Could EM 'Faraday' tensor represent 'Curvature' ??

[Widdekind]

If, in EM, the Faraday tensor [math]F^{\mu \nu} \equiv \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}[/math] (Wiki); and if, in GR, the Riemann curvature tensor is [math]R(u,v) \equiv \nabla_u \nabla_v - \nabla_v \nabla_u[/math] (Wiki); then, could one 'visually identify' the EM 4-vector [math]A^{\mu}[/math] as the 'covariant derivative' of 'something else', as [math]A^{\mu} \equiv \partial^{\mu} T[/math], and then view the Faraday tensor, [math]F^{\mu \nu} \equiv \left( \partial^{\mu} \partial^{\nu} - \partial^{\nu} \partial^{\mu} \right) T[/math], as a kind of 'curvature' tensor ?

[ajb]

It is a curvature tensor. It is the curvature of a connection on a principle U(1) bundle.

 

A report a wrote a couple of years ago maybe be of interest: http://www.angelfire.com/geek/susy/FirstYear.html

Edited March 5, 2011 by ajb