Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
This question on MSE is relevant: math.stackexchange.com/questions/147668/… –  m_t May 29 '13 at 19:32
add comment

2 Answers

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.

share|improve this answer
    
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 '10 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 '10 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 '10 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 '10 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 '10 at 18:57
add comment

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 $\chi\smile\chi$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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