Low Dimensional Topology

August 15, 2011

Mineyev and the Hanna Neumann Conjecture

Filed under: Free groups,Geometric Group Theory — Henry Wilton @ 4:14 pm

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

$\mathrm{rank}~H\cap K-1\leq (\mathrm{rank}~H-1) (\mathrm{rank}~K-1)$ .

She also proved that $\mathrm{rank}~H\cap K-1\leq 2(\mathrm{rank}~H-1) (\mathrm{rank}~K-1)$ .

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 $\chi(X)=1-\mathrm{rank} F=-1$, where $\chi$ is Euler characteristic.  The subgroups H and K correspond to (based) covering spaces $X^H$ and $X^K$ 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 $X^H\times_X X^K$.  Specifically,

$Y=\{ (x_1,x_2)\in X^H\times X^K\mid \pi_H(x_1)=\pi_K(x_2)\}$

where $\pi_H:X^H\to X$ and $\pi_K:X^K\to X$ 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 $X^H$ and $X^K$.

Remarks.

• The maps $Y\to X^H$ and $Y\to X^K$ are covering maps.
• $\mathrm{degree} (Y\to X^H)=\mathrm{degree} (X^K\to X)$ (and, symmetrically, $\mathrm{degree} (Y\to X^K)=\mathrm{degree} (X^H\to X)$).

At this point, it’s easy to prove the Hanna Neumann Conjecture for subgroups of finite index.

Proof of the HNC for $|F:H|,|F:K|<\infty$.

It follows immediately from the fact that Euler characteristic is multiplicative that

$-\chi(Y)=\chi(X^H)\chi(X^K)$ .

Let Y’ be the component of Y with $\pi_1 Y'= H\cap K$.  Because $Y\to X$ is a covering map of finite degree, each component of Y has non-positive Euler characteristic.  Therefore

$-\chi(Y')\leq \chi(Y)$ ,

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 $H\cap gKg^{-1}$.  The above proof naturally gave the following.

$\sum_{HgK\in H\backslash F/K} \mathrm{rank}~H\cap gKg^{-1} -1 = (\mathrm{rank}~H-1) (\mathrm{rank}~K-1)$ .

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 $H\cap K$ 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

$\sum_{HgK\in H\backslash F/K} \max\{\mathrm{rank}~H\cap gKg^{-1} -1,0\} \leq (\mathrm{rank}~H-1) (\mathrm{rank}~K-1)$ .

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.

1. There’s a copy of Dicks’ one page proof that’s been circulating (I received a copy from Jason Manning), although it’s a bit dense.

The only application of HNC that I know of is a result of Wise on coherence of certain groups. Are there any other applications?

I’ve wondered whether there may be a strengthening of the strengthened Hanna Neumann Conjecture, which takes into account the ranks of infinite intersections of the groups with several other conjugates (by the finiteness of height, this is still a finite list of groups)? In particular, can one say something about the intersections of a fixed subgroup with its conjugates? See http://www.ams.org/mathscinet-getitem?mr=1389776

Also, is there any version for virtually free groups?

Comment by ianagol — August 15, 2011 @ 4:59 pm

• I don’t know of any other applications of (S)HNC . Also, I believe there is a gap in Wise’s coherence argument.

Regarding your third paragraph, are you suggesting that there might be some sort of correction term to the SHNC inequality, which depends on the ranks of intersections of more than two conjugates? Do you have any idea what the statement should be? Certainly there will always be an error: if H=K is malnormal then the left hand side is always equal to rank H-1 and the right hand side is always (rank H-1)^2.

The virtually free group question is very nice! One way to proceed might be to pull back to a free subgroup F of finite index, in which case the subgroups H and K pull back to families of subgroups of F.

Comment by Henry Wilton — August 16, 2011 @ 9:50 am

2. Just to clarify, Igor has two papers on this topic, both of which are available on his webpage. I believe they use the same basic idea, but the first works in the context of L^2 cohomology and the second one is very elementary, and only 7 pages long.

Comment by Nathan Dunfield — August 15, 2011 @ 9:30 pm

• To clarify my clarification, Igor actually has three papers on the HNC, rather than two as I implied; it’s the 2nd and 3rd papers I was referring to before, which are the ones which contain proofs of the HNC. The short one is here.

Comment by Nathan Dunfield — August 15, 2011 @ 10:46 pm

3. Dicks’ one-page proof is available at

http://mat.uab.cat/~dicks/SimplifiedMineyev.pdf

It is not a draft article. It is just a variation of the 17 May 2011 e-mail from Dicks to Mineyev explaining how one could convert his brilliant 20-page Hilbert-module proof into a 2-page Bass-Serre-theory proof, which Mineyev did, expanding it into a 7-page proof which he posted on 22 May 2011.

Comment by Warren Dicks — August 16, 2011 @ 11:26 am

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