# Why do we require the generators of $\mathrm{SU(N)}$ gauge theories to be $N \times N$ matrices?

I have often read that the generators for $\mathrm{SU(N)}$ gauge theories must be $N \times N$ matrices; see for instance these notes at the top of page 3: http://www.staff.science.uu.nl/~wit00103/ftip/Ch12.pdf‎. Why is this?

I don't think this is necessary from a mathematical point of view. For instance, for $\mathrm{SU(2)}$ we can consider the $2 \times 2$ generators: $$\begin{array}{ccc} \displaystyle J_{1/2}^1=\frac{1}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \; ,& \displaystyle J_{1/2}^2=\frac{1}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \; ,& \displaystyle J_{1/2}^3=\frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{array}$$ However, also the following $3 \times 3$ generators satisfy the Lie algebra: $$\begin{array}{ccc} \displaystyle J_{1}^1=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \; ,& \displaystyle J_{1}^2=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} \; ,& \displaystyle J_{1}^3=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \end{array}$$

You should also specify the Representation.

The Representation requires SU(N) Lie group with N×N matrix is called Fundamental Representation. Which is used in Standard model U(1) x SU(2) x SU(3).

You can surely have SU(N) Lie groups with other Representation, such as Adjoint Representation, then in this case SU(N) are represented by a matrix with a rank of the number of its group element $d(G)$, which is rank-$(N^2-1)$. i.e. $(N^2-1)$ x $(N^2-1)$ matrix. In this case, Adjoint Representation for a Lie group G is a way of representing the elements of the group as linear transformations of the group's Lie algebra. So the generator $(T^a)_{bc}=-if_{ab}{}^c$, takes its value as the structure constant. (here $[T^a,T^b]=i f_{ab}{}^c T^c$.)

So there is no puzzle at all.

• Thanks for your reply. So if I understand it correctly, we have found a model (i.e. the Standard Model) which "happens" to work very well if we use the fundamental representation? There is no specific reason to use the fundamental representation, except that it "happens" that nature seems to work that way? – Hunter Mar 9 '14 at 2:44
• Hi Hunter, I don't think particle physics provides a deep reason. But, we know there is a shallow reason to have rank-$N$ matrix, for the reason that the color charge is 3 for SU(3) QCD. – wonderich Mar 9 '14 at 2:53
• btw. I know in condensed matter viewpoint, there are topological reason why there are 3 generations of 16 Weyl fermions and 3 generations of neutrinos. It is from an argument like Kitaev's 1D chain, generalized to a fluctuating 1D string in 3+1D. – wonderich Mar 9 '14 at 2:54
• Ok, thanks, I guess I was hoping there was some deeper reason. – Hunter Mar 9 '14 at 2:57
• I think there is, it is your chance to become a great physicist sorting this deep reason out, let me know your progress. :-) – wonderich Mar 9 '14 at 2:58

Every Lie group has a set of generators, and typically a group element is found by exponentiate (linear combinations) of these generators.

Since the fundamental definition of say $SU(N)$ [similarly $SO(N)$] is something like

The group of unitary (orthogonal), $n$ by $n$ matrices with unit determinant

then, the fundamental representation is given by $n$ by $n$ matrices.

However, the group IS generated by the algebra. Therefore, any set of "matrices" satisfying the commutation relations that defines the algebra generates a new (probably inequivalent) representation of the group.

## Examples

Trivial representation: If all generators are zero, $T^a = 0, \; \forall \; a$, every Lie algebra is satisfied. And the group representation is the scalar unit $1$. This is called the trivial representation.

Adjoint representation: A representation of dimension equal to the dimension of the group, i.e., $n$ by $n$ matrices where $n=\mathrm{dim}(G)$, can be defined by identifying the generators with the structure constants of the group (as pointed out by @Idear in other response of this question, $$(T^a)_{bc} = -\imath f_{bc}{}^a.$$

Additionally, one can "create" more representations of the group by multiplying the known representations (via the tensor product) or by adding them (via the direct sum). Nonetheless, the result of this operation are not necessary irreducible representations (or irreps), but that is subject for another post!

;-)