Lie algebras with non-degenerate invariant bilinear forms are semisimple #
In this file we prove that a finite-dimensional Lie algebra over a field is semisimple
if it does not have non-trivial abelian ideals and it admits a
non-degenerate reflexive invariant bilinear form.
Here a form is invariant if it invariant under the Lie bracket
in the sense that ⁅x, Φ⁆ = 0 for all x or equivalently, Φ ⁅x, y⁆ z = Φ x ⁅y, z⁆.
Main results #
LieAlgebra.InvariantForm.orthogonal: given a Lie submoduleNof a Lie moduleM, we define its orthogonal complement with respect to a non-degenerate invariant bilinear formΦas the Lie ideal of elementsxsuch thatΦ n x = 0for alln ∈ N.LieAlgebra.InvariantForm.isSemisimple_of_nondegenerate: the main result of this file; a finite-dimensional Lie algebra over a field is semisimple if it does not have non-trivial abelian ideals and admits a non-degenerate invariant reflexive bilinear form.
References #
We follow the short and excellent paper [dieudonne1953].
A bilinear form on a Lie module M of a Lie algebra L is invariant if
for all x : L and y z : M the condition Φ ⁅x, y⁆ z = -Φ y ⁅x, z⁆ holds.
Instances For
The orthogonal complement of a Lie submodule N with respect to an invariant bilinear form Φ is
the Lie submodule of elements y such that Φ x y = 0 for all x ∈ N.
Equations
- LieAlgebra.InvariantForm.orthogonal Φ hΦ_inv N = { toSubmodule := Φ.orthogonal ↑N, lie_mem := ⋯ }
Instances For
Alias of LieAlgebra.InvariantForm.orthogonal_isCompl_toSubmodule.
A finite-dimensional Lie algebra over a field is semisimple
if it does not have non-trivial abelian ideals and it admits a
non-degenerate reflexive invariant bilinear form.
Here a form is invariant if it is compatible with the Lie bracket: Φ ⁅x, y⁆ z = Φ x ⁅y, z⁆.