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 submoduleN
of a Lie moduleM
, we define its orthogonal complement with respect to a non-degenerate invariant bilinear formΦ
as the Lie ideal of elementsx
such thatΦ n x = 0
for 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 = let __spread.0 := Φ.orthogonal ↑N; { toSubmodule := __spread.0, lie_mem := ⋯ }
Instances For
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⁆
.