Let $M$ be a Riemannian manifold equipped with a free and proper isometric action by a Lie group $G$. Suppose that $M$ has positive injectivity radius.

Question: Is it possible for the quotient manifold $M/G$, equipped with the quotient Riemannian metric, to have injectivity radius $0$?

Thoughts: For this to happen, the action must be such that the quotient $M/G$ is non-compact. I think one way to construct an example would involve a $G$-action where the $G$-orbits in $M$ "shrink in size" as one approaches a particular point of $M$, similar to the rotational action of $S^1$ on $\mathbb{R}^2$ (except this is not free). I suspect there is a simple example, but can't quite see it.


1 Answer 1


The simplest example is when $(M, g)$ is the upper half plane $\mathbb H$ with the hyperbolic metric

$$g = \frac{1}{y^2} (dx^2+ dy^2),$$

while $G = \mathbb Z$ is the action of the translation

$$ (x, y) \mapsto (x+n, y).$$

The quotient has injectivity radius equals zero.


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