📂Representation Theory

Adjoint map

Definition¹

Let $\mathfrak{g}$ be a Lie algebra, and denote its elements by $X \in \mathfrak{g}$. Define the linear map $\ad_{X} : \mathfrak{g} \to \mathfrak{g}$ as follows.

$$ \ad_{X}(Y) = [X, Y] $$

The map $X \mapsto \ad_{X}$ is called the adjoint map or the adjoint representation.

Explanation

Note that the adjoint map refers to $X \mapsto \ad_{X}$, not the linear map $\ad_{X}$. Thus the domain and codomain of the adjoint map are as follows.

$$ \text{adjoint map : } \mathfrak{g} \to \operatorname{End}(\mathfrak{g}) $$

Here $\operatorname{End}(\mathfrak{g})$ denotes the endomorphism space. $\operatorname{End}(\mathfrak{g})$ itself is also a Lie algebra, and its bracket is defined by the commutator as follows.

$$ [\cdot, \cdot] : \operatorname{End}(\mathfrak{g}) \times \operatorname{End}(\mathfrak{g}) \to \operatorname{End}(\mathfrak{g}) $$

$$ [\ad_{X}, \ad_{Y}] = \ad_{X} \circ \ad_{Y} - \ad_{Y} \circ \ad_{X} $$

The notation $[X, [X, [X, Y]]]$ can be abbreviated simply as $(\ad_{X})^{3}(Y)$.

(b) indicates that $\ad : \mathfrak{g} \to \operatorname{End}(\mathfrak{g})$ is a Lie algebra homomorphism. In particular, (a) and (b) mean respectively that $\ad_{X}$ and $\ad$ act as if they satisfy a distributive law with respect to $[\cdot, \cdot]$.

Properties

(a) $\ad_{X}([Y, Z]) = [\ad_{X}(Y), Z] + [Y, \ad_{X}(Z)]$

(b) $\ad_{[X, Y]} = [\ad_{X}, \ad_{Y}] = \ad_{X} \ad_{Y} - \ad_{Y} \ad_{X}$

Proof

(a)

Since $[\cdot, \cdot]$ satisfies the Jacobi identity (see ../3772),

$$ \begin{align*} && [X, [Y, Z]] + [Z, [X, Y]] + [Y, [Z, X]] = 0 \\ \implies && [X, [Y, Z]] - [[X, Y], Z] - [Y, [X, Z]] = 0 \\ \implies && \ad_{X} ([Y, Z]) - [\ad_{X}(Y), Z] - [Y, \ad_{X}(Z)] = 0 \\ \implies && \ad_{X}([Y, Z]) = [\ad_{X}(Y), Z] + [Y, \ad_{X}(Z)] \end{align*} $$

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(b)

Since $[\cdot, \cdot]$ satisfies the Jacobi identity (see ../3772),

$$ \begin{align*} \ad_{[X, Y]}(Z) &= [[X,Y], Z] \\ &= -[[Z,X], Y] - [[Y,Z], X] \\ &= [Y, [Z,X]] + [X, [Y,Z]] \\ &= [X, [Y,Z]] + [Y, [Z,X]] \\ &= [X, [Y,Z]] - [Y, [X,Z]] \\ &= \ad_{X}([Y, Z]) - \ad_{Y}([X, Z]) \\ &= \ad_{X}(\ad_{Y}(Z)) - \ad_{Y}(\ad_{X}(Z)) \\ &= \left( \ad_{X}\ad_{Y} - \ad_{Y}\ad_{X} \right)(Z) \\ &= [\ad_{X}, \ad_{Y}] (Z) \end{align*} $$

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