What does Robert Stong mean when he says $H^*(MO(k))$ is a free Steenrod algebra in dimension less than $2k$?

Question

$\newcommand{\Z}{\mathbb Z}\newcommand{\a}{\mathfrak a}\newcommand\widetildeH{\smash{\widetilde H}}$In Robert Stong's notes on Cobordism Theory, on page 95 he asserts the following:

$\widetildeH^* (MO(r),\Z_2)$ is a free $\a_2$ module in dimension less than or equal to $2r$

where $MO(r)$ is the Thom space $\operatorname{Th}(\gamma_r)$ for the tautological bundle sitting over the classifying space $BO_k$, and $\a_2$ is the mod $2$ Steenrod algebra. My question is what does this mean/how is this possible? In what follows I will assume all cohomology is taken with $\Z_2$ coefficients.

In particular, he has just shown that $\widetildeH^*(MO)$, which is defined as the direct sum of groups: $$H^n(MO)=\lim_{k\rightarrow \infty}H^{n+k}(MO(k))$$ is a free $\mathfrak a_2$ module. This makes sense to me, as $\widetildeH^*(MO)$ should very much be an infinite dimensional vector space over $\Z_2$ because for any $k>n$ we have that by the Thom isomorphism: \begin{align} \widetildeH^{n+k}(MO(k))\cong H^n(BO_k)=(\Z_2[x_1,\dots, x_k])_n \end{align} where $\Z_2[x_1,\dots, x_k]$ has grading $|x_i|=i$. So since for all $k>n$ this is the same group we have that: \begin{align} \widetildeH^{n}(MO)\cong \widetildeH^{n+k}(MO(k))\cong H^n(BO_k)=(\Z_2[x_1,\dots, x_k])_n. \end{align} $\widetildeH^*(MO)$ is then given the structure of $\a_2$ module via the algebra acting on $\widetildeH^{n+k}(MO(k))$, but I do not believe that this makes $\widetildeH^n(MO)$ an $\a_2$ module, so how can $\widetildeH^{i}(MO(k)$ be an $\a_2$ module, let alone a free $\a_2$ module in dimension $i\leq 2k$?

Moreover, even if I made some mistake and somehow $\widetildeH^n(MO)$ is an $\a_2$ module, I do not see how it can be free. We have that $\widetildeH^n(MO)$ is a finite rank $\Z_2$ vector space, and if $\widetildeH^n(MO)$ is free of any rank it is a direct sum of $\a_2$s which is an infinite rank $\Z_2$ vector space; these two facts seem to be in complete contradiction with one another.

Perhaps, Robert Stong actually means that we can take some finite amount of generators $\{x_1,\dots, x_n\}\subset \widetildeH^*(MO(r))$ whose span contains $\widetildeH^i(MO(r))$ for all $i\leq 2r$, and these generators are linearly independent over $\a_2$? This I would believe.

The fact that these groups are somehow "free" is essential to his proof of determining the ranks of $\pi_n(MO)$, so any help would be greatly appreciated.

ETA:

Oh I forgot to mention, he justifies this as follows:

The free module structure follows from the stability $\widetildeH^{r+i}(MO(r))\cong \widetildeH^{r+i+1}(MO(r+1))$ for $i\leq r$.

I agree with this isomorphism, I fail to see how this makes this a Steenrod module, let alone a free one.

Answer 1

First, let's clarify what the statement means: The Steenrod algebra is a graded algebra, and $H^*(X)$ for any spectrum $X$ is a graded module over it. This means that the action $\mathcal{A}_2\otimes H^*(X)\to H^*(X)$ takes $\mathcal{A}_2^n \otimes H^m(X) \to H^{n+m}(X)$. So yes, it does not make sense to consider any individual $H^m(X)$ as module over the Steenrod algebra.

What does it mean to be a free module, in this graded world? You can write down the "free module over $\mathcal{A}_2$ on a generator in degree $n$", this is just the $\mathcal{A}_2$-module which is given in degree $m$ by $\mathcal{A}_2^{m-n}$. Let's write that as $\mathcal{A}_2(n)$. Now we call a graded $\mathcal{A}_2$-module $M$ free if it is isomorphic to a direct sum $\bigoplus_i \mathcal{A}_2(n_i)$.

This is for example the case for $H^*(MO)$ (this is due to Thom originally, I believe). It is not quite the case for $H^*(MO(r))$, but you get that it "looks like a free module in degrees $\leq 2r$". What does this mean? It means you find a map $\bigoplus_i \mathcal{A}_2(n_i) \to H^*(MO(r))$ of graded modules, which is an isomorphism in degrees $\leq 2r$.

Now as for how this is justified (without having double-checked that this is the way Stong does it): In principle, all of this can be verified explicitly, by using the Thom isomorphism, a description of the cohomology of $BO(r)$ with Steenrod action, and the fact that the Thom isomorphism twists the Steenrod action by Stiefel-Whitney classes. But the quote at the end of your question leads me to believe that Stong deduces the unstable statement (for $MO(r)$) from the stable one (for $MO$). The stable case is originally done by Thom through explicit computations, but there is a more conceptual argument based on the Milnor-Moore theorem. This is explained nicely in the Blog of Akhil Mathew, for example. And the stable statement implies the quoted statement for $MO(r)$, since $H^*(MO)$ in the range of degrees $[0,r]$ is isomorphic to $H^*(MO(r))$ in the range of degrees $[r,2r]$.