Jacobi fields on a "bump surface" - MathOverflow

Jacobi fields on a "bump surface"

2010-01-19T21:51:00Z

Consider a "bump surface" which looks like the following:

Such a surface is rotationally symmetric, $C^2$-smooth, has positive curvature in the middle and negative curvature along the ring (the orange region in the picture). I don't really care what happens past that (it could flatten out, or oscillate, etc.)

Here are two examples, as surfaces of revolution in $\mathbb R^3$ in cylindrical coordinates:

$z(r) = e^{-r^2/2}$ and $z(r) = \tfrac{2}{\pi} \cos(\tfrac{\pi}{2} r)$.

I need to do some Riemannian geometry on a bump surface; in particular, analyze a Jacobi field along a radial geodesic $\gamma$. I don't care what bump surface I use; it only has to feature both positive and negative curvature. For any surface of revolution, it's easy to write down a formula for the scalar curvature $K$ (see p. 142 of McCleary's Geometry from a differentiable viewpoint), and the Jacobi equation takes the form $J'' + KJ|\dot\gamma|^2 = 0$. Thus, if the scalar curvature has a simple form, then the Jacobi equation should be easy to solve. In the case of these two examples, the scalar curvature isn't particularly pretty, hence analyzing the Jacobi equation is difficult (though not intractable).

My question to the MathOverflow community: is there a better bump surface than the two examples I gave above, for which the scalar curvature has a particularly simple form?

Edit: The curvatures for the surfaces given above are

$K(r) = \frac{2 (1 - r)}{(e^{r^2/2} + r^2 e^{-r^2/2})^2}$ and $K(r) = \frac{\pi \sin(\pi r)}{2 r (1 + \sin^2(\pi r/2))^2}$,

respectively. As you can see, they're not the worst expressions possible, but they're also not as simple as I'd like them to be.

Answer by Steve Huntsman for Jacobi fields on a "bump surface"

2010-01-19T22:42:50Z

Take a portion of a pseudosphere and cap it with a portion of a sphere in such a way that the surface is ($C^1$)-smooth.

OLD (BAD) ANSWER:

Take a portion of a hyperboloid and cap it with a portion of a sphere in such a way that the surface is smooth.

Answer by Deane Yang for Jacobi fields on a "bump surface"

2010-01-19T23:15:15Z

ORIGINAL ANSWER DELETED

EDIT: I neglected to account for the need to parameterize by arclength. And I think I also misunderstood and thought that you wanted only the Jacobi field that fixes the center. You want to solve for an Jacobi field, given a point (away from the center) and a vector at that point, right?

So that's definitely not as easy as I thought. Here are my thoughts:

1) I think the already proposed surface given by a spherical cap glued to a pseudosphere is already a good enough question. In my experience you never really need a $C^2$ surface, and something with piecewise continuous curvature is almost always enough. I encourage you to try it.

2) As for the more general approach, I no longer have any easy answer, but here are some thoughts:

Let the surface be given by $(r,\theta) \mapsto X(r,\theta) = (r\cos\theta, r\sin\theta, f(r))$. If $s$ be the arclength parameter along a radial geodesic, then $s'(r) = \sqrt{1 + f'(r)^2}$. One Jacobi field $J_1(r,\theta)$ is given simply by

$J_1(r,\theta) = \partial X/\partial\theta = re_\theta$, where $e_\theta = (-\sin\theta, \cos\theta, 0)$ is a unit vector field that is orthogonal to and parallel along any radial geodesic.

If we view $r$ as a function of $s$, then the Jacobi equation says that $r'' + Kr = 0$, where $K$ is the Gauss curvature. It suffices to solve for one more Jacobi field $J_2 = h(s)e_\theta$ independent of $J_1$. The Jacobi equation for $J_2$ is given by $h'' + Kh = 0$. Since $r$ is already a solution, we can try to solve for $h$ using variation of parameters.

So the goal is to find an even function $f$ with an inflection point such that the function

$s(r) = \int_0^r \sqrt{1 + f'(t)^2} dt$

can be explicitly integrated and inverted. I suggest trying something like $f(r) = 1/(1+r^2)$.