I just have the feeling that there must be some relation between Alexander duality and linking numbers, but I don't know what is that. Will anyone tell me anything about that? Or could anyone give some references? Thanks.

  • $\begingroup$ @anomaly Usually closed curves don't intersect in $S^3$. $\endgroup$ – PVAL-inactive Feb 14 '16 at 4:22
  • $\begingroup$ @PVAL: Oops, of course not. (The linking number in $S^3$ can be described as an intersection number between one curve and a surface bounded by another, but it's easier just to compute it from a crossing diagram.) $\endgroup$ – anomaly Feb 14 '16 at 5:33

Not only there is a relation between them, but in fact the linking number may be defined (and generalized to higher dimensions) via Alexander duality as follows:

Let $M,N\subset \mathbb{R}^{m}$ be $p$ and $q$ dimensional manifolds (compact, connected, oriented and without boundary) embedded in $\mathbb{R}^{m}$ with $m=p+q+1$. Alexander duality tells us

$$ H^{p}(M)\cong H_{q}(\mathbb{R}^{m}\setminus M) \quad (\cong \mathbb{Z})$$

(in particular the homology of $M$ does not depend on the embedding). Now consider the inclusion $i\colon N \to \mathbb{R}^{m}\setminus M$, which induces a map on homology

$$i_{*}\colon H_{q}(N)\to H_{q}(\mathbb{R}^{m}\setminus M)\cong \mathbb{Z}$$

This map sends then the fundamental class of $N$ (a fixed generator of the infinite cyclic group $H_{q}(N)$ defining the orientation of $N$) to some integer times the generator of the infinite cyclic group $H_{q}(\mathbb{R}^{m}\setminus M)$ which is the image under the previous isomorphism of the fundamental class of $M$.

This integer is the linking number of $M$ and $N$.

You may want to check that this definition agrees with the classical one for the lower dimensions where it is defined. Namely, you can check that for two knots in space ($p=q=1$) you get the same integer (if you choose the sign carefully, and up to a sign in any case) with the classical definition:

Project both knots onto the plane in a nice way and for every time the first one passes under the second one, sum $+1$ or $-1$ depending on the orientation of the crossing.

You can check that these definitions agree using for example the Wirtinger presentation of $\pi_{1}(\mathbb{R}^{3}\setminus K)$, which then gives you a presentation of the homology on degree 1 (its abelianization).

Finally, a reference as requested:

Knots and links by Dale Rolfsen (see page 132).

  • $\begingroup$ That's quite useful. Thanks a lot! $\endgroup$ – Ivy Nov 16 '17 at 8:02

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