9
$\begingroup$

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

$\endgroup$

2 Answers 2

15
$\begingroup$

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 $F_X$.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks, Neil! What is a good place to read about these basic things? When I search for papers on unstable Steenrod modules, many of them don't even give the definitions (regarding them as well-known, I suppose) before jumping into doing much fancier things with them. $\endgroup$
    – Tim Campion
    Jan 14 at 15:40
4
$\begingroup$

Your question seems to be equivalent to asking for a description of the unstable condition in terms of the Milnor basis. This is easy to do: Given $r = (r_1,\dots, r_s)$, let $P(r)$ be dual to $\xi_1^{r_1}\cdots \xi_s^{r_s}$, and let $e(r)=r_1+\dots +r_s$. Then an $A$--module $M$ is unstable iff $P(r)x = 0$ if $e(r)>|x|$.

This is pretty standard stuff, I think. This would be in Lionel Schwartz' book on unstable modules, and very possibly also in Margolis' earlier book. I use this characterization in section 6 of my first generic representation theory paper [Amer.J.Math. 116 (1994)].

If you are curious about such things, you should be aware that in the 1980's and 90's, the algebra related to the unstable condition was studied quite completely, with application to both the Sullivan conjecture (and then numerous other remarkable topological consequences), and also to representation theory. Schwartz' book is a good place to start learning about this.

$\endgroup$
1
  • $\begingroup$ Thanks, this is helpful! $\endgroup$
    – Tim Campion
    Jan 18 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.