Return to Revisions

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, $H^*(Q_0S^0;Z/2)$ is polynomial. This gives the algebra structure on $H^*(Q(S^m\vee S^n);Z/2)$ for any $m,n$.