Structure constants

Previously on Visual Lie Theory it was mentioned, that the commutator of two basis elements of a Lie algebra $\cal{L}$ is again an element of $\cal{L}$. I.e., it can be written as a linear combination of the basis vectors $t_k, k=1,\ldots,d$; i.e.,
$$[t_i, t_j] = \sum_{k=1}^d f^k_{ij}\,t_k$$ From the literature I learn, that the parameters $f^k_{ij}$ ("structure constants") (almost?) completely describe the Lie algebra $\cal{L}$.

What's interesting about the structure constants is, that once the $f^k_{ij}$ are known, one immediately can write down a $d$-dimensional matrix representation of $\cal{L}$, the adjoint representation. Concretely, the adjoint representation of $t_i$, the $i$-th generator, is given by
$$(T_i)_{k,j}  =  f^k_{ij}$$ where $j = 1,\ldots,d$ and $k = 1,\ldots,d$ denote the row and column index of the matrix $T_i$.

Clearly,  the structure constants depend on the chosen basis vectors $t_k, k=1,\ldots,d$. Viewed from another basis $$t_k \rightarrow \tilde{t}_k \equiv \sum_{l=1}^d a^l_k\,t_l$$ the structure constants will change as well;
$$[\tilde{t}_i, \tilde{t}_j] = \sum_{k=1}^d \tilde{f}^k_{ij}\,\tilde{t}_k$$ It turns out, that there is a particular basis, known as the Chevalley basis, in which the $f^k_{ij}$ are integers and assume only values between $-3$ and $+3$; in addition, the sum in $$[t_i, t_j] = \sum_{k=1}^d f^k_{ij}\,t_k$$ contains at most one non-zero element.

In general, the generators $t_i, i=1,\ldots,d$ can be divided in $d-r$ generators $e_\alpha$ corresponding to non-zero roots and $r$ generators $h_\alpha$ corresponding to zero roots; $d$ being the dimension and $r$ being the rank of $\cal{L}$. Here, the label $i$ of the generator $t_i$ is replaced by the root $\alpha$, which we may regard as a $r$-dimensional index of $e_\alpha$ and $h_\alpha$. For the generators $e_\alpha$ we find
$$[e_{\alpha_i}, e_{\alpha_j}] = K_{ij}\,e_{\alpha_i+\alpha_j}$$ where $K_{ij}$ is non-zero if the sum $\alpha_i+\alpha_j$ is a non-zero root. If $\alpha_i+\alpha_j$ is not a root, the commutator vanishes ($K_{ij}=0$). If $\alpha_i+\alpha_j$ is a zero-root (i.e. $\alpha_j = -\alpha_i$) the commutator is
$$[e_{\alpha_i}, e_{-\alpha_i}] = h_{\alpha_i}$$ where $h_{\alpha_i}$ is an element of the Cartan subalgebra. All generators contained within the Cartan subalgebra commute, i.e.
$$[h_{\alpha_i}, h_{\alpha_j}] = 0$$ . Furthermore,
$$[h_{\alpha_i}, e_{\alpha_j}] = N_{ij}\,e_{\alpha_j}$$ with $N_{ij}$ being the entries of the Cartan matrix.

Since all information on $\cal{L}$ supposedly is encoded in the $f^k_{ij}$, naively I had expected that there existed tables of  $f^k_{ij}$ (or $K_{ij}$) for all (classical and exceptional) Lie algebras. My web search, however, turned up surprisingly little - perhaps I looked at all the wrong places.

So how does one (or a computer) then actually calculate the $f^k_{ij}$? The following two papers I found most useful for finding an answer :

1. V. K. Agrawala and Johan G. Belinfante: Weight diagrams for lie group representations: A computer implementation of Freudenthal's algorithm in ALGOL and FORTRAN, BIT (1969) 9(4):301-314. doi: 10.1007/BF01935862
2. R.B Howlett, L.J Rylands, D.E Taylor: Matrix Generators for Exceptional Groups of Lie Type, Journal of Symbolic Computation, volume 31, issue 4, April 2001, pages 429-445, doi: 10.1006/jsco.2000.0431

Based on algorithms and source code contained therein I wrote some MATLAB programs (the only computer language I speak somewhat fluently) for the calculation of the coefficients $K_{ij}$; these tools ("LieTools") are available for download here. (At this point Octave is not (yet) supported due to Octave's missing implementation of nested functions.)

Fig. 1: Structure constants $K_{ij}$ for the Lie algebra F4 (52 dimensions, i.e. 52-4=48 non-zero roots)

As an example figure 1 shows a graphical representation of (what "LieTools" thinks are) the coefficients $K_{ij}$ for the exceptional Lie algebra F4. In the Chevalley basis the $K_{ij}$ assume one of five possible value, $-2,-1,0,+1,+2$, here plotted in 5 different colours. (You've probably noticed that I used part of this figure for the blog header.)