5 Representations of Lie groups and Lie algebras - generalities 5.3 The adjoint representation 6 SL₂

5.4 Representations of U(1) and Maschke’s theorem

We now discuss the representation theory of the unitary group $\operatorname{U}(1)=\{e^{it}\,:\,t\in\mathbb{R}\}$ , which is isomorphic to $\mathrm{SO}(2)$ , the circle group. Its Lie algebra is $i\mathbb{R}\subset\mathbb{C}$ with trivial Lie bracket, which is isomorphic to $\mathbb{R}$ .

Proposition 5.15.

All irreducible finite-dimensional representations of $\operatorname{U}(1)=\{z\in\mathbb{C};\,|z|=1\}$ are one-dimensional. They are given by

z=e^{it}\mapsto e^{int}=z^{n}

for $n\in\mathbb{Z}$ .

Proof.

It follows from Schur’s lemma that all irreducible finite-dimensional representations of $\operatorname{U}(1)$ are one-dimensional, so are homomorphisms $\operatorname{U}(1)\rightarrow\mathbb{C}^{\times}$ . Since $\operatorname{U}(1)$ is connected, such a homomorphism is determined by the derivative $\mathfrak{u}_{1}\rightarrow\mathfrak{gl}_{1}(\mathbb{C})=\mathbb{C}$ , which has the form $it\mapsto\lambda t$ for some $\lambda\in\mathbb{C}$ . As in Example 4.52, this exponentiates to a map $\operatorname{U}(1)\rightarrow\mathbb{C}^{\times}$ if and only if $\lambda=in$ for some $n\in\mathbb{Z}$ , giving the homomorphism $z\mapsto z^{n}$ . ∎

Theorem 5.16.

1.

All representations of $\operatorname{U}(1)$ are unitary.
2.

All representations of $\operatorname{U}(1)$ are completely reducible, that is, decompose into irreducible representations.

Proof.

1.

Exercise (see Maschke’s theorem below).
2.

This follows from (1) as in the proof of Maschke’s theorem for finite groups. We sketch another proof. So take $(\rho,V)$ , a finite dimensional representation of $\operatorname{U}(1)$ . Consider its differential $D\rho:\mathfrak{u}(1)=i\mathbb{R}\to\mathfrak{gl}_{n,\mathbb{C}}$ . Let $A=D\rho(2\pi i)$ . We may write $A=D+N$ for some strictly upper triangular matrix $N$ and diagonal matrix $D$ such that $D$ and $N$ commute. Then

$\rho(\exp(A))=\rho(e^{2\pi i})=\rho(1)=I.$

On the other hand,

$\exp(A)=\exp(D+N)=\exp(D)\exp(N),$

since $D$ and $N$ commute. It follows that $\exp(N)=\exp(D)=I$ (why?), whence $N=0$ (why?).

∎

Remark 5.17.

Complete irreducibility does not hold for representations of a general Lie group. For example, the standard representation of

N=\left\{\begin{pmatrix}1&x\\ 0&1\end{pmatrix}:x\in\mathbb{R}\right\}

does not decompose into a direct sum of two one-dimensional invariant subspaces (otherwise we would diagonalize $\begin{pmatrix}1&x\\ 0&1\end{pmatrix}$ , which is impossible). Furthermore, it is not unitary (as unitary matrices are diagonalizable).

Some of this actually generalizes substantially:

Theorem 5.18.

(Maschke’s theorem for compact Lie groups) Let $G$ be a compact Lie group.

1.

Any finite-dimensional representation of $G$ is unitarizable.
2.

(complete reducibility) Any finite-dimensional representation of $G$ is a direct sum of irreducible representations.

Proof.

The second part is proved exactly as for finite groups in section 1.5.3. Let $(\rho,V)$ be a finite-dimensional representation and let $W$ be a subrepresentation. Let $\left\langle\,,\,\right\rangle$ be the $G$ -invariant Hermitian inner product on $V$ guaranteed by the first part. Then the orthogonal complement $W^{\perp}$ is also a subrepresentation, and $V=W\oplus W^{\perp}$ . Iterating, we obtain that $V$ is a direct sum of irreducible representations.

The proof of the first part also uses the same idea as for finite groups. Take $(\,,\,)$ to be any Hermitian inner product on $V$ . Then define

\left\langle v,w\right\rangle=\int_{g\in G}(gv,gw)dg.

This is also a Hermitian inner product, and

$\displaystyle\left\langle hv,hw\right\rangle$	$\displaystyle=\int_{g\in G}(ghv,ghw)dg$
	$\displaystyle=\int_{k\in G}(kv,kw)d(kh^{-1})$	(putting $k=gh$ )
	$\displaystyle=\int_{k\in G}(kv,kw)dk$	(since $dk=d(kh^{-1})$ )
	$\displaystyle=\left\langle v,w\right\rangle.$

The challenge here is to show that there is an appropriate notion of $\int_{g\in G}f(g)dg$ for which the step “ $dk=d(kh^{-1})$ ” is valid — this goes by the name of ‘existence of Haar measure’. For $U(1)$ you can do it by hand, see problem 68. ∎