$\newcommand\Sq{\mathit{Sq}}$Recall that a (graded) module $V^\ast$ over the Steenrod algebra $\mathcal A^\ast$ is said to be unstable if $\Sq^i v = 0$ for $i > |v|$. The motivating example, of course, is that if $V^\ast = H^\ast(X)$ for a space $X$ with its natural $\mathcal A^\ast$ structure, then $V^\ast$ is unstable.
The category of $\mathcal A^\ast$-modules which are finite-dimensional over $\mathbb F_p$ is dual to the category of $\mathcal A_\ast$-comodules which are finite-dimensional over $\mathbb F_p$, where $\mathcal A_\ast$ is the dual Steenrod algebra. So the instability condition should be expressible from this dual perspective.
Question 1: Let $V_\ast$ be a finite-dimensional graded $\mathbb F_p$-vector space equipped with the structure of an $\mathcal A_\ast$-comodule. Under what conditions is the dual vector space $V^\ast$ unstable (with its natural $\mathcal A^\ast$-module structure)?
Ideally, the condition would be expressed in terms of Milnor's $\{\xi_m, \tau_n\}$ generators. In principle it should be straightforward to do the translation, but I am a bit intimidated by the Adem relations. I am also stuck already because even before dualizing, I don't know how to express the instability condition in terms of the generating set $\{\Sq^{2^k} \Sq^{2^{k-1}}\dotsm \Sq^1\}$, which I suppose leads to a subsidiary question:
Question 2: Let $V^\ast$ be a finite-dimensional $\mathcal A^\ast$-module. In terms of the generating set $\{\Sq^{2^k} \Sq^{2^{k-1}} \dotsm \Sq^1\}$, when is $V^\ast$ unstable?
I'd be happy to know the answer for $p=2$, $p>2$, or both.
Best Answer
Normally people think about Steenrod comodules as graded $\mathbb{Z}/2$-modules equipped with a graded coaction $\psi\colon M_*\to M_*[\xi_1,\xi_2,\dotsc]$. However, it is equivalent to consider ungraded modules with coaction $\psi\colon M\to M[\xi_0^{\pm 1},\xi_1,\xi_2,\dotsc]$; the grading is recovered by the formula $$ M_d = \{m : \psi(m)=\xi_0^dm \pmod{\xi_k:k>0}\}, $$ and the original coaction is recovered by setting $\xi_0=1$.
Now the action is unstable iff $\psi(M)\leq M[\xi_0,\xi_1,\dotsc]$ (so no negative powers of $\xi_0$ are involved).
If $M$ has a compatible product then we can reformulate things in terms of the scheme $X=\text{spec}(M)$. A Steenrod action is equivalent to an action of the group scheme $\text{Aut}(G_a)$ on $X$ (where $G_a$ is the additive formal group). The main instability condition (that certain Steenrod operations should vanish) is equivalent to saying that the action extends over the monoid $\text{End}(G_a)$. (It is easy to see that there is at most one extension.) In the multiplicative context one also wants to impose the condition $\text{Sq}^k(x)=x^2$ when $|x|=k$. This is equivalent to saying that the Frobenius endomorphism $F_{G_a}\in\text{End}(G_a)$ should act on $X$ as the Frobenius endomorphism $F_X$.