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