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

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?

• But am I not supposed to start from the definition $\textbf{D}_\mu \phi = (\partial_\mu + i \textbf{A}_\mu ) \phi$ an then manipulate assuming $\phi = \phi^a T^a$ to get to $\textbf{D}_\mu \phi = \partial_\mu \phi + i[\textbf{A}\mu, \phi]$? – Slayer147 Apr 10 at 11:55
• Did you fill in, the attempted solution, $\left( T^e\right )_db=-if^{edb}$? – descheleschilder Apr 10 at 13:27
• Yes, and it is done in my edit too. – Slayer147 Apr 10 at 13:39

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.

• Thanks for the answer. I just don't understand your claim that "if you choose $r$ to be the adjoint, this does need the Jecobi". From what I see I can get the same answer doing the same steps above even if $r$ is the adjoint. – Slayer147 Apr 10 at 18:17
• Probably I shouldn't have said that. What I meant is that the statement $[T^a,T^b]= i f^{abc} T^c$ is the Jacobi identity if expressed in the adjoint representation. – MannyC Apr 10 at 18:27

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.

• I edited the question. Following your comment, I got to the point above. – Slayer147 Apr 10 at 2:00