Adjoint representation

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In mathematics, the adjoint representation (or adjoint action) of a Lie group G is the natural representation of G on its own Lie algebra. This representation is the linearized version of the action of G on itself by conjugation.

1 Formal definition
- 1.1 Adjoint representation of a Lie algebra
2 Examples
3 Properties
4 Roots of a semisimple Lie group
5 Variants and analogues

[edit] Formal definition

Let G be a Lie group and let $\mathfrak g$ be its Lie algebra (which we identify with T_eG, the tangent space to the identity element in G). Define a map

$\Psi : G \to \mathrm{Aut}(G)\,$

by the equation $Ψ(g) = Ψ g$ for all g in G, where the automorphism $Ψ g$ is defined by

$\Psi_g(h) = ghg^{-1}\,$

for all h in G. It follows that the derivative of Ψ_g at the identity is an automorphism of the Lie algebra $\mathfrak g$ . We denote this map by Ad_g:

$\mathrm{Ad}_g\colon \mathfrak g \to \mathfrak g.$

To say that Ad_g is a Lie algebra automorphism is to say that Ad_g is a linear transformation of $\mathfrak g$ that preserves the Lie bracket. The map

$\mathrm{Ad}\colon G \to \mathrm{Aut}(\mathfrak g)$

which sends g to Ad_g is called the adjoint representation of G. This is indeed a representation of G since $\mathrm{Aut}(\mathfrak g)$ is a Lie subgroup of $\mathrm{GL}(\mathfrak g)$ and the above adjoint map is a Lie group homomorphism. The dimension of the adjoint representation is the same as the dimension of the group G.

[edit] Adjoint representation of a Lie algebra

One may always pass from a representation of a Lie group G to a representation of its Lie algebra by taking the derivative at the identity. Taking the derivative of the adjoint map

$\mathrm{Ad}\colon G \to \mathrm{Aut}(\mathfrak g)$

gives the adjoint representation of the Lie algebra $\mathfrak g$ :

$\mathrm{ad}\colon \mathfrak g \to \mathrm{Der}(\mathfrak g).$

Here $\mathrm{Der}(\mathfrak g)$ is the Lie algebra of $\mathrm{Aut}(\mathfrak g)$ which may be identified with the derivation algebra of $\mathfrak g$ . The adjoint representation of a Lie algebra is related in a fundamental way to the structure of that algebra. In particular, one can show that

$\mathrm{ad}_x(y) = [x,y]\,$

for all $x,y \in \mathfrak g$ . For more information see: adjoint representation of a Lie algebra.

[edit] Examples

If G is abelian of dimension n, the adjoint representation of G is the trivial n-dimensional representation.
If G is a matrix Lie group (i.e. a closed subgroup of GL(n,C)), then its Lie algebra is an algebra of n×n matrices with the commutator for a Lie bracket (i.e. a subalgebra of $\mathfrak{gl}_n(\mathbb C)$ ). In this case, the adjoint map is given by Ad_g(x) = gxg⁻¹.
If G is SL₂(R) (real 2×2 matrices with determinant 1), the Lie algebra of G consists of real 2×2 matrices with trace 0. The representation is equivalent to that given by the action of G by linear substitution on the space of binary (i.e., 2 variable) quadratic forms.

[edit] Properties

The following table summarizes the properties of the various maps mentioned in the definition

$\Psi\colon G \to \mathrm{Aut}(G)\,$	$\Psi_g\colon G \to G\,$
Lie group homomorphism: $Ψ g h = Ψ g Ψ h$	Lie group automorphism: $Ψ g (a b) = Ψ g (a)Ψ g (b)$ $(\Psi_g)^{-1} = \Psi_{g^{-1}}$
$\mathrm{Ad}\colon G \to \mathrm{Aut}(\mathfrak g)$	$\mathrm{Ad}_g\colon \mathfrak g \to \mathfrak g$
Lie group homomorphism: $Ad g h = Ad g Ad h$	Lie algebra automorphism: $Ad g$ is linear $(\mathrm{Ad}_g)^{-1} = \mathrm{Ad}_{g^{-1}}$ $Ad g [x, y] = [Ad g (x),Ad g (y)]$
$\mathrm{ad}\colon \mathfrak g \to \mathrm{Der}(\mathfrak g)$	$\mathrm{ad}_x\colon \mathfrak g \to \mathfrak g$
Lie algebra homomorphism: $ad$ is linear $ad [x, y] = [ad x,ad y]$	Lie algebra derivation: $ad x$ is linear $ad x [y, z] = [ad x (y), z] + [y,ad x (z)]$

The image of G under the adjoint representation is denoted by Ad_G. If G is connected, the kernel of the adjoint representation coincides with the kernel of Ψ which is just the center of G. Therefore the adjoint representation of a connected Lie group G is faithful if and only if G is centerless. More generally, if G is not connected, then the kernel of the adjoint map is the centralizer of the identity component G₀ of G. By the first isomorphism theorem we have

$\mathrm{Ad}_G \cong G/C_G(G_0).$

[edit] Roots of a semisimple Lie group

If G is semisimple, the non-zero weights of the adjoint representation form a root system. To see how this works, consider the case G=SL_n(R). We can take the group of diagonal matrices diag(t₁,...,t_n) as our maximal torus T. Conjugation by an element of T sends

$\begin{bmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\ \end{bmatrix} \mapsto \begin{bmatrix} a_{11}&t_1t_2^{-1}a_{12}&\cdots&t_1t_n^{-1}a_{1n}\\ t_2t_1^{-1}a_{21}&a_{22}&\cdots&t_2t_n^{-1}a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ t_nt_1^{-1}a_{n1}&t_nt_2^{-1}a_{n2}&\cdots&a_{nn}\\ \end{bmatrix}.$

Thus, T acts trivially on the diagonal part of the Lie algebra of G and with eigenvectors t_it_j^-1 on the various off-diagonal entries. The roots of G are the weights diag(t₁,...,t_n)→t_it_j^-1. This accounts for the standard description of the root system of G=SL_n(R) as the set of vectors of the form e_i−e_j.

[edit] Variants and analogues

The adjoint representation can also be defined for algebraic groups over any field.

The co-adjoint representation is the contragredient representation of the adjoint representation. Alexandre Kirillov observed that the orbit of any vector in a co-adjoint representation is a symplectic manifold. According to the philosophy in representation theory known as the orbit method (see also the Kirillov character formula), the irreducible representations of a Lie group G should be indexed in some way by its co-adjoint orbits. This relationship is closest in the case of nilpotent Lie groups.

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Category: Representation theory of Lie groups