# Injectivity radius of manifolds with boundary

This question stems from the discussion in:

how to define the injectivity radius of manifolds with boundary?

Suppose $$(M,g)$$ is a compact Riemannian manifold with boundary. In this context, let the injectivity radius of a point $$x$$ be the minimum distance from $$x$$ at which there is a point $$y$$ with more than one length-minimizing geodesic connecting $$x$$ to $$y$$.

Is it true that the injectivity radius as defined this way is bounded below by some nonzero value? If so, is there a standard reference for this fact?

In the discussion linked above, the following theorem is referenced:

Corollary 2. If for a complete Riemannian manifold with boundary, M, the sectional curvatures of the interior and the outward sectional curvatures of the boundary are no greater than $$K$$, then $$N(p,\frac{\pi}{2K})$$ is open in M and the distance function from p is convex on $$N(p,\frac{\pi}{2K})$$.

Where $$N(p,\frac{\pi}{2K})$$ is the set of points connected to $$p$$ by a unique geodesic of length $$\frac{\pi}{2K}$$ or less.

This seems like it is close to the result I am looking for. Earlier in the paper, it is also stated that there are no conjugate points in $$N(p,\frac{\pi}{K})$$.

Is there a simple step from this result that proves that the injectivity radius is nonzero?

Yes, they show that any compact Riemannian manifold with boundary is locally $$\mathrm{CAT}(\kappa)$$ for some $$\kappa\in\mathbb{R}$$. In particular the injectivity radius is positive.
• This definitely works but is it necessary to go the generality of $CAT$-spaces? This should be proveable purely in Riemannian geometry language. – quarague Aug 21 '19 at 6:37
• You definitely need curvature, I just think you don't need $CAT$-spaces. A $CAT$-space is by definition a metric space that satisfies a triangle comparison inequality. Historically this is a generalisation from Riemannian manifolds with bounded curvature. One can show that a Riemannian manifold with curvature bounded from above is a $CAT$-space but this is why $CAT$-spaces are defined the way they are. The question of OP was known before $CAT$-spaces where invented/defined. So one should be able to prove it directly with Riemannian geometry machinery. – quarague Aug 22 '19 at 6:46