Consider the space $X=\mathbf{R}^3\setminus (S^1\sqcup S^1\sqcup S^1)$, where the circles are arranged in form of the Borromean rings. Let $a, b, c\in H^1(X)$ be the cohomology classes corresponding to the circles. My goal is to compute the Massey product $\langle a,b,c\rangle$.
Since I cannot draw these properly, I use the following (hopefully) homotopic replacement for $X$: first, let's shrink $\mathbf{R}^3$ as much as possible, so the new $X$ should be three (hollow) tori with the triangles enclosed by the circles appropriately filled.
Since I cannot draw the circles properly, I have tried to depict the $\Delta$-complex I want to use as a model $X$ as follows:
Explanation of the depiction. The bottoms of the three strands of the braid are to be connected to the top, forming the three tori of the link. The ambient $\mathbf R^3$ has been shrinked to sheets in the voids between the tori. These correspond to the voids between the strands of the braid, plus the triangle in the middle of the circles, which corresponds to the cell in the right.
The points in which the strands touch, together with the boundaries of the coloured triangles, are used as 0- and 1-cells of the $\Delta$-complex. In the following picture, you see the structure taken apart into the strands and triangles:
Note that this is not a simplicial complex; rather, there may be multiple distinct cells with the same list of vertices, and cells may have degenerate vertex lists without being degenerate themselves (the latter does not seem to occur here). I denote simplices by their vertices (in a fixed order) and the color ($r, g, b, k, v$ stand for red, green, blue, black, violet).
Here starts the mess
Representatives for cohomology The circles correspond to three classes $\alpha, \beta, \gamma\in H^1(X)$. What are their representing cocycles? If there weren't the triangles in the voids, I'd guess it's:
$$\begin{aligned}\alpha &= [14_g - 45_b + 53_r -31_k]\\ \beta &= [61_b - 12_k + 25_r - 56_g]\\ \gamma &= [42_g - 23_k + 36_v - 64_r]\end{aligned}$$
But now, alas, these aren't cocycles anymore. What are correct representatives for $H^1$ now? If I haven't miscalculated, with $\mathbf Z/2\mathbf Z$-coefficients, the three cohomology classes each are represented by the sum of all non-vertical edges of two of the three strands. This irritates me, because these only span two dimensions.
An additional problem is that these are quite lengthy; it seems tedious to compute cup products of these representatives. Are there any better ones?