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Can the fundamental group of an open connected subset $X$ of $\mathbb{R}^n$ have a torsion element?

Robert
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    What happens if we embed, say $SO(3)$ into $\mathbb{R}^9$ and thicken it by a small epsilon to make it open? Is the unthickened core a deformation retract of the thickened version? – Jyrki Lahtonen Jan 25 '13 at 23:06

2 Answers2

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The real projective plane $\Bbb R P^2$ embeds into $\Bbb R^4$. Taking a tubular neighborhood $N$ of this embedding, we get for all $n \geq 4$ an open subset $N \subset \Bbb R^n$ such that $\pi_1(N) \cong \Bbb Z/2$, so the fundamental group of this open subset has torsion.

For $n = 2$, there is no such subset; see the following article:

Eda, K. Free $\sigma$-products and fundamental groups of subspaces of the plane. Topology and its Applications Volume 84, Issues 1-3, 24 April 1998, pp. 283-306.

This is actually an open problem for $n = 3$.

Mizar
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Henry T. Horton
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    I think it's an open problem for compact subsets of $\mathbf{R}^3$, but the answer is certainly known for open subsets (there's no torsion). Indeed start from a torsion loop, and find in your open subset a compact 3-fold containing this loop as well as a homotopy from a power of this loop to the constant. Then you get an orientable compact 3-fold with nonempty boundary with torsion in the fundamental group. From Thurston's geometrization it should be easy to conclude, and probably there's an argument using much less machinery. – YCor Feb 01 '13 at 20:08
  • Could somebody tell me how the 2-dimensional case follows from the results in Eda's paper? It seems to deal only with spaces which are the complement of a subset of the x-axis. – Mizar May 16 '13 at 10:12
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As you can see here, the real projective plane $\mathbb{R}P^2$ can be embedded in to $\mathbb{R}^4$ and so, taking a sufficiently small $\epsilon$ open neighbourhood of the image of our embedding, gives a homotopy equivalent surface with fundamental group $\mathbb{Z}/2\mathbb{Z}$. This is probably the most simple case.

Dan Rust
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