The central amenability constant of a finite group: Part 3 of n
OK, back to the story of the central amenability constant. I’ll take the opportunity to re-tread some of the ground from the first post.
Given a finite group G, denotes the usual complex group algebra: we think of it as the vector space equipped with a suitable multiplication. This has a canonical basis as a vector space, indexed by group elements: we denote the basis vector corresponding to an element x of G by . Thus for any function , we have .
(Aside: this is not really the correct “natural” way to think of the group algebra if one generalizes from finite groups to infinite groups; one has to be more careful about whether one is thinking “covariantly or contravariantly”. is naturally a contravariant object as G varies, but the group algebra should be covariant as G varies. However, our approach allows us to view characters on G as elements of the group algebra, which is a very convenient elision.)
The centre of , henceforth denoted by , is commutative and spanned by its minimal idempotents, which are all of the form
for some irreducible character . Moreover, is a bijection between the set of irreducible characters and the set of minimal idempotents in .
and, equipping with the natural -norm, defined by
we define to be . Explicitly, if we use the convention that the value of a class function on any element of a conjugacy class C is denoted by , we have
the formula stated in the first post of this series.
2. Moving onwards
As I am writing these things up, it occurs to me that “philosophically speaking”, perhaps one should regard as an element of the group algebra , where Gop denotes the group whose underlying set is that of G but equipped with the reverse multiplication. It is easily checked that a function on is central as an element of if and only if it is central as an element of the algebra , so we can get away with the definition chosen here. Nevertheless, I have a suspicion that the picture is somehow the “right” one to adopt, if one wants to put the study of into a wider algebraic context.
is a non-zero idempotent in a Banach algebra, so it follows from submultiplicativity of the norm that . When do we have equality?
Theorem 2 (Azimifard–Samei–Spronk) if and only if G is abelian.
The proof of necessity (that is, the “only if” direction) will go in the next post. In the remainder of this post, I will give two proofs of sufficiency (that is, the “if” direction).
In the paper of Azimifard–Samei–Spronk (MR 2490229; see also arXiv 0805.3685) where I first learned of , this direction is glossed over quickly, since it follows from more general facts in the theory of amenable Banach algebras. I will return later, in Section 2.2, to an exposition of how this works for the case in hand. First, let us see how we can approach the problem more directly.
2.1. Proof of sufficiency: direct version
Suppose G is abelian, and let . Then G has exactly n irreducible characters, all of which are linear (i.e. one-dimensional representations, a.k.a. multiplicative functionals). Denoting these characters by , we have
This sum can be evaluated explicitly using some Fourier analysis — or, in the present context, the Schur column orthogonality relations. To make this a bit more transparent, recall that for all characters and all y in G. Hence by a change of variables in the previous equation, we get
For a fixed element x in G, the n-tuple is a column in the character table of G. We know by general character theory for finite groups that distinct columns of the character table, viewed as column vectors with complex entries, are orthogonal with respect to the standard inner product. Hence most terms in the expression above vanish, and we are left with
which equals , since each takes values in . This completes the proof.
The following argument is an expanded version of the one that is outlined, or alluded to, in the paper of Azimifard–Samei–Spronk. It is part of the folklore in Banach algebras — for given values of “folk” — but really the argument goes back to the study of “separable algebras” in the sense of ring theory.
Lemma 3 Let A be an associative, commutative algebra, with identity element 1A. Let be the linear map defined by . Then there is at most one element m in that simultaneously satisfies =1A and for all a in A.
Proof: Let us first omit the assumption that A is commutative, and work merely with an associative algebra that has an identity.
Define the following multiplication on :
Then is an associative algebra — the so-called enveloping algebra of A. If m satisfies the conditions mentioned in the lemma, then
and so, by taking linear combinations, for every w in . If n is another element of satisfying the conditions of the lemma, we therefore have nm=m, and by symmetry, mn=n.
Now we use the assumption that A is commutative. From this assumption, we see that is also commutative. Therefore
Now let G be a finite group and let A= . Because A is spanned by its minimal idempotents , and because minimal idempotents in a commutative algebra are mutually orthogonal, satisfies the two conditions mentioned in Lemma 3. On the other hand, if G is abelian, consider
Clearly =1A, and a direct calculation shows that for all g in G, so by linearity also satisfies both conditions mentioned in Lemma 3. Applying the lemma tells us that , and in particular