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?



Cartan-Hadamard Theorem
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.


