Homology of the product of spaces with integer coefficients and the Massey products

Question

Consider $H_*(X\wedge Y;Z)$, where $X=Y=BZ/2$ for concreteness' sake. If we write $e_i$ the generator of $H_i(BZ/2;Z/2)$., we see that the $E_2=E_{\infty}$ term of the Bockstein spectral sequence for $X\wedge Y$ is trivial, thus the permanent cycles are the image of $\beta$. As we have $$\beta e_{2j}=e_{2j-1}$$ one can choose a basis of the set of permanent cycles as $$\{e_{2i-1}\otimes e_{2j-1}\}\cup \{e_{2i}\otimes e_{2j-1}+e_{2i-1}\otimes e_{2j}\}.$$ If we still denote by $e_{2i-1}$ the lift of $e_{2i-1}\in H_{2i-1}(BZ/2;Z/2)$ to $H_{2i-1}(BZ/2;Z)$, then we see that the elements of the first set lifts simply to the "products $e_{2i-1}\otimes e_{2j-1}$" of integral homology classes, whereas those in the second lift to "some sort of Massey products $\langle e_{2i-1} , 2, e_{2j-1}\rangle $".

Now, my questions are

1. Is there any reference for this kind of facts, that is the description of the homology of the product of spaces using "Massey products"?

2. Is there a setting in which one can "really" consider the obvious lifts of the elements $e_{2i}\otimes e_{2j-1}+e_{2i-1}\otimes e_{2j}$ as Massey product?

Answer 1

Here's a very general form. Suppose that we have six chain complexes $A_0, A_1, A_2, A_{01}, A_{12}, A_{012}$, with bilinear "multiplication" pairings of chain complexes: $$ \begin{align*} A_0 \otimes A_1 &\to A_{01}\\ A_1 \otimes A_2 &\to A_{12}\\ A_{01} \otimes A_2 &\to A_{012}\\ A_0 \otimes A_{12} &\to A_{012} \end{align*} $$ Let's assume that we also have a chain homotopy $H$ between the two composite maps $A_0 \otimes A_1 \otimes A_2 \rightrightarrows A_{012}$. Given cycles $x \in A_0, y \in A_1, z \in A_2$ of degrees $p$, $q$, and $r$ respectively such that $[x \cdot y] = 0 \in H_{p+q}(A_{01})$ and $[y \cdot z] = 0 \in H_{q+r}(A_{12})$, then we can form a Massey product: choose elements $u$ and $v$ such that $\partial u = x \cdot y$ and $\partial v = y \cdot z$, and form $$ \langle x,y,z\rangle = u \cdot z - (-1)^p x \cdot v + H(x\otimes y \otimes z) $$ which represents an element in $H_{p+q+r+1}(A_{012})$. This depends in the usual way on the choices of $u$ and $v$, so there is indeterminacy. (If you use cohomological indexing then for elements in cohomological degrees $p$, $q$, and $r$ you get a Massey product in $H^{p+q+r-1}(A_{012})$.)

The "usual" definition of a Massey product is when all six chain complexes are equal to a single complex $A$ equipped with a strictly associative multiplication (allowing us to choose $H = 0$) that makes it into a differential graded algebra.

In the case you describe, we can take

• $A_0 = A_{01} = C_*(X,pt)$ the relative singular chain complex of $X$,
• $A_2 = A_{12} = C_*(Y,pt)$ the relative singular chain complex of $Y$,
• $A_{012} = C_*(X \wedge Y, pt)$ the relative singular chain complex of $X \wedge Y$, and
• $A_1 = \Bbb Z$, the constant chain complex $\Bbb Z$ in degree 0.

Then two of the multiplications are just the unit isomorphism and the other two multiplications $C_*(X,*) \otimes C_*(Y,*)$ are induced by the Eilenberg-Zilber shuffle map. The unit isomorphism is associative and so we can take $H = 0$ in this case, recovering your formula.

If you like, the entire framework of these six objects, pairings, homotopy, and chosen elements asks for a DG-category (really an $A_\infty$ DG-category) with four objects and chosen maps $$ a \stackrel{z}{\to} b \stackrel{y}{\to} c \stackrel{x}{\to} d $$ so that the double composites are null; in this description we can also think of this secondary operation $\langle x,y,z\rangle$ as related to Toda's "bracket" construction. The case of a differential graded algebra is the one-object DG-category case.