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Let $G$ be a connected Lie group with identity $e$ and let $g\in G$. Must there exist a left-invariant vector field $X_g = dL_g(v)$ (for some $v \in \mathfrak g$) such that the flow $\phi_t(e)$ of $X$ through $e$ passes through $g$? I believe this is equivalent to the question of whether the exponential map from $\mathfrak g$ to $G$ is surjective.

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up vote 6 down vote accepted

No. The equivalence you describe holds, and it's well-known that the exponential map fails to be surjective for $\text{SL}_2(\mathbb{R})$. Any $g$ with this property necessarily has a square root, which $\left[ \begin{array}{cc} -1 & 1 \\\ 0 & -1 \end{array} \right]$ doesn't. (Since any square root in $\text{SL}_2(\mathbb{R})$ has determinant $1$, it must have eigenvalues $\pm i$, so in particular it is diagonalizable.)

The exponential map is surjective if $G$ is either compact or nilpotent.

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That's a neat fact. Where could one find proof of this? In, say, Helgason? – Neal Jun 26 '12 at 4:19
@Neal If $G$ is a compact connected Lie group, then you can use Riemannian geometry to prove that the exponential map $\text{exp}:\mathfrak{g}\to G$ is surjective. Indeed, if you equip $G$ with a bi-invariant Riemannian metric (which you can do by Weyl's unitary trick combined with the compactness of $G$), then you can prove that the one-parameter subgroups of $G$ are precisely the geodesics of $G$ passing through the origin. The Hopf-Rinow theorem now establishes the surjectivity of $\text{exp}:\mathfrak{g}\to G$ ... – Amitesh Datta Jun 27 '12 at 2:15
@Neal ... Of course, if you do not wish to explicitly use Riemannian geometry, then there is an alternative proof if you assume Cartan's theorem on the maximal tori in a compact connected Lie group (Cartan's theorem can be proved using the Lefschetz fixed point theorem in algebraic topology). Indeed, each element of $G$ is an element of some maximal torus $T\subseteq G$ which means that we need only prove the surjectivity of the exponential map $\text{exp}:\mathfrak{t}\to T$ where $T$ is a torus. Of course, this is very easy to do. – Amitesh Datta Jun 27 '12 at 2:18
Thank you @AmiteshDatta. So using Riemannian geometry, we require the compactness of $G$ to equip it with a bi-invariant metric and using Lie groups we require compactness to use Cartan's theorem on maximal tori. – Neal Jun 27 '12 at 13:12

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