Gauss Curvature of a Surface Find the Gauss curvature of a surface with the first quadratic form:
$$\mathrm{d}s^2 = \mathrm{d}u^2 + 2\cos a(u,v)\mathrm{d}u\,\mathrm{d}v + \mathrm{d}v^2.$$
I have found $E$, $F$, and $G$.  $E = 1$, $F = \cos a(u,v)$, and $G = 1$.  I know that Gauss curvature can be calculated using the equation $K = (LN - M^2)/(EG - F^2)$, where $E$, $F$, and $G$ are from the first quadratic form, and $L$, $M$, and $N$ are from the second quadratic form.  Does anyone know how to find $L$, $M$, and $N$ from the information given in the problem, or perhaps a different formula that can calculate Gauss curvature directly from the first quadratic form. Thank you for any help!
 A: The first fundamental form determines the Gaussian curvature but definitely does not determine the second fundamental form. Indeed, the first fundamental form can be defined on an abstract surface (say, the hyperbolic plane) that does not even live inside $\Bbb R^3$.
You need the Gauss equations in order to compute the Gaussian curvature. That is, you need to calculate the Christoffel symbols and then various combinations of them and their derivatives.
Here you go (have fun):
\begin{align*}
\Gamma_{uu}^u &=\frac{\tfrac12GE_u+F(\tfrac12E_v-F_u)}{EG-F^2} \\
\Gamma_{uu}^v &=\frac{-\tfrac12FE_u+E(F_u-\tfrac12E_v)}{EG-F^2}\\
\Gamma_{uv}^u &=\frac{GE_v-FG_u}{2(EG-F^2)} \\
\Gamma_{uv}^v &=\frac{-FE_v+EG_u}{2(EG-F^2)}\\
\Gamma_{vv}^u &=\frac{G(F_v-\tfrac12G_u)-\tfrac12FG_v}{EG-F^2} \\
\Gamma_{vv}^v &=\frac{F(\tfrac12G_u-F_v)+\tfrac12EG_v}{EG-F^2} \\
\end{align*}
and
$$EK = \big(\Gamma_{uu}^v\big)_v - \big(\Gamma_{uv}^v\big)_u
+\Gamma_{uu}^u\Gamma_{uv}^v+\Gamma_{uu}^v\Gamma_{vv}^v-\Gamma_{uv}^u\Gamma_{uu}^v-\big(\Gamma_{uv}^v\big)^2.$$
A: There's a remarkable theorem by Gauss (that's actually what it's called: the Theorema Egregerium) that says Guassian curvature $K$ can be found using only the first fundamental form. There are many different ways to then compute it, one of which (perhaps the most straightforward) is via the Brioschi formula.
