Can a manifold have zero curvature but non-zero torsion?

Question

I am trying to work out whether it is possible for a manifold to have zero curvature but a non-zero torsion, and what restrictions this would put on the connection.

If we have a manifold $M$ equipped with a veilbein basis $\{ e_a \}$ and spin connection $\{ \omega^a_{\ b} \}$, then the curvature and torsion 2-forms are given by

$$ R^a_{\ b} = \mathrm{d}\omega^a_{\ b} + \omega^a_{\ c} \wedge \omega^c_{\ b},$$

$$ T^a = \mathrm{d} e^a + \omega^a_{\ b} \wedge e^b. $$

It is a fact that every metric connection can be split up as

$$ \omega^a_{\ b} = \tilde{\omega}^a_{\ b} + K^a_{\ b} $$

where $\tilde{\omega}^a_{\ b}$ is a connection with zero torsion (Levi-Civita) and $K^a_{\ b}$ is the contorsion one-form (reference: Nakahara's book), so the curvature and torsion take the forms

$$ R^a_{\ b} = \tilde{R}^a_{\ b} + \mathrm{d}K^a_{\ b}+ \tilde{\omega}^a_{\ c} \wedge K^c_{\ b} + K^a_{\ c} \wedge \tilde{\omega}^{c}_{ \ b} + K^a_{\ b} \wedge K^c_{\ b} $$

$$ T^a = K^a_{\ b} \wedge e^b $$

where I have used the fact that the torsion of $\tilde{\omega}^a_{\ b}$ is zero. I am looking for any possible solution that gives $R^a_{\ b} = 0$ and $T^a \neq 0$. If I set $\tilde{\omega}^a_{\ b} = 0$, my curvature and torsion reduces to

$$ R^a_b = \mathrm{d}K^a_{\ b} + K^a_{\ c} \wedge K^c_{\ b}$$

$$ T^a = K^a_{ \ b} \wedge e^b $$

From this point onwards I was able to make a suitable choice of $K^{a}_{\ b}$ to ensure that $R^a_{\ b}$ vanishes but I have some questions about my analysis:

My questions

1. Is it possible to have a manifold that has a zero torsion but a non-zero curvature?
2. Is it possible that for a general connection $\omega^a_{\ b} = \tilde{\omega}^a_{\ b} + K^a_{ \ b}$ that I can set $\tilde{\omega}^a_{\ b} = 0$ while leaving $K^a_{\ b}$ alone, in other words, are $\tilde{\omega}^a_{\ b}$ and $K^a_{\ b}$ independent?

Answer 1

1. For the Levi-Civita connection on a Riemannian manifold, the torsion is zero and most often the curvature is nonzero. Any compact orientable manifold with nonzero Euler characteristic must have nonzero curvature.

2. The easiest way to construct a connection with zero curvature and nonzero torsion is to take a (non-abelian) Lie group and declare $\nabla_X Y = 0$ for all left-invariant vector fields $X$ and $Y$. The structure equations then tell you that all the Lie algebraic structure is in the torsion. (In other words, in your notation, take the left-invariant Maurer-Cartan forms $e^a$ and set $\omega^a_b = 0$. Then, letting $\gamma^a_{bc}$ denote the structure constants of the Lie group, the structure equations $de^a = -\frac12\sum \gamma^a_{bc}e^b\wedge e^c$ tell us that this quantity is all torsion. Obviously, the curvature is dead zero.)