Let $\lag$ be a Lie algebra. For every $a\in\lag$ we define the adjoint endomorphism, a.k.a. the adjoint action, $$\ad(a):\lag\rightarrow\lag$$ to be the linear transformation with action $$\ad(a): b\mapsto [a,b],\quad b\in\lag.$$

For any vector space $V$ we use $\mathfrak{gl}(V)$ to denote the Lie algebra of $\End V$ determined by the commutator bracket. So $\mathfrak{gl}(V)=\End V$ as vector spaces, only the multiplications are different.

In this notation, treating $\mathfrak{g}$ as a vector space, the linear mapping $\ad:\lag\rightarrow \mathfrak{gl}(\lag)$ with action $$a\mapsto \ad(a),\quad a\in\lag$$ is called the adjoint representation of $\lag$ The fact that $\ad$ defines a representation is a straight-forward consequence of the Jacobi identity axiom. Indeed, let $a,b\in \lag$ be given. We wish to show that $$\ad([a,b]) = [\ad(a),\ad(b)],$$ where the bracket on the left is the $\lag$ multiplication structure, and the bracket on the right is the commutator bracket. For all $c\in\lag$ the left hand side maps $c$ to $$[[a,b],c],$$ while the right hand side maps $c$ to $$[a,[b,c]]+[b,[a,c]].$$ Taking skew-symmetry of the bracket as a given, the equality of these two expressions is logically equivalent to the Jacobi identity: $$[a,[b,c]] +[b,[c,a]] + [c,[a,b]] = 0.$$