# adjoint representation

\DeclareMathOperator\ad

ad \DeclareMathOperator\EndEnd

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

$$\text{ad}(a):\text{mathfrak}g\to \text{mathfrak}g$$ |

to be the linear transformation with action

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

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

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

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

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

$$\text{ad}([a,b])=[\text{ad}(a),\text{ad}(b)],$$ |

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

Canonical name | AdjointRepresentation |

Date of creation | 2015-10-05 17:38:19 |

Last modified on | 2015-10-05 17:38:19 |

Owner | rmilson (146) |

Last modified by | rmilson (146) |

Numerical id | 9 |

Author | rmilson (146) |

Entry type | Definition |

Classification | msc 17B10 |

Related topic | IsotropyRepresentation |

Defines | adjoint action |

Defines | gl |

Defines | general linear Lie algebra |