The exponential map associated with the (-) - connection on a Lie group is generally not surjective. This is because, for this connection, the one-parameter subgroups and geodesics coincide. If we drop the requirement that one-parameter subgroups are geodesics, is it possible to put a left invariant connection on a Lie group for which the geodesic exponential map is surjective?
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1 Answer
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Yes. Just take the Levi-Civita connection of any left-invariant Riemannian metric on the Lie group. The metric is complete, so any two points can be joined by a geodesic (Hopf-Rinow). Thus, the geodesic exponential map of that connection starting from the identity is surjective.
answered Jul 28, 2018 at 16:55
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$\begingroup$ Thanks for the answer, Robert. One implication of this is that the $(0)$-connection on a non-compact Lie group will generally not be metric. $\endgroup$ Commented Jul 29, 2018 at 23:15
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3$\begingroup$ True, its holonomy is non-compact, so it is not the Levi-Civita connection of any Riemannian metric. On the other hand if the group is semi-simple, then the (0)-connection does preserve the bi-invariant pseudo-Riemannian metric. $\endgroup$ Commented Jul 29, 2018 at 23:18