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Compute the homology groups of $S^3 \setminus K$, where $K$ is two linked copies of circles in $\mathbb R^3$. How about the homology group of $S^3 \setminus K'$ where $K'$ is just one copies of circles in $\mathbb R^3$?

What is really important and interesting is how to compute the homology group. To compute $H_*(S^3 \setminus K)$ I guess we should use Mayer-Vietoris sequence to $A$ and $B$, two copies of $S^3\setminus S^1$.

However, I even don's know how to compute $H_*(S^3 \setminus S^1)$. To do this, I try Mayer-Vietoris theorem again.

Because I have few intuition about $S^3$, I do not know what the maps are in there Mayer-Vietoris long exact sequences.

Please help!

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    $\begingroup$ By "one linked copies of circles", do you mean just one circle :-) $\endgroup$
    – Stefan Hamcke
    Jun 30, 2015 at 17:06
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    $\begingroup$ For homology, are you familiar with Alexander duality? $\endgroup$
    – Qiaochu Yuan
    Jun 30, 2015 at 19:13
  • $\begingroup$ I assume you're referring here to the Hopf link. Represent the linked circles as the z-axis $\cup \{\infty\}$ and the unit circle in the xy-plane in $\mathbb R^3 \cup \{\infty\}$. Delete these. Show that this deformation retracts onto a torus neighborhood of the circle in the xy-plane. If your K' just means a single unknotted circle, try to do something similar. The process will be simpler. If you don't mean these things to be unknotted, Alexander duality is easiest. $\endgroup$
    – user98602
    Jun 30, 2015 at 21:10

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For basic Mayer-Vietoris usage you need to thicken your link/knot a little in such way that you won't create new intersections in obtained in such way "link" of solid tori.

Namely: let $K$ be your link of $n$ circles and $U_1 = K \times D^2_{3 \varepsilon}$, where $D_{3 \varepsilon}^2$ is disk of "small" radius $3 \varepsilon$ such that you can retract $U_1$ to $K$. Then let's denote by $U_2 = S^3 \setminus \{K \times D^2_{ \varepsilon}\}$ it's thickened complement. MV sequence gives us: $$\ldots\rightarrow H_2(U_1 \cup U_2) \rightarrow H_1(U_1 \cap U_2) \rightarrow H_1(U_1) \oplus H_1(U_2) \rightarrow H_1(U_1 \cup U_2) \rightarrow \ldots$$ $$\ldots\rightarrow H_2(S^3) \rightarrow H_1(U_1 \cap U_2) \rightarrow H_1(U_1) \oplus H_1(U_2) \rightarrow H_1(S^3) \rightarrow \ldots$$ $$\ldots0 \rightarrow H_1(U_1 \cap U_2) \rightarrow H_1(U_1) \oplus H_1(U_2) \rightarrow 0 \rightarrow \ldots.$$ Observe now that:

  1. $U_1 \cap U_2$ is homologically disjoint sum of $n$ tori and hence $H_1(U_1 \cap U_2) = \bigoplus_{i=1}^n \mathbb{Z}\oplus \mathbb{Z}$

  2. $H_1(U_1) = \bigoplus_{i=1}^n \mathbb{Z}$

  3. $H_1(U_2) = H_1(S^3 \setminus K)$

hence $H_1(S^3 \setminus K) = \bigoplus_{i=1}^n \mathbb{Z}$.

The problem with this approach is that it depends on the embedding of $K$ in $S^3$. It may not be possible to find described small $\varepsilon$ (search for "wild knots"). However Qiaochu already mentioned the workaround - Alexander Duality which gives the result immediately.

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  • $\begingroup$ Regarding your last paragraph, Alexander duality needs the subspace to be locally contractible, which I suspect also depends on the embedding of $K$ in $S^3$. $\endgroup$
    – Najib Idrissi
    Jul 30, 2015 at 11:13
  • $\begingroup$ Ah, you are right. $\endgroup$
    – Adam Baranowski
    Jul 30, 2015 at 19:26

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