18
$\begingroup$

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!

$\endgroup$
5
  • $\begingroup$ 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 $\endgroup$ Jan 26, 2011 at 19:01
  • $\begingroup$ 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... $\endgroup$ Jan 26, 2011 at 19:03
  • 1
    $\begingroup$ 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) $\endgroup$ Jan 26, 2011 at 19:16
  • $\begingroup$ Thank you a lot!! I should ask the question earlier, it had troubled me for one month... $\endgroup$ Jan 27, 2011 at 16:19
  • 1
    $\begingroup$ 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: mathoverflow.net/questions/335032/…. $\endgroup$ Jun 28, 2019 at 21:53

1 Answer 1

14
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.