Umbilical points of the ellipsoid.

Question

The following is Exercise (11) of Chapter 3 of Curves and Surfaces, 2nd edition, by Montiel and Ros:

Determine the umbilical points of the ellipsoid of equation $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$ where $0 < a < b < c$.

My question is:

Why the umbilical points must have $y = 0$? Why can't it happen that $x = 0$ or $z = 0$?

The book provides a solution, that I worked out as follows (without quite understanding the conclusion):

Denote the ellipsoid by $S$. We begin by noting that $S$ is the inverse image of a regular value of the function $$ f(x, y, z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}. $$ Indeed, if $f(x, y, z) = 1$ then $$ \nabla f(x, y, z) = 2 \left (\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right) \neq 0. $$ It is clear that $S = f^{-1}(\{1\})$.

The gradient gives us a Gauss map defined on $S$: $$ N(x, y, z) = \frac{\nabla f(x, y, z)}{|\nabla f (x, y, z)|}. $$ Then, there exists a nonvanishing differentiable function $h$ defined on $S$ such that $$ h(x, y, z) N(x, y, z) = \left(\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right). $$ Using the product rule and the fact that the right-hand side is linear, for all $p = (x, y, z) \in S$ and $v = (v_1, v_2, v_3) \in T_pS$ we have that $$ (dh)_p(v)N(p) + h(p)(dN)_p(v) = \left(\frac{v_1}{a^2}, \frac{v_2}{b^2}, \frac{v_3}{c^2}\right). $$

The point $p$ is umbilical if and only if $(dN)_p$ is a multiple of the identity. We claim that this happens if and only if the left-hand side vanishes after scalar multiplication by $N(p) \wedge v$. It is clear that the first term vanishes, since $(N(p) \wedge v) \perp N(p)$. Suppose $(dN)_p$ is a multiple of the identity. Then $h(p)(dN)_p(v) \in \langle v \rangle$ and the second term vanishes as well. On the other hand, assume that $$ \left(h(p) (dN)_p(v)\right) \cdot (N(p) \wedge v) = 0. $$ Then $(dN)_p(v)$ lies in the plane spanned by $N(p)$ and $v$. But since $(dN)_p(v) \in T_pS$, if follows that it is a multiple of $v$, and therefore the claim follows.

From the previous paragraph, we conclude that
$$ \left( \frac{v_1}{a^2}, \frac{v_2}{b^2}, \frac{v_3}{c^2}\right) \cdot (N(p) \wedge v) = 0, $$ that is, $\left(\frac{v_1}{a^2}, \frac{v_2}{b^2}, \frac{v_3}{c^2}\right)$ lies on the plane spanned by $N(p)$ and $v$. Since these three vectors are linearly dependent, we have that $$ \begin{vmatrix} \displaystyle \frac{v_1}{a^2} & \displaystyle \frac{v_2}{b^2} & \displaystyle \frac{v_3}{c^2} \\ v_1 & v_2 & v_3 \\ \displaystyle \frac{x}{a^2} & \displaystyle \frac{y}{b^2} & \displaystyle \frac{z}{c^2} \end{vmatrix} = 0, \quad \text{ if } \quad N(p)\cdot v = 0. $$ But the determinant is a quadratic form in the variables $v_i$ which vanishes on a plane: $$ q(v) = \frac{z}{c^2}\left(\frac{1}{a^2} - \frac{1}{b^2}\right)v_1 v_2 + \frac{x}{a^2} \left(\frac{1}{b^2}- \frac{1}{c^2}\right)v_2 v_3 + \frac{y}{b^2} \left(\frac{1}{c^2} - \frac{1}{a^2}\right)v_1 v_3 $$ Hence, the matrix of this quadratic form is $$ A = \frac12 \begin{bmatrix} 0 & \displaystyle \frac{z}{c^2}\left(\frac{1}{a^2} - \frac{1}{b^2}\right) & \displaystyle \frac{y}{b^2} \left(\frac{1}{c^2} - \frac{1}{a^2}\right)\\ \displaystyle \frac{z}{c^2}\left(\frac{1}{a^2} - \frac{1}{b^2}\right) & 0 & \displaystyle \frac{x}{a^2} \left(\frac{1}{b^2}- \frac{1}{c^2}\right)\\ \displaystyle \frac{y}{b^2} \left(\frac{1}{c^2} - \frac{1}{a^2}\right) & \displaystyle \frac{x}{a^2} \left(\frac{1}{b^2}- \frac{1}{c^2}\right) & 0 \end{bmatrix}. $$

So far, so good. Now, the book argues that since this quadratic form vanishes on a plane, the determinant of $A$ must be zero.

Why is this so?

Moving forward, $\det A = 0$ if $x =0$ or $y = 0$ or $z = 0$. The book claims that it can only be $y = 0$. Why?

I have found this similar question, but I did not quite understand what is going on: Umbilical points of Ellipsoid alternate method

Thanks in advance and kind regards

Answer 1

I think I got this.

Suppose $x = 0$. From $N(p) \cdot v = 0$ we have that $$ v_3 = - \frac{y}{b^2} \frac{c^2}{z}v_2. $$ Then $$ q(v) = \left( \frac{z}{c^2} \left( \frac{1}{a^2} - \frac{1}{b^2} \right) - \frac{y}{b^2} \frac{c^2}{z} \left( \frac{1}{c^2} - \frac{1}{a^2}\right) \right) v_1 v_2 = 0 \quad \forall v_1, v_2 \in \Bbb{R}. $$ Then $$ \frac{z}{c^2} \left( \frac{1}{a^2} - \frac{1}{b^2} \right) = \frac{y^2}{b^2} \frac{c^2}{z} \left( \frac{1}{c^2} - \frac{1}{a^2}\right) \implies \frac{z^2}{c^2} = \frac{y^2}{b^2} \left( \frac{a^2 - c^2}{b^2 - a^2} \right) < 0, $$ which is absurd.

If $y = 0$, by the same argument we arrive at $$ \frac{x^2}{a^2} = \frac{z^2}{c^2} \left( \frac{b^2 - a^2}{c^2-b^2} \right), $$ which is feasible.

If in turn $z = 0$, then $$ \frac{x^2}{a^2} = \frac{t^2}{b^2} \left( \frac{a^2 - c^2}{c^2-b^2} \right) < 0, $$ which is also absurd. Hence, the only possible option is $y = 0$.

Any comments will be appreciated.