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In some Yang-Mills theory with gauge group $G$, the gauge fields $A_{\mu}^{a}$ transform as $$A_{\mu}^{a} \to A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c}$$ $$A_{\mu}^{a} \to A_{\mu}^{a} \pm \left(\partial_{\mu}\theta^{a}-A_{\mu}^{b}f^{bac}\theta^{c}\right)$$ $$A_{\mu}^{a} \to A_{\mu}^{a} \pm \left(\partial_{\mu}\theta^{a}-iA_{\mu}^{b}(T^{b}_{\text{adj}})^{ac}\theta^{c}\right),$$

where $T^{a}_{\text{adj}}$ is the adjoint representation of the gauge group $G$ and the gauge parameters $\theta^{a}$ are seen to transform in the adjoint representation of the gauge group $G$.


Why does this mean that the gauge fields $A_{\mu}^{a}$ transform in the adjoint representation?

Should the transformation of the gauge fields $A_{\mu}^{a}$ in the adjoint representation not be given by

$$A_{\mu}^{a} \to A_{\mu}^{a} \pm i\theta^{b}(T^{b}_{\text{adj}})^{ac}A_{\mu}^{c}?$$

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  • $\begingroup$ Except for the $\partial_\mu\theta$ term in the last expression, and perhaps up to a sign, you wrote the exact same thing four times. Are you asking how to show that the four expressions are the same thing? Or you already know that? $\endgroup$ – AccidentalFourierTransform Mar 23 '17 at 20:19
  • $\begingroup$ I would like to prove that the third term is equal to the fourth term. $\endgroup$ – nightmarish Mar 23 '17 at 20:50
  • $\begingroup$ You can't because they are not equal. The $\partial_\mu\theta$ term is missing in the fourth expression. It should be there. $\endgroup$ – AccidentalFourierTransform Mar 23 '17 at 21:21
  • $\begingroup$ I know! But it is said that both $\theta$ and $A_{\mu}^{a}$ transform in the adjoint representation. But it appears that their precise mathematical transformation is different. $\endgroup$ – nightmarish Mar 23 '17 at 22:12
  • $\begingroup$ Ah, that's the problem: $\theta$ does not transform in the adjoint representation. In fact, it does not transform at all: it is this function what parametrises the transformation itself. This is similar to when you perform a translation $x\to x+a$ - the vector $a$ does not transform. $\endgroup$ – AccidentalFourierTransform Mar 23 '17 at 22:27

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