MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

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.

-
 This question on MSE is relevant: math.stackexchange.com/questions/147668/… – mt May 29 at 19:32

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 $\chi\smile\chi$.

-

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.

-
 Hi Tyler, Is there a good reference for your assertion 'they are organized under the action of an $E_{\infty}$ operad'? – Minhyong Kim Feb 15 2010 at 23:56 I believe that one of the first sources is May's "A general algebraic approach to Steenrod operations", but the operad itself is more visible in Hinich-Schechtman's "On homotopy limit of homotopy algebras". I'd recommend Mandell's "$E_\infty$-algebras and p-adic homotopy theory". Unfortunately, I don't think many of these will give many cochain-level formulas. I can look up more references if you're interested. – Tyler Lawson Feb 16 2010 at 3:36 I am interested, but I'll first chase down the references you give and see if I can understand them. My own motivation is described somewhat in a paper on Massey products you can find on my webpage. I know you have many arithmetic interests, so if you have any comments on that paper, they will be very gratefully received. – Minhyong Kim Feb 16 2010 at 8:51 Interesting -- I had not seen the trivial Massey product calculation via $d(f^3)$. My interests in this problem as well lie in the construction of certain Massey products, hence the focus on explicit cohains from cup products. Am I right in understanding that the hierarchy of products you mention all involve operations beyond cup products? – Cam McLeman Feb 17 2010 at 7:17 They all involve operations beyond cup products, such as cup-$i$ products (as in Mosher-Tangora), but in some sense all these operations are supposed to show that the cup product is coherently nice. – Tyler Lawson Feb 17 2010 at 18:57