7
$\begingroup$

I'm a student taking my first course in algebraic topology. I've stumbled across this exercise: calculate the fundamental group of $S^3-\gamma$, where $\gamma$ is a circumference in $\mathbb{R}^3$ (i.e. $\gamma=S^1$ in $\mathbb{R}^3$) and $S^3=\mathbb{R}^3\cup\lbrace\infty\rbrace$ is the one-point compactification of $\mathbb{R}^3$. I thought of $\mathbb{R^3}$ as $\mathbb{R}^1\times\mathbb{R}^2$, then I can think of $\gamma$ as an $S^1$ in $\mathbb{R}^2$ which has the same homotopy type as $\mathbb{R}^2\setminus\lbrace (0,0)\rbrace$. So $\mathbb{R^3}\cup\lbrace\infty\rbrace$ without $\gamma$ has the same homotopy type as $\mathbb{R}\cup\lbrace\infty\rbrace$, and its one-point compactification gives $S^1$.

In the end, I would get $\pi_1(S^3\setminus\gamma)=\pi_1(S^1)=\mathbb{Z}$, but I don't know if it's correct.

Maybe I could think of $S^3$ as $S^1\ast S^1$, but I don't know if that could help.

Thank you in advance

  • 2
    What do you mean "circumference"? You mean a knot?2012-12-26
  • 0
    If you look in OP's question you gonna see that $\gamma$ is a $S^1$ @user322402012-12-26
  • 0
    Yes yes exactly, $\gamma$ is $S^1$2012-12-26
  • 0
    @Francesca: you should edit your post to say what $\gamma$ is. (the "edit" button is right under the tag, under the post).2012-12-26
  • 0
    Also, what does $S^1\ast\S^1$ mean?2012-12-26
  • 0
    @Tomás: My point is, there can be different embeddings of $\mathbb{S}^{1}$ in $\mathbb{R}^{3}$, and the result on the fundamental groups are quite different.2012-12-26
  • 0
    @user32240 Are they? At least I know that the first dimension homology groups are the same no matter the embedding (chapter 2B of Hatcher states the result), so the fundamental groups cannot be _that_ different.2012-12-27
  • 0
    @Arthur: Check http://en.wikipedia.org/wiki/Knot_group. The computation of the knot group is very difficult, and it is $\mathbb{Z}$ if and only if the knot is the unknot.2012-12-27
  • 0
    @user32240 I didn't know that. But in that case, I would assume that $\gamma$ is the boundary of an embedded disc. That also fits with OP's description of $\gamma$ as a "circumference".2012-12-27

1 Answers 1

2

If you "rotate" your $S^3 = \mathbb{R}^3\cup\lbrace\infty\rbrace$ so that one of the points of $\gamma$ is at $\infty$, you get $S^3 \setminus \gamma$ homeomorphic to $\Bbb R^3 \setminus \text{a line}$, which can be deformation retracted to a circle, so $\Bbb Z$ it is!

Also, your reasoning seems sound enough, and you get the same answer, so I would assume it is correct.

  • 0
    That's a very nice point of view and you get the same result. My only doubt is that the text says that $\gamma$ is in $\mathbb{R}^3$ so it might want to avoid the case with a point of $\gamma$ at $\infty$. But thanks a lot anyway2012-12-26
  • 0
    @Francesca: the difference is just a rotation of $S^3$. It's homogeneous (like any sphere), so you can move any point anywhere.2012-12-26
  • 0
    Oh ok. I thought it was more like removing $S^1$ from $\mathbb{R}^3$ and then compactify with the point at $\infty$, but I guess your point of view is more correct.2012-12-26