# Dual Steenrod squares

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 $$\mathrm{Sq}_r:H_i(X)\stackrel{\Phi_X^i}{\to} H^i(X)^*\stackrel{(\mathrm{Sq}^r)^*}{\to} H^{i-r}(X)^*\stackrel{(\Phi^{i-r}_X)^{-1}}{\to} H_{i-r}(X).$$ 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 $$\Delta_{i-r}\circ \mathrm{Sq}_r = \sum_{p+q=r}(\mathrm{Sq}_p\otimes\mathrm{Sq}_q)\circ \Delta_i.$$ 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 $$\alpha\in H^n(X)$$ $$\langle \alpha,a\,\#\,b\rangle = \langle\alpha^{\times 2},a\times b\rangle.$$

• What exactly do you mean by "the diagonal"? $a\to a\otimes a$? – user43326 Apr 23 at 10:04
• Yes, exactly. Is there any problem with this map? – FKranhold Apr 23 at 11:08
• It seams that the diagonal $\nabla$ imposes a coalgebra structure on $H^*(X)$ (which is not graded) and that $D$ is its dual a (non-graded) algebra structure on $H_*(X)$. Is there anything known about it? – FKranhold Apr 23 at 13:04
• Like, not being additive? – user43326 Apr 24 at 8:05