differential geometry - Covariant derivative applied to a vector vs. applied to a matrix?

I know there are (say) two different definitions/representations of the covariant derivative:

• one is the covariant derivative applied to a vector $F$, which reads as

$$DF=\partial F+iAF$$

(adapted to the coordinate system of course, $D\rightarrow D_{\mu}$ and so on)

• whereas an other one applies to a matrix $F$ and reads as

$$DF=\partial F+i\left[A,F\right]$$

where this time a commutator appears.

In both case $A$ is the gauge-potential, whereas the $F$ are not the same of course.

I wonder if there is a link between these two representations, and what kind of use they have. I know about the first representation (could someone tell me if the name "representation" is even well adapted?) but clearly the second one does not lead to the usual definition of the gauge field, say $B_{\alpha\beta}=\left[D_{\alpha},D_{\beta}\right]$, so I'm pretty much struggling to understand if it has a geometric interpretation. Most of all: What are the different names of these two representations (if one can use this terminology)?

Good references about that are warmly welcomed.

Answer

(I am dropping the bothersome factors of $\mathrm{i}$ in this answer, they contribute nothing to understanding what is going on)

The gauge covariant derivative exists for all forms on the spacetime manifold $\mathcal{M}$ taking value in a representation of the gauge group. (Formally, these are sections of associated vector bundles to the gauge principal bundle)

Given a representation $\rho : G \to \mathrm{GL}(V_\rho)$ with $V_\rho$ some vector space, $V_\rho$-valued $n$-forms are elements of $\Omega^n(\mathcal{M}) \otimes V_\rho$. The gauge covariant derivative associated to a (Lie-algebra valued) gauge field $A$ is

$$ \mathrm{d}_A = \mathrm{d} + \mathrm{d}\rho(A)$$

where $\mathrm{d}$ is the ordinary exterior derivative and $\mathrm{d}\rho$ is the induced representation of the Lie algebra.

Therefore, the explicit form of the gauge covariant derivative depends on the representation $\rho$.

For your "matrix", it is implicit that the matrix is actually taking values in the Lie algebra, i.e. the adjoint representation, for which $\mathrm{d}\rho(A)X = [A,X]$ (essentially a consequence of $\rho(g)X = gXg^{-1}$ and the Baker-Campbell-Hausdorff formula).

For your "vector", it is implied that it transforms in the fundamental representation, where, if we initially present $G$ as a matrix group, the representation is actually the identity, leading to $\mathrm{d}\rho(A)X = AX$ (Caveat: Not all Lie group are matrix groups, but we physicists don't care, ours always are)