3
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

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 $V([0])=[0]$, so $$V(Q^{2i_1,2i_2,\ldots 2i_m})[0]=Q^{i_1,i_2,\ldots i_m}[0].$$ 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 $m,n$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks, Prof. Could you explain more? Which page could I find the results? $\endgroup$
    – QSR
    Commented Sep 3, 2015 at 12:47
  • 1
    $\begingroup$ 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. $\endgroup$
    – user43326
    Commented Sep 3, 2015 at 14:47
  • $\begingroup$ Dear Prof., you mentioned that "since the fundamental class is primitive, so all the classes obtained by Dyer-Lashof operations are primitive as well". But if this is true for the homology, the cup product of cohomology would be trivial. $\endgroup$
    – QSR
    Commented Sep 5, 2015 at 12:44
  • 1
    $\begingroup$ No, in homology there are not only the elements $Q^I(\iota _n)$'s but their products. And they are not primitive. $\endgroup$
    – user43326
    Commented Sep 5, 2015 at 12:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .