Can someone point out the gap in this argument. Consider a simplyconnected Lie group with the ()connection. This connection is flat and so the sectional curvatures are zero. Then, by the CartanHadamard theorem and simpleconnectedness, the Lie group must be diffeomorphic to ${\Bbb R}^n$. However, I don't think that this is correct without an addition assumption of solvability or nilpotency. What's wrong here?

1$\begingroup$ Which flat connection do you have in mind? E.g. if $G = SU(2)$, which is $S^3$ topologically, then the $G$invariant metric is just the standard spherical metric, which is not flat. More generally, if $G$ is not abelian, then nontrivial commutators in $G$ will correspond to nontrivial curvature for the LeviCivita connection on $G$. $\endgroup$– EmertonJan 4, 2012 at 21:23

$\begingroup$ Emerton, the connection I had in mind was the ()connection (the minus connection). However, as you've all pointed out, the theorem applies to a LeviCivita connection. Thanks. $\endgroup$– Oliver JonesJan 4, 2012 at 22:59
2 Answers
As Emerton pointed out, you need to be careful about the connection. CartanHadamard theorem is a statement involving the curvature of the LeviCivita connection determined by some metric.
If $G$ is a Lie group equipped with a biinvariant metric $h$, then this metric induces a metric $\langle,\rangle$ on the Lie algebra $T_1G$. Given two unit orthogonal vectors $X,Y\in T_1G$ that span a plane $\pi\subset T_1G$, then the sectional curvature of $h$ at $1$ along the plane $\pi$ is given by
$$ K_1(\pi)=\frac{1}{4}\langle\;[x,y],[x,y]\;\rangle$$.
We see from the above equality that the sectional curvature everywhere $\leq 0$ iff the Lie algebra is Abelian.
For metrics on $G$ which are only left invariant the computations of curvatures are a bit more complicated and you can find more details in John Milnor's paper, Curvatures of left invariant metrics on Lie groups, Adv. in Math., vol. 121(1976), p. 293329.

$\begingroup$ Thanks Liviu for your helpful response. I suspected that it wasn't enough to just have a connection but didn't know what was missing. $\endgroup$ Jan 4, 2012 at 23:10
Your connection is not Riemannian; it has torsion, so cannot be the LeviCivita connection of any Riemannian metric. The CartanHadamard theorem isn't even true in Lorentzian geometry, and so you wouldn't expect it for a flat connection which isn't torsion free.

$\begingroup$ Thanks Ben. I didn't look at the CartanHadamard theorem close enough to see that it refers to a LeviCivita connection. $\endgroup$ Jan 4, 2012 at 22:55