Lol I think you became too used to Unitary and orthogonal groups. Joigus
Would it help to know SO(3.1) and SU(n) are both subgroups of SL(2,c)/Z_2 ?
lets start with the following
\[sl(2,\mathbb{C})=su(2)\oplus isu(2)\]
generators denoted e,f,h
[e,f=h]
[h,e]=2e
[h,f]=-2f
the 2C is the linear combination of e,f,h
\[\pi (h)=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\]
\[\pi( e)=\begin{pmatrix}0&1\\0&0\end{pmatrix}\]
\[\pi h=\begin{pmatrix}0&0\\-1&0\end{pmatrix}\]
\[f_i,h_i,e_i\] i=1,2,3....r
however the set of complex cannot all commute so you need commutations
\[[h_ih_j]=0\]
\[[h_i,e_j]=A_{ji}e_j\]
\[h_i,f_i]=-A_{ji}f_j\]
\[[e_i,f_j]=\delta_{ij}h_{ij}\]
where \(A_{ij} \) is the Cartan matrix ( I won't go through the ladder operators as they are fairly lengthy) however it can be expressed as
\[[h_i,e_i]=\langle\alpha_j\rangle=\frac{2}{\langle\alpha_j,\alpha_j}\langle\alpha_j,\alpha_i\rangle_j=A_{ji}e_j\]
\[\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\]
the above is for SL(2C) for sl(3,C) the Cartan matrix is an 8 dimensional algebra of rank 2 which means it has a 2 dimensional Cartan sub algebra given as follows
\[\pi(t_1)= \begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}\]
\[\pi(t_2)= \begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}\]
\[\pi(t_3)= \begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\]
\[\pi(t_4)= \begin{pmatrix}0&0&1\\0&-1&0\\0&0&0\end{pmatrix}\]
\[\pi(t_5) =\begin{pmatrix}0&0&-i\\0&0&0\\i&0&0\end{pmatrix}\]
\[\pi(t_6)= \begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}\]
\[\pi(t_7)= \begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix}\]
\[\pi(t_1)=\frac{1}{\sqrt{3}} \begin{pmatrix}0&1&0\\1&0&0\\0&0&-2\end{pmatrix}\]
You may note the last is the Gell-Mann matrices
if we take the commutator between \(\pi(t_1)\) and \(\pi(t_2)\) we get \([\pi(t_1),\pi(t_2)]=2i\pi(t_3)\) which is familiar in the su(2) algebra. Thus we can define the following
\[x_1=\frac{1}{2}t_1\]
\[x_2=\frac{1}{2}t_1\]
\[x_3=\frac{1}{2}t_3\]
\[y_4=\frac{1}{2}t_4\]
\[y_5=\frac{1}{2}t_5\]
\[z_6=\frac{1}{2}t_6\]
\[z_7=\frac{1}{2}t_7\]
\[z_8=\frac{1}{\sqrt{3}}t_8\]
with change in basis
\[e_1=x_1+ix_2\]
\[e_2=y_4+iy_5\]
\[e_3=z_6+iz_7\]
\[f_1=x_1+ix_2\]
\[f_2=y_4+iy_5\]
\[f_3=z_6+iz_7\]
Now I should inform everyone that the basis and coordinates I am describing apply to Dynken diagrams and what I am describing apply to the root diagrams...
https://en.wikipedia.org/wiki/Dynkin_diagram
the basis above in matrix form is
\[\pi(e_1)=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}\]
\[\pi(e_2)=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\]
\[\pi(e_1)=\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}\]
\[\pi(f_1)=\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix}\]
\[\pi(f_2)=\begin{pmatrix}0&0&0\\0&0&0\\1&0&0\end{pmatrix}\]
\[\pi(f_3)=\begin{pmatrix}0&0&0\\0&0&0\\0&1&0\end{pmatrix}\]
\[\pi(x_3)=\frac{1}{2}\begin{pmatrix}1&1&0\\0&-1&0\\0&0&0\end{pmatrix}\]
\[\pi(z_8)=\frac{1}{3}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}\]
@joigus That should help better understand the special linear group of Real as well as complex. Now knowing the above applies to Dynken diagrams will also help better understand the validity of the OPs link as well as the methodology.
@TheoM Hope this answers your question as well on the validity behind the LCT's and where they are applied in particle physics so yes the link overall you provided is valid