Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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 $$K=-d\omega_{12}(e_1,e_2) =-\frac{(\sqrt{G})_{uu}}{\sqrt{G}}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.