Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there a simple proof that 3-dimensional graph manifolds have residually finite fundamental groups?

By "simple" I mean the proof that does not use any hard 3d topology. I care because I wish to generalize this to higher-dimensional analogs of graph manifolds.

share|improve this question
    
A question that begs for Henry Wilton's attention. –  Greg Kuperberg Dec 15 '09 at 19:40
1  
Attention duly provided. –  HJRW Dec 15 '09 at 20:01
    
I've just rolled-back Igor's edit to the questionthat changed it to ask about residual hyperbolicity (see the edit history), and asked him to post that as a new question. –  Scott Morrison Dec 15 '09 at 21:59
add comment

1 Answer

up vote 6 down vote accepted

As far as I'm aware, every proof of this fact is essentially the same as Hempel's original proof. I don't know whether it's "simple" enough for you! The key point is that the fundamental group G of a Seifert-fibred piece has the following property.

Property. There exists an integer K such that for any positive integer n there is a finite-index normal subgroup Gn of G such that any peripheral subgroup P intersects Gn in KnP.

It's not too hard to prove. There's a nice account in a paper by Emily Hamilton (which generalizes Hempel's result).

The other important fact is that peripheral subgroups in Seifert-fibred manifold groups are separable (ie closed in the profinite topology, for any non-experts out there).

Using these two pieces of information, you can piece together finite quotients of Seifert-fibred pieces into a virtually free quotient of π1 of the graph manifold in which your favourite element doesn't die.

Note on separability of peripheral subgroups. Of course, Scott proved that Seifert-fibred manifold groups are LERF. But, by a pretty argument of Long and Niblo, a subgroup is separable if and only if the double along it is residually finite. In particular, you can deduce peripheral separability from the easier fact that Seifert-fibred manifold groups are residually finite.

share|improve this answer
    
Thanks, Henry! Unfortunately, this is too hard to generalize to higher-dimensional case. I will edit the question to reflect downgraded expectations. –  Igor Belegradek Dec 15 '09 at 20:28
    
To add a little, the key point of Henry's "Property" is that the induced subgroups are characteristic for the peripheral subgroups, and you can find common cover of each component of the JSJ decomposition which induces the same characteristic cover of each peripheral subgroup. I don't think you really need peripheral subgroup separability. You just need to know that that as K goes to infinity, the groups G_n intersect in the trivial group. –  Ian Agol Dec 15 '09 at 20:35
    
Ian, without peripheral separability it's conceivable that there's a g not in P such that g is in PG_n for every n. How do you envisage getting round this? –  HJRW Dec 15 '09 at 21:13
    
Hey Henry, I think you're right. I guess Hempel proves peripheral subgroup separability in the course of his argument, even though he doesn't phrase it in such terms. –  Ian Agol Dec 15 '09 at 22:32
    
Hi Ian, I suppose the point of the Long--Niblo argument is that you will have peripheral separability if the double is residually finite. So if you don't have it then you're hosed anyway. –  HJRW Dec 15 '09 at 23:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.