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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

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What do you mean "circumference"? You mean a knot? –  Bombyx mori Dec 26 '12 at 11:29
    
If you look in OP's question you gonna see that $\gamma$ is a $S^1$ @user32240 –  Tomás Dec 26 '12 at 11:37
    
Yes yes exactly, $\gamma$ is $S^1$ –  Francesca Dec 26 '12 at 12:25
    
@Francesca: you should edit your post to say what $\gamma$ is. (the "edit" button is right under the tag, under the post). –  tomasz Dec 26 '12 at 12:30
    
Also, what does $S^1\ast\S^1$ mean? –  tomasz Dec 26 '12 at 12:31

1 Answer 1

up vote 2 down vote accepted

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.

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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 anyway –  Francesca Dec 26 '12 at 12:27
    
@Francesca: the difference is just a rotation of $S^3$. It's homogeneous (like any sphere), so you can move any point anywhere. –  tomasz Dec 26 '12 at 12:33
    
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. –  Francesca Dec 26 '12 at 12:45

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