Let $K$ be an $n$-manifold with boundary and let $x,y,z \in H^*(K)$ be cohomology classes with $x\cup y=y\cup z=0$.

The Massey product $\langle x,y,z \rangle$ is defined as the set of cohomology classes $[a\cup \tilde z - (-1)^{\deg (x)}\tilde x \cup b]$, where $a,b$ are cochains with $\partial a=\tilde x\cup\tilde y$, $ \partial b=\tilde y \cup\tilde z$ and $\tilde x,\tilde y, \tilde z$ are representatives of the cohomology classes $x,y,z$.

In his paper "higher order linking numbers" Massey uses another approach to compute elements of Massey products using the Poincaré-duality of cohomology+cup-products and homology+intersection of manifolds:

Let the fundamental classes $[M],[N],[P] \in H_*(K,\partial K)$ be the dual classes of the cohomology classes $x,y,z$, such that $X\cap P$ and $M \cap Y$ intersect transversally, for $X,Y$ manifolds with $M⫛ N= \partial X$ and $N ⫛ P= \partial Y$.

Then the sum $X \cap P - (-1)^{n-\deg(x)}M\cap Y$ obviously represents the Poincaré-dual of a triple product.

How exactly does this obvious duality work?

In my attempt I didnt get very far...

Using Bredon's intersection theory I get $[M \cap N]= D(x \cup y)$, for $D:H^*(K,\partial K) \to H_{n-*}(K)$ the duality isomorphism. This implies, that the cap product of a representative of $[K]$ and $\tilde x \cup \tilde y$ will be send to a representative of $[M \cap N]$, which we call $r_{mn}$. Since $x\cup y = 0$ it follows that $[M\cap N]=0$ and therefore $r_{m,n}$ is the boundary of a non-cyclic chain $r$.

Since this chain is non-cyclic, it is not a representative of any homology class. Thus I dont see any way to continue, using the duality $[M \cap N]= D(x \cup y)$.


1 Answer 1


While not quite an answer from the perspective you're thinking of, one approach to this (at least if you're in a smooth manifold), would be to use a model of cohomology in which cocycles are represented by maps from other smooth manifolds. One such model is "geometric cohomology," which uses manifolds with corners. In this case, for a closed oriented manifold, homology and cohomology are tautologically the same and the cup product is the intersection product (at least for transverse things).

Geometric cohomology was introduced by Lipyanskiy in https://arxiv.org/abs/1409.1121 and is being further developed by me, Medina, and Sinha. We have one preprint recently posted that included an outline of the main results at https://arxiv.org/abs/2106.05986. More details will be forthcoming in a more foundational paper.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.