Math 202B: Lecture 9

The convolution and class algebras $\mathcal{C}(G)$ and $\mathcal{Z}(G)$ of a group $G$ are apparently quite different from the other types of algebras we have seen, namely function algebras $\mathcal{F}(X)$ and linear algebras $\mathcal{L}(V).$ However, they are in fact closely related.

Theorem 9.1. The convolution algebra $\mathcal{C}(G)$ is isomorphic to a subalgebra of a linear algebra.

Proof: Since $\mathcal{C}(G)$ has a natural scalar product, the $\ell^2$ -scalar product

$\langle A,B \rangle = \sum\limits_{g \in G} \overline{A(g)}B(g),$

it is a Hilbert space as well as an algebra. For each $A \in \mathcal{C}(G)$ , consider the function $\mathsf{L}(A) \colon \mathcal{C}(G) \to \mathcal{C}(G)$ defined by

$\mathsf{L}(A)B = AB, \quad B \in \mathcal{C}(G).$

Observe that $\mathsf{L}(A)$ is a linear operator on $\mathcal{C}(G),$ i.e.

$\mathsf{L}(A)(\beta B + \gamma C) = A(\beta B + \gamma C) = \beta AB + \gamma AC = \beta \mathsf{L}(A)B + \gamma \mathsf{L}(A)C.$

Thus, $\mathsf{L}$ is a function from the convolution algebra $\mathcal{C}(G)$ to the linear algebra $\mathcal{L}(\mathcal{C}(G)).$ Moreover, the map $\mathsf{L}$ itself is a linear transformation, i.e.

$\mathsf{L}(\alpha A + \beta B) C= (\alpha A +\beta B)C=\alpha AB +\beta AC = \alpha \mathsf{L}(A)B + \gamma \mathsf{L}(B)C.$

We claim that $\mathsf{L}$ is an injective linear transformation. Since $\mathsf{L}$ is a linear transformation of $\mathcal{C}(G)$ , it is completely determined by its action on the group basis $\{E_g \colon g \in G\}$ of $\mathcal{C}(G)$ . Thus, it suffices to show that the linear operators $\mathsf{L}(E_g)$ and $\mathsf{L}(E_h)$ on $\mathcal{C}(G)$ are distinct unless $g=h.$ Since $\mathsf{L}(E_g)$ and $\mathsf{L}(E_h)$ are linear operators on $\mathcal{C}(G),$ they are uniquely determined by their actions on the group basis $\{E_g \colon g \in G\}$ of $\mathcal{C}(G)$ . We have

$\mathsf{L}(E_g)E_k = E_gE_k = E_{gk} \quad\text{and}\quad \mathsf{L}(E_h)E_k = E_hE_k = E_{hk},$

hence

$\mathsf{L}(E_g)E_k=\mathsf{L}(E_h)E_k \iff gk = hk \iff g=h.$

From this we can conclude that the image of $\mathcal{C}(G)$ in $\mathcal{L}(\mathcal{C}(G))$ under $\mathsf{L}$ is a subspace isomorphic to $\mathcal{C}(G)$ . It remains to show that $\mathsf{L}$ is not just a linear transformation, but an algebra homomorphism, and we leave this as an exercise (that you really should do).

-QED

The proof of Theorem 8.1 is the linear version of Cayley’s theorem from group theory: instead of representing $G$ as a subgroup of the group of permutations of $G,$ it represents $\mathcal{C}(G)$ as a subalgebra of the algebra of linear operators on $\mathcal{C}(G).$ This is called the left regular representation of $\mathcal{C}(G).$

Corollary 9.2. The class algebra $\mathcal{Z}(G)$ is isomorphic to a function algebra.

Proof: Since $\mathcal{Z}(G)$ is a subalgebra of $\mathcal{C}(G)$ , and since $\mathcal{C}(G)$ is isomorphic to a subalgebra of a linear algebra, $\mathcal{Z}(G)$ is isomorphic to a subalgebra of a linear algebra. Thus, $\mathcal{Z}(G)$ is isomorphic to a commutative subgalgebra of a linear algebra, and we have previously shown that all commutative subalgebras of linear algebras are isomorphic to function algebras.

-QED

For an example illustrating how the proof of Theorem 8.1 works, let us take $G=\{g^0,g^1,g^2,g^3\}$ to be a cyclic group of order four with generator $g$ , so that $g^0=g^4=\dots=e$ is the group unit. The group basis of $\mathcal{C}(G)$ is thus $E_{g^0},E_{g^1},E_{g^2},E_{g^3},$ and we denote these vectors by $E_0,E_1,E_2,E_3$ for brevity. Let $A \in \mathcal{C}(G)$ be any function on $G,$ and let

$A = \alpha_0 E_0 + \alpha_1 E_1 + \alpha_2 E_2 + \alpha_3 E_3$

be its expansion in the group basis, so that

$\alpha_0 = A(g^0),\ \alpha_1=A(g),\ \alpha_2=A(g^2),\ \alpha_3=A(g^3).$

Problem 9.1. Show by direct calculation that the matrix of $\mathsf{L}(A) \in \mathcal{L}(\mathcal{C}(G))$ with respect to the ordered basis $E_0,E_1,E_2,E_3$ of $\mathcal{C}(G)$ is

$\mathsf{L}(A) = \begin{bmatrix} \alpha_0 & \alpha_1 & \alpha_2 & \alpha_3 \\ \alpha_3 & \alpha_0 & \alpha_1 & \alpha_2 \\ \alpha_2 & \alpha_3 & \alpha_0 & \alpha_1 \\ \alpha_1 & \alpha_2 & \alpha_3 & \alpha_0 \end{bmatrix}.$

In the case where $G$ is an abelian group, as in the example above, we have that $\mathcal{C}(G) = \mathcal{Z}(G)$ . Moreover, it is possible to establish Corollary 8.1 directly, without appealing to Theorem 8.1. This is done constructively, by finding an explicit basis of orthogonal projections in the commutative convolution algebra $\mathcal{C}(G)$ called its Fourier basis. The advantage of this direct approach is that it also gives us an explicit description of the spectrum of all matrices in the left regular representation of $\mathcal{C}(G).$ This is very useful in applications – in particular, matrices in the left regular representation of a cyclical group are called circulant matrices and they are important in engineering.

To see begin to see where a projection basis of $\mathcal{C}(G)$ might come from, recall that we previously showed the non-existence of an algebra homomorphism $\mathsf{T} \colon \mathcal{L}(V) \to \mathbb{C}$ for $V$ a Hilbert space of dimension at least two. This reflects the fact that linear algebras $\mathcal{L}(V)$ are maximally noncommuative. But we have also seen that for $G$ a group with at least two elements, the dimension of $\mathcal{Z}(G)$ is at least two, so convolution algebras are always at least one degree more commutative than linear algebras, and therefore might admit homomorphisms to the complex numbers.

Theorem 9.3. A linear transformation $\mathsf{T} \colon \mathcal{C}(G) \to \mathbb{C}$ is an algebra homomorphism if and only if the function $\chi \in \mathcal{C}(G)$ defined by

$\chi(g) = \mathsf{T}(E_g), \quad g \in G,$

is a group homomorphism taking values in $U(\mathbb{C}),$ the unitary group of the complex numbers.

Problem 9.2. Prove Theorem 8.3.

Definition 9.1. A group homomorphism $\chi \colon G \to U(\mathbb{C})$ is called a character of $G.$ The set of all characters of $G$ is denoted $\widehat{G}$ .

Observe first of all that $\widehat{G}$ is a set of functions on $G,$ is a subset of $\mathcal{C}(G),$ the space of all $\mathbb{C}$ -valued functions on $G.$ It is this special subset of functions tethered to the group law in $G$ that we look to for orthogonal projections. No matter what the group $G$ is, the set $\widehat{G}$ is nonempty: it contains at least the trivial character defined by $\chi(g)=1,$ $g \in G.$ However, for highly noncommutative groups there may not be many nontrivial characters.

Problem 9.3. Determine the Fourier dual of the symmetric group $S_n.$

For abelian groups the situation is much better and in fact we always have $|G|=|\widehat{G}|.$ To begin, recall that every finite abelian group is isomorphic to a product of cyclic groups. We thus fix a positive integer $r \in \mathbb{N},$ and $r$ positive integers $n_1,\dots,n_r \in \mathbb{N},$ and consider the group

$G = G_1 \times \dots \times G_r,$

where $G_i$ is a cyclic group of order $n_i$ with generator $g_i.$ Define the dual group of $G$ to be

$\Lambda = \{\alpha=(\alpha_1,\dots,\alpha_r) \colon \alpha_k \in \{0,1,\dots,n_k-1\},\ 1 \leq k \leq r\}.$

That is,

$\Lambda = \mathbb{Z}_{n_1} \times \dots \times \mathbb{Z}_{n_r},$

where $\mathbb{Z}_n=$ is the additive group of integers modulo $n$ . We can parameterize $G$ by the points of $\Lambda,$ writing

$g_\alpha=(g_1^{\alpha_1},\dots,g_r^{\alpha_r}), \quad \alpha \in \Lambda.$

Indeed, the parameterization $\alpha \mapsto g_\alpha$ is a group isomorphism $\Lambda \to G$ (Exercise: prove this, noting that because $|\Lambda|=|G|$ it is sufficient to show the parameterization is an injective group homomorphism).

Theorem 9.4. For every $\lambda \in \Lambda,$ the function $\chi^\lambda \colon G \to U(\mathbb{C})$ defined by

$\chi^\lambda(g_\alpha) = \omega_1^{\alpha_1\lambda_1} \dots \omega_r^{\alpha_r\lambda^r},$

where $\omega_k=\exp\left(\frac{2\pi i}{n_k}\right)$ is a principal $k$ th root of unity, is a character of $G,$ and every character of $G$ is of this form.

Proof: For any $\lambda \in \Lambda,$ it is clear that $\chi^\lambda(e)=1,$ because the identity element $e \in G$ has parameters $\alpha=(0,0,\dots,0).$ Moreover, for any $\alpha,\beta \in \Lambda$ we have

$\chi^\lambda(g_\alpha g_\beta) = \chi^\lambda(g_{\alpha+\beta})=(\omega_1)^{(\alpha_1+\beta_1)\lambda_1} \dots (\omega_r)^{(\alpha_r+\beta_r)\lambda_r)} = (\omega_1)^{\alpha_1\lambda_1} \dots (\omega_r)^{\alpha_r\lambda_r}(\omega_1)^{\beta_1\lambda_1} \dots (\omega_r)^{\beta_r\lambda_r}=\chi^\lambda(g_\alpha)\chi^\lambda(g_\beta),$

so $\chi^\lambda$ is indeed a group homomorphism $G \to U(\mathbb{C}).$ The fact that every homomorphism $\chi \colon G \to U(\mathbb{C})$ is $\chi^\lambda$ for some $\lambda \in \Lambda$ is left as an exercise.

-QED

We now have a special subset $\widehat{G}=\{\chi^\lambda \colon \lambda \in \Lambda\}$ of the convolution algebra $\mathcal{C}(G)$ of the finite abelian group $G$ , namely the set $\widehat{G}$ of all homomorphisms to the unitary group $U(\mathbb{C}).$ We now claim that the characters form a basis of $\mathcal{C}(G).$ Since the number of characters is $|\widehat{G}|=|\Lambda|=|G|,$ which is the dimension of $\mathcal{C}(G)$ , it is sufficient to show that $\widehat{G}=\{\chi^\lambda \colon \lambda \in \Lambda\}$ is a linearly independent set in $\mathcal{C}(G).$

Theorem 9.5. The set $\{\chi^\lambda \colon \lambda \in \Lambda\}$ is orthogonal with respect to the $\ell^2$ -scalar product on $G$ – we have

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

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

$\langle \chi^\lambda,\chi^\mu \rangle = \sum\limits_{\alpha \in \Lambda} \overline{\chi^\lambda(g_\alpha)}\chi^\mu(g_\alpha) = \sum\limits_{\alpha \in \Lambda}(\omega_1)^{\alpha_1(\mu_1-\lambda_1)} \dots (\omega_r)^{\alpha_r(\mu_r-\lambda_r)}=\left(\sum\limits_{\alpha_1=0}^{n_1-1} \zeta_1^{\alpha_1}\right) \dots \left(\sum\limits_{\alpha_r=0}^{n_r-1} \zeta_r^{\alpha_r}\right),$

where

$\zeta_1 = \omega_1^{\mu_1-\lambda_1},\ \dots,\ \zeta_r=\omega_r^{\mu_r-\lambda_r}.$

Thus if $\lambda=\mu$ we have

$\langle \chi^\lambda,\chi^\mu \rangle = n_1 \dots n_r = |\Lambda|=|G|,$

and if $\lambda \neq \mu$ we have

$\langle \chi^\lambda,\chi^\mu \rangle =\frac{1-\zeta_1^{n_1}}{1-\zeta_1} \dots \frac{1-\zeta_r^{n_r}}{1-\zeta_r},$

where the denominator of each fraction is nonzero and the numerator is zero, because $\zeta_k=\omega_k^{\mu_k-\lambda_k}$ is an $n_k$ th root of unity.

-QED

The orthogonal basis $\widehat{G}=\{\chi^\lambda \colon \lambda \in \Lambda\}$ of $\mathcal{C}(G)$ is called its character basis. It is convenient to write $\chi^\lambda_\alpha := \chi^\lambda(g_\alpha),$ since this highlights the symmetry $\chi^\lambda_\alpha = \chi^\alpha_\lambda.$ The $|G| \times |G|$ symmetric matrix $X=[\chi^\lambda_\alpha]$ is called the character table of $G,$ and Theorem 8.3 says that $\frac{1}{\sqrt{|G|}}$ is a symmetric unitary matrix. Another way to say the same thing is that the rescaled character basis

$E^\lambda = \frac{1}{\sqrt{|G|}} \chi^\lambda, \quad \lambda \in \Lambda,$

is an orthonormal basis of the convolution algebra $\mathcal{C}(G).$ In fact, the further scaling

$F^\lambda = \frac{1}{|G|}F^\lambda, \quad \lambda \in \Lambda,$

is even better, for the following reason.

Theorem 9.6. The elements of the basis $\{F^\lambda \colon \lambda \in \Lambda\}$ are orthogonal projections in $\mathcal{C}(G).$

Proof: For any $\lambda \in \Lambda,$ we have

$(F^\lambda)^* = \left( \frac{1}{|G|}\sum\limits_{g \in G} \chi^\lambda(g) E_g\right)^*=\frac{1}{|G|}\sum\limits_{g \in G} \overline{\chi^\lambda(g)} E_g^* =\frac{1}{|G|} \sum\limits_{g \in G} \chi^\lambda(g^{-1}) E_{g^{-1}} = F^\lambda.$

For any $\lambda,\mu \in \Lambda,$ we have

$F^\lambda F^\mu = \left( \frac{1}{|G|}\sum\limits_{g \in G} \chi^\lambda(g) E_g\right)\left( \frac{1}{|G|}\sum\limits_{g \in G} \chi^\mu(g) E_g\right)=\frac{1}{|G|^2}\sum\limits_{g \in G} \left( \sum\limits_{h \in G}\chi^\lambda(gh^{-1})\chi^\mu(h)\right)E_g = \frac{1}{|G|^2}\sum\limits_{g \in G} \chi^\lambda(g)\left( \sum\limits_{h \in G}\chi^\lambda(h^{-1})\chi^\mu(h)\right)E_g.$

The internal sum is

$\sum\limits_{h \in G}\chi^\lambda({h^{-1}})\chi^\mu(h)=\sum\limits_{h \in G}\overline{\chi^\lambda(h)}\chi^\mu(h)=\delta_{\lambda\mu}|G|,$

where the final equality is Theorem 8.3. Thus

$F^\lambda F^\mu = \frac{1}{|G|^2}\delta_{\lambda\mu}|G|\sum\limits_{g \in G} \chi^\lambda(g)E_g=\delta_{\lambda\mu}F^\lambda.$

-QED

Since we know that any algebra with a basis of orthogonal projections is isomorphic to the function algebra (Lecture 3), Theorem 8.4 gives the promised second proof of the fact that the convolution algebra $\mathcal{C}(G)$ of an abelian group $G$ is isomorphic to a function algebra. In this particular case, the basis $\{F^\lambda \colon \lambda \in \Lambda\}$ is known as the Fourier basis of $\mathcal{C}(G).$

Math 202B: Lecture 9

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