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.


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.

  • $\begingroup$ 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. $\endgroup$ – J. Salieri Mar 23 at 16:18
  • $\begingroup$ It's a closed subset of a complete metric space and therefore complete. $\endgroup$ – Hans Engler Mar 23 at 17:21

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