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Let $M$ be a Riemannian manifold. Let $\mathrm{inj}_M(p)$ be the injectiviy radius at a point $p\in M$, which is defined as the biggest $R>0$ such that $\mathrm{exp} \colon B_R(p) \rightarrow M$ is a diffeomorphism. Then, the injectivity radius of $M$ is defined as $$ \mathrm{inj}_M :=\mathrm{inf}\{\mathrm{inj}_M(p) : p\in M\}.$$ It is clear that if $M$ is not complete, $\mathrm{inj}_M=0$ since you can pick a sequence $\{q_n\}_{n\in \mathbb{N}}$ which converges to the point $q$ where the completeness fails and $\lim_{n\to \infty}\mathrm{inj}_M(q_n)=0$. Thus, $\mathrm{inj}_M=0$.

Is there any easy example of a complete Riemannian manifold $M$ with $\mathrm{inj}_M=0$?

This $M$ should be non compact. Because if $M$ is compact, then $\mathrm{inj}_M>0$ since the infimum turns minimum and $\mathrm{inj}_M(p)>0$ for each point $p\in M$ because the exponential map is a local diffeomorphism.

J. Salieri
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1 Answers1

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Consider a surface of revolution such as $$ M = \{x \in \mathbb{R}^3 : x_1^2 + x_2^2 = \frac{1}{1+x_3^2} \} $$ with Riemannian metric inherited from the ambient space. For points with large $|x_3|$, the injectivity radius becomes arbitrarily small.

Hans Engler
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  • Ok, I see the idea but how do you prove that surface is complete? A pseudosphere maybe will also be valid? Because it is like a cylinder but the radius of it shrinks as the component of it axis gets bigger. – J. Salieri Mar 23 '19 at 16:18
  • It's a closed subset of a complete metric space and therefore complete. – Hans Engler Mar 23 '19 at 17:21
  • @HansEngler I have a small question. By claiming that a closed subset of $\mathbb{R}^3$ (a complete metric space) is complete, we are assuming that the metric on that closed subset is inherited from the metric on $\mathbb{R}^3$. But on the other hand, on a Riemannian manifold, the metric space's metric is usually given by the distance function. Then how should we conclude that $M$ is still complete? – Hopf eccentric Mar 06 '20 at 07:57
  • @Hopfeccentric there is the Hopf-Rinow theorem, which states these two things are equivalent. https://en.wikipedia.org/wiki/Hopf–Rinow_theorem – Espace' etale Jun 20 '22 at 13:43
  • @Hopfeccentric, a better statement would be : "closed submanifold of a complete Riemannian manifold is complete". – Rabi Kumar Chakraborty Jan 22 '23 at 00:47