35

Consider the ellipsoid $M \subset \mathbb{R}^3$ defined by

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

where $0 < a < b < c$, equipped with the usual Riemannian induced metric from $\mathbb{R}^3$. Let $p \in M$ be a non-umbilic point. I've seen the claim that the set of singularities of the Riemannian exponential map

$$\exp_p\colon T_pM \rightarrow M$$

is an ellipse in this case. Note that I'm talking about conjugate points in the tangent space, not down on the manifold $M$, where this set might be quite ugly. In other words, I'm looking at the set

$$C_p = \{ u \in T_pM : d(\exp_p)(u)~\hbox{is singular} \}.$$

My question is: is there a nice description of $C_p$? Is it an ellipse, at least at most points?

I'm also interested in the case where $b = c$ or $a = b$, as long as it's not a sphere, basically. For the sphere $C_p$ is always a circle, of course.

Didier
  • 20,820
student
  • 4,027
  • Where one finds the proof for the circular case? – san Apr 25 '16 at 03:20
  • @san On a sphere, you can compute the exponential map explicitly. Take a look at, e.g., this question: http://math.stackexchange.com/questions/455683/exponential-map-on-the-the-n-sphere?rq=1 – student Apr 26 '16 at 02:41
  • You're essentially asking for a description of these images, aren't you? I realize these are drawn on the ellipsoid and you're asking for their preimage on the tangent space, but they probably aren't "nice", and certainly aren't an ellipse... Not much seems to be known, really. – Gro-Tsen Apr 29 '16 at 20:24
  • 2
    @Gro-Tsen The preimage on the tangent space turns out to be much nicer than those pictures. For instance, $C_p$ is a smooth submanifold of the tangent space, whereas its image has cusps. This is because $\exp_p$ is not a local diffeomorphism. Did you have an argument in mind to show that $C_p$ is not an ellipse? – student Apr 29 '16 at 21:26
  • I don't have a proof, but if I try to imagine even a bit of these curves being the image of an ellipse by the exponential map, it seems completely impossible. (One stupid thing, though: all of $C_p$ can't be an ellipse, because $C_p$ is unbounded: every geodesic direction will meet singular points arbitrarily far. But I assume you meant that $C_p$ had connected components that were ellipses, or at least the one closest to the origin or something like that.) – Gro-Tsen Apr 29 '16 at 22:20
  • @Gro-Tsen I don't see why it seems impossible. But yes, I mean each connected component. In fact, I'm mostly interested in the first conjugate locus, which means the component closest to the origin. – student Apr 29 '16 at 22:29

0 Answers0