Softwares to determine semi-simple types of Lie algebras generated over $\mathbb{R}$ or $\mathbb{C}$ by a set of matrices

Question

I wish to determine the type of Lie algebra generated over $\mathbb{R}$ or $\mathbb{C}$ by a set of square matrices with irrational elements. I've been using GAP and it is quite good at determining the type of Lie algebras over $\mathbb{Q}$, which seems to work fine when all the matrix elements are rational numbers. However, when the matrices contain irrational numbers like $\sqrt{2}$ or $\sqrt{3}$, although I can manage to get a Lie algebra over cyclotomic numbers, the function SemiSimpleType does not work with Lie algebras over cyclotomic numbers. Therefore I wish to know:

Is there a way of doing this with GAP? If not, are there any other softwares which can do this?

Answer 1

What I would do is to rationalize the matrices -- increase the dimension but get rational matrices.

gap> mats:=[nplus,nminus,n3];;
gap> B:=Basis(CF(24));
CanonicalBasis( CF(24) )
gap> new:=List(mats,x->BlownUpMat(B,x));;
gap> Length(new[1]);
32

So now you have $32\times32$ matrices, but the rational algebras spanned by them are isomorphic, and there. is no ambiguity what the rational algebra spanned by the new matrices should be:

gap> a:=LieAlgebra(Rationals,new);
<Lie algebra over Rationals, with 3 generators>
gap> SemiSimpleType(a);
"A1 A1 A1 A1 A1 A1 A1 A1"

You could do a rational base change here to expose the structure of eight copies of $A_1$. If you reduce this base change back down to $Q(\sqrt{2},\sqrt{3})$ you would get the same decomposition there (or over cyclotomics), thus it is the same.

Added in response to the comment:

Why is the R-structure the same as the Q-structure: The R-algebra contains the Q-algebra, we just extend the set of coefficients, but it does not change the multiplicative structure (multiplication of basis elements) of the algebra. A direct sum decomposition of the Q-algebra thus extends to a decomposition of the R-algebra, and any Q-basis of the Q-algebra will be am R-basis of the R-algebra. Now consider one summand: If it is $A_1$, it means we can find a Q-basis, such that the basis vectors multiply in a certain pattern. This will extend to R, and thus the R-type is the same.