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Calculation of Gaussian Curvation via Connection Forms

In this article, I am going to calculate the Gaussian curvature via connection forms.

Suppose that $X:D\to \Sigma$ is a local parametrization on a surface $\Sigma$ in $\mathbb{R}^{3}$ so that $X=X(u,v)$ is an orthogonal coordinate system, which means that the tangent vectors $X_{u}$ and $X_{v}$ are orthogonal at each point of $\Sigma$ . Here $D$ is an open subset of $\mathbb{R}^{2}$ . The first fundamental form on $\Sigma$ are given by the following three functions: $E=\langle X_{u},X_{u}\rangle$ , $F=\langle X_{u},X_{v}\rangle=0$ , $G=\langle X_{v},X_{v}\rangle$ . Here $\langle\cdot,\cdot\rangle$ denotes the induced metric on $\Sigma$ .

Let $\displaystyle E_{1}=\frac{X_{u}}{\sqrt{E}}$ , $\displaystyle E_{2}=\frac{X_{v}}{\sqrt{G}}$ . Then $\{E_{1},E_{2}\}$ forms an orthonormal frame on $X(D)$ . Suppose that $\{\theta^{1},\theta^{2}\}$ is the dual coframe to $\{E_{1},E_{2}\}$ . Then we know $\theta^{1}=\sqrt{E}du$ , $\theta^{2}=\sqrt{G}dv$ . Let $\omega_{1}^{2}$ denote the connection one form on $\Sigma$ . Then

$\displaystyle d\theta^{1}=\omega_{2}^{1}\wedge\theta^{2}=-\frac{\sqrt{E}_{v}}{\sqrt{G}}du\wedge\theta^{2}$ ,

$\displaystyle d\theta^{2}=-\omega_{2}^{1}\wedge\theta^{1}=-\frac{\sqrt{G}_{u}}{\sqrt{F}}du\wedge\theta^{1}$ .

We obtain that

$\displaystyle\omega_{1}^{2}=-\frac{\sqrt{E}_{v}}{\sqrt{G}}du+\frac{\sqrt{G}_{u}}{\sqrt{F}}dv$ .

The Gauss-Codazzi equation tells us that

$d\omega_{1}^{2}=-K\theta^{1}\wedge\theta^{2}$ .

Hence

$\displaystyle K=-\frac{1}{2\sqrt{EG}}\left\{\frac{\partial}{\partial u}\frac{\sqrt{G}_{u}}{\sqrt{EG}}+\frac{\partial}{\partial v}\frac{\sqrt{E}_{v}}{\sqrt{EG}}\right\}$ .

As an application, let us calculate the Gaussian curvature when the coordinate system is isothermal. Recall that a coordinate system $(u,v)$ on a surface $\Sigma$ is called isothermal if there exists a smooth function $\lambda=\lambda(u,v)$ so that $E=F=\lambda^{2}$ . Then we can show that