Adjoint representation of a Lie algebra

Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over $\mathbb{F}\in \{\mathbb{R},\mathbb{C}\}$. For $X\in \mathfrak{g}$, its adjoint endomorphism is

${\mathrm{ad}}_X\in {\mathrm{End}}_{\mathbb{F}}(\mathfrak{g}),{\mathrm{ad}}_X(Y)=[X,Y]\mathrm{.}$

Bilinearity of the bracket makes ${\mathrm{ad}}_X$ linear in $Y$ and makes the assignment $X\mapsto {\mathrm{ad}}_X$ linear in $X$. Moreover, the Jacobi identity in Finite-dimensional Lie algebra gives, for every $Z\in \mathfrak{g}$,

$[{\mathrm{ad}}_X,{\mathrm{ad}}_Y](Z)=[X,[Y,Z]]-[Y,[X,Z]]=[[X,Y],Z]={\mathrm{ad}}_{[X,Y]}(Z)\mathrm{.}$

Consequently

$\mathrm{ad}:\mathfrak{g}⟶\mathfrak{g}\mathfrak{l}(\mathfrak{g}),X⟼{\mathrm{ad}}_X,$

is a Lie-algebra homomorphism into the commutator Lie algebra of endomorphisms, where here a Lie-algebra homomorphism means a linear map that preserves the bracket. A linear homomorphism from a Lie algebra into the commutator Lie algebra of endomorphisms is called a representation, and this particular one is the adjoint representation of $\mathfrak{g}$. The later general definition of Lie-algebra representations uses exactly this convention but is not a prerequisite for the construction above.

For the zero algebra this is the unique map between zero spaces. Every one-dimensional Lie algebra over the stated fields has zero bracket, so its adjoint representation is zero. Degenerate adjoint maps are allowed; no faithfulness is asserted. The construction is algebraic, boundaryless and endpoint-free, uses no metric or choice principle, and contains no biconditional.