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I'm having serious trouble finding the umbilical points of the ellipsoid represented by

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1, \;\;\;a,b,c\neq 0.$$

My first thought was to use the parametrization

$$\mathbf{x}(u,v)=(a\sin(u)\cos(v),b\sin(u)\sin(v),c\cos(u)),$$

for $0<u<\pi$ and $0<v<2\pi$, compute the first and second fundamental forms, etc., but this is a nightmare. After doing some researching (and on the back solutions of Do Carmo) I came across I suppose what would be an alternate method which doesn't dig directly into a parametrization. It is explained slightly at the end of the pdf:

http://www.math.umn.edu/~voronov/5378/sample1.pdf

which essentially states to notice that $N_1=(\frac{x^2}{a^2},\frac{y^2}{b^2},\frac{z^2}{b^2})$ (the gradient) is such that $N_1=fN$, for some $f$ such that $|f|=|N_1|$, where $N$ is the unit normal vector to surface, as well as notice a point on a curve $\alpha(t)=(x(t),y(t),z(t))$ lying on the ellipsoid is an umbilical point iff the vector triple product \begin{equation} \left(\frac{dN_1}{dt}\wedge \alpha '\right)\cdot N_1=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1) \end{equation} which I mostly understand. Then it says use some trickery by multiplying a $\frac{z}{c^2}$ to the equation and put it in terms of $x,x',y',$ and $y$ like this...

The way I understand it you should start with equation (1) in this form $$ \left(\left(\frac{x'}{a^2},\frac{y'}{b^2},\frac{z'}{c^2}\right)\wedge\left(x',y',z'\right)\right)\cdot\left(\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2}\right)=0. $$ Then, making things more complicated, (plugging everything in and doing the computation) we have $$ \frac{xy'z}{a^2b^2}-\frac{xy'z'}{a^2c^2}+\frac{x'yz'}{b^2c^2}-\frac{x'yz'}{a^2b^2}+\frac{x'y'z}{a^2c^2}-\frac{x'y'z}{b^2c^2}=0. $$ Multiplying $\frac{z}{c^2}$ to both sides gives $$ \frac{xy'z'z}{a^2b^2c^2}-\frac{xy'z'z}{a^2c^4}+\frac{x'yz'z}{b^2c^4}-\frac{x'yz'z}{a^2b^2c^2}+\frac{x'y'z^2}{a^2c^4}-\frac{x'y'z^2}{b^2c^4}=0 $$ From here I suppose one would use the original equation for the ellipsoid as well as implicit derivative, $\frac{2zz'}{c^2}=-\frac{2yy'}{b^2}-\frac{2xx'}{a^2}$ to get rid of $z$ and $z'$. However, when I do that it starts getting pretty messy and I'm starting to believe I'm not quite understanding the method correctly. I also believe $y=0$ should satisfy this equation, but that's not quite working out, which also leads me to believe that I'm wrong in my thought.

Any opinions/suggestions would be greatly appreciated. Thank you

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2 Answers 2

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An umbilical point $x$ of a surface $S$ is characterized by $dN_x(v)=kv$ for every $v$ in $T_xS$.

If $v=dx/dt$ then $dN/dt=kdx/dt$ is equivalent to $\det(dN/dt,N,dx/dt)=0$.

If $N_1=fN$ then $N_1'=f'N+fN'$ and $\det(N_1',N_1,x')=f^2\det(N',N,x')$.
Then taking $N_1=(x_i/a_i)$, $N'=(v_i/a_i)$ and $x'=(v_i)$ for the ellipsoid $E$: $$\sum_{ i=1}^3 x_i^2/a_i=1 \quad \text{with} \quad 0 < a_1 < a_2 < a_3$$ we obtain that $$\det(N_1',N_1,x')v_3/a_3=[(a_3-a_2)x_1v_2+(a_3-a_1)x_2v_1](-x_3v_3/a_3)+[v_1v_2(a_2-a_1)]x_3^2/a_3.$$ Substituting $-x_3v_3/a_3~$ by $~x_1v_1/a_1+x_2v_2/a_2$ we deduce that the umbilical points $(x_i)$ of $E$ satisfy that $$v_1^2x_1x_2(a_3-a_1)/a_1+v_2^2x_1x_2/a_2+v_1v_2\Delta=0$$ for every $(v_1,v_2)$, i.e. $x_1x_2=0$ and $$\Delta:=-x_1^2(a_3-a_2)/a_1+x_2^2(a_3-a_1)/a_2+x_3^2(a_2-a_1)/a_3=0.$$ If $0<a_1<a_2<a_3$ then the only solutions are given by $$x_2=0~, \quad x_1^2(a_3-a_2)/a_1=x_3^2(a_2-a_1)/a_3$$ and $$x_1^2/a_1+x_3^2/a_3=1~,$$ that is $$x_1=\pm\sqrt{a_1(a_2-a_1)/(a_3-a_1)}~, \quad x_2=0~, \quad \text{and} \quad x_3=\pm\sqrt{a_3(a_3-a_2)/(a_3-a_1)}$$

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I will write a very detailed answer. Suppouse our ellipsoid is given by $f^{-1}(1)$ where $f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}$. Any curve passing through points in the preimage of $1$ are such that $f\circ \alpha(s)=1$. Differentiating with respect to $s$ yields $\nabla f(\alpha(s))\cdot \alpha'(s)=0$. As $\nabla f(\alpha(s))\not=0$ we may take normalize it to gain a Gauss Normal Map $N=\frac{(x/a^2,y/b^2,z/c^2)}{\sqrt{x^2/a^4+y^2/b^4+z^2/c^4} }$ where $h=\sqrt{x^2/a^4+y^2/b^4+z^2/c^4}$.

$$h(\alpha(s))N(\alpha(s))={\left(\frac{\alpha_1}{a^2},\frac{\alpha_2}{b^2},\frac{\alpha_3}{c^2}\right)}{}\Rightarrow $$ $$(h\circ\alpha)'(s)N(\alpha(s))+h\circ\alpha(s) DN(\alpha'(s))=\left(\frac{\alpha_1'}{a^2},\frac{\alpha_2'}{b^2},\frac{\alpha_3'}{c^2}\right) \tag{1}$$ Let $\alpha'=(v_1,v_2,v_3)$.

Claim: A point is umbilic if and only if $DN(v)\cdot(N\times v)=0$

Proof: $(\Rightarrow)$ Being umbilic, $DN(v)=\lambda v$ and the results holds.

$(\Leftarrow)$ $\{v, N, N\times v\}$ forms an orthogonal basis if $v\in T_pS$ is not equal to zero. $DN(v)=\lambda_v v+\lambda_2 N+\lambda_3 N\times v$. Furthermore $\lambda_2=0$ ( because $DN(v)\in T_pS$) and $\lambda_3=0$ ( because $DN(v)\cdot(N\times v)=0$). Thus $DN(v)=\lambda_v v$. In particular if $e_1$ and $e_2$ are principal directions:

$$\lambda_{e_1+e_2}(e_1+e_2)=DN(e_1+e_2)=DN(e_1)+DN(e_2)=k_1e_1+k_2e_2$$

Thus, as $\{e_1,e_2\}$ is a basis $k_1=k_2=\lambda_{e_1+e_2}$.

With this lemma combined with equation (1), we must have umbilic points if and only if $DN(v)\cdot(v\times N)=0$, which happens if and only if:

$$\left(\frac{\alpha_1'}{a^2},\frac{\alpha_2'}{b^2},\frac{\alpha_3'}{c^2}\right) \cdot(v\times N)=0 \:\text{ where $v=\alpha'\in T_pS$} \Leftrightarrow $$

$$ \frac{1}{h}\text{det}\begin{bmatrix}\frac{v_1}{a^2} & \frac{v_2}{b^2} & \frac{v_3}{c^2}\\ \frac{x}{a^2} & \frac{y}{b^2} & \frac{z}{c^2}\\ v_1 & v_2 & v_3 \end{bmatrix}=0 \:\text{ where $v\in T_p S$ and $(x,y,z)=p$} \Leftrightarrow$$ $$\frac{-x}{a^2}\left(\frac{v_2v_3}{b^2}-\frac{v_2v_3}{c^2}\right) +\frac{y}{b^2}\left(\frac{v_1v_3}{a^2}-\frac{v_1v_3}{c^2}\right)-\frac{z}{c^2}\left(\frac{v_1v_2}{a^2}-\frac{v_1v_2}{b^2}\right)=0\tag{I}$$ $$ v_1 \frac{x}{a^2}+v_2 \frac{y}{b^2}+v_3 \frac{z}{c^2}=0\tag{II}$$ $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 \tag{III}$$

Thus, umbilical points in an ellipsoid must satisfy these three equations. Let us suppose $0<a<b<c$.

Claim: $z=0$ yields no solutions.

Proof: by equation (III), $z=0$ implies that either $x\not=0$ or $y\not=0$. Let us take $x\not=0$. By equation (II) $v_2$ and $v_3$ may assume any value but $v_1$ is determined if $v\in T_pS$. Furthermore, by multiplying equation (I) by $x$ and taking $v_3,v_2\not =0$: $$ \frac{x^2}{a^2}v_2v_3\left(\frac{1}{b^2}-\frac{1}{c^2}\right)=\frac{yx}{b^2}v_1v_3\left(\frac{1}{a^2}-\frac{1}{c^2}\right)=\frac{-v_1}{v_2}\frac{x^2}{a^2}v_1v_3\left(\frac{1}{a^2}-\frac{1}{c^2}\right)\Rightarrow $$ $$\frac{x^2}{a^2}v_2^2\left(\frac{1}{b^2}-\frac{1}{c^2}\right)=-v_1^2\frac{x^2}{a^2}\left(\frac{1}{a^2}-\frac{1}{c^2}\right)$$ This equation implies that a number strictly larger than $0$ is also non positive. Contradiction! Similarly, if we had supposed $y\not=0$, by equation (II) $v_1$ and $v_3$ may assume any value but $v_2$ is determined if $v\in T_pS$. Furthermore, by multiplying equation (I) by $y$ and taking $v_1,v_2\not =0$: $$\frac{y^2}{b^2}v_1v_3\left(\frac{1}{a^2}-\frac{1}{c^2}\right)= \frac{xy}{a^2}v_2v_3\left(\frac{1}{b^2}-\frac{1}{c^2}\right)=\frac{-v_2}{v_1}\frac{y^2}{b^2}v_2v_3\left(\frac{1}{b^2}-\frac{1}{c^2}\right)\Rightarrow $$ $$\frac{y^2}{a^2}v_1^2\left(\frac{1}{a^2}-\frac{1}{c^2}\right)=-v_2^2\frac{y^2}{b^2}\left(\frac{1}{b^2}-\frac{1}{c^2}\right)$$ Thus, a strictly greater than $0$ number is also non positive, which is absurd.

We notice somethings. Firstly, $v_1$ and $v_2$ may be taken to be any real number, as long as we fix $v_3$ to be that for which equation (II) is satisfied. Furthermore, With $z$ being non-negative, equation (I) will hold if and only if it also holds when multiplied by $\frac{z}{c^2}$. This yields:

$$\frac{-xz}{a^2c^2}\left(\frac{v_2v_3}{b^2}-\frac{v_2v_3}{c^2}\right) +\frac{yz}{b^2c^2}\left(\frac{v_1v_3}{a^2}-\frac{v_1v_3}{c^2}\right)-\frac{z^2}{c^4}\left(\frac{v_1v_2}{a^2}-\frac{v_1v_2}{b^2}\right)=0 $$

By using equation (II) this further yields that equations (I) and (II) will hold if and only if:

$$v_2^2\frac{yx}{b^2a^2}\left(\frac{1}{b^2}-\frac{1}{c^2}\right)-v_1^2\frac{yx}{b^2a^2}\left(\frac{1}{a^2}-\frac{1}{c^2}\right)+v_1v_2\left(\frac{x^2}{a^4}\left(\frac{1}{b^2}-\frac{1}{c^2}-\right)-\frac{y^2}{b^4}\left(\frac{1}{a^2}-\frac{1}{c^2}-\right)-\frac{z^2}{c^4}\left(\frac{1}{a^2}-\frac{1}{b^2}\right)\right) =0$$

If $v_1=0$ and $v_2=1$, we must have $xy=0$. If $x=0$, we have a positive number $\frac{z^2}{c^4}\left(\frac{1}{a^2}-\frac{1}{b^2}\right)$ being non-positive $-\frac{y^2}{b^4}\left(\frac{1}{a^2}\right)$ which is absurd. Thus $y=0$ and we must have (I) (II) and (III) holding if and only if $y=0$ and:

$$\frac{x^2}{a^2}(c^2-b^2)=\frac{z^2}{c^2}(b^2-a^2)\quad \text{and} \quad \frac{x^2}{a^2}+\frac{z^2}{c^2}=1\Leftrightarrow$$

$$\begin{cases} x^2=a^2\frac{b^2-a^2}{c^2-a^2}\\ y=0\\ z^2=c^2\frac{c^2-b^2}{c^2-a^2} \end{cases}$$

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