user159870

I look at this exercise of A. Pressley:

Show that the Gaussian and mean curvatures of a surface S are smooth functions on $S$.


I did this:

We consider that $S$ is smooth, so also its parametrization $\sigma$.

Having that $\sigma$ is smooth, its partial derivatives of all orders are continuous.

So $$ E=\|\sigma_u\|^2, \ F=\sigma_u \cdot \sigma_v, \ G=\|\sigma_v\|^2, \ L=\sigma_{uu}\cdot \mathbf{N}, \ M=\sigma_{uv}\cdot \mathbf{N}, \\ N=\sigma_{vv}\cdot \mathbf{N}, \ \mathbf{N}=\frac{\sigma_u \times \sigma_v}{\|\sigma_u \times \sigma_v\|}$$ are smooth functions.

We have the ellipsoid $\frac{x^2}{p^2}+\frac{y^2}{q^2}+\frac{z^2}{r^2}=1 , p, q, r \neq 0$.

I want to show that it is a smooth surface.

$r(u,v)=(p\sin{u} \cos{v}, q\sin{u} \sin{v} , r\cos{u})$ a parametrization of the ellipsoid.

I calculated that $r_u \times r_v=(qr\sin^2{u} \cos{v}, pr\sin^2{u} \sin{v}, pq\cos{u} \sin{u})$.

So that it is a smooth surface, one has to show that $r_u \times r_v \neq 0$, or not?

This doesn't stand. For $u=0$ : $r_u \times r_v=(0,0,0)$, or?

Does one not show in this way that it is a smooth surface?

@TedShifrin
For the first question.
We can show that there is such a sequence as follows:

If there is no such $T_0$ , by definition that means that for every $T_0>0$
there exists $0< T'<T_0$ such that $\gamma$ is $T'$-periodic.
We use this to define the sequence iteratively: to get $T_1$ , we take $T_0=1$
in the above, and we let $T_1$ be the corresponding $T'$ . Then, we "take $T_0 =T_1$ in the above, and we let $T_2$ be the new "corresponding T ′
. Etc.: given $T_n$ we define $T_{n+1}$ to be the T ′ corresponding to $T_0=T_n$.