Fundamental group torsion for subsets of Euclidean 3-space

Importance: Medium ✭✭
Author(s): Ancient/folklore
Subject: Topology
Recomm. for undergrads: no
Posted by: rybu
on: November 7th, 2009

\begin{problem} Does there exist a subset of $\mathbb R^3$ such that its fundamental group has an element of finite order? % Enter your conjecture in LaTeX % You may change "conjecture" to "question" or "problem" if you prefer. \end{problem}

The corresponding problem in $\mathbb R^2$ has a negative answer. The corresponding problem in $\mathbb R^n$ for $n \geq 4$ has a positive answer. If the subset of $\mathbb R^3$ has a regular neighbourhood with a smooth boundary, the answer is negative. Similarly the homology of the subset is known to have no torsion via an Alexander duality argument. So any torsion in the fundamental group must be in the commutator subgroup.

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[E] Eda, K. Fundamental group of subsets of the plane. Topology and its Applications Volume 84, Issues 1-3, 24 April 1998, Pages 283-306

% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)

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