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Just a quick question to verify whether I'm right.

Claim: The fundamental group of the complement of $n$ lines through the origin in $\mathbb{R}^3$ is $F_n$, the free group on $n$ generators.

Proof: remove a line from $\mathbb{R}^3$. We may deformation retract the remaining space onto a cylinder radius $\epsilon$ about the line, and thence to a circle $S^1$. There is no trouble repeating this process with a second distinct line, except that then we will be a wedge union $S^1 \vee S^1$. Continue inductively, and recall that the wedge union of $n$ circles has the stated fundamental group.

I'm only just starting to really get my head around this stuff, so any feedback would be really useful!

Thanks!

Edward Hughes
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  • Thanks! I agree the details could be messy, but now at least I know I have the right idea. – Edward Hughes Apr 08 '12 at 19:41
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    Do you mean "minus $n$ lines" like in the title, or "minus $n$ lines through the origin"? There is a significant difference: if the $n$ lines are disjoint then the fundamental group is $F_{n}$ (seen by deformation retracting onto ($\mathbb{R}^2$ minus $n$ points)), but if $n\geq 2$ and they all intersect at the same point then the fundamental group is $F_{2n-1}$ (as shown by user8268) – William Apr 08 '12 at 22:48
  • @you, and in general the fundamental group will depend on the intersection pattern (but I don't see if it will only depend on it; probably not—and this should be known, by people who work on subspace arrangements) – Mariano Suárez-Álvarez Apr 08 '12 at 23:16
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    You say «There is no trouble repeating this process with a second distinct line». What process? WHat you explain in the case of one line cannot be done when there are two of them! – Mariano Suárez-Álvarez Apr 08 '12 at 23:19
  • @MarianoSuárez-Alvarez, I just mean to say that the title of the thread seems to be misleading, as it doesn't match the question in the main post, and I think there needs to be some clarification as to which question should be answered here. – William Apr 08 '12 at 23:25
  • This seems similar in many respects to Tai-Danae's example 4 at https://www.math3ma.com/blog/clever-homotopy-equivalences. Good question, thanks for asking. – DanLewis3264 May 05 '20 at 21:46

1 Answers1

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There is a deformation retraction of ($\mathbb{R}^3$ minus $n$ lines through the origin) to (the unit sphere with $2n$ points removed). The $2n$ points are the intersections of the lines with the sphere, the deformation retraction is along the rays from the origin.

As a result, the fundamental group is actually $F_{2n-1}$, not $F_n$.

user8268
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    How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks! – Edward Hughes Apr 10 '12 at 23:45
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    Have you heard of stereographic projection? It is a homeomorphism from the complement of a point on $S^n$ to $\mathbb{R}^n$. Relevant link: http://en.wikipedia.org/wiki/Stereographic_projection – John Stalfos Apr 12 '12 at 02:14
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    I was asking this question myself too ; I guess the right way to do it is to use one of the $2n$ points as a "north pole" for the stereographic projection, which leaves us with $\mathbb R^2$ with $2n-1$ points removed. Using van Kampen's theorem, we get $F_{2n-1}$. :) – Patrick Da Silva Sep 01 '13 at 12:11
  • How do I see the deformation retraction? – DanLewis3264 May 05 '20 at 21:52
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    @DanLewis3264 You may be familiar with the deformation retraction of $\mathbb R^3 \setminus {0}$ to $S^2$ given by $H(x, t) = \frac{x}{1 + t(| x| - 1)}$. Now just restrict the domain to $\mathbb R^3$ minus $n$ lines through the origin. – Alex G. Oct 30 '21 at 03:48