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The Bullet-Macdonald identity (c.f. On the Adem relations)is the following: $$P(s^2+st)P(t^2)=P(t^2+st)P(s^2).$$ where we have $P(s)=\Sigma _s s^iSq^i$. This is known to be equivalent to the Adem relations.

Now, this equality was proved with trivial action of Steenrod squares on the formal indeterminates in the paper (see also Larry Smith, An algebraic introduction to the Steenrod algebra, arXiv:0903.4997).

However, in the section 4 of the same paper, there is an alternative proof, where a different action was used in the course of the proof, but the conclusion is still about with the trivial action. There we have $Sq^1(t)=t^2$.

Another generating series identity equivalent to the Adem relations is the Bisson-Joyal identity, that is, we have $$Q(s)Q(t)=Q(t)Q(s)$$ with $Q(s)=\Sigma _s s^iSq_i$ where $Sq _i(x)=Sq^{n-i}(x)$ for $x\in H^n(X)$, and this time the Steenrod square acts non-trivially on the indeterminates by $$Q(s)(t)=t(s+t).$$

Since both Bullett-Macdonald and Bisson-Joyal identities are equivalent to Adem relations, they are equivalent to each other.

My questions are

  1. Is there a direct proof of the equivalence between the two identities?

  2. If there is any such proof, how does one go between the different actions of Steenrod squares?

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  • $\begingroup$ I was puzzled by the two actions, as well! Smith's paper is nice (though, alas, there are typos as he feared...); the "nontrivial" indeterminates appear naturally from the interpretation of H*(X,p)[t] == H*(X \times K(p,1),p), while understanding the "total pth power" seems to want a different interpretation of H*(X)[t]; this comes up in B+M when they formally invert the fundamental class of K(p,1)... but, I don't know, so this is not an answer. Good luck! $\endgroup$
    – Jesse C. McKeown
    Commented Sep 18, 2014 at 22:34
  • $\begingroup$ So would it be related to somehow the equivalence $F(X\times BZ/p,HZ/p)\cong F(X\times \vee _i S^i, HZ/2)$? (Their homotopy groups give $H^*(X)[[t]]$ with two different actions of Steenrod squares on $t$.) But I don't see how $F(X\times \vee _i S^i, HZ/2)$ gets into the picture. $\endgroup$
    – user43326
    Commented Sep 23, 2014 at 9:39
  • $\begingroup$ Actually $F(X\times \vee _i S^i,HZ/2)$ should be irrelevant since $H^*(X\times \vee _i S^i,HZ/2)$ doesn't have the correct ring structure. $\endgroup$
    – user43326
    Commented Sep 23, 2014 at 15:57
  • $\begingroup$ If I had to come up with a geometric description of steenrod-acting-trivially-on-formal-variables, it would be in terms of the fiberwise cohomology of $X\times B(Z/pZ)$. edit-add: ... as a bundle over $B(Z/pZ)$. $\endgroup$
    – Jesse C. McKeown
    Commented Sep 27, 2014 at 3:46

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