Math 202B: Lecture 22

Let G be a finite group, and let (V,\varphi) be a unitary representation of G. Recall that there is a canonically associated linear representation (V,\Phi) of the convolution algebra \mathcal{C}(G) – if

A = \sum\limits_{g \in G} A(g)E_g

then

\Phi^\lambda(A) = \sum\limits_{g \in G} A(g)\varphi^\lambda(g),

a sum of unitary operators in \mathcal{L}(V^\lambda). Recall also that the class algebra \mathcal{Z}(G), which is by definition the center of \mathcal{C}(G), is the algebra of functions on G which are constant on conjugacy classes, aka central functions.

Theorem 21.1. We have \Phi(A) \in \mathrm{Hom}_{\mathcal{C}(G)}(V,V) if and only if A \in \mathcal{C}(G).

Proof: If A \in \mathcal{Z}(G), then A is a central element in \mathcal{C}(G), i.e.

AB=BA \quad \text{ for all }B \in \mathcal{C}(G).

Since \Phi \colon \mathcal{C}(G) \to \mathcal{L}(V) is an algebra homomorphism, we have

\Phi(A)\Phi(B)=\Phi(AB)=\Phi(BA)=\Phi(B)\Phi(A),

which is exactly the condition for the operator \Phi(A) to be in \mathrm{Hom}_{\mathcal{C}(G)}(V,V).

Problem 21.1. Prove the converse.

-QED

Now suppose the unitary representation (V,\varphi) of G is irreducible. Then, so is the the linear representation (V,\Phi) of \mathcal{C}(G). Thus, Burnside’s theorem tells us that the image of \mathcal{C}(G) under \Phi is all of \mathcal{L}(V). Since the center of \mathcal{L}(V) is precisely the set of scalar operators \omega I, with \omega \in \mathbb{C} and I the identity operator in \mathcal{L}(V), for every A \in \mathcal{Z}(G) we have \Phi(A) = \omega^V(A)I for some \omega^V(A) \in \mathbb{C}. Thus, we have a map

\omega^V\colon \mathcal{Z}(G) \longrightarrow \mathbb{C}

which sends every A \in \mathcal{Z}(G) to the unique eigenvalue \omega^V(A) of the operator \Phi(A) \in \mathcal{L}(V). This map is called the central character of \mathcal{Z}(G) corresponding to the irreducible linear representation (V,\Phi), or equivalently the irreducible unitary representation (V,\varphi).

Problem 21.2. Prove that \omega^V is an algebra homomorphism \mathcal{Z}(G) \to \mathbb{C}, and that

\omega^V(A) = \frac{1}{\dim V}\mathrm{Tr}\, \Phi(A), \quad A \in \mathcal{Z}(G).

Now let \Lambda be a set parameterizing isomorphism classes of irreducible unitary representations of G. For each \lambda \in \Lambda, let (V^\lambda,\varphi^\lambda) be a representative of the isomorphism class of unitary irreps corresponding to \lambda, and let

\chi^\lambda(g) = \mathrm{Tr}\, \varphi^\lambda(g), \quad g \in G,

be the corresponding character. Then \chi^\lambda is a central function on G, and we have shown that \{\chi^\lambda \colon \lambda \in \Lambda\} is an orthogonal set in \mathcal{Z}(G), i.e.

\langle \chi^\lambda,\chi^\mu\rangle= \sum\limits_{g \in G} \overline{\chi^\lambda(g)}\chi^\mu(g) = \delta_{\lambda\mu}|G|.

We will now show that the set \{\chi^\lambda \colon \lambda \in \Lambda\} of irreducible characters of G is in fact a basis of the class algebra \mathcal{Z}(G).

Theorem 22.2. The orthogonal set \{\chi^\lambda \colon \lambda \in \Lambda\} spans \mathcal{Z}(G).

Proof: Let W=\mathrm{Span} \{\chi^\lambda \colon \lambda \in \Lambda\} be the subspace of \mathcal{Z}(G) spanned by the irreducible characters of G. We want to show that W=\mathcal{Z}(G). We will do this by showing that W^\perp = \{0\}. In other words, the only central function on G orthogonal to every irreducible character is the zero function.

Suppose

A = \sum\limits_{g \in G} A(g)E_g

is a central function on G such that

\langle A,\chi^\lambda \rangle = \sum\limits_{g \in G} \overline{A(g)}\chi^\lambda(g) =0

for all \lambda \in \Lambda. For each \lambda \in \Lambda, let \omega^\lambda be the central character of \mathcal{Z}(G) corresponding to (V^\lambda,\Phi^\lambda), and let

\overline{A} =\sum\limits_{g \in G} \overline{A(g)}E_g,

which is again a central function on G. We then have

\omega^\lambda(\bar{A}) = \frac{1}{\dim V^\lambda} \mathrm{Tr}\, \Phi^\lambda(\overline{A})=\frac{1}{\dim V^\lambda} \mathrm{Tr}\, \sum\limits_{g \in G} \overline{A(g)} \varphi^\lambda(g)=\frac{1}{\dim V^\lambda} \sum\limits_{g \in G} \overline{A(g)} \chi^\lambda(g)=0.

Thus, our hypothesis that A \in \mathcal{Z}(G) is orthogonal to all irreducible characters of G forces \Phi^\lambda(\bar{A}) to be the zero operator in \mathcal{L}(V^\lambda) for every \lambda \in \Lambda. Since every linear representation of \mathcal{C}(G) is isomorphic to a direct sum of irreducible representations (isotypic decomposition, Lecture 20), this says that the image of A is the zero operator in every linear representation of \mathcal{C}(G). This includes the regular representation of \mathcal{C}(G), which is faithful – the image of \mathcal{C}(G) in the regular representation is isomorphic to \mathcal{C}(G). We conclude that \bar{A}, and hence A itself, is the zero function on G, i.e. the zero element of \mathcal{Z}(G).

-QED

Now we know that the number of isomorphism classes of irreducible unitary representations of G is exactly equal to the number of conjugacy classes in G. Thus, our set \Lambda indexing the isomorphism classes of irreps of G also indexes the conjugacy classes in G. This gives us two orthogonal bases of the class algebra \mathcal{Z}(G).

First, let \{C_\alpha \colon \alpha \in \Lambda\} be the set of conjugacy classes of \mathcal{Z}(G). For each \alpha \in \Lambda, let K_\alpha \in \mathcal{Z}(G) be the indicator function of C_\alpha, i.e.

K_\alpha(g) = \begin{cases} 1, \text{ if } g\in C_\alpha \\ 0, \text{ if }g \not\in C_\alpha \end{cases}.

Then, for every central function A \colon G \to \mathbb{C} we have

A = \sum\limits_{\alpha \in \Lambda} A(\alpha) K_\alpha,

where A(\alpha) denotes the value of A(g) for any group element g \in C_\alpha. Thus, the functions \{K_\alpha \colon \alpha \in \Lambda\} span \mathcal{Z}(G). Moreover, we have

\langle K_\alpha,K_\beta \rangle = \langle \sum\limits_{g \in C_\alpha} E_g,\sum\limits_{g \in C_\beta}E_g \rangle = \sum\limits_{g\in C_\alpha}\sum\limits_{h \in C_\beta} \langle E_g,E_h\rangle = \delta_{\alpha\beta}|C_\alpha|,

where the final equality follows from the fact that C_\alpha and C_\beta are the same set if \alpha=\beta, and disjoint if \alpha \neq \beta.

We now have two orthogonal bases in \mathcal{Z}(G).

The character basis of \mathcal{Z}(G) is defined by

E^\lambda = \sum\limits_{\alpha \in \Lambda} \chi^\lambda_\alpha K_\alpha, \quad \lambda \in \Lambda,

and since

|\{E^\lambda \colon \lambda \in \Lambda\}|=|\{K^\alpha \colon \alpha \in \Lambda\}|,

in order to show that the character basis is indeed an orthogonal basis of \mathcal{Z}(G) we only need to prove its orthogonality.

Theorem 22.3. The set \{E^\lambda \colon \lambda \in \Lambda\} is orthogonal in \mathcal{Z}(G).

Proof: For \lambda,\mu \in \Lambda, we have

\langle E^\lambda,E^\mu\rangle = \sum\limits_{\alpha,\beta \in \Lambda} \overline{\chi^\lambda_\alpha}\chi^\mu_\beta \langle K_\alpha,K_\beta\rangle = \sum\limits_{\alpha \in \Lambda} \overline{\chi^\lambda_\alpha}\chi^\mu_\alpha|C_\alpha|=\sum\limits_{g \in G}\overline{\chi^\lambda(g)}\chi^\mu(g)=\delta_{\lambda\mu}|G|,

where the final equality is the original formulation of orthogonality of irreducible characters in \mathcal{Z}(G).

-QED

At this point, we are very close to constructing a Fourier basis of \mathcal{Z}(G) – we will see next time that the functions

F^\lambda = \frac{|C_\lambda|}{|G|}E^\lambda, \quad \lambda \in \Lambda

are orthogonal projections,

F^\lambda F^\mu = \delta_{\lambda\mu}F^\lambda, \quad \lambda,\mu \in \Lambda,

so that \mathcal{Z}(G) is isomorphic to the pointwise function algebra \mathcal{F}(\Lambda), extending our result from the case of abelian groups, where \mathcal{Z}(G)=\mathcal{C}(G).

Published by Jonathan Novak

Mathematician at UC San Diego.

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