Reference request: mod 2 cohomology of periodic KO theorySingular complex = cohomology ring + Steenrod operations?Homology-Cohomology PairingSteenrod algebra at a prime powerCohomology of the Image of J spectrumThe Segal Machine constructing spectra and topological $K$-TheoryEndomorphism ring spectrum of the Eilenberg-MacLane spectrumRealizing $mathcalA(2)//mathcalA(1)$ by a finite spectrumCohomology of $ko,tmf,MSpin,MString$ with coefficients $mathbbZ/p$ for odd primes $p$The connective $k$-theory cohomology of Eilenberg-MacLane spectraMaps from mod-$p$ Eilenberg-MacLane spectrum to connective $K$-theory spectrum
Reference request: mod 2 cohomology of periodic KO theory
Singular complex = cohomology ring + Steenrod operations?Homology-Cohomology PairingSteenrod algebra at a prime powerCohomology of the Image of J spectrumThe Segal Machine constructing spectra and topological $K$-TheoryEndomorphism ring spectrum of the Eilenberg-MacLane spectrumRealizing $mathcalA(2)//mathcalA(1)$ by a finite spectrumCohomology of $ko,tmf,MSpin,MString$ with coefficients $mathbbZ/p$ for odd primes $p$The connective $k$-theory cohomology of Eilenberg-MacLane spectraMaps from mod-$p$ Eilenberg-MacLane spectrum to connective $K$-theory spectrum
$begingroup$
The mod 2 cohomology of the connective ko spectrum is known to be the module $mathcalAotimes_mathcalA_2 mathbbF_2$, where $mathcalA$ denotes the Steenrod algebra, and $mathcalA_2$ denotes the subalgebra generated by $Sq^1$ and $Sq^2$.
Where can I find the original calculation for referencing it ?
What is the structure as module over the steenrod algebra of the mod 2 cohomology of periodic KO theory?
at.algebraic-topology kt.k-theory-and-homology stable-homotopy
asked Jul 24 at 16:32
Nicolas BoergerNicolas Boerger
1,0096 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
The mod 2 cohomology of the connective ko spectrum is known to be the module $mathcalAotimes_mathcalA_2 mathbbF_2$, where $mathcalA$ denotes the Steenrod algebra, and $mathcalA_2$ denotes the subalgebra generated by $Sq^1$ and $Sq^2$.
Where can I find the original calculation for referencing it ?
What is the structure as module over the steenrod algebra of the mod 2 cohomology of periodic KO theory?
at.algebraic-topology kt.k-theory-and-homology stable-homotopy
asked Jul 24 at 16:32
Nicolas BoergerNicolas Boerger
1,0096 silver badges13 bronze badges
$endgroup$
2
$begingroup$
Unless I'm mistaken $pi_*(HmathbbF_2wedge KO)=0$ (it is a ring of positive characteristic containing an isomorphism between the additive and the multiplicative formal group law), so the answer to your second question is rather trivial
$endgroup$
– Denis Nardin
Jul 24 at 16:45
1
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
Jul 24 at 16:56
4
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
Jul 24 at 19:57
add a comment |
$begingroup$
The mod 2 cohomology of the connective ko spectrum is known to be the module $mathcalAotimes_mathcalA_2 mathbbF_2$, where $mathcalA$ denotes the Steenrod algebra, and $mathcalA_2$ denotes the subalgebra generated by $Sq^1$ and $Sq^2$.
Where can I find the original calculation for referencing it ?
What is the structure as module over the steenrod algebra of the mod 2 cohomology of periodic KO theory?
at.algebraic-topology kt.k-theory-and-homology stable-homotopy
asked Jul 24 at 16:32
Nicolas BoergerNicolas Boerger
1,0096 silver badges13 bronze badges
$endgroup$
The mod 2 cohomology of the connective ko spectrum is known to be the module $mathcalAotimes_mathcalA_2 mathbbF_2$, where $mathcalA$ denotes the Steenrod algebra, and $mathcalA_2$ denotes the subalgebra generated by $Sq^1$ and $Sq^2$.
Where can I find the original calculation for referencing it ?
What is the structure as module over the steenrod algebra of the mod 2 cohomology of periodic KO theory?
at.algebraic-topology kt.k-theory-and-homology stable-homotopy
at.algebraic-topology kt.k-theory-and-homology stable-homotopy
asked Jul 24 at 16:32
Nicolas BoergerNicolas Boerger
1,0096 silver badges13 bronze badges
asked Jul 24 at 16:32
Nicolas BoergerNicolas Boerger
1,0096 silver badges13 bronze badges
asked Jul 24 at 16:32
Nicolas BoergerNicolas Boerger
1,0096 silver badges13 bronze badges
asked Jul 24 at 16:32
Nicolas BoergerNicolas Boerger
1,0096 silver badges13 bronze badges
asked Jul 24 at 16:32
Nicolas BoergerNicolas Boerger
1,0096 silver badges13 bronze badges
1,0096 silver badges13 bronze badges
2
$begingroup$
Unless I'm mistaken $pi_*(HmathbbF_2wedge KO)=0$ (it is a ring of positive characteristic containing an isomorphism between the additive and the multiplicative formal group law), so the answer to your second question is rather trivial
$endgroup$
– Denis Nardin
Jul 24 at 16:45
1
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
Jul 24 at 16:56
4
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
Jul 24 at 19:57
add a comment |
2
$begingroup$
Unless I'm mistaken $pi_*(HmathbbF_2wedge KO)=0$ (it is a ring of positive characteristic containing an isomorphism between the additive and the multiplicative formal group law), so the answer to your second question is rather trivial
$endgroup$
– Denis Nardin
Jul 24 at 16:45
1
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
Jul 24 at 16:56
4
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
Jul 24 at 19:57
2
2
$begingroup$
Unless I'm mistaken $pi_*(HmathbbF_2wedge KO)=0$ (it is a ring of positive characteristic containing an isomorphism between the additive and the multiplicative formal group law), so the answer to your second question is rather trivial
$endgroup$
– Denis Nardin
Jul 24 at 16:45
$begingroup$
Unless I'm mistaken $pi_*(HmathbbF_2wedge KO)=0$ (it is a ring of positive characteristic containing an isomorphism between the additive and the multiplicative formal group law), so the answer to your second question is rather trivial
$endgroup$
– Denis Nardin
Jul 24 at 16:45
1
1
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
Jul 24 at 16:56
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
Jul 24 at 16:56
4
4
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
Jul 24 at 19:57
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
Jul 24 at 19:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Ravenel in his Complex Cobordism and Stable Homotopy Groups of Spheres attributes this result to Stong, in Determination of $H^*(BO(k,⋯,∞),Z_2)$ and $H^∗(BU(k,⋯,∞),Z_2)$, but looking at that paper (which is concerned mainly with the "unstable" cohomology of the various constituent spaces of $ko$), he attributes this further to
Adams, J. F., On Chern characters and the structure of the unitary group, Proc. Camb. Philos. Soc. 57, 189-199 (1961). ZBL0103.16001.
There you can find indeed the required result as Lemma 4.
Regarding your second question, the cohomology $H^*(KO;mathbbF_2)$ is zero for chromatic reasons (the spectrum $KUwedge HmathbbF_2$ carries an isomorphism of the additive and multiplicative formal group law in characteristic two, so it must be trivial, and then you can use the fact that $KU=KO/eta$ to conclude, since $eta$ is nilpotent in $pi_*KO$), hence the map $H^*(KO;mathbbF_2)→H^*(ko;mathbbF_2)$ is trivial.
answered Jul 24 at 16:59
Denis NardinDenis Nardin
10.1k2 gold badges39 silver badges71 bronze badges
$endgroup$
add a comment |
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$begingroup$
Ravenel in his Complex Cobordism and Stable Homotopy Groups of Spheres attributes this result to Stong, in Determination of $H^*(BO(k,⋯,∞),Z_2)$ and $H^∗(BU(k,⋯,∞),Z_2)$, but looking at that paper (which is concerned mainly with the "unstable" cohomology of the various constituent spaces of $ko$), he attributes this further to
Adams, J. F., On Chern characters and the structure of the unitary group, Proc. Camb. Philos. Soc. 57, 189-199 (1961). ZBL0103.16001.
There you can find indeed the required result as Lemma 4.
Regarding your second question, the cohomology $H^*(KO;mathbbF_2)$ is zero for chromatic reasons (the spectrum $KUwedge HmathbbF_2$ carries an isomorphism of the additive and multiplicative formal group law in characteristic two, so it must be trivial, and then you can use the fact that $KU=KO/eta$ to conclude, since $eta$ is nilpotent in $pi_*KO$), hence the map $H^*(KO;mathbbF_2)→H^*(ko;mathbbF_2)$ is trivial.
answered Jul 24 at 16:59
Denis NardinDenis Nardin
10.1k2 gold badges39 silver badges71 bronze badges
$endgroup$
add a comment |
$begingroup$
Ravenel in his Complex Cobordism and Stable Homotopy Groups of Spheres attributes this result to Stong, in Determination of $H^*(BO(k,⋯,∞),Z_2)$ and $H^∗(BU(k,⋯,∞),Z_2)$, but looking at that paper (which is concerned mainly with the "unstable" cohomology of the various constituent spaces of $ko$), he attributes this further to
Adams, J. F., On Chern characters and the structure of the unitary group, Proc. Camb. Philos. Soc. 57, 189-199 (1961). ZBL0103.16001.
There you can find indeed the required result as Lemma 4.
Regarding your second question, the cohomology $H^*(KO;mathbbF_2)$ is zero for chromatic reasons (the spectrum $KUwedge HmathbbF_2$ carries an isomorphism of the additive and multiplicative formal group law in characteristic two, so it must be trivial, and then you can use the fact that $KU=KO/eta$ to conclude, since $eta$ is nilpotent in $pi_*KO$), hence the map $H^*(KO;mathbbF_2)→H^*(ko;mathbbF_2)$ is trivial.
answered Jul 24 at 16:59
Denis NardinDenis Nardin
10.1k2 gold badges39 silver badges71 bronze badges
$endgroup$
add a comment |
$begingroup$
Ravenel in his Complex Cobordism and Stable Homotopy Groups of Spheres attributes this result to Stong, in Determination of $H^*(BO(k,⋯,∞),Z_2)$ and $H^∗(BU(k,⋯,∞),Z_2)$, but looking at that paper (which is concerned mainly with the "unstable" cohomology of the various constituent spaces of $ko$), he attributes this further to
Adams, J. F., On Chern characters and the structure of the unitary group, Proc. Camb. Philos. Soc. 57, 189-199 (1961). ZBL0103.16001.
There you can find indeed the required result as Lemma 4.
Regarding your second question, the cohomology $H^*(KO;mathbbF_2)$ is zero for chromatic reasons (the spectrum $KUwedge HmathbbF_2$ carries an isomorphism of the additive and multiplicative formal group law in characteristic two, so it must be trivial, and then you can use the fact that $KU=KO/eta$ to conclude, since $eta$ is nilpotent in $pi_*KO$), hence the map $H^*(KO;mathbbF_2)→H^*(ko;mathbbF_2)$ is trivial.
answered Jul 24 at 16:59
Denis NardinDenis Nardin
10.1k2 gold badges39 silver badges71 bronze badges
$endgroup$
Ravenel in his Complex Cobordism and Stable Homotopy Groups of Spheres attributes this result to Stong, in Determination of $H^*(BO(k,⋯,∞),Z_2)$ and $H^∗(BU(k,⋯,∞),Z_2)$, but looking at that paper (which is concerned mainly with the "unstable" cohomology of the various constituent spaces of $ko$), he attributes this further to
Adams, J. F., On Chern characters and the structure of the unitary group, Proc. Camb. Philos. Soc. 57, 189-199 (1961). ZBL0103.16001.
There you can find indeed the required result as Lemma 4.
Regarding your second question, the cohomology $H^*(KO;mathbbF_2)$ is zero for chromatic reasons (the spectrum $KUwedge HmathbbF_2$ carries an isomorphism of the additive and multiplicative formal group law in characteristic two, so it must be trivial, and then you can use the fact that $KU=KO/eta$ to conclude, since $eta$ is nilpotent in $pi_*KO$), hence the map $H^*(KO;mathbbF_2)→H^*(ko;mathbbF_2)$ is trivial.
answered Jul 24 at 16:59
Denis NardinDenis Nardin
10.1k2 gold badges39 silver badges71 bronze badges
edited Jul 24 at 17:15
answered Jul 24 at 16:59
Denis NardinDenis Nardin
10.1k2 gold badges39 silver badges71 bronze badges
answered Jul 24 at 16:59
Denis NardinDenis Nardin
10.1k2 gold badges39 silver badges71 bronze badges
answered Jul 24 at 16:59
Denis NardinDenis Nardin
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10.1k2 gold badges39 silver badges71 bronze badges
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$begingroup$
Unless I'm mistaken $pi_*(HmathbbF_2wedge KO)=0$ (it is a ring of positive characteristic containing an isomorphism between the additive and the multiplicative formal group law), so the answer to your second question is rather trivial
$endgroup$
– Denis Nardin
Jul 24 at 16:45
1
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
Jul 24 at 16:56
4
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
Jul 24 at 19:57