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 $i$.)



What is the total square on the dual Steenrod algebra?
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:
$$
\begin{align\*}
\zeta\_n \cdot Sq &= \sum (\xi\_1^m)^\* (\zeta\_i) \zeta\_j^{2^i} = \zeta\_n + \zeta\_{n-1}^2 + \dots + \zeta\_1^{2^{n-1}} + 1\\
Sq \cdot \zeta\_n &= \sum \zeta\_i (\xi\_1^m)^\* (\zeta\_j^{2^i}) = \zeta\_n + \zeta\_{n-1} + \dots + \zeta\_1 + 1.
\end{align\*}
$$


