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Here's a problem I've found entertaining.

Is it possible to find a subset of 3-dimensional Euclidean space such that its homology groups (integer coefficients) or one of its fundamental groups contains an element of finite order?

Context: The analogous question has a negative answer in dimension 2. This is a theorem of Eda's (1998). In dimension 4 and higher, the answer is positive as the real projective plane embeds. If the subset of 3-space has a regular neighbourhood with a smooth boundary, a little 3-manifold theory says the fundamental group and homology groups are torsion-free.

edit: Due to Richard Kent's comment and the ensuing discussion, torsion in the homology has been ruled out. So any torsion in the fundamental group must belong to the commutator subgroup.

edited Nov 7 at 2:17

asked Nov 7 at 1:14

Ryan Budney
4,139●5●25

Richard Kent · Answer 1 · 2009-11-07 02:02:04Z UTC

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I'll assume that the subset is compact.

Then, if you use Cech cohomology, Alexander duality turns this into a question about the complement, which is a 3-manifold.

So, I answer with another question: Can a (wild) open submanifold of the 3-sphere have torsion in its homology? (My guess is no. But then I'm not RH Bing.)

edited Nov 7 at 2:02

answered Nov 7 at 1:51

Richard Kent
2,097●1●3●17

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Ah, that's nice. The answer to your question is no. If you had torsion in H_1 of an open 3-dimensional submanifold of R^3, you'd have it in a compact 3-dimensional submanifold of R^3. That doesn't happen -- I didn't supply the argument but it it boils down to what's known as Fox's Re-Embedding Theorem (which is an application of Dehn's Lemma), that such a 3-manifold can be re-embedded to be the complement of disjoint embedded handlebodies, another duality application says H_1 is free. So part of my question is answered. The fundamental group question remains open. – Ryan Budney Nov 7 at 2:04

Oh, very good. I knew the argument for compact submanifolds, but was spacing on whether or not you can immediately jump to a compact submanifold. Thanks. – Richard Kent Nov 7 at 2:09

Igor Belegradek · Answer 2 · 2009-11-07 02:01:52Z UTC

I think your subset of R^3 must be pretty ugly to have a fighting chance. If it is a compact subpolyhedron of R^3, then by Alexander duality its k-homology is the same as (2-k)-dimensional cohomology of an open 3-manifold. The only interesting case is k=1 because 0th (co)homology are torsion free, but if the open manifold is homotopy equivalent to a finite complex then by universal coefficients 1st cohomology is torsion free. This rules out all "nice" examples.

Torsion in homology or fundamental group of subsets of Euclidean 3-space

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