}_{21} X_{u} + Γ^{2}_{21} X_{v} + M n \\ X_{v v} & = Γ^{1}_{22} X_{u} + Γ^{2}_{22} + N n \end{aligned}

We have that $X_{u v} = X_{v u} ⟺ Γ^{a}_{a b} = Γ^{a}_{b a}$
By taking $X_{u} \cdot X_{u u} = Γ^{1}_{11} E + Γ^{2}_{11} F ⟹ \frac{1}{2} E_{u} = Γ^{1}_{11} E + Γ^{2}_{11} F$ since the LHS is simply $\frac{1}{2} (X_{u} \cdot X_{u})_{u}$
Likewise taking $X_{u u} \cdot X_{v}$ we obtain that $Γ^{1}_{11} F + Γ^{2}_{11} G = F_{u} - \frac{1}{2} E_{v}$
Thus we obtain:

\begin{aligned} Γ^{1}_{11} & = \frac{G E_{u} - 2 F F_{u} + F E_{u}}{2 (E G - F^{2})} \\ Γ^{2}_{11} & = \frac{- E_{u} F + 2 E F_{u} - E E_{u}}{2 (E_{u} - F^{2})} \end{aligned}

With the compatibility condition that $X_{u u v} = X_{u v u}$ we obtain:

\begin{aligned} (Γ^{1}_{11} X_{u} + Γ^{2}_{11} X_{v} + L n)_{v} & = (Γ^{1}_{12} X_{u} + Γ^{2}_{12} X_{v} + M n)_{u} \end{aligned}

Which gives us the Gauss-Weingarten equation:

\begin{matrix} (1) & (Γ^{1}_{11})_{v} + Γ^{2}_{11} Γ^{1}_{22} - a_{21} L = (Γ^{1}_{12})_{u} + Γ^{2}_{12} Γ^{1}_{12} - a_{11} M \end{matrix}

One can also obtain that:

\begin{aligned} (2) & a_{11} M - a_{21} L & = \frac{L N - M^{2}}{E G - F^{2}} \cdot F \\ = F \cdot K \end{aligned}

(1) + (2) gives us :

\begin{matrix} ◻ & K = \frac{1}{(E G - F^{2})^{2}} (| \begin{matrix} - \frac{1}{2} E_{u v} + F_{u v} - \frac{1}{2} G_{u u} & \frac{1}{2} E_{u} & F_{u} - \frac{1}{2} E_{v} \\ F_{v} - \frac{1}{2} G_{u} & E & F \\ \frac{1}{2} G_{v} & F & G \end{matrix} | - | \begin{matrix} 0 & \frac{1}{2} E_{v} & \frac{1}{2} G_{v} \\ \frac{1}{2} E_{v} & E & F \\ \frac{1}{2} G_{u} & F & G \end{matrix} |) \end{matrix}

Tangential Derivative

Definition

We will use the symbol $\nabla$ to denote the tangential derivative. It is the tangential component of the ordinary derivative. That is:

\nabla_{u} a = a_{u} - (n \cdot a_{u}) n

Definition

A geodesic is a vector that has the form of

\begin{aligned} \nabla_{u} X_{u} & = 0 \\ \nabla_{v} X_{v} & = 0 \end{aligned}