From here , the equation 7 gives $$\mathbf{III}-2H\mathbf{II}+K\mathbf{I}=0$$ where $\mathbf{III},\mathbf{II},\mathbf{I}$ are third, second and first fundamental forms respectively. $H$ is the mean curvature and $K$ is the Gauss curvature. I can only tackle this provlem when I use $\mathbf{III}=(d\mathbf{n},d\mathbf{n}),\mathbf{II}=(d\mathbf{r},d\mathbf{n})$,express it in $\mathbf{r}_u,\mathbf{r}_v$ and then compute the coefficient. The computation is a bit too ugly for this elegant equation. So I wonder if there is other elegant method in solving this problem? Is there geometric interpretation for this identity? Thanks!
Asked
Active
Viewed 321 times
1 Answers
7
As it so happens, I read a nice proof of this fact just two days ago.
Paraphrasing Barrett O'Neill's "Elementary Differential Geometry":
(a) Show that the characteristic polynomial of the shape operator is $$\lambda^2 - 2H \lambda + K.$$
(b) The Cayley-Hamilton Theorem says that every linear operator satisfies its characteristic polynomial. Use this fact to prove that $$\text{III} - 2H\text{II} + K\text{I} = 0.$$
Jesse Madnick
- 31,524