Lie algebra homomorphism

A Lie algebra homomorphism is a linear map $f$ satisfying $f([x,y])=[f(x),f(y)]$.

A representation of $\mathfrak{g}$ on a vector space $V$ is a Lie algebra homomorphism $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$.

For a basis $(T_a)$ of a Lie algebra, its structure constants are defined by $[T_a,T_b]=f_{ab}{}^c{T}_c$.

A Lie algebra representation is faithful when its representing homomorphism is injective.

A nonzero Lie algebra representation is irreducible when it has no proper nonzero invariant subspace.

The adjoint representation is ${\mathrm{ad}}_x(y)=[x,y]$. For a semisimple Lie algebra, its nonzero weights are the roots and its zero-weight space is the Cartan subalgebra.

The Killing form isIt is symmetric and invariant: $\kappa ([X,Y],Z)=\kappa (X,[Y,Z])$.

The radical ${\mathfrak{g}}^{\perp }=\{x:\kappa (x,\mathfrak{g})=0\}$ of the Killing form is a solvable ideal. Invariance makes it an ideal, and the Cartan solvability criterion applied to its adjoint image proves solvability.

Every invariant bilinear form on a finite-dimensional complex simple Lie algebra is a scalar multiple of its Killing form. A nondegenerate invariant form identifies the algebra with its dual; comparing this identification with the Killing form gives an endomorphism of the irreducible adjoint representation, so Schur lemma makes it scalar.

For ${\mathfrak{sl}}_n(\mathbb{C})$,

Every invariant subspace of the Adjoint representation of a Lie algebra is an ideal. Hence the adjoint representation of a Simple Lie algebra is irreducible.

For a finite-dimensional Lie algebra representation $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$, its trace form is the symmetric bilinear formIt is invariant: $([x,y],z{)}_V=(x,[y,z]{)}_V$.

For a finite-dimensional representation $d$, the trilinear form $B(X,Y,Z)=\mathrm{Tr}(d(X)d(Y)d(Z))$ is invariant under simultaneous adjoint action. This follows by writing the sum of its three infinitesimal variations as the trace of a commutator.