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.