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Is the geodesic exponential map for a Lie group with the (-)-connection a diffeomorphism? This connection is one of two flat connections introduced by Cartan and Schouten on a Lie group and has torsion $T(X,Y)=-[X,Y]$.

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Since there are groups which are not diffeomorphic to a vector space, probably no. – Mariano Suárez-Alvarez Apr 5 '13 at 23:42
$S^1$? I voted to close as "too localized". – Angelo Apr 6 '13 at 0:57
What is the/a (-)-connection? Did you mean $\Theta$? – David Roberts Apr 6 '13 at 2:07
up vote 12 down vote accepted

The $(+)$/0/$(-)$-connections on a Lie group $G$ are the left-invariant connections (I will identify connections with covariant derivatives) defined on left-invariant vector fields by $\nabla_XY=c[X,Y]$ with respectively $c=1,\frac{1}{2},0$. Their name is related to the fact that their torsion is $T(X,Y)=d[X,Y]$ with respectively $d=1,0,-1$.

The name is also, and perhaps more importantly, related to the fact that in the general theory of invariant connections on reductive homogeneous spaces (see Kobayashi-Nomizu Vol. I Chap. X.2, applied here to $(G\times G)/\operatorname{diag}(G)$) they come from a choice of $\mathfrak k$-invariant subspace of $\mathfrak g\times\mathfrak g$ supplementary to $\mathfrak k=\{(X,X)\mid X\in \mathfrak g\}$ given by $\mathfrak m_+=\{(0,X)\mid X\in\mathfrak g\}$, $\mathfrak m_0=\{(X,-X)\mid X\in\mathfrak g\}$ or $\mathfrak m_-=\{(X,0)\mid X\in\mathfrak g\}$, where $\mathfrak g=\operatorname{Lie}(G)$.

The geodesics of the $(+)$/0/$(-)$-connections on a Lie group are the one-parameter subgroups. To see that, just check the geodesic equation with $\gamma(t)=\exp(tX)$: since $\dot\gamma(t)={L_{\exp(tX)}}_*X$, we get $\nabla_{\dot\gamma(t)}\dot\gamma(t)={L_{\exp(tX)}}_*\nabla_XX={L_{\exp(tX)}}_*c[X,X]=0$.

To have one more justification, notice that if the difference $S(X,Y)=\nabla_XY-\nabla^\prime_XY$ between two connections is an antisymmetric tensor, then these connections have the same (parametrized) geodesics (to see it, write the geodesic equation $\nabla'_{\dot\gamma}\dot\gamma=0$ for $\nabla'$ in terms of that for $\nabla$ and of $S$). Since the three connections at hand differ by a multiple of the Lie bracket, they have the same geodesics.

So the exponential map of the $(-)$-connection is a diffeomorphism whenever the exponential map of the Lie group is a diffeomorphism. And this happens exactly when both

  1. the Lie group $G$ is connected and simply connected, and
  2. no eigenvalues of the adjoint action are purely imaginary (i.e., for all $X \in \operatorname{Lie}(G)$, the operator $Y\mapsto \operatorname{ad}(X)Y$ has no purely imaginary eigenvalue).

The Lie groups with these properties are called strongly exponential and are all solvable (the condition on the eigenvalues rules out the existence in $\operatorname{Lie}(G)$ of any compact semisimple subalgebra, which in turn rules out the existence of any semisimple subalgebra).

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Thanks Yannick for the comprehensive answer; very helpful. – Oliver Jones Apr 6 '13 at 21:07

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