You may have heard the big news – Igor Mineyev has announced a proof of the Hanna Neumann Conjecture. I’d like to quickly remind you what the conjecture says. It concerns the rank of a free group, by which I mean the minimal number of generators.
Let F be a non-abelian free group, which we may take to be of rank two. The starting point is an old theorem of Howson.
Theorem (Howson, 1954). If H,K are finitely generated subgroups of F then the rank of the intersection of H and K is finite.
The obvious question is: ‘What is the best bound on the rank of the intersection, in terms of the ranks of H and K?’ Shortly after Howson’s paper, Hanna Neumann conjectured the answer.
Conjecture (Hanna Neumann, 1956). If H,K are non-trivial subgroups of F, then
She also proved that .
I want to quickly convey why the Hanna Neumann Conjecture is reasonable. Specifically, I want to explain why it holds for finite-index subgroups of F, using a topological argument of Stallings.
As usual, the idea is to think of F as the fundamental group of a graph X (equipped with some basepoint). We have that , where is Euler characteristic. The subgroups H and K correspond to (based) covering spaces and respectively. Because the correspondence between subgroups and covering spaces is functorial, we can study the intersection of H and K by studying the covering space Y that is the pullback of the following diagram.
The space Y can be seen explicitly as the fibre product . Specifically,
where and are the covering maps. The intersection of H and K is the fundamental group of the component of Y picked out by the basepoints of and .
- The maps and are covering maps.
- (and, symmetrically, ).
At this point, it’s easy to prove the Hanna Neumann Conjecture for subgroups of finite index.
Proof of the HNC for .
It follows immediately from the fact that Euler characteristic is multiplicative that
Let Y’ be the component of Y with . Because is a covering map of finite degree, each component of Y has non-positive Euler characteristic. Therefore
which is exactly the statement of the conjecture. QED
If you were reading the above proof carefully, you may have been a little dissatisfied when we passed from Y to Y’, as we lost a lot of information at that point. Indeed, the the fundamental groups of the other components of Y are conjugate to subgroups of F of the form . The above proof naturally gave the following.
These sorts of considerations motivate the Strengthened Hanna Neumann Conjecture, proposed by Walter Neumann. A little care is needed, because if H and K are both of infinite index in F then will often be trivial, and so of rank 0; but we do not want such intersections to contribute negative terms to our sum, as this would make the ‘strengthened’ conjecture weaker than the original! Therefore, the strengthening takes the following form.
Strengthened Hanna Neumann Conjecture (Walter Neumann, 1989). If H,K are subgroups of F, then
The Hanna Neumann Conjecture has provided endless fun for group theorists over the past 50 years. I was told that Benson Farb used to set it as a problem to his incoming students. Much of the appeal lies in its elementary statement, and it always seemed likely that what was needed was one ingenious idea, rather than a big new piece of technology.
Well, Igor Mineyev has announced a proof of the Strengthened Hanna Neumann Conjecture. I haven’t looked at his papers on that topic, but at a recent conference in Southampton, Yago Antolin presented a simplification of Mineyev’s proof due to Warren Dicks. Slides of his talk are available here, and I heartily recommend you take a look. The proof is elementary (though also ingenious!), and can be read quite quickly.
I haven’t yet had time to absorb the proof fully. If I do, then perhaps I’ll blog about it again. But let me highlight two features.
- It uses the fact that free groups are left-orderable. I don’t think anyone saw that coming. (Feel free to boast in the comments if you did!)
- The idea of the proof is to construct a new, in some sense better, family of trees on which H, K and their intersection act. The proof then uses an estimate like the one derived from the fibre product above.