If $R:U(1)\to GL(n,\mathbf{C})$
is a smooth representation then there exists a basis of ${\mathbf{C}}^{n}$
with respect to which $R({e}^{i\theta})=\left(\begin{array}{ccc}\hfill {e}^{i{m}_{1}\theta}\hfill & \hfill \hfill & \hfill 0\hfill \\ \hfill \hfill & \hfill \mathrm{\ddots}\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill \hfill & \hfill {e}^{i{m}_{n}\theta}\hfill \end{array}\right),$
where ${m}_{1},\mathrm{\dots},{m}_{n}$
are integers called the *weights* of the representation. In other words, ${\mathbf{C}}^{n}={\oplus}_{i=1}^{n}{V}_{i}$
where each ${V}_{i}$
is a 1-dimensional subrepresentation and $R={R}_{1}\oplus \mathrm{\cdots}\oplus {R}_{n}$
with ${R}_{i}={R|}_{{V}_{i}}$
.

# Representations of U(1), part 2

## Representations of U(1), part 2

We will now prove the theorem we stated last time:

This will follow from our earlier result about complete reducibility. More precisely, we will prove the following lemmas:

Any representation of $U(1)$ admits an invariant Hermitian inner product.

This will tell us that any representation splits as a direct sum of irreducible subrepresentations. Now the theorem will follow from:

Any irreducible representation of $U(1)$ is 1-dimensional and given by $R({e}^{i\theta})={e}^{im\theta}$ for some integer $m$ .

## Proof of first lemma

Take any Hermitian inner product $\u27e8,\u27e9$ on ${\mathbf{C}}^{n}$ . We will "average" it over the group to get a new inner product: $${\u27e8u,v\u27e9}_{inv}={\int}_{0}^{2\pi}\u27e8R({e}^{i\theta})u,R({e}^{i\theta})v\u27e9\frac{d\theta}{2\pi}$$ which we will prove is invariant.

The $1/2\pi $ is just there to ensure that if $\u27e8,\u27e9$ is invariant then ${\u27e8,\u27e9}_{inv}=\u27e8,\u27e9$ .

It's an exercise to check that ${\u27e8,\u27e9}_{inv}$ is a Hermitian inner product. We'll now prove that it's invariant, that is $${\u27e8R({e}^{i\varphi})u,R({e}^{i\varphi})v\u27e9}_{inv}={\u27e8u,v\u27e9}_{inv}$$ for all $u$ and $v$ in ${\mathbf{C}}^{n}$ and ${e}^{i\varphi}\in U(1)$ .

We prove this by computing: $$\u27e8R({e}^{i\varphi})u,R{({e}^{i\varphi}v\u27e9}_{inv}={\int}_{0}^{2\pi}\u27e8R({e}^{i\theta})R({e}^{i\varphi})u,R({e}^{i\theta})R({e}^{i\varphi})v\u27e9\frac{d\theta}{2\pi}={\int}_{0}^{2\pi}\u27e8R({e}^{i(\theta +\varphi )})u,R({e}^{i(\theta +\varphi )})v\u27e9\frac{d\theta}{2\pi}$$ where we used that $R$ is a representation.

We now change variables ${\theta}^{\prime}=\theta +\varphi $ . Since $\varphi $ is just some fixed number (constant) $d{\theta}^{\prime}=d\theta $ , so the integral becomes: $${\int}_{0}^{2\pi}\u27e8R({e}^{i{\theta}^{\prime}})u,R({e}^{i{\theta}^{\prime}})v\u27e9\frac{d{\theta}^{\prime}}{2\pi}$$ Since ${\theta}^{\prime}$ is just a dummy variable we're integrating over, this agrees with the definition of ${\u27e8u,v\u27e9}_{inv}$ . This shows that ${\u27e8,\u27e9}_{inv}$ is invariant for the representation $R$ .

This works more generally for any compact group $G$
, that is a topologically closed matrix group where all the matrix entries are bounded. In this context, you can define a similar integral (called the *Haar integral*) and run the same argument. I won't prove this: you can do an in-depth project about it if you're interested.

## Proof of second lemma

Fix ${e}^{i\theta}\in U(1)$ and consider $R({e}^{i\theta})\in GL(n,\mathbf{C})$ . Because $\mathbf{C}$ is an algebraically closed field, the characteristic polynomial of $R({e}^{i\theta})$ has a root, so $R({e}^{i\theta})$ has at least one eigenvalue $\lambda \in \mathbf{C}$ for which the eigenspace ${V}_{\lambda}=\{v\in {\mathbf{C}}^{n}:R({e}^{i\theta})v=\lambda v\}$ is not zero.

${V}_{\lambda}$ is a subrepresentation of ${\mathbf{C}}^{n}$ , i.e. if ${e}^{i\varphi}$ is any element of $U(1)$ then $v\in {V}_{\lambda}$ implies $R({e}^{i\varphi})v\in {V}_{\lambda}$ .

Suppose $v\in {V}_{\lambda}$ . Apply $R({e}^{i\theta})$ to $R({e}^{i\varphi})v$ . Because $U(1)$ is abelian and $R$ is a representation, the matrices $R({e}^{i\theta})$ and $R({e}^{i\varphi})$ commute with one another, and we get $$R({e}^{i\theta})(R({e}^{i\varphi})v)=R({e}^{i\varphi})R({e}^{i\theta})v=R({e}^{i\varphi})\lambda v=\lambda (R({e}^{i\varphi})v),$$ Therefore $R({e}^{i\varphi})v$ is an eigenvector of $R({e}^{i\theta})$ with eigenvalue $\lambda $ , that is $R({e}^{i\varphi})v\in {V}_{\lambda}$ as required.

If ${\mathbf{C}}^{n}$
is irreducible then this implies ${\mathbf{C}}^{n}={V}_{\lambda}$
because ${V}_{\lambda}$
would otherwise be a proper subrepresentation. This means that $R({e}^{i\theta})=\lambda I$
because *everything* is an eigenvector with eigenvalue $\lambda $
.

We fixed a particular $\theta $ at the beginning of the proof of the second lemma, but the proof works for all $\theta $ and we get an eigenvalue $\lambda (\theta )$ that depends on $\theta $ . In other words, we can think of $\lambda $ as a map (actually a homomorphism) $\lambda :U(1)\to {\mathbf{C}}^{*}$ to the nonzero complex numbers such that $R({e}^{i\theta})=\lambda (\theta )I$ for all $\theta $ ($\lambda \ne 0$ because $R({e}^{i\theta})$ is invertible).

$\lambda (\theta )\in U(1)$ .

We have an invariant Hermitian inner product $\u27e8,\u27e9$ , so $$\u27e8v,v\u27e9=\u27e8R({e}^{i\theta})v,R({e}^{i\theta}v\u27e9=\u27e8\lambda (\theta )v,\lambda (\theta )v\u27e9=|\lambda (\theta ){|}^{2}\u27e8v,v\u27e9,$$ using the fact that the inner product is sesquilinear (we can pull out the two factors of $\lambda (\theta )$ but the first one picks up a complex conjugate sign). Therefore ${|\lambda (\theta )|}^{2}=1$ .

This tells us that $\lambda :U(1)\to U(1)$ is a homomorphism from $U(1)$ to $U(1)$ . We classified these in an earlier video: they are all of the form $\lambda (\theta )={e}^{im\theta}$ for some integer $m$ . This tells us that $$R({e}^{i\theta})={e}^{im\theta}I.$$

Since our representation is irreducible, we can now deduce that it is 1-dimensional. This is because any 1-dimensional complex line in ${\mathbf{C}}^{n}$ is preserved under the map ${e}^{im\theta}I$ (just rescales by ${e}^{im\theta}$ , which rotates every complex line inside itself by $m\theta $ ) so any complex line in ${\mathbf{C}}^{n}$ is a subrepresentation. Since ${\mathbf{C}}^{n}$ is irreducible, it must coincide with any 1-dimensional complex line inside it, and hence $n=1$ .

This completes the proof.

## Pre-class exercise

Show that if $\u27e8,\u27e9$ is a Hermitian inner product on ${\mathbf{C}}^{n}$ then $${\u27e8u,v\u27e9}_{inv}={\int}_{0}^{2\pi}\u27e8R({e}^{i\theta})u,R({e}^{i\theta})v\u27e9\frac{d\theta}{2\pi}$$ is also a Hermitian inner product on ${\mathbf{C}}^{n}$ .