# What is the total square on the dual Steenrod algebra?

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 $$i$$.)

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

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*}

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 $$Sq(\zeta_i)$$... ]