MHB What is the connection form matrix of a surface of revolution?

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Here is this week's POTW:

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Let $S$ be a surface of revolution parametrized by $x(u,v) = (f(u)\cos v, f(u)\sin v, g(u))$, where $(u,v)$ ranges over some open connected set $\Omega \subset \Bbb R^2$. Assume $f$ and $g$ are smooth, $f > 0$ on its domain, and $(f')^2 + (g')^2 = 1$. Evaluate the connection form matrix of $S$ relative to $x$.

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No one answered this week's problem. You can find my solution below.

Spoiler

We have $x_u = (f'(u)\cos v, f'(u)\sin v, g'(u))$, $x_v = (-f(u)\sin v, f(u)\cos v, 0)$, and $x_u \times x_v = (-f(u)g'(u)\cos v, -f(u)g'(u)\sin v, f(u)f'(u))$. Since $\|x_u\| = 1$ and $\|x_u \times x_v\| = f(u)$, then

$$\xi_1 = x_u, \quad \xi_3 = \frac{x_u \times x_v}{f(u)} = (-g'(u)\cos v, -g'(u)\sin v, f'(u)), \quad \xi_2 = \xi_3 \times \xi_1 = (-\sin v, \cos v, 0)$$

forms an adapted moving orthonormal frame for $S$. Now since

$$d\xi_1 = (f''(u)\cos v\, du - f'(u)\sin v\, dv, f''(u)\sin v\, du + f'(u)\cos v\, dv, g''(u)\, du)$$
$$d\xi_2 = (-\cos v\, dv, -\sin v\, dv, 0)$$
$$d\xi_3 = (g''(u)\cos v\, du - g'(u)\sin v\, dv, -g''(u)\sin v\, du - g'(u)\cos v\, dv, f''(u)\, du)$$

we have

$$d\xi_1 \cdot \xi_2 = f'(u)\, dv, \quad d\xi_1 \cdot \xi_3 = (g''(u)f'(u) - f''(u)g'(u))\, du, \quad d\xi_2\cdot \xi_3 = g'(u)\, dv.$$

Hence, the connection form matrix is

$$\begin{bmatrix}0 & -d\xi_1 \cdot \xi_2 & -d\xi_1 \cdot \xi_3\\d\xi_1 \cdot \xi_2 & 0 & d\xi_2 \cdot \xi_3\\d\xi_1 \cdot \xi_3 & d\xi_2 \cdot \xi_3 & 0\end{bmatrix} = \begin{bmatrix}0 & -f'(u)\, dv & -(g''(u)f'(u) - f''(u)g'(u)\, du\\f'(u)\, dv & 0 & -g'(u)\, dv\\(g''(u)f'(u) - f''(u)g'(u))\, dv & g'(u)\, dv& 0\end{bmatrix}.$$