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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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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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  • $\begingroup$ Can someone explain this last step, the contradiction about a connection on $S^3$? $\endgroup$
    – Ethan Dlugie
    Commented Apr 16, 2020 at 16:31
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    $\begingroup$ @EthanDlugie: Yes. The point is that, if $S^3$ admitted a torsion-free flat connection, then, because it is simply connected, it would have a global parallel coframe field, $(\theta_1,\theta_2,\theta_3)$ and the parallel condition together with the torsion-free condition would imply that the $1$-forms $\theta_i$ are closed. Since $S^3$ is simply-connected, these would have to be exact, i.e., $\theta_i = \mathrm{d}x_i$ for some functions $x_i$ on $S^3$. But these functions could not have any critical points because their differentials never vanish. Since $S^3$ is compact, this is not possible. $\endgroup$
    – Robert Bryant
    Commented Apr 16, 2020 at 18:08
  • $\begingroup$ Thanks for answering Robert. Two questions still: is torsion-free necessary to find the global parallel coframe? I would think that's just a consequence of flatness. And how to see closedness of the $\theta_i$? $\endgroup$
    – Ethan Dlugie
    Commented Apr 17, 2020 at 2:46
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    $\begingroup$ @EthanDlugie: No. The assumptions you need for global parallel coframe are flatness and simple-connectivity. The closedness of the $\theta_i$ is then equivalent to the torsion-free condition: If $X_i$ are the vector fields dual to the $\theta_i$, then they will parallel as well, so $T(X_i,X_j) = -[X_i,X_j]$. Torsion-free is then $[X_i,X_j]=0$, and then the definition of exterior derivative says $\mathrm{d}\theta_i = 0$. For the converse, read backwards. $\endgroup$
    – Robert Bryant
    Commented Apr 17, 2020 at 8:12
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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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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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I'm surprised that nobody mentioned the Tanaka-Webster connection of a strictly pseudoconvex CR spherical manifold yet.

A strictly pseudoconvex pseudo-Hermitian manifold is a triplet $(M,\theta,J)$ with $\theta$ a contact form on $M$ and $J$ an integrable almost-complex structure on $H=\ker \theta$, such that the Levi form $d\theta|_{H\times H}(\cdot,J\cdot)$ is definite positive. On such a manifold, there exists a unique connection $\nabla^{\theta}$, called the Tanaka-Webster connection, which parallelises $H$, $J$, $X_{\theta}$ the Reeb vector field of $\theta$, and whose torsion $T^{\theta}$ satisfies $$ \forall Y,Z \in H,\quad T^{\theta}(Y,Z) = d\theta(Y,Z)X_{\theta} \quad \text{and} \quad T^{\theta}(X_{\theta},JY) + JT^{\theta}(X_{\theta},Y)=0. $$ The CR-sphere $S^{2n+1}$ and the Heisenberg space $H^{2n+1}$, endowed with their standard contact forms, are examples of strictly pseudoconvex pseudo-Hermitian manifolds whose curvatures $R^{\theta}$ identically vanish, but it is clear that their torsion do not. More generally, a CR manifold which is locally CR-diffeomorphic to one of these space has the same property: they are called CR-spherical.

In this sense, the whole geometry of spherical manifolds is contained in the torsion.

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