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Everytime I hear about adding torsion to GR, something struggles me: how do you create a kinetic term for the electromagnetic field that is still gauge-invariant? One of the consequences of torsion is that covariant derivatives no more commute on scalar functions, so by looking at the $F_{\mu\nu} = \nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}$ (constrained to minimal coupling by the equivalence principle) and we perform a gauge transformation on it $A_{\mu} \to A_{\mu} + \partial_{\mu}\Lambda$ we get $F_{\mu\nu} \to F_{\mu\nu} + T^{\rho}_{\mu\nu}\partial_{\rho}\Lambda$ where $T^{\rho}_{\mu\nu}$ is the torsion tensor. So how can we handle torsion in non pure-gravity?

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The gauge invariance problem of the Maxwell field in the presence of torsion has been known for many years. There is no known perfect solution for this problem.

The importance of having the gauge invariance of the Maxwell action is that it implies charge conservation which is very well established experimentally.

One type of suggestions (please see Sabbata ) is through nonminimal coupling of the maxwell field to torsion. There are other suggestions restricting the types of connections.

However, the "solution" by Benn Dereli and Tucker Phys Lett. B. 96B 100-104 (1980) reviewed is a recent paper by Socolovsky(section 35) seems more appealing as it does not involve any change in the geometric structure of the Maxwell theory coupled to an Einstein-Cartan background, nor waives the minimal coupling. Here the Maxwell field is defined to be the exterior derivative of the vector potential (as in Pavel's answer), and the Maxwell action has its "flat space" form except for the curved space-time measure, which is manifestly gauge invariant. However, the independent variables are taken to be the tetrad components of the gauge potential $A_a =e_a^{\mu} A_{\mu}$ rather the space-time components. In this solution, the gauge noninvariance is manifested in the solution of the field equations for the torsion itself being proportional to the field's non gauge invariant spin density tensor. However, the Lorentz generators being the space integrals of the spin density zero components are still gauge invariant.

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Just to see whether I understand. The problem arises if you take a connection $\nabla:TM\rightarrow TM\otimes\Omega^1$ on the tangent bundle, use the metric to get a connection on the cotangent bundle $\nabla':\Omega^1\rightarrow\Omega^1\otimes\Omega^1$, and then define $F = \nabla' A$, where you wedge the 1-forms in the end? –  Pavel Safronov Jan 12 '12 at 14:46
    
Firstly you don't need a metric to get a connection on the cotangent bundle. Isomorphism of vector spaces induces isomorphism of dual vector spaces hence if you can parallel transport vectors you can parallel transport convectors too. Hence an affine connection yields a covariant exterior derivative operator which equals the usual exterior derivative iff the torsion vanishes. Secondly I also don't understand why we need this covariant exterior derivative in the action instead of using the ordinary exterior derivative. Perhaps it is needed because otherwise the torsion has no physical meaning. –  Squark Jan 14 '12 at 18:52
    
@Pavel, Sorry for the late response, I am afraid I didn't understand the question. Nevertheless, here are some more details , which I hope will be helpful. When formulated by means of the Vector potential, the Maxwell Lagrangian has the same form as on a Riemannian manifold. However, when written in terms of the tetrad components of the gauge field (which are locally $u(1)$ valued sections of the frame bundle, it becomes dependent on the spin connection and as a consequence one gets the correct Maxwell equations $\nabla *F = 0$ from its variation with respect to the tetrad components $A_a$. –  David Bar Moshe Jan 15 '12 at 16:04
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Your definition of $F_{\mu\nu}$ is strange. Assume the relevant $U(1)$-bundle is trivial, then $A$ is a 1-form on the base. The curvature $F=dA$ is independent of the metric. In coordinates you still use the formula without covariant derivatives: $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$.

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