Fix the ground ring $\mathbb{F}_2$ and let $X$ be a space with finite homology. Then we have an isomorphism $\Phi^i_X:H_i(X)\to H^i(X)^*,a\mapsto \langle,a\rangle$ which allows us to define the dual Steenrod squares $$\mathrm{Sq}_r:H_i(X)\stackrel{\Phi_X^i}{\to} H^i(X)^*\stackrel{(\mathrm{Sq}^r)^*}{\to} H^{ir}(X)^*\stackrel{(\Phi^{ir}_X)^{1}}{\to} H_{ir}(X).$$ I want to understand which formulae they satisfy. Many things seem easy: For example, the “dual” Cartan formula needs the comultiplication $\Delta:H_i(X)\to \bigoplus_{p+q=i}H_p(X)\otimes H_q(X)$ which is dual to the cup product, and one should get $$\Delta_{ir}\circ \mathrm{Sq}_r = \sum_{p+q=r}(\mathrm{Sq}_p\otimes\mathrm{Sq}_q)\circ \Delta_i.$$ What I am now trying to dualise is $$\mathrm{Sq}^nx = x\smile x ~~~\text{ for }x=n.$$ The right hand side is obviously the cup product precomposed with the diagonal $\nabla:H^n(X)\to H^n(X)^{\otimes 2}\subseteq \bigoplus_{p+q=2n}H^p(X)\otimes H^q(X)$. We can find a map which is dual to this diagonal and of the form $$D:\bigoplus_{p+q=2n}H_p(X)\otimes H_q(X)\to H_n(X),a\otimes b\mapsto a\,\# \,b.$$ We then should get for $a\in H_{2n}(X)$ the statement $$\mathrm{Sq}_n(a) = (D\circ \Delta)(a).$$ However, this does not look very enlightening. Does this “wrongly graded product” $D$ have any meaning? At least, it satisfies the following property for all $\alpha\in H^n(X)$ $$\langle \alpha,a\,\#\,b\rangle = \langle\alpha^{\times 2},a\times b\rangle.$$
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$\begingroup$ What exactly do you mean by "the diagonal"? $a\to a\otimes a$? $\endgroup$ – user43326 Apr 23 at 10:04

$\begingroup$ Yes, exactly. Is there any problem with this map? $\endgroup$ – FKranhold Apr 23 at 11:08

$\begingroup$ It seams that the diagonal $\nabla$ imposes a coalgebra structure on $H^*(X)$ (which is not graded) and that $D$ is its dual a (nongraded) algebra structure on $H_*(X)$. Is there anything known about it? $\endgroup$ – FKranhold Apr 23 at 13:04

$\begingroup$ Like, not being additive? $\endgroup$ – user43326 Apr 24 at 8:05