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This definitely must be well a known matter, but I don't know how to do it. So, please provide me with a reference if there is any!

Work at $p=2$. Then, it is well known that $Sq^{2^i}$ generate the Steenrod algebra, i.e. any operation $Sq^n$ can be written in terms of composition of $Sq^{2^i}$ operations with $i\geqslant 0$ (which are not necessarily admissible). What I would like to know is the possible ways to find such a decomposition.

I know that we have Adem relations and somehow one can work inductively that given $n$ look at composition of admissible non-admissible pairs $Sq^aSq^b$ with $a+b=n$ and somehow do some clever argument! For instance, $Sq^6=Sq^2Sq^4+Sq^1Sq^4Sq^1$ is something that I am interested in. But, I know of this expression, because I know the Adem relation for $Sq^2Sq^4$.

I like to know if there is any way of Using Milnor's theorem or say the anti-homomorphism of the Steenrod algebra, or the diagonal of this algebra to compute such a decomposition!!!

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    $\begingroup$ Such decompositions are not unique, as (for example) $Sq^6 = Sq^2Sq^4 + Sq^1Sq^2Sq^1Sq^2$ as well. Sage can do these types of change of basis for you. But you might also find what you want in mathweb.scranton.edu/monks/pubs/bases.pdf (Look at Wall's basis and Arnon's A basis) $\endgroup$
    – Drew Heard
    Commented Mar 3, 2018 at 10:47
  • $\begingroup$ @DrewHeard Thanks for the reference. This is an interesting example where for the same Adem relation where one uses $Sq^1Sq^4Sq^1=Sq^5Sq^1=Sq^3Sq^3$ and $Sq^3=Sq^1Sq^2$. But, there are different factors of powers of $2$ appearing in the expression! $\endgroup$
    – user51223
    Commented Mar 3, 2018 at 12:20
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    $\begingroup$ This is related: math.stackexchange.com/questions/2312884/… $\endgroup$
    – John Palmieri
    Commented Mar 3, 2018 at 16:29
  • $\begingroup$ @JohnPalmieri Thanks. I should have remembered Wood's paper. But, I am now arriving at the conclusion that despite existence of various bases, there is no clear `algorithm' to do this! $\endgroup$
    – user51223
    Commented Mar 4, 2018 at 11:30

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