An attempt at an alternative calculation of the rank of $\pi_n(MO)$

Question

$\newcommand{\a}{\mathfrak a}\newcommand{\Z}{\mathbb Z}$Let $MO$ be the Thom Spectrum, then I am trying to come up with an alternative calculation that the rank of $\pi_n(MO)$ as a $\Z_2$ vector space is the number of non dyadic partitions of $n$. The method of calculation I have a familiar source for is rather lacking in details (see here) so out of a mixture of frustration and curiosity I am wondering if the following route has any merit, while avoiding the classical route.

Let's assume for the moment that I know $H^*(MO)$ is a free $\a_2$ module, where $\a_2$ is the mod $2$ Steenrod algebra, and I know that it has a basis given by generators $\{x_i:i\neq 2^{j}-1\}$ (while I am fine assuming this as I am aware of a not too bad proof, if for some reason I do not need this fact that would be very happy to not to use this). I wish to employ the Adams spectral sequence to calculate the stable homotopy groups $\pi_n(MO)$. In particular, the Adams spectral sequence will tell me that $$E_2^{s,t}\Rightarrow \pi_*(MO)\otimes \Z_2$$ however $\pi_*(MO)$ is already a $\Z_2$ vector space so I believe that this means: $$E_2^{s,t}\Rightarrow \pi_*(MO)$$ We also have that: $$E_2^{s,t}=\operatorname{Ext}^{s,t}_{\a_2}(H^*(MO),\Z_2)$$ Since $H^*(MO)$ is free, we must have that for all $s>0$ these groups are zero, so the Adams spectral sequence stabilizes immediately, and: $$\operatorname{Hom}^n_{\a_2}(H^*(MO),\Z_2)\cong \pi_n(MO)$$ So now all we have to do is calculate the rank of $\operatorname{Hom}^n_{\a_2}(H^*(MO),\Z_2)$ as $\Z_2$ vector space. I believe as a subgroup of $\operatorname{Hom}_{\a_2}(H^*(MO),\Z_2)$, we have that: \begin{align} \operatorname{Hom}^n_{\a_2}(H^*(MO),\Z_2)=\{f\in\operatorname{Hom}_{\a_2}(H^*(MO),\Z_2):\operatorname{supp}f\subset H^n(MO) \} \end{align} Now the confusing part to me is that $f\in \operatorname{Hom}_{\a_2}(H^*(MO),\Z_2)$ means that for all $a\in \a_2$ we have that $f(a\cdot \omega)=a\cdot f(\omega)$, but $\a_2$ acts on $\Z_2$ trivially in non zero degree, so this always zero unless the degree zero part of $a$ is one, in which case $f(a\cdot \omega)=f(\omega)$. However, if $\omega$ is not of the form $x_i$ for some $i\neq 2^j-1$, then $\omega$ is a linear combination of such elements over $\a_2$, so most $\omega$ just map to zero?

Perhaps I am thinking about this incorrectly, or have set up the problem wrong, but I would appreciate any help or hints at calculating the $\Z_2$ rank of this vector space.

Answer 1

$\newcommand{\a}{\mathfrak a}\newcommand{\Hom}{\operatorname{Hom}}\newcommand{\Z}{\mathbb Z}$ This can be proven by only assuming that $H^*(MO)$ is a free module. Indeed, due to $H^*(MO)$ being a graded free module, there exists an isomorphism: \begin{align*} \bigoplus_{i=0}^n(\a_2)_i^{f(i)}\longrightarrow H^n(MO) \end{align*}where $f(i)$ is the number elements of degree $i$. We know that rank of $H^n(MO)$ is $p(n)$, the number of partitions of $n$, and that $(\a_2)_i$ has rank $p_d(i)$, the number of dyadic partitions of $n$. Therefore: \begin{align*} \sum_{i=0}^np_d(i)\cdot f(i)=p(n) \end{align*} Define a new function $g(i)=f(n-i)$, then $g(n-i)=f(n-(n-i))=f(i)$, hence: \begin{align*} \sum_{i=0}^np_d(i)\cdot g(n-i)=p(n) \end{align*} but then a statement from combinatorics forces $g(i)=p_{nd}(i)$. Hence $f(0)=g(n)=p_{nd}(n)$.

The grading of $\Hom_{\a_2}(H^*(MO),\Z_2)$ is the one induced by $H^*(MO)$, hence $\Hom^n_{\a_2}(H^*(MO),\Z_2)$ consists of $\a_2$ linear maps $H^*(MO)\rightarrow \Z_2$ which have support in $H^n(MO)$. However, $\a_2$ acts trivially on $\Z_2$, so the only elements of $H^n(MO)$ which a map $f\in \Hom^n_{\a^2}(H^*(MO),\Z_2)$ does not act trivially on, are those in the image of $(\a_2)_0^{f(0)}$. It follows that the rank of $\Hom^n_{\a^2}(H^*(MO),\Z_2)$ as a $\Z_2$ vector space, and thus $\pi_{n}(MO)$, is equal to $f(0)$, i.e. the nondyadic partitions of $n$.