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We know that the integral cohomology of $BU$, $H^*(BU) = Z[c_1,c_2,...]$ where $|c_i| = 2i$ is the $i$th Chern class, with coproduct $\Delta c_i = \Sigma c_j\otimes c_{i-j}$.

And at almost everywhere people claim that as a Hopf algebra, its dual, $H_*(BU) = Z[b_1,b_2,...]$ with $b_i = (c_1^i)^*$ and coproduct $\Delta b_i = \Sigma b_j \otimes b_{i-j}$. I tried to prove this by using the standard method in the structure of dual Steenrod algebra but I am stuck by some twisted combinatorial problems.

So may I ask for a proof or even a hint here? I think I am pretty near but I somehow need a kick...

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The proof isn't quite trivial, it is covered in section 21.4 of May-Ponto's More Concise Algebraic Topology. It forms part of a proof of Bott Periodicity, though, so its difficulty is justified. –  Justin Young Feb 19 at 5:37
    
@Young Thank you so much! I think I saw this proof(the only proof I saw) after asking and undoubtedly it is a proof but I cannot get what happened behinds it. i.e. could we have a somehow elementary proof without using concepts like universal universal Hopf algebras? And since it is a part of a proof of Bott Periodicity, could we obtain a proof by Bott Periodicity? –  Mingcong Zeng Feb 19 at 5:46
    
Indeed, it should be relatively easy (Serre spectral sequences, knowledge of $SU(n)$.) to show that $\Omega SU$ has that homology Hopf alebra, and Bott periodicity gives you a weak equivalence $BU \to \Omega SU$. –  Justin Young Feb 19 at 18:09
    
@Young Thank you so much again! I think the Serre SS is a better approach for me but I will try to get the direct proof by dual Hopf algebra. –  Mingcong Zeng Feb 20 at 2:24
    
I can't believe I put "right" for "write"...my English is deteriorating. Sure, but Bott periodicity is a much deeper theorem than the computation of the dual Hopf alegbra. For me, it was cooler the other way. –  Justin Young Feb 20 at 10:05

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