Math 202B: Lecture 13

***All problems in this lecture due Feb. 15 at 23:59***

Let G be a finite group. We have been comparing and contrasting the function algebra \mathcal{F}(G) and the convolution algebra \mathcal{C}(G). The function algebra \mathcal{F}(G) sees G as a finite set, and does not interact with its group structure. The convolution algebra does interact with the group structure of G, and in particular is commutative if and only if G is abelian.

We understand subalgebras of \mathcal{F}(G) very well: they are in bijection with partitions of G. Namely, if \mathfrak{p} is a partition of G, then we have a corresponding subalgebra of \mathcal{F}(G), the set of functions on G which are constant on the blocks of \mathfrak{p}.

Our understanding of subalgebras of \mathcal{C}(G) is less complete. We know that every subgroup H of G gives rise to a corresponding subalgebra \mathcal{A}(H) of \mathcal{C}(G), namely the set of functions on G which vanish outside H. We also used the averaging trick to construct a two-dimensional subalgebra of \mathcal{C}(G) of a different kind.

Since pointwise multiplication of functions and convolution of functions are very different operations, there is no reason to expect that \mathcal{A}(\mathfrak{p}) should be closed under convolution. But perhaps this could be the case for some “special” partition of G which arises from its group structure.

Partitions of G are the same thing as equivalence relations on G. In the context of group theory, there is a particularly important equivalence relation: conjugacy. By definition, two points g,h \in G are conjugate to one another if and only if g = khk^{-1} for some k \in G. This equivalence relation defines a partition of G whose blocks are called conjugacy classes. In group theory, the significance of this concept is that a subgroup of G is normal if and only if it is a union of conjugacy classes. We will now consider the partition of G into conjugacy classes from the algebra perspective. Functions on G which are constant on conjugacy classes are know as class functions. Could it be the case that the set of class functions on G is a subalgbra of G?

Let \{C_\alpha \colon \alpha \in \Lambda\} be the partition of G into conjugacy classes. Here \Lambda is a finite set of labels which parameterize the blocks of this partition.

Problem 13.1. Prove that |\Lambda|=|G| if and only if G is abelian, and that |\Lambda|=1 if and only if |G|=1. Next, prove that the cardinality of the set \{(g,h) \in G \times G \colon gh=hg\} of commuting pairs of elements from G is |G||\Lambda|.

We have defined class functions to be those functions on G which are constant on its conjugacy classes. A useful alternative characterization of such functions is that they are precisely those functions which are insensitive to any noncommutativity in G.

Problem 13.2. A function A \in \mathcal{C}(G) is a class function if and only if A(gh)=A(hg).

Now we return to our original question: is the set of class functions on G a subalgebra of \mathcal{C}(G)? The answer is yes, as the following theorem shows.

Theorem 13.1. The set of class functions on G is precisely the center of \mathcal{C}(G).

Proof: Suppose first that A \in \mathcal{C}(G) is a class function, and let B \in \mathcal{C}(G) be any function. We want to prove that A commutes with B. According to the definition of convolution, we have

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

where to get the second equality we used the substation h \rightsquigarrow gh^{-1} and to get the third inequality we used the commutativity of \mathbb{C}. Since A is a class function, we conclude that

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

Conversely, suppose that A \in \mathcal{C}(g) is a central function on G. Then, it commutes with each of the elementary functions E_g, which form a basis of \mathcal{C}(G) as g ranges over G. Now observe that

[AE_g](h) = \sum\limits_{k \in G} A(hk^{-1})E_g(k)=A(hg^{-1}),

whereas

[E_gA](h)=\sum\limits_{k \in G} E_g(hk^{-1})A(k)=A(g^{-1}h).

Since these are equal due to the centrality of A, Problem 13.2 shows that A is a class function. \square

The fact that the set of class functions is a subalgebra of \mathcal{C}(G) now follows from the fact that the center of any algebra is a subalgebra thereof. Let us write \mathcal{K}(G):=Z(\mathcal{C}(G)) for brevity.

Theorem 13.2. The dimension of \mathcal{K}(G) is equal to |\Lambda|, the number of conjugacy classes in G.

Proof: Let

K_\alpha = \sum\limits_{g \in C_\alpha} E_g

be the indicator function of the conjugacy class C_\alpha. Then, the set \{K_\alpha \colon \alpha \in \Lambda\} spans \mathcal{K}(G). Moreover, for the $L^2$-scalar product on \mathcal{C}(G) we have

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

Since \{C_\alpha \colon \alpha \in \Lambda\} is a partition of G, this gives

\langle K_\alpha,K_\beta\rangle = \delta_{\alpha\beta}|C_\alpha|,

hence \{K_\alpha \colon \alpha \in \Lambda\} is an orthogonal set of nonzero functions. \square

It is useful to think of \mathcal{K}(G) and \mathcal{C}(G) on the same footing, as two algebras associated to any finite group. The class algebra \mathcal{K}(G) is always commutative, while the convolution algebra \mathcal{C}(G) is commutative precisely when G is abelian, and coincides with \mathcal{K}(G) in this case. Any group G with at least two elements has at least two conjugacy classes, hence the dimension of \mathcal{K}(G) is at least two – the convolution algebra \mathcal{C}(G) of a nontrivial group can be noncommutative, but not a noncommutative as the endomorphism algebra of a Hilbert space, whose center is one-dimensional.

An interesting aspect of the class algebra \mathcal{K}(G) is that the connection coefficients of the class basis \{K_\alpha \colon \alpha \in \Lambda\} count solutions to equations in G. More precisely, consider the |\Lambda|^3 structure constants defined by

K_\alpha K_\beta = \sum\limits_{\gamma \in \Lambda} c_{\alpha\beta\gamma}K_\gamma, \quad \alpha,\beta \in \Lambda.

Problem 13.3. Let k \in C_\gamma. Show that

c_{\alpha\beta\gamma}=|\{(x,y) \in C_\alpha \times C_\beta \colon xy=k\}|.

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