Going from full non-Abelian gauge transformation to its infinitesimal version in component notation

Question

Let $A_\mu^a(x)$ be a non-Abelian gauge field, with $\mathrm{SU}(N)$ generators $T_a$. We can write the field as a Lie-algebra-valued object $$ \mathbf{A}_\mu \equiv A_\mu^a T_a.$$ The full local gauge transformation is $U(x)=\exp(i \alpha^a (x) T_a)$, parameterized by local parameters $\alpha^a(x)$. The field transforms as $$\mathbf{A}_\mu \longrightarrow U \mathbf{A}_\mu U^{-1} - \frac{i}{g} (\partial_\mu U)U^{-1}$$ In the infinitesimal version (i.e. to first order in $\alpha$), in component notation, the transformation should correspond to $$A^a_\mu(x) \longrightarrow A^a_\mu(x) + \frac{1}{g}\partial_\mu\alpha^a(x)-f^{abc} \alpha_b(x)A^c_\mu (x),$$ where $f^{abc}$ stands for the structure constants of $\mathfrak{su}(N)$, defined by $$[T^a,T^b]=if^{abc}T_c.$$

I tried going from the first transformation to the second, using the definitions of $\mathbf{A}_\mu$ and $f^{abc}$, but I just can't get the right form, the last term with $A^c_\mu(x)$ confuses me the most...

For the sake of completeness, let me just state that I also used these two relations: $$U(x)=1+i\alpha^a(x)T_a + \mathcal{O}(\alpha^2)$$ $$U^{-1}(x)=1-i\alpha^a(x)T_a + \mathcal{O}(\alpha^2)$$

Any hints? Am I missing something obvious here?

Answer 1

At first order in $\alpha$, you have \begin{eqnarray} \mathbf{A}_{\mu} &\longrightarrow& (1+i\alpha^a (x) T_a) A_{\mu}^b T_b (1-i\alpha^c (x) T_c) - \frac{i}{g} (i \partial_{\mu} \alpha^a (x) T_a) (1-i\alpha^b (x) T_b)\\ &\longrightarrow& \mathbf{A}_{\mu} + i \alpha^a (x) A_{\mu}^b T_a T_b - i \alpha^c (x) A_{\mu}^b T_b T_c + \frac{1}{g} \partial_{\mu} \alpha^a (x) T_a \\ &\longrightarrow& \mathbf{A}_{\mu} + i \alpha^a (x) A_{\mu}^b (T_a T_b-T_b T_a)+ \frac{1}{g} \partial_{\mu} \alpha^a (x) T_a \\ &\longrightarrow& \mathbf{A}_{\mu} - \alpha^a (x) A_{\mu}^b f_{abc} T^c+ \frac{1}{g} \partial_{\mu} \alpha^a (x) T_a \\ &\longrightarrow& A_{\mu}^a T_a - \alpha_b (x) A_{\mu}^c f^{abc} T_a+ \frac{1}{g} \partial_{\mu} \alpha^a (x) T_a \\ \end{eqnarray} Then identifying on both side the coefficient in front of $T^a$ you have \begin{eqnarray} A_{\mu}^a &\longrightarrow& A_{\mu}^a - \alpha_b (x) A_{\mu}^c f^{abc} + \frac{1}{g} \partial_{\mu} \alpha^a (x) \, . \end{eqnarray}

Answer 2

The term with the structure constant arises from $U \mathbf A_\mu U^{-1}$ because $\mathbf A_\mu$ and $T_a$ do not commute. If you're careful where you place $\mathbf A_\mu$ during the calculation, a commutator will come out giving you the term you're looking for.