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Let $\mathcal{M}$ be a Riemannian manifold, and let $\operatorname{inj} \mathrel\colon \mathcal{M} \to (0, \infty]$ be its injectivity radius function.

It is known that if $\mathcal M$ is connected and complete, then $\operatorname{inj}$ is a continuous function: see for example [Lee, Introduction to Riemannian Manifolds, 2018, Prop. 10.37].

What is known in the case where $\mathcal M$ is not complete? Is $\operatorname{inj}$ also continuous? If not, is there a known counter-example? Would $\operatorname{inj}$ still be semi-continuous?

This question is similar to the question "The continuity of Injectivity radius", but the discussion there focuses on compact or complete manifolds.

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  • $\begingroup$ In Chavel's 2006 book, "Riemannian geometry: a modern introduction", Theorem III.2.3 (p118) states the injectivity radius is continuous, without any conditions on the Riemannian manifold $\cal M$. However, there is no proof, and I did not find any comment about where to find a proof, nor did I find any papers referencing that theorem. On the bright side, the two answers below (by Jack Lee and Stephen M), combined, provide such a proof. Thank you both! $\endgroup$ – Nicolas Boumal Sep 12 '19 at 18:55
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I believe the following proof shows that it is lower semi-continuous, but I do not know a whether it's continuous (or a counterexample), and like OP, would be interested.

The idea is to take a ball just slightly less than the injectivity radius at a point, push everything to the unit ball, then use a bump function to change the metric near the boundary and embed it smoothly in the sphere, at which point the theorem for complete manifolds will give the result.

Let $x \in M$ and let $i(x)$ be its injectivity radius. Let $0 < \varepsilon < i(x)$. We want to show that for some $\delta > 0$, every point $y$ in $B_{\delta}(x)$ has $i(y) > i(x) - \varepsilon$, i.e., that the exponential map $\exp_y: T_y(M) \to M$ is a diffeomorphism on the ball of radius $i(x) - \varepsilon$.

The exponential map $\exp_x$ is a diffeomorphism from the closed ball of radius $r = i(x) - \varepsilon/4$ to the closed ball in $\mathbb{R}^n$ of radius $r$, say $B_r$. Let $\hat{x}$ be the origin of the latter, corresponding to $x$. Let $\hat{g} = (\exp_x)_*g$ be the pushforward of the metric to this ball. Let $\chi$ be a bump function on $B_r$ that is $1$ on $B_{r - \varepsilon/2}$ and supported on $B_{r - 3\varepsilon/4}$. Let $h = \chi\hat{g} + (1 - \chi)g_0$, where $g_0$ is the round metric on the closed upper hemisphere $S^n_+$. Then we can clearly double $B_r$ to $S^n$, extending $h$ to be the usual round metric on the lower hemisphere. So $h$ is a smooth metric on $S^n$ that coincides with $\hat{g}$ on $B_{r - \varepsilon/2}$.

Thus, by the usual result for complete manifolds (e.g. in Lee), the injectivity radius of $h$ is a continuous function -- let's call it $\hat{i}$. So there's some $\delta > 0$ so that

  • $\delta < \varepsilon/4$, and
  • for $y$ in the $h$-ball of radius $\delta$ about $\hat{x}$, $\hat{i}(y) > \hat{i}(x) - \varepsilon/4 > i(x) - \varepsilon/2$.

Observe now that for such $y$, the ball $A$ of radius $r - 3\varepsilon/4 = i(x) - \varepsilon$ about $y$ is contained in $B_{r - \varepsilon/2}$, and the exponential map at $y$ is a diffeomorphism onto $A$. Since $(B_{r - \varepsilon/2},h|_{B_{r - \varepsilon/2}})$ is isometric to $B_{r - \varepsilon/2}(x)$ in $M$, this yields the claim.

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  • $\begingroup$ Thanks for this detailed proof Stephen, this is a nice argument, giving at least semi continuity in all cases. It would be great to know also if inj should actually be continuous in all cases, or if there is a counter-example. if the latter, seeing the counter-example would surely help intuition significantly too. I'll wait a bit to accept your answer in the hope someone can provide references / pointers for this. $\endgroup$ – Nicolas Boumal Jul 1 '19 at 11:22
  • $\begingroup$ Thank you. Please do wait, as I'm also interested in the other question. $\endgroup$ – Stephen M Jul 1 '19 at 22:36
  • $\begingroup$ For (connected and) complete manifolds, Lee's Prop. 10.36 states the injectivity radius at $x$ is the distance to the cut locus of $x$ (infinite if empty). The distance function to a fixed set is always continuous. If Prop. 10.36 holds also for incomplete manifolds, then we may be able to learn something about the (lack of) continuity of the injectivity radius by looking at manifolds whose cut loci have "jumps" (as a set-valued map). $\endgroup$ – Nicolas Boumal Jul 2 '19 at 11:45
  • $\begingroup$ I wanted to accept both Stephen's and Jack's answers, as they form a whole. Since MO allows for only one accepted answer, I chose Stephen's, with the arbitrary rationale that it was posted first. $\endgroup$ – Nicolas Boumal Dec 3 '19 at 21:46
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I think that Stephen's idea can be adapted to show that the injectivity radius is also upper semicontinuous in the incomplete case. Here's my argument -- let me know if you see anything wrong.

Let $(M,g)$ be a connected Riemannian $n$-manifold, and let $i\colon M\to(0,\infty]$ be the injectivity radius function. Suppose $i$ is not upper semicontinuous at $x\in M$, and let $r = i(x)$. Then there exists a sequence of points $x_k\to x$ and a number $R>r$ such that $i(x_k)\ge R$ for all $k$. Let $\varepsilon = (R-r)/3$, $R'=r+\varepsilon$, and $R'' =r+2\varepsilon$, so that $r<R'<R''<R$. Choose $k_0$ large enough that $d(x_k,x)$ and $d(x_k,x_{k_0})$ are both less than $\varepsilon$ for all $k\ge k_0$. Since the geodesic balls $B_r(x)$, $B_{R'}(x_k)$, and $B_{R''}(x_k)$ are also metric balls, the triangle inequality implies that for each $k\ge k_0$, $$ B_r(x) \subseteq B_{R'}(x_k) \subseteq B_{R''}(x_{k_0}). $$

Using normal coordinates, we can identify $B_R(x_{k_0})$ with the Euclidean ball $B_R(0)\subset \mathbb R^n$, and then by using a bump function we can create a complete metric $\hat g$ on $\mathbb R^n$ that agrees with $g$ on $\overline B_{R''}(x_{k_0})$. The set inclusions above imply that $g$ and $\hat g$ agree on $B_{R'}(x_k)$ for each $k$, and thus $\hat i(x_k)\ge R'$ for all $k\ge k_0$. By continuity of $\hat i$, we have $\hat i(x) = \lim_k \hat i(x_k) \ge R'$.

This means that the exponential map of $\hat g$ is injective on the ball of radius $R'$ in $T_xM$. If we can show that the exponential map of $\hat g$ is equal to that of $g$ on that ball, then we have a contradiction to the assumption $i(x)=r<R'$.

Let $\gamma\colon [0,R')\to \mathbb R^n$ be a unit-speed $\hat g$-geodesic starting at $x$, and let $$t_0 = \sup\{t\in [0,R'): \gamma(t) \in B_{R''}(x_{k_0})\}.$$ Then for all $0\le t < t_0$, $\gamma(t)$ is in the set where $g=\hat g$, and thus $\gamma|_{[0,t_0)}$ is also a $g$-geodesic. We need to show that $t_0=R'$ for every such $\gamma$.

Suppose $t_0<R'$ for some such $\gamma$. Then $\gamma(t_0)\in \partial B_{R''}(x_{k_0})$, which means that $d_g(x_{k_0},\gamma(t_0)) = R''$. However, \begin{align*} d_g(x_{k_0},\gamma(t_0)) &\le d_g(x_{k_0},x) + d_g(x,\gamma(t_0))\\ &< \varepsilon + L_g(\gamma|_{[0,t_0]})\\ &= \varepsilon + t_0\\ &< \varepsilon + R' \\ &= R'', \end{align*} which is a contradiction.

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  • $\begingroup$ Thanks Jack! I had a close look at your proof, and I believe I agree with every step. The more technical step (where you build a complete metric $\hat g$ on $\mathbb{R}^n$, smoothly transitioning from $g$ to the Euclidean metric) seems to be essentially the same as in Stephen's proof, where he does so by building a metric on a sphere, necessarily complete by compactness. Combining your two proofs should give continuity of $i(x)$ for any (connected) manifold, which is great. It looks like there should be a good way to write both of your arguments with the same language. $\endgroup$ – Nicolas Boumal Jul 12 '19 at 20:55
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The proofs provided here by Stephen M and Jack Lee now appear in my book about Riemannian optimization: An introduction to optimization on smooth manifolds. See Section 10.8.

The answer is: yes, the injectivity radius is a continuous function even if the Riemannian manifold $\mathcal{M}$ is not complete.

Thank you both!

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  • $\begingroup$ I think that this should probably be part of the question, not a separate answer. $\endgroup$ – LSpice May 22 at 23:14
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    $\begingroup$ Is that the norm in this situation on mathoverflow? I'm happy to move it there if so. Should it then appear separated, visibly as an edit? $\endgroup$ – Nicolas Boumal 2 days ago
  • $\begingroup$ Well, I flagged it accordingly and the flag was disputed. I don't really know what that means, but obviously it's not universal! It makes more sense to me as part of the question (separated as an edit, as you say), but, having heard my suggestion and the fact that someone disagrees, I think it's appropriate for you to do what makes sense to you. $\endgroup$ – LSpice 2 days ago

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