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The full problem is:

Let $(x,y,z)$ be Cartesian coordinates in $\mathbb{R}^3$. Let $x,y,z$ all be a parameterization of a surface $M$ in local coordinates $(u,v)$. Let local coordinates $(u,v)$ be orthogonal: $\partial_ux\partial_vx+\partial_uy\partial_vy+\partial_uz\partial_v=0$.

Then the Euclidean metric $ds^dx^2+dy^2+dz^2$ in $\mathbb{R}^3$ induces the Riemannian metric on the surface $M$ $$ds^2=A^2(u,v)du^2+B^2(u,v)dv^2,$$ where $A^2=(\partial_ux)^2+(\partial_uy)^2+(\partial_uz)^2$ and $B^2=(\partial_vx)^2+(\partial_vy)^2+(\partial_vz)^2$. Find the Riemann curvature tensor of the induced metric of the surface $M$ (in terms of the functions $A$ and $B$).

(Hint: the Riemann curvature tensor in two dimensions has only one independent component.)

Attempt at answer:

$g=\left[ \begin{array}{cc} A^2 & 0 \\ 0 & B^2 \end{array} \right]$ implies $\sigma^1=Ad\mu$ and $\sigma^2=Bdv$. Then $\omega_{12}=\alpha(u,v)\sigma^1+\beta(u,v)\sigma^2$.

$d\sigma^1=-\omega_{12}u\sigma^2=d(Adu)=\frac{\partial A}{\partial v}du\wedge dv=-\frac{1}{AB}\frac{\partial A}{\partial v}\sigma^1\wedge\sigma^2=-(\alpha\sigma^1\wedge\sigma^2)$ Similarly for $d\sigma^2$ (but using $\beta$). After some manipulations, I get $\alpha=\frac{1}{AB}\frac{\partial A}{\partial v}$ and $\beta=-\frac{1}{AB}\frac{\partial B}{\partial u}$.

This gives me $\omega_{12}=\frac{1}{B}\frac{\partial A}{\partial u}-\frac{1}{A}\frac{\partial B}{\partial v}$.

I have no clue how to go on or if I even did the right thing. Any help/advice would be a godsend.

PS: Our textbook is Frankel, if that's at all relevent.

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Once you know $g_{ij}$ you can fix the Levi-Civitia connection, then the curvature should be immediate. – Bombyx mori May 6 '13 at 3:12
up vote 2 down vote accepted

You're on the right track, although there are small errors, presumably typos. When you finally got to $\omega_{12}$, you omitted the $du$ and $dv$. Now use $d\omega_{12}=K\sigma^1\wedge\sigma^2$ to get $K$.

You should find $\omega_{12} = \dfrac{A_v}{B}du - \dfrac{B_u}{A}dv$ and $$K = -\frac1{AB}\left(\left(\frac{B_u}A\right)_u + \left(\frac{A_v}B\right)_v\right)\,.$$

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As a final answer, I get $$\frac{1}{AB^3}\frac{\partial B}{\partial v}\frac{\partial A}{\partial v}-\frac{1}{AB^2}\frac{\partial^2A}{\partial v^2}+\frac{1}{A^3B}\frac{\partial A}{\partial u}\frac{\partial B}{\partial u}-\frac{1}{A^2B}\frac{\partial^2B}{\partial u^2}$$ – Desperate Fluffy May 6 '13 at 4:43
I get the same $\omega$, but not $K$. Can you give further explanation? – Desperate Fluffy May 6 '13 at 14:50
I think we are agreeing, my apologies. Writing out the product rule with my expression gives what you have. The form I have is more compact and, in practice, more useful :) – Ted Shifrin May 6 '13 at 16:24

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