adjoint representationAdjointRepresentation2002-05-29 09:14:592007-01-25 16:59:20Definitionadjoint actionglgeneral linear Lie algebra\usepackage{amsmath}
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\newcommand{\End}{\mathop{\mathrm{End}}\nolimits}Let $\lag$ be a Lie algebra. For every $a\in\lag$ we define the
\PMlinkescapetext{{\em adjoint endomorphism}}, a.k.a. the {\em 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 {\em 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:
$$[a,[b,c]] +[b,[c,a]] + [c,[a,b]] = 0.$$