6
$\begingroup$

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?

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you, this answers the second question. Would there be any published reference? $\endgroup$ – user43326 Jan 4 '17 at 9:23

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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