Connection, Metric and Torsion

[pervect]

In the presence of torsion, is it correct to say that the metric doesn't give rise to a unique connection? So, if we were using, say ECKS theory, which unlike GR includes torsion, in order to find the connection, we'd need not only the metric, but a specification of the torsion.

My thinking is that the metric gives the symmetric part of the Christoffel symbols of the second kind, but in the presence of torsion, there is also an asymmetric part for these symbols which is given by the torsion. However, I could use a sanity check here.

A related question: if we define a geodesic by saying that it parallel transports its own tangent vector, is the geodesic equation unaffected by the presence of the torsion? So the metric still gives us the geodesic equation, the unspecified torsion terms don't matter?

 

[Orodruin]

My thinking is that the metric gives the symmetric part of the Christoffel symbols of the second kind, but in the presence of torsion, there is also an asymmetric part for these symbols which is given by the torsion. However, I could use a sanity check here.

This is generally not correct. The symmetric part of the Christoffel symbols in a metric compatible connection with non-zero torsion are not the Christoffel symbols of the Levi-Civita connection.

A related question: if we define a geodesic by saying that it parallel transports its own tangent vector, is the geodesic equation unaffected by the presence of the torsion? So the metric still gives us the geodesic equation, the unspecified torsion terms don't matter?

No. Generally, the connection with torsion has different geodesics. As an example, consider the connection on the sphere (minus the poles) for which compass directions are parallel transported.

 

[Orodruin]

To make the example explicit. Consider the two-dimensional sphere (with the poles removed) and the vector fields
$$
X = \partial_\theta \quad \mbox{and} \quad Y = \frac{1}{\sin\theta} \partial_\varphi,
$$
where $\theta$ and $\varphi$ are the spherical coordinates. With the standard metric, these fields are orthonormal everywhere and if we define a connection such that $\nabla_Z X = \nabla_Z Y = 0$ for all $Z$, i.e., both $X$ and $Y$ are parallel, the connection is metric compatible (you can easily check that it conserves inner products under parallel transport by decomposing an arbitrary vector into a linear combination of $X$ and $Y$).

This connection has only one non-zero Christoffel symbol, $\Gamma_{\theta\varphi}^\varphi = \cot(\theta)$ and is not torsion free (in particular $\Gamma_{\varphi\theta}^\varphi = 0$. If you take the symmetric part of the Christoffel symbols, the only non-zero element is $\Gamma_{\{\theta\varphi\}}^\varphi =\Gamma_{\{\varphi\theta\}}^\varphi = \cot(\theta)/2$. This does not correctly reproduce the non-zero Christoffel symbols of the Levi-Civita connection.

The geodesics of the above mentioned connection are generally not great circles (although the equator and the meridians are geodesics and great circles), but the curves that move in a fixed compass direction. For example, the curves of constant latitude go east$\leftrightarrow$west and are geodesics in this connection.

 

[pervect]

OK, before I work on some possible conceptual confusions I might be suffering from, I need to clear up some linguistic confusions. Is a connection generally defined to be the Ricci rotation coefficients, the connenction one-forms, usually written something like $\omega_{\lambda\mu\nu}$ (Wald, pg 50, for instance)?. Or does "a connecton" sometimes mean the Christoffel symbols $\Gamma$ as well?

Additionally, are the Christoffel symbols $\Gamma$ defined in a non-coordinate basis, and if so, how? I've only seen Christoffel symbols defined and (up to now) used in the context of a coordinate basis. This if (as in your example) we have a non-coordinate basis, and we start talking about the Christoffel symbols in that basis, I get confused. The Christoffel symbols aren't tensors, so I don't think we can do the usual and just transform the basis :(.