# Representations of U(1), part 2

## Representations of U(1), part 2

We will now prove the theorem we stated last time:

Theorem:

If $R\colon U(1)\to GL(n,\mathbf{C})$ is a smooth representation then there exists a basis of $\mathbf{C}^{n}$ with respect to which $R(e^{i\theta})=\begin{pmatrix}e^{im_{1}\theta}&&{\huge 0}\\ &\ddots&\\ {\huge 0}&&e^{im_{n}\theta}\end{pmatrix},$ where $m_{1},\ldots,m_{n}$ are integers called the weights of the representation. In other words, $\mathbf{C}^{n}=\bigoplus_{i=1}^{n}V_{i}$ where each $V_{i}$ is a 1-dimensional subrepresentation and $R=R_{1}\oplus\cdots\oplus R_{n}$ with $R_{i}=R|_{V_{i}}$ .

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

Lemma (Weyl's unitarian trick):

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:

Lemma (Basically Schur's lemma plus a little bit):

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 $\langle,\rangle$ on $\mathbf{C}^{n}$ . We will "average" it over the group to get a new inner product: $\langle u,v\rangle_{inv}=\int_{0}^{2\pi}\langle R(e^{i\theta})u,R(e^{i\theta})% v\rangle\frac{d\theta}{2\pi}$ which we will prove is invariant.

Remark:

The $1/2\pi$ is just there to ensure that if $\langle,\rangle$ is invariant then $\langle,\rangle_{inv}=\langle,\rangle$ .

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

We prove this by computing: $\langle R(e^{i\phi})u,R(e^{i\phi}v\rangle_{inv}=\int_{0}^{2\pi}\langle R(e^{i% \theta})R(e^{i\phi})u,R(e^{i\theta})R(e^{i\phi})v\rangle\frac{d\theta}{2\pi}=% \int_{0}^{2\pi}\langle R(e^{i(\theta+\phi)})u,R(e^{i(\theta+\phi)})v\rangle% \frac{d\theta}{2\pi}$ where we used that $R$ is a representation.

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

Remark:

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.

Lemma (Schur's lemma):

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

Proof:

Suppose $v\in V_{\lambda}$ . Apply $R(e^{i\theta})$ to $R(e^{i\phi})v$ . Because $U(1)$ is abelian and $R$ is a representation, the matrices $R(e^{i\theta})$ and $R(e^{i\phi})$ commute with one another, and we get $R(e^{i\theta})(R(e^{i\phi})v)=R(e^{i\phi})R(e^{i\theta})v=R(e^{i\phi})\lambda v% =\lambda(R(e^{i\phi})v),$ Therefore $R(e^{i\phi})v$ is an eigenvector of $R(e^{i\theta})$ with eigenvalue $\lambda$ , that is $R(e^{i\phi})v\in V_{\lambda}$ 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) $\lambda\colon U(1)\to\mathbf{C}^{*}$ to the nonzero complex numbers such that $R(e^{i\theta})=\lambda(\theta)I$ for all $\theta$ ($\lambda\neq 0$ because $R(e^{i\theta})$ is invertible).

Lemma:

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

Proof:

We have an invariant Hermitian inner product $\langle,\rangle$ , so $\langle v,v\rangle=\langle R(e^{i\theta})v,R(e^{i\theta}v\rangle=\langle% \lambda(\theta)v,\lambda(\theta)v\rangle=|\lambda(\theta)|^{2}\langle v,v\rangle,$ 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 $\lambda\colon U(1)\to U(1)$ 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

Exercise:

Show that if $\langle,\rangle$ is a Hermitian inner product on $\mathbf{C}^{n}$ then $\langle u,v\rangle_{inv}=\int_{0}^{2\pi}\langle R(e^{i\theta})u,R(e^{i\theta})% v\rangle\frac{d\theta}{2\pi}$ is also a Hermitian inner product on $\mathbf{C}^{n}$ .