Research Updates: Boston–Shalev for conjugacy classes, growth in linear groups, and the (amazing) Kelley–Meka result

1. Boston–Shalev for conjugacy classes

Last week Daniele Garzoni and I uploaded to the arxiv a preprint on the Boston–Shalev conjecture for the conjugacy class weighting. The Boston–Shalev conjecture in its original form predicts that, in any finite simple group G, in any transitive action, the proportion of elements acting as derangements is at least some universal constant c > 0. This conjecture was proved by Fulman and Guralnick in a long series of papers. Daniele and I looked at conjugacy classes instead, and we found an analogous result to be true: the proportion of conjugacy classes containing derangements is at least some universal constant c' > 0.

Our proof depends on the correspondence between semisimple conjugacy classes in a group of Lie type and polynomials over a finite field possibly with certain restrictions: either symmetry or conjugate-symmetry. We studied these sets of polynomials from an “anatomical” perspective, and we needed to prove several nontrivial estimates, e.g., for

  • the number of polynomials with a factor of a given degree (which is closely related the “multiplication table problem”),
  • the number of polynomials with an even or odd number of irreducible factors,
  • the number of polynomials with no factors of small degree,
  • or the number of polynomials factorizing in a certain way (e.g., as f = gg^*, g irreducible, g^* the reciprocal polynomial).

For a particularly neat example, we found that, if the order of the ground field is odd, exactly half the self-reciprocal polynomials have an even number of irreducible factors — is there a simple proof of this fact?

2. Growth in Linear Groups

Yesterday Brendan Murphy, Endre Szabo, Laci Pyber, and I uploaded a substantial update to our preprint Growth in Linear Groups, in which we prove one general form of the “Helfgott–Lindenstrauss conjecture”. This conjecture asserts that if a symmetric subset A of a general linear group \mathrm{GL}_n(F) (n bounded, F an arbitrary field) exhibits bounded tripling, |A^3| \le K|A|, then A suffers a precise structure: there are subgroup H \trianglelefteq \Gamma \le \langle A \rangle such that \Gamma / H is nilpotent of class at most n-1, H is contained in a bounded power A^{O_n(1)}, and A is covered by K^{O_n(1)} cosets of \Gamma. Following prodding by the referee and others, we put a lot more work in and proved one additional property: \Gamma can be taken to be normal in \langle A \rangle. This seemingly technical additional point is actually very subtle, and I strongly doubted whether it was true late into the project, more-or-less until we actually proved it.

We also added another significant “application”. This is not exactly an application of the result, but rather of the same toolkit. We showed that if G \le \mathrm{GL}_n(F) (again F an arbitrary field) is any finite subgroup which is K(n)-quasirandom, for some quantity K(n) depending only on n, then the diameter of any Cayley graph of G is polylogarithmic in the order of |G| (that is, Babai’s conjecture holds for G). This was previously known for G simple (Breuillard–Green–Tao, Pyber–Szabo, 2010). Our result establishes that it is only necessary that G is sufficiently quasirandom. (There is a strong trend in asymptotic group theory of weakening results requiring simplicity to only requiring quasirandomness.)

The intention of our paper is more-or-less to “polish off” the theory of growth in bounded rank. By contrast, growth in high-rank simple groups is still poorly understood.

3. The Kelley–Meka result

Not my own work, but it cannot go unmentioned. There was a spectacular breakthrough in additive combinatorics last week. Kelley and Meka proved a Behrend-like upper bound for the density of a subset A \subset \{1, \dots, n\} free of three-term arithmetic progressions (Roth’s theorem): the density of A is bounded by \exp(-c (\log n)^\beta) for some constants c, \beta > 0. Already there are other expositions of the method which are also worth looking at: see the notes by Bloom and Sisask and Green (to appear, possibly).

Until this work, density 1 /\log n was the “logarithmic barrier”, only very recently and barely overcome by Bloom and Sisask. Now that the logarithmic barrier has been completely smashed, it seems inevitable that the new barometer for progress on Roth’s theorem is the exponent \beta. Kelley and Meka obtain \beta = 1/11, while the Behrend construction shows \beta \le 1/2.

Fekete’s lemma and sum-free sets

Just a quick post to help popularise a useful lemma which seems to be well known to researchers but not to undergraduates, known variously as Fekete’s lemma or the subadditive lemma. It would make for a good Analysis I exercise. Let {\mathbf{N}} exclude {0} for the purpose of this post.

Lemma 1 (Fekete’s lemma) If {f:\mathbf{N}\rightarrow\mathbf{R}} satisfies {f(m+n)\leq f(m)+f(n)} for all {m,n\in\mathbf{N}} then {f(n)/n\longrightarrow\inf_n f(n)/n}.

Proof: The inequality {\liminf f(n)/n \geq \inf_d f(d)/d} is immediate from the definition of {\liminf}, so it suffices to prove {\limsup f(n)\leq f(d)/d} for each {d\in\mathbf{N}}. Fix such a {d} and set {M=\max\{0,f(1),f(2),\ldots,f(d-1)\}}. Set {f(0)=0}. Now for a given {n} choose {k} so that {0 \leq n-kd < d}. Then

\displaystyle  \frac{f(n)}{n} \leq \frac{f(n-kd)}{n} + \frac{f(kd)}{n} \leq \frac{M}{n} + \frac{kf(d)}{kd} \longrightarrow \frac{f(d)}{d},

so {\limsup f(n)/n \leq f(d)/d}. \Box

Here is an example. For {n\in\mathbf{N}} let {f(n)} denote the largest {k\in\mathbf{N}} such that every set of {n} nonzero real numbers contains a subset of size {k} containing no solutions to {x+y=z}. We call such a subset sum-free.

Proposition 2 {f(n)/n} converges as {n\rightarrow\infty}.

Proof: Let {A} be a set of size {m} containing no sum-free subset of size larger than {f(m)}, and {B} a set of size {n} containing no sum-free subset of size larger than {f(n)}. Then for large enough {M\in\mathbf{N}}, the set {A\cup MB} is a set of size {m+n} containing no sum-free subset of size larger than {f(m)+f(n)}, so {f(m+n)\leq f(m)+f(n)}. Thus by Fekete’s lemma {f(n)/n} converges to {\inf f(n)/n}. \Box

The value of {\sigma = \lim f(n)/n} not easy to compute, but certainly {0\leq\sigma\leq\frac{1}{2}}. A famously simple argument of Erdos shows {\sigma\geq\frac{1}{3}}: Fix a set {A}, and for simplicity assume {A\subset\mathbf{N}}. For {\theta\in[0,1]} consider those {n\in A} such that the fractional part of {\theta n} lies between {\frac{1}{3}} and {\frac{2}{3}}. This set {A_\theta} is sum-free, and if {\theta} is chosen uniformly at random then {A_\theta} has size {|A|/3} on average.

The example {A=\{1,2,3,4,5,6,8,9,10,18\}}, due to Malouf, shows {\sigma\leq\frac{2}{5}}. The current record is due to Lewko, who showed by example {\sigma\leq\frac{11}{28}}. Since these examples are somewhat ad hoc, while Erdos’s argument is beautiful, one is naturally led to the following conjecture (indeed many people have made this conjecture):

Conjecture 3 {\sigma=\frac{1}{3}}.

Ben Green, Freddie Manners, and I have just proved this conjecture! See our preprint.