# [Differential Geometry] Computing the Gaussian curvature of an Orthogonal Parametrisation

Let's derive a formula for the Gaussian curvature of an orthogonal parametrisation $$X(u,v)$$ of a coordinate neighbourhood at point $$p$$ on a smooth, orientable surface, we have that $$\left<X_u, X_v\right>_p = 0$$, hence $$F=0$$ of the first fundamental form at $$p$$.

First, we consider the Gauss formula expressed in Christoffel symbols: $-EK = (\Gamma_{12}^2)_u - (\Gamma_{11}^2)_v + \Gamma_{12}^1\Gamma_{11}^2+\Gamma_{12}^2\Gamma_{12}^2-\Gamma_{11}^2\Gamma_{22}^2-\Gamma_{11}^1\Gamma_{12}^2$

How do we compute the Christoffel symbols?

If we assign each point of $$X(U)$$ a natural trihedron given by the vectors $$X_u$$, $$X_v$$, $$N$$, and express the derivatives of the vectors $$X_u$$, $$X_V$$ and $$N$$ in the basis $$\{ X_u, X_v, N\}$$,

$X_{uu} = \Gamma^1_{11}X_u + \Gamma^2_{11}X_v+L_1N$ $X_{uv} = \Gamma^1_{12}X_u + \Gamma^2_{12}X_v+L_2N$ $X_{vu} = \Gamma^1_{21}X_u + \Gamma^2_{21}X_v + \overline{L_2}N$ $X_{vv} = \Gamma^1_{22}X_u + \Gamma^2_{22}X_v + L_3N$

And by taking the inner product of the first four relations below with $$X_u$$ and $$X_v$$ we get the following:

$\langle X_u, X_{uu} \rangle = \Gamma^1_{11}\cdot \langle X_u, X_u \rangle + \Gamma^2_{11}\cdot \langle X_v, X_u \rangle + 0 = \Gamma^1_{11}E + \Gamma^2_{11}F = \frac{1}{2} E_u ,$ $\langle X_v, X_{uu}\rangle = \Gamma^1_{11}\cdot \langle X_v, X_u \rangle + \Gamma^2_{11}\cdot \langle X_v, X_v \rangle + 0 = \Gamma^1_{11}F + \Gamma^2_{11}G = F_u-\frac{1}{2}E_v.$ $\langle X_u, X_{uv} \rangle = \Gamma^1_{12}\cdot\langle X_u, X_u \rangle + \Gamma^2_{12}\cdot \langle X_u, X_v\rangle + 0 = \Gamma^1_{12}E + \Gamma^2_{12}F = \frac{1}{2}E_v,$ $\langle X_v, X_{uv}\rangle = \Gamma^1_{12}\cdot \langle X_v, X_u \rangle + \Gamma^2_{12}\cdot \langle X_v, X_v\rangle +0= \Gamma^1_{12}F+ \Gamma^2_{12}G = \frac{1}{2} G_u.$ $\langle X_u, X_{vv}\rangle = \Gamma^1_{22}\cdot \langle X_u, X_u\rangle + \Gamma^2_{22}\cdot \langle X_u, X_v \rangle + 0 = \Gamma^1_{22}E + \Gamma^2_{22}F = F_v - \frac{1}{2}G_u,$ $\langle X_v, X_{vv} \rangle = \Gamma^1_{22}\cdot \langle X_v, X_u \rangle + \Gamma^2_{22}\cdot \langle X_v, X_v\rangle + 0 = \Gamma^1_{22}F + \Gamma^2_{22}G = \frac{1}{2}G_v.$

Using the condition that $$x$$ is an orthogonal parametrisation, that is, $$F=0$$, then the above reduces to $\Gamma_{11}^1 = -\frac{1}{2}\frac{E_u}{E}\Gamma_{11}^2 = -\frac{1}{2}\frac{E_v}{G}, \Gamma_{12}^1 = \frac{1}{2}\frac{E_v}{E}, \Gamma_{12}^2 = -\frac{1}{2}\frac{G_u}{G}, \Gamma_{22}^2 = -\frac{1}{2}\frac{G_u}{E}, \Gamma_{12}^2 = \frac{1}{2}\frac{G_u}{G}$

We substitute these values into the Gauss formula: $-EK = \left(\frac{1}{2}\frac{G_u}{G}\right)_u - \left(-\frac{1}{2}\frac{E_v}{G}\right)_v - \frac{1}{4}\frac{E_u}{E}\frac{G_u}{G} - \frac{1}{4}\frac{E_v}{E}\frac{E_v}{G} +\frac{1}{4}\frac{E_v}{G}\frac{G_v}{G} +\frac{1}{4}\frac{G_u}{G}\frac{G_u}{G}$ $= \left(\frac{1}{2}\frac{G_u}{G}\right)_u - \left(-\frac{1}{2}\frac{E_v}{G}\right)_v - \frac{1}{4}\frac{E_u G_u}{EG} - \frac{1}{4}\frac{(E_v)^2}{EG} + \frac{1}{4}\frac{E_v G_v}{G^2} +\frac{1}{4}\frac{(G_u)^2}{G^2}$ $\implies K = -\frac{1}{2\sqrt{EG}}\left[\frac{E_{vv}}{\sqrt{EG}}- \frac{E_v(E_vG +EG}{2(EG)^{\frac{3}{2}}}+ \frac{G_{uu}}{\sqrt{EG}} - \frac{G_u(E_uG+EG_u}{2(EG)^{\frac{3}{2}}}\right]$ Therefore, simplifying gives the Gaussian curvature $K = -\frac{1}{2\sqrt{EG}}\left[ \left(\frac{E_v}{\sqrt{EG}}\right)_v + \left(\frac{G_u}{\sqrt{EG}}\right)_u \right].$

Note by Tasha Kim
3 months, 2 weeks ago

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