2
$\begingroup$

Considering the 2-Form Gauge Potential $B_{\mu \nu}$, we can write $dB=H$.

In a flat manifold we have that $H_{\mu \nu \rho}=\partial_{\mu}B_{\nu \rho}+\partial_{\rho}B_{\mu \nu}+\partial_{\nu}B_{\rho \mu}$, does this still applies on a curved manifold (spacetime)? Or we have to substitute the normal derivatives with the covariant derivatives? ($H_{\mu \nu \rho}=\nabla_{\mu}B_{\nu \rho}+\nabla_{\rho}B_{\mu \nu}+\nabla_{\nu}B_{\rho \mu}$).

I know that if the potential is a 1-form ($A_\mu$) the result is the same, but I want to know what is the correct definition for a generic $p$-form potential.

$\endgroup$
5
$\begingroup$

The exterior derivative $\mathrm{d}$ needs no connection $\nabla$. The two types of derivatives on $p$-forms are the same if the connection $\nabla$ is torsionfree.

$\endgroup$
4
  • $\begingroup$ Oh Thanks! And what about if the connection $\nabla$ is not TorsionFree? The correct definition should be with the partial derivative $\partial$ then? $\endgroup$ – Andrea Di Pinto Jan 30 at 1:49
  • 1
    $\begingroup$ Yes, if you mean the exterior derivative $\mathrm{d}$. $\endgroup$ – Qmechanic Jan 30 at 1:50
  • $\begingroup$ Ok, perfect. And there exist some kind of "exterior covariant derivative" $\tilde{d}$ constructed with the connection $\nabla$? Or this has no use? $\endgroup$ – Andrea Di Pinto Jan 30 at 14:54
  • 1
    $\begingroup$ Yes, it's mathematically well-defined. Whether it is relevant in physics is another matter. $\endgroup$ – Qmechanic Jan 30 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.