I recently found out from this answer that every Lie group equipped with a left-invariant Riemannian metric is a (geodesically) complete Riemannian manifold.
I wonder whether the same holds also for a semi-Riemannian group, i.e., a Lie group equipped with a bi-invariant semi-Riemannian metric. Any semisimple and any compact Lie group admits such a metric.
Question. Is every semi-Riemannian group geodesically complete?
If the answer is no, as I suspect, does it then change anything to assume that the group has dimension three?
Best Answer
To correct my earlier hasty comment:
Surprisingly (at least, to me), the answer is negative. Joe Wolf in Corollary 11.1 in
Wolf, J. A., Isotropic manifolds of indefinite metric, Comment. Math. Helv. 39, 21-64 (1964). ZBL0125.39203.
proves the following:
Let $G$ denote the unique (up to an isomorphism) noncommutative 2-dimensional simply-connected Lie group. (It is the Lie group of orientation-preserving affine transformations of ${\mathbb R}$.) Then $G$ admits an incomplete left-invariant flat metric of signature $(1,1)$.