6
$\begingroup$

I have two question:

1) Are there any examples of complete manifold with strictly positive and bounded section curvature which has zero injectivity radius?

2) Is there a sequence of non-compact complete manifolds with strictly positive and bounded section curvature with injectivity radius approach to zero?

I think one may construct these examples from Beger's spheres, but I cannot do it rigorously.

$\endgroup$
  • 2
    $\begingroup$ Here is an idea: start from a paraboloid in $\mathbb R^3$ and introduce a sequence of "pimples" that run off to infinity and each approximate a tiny cone. The easiest way to visualize this is by attaching little cones to a parabola in $\mathbb R^2$ so that the result is a convex curve, then approximate by strictly convex curve, and consider the corresponding surface of revolution. Near each attached "cone" the injectivity radius will be small. The trick will be to pick cone regions so that the curvature stays bounded above and injectivity radius tends to zero. $\endgroup$ – Igor Belegradek Apr 29 '19 at 18:41
  • 1
    $\begingroup$ Thanks for your answer, I tried to construct a exampale as you have suggested ,But when attaching some cones to parabola, I should take this cone long and thin to make sure injectivity radius tends to zero,. this seems not true Under the curvature bounded assumption. are there any point I have omitted? $\endgroup$ – Yuchen Bi May 4 '19 at 8:14
4
$\begingroup$

If by strictly positive, you mean that there is a $\epsilon>0$ so that the $Sect >\epsilon$, then there are no such examples. The reason for this is that this curvature assumption implies that a manifold is compact. Since the injectivity radius of a point is a positive and continuous function on a manifold, it has a non-zero minimum.

Igor Belegradek pointed out (thanks for this) that strictly positive also has a another meaning, which is that the sectional curvature is everywhere positive, without a positive lower bound. If we allow this as the definition of "strictly positive", there is a simple counterexample, which is just to take a disjoint union of countably many collapsing Berger spheres. Some of the sectional curvature of the Berger spheres converge to zero as the metric collapses (See here), but the curvature is always positive throughout the collapse. This also addresses your second question in terms of this definition of positive curvature.

However, it seems worth asking whether we can create a connected manifold with positive curvature and zero injectivity radius. I was unable to find a simple argument to rule out such examples, but I strongly suspect that none exist. As I mentioned before, the only spaces we need to worry about are non-compact. In this case, one can either use the Soul Theorem or results of Gromoll and Meyer to see that the manifold must be diffeomorphic to $\mathbb{R}^n$. Such a manifold also has no closed geodesics and there is a lower bound on the distance between conjugate points, which is strong evidence that there should be control on the injectivity radius. However, I'm not sure how to derive such an estimate. Maybe someone more familiar with this can help.

$\endgroup$
  • 5
    $\begingroup$ It seems likely that by "strictly positive" the OP means "$>0$". If so, then your first paragraph is incorrect, e.g. the paraboloid in $\mathbb R^3$ has positive curvture everywhere. $\endgroup$ – Igor Belegradek Apr 29 '19 at 15:36
  • 2
    $\begingroup$ How do you show that "Such a manifold also has no closed geodesics and there is a lower bound on the distance between conjugate points"? $\endgroup$ – Igor Belegradek Apr 29 '19 at 18:16
  • 2
    $\begingroup$ The bound on the conjugate points is from looking at the Jacobi equations using the sectional curvature upper bound. The result on the closed geodesics is from the paper "Closed geodesics on non-compact Riemannian manifolds" by Gudlaugur Thorbergsson. $\endgroup$ – Gabe K Apr 29 '19 at 18:33
  • 1
    $\begingroup$ I guess more accurately the closed geodesic result is due to Gromoll and Meyer but Thorbergsson states it in the introduction. $\endgroup$ – Gabe K Apr 29 '19 at 18:35
  • 2
    $\begingroup$ I see. Gromoll-Meyer prove in theorem 4 of "On complete open manifolds of positive curvature" that at any point the exponential map is proper, and in particular, there is no closed geodesics. This does not help to answer the OP question though. $\endgroup$ – Igor Belegradek Apr 29 '19 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.