For manifolds without boundary one defines the injectivity radius as the maximal radius where the exponential map is a diffeomorphism. One can then show that the injectivity radius is the maximum number that such any two points with distance less than that number have a unique geodesic length minimizer between them.

If one looks at a (smooth complete compact) manifold with boundary, using the exponential map definition, points close to the boundary have small injectivity radius and the injectivity radius of the entire manifold would be zero. Using the interpretation about unique length minimizer instead would give a nontrivial notion of injectivity radius for manifolds with boundary.

My question is mostly a reference request, is it written down somewhere how to define the injectivity radius of a manifold with boundary?

My question came from papers that have general statements of the form: Let $(M,g)$ be a manifold with boundary, such that some curvature bounds hold and the injectivity radius is bounded from below. Then bla bla bla. I assume these are not theorems about the empty set.

Edit: Maybe this is more complicated than I thought. It seems we have three alternative proposals: 1. my original idea: there exists a unique length minimizer 2. Thomas Rot's idea: use the injectivity radius of the double of the manifold 3. Schick's paper, referenced by user44172: using a collar near the boundary

I think 1. and 2. are equivalent. Whether these two are equivalent to 3. looks nontrivial. Additionally, all these definitions are usually only used in combination with some curvature bounds, so the definitions might be distinct in full generality but equivalent under some suitable curvature bounds.

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    $\begingroup$ Maybe one should define it as the injectivity radius of the double of the manifold? $\endgroup$
    – Thomas Rot
    Apr 14, 2016 at 7:22
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    $\begingroup$ The paper "Geometric curvature bounds in Riemannian manifolds with boundary" by Alexander, Berg and Bishop is related. In particular, it follows a lower bound on your unique-length-minimizer-injectivity radius from upper curvature bound and second fundamental form of the boundary. ams.org/journals/tran/1993-339-02/S0002-9947-1993-1113693-1/… $\endgroup$ Apr 14, 2016 at 10:17
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    $\begingroup$ @ThomasRot This would not work well for manifolds where the boundary is not totally geodesic. For example, if you take the double of the flat unit disk, then two points on the boundary would have two minimizing geodesics between them, one on each part. Neither would the exponential map at a boundary point be injective. So with that definition, you get $0$ again. If you just ask for a unique minimizer, there is no trouble. Schick's definition seems to work as well, but gives $1$ (coming from injectivity of the collar) unless I misinterpreted the statement. $\endgroup$ Apr 15, 2016 at 9:42

1 Answer 1


http://arxiv.org/abs/math/0001108 This paper by Schick is nicely written and contains coordinate-wise and coordinate-free definitions of bounded geometry for manifolds with boundary.

In particular, it says that in this case the condition of having injectivity radius bounded from below translates to the following conditions:

  1. Normal collar: there is $r_c>0$ such that the geodesic collar $\partial M \times [0,r_c] \to M$ sending $(x,t)$ to $K(\exp_x(tv_x))$ (where $v_x$ is the unit inward normal vector) is a diffeomorphism into its image.
  2. Positive injectivity radius in $\partial M$.
  3. Positive injectivity radius in $M\setminus K(\partial M \times [0,r_c])$.

Most theorems that assume that $M$ is a borderless manifold with positive injectivity radius should have an analogue for the case of manifolds with border satisfying the above conditions.


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