2.10. Lecture 10

2.10.1. The adjoint representation

Let $G$ be a linear Lie group and $\mathfrak{g}$ be its Lie algebra. Then the adjoint representation $\operatorname{Ad}$ of $G$ is the action on $\mathfrak{g}$ by conjugation. We usually write $\operatorname{Ad}_{g}$ instead of $\operatorname{Ad}(g)$ , so that

\operatorname{Ad}_{g}(Y)=gYg^{-1}

for $g\in G$ and $Y\in\mathfrak{g}$ . By Proposition 1.3.8(ii), $gYg^{-1}$ is indeed in $\mathfrak{g}$ . Thus the map

\operatorname{Ad}:G\longrightarrow\operatorname{GL}(\mathfrak{g})\ ;\ g% \longmapsto\operatorname{Ad}_{g},

is a Lie group homomorphism.

Remark 2.10.1.

An alternative definition, that works for general Lie groups, is to consider the conjugation by $g$ map $h\mapsto ghg^{-1}$ and then take the derivative:

\operatorname{Ad}_{g}(Y)=\frac{d}{dt}g\exp(tY)g^{-1}|_{t=0}.

The derivative of $\operatorname{Ad}$ , denoted by $\operatorname{ad}$ , is called the adjoint representation of the Lie algebra $\mathfrak{g}$ . Thus

\operatorname{ad}=D\operatorname{Ad}.

Again, we write $\operatorname{ad}_{X}$ for $\operatorname{ad}(X)$ , so we have

\operatorname{ad}:\mathfrak{g}\longrightarrow\operatorname{End}(\mathfrak{g})=% \mathfrak{gl}(\mathfrak{g})\ ;\ X\longmapsto\operatorname{ad}_{X}.

By Theorem 1.5.6 we have the formula

\operatorname{Ad}_{\exp(tX)}=\exp(t\operatorname{ad}_{X})

as elements of $\operatorname{GL}(\mathfrak{g})$ .

Theorem 2.10.2.

Let $G$ and $\mathfrak{g}$ be as above and let $X,Y\in\mathfrak{g}$ . Then

(i)

$\operatorname{ad}_{X}(Y)=[X,Y]=XY-YX.$
(ii)

The map $\operatorname{ad}$ is a Lie algebra homomorphism, so that

$\operatorname{ad}_{[X,Y]}=[\operatorname{ad}_{X},\operatorname{ad}_{Y}].$

Thus, for all $Z\in\mathfrak{g}$ ,

$\operatorname{ad}_{[X,Y]}(Z)=[\operatorname{ad}_{X},\operatorname{ad}_{Y}](Z)=% \operatorname{ad}_{X}(\operatorname{ad}_{Y}(Z))-\operatorname{ad}_{Y}(% \operatorname{ad}_{X}(Z)).$

Proof.

(i)

Since $\operatorname{Ad}_{\exp(tX)}(Y)=\exp(tX)Y\exp(-tX)$ , taking the differential at $t=0$ , we get

$\operatorname{ad}_{X}(Y)=XY-YX=[X,Y].$
(ii)

The map $\operatorname{ad}$ is a Lie algebra homomorphism because it is the differential of a Lie group homomorphism, Theorem 1.5.6(iii).

∎

Remark 2.10.3.

This explains the origin of the Jacobi identity:

	$\displaystyle[\operatorname{ad}_{X},\operatorname{ad}_{Y}](Z)$	$\displaystyle=\operatorname{ad}_{X}(\operatorname{ad}_{Y}(Z))-\operatorname{ad% }_{Y}(\operatorname{ad}_{X}(Z))$
		$\displaystyle=[X,[Y,Z]]-[Y,[X,Z]]$
		$\displaystyle=[X,[Y,Z]]+[Y,[Z,X]],$

while

	$\displaystyle\operatorname{ad}_{[X,Y]}(Z)$	$\displaystyle=[[X,Y],Z]$
		$\displaystyle=-[Z,[X,Y]].$

Equating these gives the Jacobi identity.

Remark 2.10.4.

The first formula, $\operatorname{ad}_{X}(Y)=[X,Y]$ , could have been used to define the adjoint representation for any Lie algebra, without reference to Lie groups.

Remark 2.10.5.

Warning! It is very easy to misinterpret some of the formulas concerning the adjoint representation. For example, $\operatorname{ad}_{[X,Y]}=[\operatorname{ad}_{X},\operatorname{ad}_{Y}]$ does not mean that

\operatorname{ad}_{[X,Y]}(Z)=[\operatorname{ad}_{X}(Z),\operatorname{ad}_{Y}(Z% )],

but (as already noted and proved) that

\operatorname{ad}_{[X,Y]}(Z)=[\operatorname{ad}_{X},\operatorname{ad}_{Y}](Z)=% \operatorname{ad}_{X}(\operatorname{ad}_{Y}(Z))-\operatorname{ad}_{Y}(% \operatorname{ad}_{X}(Z)).

Similarly, $\operatorname{Ad}_{\exp(tX)}=\exp(t\operatorname{ad}_{X})$ does not mean that ${\exp(tX)}Y\exp(-tX)$ is equal to $\exp(t[X,Y])$ but rather is the identity

	$\displaystyle{\exp(tX)}Y\exp(-tX)$	$\displaystyle=\exp(t\operatorname{ad}_{X})(Y)$
		$\displaystyle=\sum_{k=0}^{\infty}\frac{t^{k}(\operatorname{ad}_{X})^{k}}{k!}(Y)$
		$\displaystyle=\sum_{k=0}^{\infty}\frac{[X,[X,\dots,[X,Y]]\ldots]}{k!}t^{k}.$

Proposition 2.10.6.

If $G$ is abelian, so is $\mathfrak{g}$ . If, moreover, $G$ is connected, then the converse holds.

Proof.

Suppose that $G$ is abelian. Then, for all $g\in G$ and $Y\in\mathfrak{g}$ ,

g\exp(tY)g^{-1}=\exp(tY).

Taking the derivative at $t=0$ we see that $gYg^{-1}=Y$ . Thus $\operatorname{Ad}$ is trivial. Differentiating, we see $\operatorname{ad}$ is trivial, so $\operatorname{ad}_{X}=0$ for all $X$ . Thus $[X,Y]=0$ for all $X,Y\in\mathfrak{g}$ as required.

Conversely, suppose $G$ is connected and $\mathfrak{g}$ is abelian. Then $\operatorname{ad}$ is trivial and, since $G$ is connected, $\operatorname{Ad}$ is trivial. Thus $gYg^{-1}=Y$ for all $g\in G,Y\in\mathfrak{g}$ . Thus $g\exp(Y)g^{-1}=\exp(Y)$ for all $g\in G$ and all $Y\in\mathfrak{g}$ . Since $\exp(\mathfrak{g})$ generates $G$ , we see that $G$ is commutative. ∎

2.10.2. Exercises

Problem 26. Prove that, for $X,Y\in\mathfrak{gl}_{n}$ ,

(\operatorname{ad}_{X})^{m}(Y)=[X,[X,\ldots,[X,Y]\ldots]]=\sum_{i=0}^{m}\binom% {m}{k}X^{k}Y(-X)^{m-k}.

Hence give a direct proof that $\exp(\operatorname{ad}_{X})=\operatorname{Ad}_{\exp(X)}$ .