Is the injectivity radius (semi) continuous on a non-complete Riemannian manifold?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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!