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Prove that the ellipsoid $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$ always has at least three geodesics.

I think these three geodesics should be cross sections of ${(x,y,0)}$, ${(x,0,z)}$ and ${(0,y,z)}$ with our ellipsoid. And I want prove that the geodesic curvature is zero on these curves. The formula I want to use is $s'k_{g}=(T\times T')N$. So all I have to prove is that $T'//N$. And $N=({\frac{2x}{a^{2}}},\frac{2y}{b^{2}},\frac{2z}{c^{2}})$. But $T'$ is not very easy to get.

Jack
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2 Answers2

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Let's look at the cross-section with $z = 0$. We can parametrise the curve as $$ (x,y,z) = (a \cos t, b \sin t, 0),$$ so at a given value of $t$, the normal to the surface is $N(t) = (\frac 2 a \cos t, \frac 2 b \sin t , 0)$.

It is possible to convince yourself that this is a geodesic without doing any calculation. Notice that the curve lies in the plane $z = 0$. So, at every value of $t$, the unit-normalised velocity vector $T(t)$ and its derivative $T'(t)$ both lie in the plane $z = 0$. But at any value of $t$, the surface normal $N(t)$ also lies in the plane $z = 0$.

So all three vectors lie in the same plane! Hence the triple product $(T(t)\times T'(t)). N(t)$ is zero at all values of $t$, and the geodesic curvature vanishes.

Kenny Wong
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For posterity: Generally, if $(M, g)$ is a Riemannian manifold and $\phi:M \to M$ is an isometry, then every component of the fixed-point set is a totally-geodesic submanifold.

Here, reflection across each coordinate plane is an isometry, and the section of the ellipsoid by each coordinate plane is fixed, hence is a totally geodesic submanifold.

Andrew D. Hwang
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