Let $S^p$ and $S^q$ be disjoint spheres in $\mathbb{R}^n$ with $n=p+q+1$ and let $X= \mathbb{R}^n- (S^p\cup S^q)$. By Alexander duality, their fundamental classes yield cohomology classes in $\tilde{H}^p(X)$ and $\tilde{H}^q(X)$, respectively, say $\alpha^p$ and $\alpha^q$. The cup product $\alpha^p \cup \alpha^q$ lies in $\tilde{H}^{p+q}(X)\cong \tilde{H}_0(S^p \cup S^q) \cong \mathbb{Z}$, so there's an integer assigned to this cup product.

Now, consider the case $p=q=1, n=3$.

Can I and if yes, how can I show that this integer is the linking number of the spheres defined by the "usual" counting of signs at the crossings?

Can this be generalized to a linking number in higher dimensions?

Edit: Thanks for your answer, Aloizio Macedo.

I think saying that Alexander duality is given by capping with an orientation class is not totally correct in this case since I need compact manifolds with boundary for Lefschetz duality. But I still tried to use your ideas for the case $p=q=1, n=3$ using $S^3$ instead of $\mathbb{R}^3$:

Call the spheres $S_1, S_2$. Let $\beta_1 = \alpha^1 \cap [S^3]$, $\beta_2 = \alpha^2 \cap [S^3] \in H_2(S^3,S_1\cup S_2)$. (Then $[S_i] = \partial(\beta_i) \in H_1(S_1\cup S_1)$ for $i=1,2$.) Let $D_1$ be the inverse of $\cdot \cap [S^3]$. We have $\alpha^1 \cup \alpha^2 = D_1(\beta_1) \cup D_1(\beta_2) = D_1(\beta_2 \circ \beta_1)$, where $\circ$ is the intersection product from Bredon. Now, if $\beta_i=[N_i]$ for submanifolds $N_i$, $i=1,2$, I get $\alpha^1\cup \alpha^2 = D_1(\beta_2 \circ \beta_1) = D_1([N_1]\circ [N_2])=D_1([N_1 \cap N_2])$. But $N_i$ would have to be two-dimensional such that $N_1 \cap N_2$ is $1$-dimensional and I'm not quite sure how to get a linking number from here... (An intersection of a Seifert surface and a knot usually is $0$-dimensional, a set of points, right?)

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    $\begingroup$ This may be a stretch, but there is an intimate relationship between cup products and intersections. This can be seen in the "Intersection Theory" chapter in Bredon, for example. Specifically Theorem $11.9$, which essentially says that the dual of the orientation class of the intersection of two compact orientable submanifolds is the cup product of the duals of the orientation classes of each one. (...) $\endgroup$ – Aloizio Macedo Sep 15 '18 at 16:00
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    $\begingroup$ (...) Maybe going through the relations of cup/cap product (recall that Alexander duality is given by capping with an orientation class) and perhaps using some alternate definition of the linking number (e.g., via Seifert surfaces) can yield something along the lines you want. $\endgroup$ – Aloizio Macedo Sep 15 '18 at 16:00
  • $\begingroup$ Thanks for your answers, Aloizio, I tried to use your ideas, see the edited question. But I'm still not sure, how to get a linking number here. $\endgroup$ – schneeewittchen Sep 19 '18 at 14:47
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    $\begingroup$ Your intersection $[N_1\cap N_2]$ is an element of $H_1(S^3,S_1\cup S_2)$; you have to now apply the boundary operator to get your element of $\tilde{H}_0(S_1\cup S_2)$. $\endgroup$ – Steve D Sep 20 '18 at 4:26
  • $\begingroup$ Yes, that makes sense. Thank you, Steve. $\endgroup$ – schneeewittchen Sep 20 '18 at 9:38

The OP has already worked out an answer, but I wanted to provide how I think about this. Let $L$ be the link, made up of circles $\alpha$ and $\beta$.

We have an Alexander duality map $D: H^1(S^3\setminus L)\rightarrow H_1(L)$, given by $\gamma\mapsto\partial(\omega\frown\gamma)$, where $\omega$ is a fundamental class. This has an inverse, $D^{-1}: H_1(L)\rightarrow H^1(S^3\setminus L)$, which can be described as follows: for any 1-cycle $\sigma\subset S^3\setminus L$, it is the boundary of a 2-cycle in $S^3$, which we denote by $\partial^{-1}\sigma$. Then $$ D^{-1}(\tau)(\sigma) = [\tau\cap\partial^{-1}\sigma]\in\mathbb{Z} $$ where $[\tau\cap\partial^{-1}\sigma]$ is just the oriented intersection. I don't know a reference for this. @CheerfulParsnip mentions this in his answer here; maybe he knows a source?

Anyway, from here it is easy. We have \begin{align*} D\left(D^{-1}(\alpha)\smile D^{-1}(\beta)\right) &= \partial(\omega \frown(D^{-1}(\alpha)\smile D^{-1}(\beta)))\\ &= \partial(\omega\frown D^{-1}(\alpha))\frown D^{-1}(\beta)\\ &= \alpha\frown D^{-1}(\beta)\\ &= D^{-1}(\beta)(\alpha)\\ &= [\beta\cap\partial^{-1}\alpha] \end{align*} and this is just the intersection of $\beta$ with the Seifert surface of $\alpha$. If we choose orientations right, this is the linking number of $\alpha$ and $\beta$.

  • $\begingroup$ Hi Steve, I don't quite understand your description of $D^{-1}$. Could you explain it in more detail? Do you mean $D^{-1}(\tau)(\sigma)=[\tau \cap \partial^{-1}\sigma]$? For your computation, I'm not sure if $D^{-1}(\beta)(\alpha)$ makes sense, since $\beta$ and $\alpha$ are $1$-cycles in the link $L$, not in the complement. $\endgroup$ – schneeewittchen Sep 24 '18 at 15:02
  • $\begingroup$ Whoops, yes, that's a typo. I'll fix it. $\endgroup$ – Steve D Sep 24 '18 at 15:04
  • $\begingroup$ I'm still not sure how your definition of $D^{-1}$ works for $D^{-1}(\beta)(\alpha)$. We have $\alpha \in H_1(L)$ and $\alpha \not\in H_1(S^3\backslash L)$, right? So how can I apply $D^{-1}(\beta)$ to it? And why do we have $\alpha \cap D^{-1}(\beta)=D^{-1}(\beta)(\alpha)$? Thanks in advance... $\endgroup$ – schneeewittchen Sep 25 '18 at 13:49
  • $\begingroup$ When I have time I'll try and come back and write this out in more detail if you like. Essentially were talking about something homologous to the circle, still living in the complement. As for the cap product at the end, that's just the definition of cap product. $\endgroup$ – Steve D Sep 25 '18 at 15:00
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    $\begingroup$ @SteveD Can you get the symmetry of the linking form from this? I'm confused because the cup product on 1-forms is anticommutative. $\endgroup$ – Kyle Miller Nov 26 '18 at 6:50

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