Can a flat space have nonzero torsion?

I know that in general a curved space can have torsion or be torsion-free, however, can torsion exist in a flat space?

I'm guessing that it cannot for the reason that torsion is the antisymmetrization of the connection coefficients and they vanish in cartesian coordinates. Unless torsion, unlike the Riemann curvature, is not an intrinsic property of the space, but an artifact of the basis vectors changing independently along different paths.

The connection and the metric of a manifold are technically two independent quantities. If we allow both of those to be independent, then making a flat manifold with torsion is fairly trivial, for instance

$$\begin{eqnarray} ds^2 &=& -dt^2 + dx^idx_i\\ {\Gamma^\alpha}_{\mu\nu} &=& g^{\alpha\beta} \varepsilon_{\beta\mu\nu} \end{eqnarray}$$

This is one of the simplest connection with a torsion. it can easily be checked to have torsion, via

$$$${T^\alpha}_{\mu\nu} = {\Gamma^\alpha}_{[\mu\nu]} = g^{\alpha\beta} \varepsilon_{\beta\mu\nu}$$$$

with also (for what follows)

$$\begin{eqnarray} {T}_{\gamma\mu\nu} &=& g_{\alpha\gamma} g^{\alpha\beta} \varepsilon_{\beta\mu\nu}\\ &=& \varepsilon_{\gamma\mu\nu} \end{eqnarray}$$

and the contorsion tensor

$$\begin{eqnarray} K_{\alpha\beta\gamma} &=& \frac{1}{2} (T_{\alpha\beta\gamma} - T_{\beta\gamma\alpha} + T_{\gamma\alpha\beta})\\ &=& \frac{1}{2} T_{\alpha\beta\gamma} \end{eqnarray}$$

You can check fairly easily that despite the arbitrariness of our connection choice, it has torsion but it is still a metric connection :

$$\begin{eqnarray} \nabla_\alpha g_{\mu\nu} &=& \partial_\alpha g_{\mu\nu} - {\Gamma^\beta}_{\mu\alpha} g_{\beta\nu} - {\Gamma^\beta}_{\nu\alpha} g_{\mu\beta}\\ &=& - g^{\beta\gamma} \varepsilon_{\gamma\mu\alpha} g_{\beta\nu} - g^{\beta\gamma} \varepsilon_{\gamma\nu\alpha} g_{\mu\beta}\\ &=& - \varepsilon_{\nu\mu\alpha} - \varepsilon_{\mu\nu\alpha}\\ &=& 0 \end{eqnarray}$$

So that we have a connection that has torsion but is nethertheless metric, so that it still fits within the framework of the Einstein-Cartan theory, if we so wish. You can also check that this is flat via

$$\begin{eqnarray} {R^\alpha}_{\mu\nu\sigma} &=& 2 {\Gamma^\alpha}_{[\nu | \lambda} {\Gamma^\lambda}_{\sigma| \mu]}\\ &=& g^{\alpha\beta} g^{\lambda\beta} (\varepsilon_{\beta\nu\lambda} \varepsilon_{\gamma\sigma\mu} - \varepsilon_{\beta\mu\lambda} \varepsilon_{\gamma\sigma\nu})\\ &=& 0 \end{eqnarray}$$

As we are trying to contract a tensor symmetric in $$\lambda \beta$$ and one antisymmetric in it.

There is in fact a whole theory based on that notion called teleparallel gravity, which contains a flat connection with torsion, and it is an entirely acceptable theory of gravity.

• what is it about this theory that doesn't map onto our observations about flat space? That the geodesics aren't straight lines? – QuantumEyedea Oct 17 '19 at 14:02
• The main physical objects affected by a metric connection with torsion are spinor fields. As far as we can tell, fermions move freely in a way that is consistent with zero torsion. – Slereah Oct 17 '19 at 14:09
• Interesting. Why is this the case? Does the spinorial action have to be written down differently when there is torsion? – QuantumEyedea Oct 17 '19 at 14:31
• Check this out : physics.stackexchange.com/questions/212298/… – Slereah Oct 17 '19 at 14:32
• I still don't quite understand one thing. Is there any difference between a flat space with torsion and the one without? Since torsion is the antisymmetric part of the connection coefficients, and those can be calculated from the first derivatives of the basis, but if we choose the basis to be the Riemann normal coordinates then torsion disappear. So by choosing the right coordinates, torsion can be eliminated. Does that mean torsion is not an intrinsic property of the space, rather it only depends on the choice of basis? – Questioneer Oct 19 '19 at 8:06