Gauge covariant derivative of an adjoint action: $\psi(x) \to g \psi(x) g^{-1}$, instead of a left action $\psi(x)\to e^{iq\theta(x)} \psi(x)$

Question

In the case where the transformation on $\psi$ is applied from the left:

$$ \psi(x)\to e^{-iq\theta(x)}\psi(x). $$

The gauge covariant derivative is

$$ D_\mu = \partial_\mu - iqA_\mu \tag{1} $$

and the field is given as follows:

$$ F_{\mu\nu}=[D_\mu,D_\nu]. \tag{2} $$

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My question is; what are the equivalents to equation (1) and (2) if we have an adjoint action such as this

$$ \psi(x) \to g(x)\psi(x)g^{-1}(x) $$

where $g(x)$ could be arbitrary general linear transformations for instance. Does the use of a adjoint action transformation significantly changes (1) and (2)?

Answer 1

I find that $ \partial_\mu (g\psi g^{-1}) = g \left( \partial_\mu \psi + [ g^{-1}(\partial_\mu g), \psi ] \right) g^{-1} .$ Therefore, we set $D_\mu\psi = \partial_\mu\psi - [iq A_\mu, \psi].$ Then, $$ D_\mu(g\psi g^{-1}) = g \left( \partial_\mu g - [iq g^{-1}A_\mu g-g^{-1}(\partial_\mu g),\psi] \right) g^{-1} $$ so $A_\mu$ should transform to $g^{-1}A_\mu g-\frac{1}{iq}g^{-1}(\partial_\mu g).$

I'm no expert on this, but I think that $F_{\mu\nu}=[D_\mu,D_\nu]$ is still valid.

Answer 2

Here’s an “index free” description.

Let A be a connection on a principal $G$-bundle $P$, and $\rho:G\to\text{GL}(V)$ be a representation. Then $\rho$ induces a representation (its derivative) which we also denote $\rho:\mathfrak g \to \text{End}(V)$.

If $\psi$ is a section of an associated vector bundle $E:=P\times_{G,\rho}V$, then the covariant derivative $\nabla^A$ associated to $A$ is given by $$ \nabla^A \psi = d\psi + \rho(A)\psi $$ So for a straightforward left action, $A$ will act similarly. For the Adjoint action $$ \nabla^A \psi = d\psi + [A,\psi] $$

For curvature, $$ \rho(F_{X,Y})\psi = [\nabla_X,\nabla_Y]\psi-\nabla_{[X,Y]}\psi $$ (For coordinate derivatives $\partial_\mu$ you can ignore the last term.) This implies the defining formula for $F$ is the same, but if written out in matrices, say, $F$ will act differently on fields $\psi$ of different representations.