# Umbilical points of Ellipsoid alternate method

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 $$\left(\frac{dN_1}{dt}\wedge \alpha '\right)\cdot N_1=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$$ 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

## 2 Answers

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 $$0 then the only solutions are given by $$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)}$$

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.

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:

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

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