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I have been studying the mod 2 Steenrod Algebra (denoted $\mathcal{A}$), using Mosher & Tangora.

We have the Serre-Cartan (or Adem basis): Let $I = \{i_1,i_2,\ldots,i_n\}$ be a sequence of integers. We say this sequence is admissable if $i_k \ge 2 i_{k+1}$ for each $k$. The elements $Sq^I = Sq^{i_1} Sq^{i_2} \ldots $ where $I$ is an admissable sequence form a basis for the mod 2 Steenrod algebra (I think maybe the more correct statement is a basis as a $\mathbb{Z}/2$-module?).

The Milnor basis is more complicated - I will give a brief explanation of (my understanding of) how it arises. First we show that $\mathcal{A}$ is actually a Hopf algebra, and thus the dual $\mathcal{A}^\ast$ is also a Hopf algebra. Then for each $i > 0$ we can define an element $\xi_i$ in $\mathcal{A}^\ast_{2^i-i}$ defined via the relation $\xi_i(\theta)(x)^{2^i}=\theta(x) \in H^{2i}(K(\mathbb{Z}/2,1);\mathbb{Z}/2)$ where $\theta \in \mathcal{A}_{2^i-1}$ and $x$ is the generator of $H^1(K(\mathbb{Z}/2,1);\mathbb{Z}/2)$. We use the notation $\langle \xi_i,\theta \rangle = \xi_i(\theta)$

Let $\mathcal{R}$ denote the set of all infinite sequences of non-negative integers with only finitely many non-zero entries. Let $\mathcal{J}$ denote the subset of $\mathcal{R}$ of admissible sequences. For each $R \in \mathcal{R}$ define a sequence $\gamma:\mathcal{J} \to \mathcal{R}$ by $\gamma(\{a_1,\ldots,a_k \}) = \{ a_1 - 2a_2,a_2-2 a_3,\ldots, a_k,0,\ldots \}$. We can order sequences of $\mathcal{J}$ lexiographically from the right, e.g. $ \{ 8,4,2,0,\ldots \} > \{ 9,4,1,\ldots \}$.

For each $R \in \mathcal{R}$ we define $\xi^R \in \mathcal{A}^\ast$ by $\xi^R = \prod_{i=1}^\infty (\xi_i)^{r_i}$ where $R = \{ r_1,r_2 \ldots \}$.

We have the result, (which might help with what I am trying to do) that $$\langle \xi^{\gamma(J)}, Sq^I \rangle = \begin{cases} 0, &I<J \\ 1, &I=J \end{cases} $$ Now the beautiful result of Milnor is that the dual mod 2 Steenrod algebra $\mathcal{A}^\ast$ is the polynomial ring over $\mathbb{Z}/2$ generated by $\{ \xi_i \}$ for $i \ge 1$.

So $\{\xi^R\}$ forms a basis for $\mathcal{A}^\ast$. The dual basis elements are denoted $\{ Sq^R \}$ are are called the Milnor basis for $\mathcal{A}$. They satisfy $\langle \xi^R,Sq^{R'} \rangle = 1$ if $R = R'$ and $0$ else.

I am now wondering how to convert from the Milnor to the Serre-Cartan basis.

I played around a bit with Sage and I 'found' that $Sq^{\{i,1\}} = Sq^{i+2}Sq^1 + \alpha Sq^{2i+1}$ (LHS is Milnor basis, RHS is Serre-Cartan, and $\alpha$ is $0$ for odd $i$ and $1$ else). My basic question is:

How do I convert between these two basis'?

Update: I've been thinking about my particular case. $Sq^{\{i, 1\}}$ has degree $i+3$. So let's take $i=3$ as an example. The admissable sequences (Serre-Cartan basis) of degree $6$ are $Sq^6$, $Sq^5 Sq^1$, $Sq^4 Sq^2$. These cannot be reduced further (i.e. by using the Adem relations). So why is $Sq^{\{ 3,1 \}} = Sq^5 Sq^1$ only?

Further udpate: I'm now not convinced this is an easy problem, except in some of the simple (small) cases. See this paper

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