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This is question about result from Brown and Peterson $H^*(MO)$ as an algebra over the Steenrod algebra. Unfortunately, the paper is not available on the Internet, so I can't find the proof.

One of results of the paper is the following. Let $w_n^* \in H_{n}(BO;\mathbb F_2)$ be the dual of Stiefel-Whitney class $w_n$ with respect to the basis of Stiefel-Whitney monomials. Denote by $z_n \in H_*(MO; \mathbb F_2)$ its image under the Thom isomorphism. Then the coaction $\rho\colon H_*(MO; \mathbb F_2) \to \mathfrak A_2^*\otimes H_*(MO; \mathbb F_2)$ maps $z_{2^j-1}$ to $\sum\limits_{i=0}^j \zeta_i \otimes z_{2^{j-i}-1}^{2^i}$, where $\zeta_i$ is the conjugate of Milnor's $\xi$'s. (Thanks to John Greenwood for corrections!)

Questions:

1) How to prove this result?

2) Does the similar formula hold for the coaction of the mod $p$ Steenrod algebra on the Chern classes?

3) Not exactly about result* Is there any way to get the paper?

MR0761717 Brown, E. H., Jr.(1-BRND); Peterson, F. P.(1-MIT) H∗(MO) as an algebra over the Steenrod algebra. Conference on homotopy theory (Evanston, Ill., 1974), 11–19, Notas Mat. Simpos., 1, Soc. Mat. Mexicana, México, 1975. 55S99 (Thanks to Ben McKay!)

UPD: In this paper Brown, Davis and Peterson give quite similar description, but for the right coaction in $BO$ and $BU$: \begin{equation} \rho(\sum_{i=0} w_i^*) = \sum_{i=0} w_i^*\otimes (\sum_{j=0}\zeta_j)^{i-1} \end{equation}

\begin{equation} \rho(\sum_{i=0} c_i^*) = - 1\otimes (\sum_{j=0}\zeta_j)^{-1} + \sum_{i=1} c_i^*\otimes (\sum_{j=0}\zeta_j)^{i-1} \end{equation}

Is there any way to rewrite the latter for the left coaction?

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  • $\begingroup$ MR0761717 Brown, E. H., Jr.(1-BRND); Peterson, F. P.(1-MIT) H∗(MO) as an algebra over the Steenrod algebra. Conference on homotopy theory (Evanston, Ill., 1974), 11–19, Notas Mat. Simpos., 1, Soc. Mat. Mexicana, México, 1975. 55S99 $\endgroup$
    – Ben McKay
    May 20, 2020 at 8:25
  • $\begingroup$ @BenMcKay Yes, I know that this paper is contained there, but, unfortunately, this journal isn't available online do you have the pdf? $\endgroup$
    – user131113
    May 20, 2020 at 16:23
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    $\begingroup$ no, sorry, I just wanted readers of the question to be clear about what the problem is. Maybe someone connected to Mexico will have a solution. $\endgroup$
    – Ben McKay
    May 20, 2020 at 17:05
  • $\begingroup$ @BenMcKay Thank you! I've added this information to the question. $\endgroup$
    – user131113
    May 20, 2020 at 20:58
  • $\begingroup$ Yours \omega_i^* are exactly primitive homology classes and given by newton polynomials by s(t)=log(b(t))' $\endgroup$
    – Bad English
    Jun 10, 2020 at 22:17

1 Answer 1

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The elements $z_n$ of $(H\mathbb{F}_2)_*MO$ that you described come from classes in $(H\mathbb{F}_2)_*\mathbb{RP}^\infty$ (namely ones that are dual to powers of the first Stiefel-Whitney class) under the inclusion $\mathbb{RP}^\infty\rightarrow \Sigma MO$ so the coaction formula follows from the coaction formula on $H_*\mathbb{RP}^\infty$.

The same argument works at larger primes with $MO$ replaced by $MU$, $\mathbb{RP}^\infty$ replaced by $\mathbb{CP}^\infty$, and Stiefel-Whitney replaced by Chern.

I couldn't find anything about that paper though.

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  • $\begingroup$ This is not quite true. The dual to Stiefel-Whitney class $w_n$ with $n\geqslant 2$ don't come from classes in $H_*(\mathbb R P^\infty; \mathbb F_2)$. The latter contains only duals to degrees of the first Stiefel-Whitney class. $\endgroup$
    – user131113
    May 20, 2020 at 16:53
  • $\begingroup$ It is correct (see e.g. Adams "stable homotopy and generalised homology" page 8 for the MU version. I know it seems counterintuitive since the classes start out being dual only to powers of the first Stiefel-Whitney class. But once they arrive in $H_*MU$ they are actually polynomial generators for it. $\endgroup$
    – John Greenwood
    May 20, 2020 at 17:13
  • $\begingroup$ Classes $b_i \in H_*(MU)$ are the images of the dual to the Milnor genus classes. And the coaction maps is not the described one, but $\rho(b_i) = \sum_j (\sum_k \xi_k)^{j+1}_{2(i-j)} \otimes b_j$. $\endgroup$
    – user131113
    May 20, 2020 at 18:01
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    $\begingroup$ @SemyonAbramyan Sorry perhaps I was too brief. I don't claim that the $b_n$ are equal to your $z_n$. I am just pointing out that the coaction of the dual Steenrod algebra in $H_*MO$ is completely determined by the coaction in $H_*\mathbb{RP}^\infty$ (and as you pointed out in your comment, easily described on the $b_n$) so it is a matter of pure algebra to get the formula on the $z_n$. $\endgroup$
    – John Greenwood
    May 20, 2020 at 19:00
  • $\begingroup$ of course the coaction in $H_*(MO)$ and $H_*(MU)$ is determined by the coaction in $H_*(\mathbb R P^\infty)$ and $H_*(\mathbb C P^\infty)$, but it is not clear for me how to write down explicit formulae for $z_n$ in terms of classical polynomial generators of $H_*(\mathbb R P^\infty)$ or $H_*(\mathbb C P^\infty)$. If you know how, it will be really helpful! $\endgroup$
    – user131113
    May 20, 2020 at 20:57

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