bayakiv said:
The corresponding Lie algebra is generated by 28 rotations of an 8-dimensional Euclidean space and 6 pseudorotations of a doublet of Minkowski spaces.
It is not true. Really, using paired rotations, one can get Lie algebras ##sl_n (\mathbb{C})##. In fact, let
$$\begin{equation}
\begin{split}
& I_{ij} = \left(1_{2i-1, \,2j-1} - 1_{2j-1, \,2i-1}\right) + \left(1_{2i, \,2j} - 1_{2j, \,2i}\right)\\
& J_{ij} = \left(1_{2i-1, \,2j-1} + 1_{2j-1, \,2i-1}\right) + \left(1_{2i, \,2j} + 1_{2j, \,2i}\right) \\
\end{split}
\end{equation}$$
where ##i<j## and ##i,j = 1,\ldots,n##, and
$$\begin{equation}
D_{ii} = \left(1_{2i-1, \,2i-1} - 1_{2n-1, \,2n-1}\right) + \left(1_{2i, \,2i} - 1_{2n, \,2n}\right)
\end{equation}$$
where ##i = 1,\ldots,n-1##, and
$$\begin{equation}
I = \sum\limits_{1}^{n}\left(1_{2i-1, \,2i}-1_{2i, \,2i-1}\right)
\end{equation}$$
Then the set ##\left\{I_{ij},J_{ij},D_{ii},II_{ij},IJ_{ij},ID_{ii}\right\}## is linearly independent basis for the algebra ##sl_n(\mathbb{C})## and the set ##\left\{I_{ij},IJ_{ij},ID_{ii}\right\}## forms a basis of the algebra ##su(n)##, and ##\left\{sl_n(\mathbb{C})\right\} = \left\{su(n)\right\} + I\left\{su(n)\right\}##. The Lie algebra ##sl_n (\mathbb{C})## is implemented as the proper motions of the torus ##T^{n}=S^1 \times\cdots\times S^1## over Villarso circles (due to paired rotations of the torus in a ##2n## - dimensional Euclidean space) and as the motions of this torus over the surface of the hypersphere of a ##2n##-dimensional space with a neutral metric.
However, it seems that the isomorphic ##sl_4 (\mathbb{C})## Clifford algebra can also be realized as geometric algebras of the doublet (a direct sum) of Minkowski spaces with signatures (1,3) and (3,1). To confirm it is enough to refer to the Dirac representation.
