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

Question

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 is not torsion-free?

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).

Answer 1

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 https://www.maths.ed.ac.uk/~v1ranick/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 the paper

• Sergey A. Melikhov, Steenrod homotopy, Russ. Math. Surveys 64 (2009) 469-551; translated into Russian in: Uspekhi Mat. Nauk 64:3 (2009) 73-166, doi:10.1070/RM2009v064n03ABEH004620, arXiv: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).

Answer 2

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.)

Answer 3

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 $H_*(X,Z)$.

Answer 4

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.

Answer 5

Proposition. Suppose that a subset $X\subset R^3$ is semilocally simply connected (SLSC). Then for each $x\in X$, the group $G= \pi_1(X,x)$ is torsion-free.

Proof. I will be using the fact that if $U$ is an open connected subset of $R^3$, then $\pi_1(U)$ is torsion-free, see for instance here.

Let $c: S^1\to X$ be a loop representing an element of order $n$ in $\pi_1(X,x)$. Accordingly, let $c_n: S^1\to X$ be the precomposition of $c$ with the map $z\mapsto z^n$ and $h: D^2\to X$ be an extension of $c_n$ to the unit disk. The image $Y:= h(D^2)$ is compact and locally path-connected. Thus, there is a function $\phi(\delta)$ such that if $d(y_1, y_2)\le \delta$, $y_1, y_2\in Y$, then there is a path of diameter $\le \phi(\delta)$ in $Y$ connecting $y_1, y_2$. Furthermore, since $Y$ is compact and $X$ is assumed to be SLSC, there exists $\epsilon>0$ such that if $\alpha: S^1\to Y$ has image of diameter $\le \epsilon$, then $\alpha$ extends to a continuous map $D^2\to X$.

Now, consider the system of open $\frac{1}{i}$-neighborhoods $U_n$ of $Y$ in $R^3$. Let $r: U_i\to Y$ denote a (likely discontinuous) nearest-point projection (defined via the Axiom of Choice). Since each group $\pi_1(U_i,x)$ is torsion-free, the map $c: S^1\to Y$ extends to a (continuous) map $f_i: D^2\to U_i$.

I will now imitate the standard argument (I think, due to Borsuk), from the proof that each finite-dimensional compact locally-contractible metrizable space is ANR.

Given a triangulation $T$ of $D^2$, I define the map $g_i: T^{(0)}\to Y$ as the composition of the restriction of $f_i$ to the $T^{(0)}$ with the projection $r$. The goal is to show that for large $i$, the map $g_i$ extends to a map $D^2\to X$ which restricts to $c$ on $S^1=\partial D^2$.

First of all, if $\frac{1}{i}\le \delta$, and $T$ is such that the diameters of the images under $f_i$ of the edges of $T$ are $\le\delta$, then for each edge $e=[v,w]$ of $T$, $d(g_i(v), g_i(w))\le 3\delta$. Hence, by the local path connectivity of $Y$, we can extend $g_i$ to $e$ so that $g_i(e)\subset Y$ has diameter $\le \phi(3\delta)$. In the case of boundary edges of the disk $D^2$, we will assume that $g_i|_e=c|_e$. Note that for each 2-simplex $\Delta$ in $T$, the diameter of $g_i(\partial \Delta)$ is $\le 3\phi(3\delta)$. By taking $i$ sufficiently large and taking the triangulation $T$ sufficiently fine, we can assume that $3\phi(3\delta)\le \epsilon$, where $\epsilon$ is defined as above. Hence, for this value of $i$, the map $g_i$ extends to a map $g: D^2\to X$. It follows that $[c]=1\in G=\pi_1(X,x)$ and, hence, $G$ is torsion-free. qed

Answer 6

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