most recent 30 from http://mathoverflow.net 2013-05-22T03:55:43Z http://mathoverflow.net/feeds/question/68132 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68132/fundamental-groups-of-surfaces unkown 2011-06-18T09:48:53Z 2013-02-26T03:21:57Z

<p>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 ?? </p>

http://mathoverflow.net/questions/68132/fundamental-groups-of-surfaces/68133#68133 Mark Sapir 2011-06-18T09:51:21Z 2011-06-18T09:51:21Z

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

http://mathoverflow.net/questions/68132/fundamental-groups-of-surfaces/68135#68135 HW 2011-06-18T10:37:53Z 2011-06-18T10:37:53Z

<p>This question is very vague, but here are some thoughts to add to Mark's answer.</p> <p>First, note that <em>any</em> finitely presented group arises as the fundamental group of a closed manifold of dimension 4 (see <a href="http://mathoverflow.net/questions/15411/finite-generated-group-realized-as-fundamental-group-of-manifolds/15414#15414" rel="nofollow">this MO question</a>), which is a huge contrast to the very special case of dimension 2.</p> <p>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 <a href="http://arxiv.org/abs/0910.5501" rel="nofollow">work of Kahn and Markovic</a>.</p> <p>I don't think any closed 3-manifold has cohomological dimension 2, so that property actually does it on its own.</p>

http://mathoverflow.net/questions/68132/fundamental-groups-of-surfaces/68182#68182 Jim Conant 2011-06-19T01:27:23Z 2011-06-19T01:27:23Z

<p>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. </p>

http://mathoverflow.net/questions/68132/fundamental-groups-of-surfaces/68188#68188 Agol 2011-06-19T02:39:08Z 2011-06-19T04:32:55Z

<p>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. </p>

http://mathoverflow.net/questions/68132/fundamental-groups-of-surfaces/68209#68209 Roberto Frigerio 2011-06-19T12:59:04Z 2011-06-19T12:59:04Z

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

http://mathoverflow.net/questions/68132/fundamental-groups-of-surfaces/122947#122947 Agol 2013-02-26T03:21:57Z 2013-02-26T03:21:57Z

<p><a href="http://books.google.com/books?id=zaLQFGI-1AoC&amp;lpg=PA51&amp;dq=poincare%2520duality%2520groups%2520of%2520dimension%2520two&amp;pg=PA35#v=onepage&amp;q&amp;f=false" rel="nofollow">Poincar\'e duality groups of dimension 2 are surface groups.</a> </p>