# How are the generators of $\mathrm{SU}(3)$ represented on the gluon space?

I was watching some new lectures on QCD from Colorado and I have a few questions about what I heard:

The $\lambda^a_{ij}$ are generators of $\mathrm{SU}(3)$ in the fundamental representation so are $3 \times 3$ matrices. That is because the $ij$ indices are colour indices and they act on a $3 \times 1$ vector in colour space (the colour wavefunction of the quarks).There are 8 generators (labelled by the index '$a$') so the $\lambda$ are vectors in what the prof in the Colorado video calls 'gluon space'. This gluon space is spanned by eight independent non-trivially transforming colour octet states in a eight-dimensional real Hilbert space so each of these states can be mapped to a unit vector in the real Hilbert space, i.e. each one is a $8 \times 1$ unit basis vector.

My understanding from the video was that the representation of the $\lambda$ in colour space are the $3 \times 3$ Gell-mann matrices acting on the colour component of the quark fields embedded in the fundamental representation. What is the representation of these lambda as vectors in the gluon space and what do they act on?

In the equation $A^{\mu} = A^{\mu}_a \lambda^a/2$ are we saying that the gluons are exactly the generators of SU($3$) in the fundamental representation which give rise to non-trivial colour transformations in colour space which when acting on quark colour wavefunctions mix around the colours? This equation also tells us the gluons live in the Lie algebra of $\mathrm{SU}(3)$ since the gluon $A^{\mu}$ can be expanded in the basis of generators $T^a = \lambda^a/2$. The lie algebra is 8 dimensional but why do we say they transform under the adjoint representation of $\mathrm{SU}(3)$? I guess it makes contact with the above in that we can write down each possible gluon as a basis vector in a eight-dimensional space but what is it that they are transformed by?

The Lie algebra and the "adjoint representation" of a Lie group are, almost by definition, the same.

When we speak of "generators", we usually mean a basis of the Lie algebra, in this case denoted by $\lambda^a, a\in\{1,2,\dots,8\}$. The indices $i,j$ in $\lambda^a_{ij}$ for the Gell-Mann matrices run from 1 to 3 and $\lambda^a_{ij}$ is simply the $ij$-th entry in $\lambda^a$ *in the fundamental representation of $\mathfrak{su}(3)$ (i.e. in the representation as $3\times 3$-matrices). The quark transform in the fundamental representation so the $\lambda^a$ act on their 3d colour vectors as multiplication by these Gell-Mann matrices.

As every other gauge field, the gluon field transforms in the adjoint of the gauge group because it is Lie algebra-valued. The adjoint representation is defined by the action of the $\lambda^a$ on any other $\lambda^b$ as $[\lambda^a,\lambda^b]$, i.e. the representation map is given by the Lie bracket $\mathfrak{su}(3)\to\mathrm{End}(\mathfrak{su}(3)), \lambda^a \mapsto [\lambda^a,-]$. Since the structure constants of the Lie algebra are defined as $[\lambda^a,\lambda^b] = f^{ab}_c \lambda^c$ (with summation convention in effect), we may write the action of $\lambda^a$ on another $\lambda^b$ also as $f^{ab}_c\lambda^c$, and so a "gluon vector" $A_a\lambda^a$ transforms into $A_a f^{ab}_c \lambda^c$ when acted upon by $\lambda^b$. Therefore, if you want to write the $\lambda^b$ as a $8\times 8$ matrix, the components of that matrix are just the structure constants $f^{ib}_j$!

Note that the gluon field is Lie algebra-valued, and not inherently "in the fundamental representation" as you seem to think. When the gluon field couples to the quarks, it acts on them as the Gell-Mann matrices because the quarks are in the fundamental, but in the gluon-gluon couplings we get the structure constants/adjoint action, because a gluon field acts on another gluon field by the Lie bracket.

• Thanks for the answer! Few comments: I understand a representation to be a homomorphic map from a group $G$ to a set of operators $GL(V)$ acting on a linear vector space $V$. How is this definition consistent with the map you gave by the lie bracket? Is there a typo in $A_a f^{ab}_c \lambda^a$? Lastly, I was wondering if you could provide me a reference where I could find the comments you have written about the gluons - the content of much of the third paragraph there is new to me and I have not seen $\text{End}$ used in this context before. Thanks again! – CAF Jan 4 '17 at 14:43
• @CAF Note that I am giving the representation map $\mathfrak{g}\to\mathrm{End}(V)$ of the Lie algebra. There is a corresponding representation map of Lie groups $G\to\mathrm{GL}(V)$ (the Lie algebra of $\mathrm{GL}(V)$ is $\mathrm{End}(V)$). Yes, there was a typo. As a reference, any introduction to non-Abelian Yang-Mills theory should discuss these points, albeit not in the exact words I chose here. For example, see chapters 6 and 9 here. – ACuriousMind Jan 4 '17 at 14:48
• Ok thanks. I guess I should have asked for a reference where this map involving $\text{End}$ is discussed in the context of the adjoint action. Could you provide one? e.g I am unfamiliar with the notation $[\lambda^a, -]$ etc – CAF Jan 4 '17 at 16:19
• @CAF I suspect you are overthinking this: $\mathrm{End}$ simply denotes the endomorphisms of $V$ - which are, in plainer language, for an n-dimensional $V$ just all square matrices of dimension $n$. $[\lambda^a,-]$ is intended to denote the linear map that's given by taking the Lie bracket with $\lambda^a$, it is also often written $\mathrm{ad}_{\lambda^a}$, cf. e.g. the Wikipedia article on adjoint representations. – ACuriousMind Jan 4 '17 at 16:29
• The adjoint action at the level of the lie group I guess takes the form $\mathrm{Ad}_g: \mathrm{SU}(3) \rightarrow \mathrm{GL}(8)$ with $x \mapsto g x g^{-1}$. Given $g = 1+ \epsilon \lambda_a, x = 1 + \epsilon \lambda_b$, to first order in $\epsilon$, this mapping yields just $x$, that is $gxg^{-1} = x.$ What does this mean? – CAF Jan 7 '17 at 11:53