# Formula for Gaussian curvature in terms of unit tangent vector fields?

Let $X\in\mathbb{R}^3$ be a surface with a local geodesic polar parametrisation with first fundamental form $du^2+G(u,v)dv^2$. How do we define unit tangent vector fields $e_1$, $e_2$ on $X$, forming an orthonormal basis for the tangent plane at each point, such that the Gaussian curvature is $K=\sqrt{G}\Big(\frac{\partial}{\partial v}<\frac{\partial e_1}{\partial u},e_2> - \frac{\partial}{\partial u}<\frac{\partial e_1}{\partial v},e_2>\Big)$?

I think $e_1$ and $e_2$ should be $\frac{1}{\sqrt{G}}dv=(0,\frac{1}{\sqrt{G}})$ and $du=(1,0)$, but I can't get the required formula for $K$ from this. Also, with these $e_1$ and $e_2$ I can't see how to derive $\frac{(\sqrt{G})_{uu}}{\sqrt{G}}$ (which is well known) from the required expression!

Are the numbers $\frac{\partial e_1}{\partial u}$ and $\frac{\partial e_1}{\partial v}$ somehow related to the Christoffel symbols? Maybe the required expression comes out somehow from the Gauss formula?

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If $$dudu + Gdvdv$$ then coframe is $$\omega_1 = du,\ \omega_2=\sqrt{G}dv\ (e_1=\partial_u,\ e_2=\frac{1}{\sqrt{G}}\partial_v)$$
Then from $$d\omega_k =\omega_{kl}\wedge \omega_l$$ we have $$\omega_{12} =(\sqrt{G})_u dv$$ And $$K=-d\omega_{12}(e_1,e_2) =-\frac{(\sqrt{G})_{uu}}{\sqrt{G}}$$