Gaussian curvature $K$ of of orthogonal parametrization $X$

Question

Let $X$ be an orthogonal parametrization of some surface $S$. Prove that the Gaussian curvature $K = - \frac{1}{2 \sqrt{E G}} ((\frac{E_{v}}{\sqrt{E G}})_{v} + (\frac{G_{u}}{\sqrt{E G}})_{u})$, where subscripts denote partial differentiation of the quantity with the subscript with respect to the terms within the subscript, and where $E$, $F$, and $G$ give the first fundamental form of $S$ by $X$.

Answer 1

I found this pdf on the Gaussian formula with a full derivation - similar to Ivo's answer above but I liked that the formulae were derived and not given (other than $K=(eg-f^2)/(EG-F^2)$). I've written it out below:

If $F=0$ then the system of determining the Christoffel symbols simplifies to \begin{align*} \Gamma_{11}^1 E + \Gamma_{11}^2 F = \Gamma_{11}^1 E = X_{uu}\cdot X_u &= \frac{E_u}{2} & \implies \Gamma_{11}^1 &= \frac{E_u}{2E} \\ \Gamma_{11}^1 F + \Gamma_{11}^2 G = \Gamma_{11}^2 G = X_{uu}\cdot X_v = F_u - \frac{E_v}{2} &= - \,\frac{E_v}{2} & \implies \Gamma_{11}^2 &= -\, \frac{E_v}{2G} \\ \Gamma_{12}^1 E + \Gamma_{12}^2 F = \Gamma_{12}^1 E = X_{uv}\cdot X_u &= \frac{E_v}{2} & \implies \Gamma_{12}^1 &= \frac{E_v}{2E} \\ \Gamma_{12}^1 F + \Gamma_{12}^2 G = \Gamma_{12}^2 G = X_{uv}\cdot X_v &= \frac{G_u}{2} & \implies \Gamma_{12}^2 &= \frac{G_u}{2G} \\ \Gamma_{22}^1 E + \Gamma_{22}^2 F = \Gamma_{22}^1 E = X_{vv}\cdot X_u = F_v - \frac{G_u}{2} &= - \, \frac{G_u}{2} & \implies \Gamma_{22}^1 &= - \, \frac{G_u}{2E} \\ \Gamma_{22}^1 F + \Gamma_{22}^2 G = \Gamma_{22}^2 G = X_{vv}\cdot X_v &= \frac{G_v}{2} & \implies \Gamma_{22}^2 &= \frac{G_v}{2G} \end{align*} Also, it is clear from the above that, \begin{align*} X_{uu} &= \Gamma_{11}^1 X_u + \Gamma_{11}^2 X_v +eN & \implies X_{uu} &= \frac{E_u}{2E} X_u - \frac{E_v}{2G} X_v + eN; \\ X_{uv} &= \Gamma_{12}^1 X_u + \Gamma_{12}^2 X_v +fN & \implies X_{uv} &= \frac{E_v}{2E} X_u + \frac{G_u}{2G} X_v + fN; \\ X_{vv} &= \Gamma_{22}^1 X_u + \Gamma_{22}^2 X_v +gN & \implies X_{vv} &= - \, \frac{G_u}{2E} X_u + \frac{G_v}{2G} X_v + gN. \end{align*} The next two equations are given on pages 154-5 of do Carmo's book, and is simplified as $F=0$: \begin{align*} N_u &= \frac{fF-eG}{EG-F^2} X_u + \frac{eF-fE}{EG-F^2} X_v = - \, \frac{e}{E} X_u - \frac{f}{G} X_v; \\ N_v &= \frac{gF - fG}{EG - F^2} X_u + \frac{fF - gE}{EG - F^2} X_v = - \, \frac{f}{E} X_u - \frac{g}{G} X_v. \end{align*} Furthermore, the order of differentiation does not matter, such that $X_{uuv} = X_{uvu}$, that is \begin{align*} X_{uuv} &= \left( X_{uu} \right)_v = \left( \frac{E_u}{2E} X_u - \frac{E_v}{2G} X_v + eN \right)_v \\ &= \left( \frac{E_u}{2E} \right)_v X_u + \frac{E_u}{2E} X_{uv} - \left( \frac{E_v}{2G} \right)_v X_v - \frac{E_v}{2G} X_{vv} + e_v N + eN_v \\ &= \left( \frac{E_u}{2E} \right)_v X_u + \frac{E_u}{2E} \left( \frac{E_v}{2E} X_u + \frac{G_u}{2G} X_v + fN \right) - \left( \frac{E_v}{2G} \right)_v X_v - \frac{E_v}{2G} \left( - \, \frac{G_u}{2E} X_u + \frac{G_v}{2G} X_v + gN \right) \\ &{} \qquad {} \qquad + e_v N + e \left( - \, \frac{f}{E} X_u - \frac{g}{G} X_v \right) \\ &= \left[ \left( \frac{E_u}{2E} \right)_v + \frac{E_uE_v}{4E^2} + \frac{E_vG_u}{4EG} - \frac{ef}{E} \right] X_u + \left[ - \, \left( \frac{E_v}{2G} \right)_v + \frac{E_uG_u}{4EG} - \frac{E_vG_v}{4G^2} - \frac{eg}{G} \right] X_v \\ &{} \qquad {} \qquad + \left[ \frac{E_u f}{2E} - \frac{E_v g}{2G} + e_v \right] N. \\ X_{uvu} &= \left( X_{uv} \right)_u = \left( \frac{E_v}{2E} X_u + \frac{G_u}{2G} X_v + fN \right)_u \\ &= \left( \frac{E_v}{2E} \right)_u X_u + \frac{E_v}{2E} X_{uu} + \left( \frac{G_u}{2G} \right)_u X_v + \frac{G_u}{2G} X_{vu} + f_u N + fN_u \\ &= \left( \frac{E_v}{2E} \right)_u X_u + \frac{E_v}{2E} \left( \frac{E_u}{2E} X_u - \frac{E_v}{2G} X_v + eN \right) + \left( \frac{G_u}{2G} \right)_u X_v + \frac{G_u}{2G} \left( \frac{E_v}{2E} X_u + \frac{G_u}{2G} X_v + fN \right) \\ &{} \qquad {} \qquad + f_u N + f \left( - \, \frac{e}{E} X_u - \frac{f}{G} X_v \right) \\ &= \left[ \left( \frac{E_v}{2E} \right)_u + \frac{E_vE_u}{4E^2} + \frac{G_uE_v}{4EG} - \frac{ef}{E} \right] X_u + \left[ - \, \frac{E_v^2}{4EG} + \left( \frac{G_u}{2G} \right)_u + \frac{G_u^2}{4G^2} - \frac{f^2}{G} \right] X_v \\ &{} \qquad {} \qquad + \left[ \frac{E_ve}{2E} + \frac{G_uf}{2G} + f_u \right] N. \end{align*} Before going further, we determine the following: \begin{align*} \left( \frac{E_v}{2G} \right)_v &= \frac{2GE_{vv} - 2G_vE_v}{4G^2} = \frac{E_{vv}}{2G} - \frac{G_vE_v}{2G^2}; \\ \left( \frac{G_u}{2G} \right)_u &= \frac{2G G_{uu} - 2G_u^2}{4G^2} = \frac{G_{uu}}{2G} - \frac{G_u^2}{2G^2}. \end{align*} As $X_{uvu}-X_{uuv}=0$, the coefficients on the basis $\left\{ X_u, X_v, N \right\}$ for $X_{uvu}-X_{uuv}$ must all be zero, that is \begin{align*} 0 = X_{uvu}-X_{uuv} &= \underbrace{\left[ \left( \frac{E_u}{2E} \right)_v - \left( \frac{E_v}{2E} \right)_u \right]}_{=0} X_u \\ &{} \qquad {} \qquad + \underbrace{\left[ \frac{E_u G_u}{4EG} + \frac{E_v^2}{4EG} - \left( \frac{E_v}{2G} \right)_v - \frac{E_vG_v}{4G^2} - \frac{G_u^2}{4G^2} - \frac{eg-f^2}{G} - \left( \frac{G_u}{2G} \right)_u \right]}_{=0} X_v \\ &{} \qquad {} \qquad + \underbrace{\left[ \frac{E_u f}{2E} - \frac{E_ve}{2E} - \frac{E_vg}{2G} - \frac{G_uf}{2G} + e_v - f_u \right]}_{=0} N. \end{align*} The coefficient on $X_v$ gives us \begin{align*} K = \frac{eg-f^2}{EG} &= \frac{1}{E} \left[ \frac{E_u G_u}{4EG} + \frac{E_v^2}{4EG} - \left( \frac{E_v}{2G} \right)_v - \frac{E_vG_v}{4G^2} - \frac{G_u^2}{4G^2} - \left( \frac{G_u}{2G} \right)_u \right] \\ &= - \, \frac{1}{2\sqrt{EG}} \left[ - \, \frac{E_u G_u}{2E\sqrt{EG}} - \frac{E_v^2}{2E\sqrt{EG}} + \frac{2\sqrt{G}}{\sqrt{E}} \left( \frac{E_v}{2G} \right)_v + \frac{E_vG_v}{2G\sqrt{EG}} + \frac{G_u^2}{2G\sqrt{EG}} + \frac{2\sqrt{G}}{\sqrt{E}} \left( \frac{G_u}{2G} \right)_u \right] \\ &= - \, \frac{1}{2\sqrt{EG}} \left[ - \, \frac{E_u G_u}{2E\sqrt{EG}} - \frac{E_v^2}{2E\sqrt{EG}} + \frac{2\sqrt{G}}{\sqrt{E}} \left( \frac{E_{vv}}{2G} - \frac{G_vE_v}{2G^2} \right) \right. \\ &{} \qquad {} \qquad {} \qquad {} \qquad \left. + \frac{E_vG_v}{2G\sqrt{EG}} + \frac{G_u^2}{2G\sqrt{EG}} + \frac{2\sqrt{G}}{\sqrt{E}} \left( \frac{G_{uu}}{2G} - \frac{G_u^2}{2G^2} \right) \right] \\ &= - \, \frac{1}{2\sqrt{EG}} \left[ \left( \frac{E_{vv}}{\sqrt{EG}} - \frac{E_v \left( E_vG + EG_v \right)}{2EG\sqrt{EG}} \right) + \left( \frac{G_{uu}}{\sqrt{EG}} - \frac{G_u \left( E_u G + EG_u \right)}{2EG\sqrt{EG}} \right) \right] \\ &= - \, \frac{1}{2\sqrt{EG}} \left[ \left( \frac{E_v}{\sqrt{EG}} \right)_v + \left( \frac{G_u}{\sqrt{EG}} \right)_u \right], \end{align*} where \begin{align*} \left( \frac{E_v}{\sqrt{EG}} \right)_v &= \frac{E_{vv}\sqrt{EG} - E_v\left( E_v\sqrt{G}/\sqrt{E} + \sqrt{E}G_v/\sqrt{G} \right)/2}{EG}; \\ \left( \frac{G_u}{\sqrt{EG}} \right)_u &= \frac{G_{uu}\sqrt{EG} - G_u\left( E_u \sqrt{G}/\sqrt{E} + \sqrt{E} G_u/\sqrt{G} \right)/2}{EG}. \end{align*}

Answer 2

Orthogonal parematrization means that the first fundamental form has $F=0$. We assume sufficient niceness of the surface $S$ (so that we never divide by $0$ and all functions are infinitely differentiable in all arguments, etc.).

We first derive two related results, where $\Gamma_{i,j}^{k}$ denotes Christoffel symbols of the first kind:

$\Gamma_{1,1}^{1} F + \Gamma_{1,1}^{2} G = X_{u,u} \cdot X_{v} = (X_{u} \cdot X_{v})_{u} - \frac{(X_{u} \cdot X_{v,u})}{2} = F_{u} - \frac{1}{2} E_{v}$.

Likewise, $\Gamma_{1,2}^{1} F + \Gamma_{1,2}^{2} G = X_{u,v} \cdot X_{v} = \frac{1}{2} G_{u}$.

Next recall the formula $K = \frac{1}{\sqrt{E G - F^2}} (\frac{\partial}{\partial v} (\frac{\sqrt{E G - F^2}}{E} \Gamma_{1,1}^{2}) - \frac{\partial}{\partial u} (\frac{\sqrt{E G - F^2}}{E} \Gamma_{1,2}^{2}))$.

From these equations, constraining $F=0$, it follows immediately by substitution:

$K = \frac{1}{\sqrt{E G}} (\frac{\partial}{\partial v} (-\frac{1}{2} \sqrt{\frac{G}{E}} \frac{E_{v}}{G}) - \frac{\partial}{\partial u} (\frac{1}{2} \sqrt{\frac{G}{E}} \frac{G_{u}}{G})) = - \frac{1}{2 \sqrt{E G}} ((\frac{E_{v}}{\sqrt{E G}})_{v} + (\frac{G_{u}}{\sqrt{E G}})_{u})$. QED

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Bonus information

The isothermal formula for Gaussian curvature $K$ follows immediately. The isothermal case is a special case of orthogonal parametrization ($F=0$) in which $E = G= \lambda \dot{=} \lambda (u,v)$.

In this case: $K = -\frac{1}{2 \sqrt{\lambda^2}} ((\frac{\lambda_{v}}{\lambda})_{v} + (\frac{\lambda_{u}}{\lambda})_{u}) = -\frac{1}{2 \lambda} ((log({\lambda})_{v})_{v} + (log({\lambda})_{u})_{u}) = -\frac{1}{2 \lambda} (log({\lambda})_{v,v} + log({\lambda})_{u,u}) = -\frac{1}{2 \lambda} (\frac{\partial^2}{(\partial u)^2} + \frac{\partial^2}{(\partial v)^2})(\log({\lambda})) = -\frac{1}{2 \lambda} \Delta(\log\lambda)$, where the penultimate equation uses a formal sum of derivative (left) operators for convenience and the cleaning up of notation and where the last equation further simplifies it by denoting this formal sum as the Laplacian operator $\Delta$.