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 8 hours ago
Nicolas BoergerNicolas Boerger
9995 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 8 hours ago
Nicolas BoergerNicolas Boerger
9995 silver badges13 bronze badges
$endgroup$
1
$begingroup$
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
7 hours ago
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
7 hours ago
1
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
4 hours ago
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 8 hours ago
Nicolas BoergerNicolas Boerger
9995 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 8 hours ago
Nicolas BoergerNicolas Boerger
9995 silver badges13 bronze badges
asked 8 hours ago
Nicolas BoergerNicolas Boerger
9995 silver badges13 bronze badges
asked 8 hours ago
Nicolas BoergerNicolas Boerger
9995 silver badges13 bronze badges
asked 8 hours ago
Nicolas BoergerNicolas Boerger
9995 silver badges13 bronze badges
asked 8 hours ago
Nicolas BoergerNicolas Boerger
9995 silver badges13 bronze badges
9995 silver badges13 bronze badges
1
$begingroup$
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
7 hours ago
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
7 hours ago
1
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
4 hours ago
add a comment |
1
$begingroup$
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
7 hours ago
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
7 hours ago
1
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
4 hours ago
1
1
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
7 hours ago
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
7 hours ago
1
1
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
4 hours ago
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
4 hours ago
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 7 hours ago
Denis NardinDenis Nardin
10k2 gold badges38 silver badges70 bronze badges
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f336884%2freference-request-mod-2-cohomology-of-periodic-ko-theory%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 7 hours ago
Denis NardinDenis Nardin
10k2 gold badges38 silver badges70 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 7 hours ago
Denis NardinDenis Nardin
10k2 gold badges38 silver badges70 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 7 hours ago
Denis NardinDenis Nardin
10k2 gold badges38 silver badges70 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 7 hours ago
Denis NardinDenis Nardin
10k2 gold badges38 silver badges70 bronze badges
edited 7 hours ago
answered 7 hours ago
Denis NardinDenis Nardin
10k2 gold badges38 silver badges70 bronze badges
answered 7 hours ago
Denis NardinDenis Nardin
10k2 gold badges38 silver badges70 bronze badges
answered 7 hours ago
Denis NardinDenis Nardin
10k2 gold badges38 silver badges70 bronze badges
10k2 gold badges38 silver badges70 bronze badges
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f336884%2freference-request-mod-2-cohomology-of-periodic-ko-theory%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
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
7 hours ago
$begingroup$
Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology!
$endgroup$
– Prasit
7 hours ago
1
$begingroup$
It is more usual to use $mathcalA_1$ to denote the algebra generated by $mathrmSq^1$ and $mathrmSq^2$.
$endgroup$
– John Palmieri
4 hours ago