Let $M$ be a complete Riemannian manifold, and let $C$ be a subset of $M$. We will say $C$ is convex if for any points $p,q \in C$, there exists a unique normal minimal geodesic $\gamma$ joining $p$ and $q$, such that $\gamma \subset C$.

We will define:

  • the convexity radius of $M$ by "the infimum of positive numbers $r$ such that the geodesic open ball $B(p,r)$ is convex for every $p \in M$",

  • the injectivity radius at $p$ by "the infimum of the cut values of the various geodesics emanting from $p$" (i.e. the radius of the largest geodesic ball on which $\exp$ is a diffeomorphism)

  • the injectivity radius of $M$ as $\inf_{p \in M} \{\textrm{injectivity radius of } $p$\}$

(these definitions are taken from Berger's A Panoramic View of Riemannian Geometry, but appear to be standard across the field).

Now, we have the following theorem, bounding the convexity radius from below in terms of the injectivity radius, and sectional curvature $K$:

THM: If $M$ is compact, $\textrm{Convexity Radius}(M) \geq \min \{ \frac12 \pi/\sqrt{ \sup K}, \frac 12 \textrm{Injectivity Radius}(M)\}$

Where $\frac12 \sup \pi/\sqrt K$ is replaced by $\infty$ if $\sup K \leq 0$.

(and if $M$ is not compact, a local version of the statement still holds)

and Berger additionally notes: "Apparently there is no example in the literature with the convexity radius smaller than half of the injectivity radius. A natural conjecture is that such a bound should not be too difficult to prove."

My question is essentially about a weaker form of this conjecture; in particular, if we can assume the injectivity radius is maximal, can we conclude something similar of the convexity radius (and without considering the curvature)? Or more precisely:

Q: If the manifold is simple (meaning any two points are connected by a unique minimizing geodesic; equivalently the exponential is a global diffeomorphism at any point), is the geodesic ball $B(r,p)$ convex for any $r> 0$?

  • 1
    $\begingroup$ Proposition III.1.5 here says so. I'm surprised I couldn't find a reference in non-thesis form. $\endgroup$
    – user147263
    Mar 30 '15 at 22:02
  • $\begingroup$ @Woodface That is exactly what I want, thank you so much. I was also surprised I couldn't find this stated or even discussed (using the wrong search queries I suppose). Discussion on the result for non-positive curvature / not-necessarily-simple is more abundant (and indeed the result follows obviously from the theorem above in the case there are no conjugate points). If you want to write this as an answer, I will accept it; however brief it may be, this perfectly answers my question. $\endgroup$
    – BaronVT
    Mar 30 '15 at 22:32
  • 2
    $\begingroup$ @Woodface I may be missing something, but the proof in the thesis you linked seems to only work for (simply connected) manifolds without focal points, which is stronger than the condition of simple given in the question (which looks like it's equivalent to simply connected + no conjugate points). $\endgroup$
    – mollyerin
    Mar 31 '15 at 4:23

I thought this question was interesting; here's what some internet research has turned up:

1. When $M$ is simply connected, the convexity radius of $M$ is $\infty$ iff $M$ has no focal points.

On the one hand, if $M$ has no focal points then the argument linked to by Woodface in the comments above shows that all geodesic balls in $M$ are convex.

On the other hand, suppose $M$ has focal points. In particular there exists a geodesic $\gamma$ intersecting a geodesic $\sigma$ orthogonally at a point $p$ of $M$, such that $\sigma$ has a focal point at some $q$ along $\gamma$. The Morse Index Theorem tells us that for points $q'$ along $\gamma$ past $q$, $\gamma$ does not minimize the distance from $q'$ to $\sigma$.

That is, take some such point $q'$; let $d(p, q') = r$. The condition that $\sigma$ have a focal point between $p$ and $q'$ tells us that there are points $p_1, p_2$ close to $p$ on $\sigma$ such that $d(p_1, q') < r$ and $d(p_2, q') < r$. If $s < r$ is the larger of these two distances, we see that the geodesic ball of radius $s$ at $q'$ can't be convex.

More generally this is an argument that if $p \in M$ where $M$ is a complete Riemannian manifold, then the convexity radius $r_{\text{conv}}(p)$ at $p$ is bounded above by the focal radius $r_f(p)$ at $p$, where $r_f(p)$ is the minimum distance of a point $q$ that is a focal point of some geodesic through $p$.

2. There exist examples of manifolds with focal points but without conjugate points.

The universal covers of such manifolds will be "simple" by your definition (simply connected and without conjugate points), but, since they will also have focal points, they will have finite convexity radii. The main class of examples seem to be due to Gulliver in this paper.

Thus, the answer to your question is no: "simple" manifolds need not have infinite convexity radius.

3. This very recent paper (by Dibble) claims that even for compact manifolds, the convexity radius can be made arbitrarily small relative to the injectivity radius.

Thus Berger's conjecture that "the bound $r_{\text{conv}}(M) \geq \tfrac{1}{2} \text{inj}(M)$ should not be too difficult to prove" is quite false (since apparently the bound itself is false).

The paper (if I understand it correctly) also claims the positive result that the convexity radius is the minimum of (a) the focal radius and (b) one-quarter the length of the shortest non-trivial closed geodesic in $M$. This is a very nice result analagous to the classical result that the injectivity radius is the minimum of the conjugate radius and one-half the length of the shortest non-trivial closed geodesic. (This is due to Klingenberg. It's given, for instance, as Proposition 2.2 in Chapter 13 of do Carmo's book, and it's also cited in the above linked paper.)

  • 1
    $\begingroup$ Ahh, I see now in reading that thesis closer the gap between conjugate points and focal points. Thank you for clarifying this and providing further details. $\endgroup$
    – BaronVT
    Mar 31 '15 at 15:08
  • 1
    $\begingroup$ a follow-up question: so is my definition of simplicity ambiguous then? I.e. is the requirement that the exponential be a global diffeomorphism stronger than no conjugate points? (and in particular, does "exp is global diff" $\Leftrightarrow$ "no focal points"?) $\endgroup$
    – BaronVT
    Mar 31 '15 at 15:55
  • 2
    $\begingroup$ @BaronVT The exponential map is a global diffeomorphism for every $p \in M$ iff $M$ is simply connected and has no conjugate points. $\endgroup$
    – mollyerin
    Mar 31 '15 at 17:30
  • 1
    $\begingroup$ gotcha. thank you again. $\endgroup$
    – BaronVT
    Mar 31 '15 at 17:47

There's one thing that I should add to the nice answer of @mollyerin: The proof of the pointwise bound $r(p) \leq r_f(p)$ (Lemma 2.4) in the arXiv version of my paper contains an error; it uses the continuity of the convexity radius $r$ at $p$, which cannot be taken for granted. However, the given argument does prove the global bound $r(M) \leq r_f(M)$, which suffices to obtain all of the paper's main results.

I'll be posting an updated version on arXiv, which also corrects a couple other minor misstatements, once I receive galley proofs from the journal. But anybody who'd like a copy can email me.

  • $\begingroup$ Welcome to MathSE. In general, a statement like this should be left as a comment on the other answer and not a stand-alone answer (because it's not really an answer to the question on its own). But I don't think you have enough reputation to leave a comment, so your post should probably be left here. Also, if you are James Dibble (so who you are is relevant to the post), you might want to use your full name, have a profile picture, link to your academic webpage, etc. to provide some credibility (because people lie about who they are on the internet). Thank you. $\endgroup$ Oct 14 '15 at 14:55

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.