# Action of Steenrod algebra on Chern classes

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 $$BU$$: $$$$\rho(\sum_{i=0} w_i^*) = \sum_{i=0} w_i^*\otimes (\sum_{j=0}\zeta_j)^{i-1}$$$$

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

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

• 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 – Ben McKay May 20 at 8:25
• @BenMcKay Yes, I know that this paper is contained there, but, unfortunately, this journal isn't available online do you have the pdf? – Semyon Abramyan May 20 at 16:23
• 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. – Ben McKay May 20 at 17:05
• @BenMcKay Thank you! I've added this information to the question. – Semyon Abramyan May 20 at 20:58

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.
• 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. – Semyon Abramyan May 20 at 16:53
• 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. – John Greenwood May 20 at 17:13
• 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 $\rho(b_i) = \sum_j (\sum_k \xi_k)^{j+1}_{2(i-j)} \otimes b_j$. – Semyon Abramyan May 20 at 18:01
• @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 $z_n$. – John Greenwood May 20 at 19:00
• 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! – Semyon Abramyan May 20 at 20:57