8
$\begingroup$

The dual Steenrod algebra ($p=2$) has generators $\xi_n$ and these have conjugates that are often labeled $\zeta_n$. I am curious about the left and right actions of the Steenrod algebra on its dual, and in particular, what the total square is. I have seen in papers that $(\xi_n)Sq = \xi_n + \xi_{n-1}$ and $Sq(\xi_n) = \xi_n + \xi_{n-1}^2$ [1]. On the other hand, I have seen that $(\zeta_n)Sq = \zeta_n + \zeta_{n-1}^2 + \dots + \zeta_1^{2^{n-1}} + 1$ [2]. I can't find a reference anywhere for the left total square on $\zeta_n$. I am not sure how to prove these actions, although it seems to me that it should follow from fairly elementary Kronecker product arithmetic along with duality knowledge.

I am interested in either a reference for the left total square, or a way to prove it.

[1] See, for example, Mahowald -- bo-resolutions, page 369.

[2] Bruner, May, McClure, Steinberger -- $H_\infty$ Ring Spectra and their Applications, page 78. (There is a typo: 1 should be $i$.)

$\endgroup$
2
  • $\begingroup$ I assume you're working at the prime 2? $\endgroup$ Mar 17 '19 at 7:31
  • $\begingroup$ Yeah, working at p=2. $\endgroup$
    – Ekie
    Mar 17 '19 at 13:52
3
$\begingroup$

I don't have a reference for you, but here is a comment on how to prove these formulas using the Kronecker pairing that you alluded to.

The Steenrod operation $Sq^m$ is dual to the element $\xi_1^m$ in the monomial basis of the dual Steenrod algebra; the left and right actions of the Steenrod algebra on $\mathcal{A}_*$ are composites of the coproduct in the dual Steenrod algebra and the action on the right or left side. If the coproduct satisfies $\Delta x = \sum x' \otimes x''$, we then get $$ \begin{align*} x \cdot Sq^m &= \sum (\xi_1^m)^*(x') x'',\\ Sq^m \cdot x &= \sum x' (\xi_1^m)^* (x''). \end{align*} $$ (The apparent order reversal is necessary to make this into a left/right action.) We'd like to apply this to the comultiplication formulas $\Delta \xi_n = \sum_{i+j=n} \xi_i^{2^j} \xi_j$ and $\Delta \zeta_n = \sum_{i+j=n} \zeta_i \zeta_j^{2^i}$. Here by convention $\xi_0 = \zeta_0 = 1$.

To apply this to the $\xi_n$, we first remark that $$ \sum_m (\xi_1^m)^*(\xi_i^{2^j}) = \begin{cases}1 &\text{if }i=0,1,\\0&\text{otherwise.}\end{cases} $$ Therefore: $$ \begin{align*} \xi_n \cdot Sq &= \sum (\xi_1^m)^* (\xi_i^{2^j}) \xi_j = \xi_n + \xi_{n-1}\\ Sq \cdot \xi_n &= \sum \xi_i^{2^j} (\xi_1^m)^* (\xi_j) = \xi_n + \xi_{n-1}^2. \end{align*} $$ To figure out the corresponding result for the $\zeta_n$, we have to figure out what the coefficient of $\xi_1^{2^n-1}$ is in the formula for $\zeta_n$. The $\zeta_i$ are defined inductively, for $n > 0$, using the formula $$ \sum_{i+j=n} \xi_i^{2^j} \zeta_j = 0. $$ If we take the quotient by the ideal generated by $\xi_2, \xi_3, \dots$ we find that this formula reduces to $$ \zeta_n + \xi_1^{2^{n-1}} \zeta_{n-1} \equiv 0 $$ and so inductively $\zeta_n \equiv \xi_1^{2^n - 1}$ mod the higher $\xi_i$. This means $$ \sum_m (\xi_1^m)^*(\zeta_j^{2^i}) = 1 $$ for any $i$ and $j$.

Therefore: $$ \begin{align*} \zeta_n \cdot Sq &= \sum (\xi_1^m)^* (\zeta_i) \zeta_j^{2^i} = \zeta_n + \zeta_{n-1}^2 + \dots + \zeta_1^{2^{n-1}} + 1\\ Sq \cdot \zeta_n &= \sum \zeta_i (\xi_1^m)^* (\zeta_j^{2^i}) = \zeta_n + \zeta_{n-1} + \dots + \zeta_1 + 1. \end{align*} $$

$\endgroup$
6
$\begingroup$

We bothered to write it down in our paper. Look at pg 6, we give some of references that we know of.

I did not find a formula for the left action of the $Sq$ on $\zeta_i$s in the literature. But from the formula for left action of $Sq$ on $\xi_i$ and formulas relating $\xi_i$s and $\zeta_i$s one can do an extensive combinatorial argument to see that $$ Sq(\zeta_i) = \zeta_i + \zeta_{i-1} + \dots + \zeta_1 + 1$$.

(In my experience the combinatorial inductive argument was tedious but straightforward!)

[ For example, let's consider the first nontrivial case, ie calculate $Sq(\zeta_2)$. Keep in mind that $\zeta_2 = \xi_2 + \xi_1^3$ and $\zeta_1 = \xi_1$. Then $$Sq(\zeta_2) = Sq(\xi_2 + \xi_1^3) = (\xi_2 + \xi_1^2) + (\xi_1 +1)^3 = \zeta_2 + \zeta_1 + 1.$$ Keep going inductively to get the formulas for $Sq(\zeta_i)$... ]

$\endgroup$

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.