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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?

flag	asked Jan 4 2012 at 21:14 Oliver Jones 227●6

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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 non-trivial commutators in $G$ will correspond to non-trivial curvature for the Levi--Civita connection on $G$. – Emerton Jan 4 2012 at 21:23

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 Levi-Civita connection. Thanks. – Oliver Jones Jan 4 2012 at 22:59

Liviu Nicolaescu · Accepted Answer · 2012-01-04 21:59:08Z UTC

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.

Ben McKay · Answer 2 · 2012-01-04 21:58:31Z UTC

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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.

answered Jan 4 2012 at 21:58

Ben McKay
2,411●1●7●9

Thanks Ben. I didn't look at the Cartan-Hadamard theorem close enough to see that it refers to a Levi-Civita connection. – Oliver Jones Jan 4 2012 at 22:55

Cartan-Hadamard Theorem

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