Mixing
Milnor basis of Steenrod Algebra
$begingroup$
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.
algebraic-topology homotopy-theory
asked Feb 26 '17 at 9:37
happymathhappymath
3,83011439
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add a comment |
$begingroup$
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.
algebraic-topology homotopy-theory
asked Feb 26 '17 at 9:37
happymathhappymath
3,83011439
$endgroup$
add a comment |
$begingroup$
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.
algebraic-topology homotopy-theory
asked Feb 26 '17 at 9:37
happymathhappymath
3,83011439
$endgroup$
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.
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
asked Feb 26 '17 at 9:37
happymathhappymath
3,83011439
asked Feb 26 '17 at 9:37
happymathhappymath
3,83011439
asked Feb 26 '17 at 9:37
happymathhappymath
3,83011439
asked Feb 26 '17 at 9:37
happymathhappymath
3,83011439
asked Feb 26 '17 at 9:37
happymathhappymath
3,83011439
3,83011439
add a comment |
add a comment |
1 Answer
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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.
answered Jan 15 at 1:00
Weinan LinWeinan Lin
111
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1 Answer
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1 Answer
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active
oldest
votes
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active
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votes
$begingroup$
$xi_i$ 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.
answered Jan 15 at 1:00
Weinan LinWeinan Lin
111
$endgroup$
add a comment |
$begingroup$
$xi_i$ 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.
answered Jan 15 at 1:00
Weinan LinWeinan Lin
111
$endgroup$
add a comment |
$begingroup$
$xi_i$ 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.
answered Jan 15 at 1:00
Weinan LinWeinan Lin
111
$endgroup$
$xi_i$ 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.
answered Jan 15 at 1:00
Weinan LinWeinan Lin
111
answered Jan 15 at 1:00
Weinan LinWeinan Lin
111
answered Jan 15 at 1:00
Weinan LinWeinan Lin
111
answered Jan 15 at 1:00
Weinan LinWeinan Lin
111
111
add a comment |
add a comment |
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