Compact surface with constant strictly positive curvature is a sphere

Question

I'm following Cartan's Differential forms. I'm trying to do exercise 8 on page 161. The chapter is about moving frames and differential forms in surface theory.

Consider the frame of Ex. 2 (principal frame), show that if $dk_1 = dk_2 = 0$ at the point M, then at M $k_1 = k_2$ or $\omega_{12} = 0$. Deduce that on a surface S which has constant strictly positive gaussian curvature K, the principal curvature cannot have a relative maximum or minimum at a point which is not umbilical.

For the first part all ok. In fact we have

$\omega_{13} = k_1\omega_1 \\ \omega_{23} = k_2\omega_2 \\ d\omega_1 = -\omega_2\wedge\omega_{12} \\ d\omega_2 = \omega_1\wedge\omega_{12} \\ d\omega_{12} = -k_1k_2\omega_1\wedge\omega_2 \\ d\omega_{13} = k_2d\omega_1 \\ d\omega_{23} = k_1d\omega_2$

Differentiating the first two and substituting the last two

$ d\omega_{13} = dk_1\wedge\omega_1 + k_1d\omega_1 = k_2d\omega_1 \\ d\omega_{23} = dk_2\wedge\omega_2 + k_2d\omega_2 = k_1d\omega_2$

we obtain

$dk_1\wedge\omega_1 = (k_2 - k_1)d\omega_1 \\ dk_2\wedge\omega_2 = (k_1 - k_2)d\omega_2 $

So if $dk_1 = dk_2 = 0$ we have or $k_1 = k_2$ or $d\omega_1 = d\omega_2 = 0$ and so or $k_1 = k_2$ or $\omega_{12} = 0$.

Now suppose that M is not umbilical, so $k_1 \neq k_2$ and $\omega_{12} = 0$. The frame becomes at M

$\omega_{12} = 0 \\ \omega_{13} = k_1\omega_1 \\ \omega_{23} = k_2\omega_2 \\ d\omega_1 = 0 \\ d\omega_2 = 0 \\ d\omega_{12} = -k_1k_2\omega_1\wedge\omega_2 \\ d\omega_{13} = 0 \\ d\omega_{23} = 0$

Now I don't know how to continue to get a contradiction. I know I have to show that $k_1k_2 \leq 0$, against the hypothesis. I also know the solution working in local coordinates, but I don't know how can I translate this in the language of differential forms. The proof here is from Shifrin's book

I don't know how to get second derivatives with differential forms (because $d^2\omega = 0$), but I suppose (and also I prefer) I have to work avoiding local coordinates.

Thanks in advance

Answer 1

Even with the moving frames computation, you're going to have to do something analogous to the local computation with second-order partial derivatives. How else can we check that a critical point is a local maximum/minimum?

Here's how you should start: Write $dk_i = \sum\limits_j k_{ij}\omega_j$ (so we know that $k_{ij} = 0$ at $M$ for $i,j=1,2$). Then write $dk_{ij} = \sum\limits_\ell k_{ij\ell}\omega_\ell$. If $k_1>k_2$ locally, then we know that $k_{1jj} \le 0$ and $k_{2jj}\ge 0$ at $M$ for $j=1,2$.

I would rather write your third displayed equations as \begin{align*} dk_1\wedge\omega_1 &= (k_2-k_1)\omega_{12}\wedge\omega_2 \\ dk_2\wedge\omega_2 &= (k_1-k_2)\omega_1\wedge\omega_{12}. \end{align*} Solve these to obtain $(k_1-k_2)\omega_{12} = A\omega_1+B\omega_2$. Now can you proceed?