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

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 Autumn Kent's comment and the ensuing discussion, torsion in the homology has been ruled out provided the subset of $$\mathbb R^3$$ is compact and has the homotopy-type of a CW-complex (more precisely, if Cech and singular cohomologies agree).

I don't think that

torsion in the homology has been ruled out

Certainly, torsion in Cech cohomology has been ruled out for a compact subset. The "usual" universal coefficient formula, relating Cech cohomology to $\operatorname{Hom}$ and $\operatorname{Ext}$ of Steenrod homology, is not valid for arbitrary compact subsets of $\Bbb R^3$ (although it is valid for ANRs, possibly non-compact). The "reversed" universal coefficient formula, relating Steenrod homology to $\operatorname{Hom}$ and $\operatorname{Ext}$ of Cech cohomology is valid for compact metric spaces, but it does not help, because $\operatorname{Ext}(\Bbb Z[\frac1p],\Bbb Z)\simeq\Bbb Z_p/\Bbb Z\supset\Bbb Z_{(p)}/\Bbb Z$, which contains $q$-torsion for all primes $q\ne p$. (Here $\Bbb Z_{(p)}$ denotes the localization at the prime $p$, and $\Bbb Z_p$ denotes the $p$-adic integers. The two UCFs can be found in Bredon's Sheaf Theory, 2nd edition, equation (9) on p.292 in Section V.3 and Theorem V.12.8.)

The remark on $\operatorname{Ext}$ can be made into an actual example. The $p$-adic solenoid $\Sigma$ is a subset of $\Bbb R^3$. The zeroth Steenrod homology $H_0(\Sigma)$ is isomorphic by the Alexander duality to $H^2(\Bbb R^3\setminus\Sigma)$. This is a cohomology group of an open $3$-manifold contained in $\Bbb R^3$, yet it is isomorphic to $\Bbb Z\oplus(\Bbb Z_p/\Bbb Z)$ (using the UCF, or the Milnor short exact sequence with $\lim^1$), which contains torsion. Of course, every cocycle representing torsion is "vanishing", i.e. its restriction to each compact submanifold is null-cohomologous within that submanifold.

By similar arguments, $H_i(X)$ (Steenrod homology) contains no torsion for $i>0$ for every compact subset $X$ of $\Bbb R^3$.

It is obvious that "Cech homology" contains no torsion (even for a noncompact subset $X$ of $\Bbb R^3$), because it is the inverse limit of the homology groups of polyhedral neighborhoods of $X$ in $\Bbb R^3$. But I don't think this is to be taken seriously, because "Cech homology" is not a homology theory (it does not satisfy the exact sequence of pair). The homology theory corresponding to Cech cohomology is Steenrod homology (which consists of "Cech homology" plus a $\lim^1$-correction term). Some references for Steenrod homology are Steenrod's original paper in Ann. Math. (1940), Milnor's 1961 preprint (published in http://www.maths.ed.ac.uk/~aar/books/novikov1.pdf), Massey's book Homology and Cohomology Theory. An Approach Based on Alexander-Spanier Cochains, Bredon's book Sheaf Theory (as long as the sheaf is constant and has finitely generated stalks) and this paper http://front.math.ucdavis.edu/math/0812.1407

As for torsion in singular $4$-homology of the Barratt-Milnor example, this is really a question about framed surface links in $S^4$ (see the proof of theorem 1.1 in the linked paper).

• Are you saying that there is torsion in $H_4$ of Barratt-Milnor ? – BS. Nov 14 '10 at 10:13
• I'm only saying that one who is interested in whether there is torsion in singular $H_4$ of the $2$-dimensional Barratt-Milnor example (which I think is conceivable) might find it helpful to look at my geometric proof of the Barratt-Milnor original result that its singular $H_3$ is nonzero (and in fact uncountable). Personally I am not interested at all, because I don't know of any single true application of singular homology/homotopy beyond the case where it coincides with Steenrod homology/homotopy. (They coincide on spaces that are homotopy equivalent to ANRs, including CW-complexes.) – Sergey Melikhov Nov 15 '10 at 19:38
• Perhaps I should clarify that the Barratt-Milnor example, as I see it, shows precisely that singular homology is pathological on compact metric spaces (because it doesn't feel dimension). I wanted to understand why this is so, but haven't signed up for all pathology. There are other ways in which singular homology is pathological on compact metric spaces: it fails the strong excision axiom and the Alexander duality; it doesn't understand in any way the operation of inverse limit; K(Z,n) doesn't represent singular cohomology. Steenrod homology is free of all these deficiencies. Hope this helps. – Sergey Melikhov Nov 16 '10 at 11:10
• Dear Sergey, sorry for taking so long to get back on this. Yes I believe you are correct and I stated too much in ruling out torsion. I'll correct that statement. – Ryan Budney Nov 19 '10 at 19:43
• Hi Ryan, sorry that I messed up dimensions in the original response. – Sergey Melikhov Nov 19 '10 at 20:24

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.)

• 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 '09 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. – Autumn Kent Nov 7 '09 at 2:09
• If a manifold is aspherical (a $K(G,1)$) then its fundamental group cannot have an element of order $n>0$ because then the manifold would have a covering space which is both a manifold and a $K(\mathbb Z/n,1)$. Thus a noncompact $3$-manifold with torsion in $\pi_1$ would have to have $\pi_2\ne 0$. Does that help? – Tom Goodwillie Jul 13 '10 at 18:35
• n>1, I meant of course – Tom Goodwillie Jul 13 '10 at 23:34
• Tom, I think the Sphere Theorem (that any 3-manifold containing an essential sphere contains an essential embedded sphere) works for any 3-manifold, so a submanifold of $S^3$ with connected complement would be aspherical and so wouldn't have torsion in $\pi_1$. For the problem at hand, you can of course assume the subset of interest, X say, is connected, but it would seem that the issue is that you're interested in the fundamental group of X, and not the complementary 3-manifold (and you don't have duality at your disposal for $\pi_1$. – Autumn Kent Jul 14 '10 at 0:23

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.

There is the Barratt-Milnor 1962 example of "anomalous (singular) homology", showing that the rational singular homology of the one point union $X$ of countably many spheres $S^2$ whith radius tending to $0$ is non zero in all dimensions $>1$ (and is in fact uncountable). They use Hurewicz maps and infinite sums of Whitehead products of elements of homotopy groups of spheres, but I don't see if torsion in higher $\pi_i(S^2)$ could give torsion in $H_*(X,Z)$.

The answer in this post shows that if $$U$$ is open and connected, its fundamental group must be torsion-free.

• I put this information in my original question, 10 years ago. – Ryan Budney Jun 18 at 23:06