In Michel Lazard's "Commutative Formal Groups" Springer Lecture Notes, he defines an operator on a polynomial 3-cochain $f$ denoted $\Gamma_n(f)$, which defines as the $n^{th}$ homogeneous piece of the difference $f(f(x,y),z)-f(x,f(y,z))$. He later notes (in equation 5.10) that $\Gamma_n(f)$ is always a cocycle, that is, $\delta\Gamma_n(f)=0$ for any $n$ and any symmetric $f$, where $\delta(g)(x,y,z,w)=g(y,z,w)-g(x+y,z,w)+g(x,y+z,w)-g(x,y,z+w)+g(x,y,z).$ He claims that proving that such a thing is a cocycle is easy (even for non-symmetric polynomials) and says he will leave the proof to the reader. I have tried to prove it repeatedly and failed, and am worried I'm just being stupid here. I've basically been writing down a lot of polynomials and trying to show things cancel, which is perhaps a bit naive. Does anyone know how this proof goes? We're basically showing that a certain obstruction is always a cocycle.
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asked Jul 20, 2013 at 22:57
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$\begingroup$ You have to use the hypothesis that $f$ is an $(n-1)$-bud (in Lazard's terminology); not sure what you mean by "polynomial 3-cochain", but if you don't use the $(n-1)$-bud hypothesis then you're trying to prove something false. $\endgroup$– user36938Commented Jul 20, 2013 at 23:30
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$\begingroup$ Sorry, yes, I forgot to mention that as part of the hypothesis. I certainly have been using that. $\endgroup$– Jonathan BeardsleyCommented Jul 21, 2013 at 4:29
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$\begingroup$ @user36938 I probably shouldn't say polynomial 3-cochain, I just meant that it's not yet clear that it's a cocycle, so it's a cochain. $\endgroup$– Jonathan BeardsleyCommented Jul 21, 2013 at 4:31
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$\begingroup$ But you put the word "cochain" in a place that refers to $f$, not $\Gamma_n(f)$, so it is confusing. Do you understand why it works when $f$ is an $n$-bud? $\endgroup$– user36938Commented Jul 21, 2013 at 9:23
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$\begingroup$ @user36938 Hm okay, yeah, no I agree, I have some sort of strange terminology there. $\endgroup$– Jonathan BeardsleyCommented Jul 21, 2013 at 15:37
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