Lie Algebras

We create the Lie algebra as a structure constant algebra. First, we construct from the full matrix algebra and get as the derived algebra of .

> gl3 := LieAlgebra(Algebra(MatrixRing(Rationals(), 3)));
> sl3 := gl3 * gl3;
> sl3;
Lie Algebra of dimension 8 with base ring Rational Field

Let's see how the first basis element acts.

> for i in [1..8] do
>     print sl3.i * sl3.1;
> end for;
(0 0 0 0 0 0 0 0)
( 0 -1  0  0  0  0  0  0)
( 0  0 -2  0  0  0  0  0)
(0 0 0 1 0 0 0 0)
(0 0 0 0 0 0 0 0)
( 0  0  0  0  0 -1  0  0)
(0 0 0 0 0 0 2 0)
(0 0 0 0 0 0 0 1)

Since it acts diagonally, this element lies in a Cartan subalgebra. The next candidate seems to be the fifth basis element.

> for i in [1..8] do
>     print sl3.i * sl3.5;
> end for;
(0 0 0 0 0 0 0 0)
(0 1 0 0 0 0 0 0)
( 0  0 -1  0  0  0  0  0)
( 0  0  0 -1  0  0  0  0)
(0 0 0 0 0 0 0 0)
( 0  0  0  0  0 -2  0  0)
(0 0 0 0 0 0 1 0)
(0 0 0 0 0 0 0 2)

This also acts diagonally and commutes with sl3.1, hence we have luckily found a full Cartan algebra in . We can now easily work out the root system. Obviously the root spaces correspond to the pairs (sl3.2, sl3.4), (sl3.3, sl3.7) and (sl3.6, sl3.8). The product of a positive root with its negative should lie in the Cartan algebra.

> sl3.2*sl3.4;
( 1  0  0  0 -1  0  0  0)
> sl3.3*sl3.7;
(1 0 0 0 0 0 0 0)
> sl3.6*sl3.8;
(0 0 0 0 1 0 0 0)

Clearly some choices have to be made and we fix sl3.3 as the element corresponding to the first fundamental root , sl3.7 as and get sl3.1 as . For the other fundamental root we have to find an element such that is non-zero.

> sl3.3*sl3.2;
(0 0 0 0 0 0 0 0)
> sl3.3*sl3.4;
( 0  0  0  0  0 -1  0  0)
> sl3.3*sl3.6;
(0 0 0 0 0 0 0 0)
> sl3.3*sl3.8;
(0 1 0 0 0 0 0 0)

We choose sl3.8 as , sl3.6 as and consequently -sl3.5 as . This now determines to be sl3.2 and to be sl3.4.

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