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Dear all,

when reading a book of M. Berger, I learned that the injectivity radius Inj(x) on a compact Riemannian manifold depends continuously on the point x.

When the manifold is complete and non-compact, Inj may not be continuous. For example, Inj(x) decreases to zero when x moves to the most curved point on a paraboloid. However, it could be infinity at that point.

My question is, can we prove the continuity of Inj on a non-compact manifold under some conditions?

(I think that the weakest condition is to assume the finiteness of Inj.)

ps. I must admit that I don't know how to prove the continuity of Inj even on a compact manifold. I think that the argument should involve the stability of ODEs (the geodesic equation and Jacobi equation). If one of you have a reference about this, could you please tell me? thanks a lot!

Chih-Wei Chen
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  • Are you sure about your paraboloid example? Take a look at Proposition 2.1.10 on Page 131 of W. Klingenberg's book Riemannian Geometry – Willie Wong Jan 26 '11 at 19:01
  • You write: "Inj(x) decreases to zero when x moves to the most curved point on a paraboloid." this is not true, InjRad does not not go to zero... – Anton Petrunin Jan 26 '11 at 19:03
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    In fact, on any compact region of a smooth Riemannian manifold, you have that the injectivity radius is bounded below by a strictly positive number... (see the same reference that I gave above) – Willie Wong Jan 26 '11 at 19:16
  • Thank you a lot!! I should ask the question earlier, it had troubled me for one month... – Chih-Wei Chen Jan 27 '11 at 16:19
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    It is known that if M is connected and complete, then inj is a continuous function: see for example [Lee, Introduction to Riemannian Manifolds, 2018, Prop. 10.37]. See also this related question: https://mathoverflow.net/questions/335032/is-the-injectivity-radius-semi-continuous-on-a-non-complete-riemannian-manifol. – Nicolas Boumal Jun 28 '19 at 21:53

1 Answers1

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The compactness is irrelevant; i.e if it is true for compact manifolds then the same is true for complete ones. (The same proof as in comact case works, but it is easier to do this way.)

If $R<\mathrm{InjRad}_p$ then one can construct a smooth metric on a sphere with an isometric copy of $B_R(p)$ inside.

If there is a sequence of points $x_n\to x$ such that $$\lim\ \mathrm{InjRad}_{x_n}< \mathrm{InjRad}_x,$$ apply above consruction for $R$ slightly smaller than $\mathrm{InjRad}_x$. You get a compact manifold with non-continuous InjRad.

If there is a sequence of points $x_n\to x$ such that $$\lim\ \mathrm{InjRad}_{x_n}> \mathrm{InjRad}_x$$ then apply above construction for $p=x_n$ for large enough $n$ and $R>\mathrm{InjRad}_x$. That leads to a contradiction again.

Anton Petrunin
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