$\newcommand\Sq{\mathit{Sq}}$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.



What is an unstable dual-Steenrod comodule?
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 $F\_X$.


