# Adjoint transformation of gauge fields

1. Aug 19, 2014

### spookyfish

A gauge field $W_\mu$ is known to transform as
$$W_\mu\to W'_\mu=UW_\mu U^{-1} +(\partial_\mu U)U^{-1}$$
under a gauge transformation $U$, where the first term $UW_\mu U^{-1}$ means it transforms under the adjoint representation. Can anyone explain to me why it means a transformation under the adjoint representation? all I know is the definition of the adjoint representation
$$[T_i]_{jk}=-if_{ijk}$$

Last edited: Aug 19, 2014
2. Aug 19, 2014

### ChrisVer

Because the gauge field $W_{\mu}$ belongs to the adjoint representation of the gauge group.

3. Aug 19, 2014

### spookyfish

How can you see that?

4. Aug 19, 2014

### ChrisVer

Well in fact you need to check the adjoint representation. The definition of the adj. repr is that of an automorphism.
In particular it's because the transformation $W \rightarrow U^{-1} W U$ preserves the Lie Bracket.
The above transformation in practice means $W^{a}_{\mu} T^{a}_{ij} \rightarrow W_{\mu} (U^{-1} T^{a} U)_{ij}$.

Actually are we talking in particular for SU(2)?
In the SU(2) case you have the (dim) reprs:
$2 \equiv \bar{2}, 2 \otimes \bar{2} = 3 \oplus 1$
the $3$ is where the gauge bosons (spin=1) belong. and that's the adjoint repr.

Last edited: Aug 19, 2014
5. Aug 19, 2014

### spookyfish

I understand. But I don't see how this transformation rule is consistent with the definition I know of the adjoint rep: Is it possible to assume that T transforms as $GT_a G^{-1}$ and then prove that it is given by the adjoint representation $[T_a]_{bc}=-if_{abc}$?
where $f_{abc}$ are determined from:
$$[T_a,T_b]=if_{abc}T_c$$
and I don't look specifically for SU(2) reps. also SU(3), and generally any Lie group.

6. Aug 19, 2014

### The_Duck

An infinitesimal symmetry transformation can be parametrized by some numbers $\omega^a$, where $a$ runs over the generators of the symmetry group. Then an object $A_i$ is said to transform in the representation $R$ if, under an infinitesimal transformation,

$A_i \to A_i + i \omega^a (T^a_R)_{ij} A_j$.

where the $T_R^a$'s are the representations of the generators in the representation $R$.

Let's look at how the vector potential $W_\mu^a$ transforms under a global gauge transformation. I'll drop the Lorentz index $\mu$ because it's irrelevant. We have

$W \to U W U^{-1}$

where $U$ is the gauge transformation matrix (we will look at a global transformation, so $\partial_\mu U = 0$). For an infinitesimal gauge transformation $U$ can be written

$U = 1 + i \omega^a T^a_F$

where the $T^a_F$ are the generators in the fundamental representation. Similarly $W$ can be written in terms of the fundamental generators:

$W = W^a T^a_F$.

So we can rewrite the transformation rule, for an infinitesimal gauge transformation, as

$W^a T^a_F \to (1 + i \omega^a T^a_F) W^b T^b_F (1 - i \omega^c T^c_F)$

or, dropping negligible terms of order $\omega^2$,

$W^a T^a_F \to W^a T^a_F + i \omega^a W^b [T^a_F, T^b_F]$.

But we know from the commutation rules that $[T^a_F, T^b_F] = i f^{abc} T^c_F$. So the transformation rule becomes

$W^a T^a_F \to W^a T^a_F - \omega^a f^{abc} W^b T^c_F$

By renaming indices this can be rewritten

$W^a T^a_F \to (W^a + \omega^c f^{cab} W^b) T^a_F$

or just

$W^a \to W^a + \omega^c f^{cab} W^b$

Looking back at the first equation, this is the transformation rule for an object that lives in a representation $R$ where the generators are given by

$(T_R^c)^{ab} = -i f^{cab}$.

This is exactly the adjoint representation.

7. Aug 19, 2014

### spookyfish

Hi. Thank you very much! This is exactly what I was looking for. the explanation is very clear. I only have one question (that I think does not affect the proof): Why did you assume that $U$ is given in the fundamental representation $$U=1+i\omega^aT_F^a$$

8. Aug 20, 2014

### haushofer

I guess because we are talking about matrix Lie groups (their elements are matrices), and their algebra elements correspond to the fundamental representation.

9. Aug 20, 2014

### The_Duck

It doesn't matter; you can pick any representation $R$ and think of $W$ as the matrix $W^a T_R^a$ and $U$ as the matrix $1 + i \omega^a T_R^a$.