# 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 from U(1) to G L n C is a smooth representation then there exists a basis of C n with respect to which R of e to the i theta is the diagonal matrix with entries e to the i m_1 theta down to e to the i m_n theta, where m_1 up to m_n are integers called the weights of the representation. In other words, C n equals the direct sum from i = 1 to n of V_i where each V_i is a 1-dimensional subrepresentation and R equals R_1 direct sum R_2 direct sum dot dot dot R_n with R_i equals R restricted to 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 of e to the i theta equals e to the i m theta for some integer m.

## Proof of first lemma

Take any Hermitian inner product angle brackets on C n. We will "average" it over the group to get a new inner product: u angle brackets v subscript inv equals the integral from 0 to 2 pi of the angle bracket between R of e to the i theta applied to u and R of e to the i theta applied to v, d theta over 2 pi which we will prove is invariant.

Remark:

The one over 2 pi is just there to ensure that if angle brackets is invariant then angle brackets subscript inv equals angle brackets.

It's an exercise to check that angle brackets subscript inv is a Hermitian inner product. We'll now prove that it's invariant, that is the angle bracket (subscript inv) between R of e to the i phi applied to u with R of e to the i phi applied to v, equals the angle bracket (subscript inv) of u with v for all u and v in C n and e to the i phi in U(1).

We prove this by computing: the angle bracket (subscript inv) between R of e to the i phi applied to u and R of e to the i phi applied to v, equals the integral from 0 to 2 pi of (the original angle bracket between R of e to the i theta times R of e to the i phi applied to u and R of e to the i theta times R of e to the i phi applied to v) d theta over 2 pi, which equals the integral of (the original angle bracket between R of e to the i (theta plus phi) applied to u and R of e to the i (theta plus phi) applied to v) d theta over 2 pi where we used that R is a representation.

We now change variables theta prime equals theta plus phi. Since phi is just some fixed number (constant) d theta prime equals d theta, so the integral becomes: the integral from 0 to 2 pi of the original angle bracket between R of e to the i theta prime u and R of e to the i theta prime v d theta prime over 2 pi Since theta prime is just a dummy variable we're integrating over, this agrees with the definition of angle bracket (subscript inv) of u and v. This shows that angle bracket subscript 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 to the i theta in U(1) and consider R of e to the i theta in G L n C. Because C is an algebraically closed field, the characteristic polynomial of R of e to the i theta has a root, so R of e to the i theta has at least one eigenvalue lambda in C for which the eigenspace V lambda equals the set of v in C n such that R of e to the i theta v equals lambda v is not zero.

Lemma (Schur's lemma):

V lambda is a subrepresentation of C n, i.e. if e to the i phi is any element of U(1) then v in V lambda implies R of e to the i phi v is in V lambda.

Proof:

Suppose v is in V lambda. Apply R of e to the i theta to R of e to the i phi v. Because U(1) is abelian and R is a representation, the matrices R of e to the i theta and R of e to the i phi commute with one another, and we get R of e to the i theta applied to (R of e to the i phi v equals R of e to the i phi times R of e to the i theta v, which equals R of e to the i phi lambda v, which equals lambda times R of e to the i phi v Therefore R of e to the i phi v is an eigenvector of R of e to the i theta with eigenvalue lambda, that is R of e to the i phi v is in V lambda as required.

If C n is irreducible then this implies C n equals V lambda because V lambda would otherwise be a proper subrepresentation. This means that R of e to the i theta equals 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 of theta that depends on theta. In other words, we can think of lambda as a map (actually a homomorphism) lambda from U(1) to C star to the nonzero complex numbers such that R of e to the i theta equals lambda of theta times the identity for all theta (lambda is nonzero because R of e to the i theta is invertible).

Lemma:

lambda of theta is a unit complex number.

Proof:

We have an invariant Hermitian inner product angle brackets, so v angle brackets v equals the angle brackets of R of e to the i theta v with itself, which equals the angle brackets of lambda of theta times v with itself, which equals the absolute value squared of lambda of theta times v angle brackets v using the fact that the inner product is sesquilinear (we can pull out the two factors of lambda of theta but the first one picks up a complex conjugate sign). Therefore the absolute value squared of lambda of theta equals 1.

This tells us that lambda from 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 of theta equals e to the i m theta for some integer m. This tells us that R of e to the i theta equals e to the i m theta times the identity

Since our representation is irreducible, we can now deduce that it is 1-dimensional. This is because any 1-dimensional complex line in C n is preserved under the map e to the i m theta times the identity (just rescales by e to the i m theta, which rotates every complex line inside itself by m theta) so any complex line in C n is a subrepresentation. Since 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 \CC^n then u angle brackets v subscript inv equals the integral from 0 to 2 pi of the angle bracket between R of e to the i theta applied to u and R of e to the i theta applied to v, d theta over 2 pi is also a Hermitian inner product on \CC^n.