How to prove that an ellipsoid is an ovaloid?

Question

An ovaloid is a connected and compact surface whose Gauss curvature is always positive. And the ellipsoid is given by the equation $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} +\frac{z^2}{c^2} - 1 = 0, $$ where $a,b,c$ are constants.

How can I prove that an ellipsoid is an ovaloid?

Answer 1

Following the advice I wrote, one can proceed as follows:

1. Build a map from the sphere to the ellipsoid:

$f: \mathbb{S}^2 \to E$ given by $(x,y,z) \mapsto (ax,by,bz)$

you can check this is well defined and has inverse:

$f^{-1}: E \to \mathbb{S}^2$ given by $(x,y,z) \mapsto \Big(\frac x a,\frac y b,\frac z c\Big)$

here you need $a,b,c \in \mathbb{R} \setminus \{0\}$ because otherwise you get a degenerated ellipsoid.

You can further show that these are continuous so $f$ is indeed an isomorphism (topologically). From here, you derive that since compactness and connectedness are topological invariants, then $E$ has to be compact and connected.

2. You can choose the fourth formula in Wikipedia -> Alternative formulas. This will yield a positive value since $\det\begin{pmatrix} 2/a² & 0 & 0 & 2x/a² \\ 0 & 2/b² & 0 & 2y/b² \\ 0 & 0 & 2/c² & 2z/c² \\ 2x/a² & 2y/b² & 2z/c² & 0\end{pmatrix} = -\frac{16}{a^4b^4c^4}(a^2b²z^2+a^2c^2y^2+b^2c^2x^2) $ which is always negative (then you need to change the sign!!).

This shows that the gaussian curvature is positive and your done.

I would be interested if you check the references for the fourth formula.