# Gauge covariant derivative for fields in tensor representations with multiple indices

In QFT, for fields transforming under some Gauge group, one defines the covariant derivative as

$$(1)\qquad D_{\mu} \phi = \partial_{\mu}\phi -igA_{\mu}^k \rho(t_k)_{ab}\phi_b$$

If $$dim\rho=dim(\text{Lie algebra})$$, one can decompose the fields on the Lie algebra and write them as matrices $$\Phi(x):=\phi(x)_a t_a$$ instead of multiplets.

Then, taking $$\rho$$ as the adjoint repr. it's possible to write (1) as a formula for the matrix field $$\Phi$$ as $$D_{\mu}=\partial_{\mu}\Phi+ig[A_{\mu},\Phi]$$. A field in the adjoint rep can also be written as a field carrying 2 indices $$\phi_{a\bar{b}}$$ where $$a$$ transforms in the fundamental of $$G$$ and $$\bar{b}$$ in the antifundamental. This corresponds to the double sided transformation: $$U\phi U^{\dagger}$$

Now, In my lecture notes i have a generalization of the above situation that I don't understand. It is stated that if I have a field transforming as $$\Psi \rightarrow U_{L} \Psi U_{R}^{\dagger}$$ where $$U_L$$ and $$U_R$$ are two arbitrary irreps of G, (not necessarily the fundamental and antifundamental) i can write the covariant derivative in matrix notation as

$$(2)\qquad D_{\mu}\Psi = \partial_{\mu}\Psi+ig_{L}\left(A_{L\mu}^{a}t_{L}^{a}\right)\Psi-ig_{R}\Psi\left(A_{R\mu}^{a}t_{R}^{a}\right)$$

I don't understand how this can be shown. My notes say that it is a generalization by analogy looking at what happens in in the adjoint representation. This makes sense to me since if we put $$U_R=U_L$$ and declare $$U_L$$ to be the fundamental repr, then (2) reduces to $$\partial_\mu \Psi + ig [A_\mu,\Psi]$$, however i want to see the details on how to derive this.

What i did so far is notice that such field must carry 2 indices $$a$$ , $$b$$ and then try to apply the definition of covariant derivative (1) to each index separately but i can't get anywhere.

• If you have the same group for both L and R action, why are you contrasting the two gauge fields? They should be the same field with independent components ranging over the dimension of the Lie algebra of the group, no? Jun 7, 2023 at 16:39
• You are right, G is has 2 factors: $G=G_L \times G_R$ with corresponding gauge fields. I hadn't notice this as it was not clear to me that you should have fields for every factor of G in $D_\mu$. I guess this solves the problem because if I put a term in $D_\mu$ for every factor of G and then use the fact that for the complex conjugate rep the generators are: $\rho (\bar{t_k}) = -t_k^{*} = -t_k^{T}$. Jun 8, 2023 at 7:42
• This solves the problem, but now i have 2 more questions. (1) If I have a field in a tensor representation of a group G (that is with multiple indices, right?), how do I write down the Covariant derivative? (2) Why should i have a connection term in $D_\mu$ for every factor of G? is it obvious? Jun 8, 2023 at 7:49
• Yeah, it is obvious. Expand all U s everywhere to lowest order in the "angles" (parameters), $U=\exp (i \theta_k t_k)$ for every group, and representation $t_k$, acting on the left or right, and ensure the covariant derivative cancels any and all gradients of all angles $\theta_k$ in all terms/representations, etc... upon all infinitesimal transformations... Jun 8, 2023 at 14:05

A Lie group $$G$$ representation $$\rho:G\to GL(V)$$ naturally induces a representation of its Lie algebra $$\mathfrak g$$. Remember that the associated Lie algebra is the tangent space at unity of the Lie group. Abstractly, you can define a Lie algebra representation by considering the differential of $$\rho$$ at the identity. More explicitly, using the exponential map, for any $$x\in \mathfrak g$$, $$e^x\in G$$, so you can define (the extra $$i$$ is for the physicists’ convention): $$\hat \rho (x)=\frac{d}{dt}_{t=0}e^{itx}$$
In your case, you can check that the corresponding representation of: $$\Psi\to U_L\Psi U_R^{-1}$$ is: $$\Psi\to iA_L\Psi -i\Psi A_R$$ from Leibnitz’ rule. Note that if you truly want the representation: $$\Psi\to U_L\Psi U_R^\dagger$$ then the corresponding representation is: $$\Psi\to iA_L\Psi +i\Psi A_R^\dagger$$ In general the two representations are different if the $$U_R$$ are not unitary (equivalently $$A_R$$ are hermitian). Typically when $$G$$ is not compact this is not the case.
Once you know this, you can simply apply this to the covariant derivative: $$D\Psi =\partial \Psi -igA\psi$$ remembering that in general the second term stands for the action of the Lie algebra.