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.

  • $\begingroup$ 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? $\endgroup$ Commented Jun 7, 2023 at 16:39
  • $\begingroup$ 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: $\rho (\bar{t_k}) = -t_k^{*} = -t_k^{T}$. $\endgroup$
    – geodesic
    Commented Jun 8, 2023 at 7:42
  • $\begingroup$ 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? $\endgroup$
    – geodesic
    Commented Jun 8, 2023 at 7:49
  • $\begingroup$ 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... $\endgroup$ Commented Jun 8, 2023 at 14:05

1 Answer 1


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.

Hope this helps.


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