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Hi all,

I am beginning to learn about geometric group theory. I would like to write a little exposé intended to be read by the uninitated, so it would be nice to talk about (preferably simple) results which were inaccessible until geometric methods had been applied. Do you have suggestions?

Edit: I would also be interested to hear about results which aren't necessarily inaccessible otherwise, but admit simpler proofs within a geometric approach.

Best to all

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    $\begingroup$ This should be Community Wiki (you'll have to edit your post to make this change) as there's no "right" answer. Also, your question is a little vague, in that you're basically asking people to tell you about geometric group theory. Could you narrow it down a bit? For example, the Wikipedia "Geometric Group Theory" page must partially answer your question -- if so, what else are you looking for? $\endgroup$
    – Ryan Budney
    Commented Sep 4, 2011 at 3:30
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    $\begingroup$ What about the subgroup of a free group is free. $\endgroup$
    – Spice the Bird
    Commented Sep 4, 2011 at 3:49
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    $\begingroup$ @Birdman, if you're thinking about the covering space argument, that specific proof technically predates geometric group theory. Moreover, there were earlier (essentially equivalent) proofs that predated covering space arguments, so such an argument certainly wasn't "inacessible" before geometric group theory. $\endgroup$
    – Ryan Budney
    Commented Sep 4, 2011 at 3:56
  • $\begingroup$ Perhaps I have misunderstood the question, but it seems to veer uncomfortably close to "please write an enyclopaedia entry for me", which I thought we had decided was not MO's metier $\endgroup$
    – Yemon Choi
    Commented Sep 4, 2011 at 4:45
  • $\begingroup$ (On 2nd thoughts I am being a bit uncharitable. Nevertheless, I second Ryan's suggestion/request for a more focused question. Maybe something on Bass-Serre theory?) $\endgroup$
    – Yemon Choi
    Commented Sep 4, 2011 at 4:46

4 Answers 4

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Gromov's theorem that a group of polynomial growth is virtually nilpotent. This still has no algebraic proof to the best of my knowledge.

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    $\begingroup$ This is a good example, but since the definition of "polynomial growth" is geometric, maybe it's not surprising that geometry is an ingredient in the solution. $\endgroup$
    – Ian Agol
    Commented Sep 4, 2011 at 17:33
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    $\begingroup$ @Agol, although one can, and should, think of growth in terms of balls in the Cayley graph the definition of growth is purely combinatorial. You are counting the number of elements that have a shortest length representative of size at most n. People consider growth of other algebraic structures like semigroups and algebras where one doesn't have the corresponding geometry. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 5, 2011 at 0:33
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    $\begingroup$ @Ben: I agree it's combinatorial, but since you used the term "length", you (and others) are clearly thinking of it in a geometric fashion. In fact, I think Milnor originally introduced the notion of growth of groups with geometric applications in mind. A choice of generating set (and therefore Cayley graph) is in some sense a geometric choice, not intrinsic to the group (like the choice of a Riemannian metric on a manifold). The answer "virtually nilpotent" is free of notions of length, which makes the theorem interesting. $\endgroup$
    – Ian Agol
    Commented Sep 5, 2011 at 15:20
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    $\begingroup$ @Agol, I'm pretty sure that people have used length to describe the number of letters in a word since the time of Post, Thue and Dyck, if not earlier. Length of a word is just a combinatorial notion. But I agree nobody considered growth before Milnor and Svarc, who were geometrically motivated. But people could have, and the notion is considered for finitely generated algebras. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 5, 2011 at 20:56
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Dani Wise has recently proven a conjecture of Baumslag, that 1-relator groups whose relator is a proper power are residually finite (in fact, linear). The proof makes use of techniques from geometric group theory, in particular using techniques of hyperbolic groups and CAT(0) cube complexes.

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    $\begingroup$ I think this fails at the "preferably simple" stipulation... $\endgroup$
    – ADL
    Commented Sep 5, 2011 at 9:47
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The proof of Stallings's Ends Theorem is topological. Note that the set of ends of a group $\Gamma$ can be identified with $H^1(\Gamma,\mathbb{Z}_2\Gamma)$, so you don't have to define ends geometrically.

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  • $\begingroup$ Perhaps an even better application is the corollary that groups of cosmological dimension one are free since this is a purely algebraic statement. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 5, 2011 at 3:49
  • $\begingroup$ Cosmological should be cohomological. Somebody should tell me how to remove the autocorrect on iPad. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 5, 2011 at 3:49
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    $\begingroup$ I'm not sure why you think one cohomologically-defined invariant (cohomological dimension) is more algebraic than another (ends). But 'cosmological dimension' is brilliant! $\endgroup$
    – HJRW
    Commented Sep 5, 2011 at 9:32
  • $\begingroup$ As you say in your answer ends don't have to be defined geometrically, but they are naturally defined that way. The cohomological description is not really the natural definition. On the other hand Stallings original paper on ends was precisely targeted at proving the Eilenberg-Ganea conjecture that groups of cohomological dimension 1 are free. In other words, a geometric proof is given that the augmentation ideal of a group algebra is projective iff the group is free. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 5, 2011 at 20:52
  • $\begingroup$ As I think of the cohomology of a group as the cohomology of an Eilenberg--Mac Lane space, I don't really think that the algebraic definition of cohomological dimension is the 'natural' one either! $\endgroup$
    – HJRW
    Commented Jan 4, 2012 at 12:31
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I gave an example here of a topological proof that a product of two commutators in a free group is not itself always a commutator. In answer to the same question, Arturo Magidin indicated how to give a combinatorial proof. I think it's fair to say that the two proofs have completely different flavours, although you can judge for yourself which is 'simpler'.

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