# Lie algebra decomposition of the gluon field

It is commonly written in the literature that due to it transforming in the adjoint representation of the gauge group, a gauge field is lie algebra valued and may be decomposed as $A_{\mu} = A_{\mu}^a T^a$. For $\mathrm{SU}(3)$ the adjoint representation is 8 dimensional so objects transforming under the adjoint representation are $8 \times 1$ real Cartesian vectors and $3 \times 3$ traceless hermitean matrices via the lie group adjoint map. The latter motivates writing $A_{\mu}$ in terms of generators, $A_{\mu} = A_{\mu}^a T^a$.

My first question is, this equation is said to be valid independent of the representation of $T^a$ - but how can this be true? In some representation other than the fundamental representation, the $T^a$ will not be $3 \times 3$ hermitean traceless matrices and thus will not contain $8$ real parameters needed for transformation under the adjoint representation. But we know the gluon field transforms under the adjoint representation so what is the error in this line of reasoning which is suggestive of constraining the $T^a$ to be the Gell mann matrices?

Consider the following small computation: $$A_{\mu}^a \rightarrow A_{\mu}^a D_b^{\,\,a} \Rightarrow A_{\mu}^a t^a \rightarrow A_{\mu}^b D_b^{\,\,a}t^a$$ Now, since $Ut_bU^{-1} = D_b^{\,\,a}t^a$ we have $A_{\mu}^a t^a \rightarrow A_{\mu}^b (U t^b U^{-1}) = U A_{\mu}^b t^b U^{-1}$. The transformation law for the $A_{\mu}^a$ is in fact $A_{\mu} \rightarrow UA_{\mu}U^{-1} - i/g (\partial_{\mu} U) U^{-1}$.

1. What is the error that amounts to these two formulae not being reconciled?

2. The latter equation doesn't seem to express the fact that the gluon field transforms in the adjoint representation. I was thinking under $\mathrm{SU}(3)$ colour, since this is a global transformation, $U$ will be independent of spacetime so the derivative term goes to zero but is there a more general argument?

The Lie algebra is an object that exists independent of any representations. It carries, naturally, on it the adjoint representation, but the Lie algebra as such is agnostic about representations. When we say the gauge field is Lie algebra-valued, this statement is independent of the gauge field acting on anything. It is an axiom of gauge theory that the gauge field $A$ is a Lie algebra-valued 1-form (or, more precisely, locally such a field associated to a connection on a principal bundle, but this does not concern us here).

The Lie algebra is a vector space and as such it has a basis. We fix one such basis, call it $T^a$, and can then of course expand the 1-form as $A = A_\mu \mathrm{d}x^\mu = A^a_\mu T^a\mathrm{d}x^\mu$ (summation over all repeated indices implied). In this abstract setting, the $T^a$ are not to be thought of as matrices. While many Lie groups are matrix groups, not all are, and one should not, for clarity of the theory, impose that the objects in question are matrices at this level.

What both the group and the algebra elements have are representations as matrices, namely, for the Lie algebra, maps $\rho : \mathfrak{g}\to\mathfrak{gl}(n)$, where $\mathfrak{gl}(n)$ denotes all $n\times n$-square matrices, which are the Lie algebra of the matrix group $\mathrm{GL}(n)$. The image of such a representation map is always a subalgebra of dimension eight or less. Again, the $T^a$ themselves are not matrices, but their image under $\rho_\text{fund} : \mathfrak{g}\to \mathfrak{gl}(3)$ is (it's of course the Hermitian traceless matrices for $\mathfrak{g}\to \mathfrak{su}(3)$), and so is their image under $\rho_\text{ad}: \mathfrak{g}\to \mathfrak{gl}(\mathfrak{g}), h \mapsto \mathrm{ad}_h$ where $\mathrm{ad}_h$ is the linear map/matrix on $\mathfrak{g}$ defined by $\mathrm{ad}_h k = [h,k]$, where the bracket is the Lie bracket on $\mathfrak{g}$. In the basis $T^a$, the components of $\mathrm{ad}_{T^a}$ are precisely the structure constants $f^{ia}_j$. It is crucial to realize that these are not the components of the $\rho_\text{fund}(T^a)$ as 3-by-3 matrices in the fundamental representation. In the trivial (colourless) representation, all the $\rho_\text{trivial}(T^a)$ are just the identity! Mathematicians are usually careful about distinguishing abstract elements of the algebra with their values as matrices in specific representations, physicists much less so and rely on the reader gathering the correct meaning from context.

The error in your "computation" is that you are confusing the gauge group $G$ with the group of gauge transformations $\mathcal{G} := \{ g: M\to G\mid g \text{ smooth}\}$, where $M$ is the spacetime (it does not help that some call the latter group the gauge group and the former the global gauge group). The gauge field transforms in the adjoint of the gauge group but in a non-linear representation of the group of gauge transformations. Indeed, the reconciliation of the formulae is just that they become the same for a globally constant gauge transformation, which is just an element of the gauge group, and which is the only thing we ever claimed the gauge field transforms in the adjoint representation of.

• Thanks! Do you mean to say there is an error in my 'computation' or just an error in the conclusion I made from it? What is the difference between the gauge group and the group of gauge transformations? I thought the gauge group in question was $\mathrm{SU}(3)_C$ which is an internal symmetry group of gauge transformations of the QCD lagrangian? – CAF Jan 8 '17 at 20:51
• @CAF Using the terminology of ACuriousMind, the group of gauge transformations $\mathcal{G}$ is the group of smooth mappings $M\rightarrow G$ where $G$ is the gauge group. So $G$ and $\mathcal{G}$ are different. – coconut Jan 8 '17 at 22:22
• Ok thanks - does it mean that elements in both groups are of the form $\exp(i\theta^a t^a)$, just that in $G$ the parameters $\theta^a$ are restricted to be constants? – CAF Jan 9 '17 at 8:07
• @CAF $\mathrm{SU}(3)$ is a gauge group. When we write $\mathrm{SU}(3)_c$ we mean it's an $\mathrm{SU}(3)$ that acts on the colours, but it's still the same group. There is no shorthand notation for the group of gauge transformations (in flat space, it would be the set of all smooth maps $\mathbb{R}^4 \to \mathrm{SU}(3)$). – ACuriousMind Jan 9 '17 at 18:26
• @CAF Almost, but $\mathrm{SU}(3)_c$ really just means it's the $\mathrm{SU}(3)$ of QCD and not an $\mathrm{SU}(3)$ from some other gauge theory. $\mathrm{SU}(3)_c$ is a name for one specific copy of $\mathrm{SU}(3)$, nothing more. – ACuriousMind Jan 9 '17 at 19:08