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
I am a bit suprised and disappointed to see that the online maths communities I lurk around seem largely oblivious to this recent preprint 1306.3969. Here is the abstract: the added emphasis is mine.
We use the method of interlacing families of polynomials to prove Weaver’s conjecture KS2, which is known to imply a positive solution to the Kadison-Singer problem via Anderson’s Paving Conjecture. Our proof goes through an analysis of the largest roots of a family of polynomials that we call the “mixed characteristic polynomials” of a collection of matrices.
(A few years ago, the 2nd and 3rd authors of that preprint recently made a dramatic improvement in our understanding of a theorem of Bourgain and Tzafriri, see arXiv 0911.1114. So this paper is certainly worth taking seriously at the very least.)
Over on G+, Willie Wong quite sensibly asked for some brief explanation of what the problem said, and why people care(d). I must confess that the full background to the Kadison-Singer conjecture/problem is well outside my area of technical expertise, possibly outside my area of competence. Nevertheless, I can at least link to this article by Casazza and Tremain, which mentions some other conjectures in functional analysis that are known to be equivalent to the Paving Conjecture, and hence (by work of Anderson) to the Kadison-Singer conjecture.
P. G. Casazza, J. C. Tremain. The Kadison–Singer Problem in mathematics and engineering. PNAS vol. 103 (2006) no. 7, 2032–2039
Here is a link to some web material for an AIM workshop on the Kadison-Singer problem, which may give the general audience some idea of work in recent years.
The paper of Weaver which the preprint refers to is:
The MathReview of Weaver’s paper, by P. J. Stacey, is short enough that it can be reproduced here:
In [Amer. J. Math. 81 (1959), 383–400; MR0123922 (23 #A1243)], R. V. Kadison and I. M. Singer asked if every pure state on an atomic maximal abelian subalgebra of B(H), the algebra of bounded operators on a separable Hilbert space H, extends uniquely to a pure state on B(H). Developing the approach in [C. A. Akemann and J. Anderson, Mem. Amer. Math. Soc. 94 (1991), no. 458, iv+88 pp.; MR1086563 (92e:46113)], the author formulates a combinatorial version of the Kadison-Singer problem, in terms of unit vectors in Ck. Some positive partial results are then obtained using discrepancy theory.
Perhaps I will keep this blog post updated with some more links, if anyone has suggestions. Though really it should be left to the operator theorists, operator algebraists, and combinatorists to write some expositions in the weeks to come.
I see there is some attention now that Terence Tao has mentioned this on G+ and thence on the Selected Papers Network. (I admit that when I mentioned the paper on G+, I didn’t tag it with #spnetwork, mainly because I didn’t feel I had anything intelligent to say at the time; and if this #spnetwork is to become useful to the community of research mathematicians, it needs less noise from spectators, and more commentary from people who understand some ideas in the papers under discussion!)
Gil Kalai has a blogpost which says a little more about how the paper of Marcus, Spielman and Srivastava relates to the previous results of Bourgain and Tzafriri, and mentions that Spielman and Srivastava had previously given a new proof – an improved proof? – of Bourgain-Tzafriri’s restricted invertibility theorem.
Orr Shallit has also picked up on this, and offers some thoughts from the perspective of an operator algebraist/operator theorist.
As promised in the previous blogpost, here is some finite group theory. (Those of you familiar with the British TV show “Faking It” will appreciate that you don’t have to fool all of the people all of the time, just some of the people at the right moments.)
Some preliminary terminology is useful, since we will need it in later posts. We say that H is (isomorphic to) a proper quotient of a group G if there is a homomorphism from G onto H which has non-trivial kernel. A group G is said to be just non-abelian (or JNA or short) if it is non-abelian, yet every proper quotient is abelian. A little thought shows that this is equivalent to the condition that the derived subgroup [G,G] is contained in each non-trivial normal subgroup of G.
(I don’t remember explicitly hearing about JNA groups in past courses or talks, but about seven years ago in Newcastle I heard a visiting speaker talk about families of groups that were “just-” for some property .)
Doing some digging in the literature: in the case where the group G is JNA and [G,G] is abelian, there is a classification or structure theorem available, through work of M. F. Newman in two papers that appear in the Proceedings of the London Mathematical Society (both in volume 10, 1960). Later on, when we resume the story of the central amenability constant, JNA groups will turn up very naturally. However, none of that is needed to follow the proof of the following result:
Theorem 1. Let G be a finite JNA group which has trivial centre and which has a conjugacy class of size 2. Then G contains an involution and an element of odd prime order p, such that G has order 2p and .
Up to isomorphism, the only group satisfying the conclusions of the theorem is the dihedral group of order 2p. (Checking that this group actually is JNA is not too difficult, but I won’t go into the details here.)
Remark. In any finite group with a conjugacy class of size 2, both elements in the conjugacy class have the same centralizer — this point will be reiterated below in a little more detail — and this centralizer has index 2 in the parent group, hence is normal. This is encouraging in our setting, since the JNA condition now gives us extra information.
A leisurely proof
The purpose of this post is to provide a proof of Theorem 1 that uses only basic facts of finite group theory, such as may be found in a first course that covers notions such as conjugacy classes and normal subgroups.
We recall that the derived subgroup of G, sometimes called the commutator subgroup of G, is the subgroup of G generated by all elements of the form as vary over G. (For those who prefer to think categorically, [G,G] is uniquely determined by the property that it is contained in the kernel of every homomorphism from G to any abelian group, i.e. it is the kernel of the abelianization homomorphism.)
Now in Theorem 1, the hypotheses on G are as follows:
- there exists G with exactly one other conjugate,
- If N is a non-trivial normal subgroup of G, then every commutator belongs to N.
It is immediate from Condition 2 that any given G either centralizes both and , or else swaps them (this is true for any group action on any 2-point set, of course). More formally:
Lemma 2. Let G. Either and , or and .
In particular, this lemma implies that , i.e. and commute. Applying the lemma to shows that for all in G, and so by Condition 1 we must have
Let N be the order of : this is an integer . If N were even, say N=2m, then , and applying Lemma 2 we would find that Z(G), which contradicts Condition 1. Therefore N is odd.
Because has odd order, this implies that , and so [G,G] contains the subgroup of G generated by , which we denote by . On the other hand, observe that since , Lemma 2 implies that is a normal subgroup of G. Therefore, by Condition 3,
Let H be the centralizer in G of the element (and hence, as remarked above, of the element ). We know that H has index 2 in G (by applying the orbit-stabilizer theorem to the conjugation action of G on ). At this point we could now invoke the general fact that index 2 subgroups of any group are necessarily normal subgroups. To keep things self-contained, we instead use some ad hoc arguments (although this admittedly leaves out the bigger picture which motivates our calculations).
Lemma 3. Let G. Then either or belongs to H.
Proof: If belongs to G but not H, then does not lie in H, and so by Lemma 2. So and rearranging shows that centralizes .
Moreover, by definition H contains . Hence it contains [G,G], by Equation (2). The next step in our argument is to show that in fact H=[G,G].
Digression. At this point, if we just wanted to proceed as quickly as possible, we could appeal to Theorem 3.4 of
M. F. Newman, On a class of metabelian groups. Proc. London Math. Soc. (3) 10 1960 354–364. MR0117293 (22 #8074)
Indeed, this point is where I’d originally got stuck on my first attempt to prove the theorem, and I has to resort to Newman’s paper to check that what I was hoping to prove was actually true. Nevertheless, it seems worth giving an ad hoc argument using only “bare-hands techniques”, since we are in a much more specialized setting than that covered by Newman’s theorem. What follows is my own argument, found by some trial and error after I had used Newman’s paper to check I was on the right lines.
Proof that H=[G,G]
Consider the function θ : H [G,G] defined by . Since [G,G]= is contained in Z(H), for all H we have
in other words, θ is a homomorphism. Our goal is to show θ is a bijection, which will force H and [G,G] to have the same cardinality. Since we already know that [G,G] is contained in H, we can conclude that [G,G]=H as required.
First, we show θ is surjective. Note that , so if we let m=(N+1)/2 and k be any integer, induction (or the fact θ is a homomorphism) implies that . Since [G,G]=, this shows θ(H)=[G,G].
Secondly, we show θ is injective. Let K=, which is a normal subgroup of H (being the kernel of a homomorphism). Note that K=. It now follows from Lemma 3 that K is normal as a subgroup of G. Suppose K is not the trivial subgroup; then by Condition 3, K contains [G,G], and in particular belongs to K. But since (Equation (1)) we have
(since ) and we get a contradiction. Therefore K is the trivial subgroup, and θ:H[G,G] is indeed injective.
Continuing the proof of Theorem 1
Let us take stock. We have shown that there exist elements G such that:
- (this follows from Equation (1) and the choice of );
- has odd order, say N (this follows from the remarks after Equation (1));
- is an index 2 subgroup in G (this follows from Equation (2) and the result just proved).
This is close to what is needed: it only remains to show that and that N is prime.
We will show that belongs to the centre of G, which combined with Condition 1 forces . To do this, first observe that by Lemma 2, centralizes , and hence centralizes . But since is an index 2 subgroup of G and , every element of G belongs to either or , from which it follows that centralizes everything in G, as required.
Remark. The argument just given seems the quickest way to do things, given what we have already shown to date. However, it relies on knowing that has index 2 in G. It may be of some interest to note that one can show centralizes G more directly, using only Equations (1) and (2). Thus, let G}. By Equation (2), there exists some integer k such that
Since , it follows that
and so . Rearranging gives
so that . Since is arbitrary in G, belongs to the centre of G.
Finally, to finish things off, it suffices to show that N is prime. Let p be a prime factor of N. Then is a proper subgroup of , and it is normal in G by Equation (1), since . Condition 3 therefore forces , and so N=p. This completes the proof of Theorem 1.
Well, having already been drenched on the way into work today, I am now stuck in my office as I wait for a lull in the rainstorm. So I guess I should put the time to use, and a blog post is as good as any right now.
This continues from the previous post. I’ll also continue the experiment of writing things up in an anecdotal style, rather than in the mode of a presentation or paper — the aim is not to present the outcome as quickly or cleanly as possible, but to give an account of its evolution. Apologies if it proves tedious, irksome, or outright Pooterish.
This is also my first attempt at using the LaTeX-to-Wordpress script written by Luca Trevisan, which has saved me a certain amount of teeth-grinding when fighting WP’s limited LaTeX support.
A few weeks after writing the previous post, I was at a rather enjoyable conference in Granada where I spent some time, on and off during lulls in the conference schedule, reconstructing the proof of the lower bound 4/3, from memory while away from my books. Annoyingly, the 4/3 comes as the limiting value of a weak lower bound, in one branch of a case-by-case analysis, but in a branch where known results suggest the true value of the amenability constant should be at least 2. (In the other branch of the analysis, one is led to 7/4 and knows that this is sharp.)
Without going into details about how the case-by-case analysis works, let’s just say that for all finite groups G in a certain class , the method in question yields a lower bound
However, for various speculative reasons (well, guesswork) it seemed likely that for all
In fact, the existing method already implied that whenever and all non-central conjugacy classes have size at least 3. This got me thinking about what one could say about those which had a conjugacy class of size 2 in the group, since — as part of the small amount of finite group theory that I remembered from my undergraduate days — this ensures G has a normal subgroup of index 2, from which one might try to play games with induction or restriction of characters.
Fast-forward to June. About a week ago, shortly after getting back to Saskatoon, some idle tweaking done while waiting for food showed that the method used to get the 4/3 bound could be modestly improved by a slightly more careful use of various inequalities. The improved method still only gave
but it did now show that whenever and all its non-central conjugacy classes have size at least 3.
The upshot? To prove that for all , one now only needed to do it for those which had a conjugacy class of size 2. Moreover, the calculations I’d been doing in Granada hinted strongly that any such group would be forced to be something like a dihedral group of order 2 mod 4, and in my earlier paper with Alaghmandan and Samei, we had already calculated the exact values of for dihedral groups, showing that in the 2 mod 4 case this constant is at least 7/3.
So, at this point, you naturally go and ask a group theorist. On the other hand, since none were at hand, and since the question seemed a little basic to throw on MathOverflow without having a more serious go on one’s own…
… and it works. To be precise for a moment, here is what I was able to show last week.
Theorem 1 Let G be a finite group which has trivial centre, and where every proper quotient is abelian. If G has a conjugacy class of size 2, then it is isomorphic to the dihedral group of order 2p for some odd prime p.
Well, assuming there isn’t a mistake, for many of you this is probably like watching a young child come up to you proudly to show you their latest kindergarten handiwork. I have a strong suspicion that this result is already known, perhaps implicitly, or as one of those folklore exercises that is set in Proper Courses for Proper Students. Certainly, in the argument I have right now, one shows en route that G is solvable, whereupon one can invoke a classification of “just non-abelian, metabelian groups” due to M. F. Newman.
On the other hand, since it turns out that one can prove the theorem using only what one learns in a first course on finite group theory, my plan is to write up the proof in the next blog post in this series.
(The rain has stopped.)
Update 06/06/2013: see the bottom of the post (some typos also fixed)
As I am currently too besieged by a jumped-up virus to get any proper work done, I suppose I might as well put this blog to one of its nominal uses and actually talk about some mathematics I did recently (last Monday and Tuesday, in fact). Well, I say talk, I really mean “present” (and I am not at all sure I will manage to “explain”). Consider this my first experiment in using the blog for channeling rainwater…
This post will just try to set up enough to state the result, and then if there is ever a follow-up post I will give more of the background story and the necessary details.
Fix a finite group G and look at the centre of its complex group algebra . This algebra, which we call A for now, is commutative and spanned by its minimal idempotents. What are these minimal idempotents? they are scalar multiples of the irreducible characters of G. (A character of G is, for us, the trace of some representation over ; we say the character is irreducible if the corresponding representation is.) If is such a character then it is not hard to show that the corresponding idempotent in A is .
Now an algebra such as A admits a unique so-called separating idempotent, that is, an element satisfying for all and , where is the linearization of the multiplication map. (This separating idempotent is a witness to the fact that A has trivial Hochschild cohomology, or trivial Ext if you prefer to think of it like that.) In fact it is not at all hard to work out what has to be:
Let us accept for now that this element should be worth studying, and introduce norms. We can always equip the complex group algebra with the -norm, that is . The resulting Banach algebra is usually studied only for infinite G, but if one is interested in quantitative phenomena then the finite case still holds some mysteries and — I claim — some interest. In any case, since we can identify with the centre of the complex group algebra of , it inherits a natural norm as a subalgebra of .
We now define the central amenability constant of G to be the -norm of the separating idempotent , and denote this number by . If we wish to write things out explicitly, some more notation will be useful. Let denote the set of conjugacy classes in G, and if C is a conjugacy class in G we write for the value takes on one (hence on every) element of C. Then
This formula can be found1 as Theorem 1.8 of a 2009 JFA paper of Ahmadreza Azimifard, Ebrahim Samei and Nico Spronk (and before you complain about paywalls, this one is on the arXiv). At this point I should acknowledge that I first got interested in this invariant of G while listening to some talks given by Ebrahim about 5-6 years ago, and have since had a few enjoyable and educative discussions with both him and Nico on this topic.
A little patience (or some prior knowledge of Banach algebras) shows that if G is abelian then . Moreover, by an argument that will be outlined in the next installment, whenever G is non-abelian. What is not all obvious, at least to me, is that 1 is an isolated value of this invariant. The following is extracted from the proof of Theorem 1.10 in the aforementioned paper.
There exists δ > 0 such that whenever G is non-abelian.
The authors obtain the existence of such a δ as a special case of a theorem of D. A. Rider (Transactions AMS, 1973). As an aside: for those of you who frequented MathStackExchange, this application is what ultimately prompted me to raise this question. What is rather unsatisfactory is how small the δ given by Rider’s proof is — Rider’s bounds yield δ = 1/300, although in some back of the envelope calculations I thought I could bring it down to around 1/90 — while the smallest value known for when G is nonabelian is 7/4.
After some failed attempts to get a better lower bound, I put the problem on the backburner, although from time to time it would come back to haunt me. More recently, in some joint work with Ebrahim and our PhD student Mahmood Alaghmandan, we had a closer look at the problem of calculating explicitly when you don’t know how to use GAP, and don’t have anyone at hand who does. See this preprint if you want to know more. In the process, I found myself getting interested again in the original problem of getting a lower bound closer to 7/4 than to 301/300…
Theorem (yours truly, some time last Monday)
whenever G is non-abelian.
This bound is still probably not sharp, but it does at least approach what seems to be the right value. However, more important than the improvement in the constant is that the argument now bypasses Rider’s theorem, which is much more general and whose proof I find somewhat hard to understand2. Instead of using general non-abelian Fourier analysis like Rider, who was in any case working on compact groups, we can get by with some basic arguments from character theory, together with a little structure theory for finite groups.
Well, this post looks long enough, so further explanation will have to wait for the next post…
- I am cheating a little here, in that those authors have a slightly more conceptual definition of , and have to do a little work to show it gives the same number as the formula above.
- One can follow the proof through line by line, but it leaves me rather baffled about what is conceptually going on, and what the intuition is.
Update 6th June 2013: it turns out that with a very modest tweak to some inequalities I was using, one can get a slight improvement from 4/3 to 13/9. More interestingly, some calculations in progress suggest that by using some group-theory and elbow-grease to deal with certain “edge cases”, one might be able to prove a lower bound of 7/4 (which we know would be sharp). Details to follow, perhaps …
Possibly NSFW for language, depending on your melon-farming place of employment, I suppose:
Anyway, a well-meant correction made to proofs of a recent paper has reminded me of the following nicety.
The following information is provided to meet the requirements of HEFCE:
The legal name and address of the University are:
The University of Newcastle upon Tyne
Newcastle upon Tyne
Since February 2006, the University has adopted the trading title of Newcastle University.
This was first pointed out to me by Antony Mee, a contemporary at Newcastle. (I’ve never understood why the original name was deemed a “defective brand”, and in any case, it just invites confusion for some of us with this University of Newcastle.)
I suspect I must have blogged this quote already, but since I can’t find such a post, I may as well repeat it.
Dr Blind (pronounced ‘Blend’) was about ninety years old and had taught, for the past fifty years, a course called ‘Invariant Subspaces’ which was noted for its monotony and virtually absolute unintelligibility, as well as for the fact that the final exam, as long as anyone could remember, had consisted of the same single yes-or-no question. The question was three pages long but the answer was always ‘Yes.’ That was all you needed to know to pass Invariant Subspaces.
He was, if possible, even a bigger windbag than Dr Roland. Together, they were like one of those superhero alliances in the comic books, invincible, an unconquerable confederation of boredom and confusion. I murmured an excuse and slipped away, leaving them to their own formidable devices.
(From Donna Tartt’s excellent novel The Secret History, which has nothing to do with mathematics.)
Update 25th Jan 2013: Professor Blind, meet Professors Cowen and Gallardo.
further update 5th Feb 2013: As you were; nothing to see here; move along.