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
Best Answer
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?