8
$\begingroup$

I'm trying to understand an example in "Elements of Homology Theory" from V.V. Prasolov (p. 85-88) where he shows that the Borromean rings represented by three spheres $S_1, S_2, S_3$ in $S^3$ are linked.

He does that by showing that the Massey triple product $\langle\alpha^1, \beta^1,\gamma^1\rangle$ with $\alpha^1, \beta^1, \gamma^1$ being the cohomology classes corresponding to $[S_1], [S_2], [S_3] \in H_1(S_1 \cup S_2 \cup S_3)$ doesn't vanish.

I guess I'm trying to formalize the proof, but need some help with it.

In the end, the triple product is represented by a cycle $c \cap w_3 + w_1 \cap d$ in $H_1(S^3, S_1 \cup S_2 \cup S_3)$ where $w_1, w_3$ are chains corresponding to disks spanned by $S_1$ and $S_3$, respectively, and $c,d$ are certain half disks such that $w_1 \cap w_2 = \partial c$, $w_2 \cap w_3 = \partial d$ ($w_2$ corresponds to the disk spanned by $S_2$).

I'm not quite sure how to understand the intersections $w_1 \cap w_2$ and $w_2 \cap w_3$. Identifying the disks with their fundamental classes, I would like to use something like $[w_1] \circ [w_2] = [w_1 \cap w_2]$ (VI.11.9 from Bredon, Topology and Geometry).

The problem now is that this equality holds only for fundamental classes of submanifolds in a manifold with boundary (as far as I know), but the the homology classes of $w_1, w_2, w_3$ live in the relative homology group $H_2(S^3, S_1 \cup S_2 \cup S^3)$. I can construct a manifold $M$ with boundary $B=\partial M$ such that $H_2(S^3, S_1 \cup S_2 \cup S^3) \cong H_2(M, B)$, namely the complement in $S^3$ of the "fattened up versions" of $S_1, S_2, S_3$.

But after these identifications I'm not quite sure how the intersection $w_1 \cap w_2$ looks like. Can I think of the element in $H_2(M,B)$ corresponding to $w_1$ as just a shrunk version of the same chain?

Maybe it would be easier to define the intersection product directly in $H_2(S^3, S_1 \cup S_2 \cup S^3)$. Is there an easier way to do this than using the identification $H_2(S^3, S_1 \cup S_2 \cup S^3) \cong H_2(M, B)$ and the intersection product in $H_{\ast}(M, B)$ defined by the cup product and Poincare duality?

I'm a bit confused by all the identifications made in the proof. In the end, $\alpha^1 \in H^1(S^3\backslash (S_1 \cup S_2 \cup S_3))$ corresponds to $[w_1] \in H_2(S^3, S_1 \cup S_2 \cup S_3)$ and to $[S_1] \in H_1(S_1 \cup S_2 \cup S_3)$. I guess I can use a long exact sequence to get from $[w_1]$ to $[S_1]$, but how is $\alpha^1$ related to these two homology classes? I know, the key is Alexander duality, but I don't have the intuition how to think of this isomorphism.

Thank you in advance.

$\endgroup$
  • $\begingroup$ By now, I'm pretty convinced that for a rigorous proof I should define the intersection product on (relative) chain level. Do you know a reference for this? $\endgroup$ – schneeewittchen Oct 1 '18 at 10:43
  • $\begingroup$ Here's a reference: A Combinatorial Algorithm for Computing Higher Order Linking Numbers by Chun-Chung Hsieh, Louis Kauffman, Chichen M. Tsau (arxiv.org/abs/1407.5027) $\endgroup$ – schneeewittchen Dec 18 '18 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.