Breuillard’s ICM talk: uniform expansion, Lehmer’s conjecture, tauhat

Emmanuel Breuillard is in Korea talking at the ICM; here’s his paper, a very beautiful survey of uniformity results for growth in groups, by himself and others, and of the many open questions that remain.

He starts with the following lovely observation, which was apparently in a 2007 paper of his but which I was unaware of.  Suppose you make a maximalist conjecture about uniform growth of finitely generated linear groups.  That is, you postulate the existence of a constant c(d) such that, for any finite subset S of GL_d(C),  you have a lower bound for the growth rate

\lim |S^n|^{1/n} > c(d).

It turns out this implies Lehmer’s conjecture!  Which in case you forgot what that is is a kind of “gap conjecture” for heights of algebraic numbers.  There are algebraic integers of height 0, which is to say that all their conjugates lie on the unit circle; those are the roots of unity.  Lehmer’s conjecture says that if x is an algebraic integer of degree n which is {\em not} a root of unity, it’s height is bounded below by some absolute constant (in fact, most people believe this constant to be about 1.176…, realized by Lehmer’s number.)

What does this question in algebraic number theory have to do with growth in groups?  Here’s the trick; let w be an algebraic integer and consider the subgroup G of the group of affine linear transformations of C (which embeds in GL_2(C)) generated by the two transformations

x -> wx

and

x -> x+1.

If the group G grows very quickly, then there are a lot of different values of g*1 for g in the word ball S^n.  But g*1 is going to be a complex number z expressible as a polynomial in w of bounded degree and bounded coefficients.  If w were actually a root of unity, you can see that this number is sitting in a ball of size growing linearly in n, so the number of possibilities for z grows polynomially in n.  Once w has some larger absolute values, though, the size of the ball containing all possible z grows exponentially with n, and Breuillard shows that the height of z is an upper bound for the number of different z in S^n * 1.  Thus a Lehmer-violating sequence of algebraic numbers gives a uniformity-violating sequence of finitely generated linear groups.

These groups are all solvable, even metabelian; and as Breuillard explains, this is actually the hardest case!  He and his collaborators can prove the uniform growth results for f.g. linear groups without a finite-index solvable subgroup.  Very cool!

One more note:  I am also of course pleased to see that Emmanuel found my slightly out-there speculations about “property tau hat” interesting enough to mention in his paper!  His formulation is more general and nicer than mine, though; I was only thinking about profinite groups, and Emmanuel is surely right to set it up as a question about topologically finitely generated compact groups in general.

 

 

 

 

 

 

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2 thoughts on “Breuillard’s ICM talk: uniform expansion, Lehmer’s conjecture, tauhat

  1. arithmetica says:

    A small correction: “algebraic number” in the third paragraph should be “algebraic integer”.

  2. JSE says:

    Thanks, fixed!

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