# Showing that Yang-Mills equations transform homogeneousy

1. ### center o bass

540
Hi! I'm trying to show that the differential from equation
$$D \star F = 0$$ transform homogeneously under the adjoint action ##F \mapsto gFg^{-1}## of the lie group ##G##, where ##D## denotes the covariant exterior derivative ##D\alpha = d \alpha + A \wedge \alpha## for some lie algebra valued form ##\alpha##. Since
$$A \mapsto gAg^{-1} + gdg^{-1}$$
we get
$$D\star F \mapsto d(g\star Fg^{-1}) + (A \mapsto gAg^{-1} + gdg^{-1})\wedge \star F = dg \wedge \star F g^{-1} + g d\star F g^{-1} + g \star F \wedge dg^{-1} + g A \wedge \star F g^{-1} + g dg^{-1} \wedge g^{-1} \star F g^{-1}$$
where if we use that ##gdg^{-1} = - (dg) g^{-1}##, then the last term seem to cancel the first. However, if this is correct, then the third term has to be zero.

Is the third term zero, and if so why?

Am I going wrong somewhere else?

2. ### Greg Bernhardt

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. ### tom.stoer

5,489
I overlooked this question completely; I'll come back to you asap

4. ### ChrisVer

Let me try it out :)
$D \star F=0$
or
$d \star F + A \wedge \star F = 0$

Now make the mappings of $D \star F$....

$d(g \star Fg^{-1}) + (gAg^{-1} + g dg^{-1}) \wedge \star gFg^{-1}$

I think here we reach different results... ?
I can't continue because I am not sure what star means.... and I am not really familiar with the wedges...

5. ### tom.stoer

5,489
I'll try to answer using very very old notes; hope the signs and details a correct. I start w/o using forms. I think the problem is not in the exterior derivative but in the su(n) part. In the following F and A are su(N) matrices, g is an SU(N) group element and the commutators [.,.] always refer to the su(N) matrices.

We have the following transformations:

$$A_\mu \to A_\mu^\prime = g\,(A_\mu + i\partial_\mu)\,g^\dagger$$
$$F_{\mu\nu} \to F_{\mu\nu}^\prime = g\,F_{\mu\nu}\,g^\dagger$$

As far as I remember the covariant derivative of F is defined as

$$(DF)^\nu = \partial_\mu F^{\mu\nu} - i [A_\mu,F^{\mu\nu}]$$

Now we have to check the transformation properties of DF:

$$(DF)^\nu \to {(DF)^\prime}^\nu = \partial_\mu (g\,F^{\mu\nu}\,g^\dagger) -i[g\,A_\mu\,g^\dagger + ig\,\partial_\mu\,g^\dagger, g\,F^{\mu\nu}\,g^\dagger]$$

I hope this Ansatz helps ...

Last edited: Aug 9, 2014