Definition. The character of a continuous G-module V is defined as:
This is a continuous map since it is an example of a matrix coefficient.
Clearly for any . The following are quite easy to show:
The last equality, that is the complex conjugate of , follows from the following:
Lemma 1. For , the linear map has all eigenvalues on the unit circle T.
Suppose the eigenvalues are with the right multiplicity. Then has trace But is continuous; since G is compact the image is bounded. It remains to show: if is bounded for all integers m we must have all .
Suppose ; replacing g by its inverse, we assume and If , increases much faster than the other and we get a contradiction. On the other hand, if , write for where R>1 and are real. Consider the continuous homomorphism
Its image lies in the torus group which is compact. Hence, there are arbitrarily large m such that (see exercise). So the real part of exceeds and increases much faster than the remaining ♦
Prove that if is a continuous homomorphism to a compact Hausdorff group, then is infinite for any open neighbourhood U of e.
Lemma 2. For , the linear map is diagonalizable.
Write in block Jordan form. Suppose the matrix is not diagonal, say entry (1, 2) is 1 with corresponding eigenvalue . The matrix for has entry (1, 2) equal to which is unbounded since However, G is compact so taking entry (1, 2) of the representation has bounded image, which is a contradiction. ♦
Orthogonality of Characters
The space of G-invariant elements of a G-module V is denoted by:
Proposition. The dimension of is:
Left-invariance then implies that the image of π lies in Also, if then Putting these two facts together gives us so π is the projection map onto Hence:
as desired. ♦
In particular, replacing V by , note that:
So the dimension of this space is the integral of:
In other words:
which we denote by Schur’s lemma then gives us:
Orthonormality of Irreducible Characters.
If V, W are irreps of G, then
Suppose now G is compact Hausdorff and abelian.
Theorem. Each irreducible representation V of G is 1-dimensional.
First we show that every acts as a scalar. By lemma 2 above, V decomposes as a direct sum of eigenspaces for Each is G-invariant since:
Since V is irreducible, we have so g acts as a scalar.
Since every element of G acts as a scalar, every vector subspace is G-invariant; so dim(V) = 1. ♦
Hence by lemma 1, every irrep of G is a continuous homomorphism , where is the unit circle. The following special case is of interest.
Definition. The n-torus is the topological group
[In fact, any compact Hausdorff abelian G must be of the form (finite group) but we won’t be needing this fact.]
For i = 1,…,n, let be the homomorphism taken by projecting to the i-th coordinate. More generally, for any , we define:
Thus it takes to
Theorem. Any irrep of is of the form for some
Suppose n=1: for a continuous write T as and compose The subgroup lies in the kernel; since the kernel is a closed subgroup it is either or for some m>0. So is of the form for some
For n>1 consider the homomorphism for i = 1,…,n, by taking where the i-th component is v. By the previous case, is of the form for some Since for all we get:
Corollary. A character of is a finite linear combination of whose coefficients are positive integers.
Replacing with variables , this gives us a polynomial in , also known as a Laurent polynomial in Hence we have a correspondence: