Let $\mathfrak{g}$ be a Lie algebra. For every $a\in\mathfrak{g}$ we define the , a.k.a. the adjoint action,

 $\ad(a):\mathfrak{g}\rightarrow\mathfrak{g}$

to be the linear transformation with action

 $\ad(a):b\mapsto[a,b],\quad b\in\mathfrak{g}.$

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 $\ad:\mathfrak{g}\rightarrow\mathfrak{gl}(\mathfrak{g})$ with action

 $a\mapsto\ad(a),\quad a\in\mathfrak{g}$

is called the adjoint representation of $\mathfrak{g}$. The fact that $\ad$ defines a representation is a straight-forward consequence of the Jacobi identity axiom. Indeed, let $a,b\in\mathfrak{g}$ be given. We wish to show that

 $\ad([a,b])=[\ad(a),\ad(b)],$

where the bracket on the left is the $\mathfrak{g}$ multiplication structure, and the bracket on the right is the commutator bracket. For all $c\in\mathfrak{g}$ 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:

 $[a,[b,c]]+[b,[c,a]]+[c,[a,b]]=0.$
Title adjoint representation AdjointRepresentation 2015-10-05 17:38:19 2015-10-05 17:38:19 rmilson (146) rmilson (146) 9 rmilson (146) Definition msc 17B10 IsotropyRepresentation adjoint action gl general linear Lie algebra