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Jul 23 at 14:34 comment added Antoine Now I see your point, thank you!
Jul 23 at 12:14 comment added Dave Benson For the linear part, you can use the "carry digit" cocycle familiar from your youth.
Jul 23 at 11:49 comment added Dave Benson Cocycles form an abelian group. Just add them.
Jul 23 at 11:44 comment added Antoine Yes, there are $B: V\times V\to \mathbb{F}_p$ and $q: V\to \mathbb{F}_p$, and q is non trivial for $E=p^{1+2n}_{-}$ (p odd). In what sense can I add them?
Jul 23 at 11:05 comment added Dave Benson For $p$ odd, you don't have a quadratic form. You have a symplectic form giving the commutators, and independently a linear form giving the $p$th powers. There's no connection between them. So just treat the two cocycles separately and add.
Jul 23 at 10:54 comment added Antoine Btw, regarding $E=p^{1+2n}_{-}$, I was thinking that $\beta(v,w)=\frac{1}{2}([v,w]+q(v)+q(w))$ is the natural choice, but it fails to be a cocycle. Do you suggest another candidate?
Jul 17 at 4:23 comment added Antoine now it is clear, thank you again!
Jul 16 at 16:39 vote accept Antoine
Jul 16 at 16:32 comment added Dave Benson I guess the point is that if you lift elements of $V$ to $E$ then $\beta(x,y)$ tells you how the lift of the product differs from the product of the lifts. So $\beta(x,y)-\beta(y,x)$ tells you the commutator of the lifts because $V$ is commutative. The formula for the commutator in terms of the square tells you this is $q(x-y)-q(x)+q(y)$. And as mentioned before, minus is plus. I hope that helps.
Jul 16 at 16:28 comment added Antoine Thanks a lot! I know where the condition $\beta(x,x)=q(x)$ comes from, but the other condition is not evident to me, does it follow from the cocycle equation?
Jul 16 at 15:30 comment added Dave Benson It might have made it clearer what was going on if I had written some of the plusses as minuses. But then people would complain that it makes no difference, so why do it? You can't win, you know.
Jul 16 at 14:47 history answered Dave Benson CC BY-SA 4.0