Cartan-Hadamard Theorem Can someone point out the gap in this argument. Consider a simply-connected Lie group with the (-)-connection. This connection is flat and so the sectional curvatures are zero. Then, by the Cartan-Hadamard theorem and simple-connectedness, 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?
 A: As Emerton pointed out, you need to be careful about the connection.  Cartan-Hadamard theorem is a statement  involving the curvature of the Levi-Civita connection  determined by some metric.
If $G$ is a Lie group equipped with a bi-invariant 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. 293-329.
A: Your connection is not Riemannian; it has torsion, so cannot be the Levi-Civita connection of any Riemannian metric. The Cartan-Hadamard theorem isn't even true in Lorentzian geometry, and so you wouldn't expect it for a flat connection which isn't torsion free.
