2.8. Lecture 8

2.8.1. Basics

Definition 2.8.1.

A finite-dimensional (complex) representation $(\rho,V)$ of a Lie group $G$ is a Lie group homomorphism

\rho:G\longrightarrow\operatorname{GL}(V),

where $V$ is a finite-dimensional complex vector space.

Remark 2.8.2.

Infinite dimensional representations are important, but subtle. In general one must equip $V$ with some topology and add topological conditions to all notions which follow. For instance, one might take $V$ to be a Hilbert space.

An example of such a representation arises naturally if you attempt to generalise the regular representation! One must take $V$ to be something like the space of square-integrable functions on the group, rather than the space of arbitrary functions, to get a pleasant theory.

If $\rho$ is a finite-dimensional representation of $G$ as above, then we can take its derivative:

D\rho:\mathfrak{g}\longrightarrow\mathfrak{gl}(V)=\operatorname{End}(V),

mapping from the Lie algebra $\mathfrak{g}$ of $G$ to the space of endomorphisms of $V$ . Note that

\operatorname{End}(V)

is a Lie algebra with bracket

[S,T]=ST-TS.

The map $D\rho$ is a Lie algebra homomorphism, by Theorem 1.5.6(iii). Often we write, abusively, $\rho$ instead of $D\rho$ .

Note that choosing an isomorphism $V\cong\mathbb{C}^{n}$ induces isomorphisms $\operatorname{GL}(V)\cong\operatorname{GL}_{n}(\mathbb{C})$ and $\mathfrak{gl}(V)\cong\mathfrak{gl}_{n,\mathbb{C}}$ .

Definition 2.8.3.

A (complex) representation $(\rho,V)$ of a Lie algebra $\mathfrak{g}$ is a Lie algebra homomorphism

\rho:\mathfrak{g}\longrightarrow\mathfrak{gl}(V),

where $V$ is a complex vector space. That is,

(i)

$\rho$ is $\mathbb{R}$ -linear;
(ii)

$\rho([X,Y])=[\rho(X),\rho(Y)]$ .

By our above discussion the differential of a Lie group representation is a Lie algebra representation.

Remark 2.8.4.

Warning! It is not the case that, if $\rho$ is a Lie algebra representation, then

\rho(XY)=\rho(X)\rho(Y).

Indeed, in general $XY$ need not be an element of the Lie algebra at all, and even if it is the displayed equation will not usually hold.

The notions of $G$ -homomorphism (or $\mathfrak{g}$ -homomorphism, or intertwiner), isomorphism, subrepresentation, and irreducible representation stay the same as for finite groups. For example, a $\mathfrak{g}$ -homomorphism from $(\rho,V)$ to $(\rho^{\prime},V^{\prime})$ is a linear map $\phi:V\rightarrow V^{\prime}$ such that

\phi(\rho(X){\bf v})=\rho^{\prime}(X)\phi({\bf v})

for all ${\bf v}\in V$ and $X\in\mathfrak{g}$ .

Definition 2.8.5.

A $\mathbb{C}$ -linear representation of $\mathfrak{g}$ is a complex representation $\rho$ of $\mathfrak{g}$ such that

\rho(\lambda g)=\lambda\rho(g)

for all $\lambda\in\mathbb{C}$ .

If $G$ is a complex Lie group, then a holomorphic representation of $G$ is a complex representation whose derivative is $\mathbb{C}$ -linear; equivalently, the map $G\to\operatorname{GL}(V)$ is holomorphic.

Theorem 2.8.6.

Let $G$ be a Lie group, $\mathfrak{g}$ be a Lie algebra.

(i)

If $V_{1}$ and $V_{2}$ are irreducible finite-dimensional representations of $G$ (or $\mathfrak{g}$ ), then

$\dim\operatorname{Hom}_{G\text{ (or }\mathfrak{g})}(V_{1},V_{2})=\begin{cases}% 1&\text{if $V_{1}\cong V_{2}$}\\ 0&\text{otherwise}.\end{cases}$

If $V_{1}=V_{2}$ , then any $G$ - (or $\mathfrak{g}$ -)homomorphism $T:V\rightarrow V$ is scalar.
(ii)

Any irreducible finite-dimensional representation of an abelian Lie group or Lie algebra is one-dimensional.
(iii)

If $(\rho,V)$ is an irreducible finite-dimensional representation of $G$ (or $\mathfrak{g}$ ) and $Z$ (or $\mathfrak{z}$ ) is the center of $G$ (or $\mathfrak{g}$ ) then there is a homomorphism $\chi:Z\rightarrow\mathbb{C}^{\times}$ (or $\chi:Z\rightarrow\mathbb{C}$ ) such that

$\rho(z){\bf v}=\chi(z){\bf v}$

for all $z\in Z$ (or $\mathfrak{z}$ ) and ${\bf v}\in V$ . We call this the central character.

Proof.

The proofs are all the same as in the finite group case! ∎

Proposition 2.8.7.

Let $(\rho,V)$ be a finite-dimensional representation of a Lie group $G$ . Let $D\rho$ be its derivative.

(i)

If $W\subseteq V$ is invariant under $\rho(G)$ , then $W$ is invariant under $D\rho(\mathfrak{g})$ .
(ii)

If $D\rho$ is irreducible, then $\rho$ is irreducible.
(iii)

If $\rho$ is unitary, that is, there is a basis for $V$ such that $\rho(g)\in\operatorname{U}(n)$ for all $g\in G$ , then $D\rho$ is skew-Hermitian, that is, $D\rho(X)\in\mathfrak{u}(n)$ for all $X\in\mathfrak{g}$ (using the same basis for $V$ ).
(iv)

Let $(\rho^{\prime},V^{\prime})$ be another finite-dimensional representation of $G$ . If $\rho\cong\rho^{\prime}$ , then $D\rho\cong D\rho^{\prime}$ .

If $G$ is connected, then the converses to these statements hold.

So, for connected Lie groups, we can test irreducibility and isomorphism at the level of Lie algebras.

Proof.

For (i), we know that $\rho(\exp(tX))({\bf w})\in W$ for any $X\in\mathfrak{g}$ and ${\bf w}\in W$ . Taking the derivative at $t=0$ , it follows that $D\rho(X)({\bf w})\in W$ as required. Part (ii) follows from (i).

For (iii), if $\rho$ is unitary, then after choosing a basis appropriately it is a Lie group homomorphism $\rho:G\to U(n)$ . The derived homomorphism therefore lands in the Lie algebra $\mathfrak{u}(n)$ of $\operatorname{U}(n)$ .

For part (iv), let $T$ be a $G$ -isomorphism, so that in particular,

T\rho_{1}(\exp(tX))T^{-1}=\rho_{2}(\exp(tX))

for all $X\in\mathfrak{g}$ and $t\in\mathbb{R}$ . Taking the derivative at $t=0$ gives

TD\rho_{1}(X)T^{-1}=D\rho_{2}(X)

so that $T$ is a $\mathfrak{g}$ -isomorphism as required.

If $G$ is connected, then $G$ is generated by $\exp(\mathfrak{g})$ . Hence all proofs above can be reversed. For example, for (i), suppose that $W$ is preserved by $D\rho(\mathfrak{g})$ . If ${\bf w}\in W$ and $X\in\mathfrak{g}$ , then

\rho(\exp(X)){\bf w}=\exp(D\rho(X)){\bf w}=\sum_{n=0}^{\infty}\frac{(D\rho(X))% ^{n}}{n!}{\bf w}\in W

as $W$ is preserved by $D\rho(X)$ and also closed. Since every $g\in G$ can be written as a finite product of $\exp(X_{i})$ for $X_{i}\in\mathfrak{g}$ , we see that $W$ is preserved by $\rho(G)$ as required. The converse of (iii) and (iv) are left as exercises. ∎

2.8.2. Exercises

Problem 22. Let $G$ be connected. Prove the converse of Proposition 2.8.7(iii) and (iv).