What are some properties that hold for the fundamental group of a surface and do not necessarily hold for the fundamental groups of manifolds of higher dimensions?
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18$\begingroup$ "fundamental groups of manifolds of higer dimensions": every finitely presentable group is the fundamental group of some manifold of dim $\ge 4$. $\endgroup$– André HenriquesCommented Jun 18, 2011 at 10:39
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1$\begingroup$ 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. $\endgroup$– Alain ValetteCommented Jun 19, 2011 at 5:17
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$\begingroup$ If fundamental groups cannot differentiate manifolds of dimension $\ge 4$ then what sort of topological invariants DO differentiate those manifolds? $\endgroup$– Sidharth GhoshalCommented Jul 10 at 16:11
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$\begingroup$ @SidharthGhoshal This question is about characterizing fundamental groups of various classes of spaces, not whether the fundamental groups distinguish the spaces in question. $\endgroup$– Sam HopkinsCommented Jul 10 at 17:46
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$\begingroup$ @SidharthGhoshal: To very briefly address your comment, there is a giganto-enourmous apparatus of homotopy theory, and surgery theory, and this theory, and that theory, which is brought to bear on the classification of higher dimensional manifolds. $\endgroup$– Lee MosherCommented Jul 10 at 17:49
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6 Answers
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It's a conjecture that surface groups are characterized (among infinite groups, with the exception of $\mathbb{Z}/2\mathbb{Z}$) by being the only 1-relator groups such that every finite-index subgroup is also 1-relator and every infinite index subgroup is free.
Addendum July 2024: This conjecture has been proved for 2-generator groups:
- Giles Gardam, Dawid Kielak, Alan D. Logan, The Surface Group Conjectures for groups with two generators, Math. Res. Lett. 30 Number 1 (2023) pp 109–123, doi:10.4310/MRL.2023.v30.n1.a5, arXiv:2202.11093.
Further addendum July 2024: Recently, this conjecture has been proved in full generality. More precisely, every one-relator group such that every subgroup of infinite index is free is either a free group or a surface group. (The related conjecture, that every residually finite one-relator group with every finite-index subgroup one-relator must be either free, a surface group or a Baumslag--Solitar group, remains open.)
- Henry Wilton, Surface groups among cubulated hyperbolic and one-relator groups, arXiv:2406.02121.
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2$\begingroup$ Neat. Whose conjecture is that? $\endgroup$ Commented Jun 19, 2011 at 2:40
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$\begingroup$ oops, I forgot a condition. Ben Fine: sci.ccny.cuny.edu/~shpil/gworld/problems/probone-rel.html $\endgroup$– Ian AgolCommented Jun 19, 2011 at 4:33
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1$\begingroup$ FYI, I recently proved this for 'cyclically pinched' one-relator groups: arxiv.org/abs/1102.2866 . $\endgroup$– HJRWCommented Jun 19, 2011 at 6:33
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3$\begingroup$ 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). $\endgroup$– HJRWCommented Jun 20, 2011 at 6:01
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1$\begingroup$ Note: finite cyclic group satisfy this (and are the only finite groups satisfying this). So one needs to add "infinite" in the conjecture. $\endgroup$– YCorCommented Jul 10 at 9:17
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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:
- Immersing almost geodesic surfaces in a closed hyperbolic three manifold, Annals of Mathematics, 175 Issue 3 (2012) pp 1127–1190, doi:10.4007/annals.2012.175.3.4, arXiv.
I don't think any closed 3-manifold has cohomological dimension 2, so that property actually does it on its own.
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Every subgroup of infinite index is free, the group is residually finite, and the cohomological dimension is 2.
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$\begingroup$ how do you prove thay every subgroup of infinite index is free ? $\endgroup$– unkownCommented Jun 18, 2011 at 9:57
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1$\begingroup$ It is well known: A. Hoare, A. Karrass and D. Solitar, Subgroups of infinite index in Fuchsiangroups, Math. Z. , 125, 1972, 59–69 $\endgroup$– user6976Commented Jun 18, 2011 at 10:11
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1$\begingroup$ unknown - consider the corresponding covering space, and convince yourself that it deformation retracts onto a graph. $\endgroup$– HJRWCommented Jun 18, 2011 at 10:38
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3$\begingroup$ @HW: your statement is actually a nontrivial theorem of Whitehead, so you might be overestimating the OP's mathematical prowess. $\endgroup$ Commented Jun 19, 2011 at 0:39
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$\begingroup$ Igor - I suppose I really had the finitely generated case in mind. $\endgroup$– HJRWCommented Jun 19, 2011 at 6:35
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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
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2$\begingroup$ Not really special to two dimensions, since also true in three dimensions. $\endgroup$ Commented Jun 19, 2011 at 2:18
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$\begingroup$ @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. $\endgroup$ Commented Jun 19, 2011 at 12:22
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2$\begingroup$ @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"? $\endgroup$ Commented Jun 20, 2011 at 0:49
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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.
answered Jun 19, 2011 at 12:59
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- Beno Eckmann, Poincaré duality groups of dimension 2 are surface groups, In: Combinatorial Group Theory and Topology. (AM-111), 1987, pp. 35-52, JSTOR, Google Books.
Also available in: Mathematical Survey Lectures 1943–2004. Springer, Berlin, Heidelberg, 2006. https://doi.org/10.1007/978-3-540-33791-1_9