Math 202B: Lecture 12

Let $G$ be a group with group law written multiplicatively: for any elements $g,h \in G$ , their product is $gh \in G$ . Since $G$ is a finite set, we can form the algebra $\mathcal{F}(G)$ of functions $A \colon G \to \mathbb{C}$ . As we know, this function algebra has a natural basis consisting of the elementary functions

$E_g(h) = \delta_{gh},\quad g,h \in G$

which are selfadjoint orthogonal idempotents in $\mathcal{F}(G)$ . However, the function algebra $\mathcal{F}(G)$ has nothing to do with the group structure of $G,$ so from the point of view of this construction the fact that $G$ is a group and not just a set is irrelevant.

We can use the underlying group structure to define a different algebra $\mathcal{C}(G)$ whose elements are functions $A \colon G \to \mathbb{C}$ , but whose operations differ from those of $\mathcal{F}(G)$ and are defined using the group structure of $G$ . In $\mathcal{C}(G)$ , multiplication of functions not defined pointwise, but rather by convolution: we declare

$E_gE_h = E_{gh}, \quad g,h \in G.$

Thus, the multiplication tensor of $\mathcal{C}(G)$ is not the three-dimensional identity matrix, but rather a two-dimensional object: the multiplication table of $G.$ Thus for any two functions

$A = \sum\limits_{g \in G} \alpha_g E_g \quad\text{and}\quad B=\sum\limits_{g \in G}\beta_gE_g,$

their convolution product is given by

$AB=\sum\limits_{g,h \in G} \alpha_g\beta_h E_{gh},$

which can equivalently be written

$AB = \sum\limits_{k \in G} \left(\sum\limits_{gh=k}\alpha_g\beta_h \right)E_k,$

$AB= \sum\limits_{g \in G} \left(\sum\limits_{h \in G}\alpha_{gh^{-1}}\beta_h \right)E_g$

$AB= \sum\limits_{g \in G} \left(\sum\limits_{h \in G}A(gh^{-1})B(h) \right)E_g.$

This is very different from the product of $A$ and $B$ viewed as elements of $\mathcal{F}(G)$ , which is

$AB=\sum\limits_{g \in G} A(g)B(g)E_g.$

Similarly, instead of defining conjugation in $\mathcal{C}(G)$ using pointwise using conjugation in $\mathbb{C}$ , we define it using the the natural involution in $G$ , which is taking inverses,

$E_g^* = E_{g^{-1}}, \quad g \in G,$

and extend this antilinearly,

$A^* = \left(\sum\limits_{g \in G} \alpha_g E_g\right)^*=\sum\limits_{g \in G}\bar{\alpha}_g E_{g^{-1}} = \sum\limits_{g \in G} \bar{\alpha}_{g^{-1}}E_g.$

This group-theoretic conjugation is indeed antimultiplicative with respect to convolution,

$(E_g*E_h)^* = E_{gh}^*=E_{(gh)^{-1}}=E_{h^{-1}g^{-1}} = E_{h^{-1}}*E_{g^{-1}}=E_h^**E_g^*,$

as well as involutive,

$(E_g^*)^* = E_{g^{-1}}^*=E_{(g^{-1})^{-1}}= E_g.$

Finally, the multiplicative unit in $\mathcal{C}(G)$ is $I= E_e$ , where $e \in G$ is the group unit, as opposed to being the constant function equal to $1$ on every $g \in G$ , which is the multiplicative unit in $\mathcal{F}(G).$

Let us record the above as an official definition.

Definition 12.1. The convolution algebra (aka a group algebra) of a group $G$ is the vector space

$\mathcal{C}(G) = \{A \colon G \to \mathbb{C}\}$

of complex-valued functions on $G$ with multiplication defined by

$[AB](g) = \sum\limits_{h \in G}A(gh^{-1})B(h)$

and conjugation defined by

$[A^*](g) = \overline{A(g^{-1})}.$

In summary, $\mathcal{F}(G)$ and $\mathcal{C}(G)$ are exactly the same as vector spaces, but they are different as algebras. In particular, the elementary functions $\{E_g \colon g \in G\}$ are a basis of $\mathcal{F}(G)$ consisting of orthogonal projections, but in $\mathcal{C}(G)$ they form a basis consisting of unitary elements.

Problem 12.1. Prove that if $G,H$ are isomorphic groups then their convolution algebras $\mathcal{C}(G)$ and $\mathcal{C}(H)$ are isomorphic algebras (the converse is false, bonus points if you give a counterexample). Also show that $\mathcal{C}(G)$ is commutative if and only if $G$ is abelian.

To further highlight the difference between $\mathcal{C}(G)$ and $\mathcal{F}(G)$ , let us consider their selfadjoint elements. To say that a function $A \colon G \to \mathbb{C}$ is selfadjoint as an element of $\mathcal{F}(G)$ simply means that it is real-valued:

$A^*(g) = A(g) \iff \overline{A(g)} = A(g).$

However, $A$ being selfadjoint in $\mathcal{C}(G)$ means something else:

$A^*(g)=A(g) \iff \overline{A(g^{-1}})=A(g).$

To further compare and contrast the function algebra $\mathcal{F}(G)$ and the convolution algebra $\mathcal{C}(G)$ , let us consider their subalgebras. From Lecture 3 we know that the lattice of subalgebras of $\mathcal{F}(G)$ is isomorphic to the lattice of partitions of $G$ , which has nothing to do with the group structure on $G$ . On the other hand, we have a subalgebra of $\mathcal{C}(G)$ associated to every subgroup of $H$ of $G$ , defined by

$\mathcal{A}(H) = \{A \in \mathcal{C}(g) \colon A(g) = 0 \text{ unless }g \in H\}$ .

In lecture, we also gave a characterization of unitary elements in $\mathcal{C}(G)$ .

Problem 12.2. Prove that $\mathcal{A}(H)$ is a subalgebra of $\mathcal{C}(G)$ , and moreover that $\mathcal{A}(H)$ is isomorphic to $\mathcal{C}(H).$

This raises the question of whether subalgebras of $\mathcal{C}(G)$ are in bijection with subgroups of $G$ , i.e. whether the lattice of subalgebras of $\mathcal{C}(G)$ is isomorphic to the lattice of subgroups of $G$ , which would be analogous to the situation with function algebras. This is not true in general. Consider the following element of $\mathcal{C}(G)$ ,

$P = \frac{1}{|G|} \sum_{g \in G} E_g,$

which is the average of the elementary functions. Equivalently, $P\colon G \to \mathbb{C}$ is the constant function

$P(g) = \frac{1}{|G|}, \quad g \in G.$

Then $P$ is selfadjoint (make sure you understand why), and also

$P^2 = \frac{1}{|G|^2} \sum_{h \in G} \sum_{g \in G} E_gE_h=\frac{1}{|G|^2} \sum_{h \in G} \sum_{g \in G} E_{gh}.$

Now, the internal sum is just

$\sum_{g \in G} E_{gh} = \sum_{g \in G} E_g,$

since multiplying each element $g \in G$ by a fixed group element $h \in G$ just permutes the points of the group (make sure you understand why). Therefore, $P^2=P$ and we conclude that the average of all elementary functions $E_g$ is a projection in $\mathcal{C}(G)$ . Now consider the subalgebra of $\mathcal{C}(G)$ generated by the the multiplicative identity $I=E_e$ and $P$ ,

$\mathcal{A}=\mathrm{alg}\{I,P\} = \{\alpha I + \beta P \colon \alpha,\beta \in \mathbb{C}\}$ .

Since $I,P$ are linearly independent (make sure you understand why), we have located a two-dimensional subalgebra inside $\mathcal{C}(G)$ , for any group $G$ . If $G$ is a group of odd order, Lagrange’s theorem tells us that any subgroup $H \leq G$ must also have odd order, and consequently the subalgebra $\mathcal{A}=\mathrm{alt}\{I,P\}$ cannot be the algebra of functions vanishing outside some subgroup $H$ of $G$ , for then $\mathcal{A}$ would be isomorphic to $\mathcal{C}(H)$ and its dimensions would be the cardinality of $H$ , which is impossible since $|H|$ cannot be even.

Math 202B: Lecture 12

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