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Does there exist an $X\subseteq\Bbb R^3$ such that $\pi_1(X)$ is finite but nontrivial? (note that $X$ doesn't have to be a manifold or anything nice).

More generally let $f:\Bbb N\to\Bbb N$ be the function that maps $n$ to the smallest $m$ such that there is a subspace of $\Bbb R^m$ whose $\pi_n$ is finite but nontrivial, is $f(n)=n+3$? If not what can be said about this function?

Alessandro Codenotti
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    https://math.stackexchange.com/questions/36279/is-the-fundamental-group-of-every-subset-of-mathbbr2-torsion-free – Moishe Kohan Dec 21 '18 at 22:51
  • Thanks @MoisheCohen, I wish I had found that question before asking mine! I suppose there's not hope for the general case then and I should ask about which results are known for $\pi_1$ and $\Bbb R^3$ instead maybe – Alessandro Codenotti Dec 21 '18 at 22:53
  • However as @PaulPlummer pointed out, it could be possible that it can be proved that there are no subspaces with finite nontrivial $\pi_1$, while the general question about torsion remains open – Alessandro Codenotti Dec 21 '18 at 23:07
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    The standard algebraic topology invariants (like homotopy groups and singular co/homology groups) were designed for reasonably "nice" topological spaces (originally, smooth manifolds, later on - CW complexes). For more pathological spaces (such as arbitrary subsets of $R^n$) one should look for other invariants, for instance, Chech homotopy and cohomology groups, Sitnikov homology, etc. – Moishe Kohan Dec 23 '18 at 16:59
  • @MoisheCohen Sp perhaps the first step should be to ask whether there are finite polyhedra $P \subset \mathbb{R}^3$ having finite fundamental groups. Since the fundamental group of $P$ agrees with the fundamental group of the $2$-skeleton of $P$, it suffices to consider $2$-dimensional polyhedra $P \subset \mathbb{R}^3$. Perhaps something is known about these spaces? – Paul Frost Dec 27 '18 at 16:01
  • @PaulFrost: No, $\pi_1(P)$ will be torsion-free, see my answer here: https://math.stackexchange.com/questions/680998/is-there-a-domain-in-mathbbr3-with-finite-non-trivial-pi-1-but-h-1-0 – Moishe Kohan Dec 27 '18 at 16:42

1 Answers1

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I do not have an answer to your first question (and I also do not find it very natural since you are considering arbitrary subsets of $R^3$, see my comment above). I will answer the second question.

No, $f(n)\ne n+3$. For instance, $f(4)=3$ since $\pi_4(S^2)\ne 0$ and every subset of $R^2$ has trivial higher homotopy groups. For the same reason, $f(n)=3$, $n\ge 4$. Indeed, $\pi_i(S^2)$ is finite for every $i\ge 4$, and is nonzero, see

S. Ivanov, R. Mikhailov, J. Wu, "On nontriviality of homotopy groups of spheres", 2015.

Read more on homotopy groups of spheres here.

"The difference between general topology and algebraic topology is that the former asks simple questions about complicated spaces while the latter asks difficult questions about simple spaces."

Edit. Just for completeness:

  1. For every open connected subset $U\subset R^3$, $\pi_1(U)$ is torsion-free, see my answer here.

  2. Consequently, for every compact connected subset $K\subset R^3$ its Chech fundamental group $\check{\pi}_1(K)$ is also torsion-free. This is because $$ \check{\pi}_1(K)\cong colim_{U} \pi_1(U) $$ where the inverse limit is taken over all open neighborhoods of $K$ in $R^3$.

  3. Therefore, for every connected CW complex $C$ (more generally, an ENR) which admits a (topological) embedding $C\to R^3$, $\pi_1(C)$ is torsion-free.

Moishe Kohan
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