Coboundary Representations for Trivial Cup Products Suppose $G$ is a pro-$p$-group, $p$ odd, and $\mathbb{F}_p$ is given the trivial $G$-action.  By skew-symmetry of the cup-product in degree 1, given $\chi\in H^1(G,\mathbb{F}_p)$, we have $\chi\cup\chi=0\in H^2(G,\mathbb{F}_p)$.  In fact, in this case, it's even possible to explicitly write $\chi\cup\chi$ as a coboundary -- $\chi\cup\chi=d\left(\binom{\chi}{2}\right)$, the coboundary of "$\chi$ choose 2".  
In any case, my question is whether or not there anyone has seen any other tricks of this sort, i.e., for the explicit realization of a trivial cup product as a coboundary.  In my specific case, I know a particular cup product is zero since I can force it, via the $G$-equivariance of the cup-product, to land in a known-to-be-trivial eigenspace of $H^2$.  I was hoping there was some "eigenspace-averaging" trick similar to the construction of orthogonal idempotents to get my hands on an explicit pre-image, but really, I'd just like to be aware of any tricks for doing this.
 A: These kinds of things appear somewhat regularly in studying power operations in algebraic topology.  There is a whole hierarchy of such products and they are organized under the action of an $E_\infty$ operad.
In this group cohomology example, for a 1-cocycle $f$ there is a sequence of 1-cochains $f^k: g \mapsto f(g)^k$.  The first one satisfies $d(f^2) = -2(f \cup f)$, and so you can divide off the front $2$ if $p$ is an odd prime.  The next one satisfies $d(f^3) = \pm 3f \cup f^2 \pm 3 f^2 \cup f$ (I cannot remember the sign, my apologies) and expresses the triviality of a "Massey product" $3 \langle f, -2f, f\rangle$; if $3$ is invertible you can divide off and get a genuine relation.
These generalize to higher cocycles and products of more elements, and particularly give rise to Steenrod operations.
A: You can make explicit an homotopy showing that the map $(\chi,\xi)\mapsto \chi\smile\xi\pm\xi\smile\chi$ is homotopic to zero. Using it, you can generalize your formula for $\chi\smile\chi$.
