Date: Mon, 6 Mar 95 11:32:26 CST From: rusin@math.niu.edu (Dave Rusin) To: povl@euklid.euromath.dk Subject: Re: fundamental group for subset of R^2 In article you write: >Does anyone have a counterexample to (or a proof of) this conjecture: > >The fundamental group of a subset of R^n is torsion free for n<=3. I suspect it's true but I have no proof. I would appreciate a copy of any responses you get. You could analyze the problem like this: If X in R^n has torsion in its fundamental group, let f: S^1 -> X be a path of finite order p in pi_1(X). Without loss of generality we may assume p is prime. Then the statement that f^p=e implies a homotpoy between f^p and a constant map, and thus a F: D --> X where D is the disk in R^2 and F | boundary(D) is f^p. That is, there is a map which I would also call F : D/Z_p -> X, where in the quotient space we identity each z in the boundary of D with z.w for each p-th root of unity w. The image of F (inside X ) is a subset of R^n in which the loop F|S^1 is not contractible. Conversely, given any map F : D/Z_p --> R^n let X be the image of F. Then F|S^1 is a loop f in pi_1(X) with f^p=e, so as long as f itself is not contractible, then f has order precisely p in pi_1(X). So the answer to your question depends on the study of all such maps F. I cannot imagine a map F in which F|S^1 is not contractible in X, but as we all know that's doesn't mean one cannot exist :-) dave ============================================================================== Newsgroups: sci.math.research From: greg@dent.uchicago.edu (Greg Kuperberg) Subject: Re: fundamental group for subset of R^2 Date: Tue, 14 Mar 1995 04:42:07 GMT In article , Povl AAge wrote: >Does anyone have a counterexample to (or a proof of) this conjecture: >The fundamental group of a subset of R^n is torsion free for n<=3. Here is a partial answer inspired by some comments by Danny Ruberman. If the subset of R^3 is a 3-manifold with boundary, then pi_1 certainly is torsion-free. You can prove that using the sphere theorem, or using Haken decomposition, or if you want to go overboard, using Thurston geometry. The last approach is not kosher, but it is the easiest to summarize: Such a 3-manifold has geometric pieces by Thurston's theorem, and the only type of geometric piece that yields torsion, a spherical piece, can't appear. It follows that any torsion in pi_1 for an arbitrary subset lies in the kernel of the map from pi_1 to shape-theoretic pi_1 (which came up recently in another thread.) Therefore a simplicial complex in R^3, for example, has torsion-free pi_1. I bet that you can't have any torsion, period.