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

enter image description here

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.

  • $\begingroup$ Following Mariano's comment, I guess it's important to you that the surface be $C^2$, right? $\endgroup$ Jan 19, 2010 at 23:50
  • $\begingroup$ Steve: You're correct. Thanks for pointing that out. I'll edit the post. $\endgroup$ Jan 20, 2010 at 0:05
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    $\begingroup$ FYI: jstor.org/stable/2162371 $\endgroup$ Jan 20, 2010 at 0:44
  • $\begingroup$ Steve, that looks like an extremely useful paper. Thanks for the link! $\endgroup$ Jan 20, 2010 at 0:56
  • $\begingroup$ One thing you can do is pick a curvature function of your liking and integrate it to a surface; then you'll like the curvature but probably not the surface :) For a surface of revolution, the integrability conditions might not be too hard to satisfy---but I haven't checked... $\endgroup$ Jan 20, 2010 at 1:12

3 Answers 3


Actually, I think you'd be better off just starting with an abstract surface of revolution and doing the calculation you want. An abstract surface of revolution in polar coordinates looks like $$ g = \mathrm{d}r^2 + \bigl(f(r)\,\mathrm{d}\theta\bigr)^2 $$ where $f$ is an odd function of $r$ that satisfies $f'(0)=1$. Then the Gauss curvature is $$ K = -\frac{f''(r)}{f(r)}, $$ so $J_1 = f(r)$ is the odd solution of the Jacobi equation $J''+K J = 0$. (Note that $r$ is already arclength in the radial direction.) The even solution satisfying $J_0(0)=1$ is then given by $$ J_0(r) = -f(r)\int_0^r \frac{\mathrm{d}\rho}{f(\rho)^2}. $$ Now you just have to choose $f$ so that $K$ starts out positive and then goes negative while the integral for $J_0$ can be performed explicitly.

It turns out that, taking $$ f(r) = \frac{r}{1{+}r^2} $$ gives $$ K = \frac{2(3{-}r^2)}{(1{+}r^2)^2},\qquad\text{and}\qquad J_0(r) = \frac{3{-}6r^2{-}r^4}{3(1{+}r^2)},\qquad J_1(r) = \frac{r}{1{+}r^2},\quad $$ which has all of your desired properties.

Now, the question is whether you want to see this isometrically embedded into $\mathbb{R}^3$ as a surface of revolution. This you can do in the form $$ (x,y,z) = \left(\frac{r\,\cos\theta}{1{+}r^2},\ \frac{r\,\sin\theta}{1{+}r^2},\ b(r^2)\right) $$ where $b$ is the function of $t\ge0$ given by the elliptic integral $$ b(t) = \frac12\int_0^{t}\frac{\sqrt{6{+}5\rho{+}4\rho^2{+}\rho^3}}{1+\rho}\,\mathrm{d}\rho. $$ Thus, this surface has the shape of a 'bubble' with a long tail.

If you want a surface that is asymptotic to a plane with a bump of positive curvature in the middle, take, for example, $$ f(r) = \frac{r(1{+}r^2)^2}{1{+}6r^2{+}r^4}, $$ which gives $J_1(r) = f(r)$ while $$ K = \frac{8(3{-}6r^2{-}12r^4{-}18r^6{+}r^8)}{(1{+}r^2)^2(1{+}6r^2{+}r^4)^2}\quad\text{and}\quad J_0(r) = \frac{3{-}23r^4{-}12r^6-15r\arctan(r)(1{+}r^2)^3}{3(1{+}r^2)(1{+}6r^2{+}r^4)}, $$ and has all of your desired properties.


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.


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

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    $\begingroup$ But Tom wants a surface on which the curvature has a nice form... your surface has the correct signs, but depending on how artfully (or not) you do the capping, the curvature may be quite ugly! $\endgroup$ Jan 19, 2010 at 22:49
  • $\begingroup$ You're right, I was confused about hyperbolic space vs. a hyperboloid. $\endgroup$ Jan 19, 2010 at 23:09
  • $\begingroup$ Now, capping the pseudosphere with a sphere cannot be made $C^2$-smoothly, for if you could the curvature function would not be continuous. $\endgroup$ Jan 19, 2010 at 23:25
  • $\begingroup$ Mariano's right: while it's easy to write down many such bump surfaces, I'm struggling with finding one that has a really simple expression for the curvature. $\endgroup$ Jan 19, 2010 at 23:33
  • $\begingroup$ Yeah, there's a singularity at the gluing circle. $\endgroup$ Jan 19, 2010 at 23:39


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 $f(r) = 1/(1+r^2)$.

  • $\begingroup$ Deane: Thanks for the response. Let me clarify why I care about this this surface in particular. I am interested in length-minimizing geodesics in random Riemannian manifolds. My strategy is to show that the radial geodesic has conjugate points, and perturb the space so that the "radial" geodesic in the perturbed space still has a conjugate point. Thus I really am trying to study Jacobi fields on a space like this. $\endgroup$ Jan 19, 2010 at 23:31
  • $\begingroup$ That's fine. I'm just pointing out that you already have simple explicit formulas for both the Jacobi field and the curvature for the specific examples you give above. So there's no need to look any further. $\endgroup$
    – Deane Yang
    Jan 19, 2010 at 23:45
  • $\begingroup$ Also, that if it's the Jacobi field you really care about, there's no need to compute curvature first. $\endgroup$
    – Deane Yang
    Jan 19, 2010 at 23:53
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    $\begingroup$ Could you explain further? I don't quite understand. What's the simple explicit formula for Jacobi fields along a radial geodesic in a surface of revolution? Also, this might not be clear from the above: I am starting the geodesic at an arbitrary point, with initial velocity pointing directly toward the center. $\endgroup$ Jan 20, 2010 at 0:27

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