PlanetMath: adjoint representation

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adjoint representation

(Definition)

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

$\displaystyle \mathop{\mathrm{ad}}\nolimits (a):\mathfrak{g}\rightarrow\mathfrak{g}$

to be the linear transformation with action

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

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

In this notation, treating $\mathfrak{g}$ as a vector space, the linear mapping $\mathop{\mathrm{ad}}\nolimits :\mathfrak{g}\rightarrow \mathfrak{gl}(\mathfrak{g})$ with action

$\displaystyle a\mapsto \mathop{\mathrm{ad}}\nolimits (a),\quad a\in\mathfrak{g}$

is called the adjoint representation of $\mathfrak{g}$ . The fact that $\mathop{\mathrm{ad}}\nolimits$ 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

$\displaystyle \mathop{\mathrm{ad}}\nolimits ([a,b]) = [\mathop{\mathrm{ad}}\nolimits (a),\mathop{\mathrm{ad}}\nolimits (b)],$

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

$\displaystyle [[a,b],c],$

while the right hand side maps

$\displaystyle [a,[b,c]]+[b,[a,c]].$

Taking skew-symmetry of the bracket as a given, the equality of these two expressions is logically equivalent to the Jacobi identity:

$\displaystyle [a,[b,c]] +[b,[c,a]] + [c,[a,b]] = 0.$

"adjoint representation" is owned by rmilson. [ full author list (2) ]

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