Computing a Massey product. Here is the question I am trying to solve:



Can anyone help me in showing me how to compute this Massey Product?
 A: In order to consider the length 4 Massey product $\langle h_{01}, h_{12}, h_{23}, h_{12} \rangle$, you have to show that the length 3 Massey products $\langle h_{01}, h_{12}, h_{23} \rangle$ and $\langle h_{12}, h_{23}, h_{12} \rangle$ are defined and contain zero. So as a start, let's look at the length 3 cases. First, look at $\langle h_{01}, h_{12}, h_{23} \rangle$: since $h_{02} \mapsto h_{01} h_{12}$ and $h_{13} \mapsto h_{12} h_{23}$, then $h_{02} h_{23}$ and $h_{01} h_{13}$ both go to the product $h_{01} h_{12} h_{13}$, so their sum goes to zero and hence is a cycle. (We're working in characteristic two, so we don't have to worry about the many signs that often come up in Massey product computations.) Therefore $h_{02} h_{23} + h_{01} h_{13}$ represents the Massey product $\langle h_{01}, h_{12}, h_{23} \rangle$. It may not be the unique representative: the important part is what the differential does to, for example, $h_{02} h_{23}$. We would get the same effect if we used $(h_{02} + \alpha) h_{23}$ for any cycle $\alpha$. I think that in this case, the relevant bidgree is spanned by $h_{02}$, so the only cycle in that bidegree is $\alpha=0$, but I haven't checked that.
Now observe that $h_{03} \mapsto h_{02} h_{23} + h_{01} h_{13}$. This says that the length three Massey product contains zero — one of its representatives (I think its only representative) is a boundary.
A similar computation should show that the other length 3 Massey product, $\langle h_{12}, h_{23}, h_{12} \rangle$, is represented by $h_{12} h_{13} + h_{13} h_{12} = 0$.
For the length 4 Massey product, find elements $\beta$ and $\gamma$ whose boundaries are representatives for the two length 3 Massey products; then the length 4 Massey product is represented by $\beta h_{12} + h_{01} \gamma$. As in the length 3 cases, there might be different choices, because we could modify $\beta$ and $\gamma$ by adding cycles to either or both. But at this point we already have one choice: $\beta = h_{03}$ and $\gamma = 0$, so one choice of representative is $h_{03} h_{12}$. Are there others? That just depends on whether there are cycles in the relevant bidgrees. The Massey product is defined to be the set of all elements obtained by choosing representatives like this.
