# cohomology ring of infinite iterated loop space

What is the cohomology ring $$H^*(\Omega^\infty \Sigma^\infty (S^m\vee S^n);\mathbb{Z}_2)?$$ I already write out the graded-vector-space basis using Dyer-Lashof operations, but I do not know how to compute the cup product...

Any references? thanks.

We have $Q(X\vee Y)\cong QX\times QY$ so, as a ring, we have $$H^*(Q(S^m\vee S^n);Z/2)\cong H^*(QS^m;Z/2)\otimes H^*(QS^n;Z/2).$$ Furthermore, if $k$ is positive, $H^*(Q(S^k;Z/2)$ is an exterior algebra (for example you can find this in Wellington's AMS memoirs that I mentioned before), since the fundamental class is primitive, so all the classes obtained by Dyer-Lashof operations are primitive as well). If $k$ is zero, since in $H_*(Q_0S^0;Z/2)$ we have $V()=$, so $$V(Q^{2i_1,2i_2,\ldots 2i_m})=Q^{i_1,i_2,\ldots i_m}.$$ Thus Verschiebung is surjective in $H_*(Q_0S^0;Z/2)$, dually the square is injective in $H^*(Q_0S^0;Z/2)$, so $H^*(Q_0S^0;Z/2)$ is polynomial. This gives the algebra structure on $H^*(Q(S^m\vee S^n);Z/2)$ for any $m,n$.
• The first fact is the consequence of the fact that $\Omega ^{\infty}$ is a right adjoint, the second, as is explained, follows from the fact that all polynomial generators of $H_*(QS^n)$ with $n>0$ are primitive. I presume you can find all of these (except the first one ) in somewhere section 3 of Wellington's book, but unfortunately I don't have it on the hand. – user43326 Sep 3 '15 at 14:47
• No, in homology there are not only the elements $Q^I(\iota _n)$'s but their products. And they are not primitive. – user43326 Sep 5 '15 at 12:47