Showing the form of the covariant derivative of $\phi$, if $\phi$ transforms as the adjoint representation of $SU(n)$

Question

I want to show that if $\phi$ transforms as the adjoint representation of SU(n), its covariant derivative is given by $\textbf{D}_\mu \phi = \partial_\mu \phi + i [\textbf{A}_\mu, \phi]$. (Exercise in chapter 6 of P. Ramond QFT book)

The covariant derivative is defined as

$$\textbf{D}_\mu= \partial_\mu + i\textbf{A}_\mu$$

If $\phi$ have the transformation properties assumed, we know that $\phi= \phi^a T^a$, where $T^a$ are the generators of SU(n) and in the adjoint representation $(T^b)_{cd}= -if^{bcd} $. Besides that, we have $\textbf{A}_\mu=A_\mu^a T^a$, etc.

The attempted solution:

$$\begin{split}(\textbf{D}_\mu \phi)_{ab} & = (\partial_\mu \phi)_{ab} + i (\textbf{A}_\mu \phi)_{ab} \\ & = (\partial_\mu \phi)_{ab} + i A_\mu^c \phi^e (T^c)_{ad}(T^e)_{db} \end{split}$$

And now I can use the relation $(T^b)_{cd}= -if^{bcd} $, but it doesn't get me to the right answer because there isn't one $f$ that contracts with both $A$ and $\phi$ as that should be to get the correct result. We can see this writing the answer with explicit components

$$(\textbf{D}_\mu \phi)_{ab} = (\partial_\mu \phi)_{ab} - A_\mu^d \phi^e f^{dec}(T^c)_{ab}.$$

How should I proceed? Any help will be appreciated.

$\textbf{Edit}$

Following the comments, I got to a point and now I am stuck:

$$\begin{split}(\textbf{D}_\mu \phi)_{be} & = (\partial_\mu \phi)_{be} + i A_\mu^a \phi^d (T^a)_{bc}(T^d)_{ce} \\ & = (\partial_\mu \phi)_{be} + i A_\mu^a \phi^d (-if^{abc})(-if^{dce}) \\ & = (\partial_\mu \phi)_{be} + i A_\mu^a \phi^d (-f^{adc}f^{ceb}-f^{aec}f^{cbd}) \\ & = (\partial_\mu \phi)_{be} - A_\mu^a \phi^d f^{adc}(T^c)_{be}-A_\mu^a \phi^d f^{aec}f^{cbd} \\ & =(\partial_\mu \phi)_{be} +i A_\mu^a \phi^d [T^a,T^d]_{be}-A_\mu^a \phi^d f^{aec}f^{cbd}\end{split} $$

Which shows that I got the right answer plus $-A_\mu^a \phi^d f^{aec}f^{cbd}$. Is this right so far?

Answer 1

I see now the problem. The definition of the covariant derivative in a general representation $r$ for a field $\phi_\alpha$, with $\alpha = 1,\ldots \mathrm{dim}\,r$, is $$ (\mathbf{D}_\mu\phi)_\alpha = \partial_\mu\phi_\alpha + i A_\mu^a (T^a_r)_{\alpha\beta}\phi_\beta\,. $$ Where $T^a_r$ are generators in the representation $r$.

A field in the adjoint representation of an algebra $\mathfrak{g}$ can be seen as a vector with an index $a = 1,\ldots,\mathrm{dim}\,\mathfrak{g}$, or as a matrix with two indices belonging to any representation. If we denote it as a matrix the definition of the covariant derivative is via the commutator.

This is what the exercise asks for:

Show that these two definitions of the covariant derivative in the adjoint representation are equivalent: $$ (\mathbf{D}_\mu\phi)_{a} = \partial_\mu\phi_a + i A_\mu^c (T^c_\mathfrak{g})_{ab}\phi_b\,, $$ $$ (\mathbf{D}_\mu\phi)_{\alpha\beta} = \partial_\mu\phi_{\alpha\beta} + i [\mathbf{A}_\mu, \phi]_{\alpha\beta}\,. $$

The indices in the second equation can be any representation. As you correctly stated $$ (T_\mathfrak{g}^c)_{ab} = -i f^{cab}\,. $$ In order to show the claim one needs to multiply the first equation by $T^a_r$.

$$ \partial_\mu\phi_{\alpha\beta} + A_\mu^c (T^a_r)_{\alpha\beta}f^{cab}\phi_b = \partial_\mu \phi_{\alpha\beta} + i A_\mu^c [T^c_r,T^b_r]\phi^b\,. $$

This is true and one does not even need the Jacobi identity. If you choose $r$ to be the adjoint, this does need the Jacobi, but only because in the adjoint representation the Jacobi plays the role of the commutation rule.

Answer 2

Remember that the structure constants satisfy the Jacobi-relation

$$ f^{abc}f^{cde} + f^{adc}f^{ceb} + f^{aec}f^{cbd} = 0$$

and that the gauge field also transforms under the adjoint representation. Using this, i think it should be straightforward to prove the statement.