I am trying to understand the Milnor basis for the Steenrod Algebra. For this first we define $\xi_i=(Sq^{2^{i-1}}....Sq^2Sq^1)^*$. Then he proves that these elements freely generate the dual algebra. Then he takes the dual of the monomial basis to obtain another basis of the Steenrod Algebra. But the dual of $\xi_i$ is just $Sq^{2^{i-1}}....Sq^2Sq^1$. But in these notes he seems to claim that it is $[Sq^{2^{i-1}},\xi_{i-1}^*]$. So where am I going wrong. Thank you.
Milnor basis of Steenrod Algebra
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algebraic-topology
homotopy-theory
1 Answers
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$\xi_i$ is dual to $Sq^{2^{i-1}}\cdots Sq^2Sq^1$ with respect to the admissible basis for the Steenrod algebra, while $[Sq^{2^{i-1}}, \xi_{i-1}^*]$ is dual to $\xi_i$ with respect to the monomial basis for the dual Steenrod algebra. These two basis are NOT dual. So there is no contradiction here.