# Solving the Lie algebra of generators: path from algebra to matrix representation

Given the Lie algebra, what is the systematic way to construct the matrix representation of the generators of the desired dimension? I ask this question here because it is the physicists for whom representation of groups is more important than mathematicians.

Let us, for example, take $SU(2)$ for concreteness. Starting from the generic parametrization of a $3\times 3$ unitary matrix $U$ with $\det U=1$, and using the formula of generators $$J^i=-i\Big(\frac{\partial U}{\partial \theta_i}\Big)_{\{\theta_i=0\}}$$ one can find the $3\times 3$ matrix representation of the generators.

However, I'm looking for something else.

1. Given the Lie algebra $[J^i, J^j]=i\epsilon^{ijk}J^k$, is there a way that one can explicitly construct (not by guess or trial) the $3\times 3$ representations of $\{J^i\}$?

2. Will the same procedure apply to solve other Lie algebras appearing in physics such as that of $SO(3,1)$ (or $SL(2,\mathbb{C})$)?

• $(t^\mathrm{adj}_i)^j{}_k=i\epsilon^{ijk}$. – AccidentalFourierTransform Jan 1 '18 at 15:41
• Probably you mean SU(3) instead of SU(2). And if you are interested in the general procedure (rather than the adjoint representation mentioned by @AccidentalFourierTransform), there are even software packages such as LiE and the group theory part of SUSYNo. – marmot Jan 1 '18 at 15:48

If you have the structure constants, i.e. the coefficients $f_{ab}^c$ in the commutation relations $[T_a,T_b]=\sum_c f_{ab}^cT_c$ with $a,b,c\in \{1,\ldots p\}$ then you can construct $p$ matrices $M_a$ (labelled by $a$) of size $p\times p$ with entries $(M_a)_{cb}$. These matrices will be a $p\times p$ representation of the algebra (in fact, the adjoint representation.)

This will work for every algebra. However, in the case of non-compact algebras, the resulting representation is obviously finite dimensional and thus cannot be make hermitian (i.e. it cannot exponentiate to a unitary, finite dimensional representation of the group.)

Be aware that in the case of $so(3,1)$, there is a subtle point that comes in going from the complex back to the real form. Over $\mathbb{C}$, the adjoint of $so(3,1)$ is reducible into $su(2)\oplus su(2)$ but over the reals the adjoint is irreducible. In other words, for non-compact real forms, there are issues with reducibility and unitary.

More generally, given the adjoint representation

$${\rm ad}(x)y ~:=~ [x,y], \qquad x,y~\in~L, \tag{1}$$

of an $n$-dimensional Lie algebra $L$ over a field $\mathbb{F}$, with a basis of generators

$$(t_1, \ldots, t_n ),\tag{2}$$

satisfying

$$[t_i,t_j]~=~\sum_{k=1}^n t_k~f^k{}_{ij} ,\qquad i,j~\in~\{1,\ldots, n\},\tag{3}$$

where

$$f^k{}_{ij}~\in~ \mathbb{F},\qquad i,j,k~\in~\{1,\ldots, n\},\tag{4}$$

are structure constants, then an $n\times n$ matrix representation of the adjoint representation

$${\rm ad}: L ~~\to~~ {\rm End}(L)\tag{5}$$

is given by the structure constants

$${\rm ad}(t_i)^k{}_j~=~f^k{}_{ij},\qquad i,j,k~\in~\{1,\ldots, n\}.\tag{6}$$

Proof of eq. (6): If a Lie algebra element

$$x~=~\sum_{j=1}^n t_j~x^j~\in ~L \tag{7}$$

has components

$$x^j~\in~\mathbb{F}, \qquad j~\in~\{1,\ldots, n\},\tag{8}$$

then for fixed $i \in~\{1,\ldots, n\}$, the endomorphism $${\rm ad}(t_i)~\in~{\rm End}(L),\tag{9}$$ which takes Lie algebra elements in Lie algebra elements $$x ~~\mapsto ~~\sum_{j=1}^n ~t_k~x^{\prime k} ~=~x^{\prime}~:=~{\rm ad}(t_i)x ~\stackrel{(1)}{=}~[t_i,x] ~\stackrel{(7)}{=}~\sum_{j=1}^n [t_i,t_j]~x^j ~\stackrel{(3)}{=}~\sum_{j,k=1}^n t_k~ f^k{}_{ij}~x^j, \tag{10}$$ is represented by the matrix $$\left({\rm ad}(t_i)^k{}_j\right)_{1\leq j,k\leq n},\tag{11}$$ such that the components transform as $$x^{\prime k}~=~\sum_{j=1}^n {\rm ad}(t_i)^k{}_j~x^j , \qquad k~\in~\{1,\ldots, n\}.\tag{12}$$ Finally combine eqs. (10) & (12) to derive eq. (6). $\Box$