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What are the properties that hold for a fundamental group of a surface and does not hold necessary for the fundamental groups of manifolds of higer dimensions ??

flag	asked Jun 18 2011 at 9:48 unkown 91●2

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"fundamental groups of manifolds of higer dimensions": every finitely presentable group is the fundamental group of some manifold of dim $\ge 4$. – André Henriques Jun 18 2011 at 10:39

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In some sense, André's comment is the answer (see also Jim's answer below): closed 4-dimensional manifolds are "as bad" as finitely presented groups: so their fundamental group can be non-linear, non-residually finite, non-hopfian,etc...; it can have Kazhdan's property (T), it can have all $L^2$-Betti numbers vanishing, etc... I share Henri's opinion that the OP is vague. – Alain Valette Jun 19 2011 at 5:17

Agol · Answer 1 · 2013-02-26 03:21:57Z UTC

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Poincar\'e duality groups of dimension 2 are surface groups.

answered Feb 26 at 3:21

Agol
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Roberto Frigerio · Answer 2 · 2011-06-19 12:59:04Z UTC

A surface group is either virtually abelian, or word hyperbolic (or both, when it is finite).

In some sense, this reflects the fact that every surface admits a Riemannian metric of constant curvature, and that the sign of the curvature is detected by the fundamental group.

In dimension 3, Perelman's uniformization implies that compact manifolds can be decomposed into "geometric pieces" (that are again detected in a suitable sense by their fundamental groups), while in higher dimension there is no hope for a simple result of this type.

Agol · Answer 3 · 2011-06-19 04:32:55Z UTC

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It's a conjecture that surface groups are characterized by being the only 1-relator groups such that every finite-index subgroup is also 1-relator and every infinite index subgroup is free.

edited Jun 19 2011 at 4:32

answered Jun 19 2011 at 2:39

Agol
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Neat. Whose conjecture is that? – Ryan Budney Jun 19 2011 at 2:40

oops, I forgot a condition. Ben Fine: sci.ccny.cuny.edu/~shpil/gworld/problems/… – Agol Jun 19 2011 at 4:33

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FYI, I recently proved this for 'cyclically pinched' one-relator groups: arxiv.org/abs/1102.2866 . – HW Jun 19 2011 at 6:33

Hey Henry, I haven't thought about this much - why is the condition on infinite index subgroups being free necessary? Is there a (residually finite) 1-relator group with finite-index subgroups 1-relator, but non-free infinite-index subgroups? – Agol Jun 19 2011 at 23:34

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Ian - The Baumslag--Solitar group BS(1,m) has this property for any m. In fact, this question goes back to Melnikov in the Kourovka notebook (though BS(1,m) is a counterexample to Melnikov's original question). – HW Jun 20 2011 at 6:01

Jim Conant · Answer 4 · 2011-06-19 01:27:23Z UTC

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The word problem for the fundamental group of a closed surface is solvable, using Dehn's algorithm. Since any finitely presented group appears as the fundamental group of some closed $4$-manifold, and there are such groups for which the word problem is unsolvable, this is indeed a special property for two dimensions.

answered Jun 19 2011 at 1:27

Jim Conant
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Not really special to two dimensions, since also true in three dimensions. – Igor Rivin Jun 19 2011 at 2:18

@Igor: I didn't interpret the question that way. I interpreted it to mean, what's true in two dimensions that's not true in general. – Jim Conant Jun 19 2011 at 12:22

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@Jim: I personally think the question is rather poor, so any answer makes more sense than the question itself. To increase the silliness slightly: "what is true of free groups which is not true of general groups"? – Igor Rivin Jun 20 2011 at 0:49

HW · Answer 5 · 2011-06-18 10:37:53Z UTC

This question is very vague, but here are some thoughts to add to Mark's answer.

First, note that any finitely presented group arises as the fundamental group of a closed manifold of dimension 4 (see this MO question), which is a huge contrast to the very special case of dimension 2.

The properties of the fundamental groups of 3-manifolds are a subject of very active research, much aided by Perelman's solution to the Geometrisation Conjecture. Like the 2-dimensional case, 3-manifold groups are residually finite (a theorem of Hempel). The fact that there is no closed 3-manifold with every infinite-index subgroup free is only very recent known, as a result of work of Kahn and Markovic.

I don't think any closed 3-manifold has cohomological dimension 2, so that property actually does it on its own.

Mark Sapir · Answer 6 · 2011-06-18 09:51:21Z UTC

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Every subgroup of infinite index is free, the group is residually finite, and the cohomological dimension is 2.

answered Jun 18 2011 at 9:51

Mark Sapir
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how do you prove thay every subgroup of infinite index is free ? – unkown Jun 18 2011 at 9:57

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It is well known: A. Hoare, A. Karrass and D. Solitar, Subgroups of infinite index in Fuchsiangroups, Math. Z. , 125, 1972, 59–69 – Mark Sapir Jun 18 2011 at 10:11

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unknown - consider the corresponding covering space, and convince yourself that it deformation retracts onto a graph. – HW Jun 18 2011 at 10:38

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@HW: your statement is actually a nontrivial theorem of Whitehead, so you might be overestimating the OP's mathematical prowess. – Igor Rivin Jun 19 2011 at 0:39

Igor - I suppose I really had the finitely generated case in mind. – HW Jun 19 2011 at 6:35

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