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Suppose I have a smooth manifold with a tangent bundle, and I have a connection. If this connection is curvature-free, is it guaranteed to be torsion-free? (I am not assuming a metric, just a finite-dimensional smooth manifold.)

I know that in general curvature-free connections do not exist, and that in general torsion-free connections do. But is the existence of a curvature-free connection sufficient to prove the existence of a torsion-free connection?

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Torsion is defined only for connections on the tangent bundle. – Tom Goodwillie Nov 21 '13 at 1:51
Typo. I just fixed it. – pavpanchekha Nov 21 '13 at 1:52
up vote 21 down vote accepted

Many manifolds have curvature-free (i.e., flat) connections on their tangent bundles. For example, any orientable $3$-manifold $M$ is parallelizable, i.e., its tangent bundle is trivial, so it carries a flat connection (in fact, many flat connections). However, nearly all of these connections will have torsion. In fact, they all will unless $M$ is very special; essentially, $M$ has to be what is known as a flat affine manifold.

I don't understand your last question. For example, the $3$-sphere has a flat connection (just regard it as a Lie group and take the connection that makes the left-invariant vector fields parallel). However, the $3$-sphere certainly does not admit a torsion-free flat connection, since it is compact and simply-connected.

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Special examples of Robert Bryants answer are Lie groups. On any Lie group, the left trivialization of the tangent bundle corresponds to a flat connection whose torsion is essentially the Lie bracket, whereas the right trivialization corresponds to another connection without curvature whose torsion is essentially the negative of the Lie bracket.

On the other hand any inner product on the Lie algebra gives rise to a right invariant metric (biinvariant if the inner product is invariant under the adjoint representation). Its Levi-Civita connections is torsion free and it has a very interesting curvature (See a paper by Milnor for the finite dimensional case or the famous paper by Arnold on volume preserving diffeomorphisms).

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I believe that this is the Milnor's paper: J. Milnor, Curvatures of Left Invariant Metrics on Lie Groups, Advances in Math ( – Vít Tuček Nov 21 '13 at 21:05

A bundle $E\rightarrow M$ will have a curvature free connection if it is trivial on some covering space $\pi:\tilde{M}\rightarrow M$ (one can take the universal cover) of $M$.

Vanishing curvature form $\Omega=d\omega+[\omega,\omega]$, $\omega$ is the connection form, means the horizontal spaces on the principal bundle $L_{GL(n)} \rightarrow M$ are tangent to a foliation.

The torsion is only defined when $E=TM$. A flat connection will usually not be torsion free. Locally, a flat connection is given by a trivializing frame field $(X_1,\ldots,X_n)$, a parallel frame. In this case $X_i$ are vector fields. But the connection will only be torsion free if $[X_i,X_j]=0,\ \forall i,j$.

The answer to your first question is a strong negative. Since any local frame field locally defines a flat connection, which has torsion if they don't commute.

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