# Gauge transformations in differential forms

I am aware of gauge transformations and covariant derivatives as understood in Quantum Field Theory and I am also familiar with deRham derivative for vector valued differential forms.

I thinking of the gauge field A of the gauge group G as a Lie(G) valued 1-form on the manifold.

But I can't see why under a gauge transformation on A by an element $g\in G$ amounts to the following change, $A \mapsto A^g = gAg^{-1} -dgg^{-1}$ (if G is thought of as a matrix Lie Group) or in general $A_g = Ad(g)A + g^* \omega$ (where $\omega$ is the left invariant Maurer-Cartan form on G and I guess $g^*$ is pull-back of $\omega$ along left translation map by $g$).

Curvature is defined as $F = dA + \frac{1}{2}[A,A]$ and using this one wants to now see why does $F \mapsto F_g = gFg^{-1}$.

Firstly is the expression for $A_g$ a definition or is there a derivation for that?

When I try proving this (assuming matrix Lie groups) I am getting stuck in multiple places like what is $dA_g$ ?

I would be happy if someone can explain the explicit calculations and/or give a reference where such things are explained. Usual books which explain differential forms or connections on principal bundles don't seem to help with such calculations.

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I'm treating that question in my bachelor's thesis in the section "principal bundles": (enigmage.de/bachelor.pdf) –  Turion Jul 5 '11 at 8:53

The method is to use the Leibniz rule in the differentiation and and change the sign whenever the exterior derivative moves past an odd form. In addition the following identity must be used $dgg^{-1} + g dg^{-1} = 0$, and remember that the commutator is is between odd forms thus it is with a plus sign.

Here are the intermediate results:

$\frac{1}{2}[A_g, A_g] = \frac{1}{2}g[A, A] g^{-1} -[dgg^{-1}, g A g^{-1}] + dgg^{-1}\wedge dgg^{-1}$

$dA_g = g dA g^{-1} + [dgg^{-1}, g A g^{-1}] - dgg^{-1}\wedge dgg^{-1}$

Here are the required details:

$d(gAg^{-1})$

$d(gAg^{-1}) = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1}$

1. Using the identities $g g^{-1} = 1$ in the first term and $dg^{-1} = - g^{-1}dg g^{-1}$ in the last term

$= dg g^{-1} g \wedge A g^{-1} + g dA g^{-1} +g A g^{-1} \wedge dg g^{-1}$

1. Collection of the first and last term into a commutator:

$= g dA g^{-1} +[ dg g^{-1}, g A g^{-1} ]$

$d(dg g^{-1})$

$d(dg g^{-1}) = ddg g^{-1} - dg \wedge dg^{-1}$
1. Using the identities $dd = 0$ and again $dg^{-1} = - g^{-1}dg g^{-1}$, we obtain:
$d(dg g^{-1}) = + dg g^{-1}\wedge dg g^{-1}$
Thanks for your reply but I am still quite in the dark. I suppose when you take $dA_g$ one is keeping $g$ fixed? I don't seem to understand how you got the last two terms in your expression for $dA_g$ Can you sort of separately write what does $d(gAd^{-1})$ and $d(dgg^{-1})$ evaluate to? –  Anirbit Aug 24 '10 at 18:14
I meant $d(gAg^{-1})$ –  Anirbit Aug 24 '10 at 18:22