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!


  • 1
    $\begingroup$ Sounds right : ) $\endgroup$ – Rudy the Reindeer Apr 8 '12 at 19:33
  • 1
    $\begingroup$ This is basically right, though writing down all the details could be messy. $\endgroup$ – Cheerful Parsnip Apr 8 '12 at 19:39
  • $\begingroup$ Thanks! I agree the details could be messy, but now at least I know I have the right idea. $\endgroup$ – Edward Hughes Apr 8 '12 at 19:41
  • 1
    $\begingroup$ 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) $\endgroup$ – William Apr 8 '12 at 22:48
  • 3
    $\begingroup$ 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! $\endgroup$ – Mariano Suárez-Álvarez Apr 8 '12 at 23:19

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 $F_n$.

  • 1
    $\begingroup$ How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks! $\endgroup$ – Edward Hughes Apr 10 '12 at 23:45
  • 2
    $\begingroup$ 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: en.wikipedia.org/wiki/Stereographic_projection $\endgroup$ – John Stalfos Apr 12 '12 at 2:14
  • 4
    $\begingroup$ 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 $F_{2n-1}$. :) $\endgroup$ – Patrick Da Silva Sep 1 '13 at 12:11
  • $\begingroup$ How do I see the deformation retraction? $\endgroup$ – bounceback May 5 '20 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.