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.


2 Answers 2


Actually, they are one and the same thing.

Before I delve into the question you asked, let me quickly describe a closely related analogy - the covariant derivative in GR. This is a quantity $\nabla_\mu$ that acts differently on different objects. In particular $$ \nabla_\mu \phi = \partial_\mu \phi,~~~ (\nabla_\mu V)_\nu = \partial_\mu V_\nu - \Gamma^\lambda_{\mu\nu} V_\lambda,~~~ \cdots $$ Here, $\phi$ is a scalar and $V$ is a vector, and so on.

Exactly as above, the covariant derivative on a gauge group acts differently depending on the representation of the field it acts on (incidently the same is true in GR, since $\phi$, $V$, etc. are classified as scalars and vectors under representations of $GL(n,{\mathbb R})$ which is the diffeomorphism group). For instance, if the field $F$ is a vector in representation $R_k$ of the gauge group $G$, then the covariant derivative acts as $$ (D_\mu F)_i = \partial_\mu F_i - i A_\mu^a (T^a_k)_{ij} F_j ~~~~~~~...... (1) $$ where $T^a_k$ is the generator of the Lie algebra ${\mathfrak g}$ corresponding to the representation $R_k$.

In particular, when one talks about a "matrix" in a representation one is really talking about a quantity in the adjoint representation. The adjoint representation is special because it can be described as a vector in the adjoint representation or as a matrix in any other representation $R_k$ as follows - A field in the adjoint representation has index $a$, i.e. $F^a$ (think gauge field). In this notation, it is a vector. However, it can analogously be described as a matrix as $F_{ij} = F^a (T^a_k)_{ij}$ in a representation $R_k$.

Now, let us apply our earlier definition of covariant derivative on the vector $F^a$ and see what it means for the matrix $F_{ij}$. First, since $F^a$ is a vector in the adjoint $$ (D_\mu F)^a = \partial_\mu F^a - i A_\mu^b (T_{adj}^b)^{ac} F^c $$ But, the generators in the adjoint representation are given by $$ (T_{adj}^b)^{ac} = - i f^{bac} $$ which implies $$ (D_\mu F)^a = \partial_\mu F^a - A_\mu^b f^{bac} F^c = \partial_\mu F^a + A_\mu^b F^c f^{bca} $$ Let us now go to the matrix notation by first contracting the above equation with $T^a_k$ $$ (D_\mu F)_{ij} = \partial_\mu F_{ij} + A_\mu^b F^c f^{bca} (T^a_k)_{ij} $$ Now, we use the Lie algebra to write $$ f^{bca} (T^a_k)_{ij} = - i [ T^b_k , T^c_k ] _{ij} $$ which implies $$ (D_\mu F)_{ij} = \partial_\mu F_{ij} - i A_\mu^b F^c [ T^b_k , T^c_k ] _{ij} = \partial_\mu F_{ij} - i [ A_\mu , F ]_{ij} $$ or in simple matrix notation $$ D_\mu F = \partial_\mu F - i [ A_\mu , F] $$

Thus, we see here that the covariant derivative on a matrix in a representation $R_k$ is really the covariant derivative on a vector in the adjoint representation whose action is universally given by (1). They are therefore, one and the same thing.


(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)

  • $\begingroup$ Thanks a lot for this clear answer. Just two remarks: $\Omega$ is not defined (I guess it is the gauge principal bundle, but I might well be wrong). Also, what is the terminology ? Does one say: the fundamental/adjoint representation of the gauge-covariant derivative, or the gauge-covariant derivative of the fundamental/adjoint representation of $F$ ? Thanks again. $\endgroup$
    – FraSchelle
    Nov 3, 2014 at 15:27
  • 1
    $\begingroup$ @FraSchelle: $\Omega^n(\mathcal{M})$ is just the ordinary space of $\mathbb{R}$-valued n-forms on $\mathcal{M}$ (I use that notation because it is quite common). The representation is on the space $F$ lives in ($V_\rho$ in my notation) of the gauge group, and has, as such, little to do with the covariant derivative itself - the representation dictates how every group/algebra element acts upon the elements of the vector space it is on. $\endgroup$
    – ACuriousMind
    Nov 3, 2014 at 15:38

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