The structure of complex cobordism cohomology of the Eilenberg-Maclane spectrum











up vote
6
down vote

favorite
2












Let $MU$ be the complex bordism spectrum and let $Hmathbb{Z}$ be the Eilenberg-Maclane spectrum.




Is it know what the structure of the complex cobordism cohomology $MU^{*}(Hmathbb{Z})$ is?




EDIT: What if instead $Hmathbb{Z}$, one consider $Hmathbb{Z}/(p)$ for a prime $p$?










share|cite|improve this question
























  • Hrmm.. we know $MU_*Hmathbb{Z}=Hmathbb{Z}_*MU=mathbb{Z}[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
    – Denis Nardin
    yesterday








  • 2




    If I remember correctly, it is $0$, but I don't remember a reference off the head.
    – user43326
    yesterday















up vote
6
down vote

favorite
2












Let $MU$ be the complex bordism spectrum and let $Hmathbb{Z}$ be the Eilenberg-Maclane spectrum.




Is it know what the structure of the complex cobordism cohomology $MU^{*}(Hmathbb{Z})$ is?




EDIT: What if instead $Hmathbb{Z}$, one consider $Hmathbb{Z}/(p)$ for a prime $p$?










share|cite|improve this question
























  • Hrmm.. we know $MU_*Hmathbb{Z}=Hmathbb{Z}_*MU=mathbb{Z}[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
    – Denis Nardin
    yesterday








  • 2




    If I remember correctly, it is $0$, but I don't remember a reference off the head.
    – user43326
    yesterday













up vote
6
down vote

favorite
2









up vote
6
down vote

favorite
2






2





Let $MU$ be the complex bordism spectrum and let $Hmathbb{Z}$ be the Eilenberg-Maclane spectrum.




Is it know what the structure of the complex cobordism cohomology $MU^{*}(Hmathbb{Z})$ is?




EDIT: What if instead $Hmathbb{Z}$, one consider $Hmathbb{Z}/(p)$ for a prime $p$?










share|cite|improve this question















Let $MU$ be the complex bordism spectrum and let $Hmathbb{Z}$ be the Eilenberg-Maclane spectrum.




Is it know what the structure of the complex cobordism cohomology $MU^{*}(Hmathbb{Z})$ is?




EDIT: What if instead $Hmathbb{Z}$, one consider $Hmathbb{Z}/(p)$ for a prime $p$?







at.algebraic-topology homotopy-theory cobordism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked yesterday









user438991

1216




1216












  • Hrmm.. we know $MU_*Hmathbb{Z}=Hmathbb{Z}_*MU=mathbb{Z}[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
    – Denis Nardin
    yesterday








  • 2




    If I remember correctly, it is $0$, but I don't remember a reference off the head.
    – user43326
    yesterday


















  • Hrmm.. we know $MU_*Hmathbb{Z}=Hmathbb{Z}_*MU=mathbb{Z}[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
    – Denis Nardin
    yesterday








  • 2




    If I remember correctly, it is $0$, but I don't remember a reference off the head.
    – user43326
    yesterday
















Hrmm.. we know $MU_*Hmathbb{Z}=Hmathbb{Z}_*MU=mathbb{Z}[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
– Denis Nardin
yesterday






Hrmm.. we know $MU_*Hmathbb{Z}=Hmathbb{Z}_*MU=mathbb{Z}[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
– Denis Nardin
yesterday






2




2




If I remember correctly, it is $0$, but I don't remember a reference off the head.
– user43326
yesterday




If I remember correctly, it is $0$, but I don't remember a reference off the head.
– user43326
yesterday










1 Answer
1






active

oldest

votes

















up vote
11
down vote













One can prove that $mathrm{Map}(Hmathbf{F}_p,MU)$ is contractible. We know that $Hmathbf{F}_p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories"), but $MU$ is harmonic (Theorem 4.2 of that paper). Since dissonant spectra (resp. harmonic spectra) are by definition $E$-acyclic (resp. local) for the spectrum $E = bigvee_p E_p$, where $E_p = bigvee_{0leq n<infty} K(n)$, the claim follows. (I just realized that this is Corollary 4.10 of Ravenel's paper.)



It is, however, not the case that $mathrm{Map}(Hmathbf{Z},MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $Hmathbf{Z}$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrm{Map}(Hmathbf{Z},MU)$ and $mathrm{Map}(L_E Hmathbf{Z},MU)$. We therefore need to understand $L_E Hmathbf{Z}$. By the discussion at this question, we can conclude that $L_{E_p} Hmathbf{Z} simeq Hmathbf{Q}_p$. It therefore suffices to understand $MU^ast(Hmathbf{Q})$. But $Hmathbf{Q}$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(Hmathbf{Q})$ admits a description in terms of $lim^0$ and $lim^1$ of multiplication by $2,3,5,7,cdots$ on $pi_ast MU$. In particular, the $lim^1$ term is $mathrm{Ext}^1_mathbf{Z}(mathbf{Q,Z}) cong widehat{mathbf{Z}}/mathbf{Z}$.






share|cite|improve this answer



















  • 1




    I don't think that the last bit of this is right. $MU^0(Hmathbb{Q}_p)$ is not a ring. We can write $Hmathbb{Q}=Smathbb{Q}$ as the telescope of multiplication by $2,3,4,5,dotsc$ on $H$ or on $S$. From the first description together with $F(H/n,MU)=0$ we get $F(H,MU)=F(Hmathbb{Q},MU)=F(Smathbb{Q},MU)$. The second description relates $[Smathbb{Q},MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $text{Ext}(mathbb{Q},mathbb{Z})=widehat{mathbb{Z}}/mathbb{Z}$.
    – Neil Strickland
    yesterday










  • @NeilStrickland you're right, thanks! I'll edit my answer.
    – skd
    yesterday






  • 4




    Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $Hmathbb{F}_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcal{A}^*$; this is already enough to show that $MU^*(Hmathbb{F}_p) = 0$. Then the sequence $mathbb{Z} to mathbb{Q} to mathbb{Q}/mathbb{Z}$ tells you that $Y^*mathrm{H}mathbb{Z}$ will always be a humongous sum of $mathrm{Ext}(mathbb{Q}, ?)$'s if $Y^*(Hmathbb{F}_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
    – Dylan Wilson
    yesterday










  • Good to know, thanks @DylanWilson!
    – skd
    yesterday











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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',
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
});


}
});














 

draft saved


draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f315320%2fthe-structure-of-complex-cobordism-cohomology-of-the-eilenberg-maclane-spectrum%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








up vote
11
down vote













One can prove that $mathrm{Map}(Hmathbf{F}_p,MU)$ is contractible. We know that $Hmathbf{F}_p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories"), but $MU$ is harmonic (Theorem 4.2 of that paper). Since dissonant spectra (resp. harmonic spectra) are by definition $E$-acyclic (resp. local) for the spectrum $E = bigvee_p E_p$, where $E_p = bigvee_{0leq n<infty} K(n)$, the claim follows. (I just realized that this is Corollary 4.10 of Ravenel's paper.)



It is, however, not the case that $mathrm{Map}(Hmathbf{Z},MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $Hmathbf{Z}$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrm{Map}(Hmathbf{Z},MU)$ and $mathrm{Map}(L_E Hmathbf{Z},MU)$. We therefore need to understand $L_E Hmathbf{Z}$. By the discussion at this question, we can conclude that $L_{E_p} Hmathbf{Z} simeq Hmathbf{Q}_p$. It therefore suffices to understand $MU^ast(Hmathbf{Q})$. But $Hmathbf{Q}$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(Hmathbf{Q})$ admits a description in terms of $lim^0$ and $lim^1$ of multiplication by $2,3,5,7,cdots$ on $pi_ast MU$. In particular, the $lim^1$ term is $mathrm{Ext}^1_mathbf{Z}(mathbf{Q,Z}) cong widehat{mathbf{Z}}/mathbf{Z}$.






share|cite|improve this answer



















  • 1




    I don't think that the last bit of this is right. $MU^0(Hmathbb{Q}_p)$ is not a ring. We can write $Hmathbb{Q}=Smathbb{Q}$ as the telescope of multiplication by $2,3,4,5,dotsc$ on $H$ or on $S$. From the first description together with $F(H/n,MU)=0$ we get $F(H,MU)=F(Hmathbb{Q},MU)=F(Smathbb{Q},MU)$. The second description relates $[Smathbb{Q},MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $text{Ext}(mathbb{Q},mathbb{Z})=widehat{mathbb{Z}}/mathbb{Z}$.
    – Neil Strickland
    yesterday










  • @NeilStrickland you're right, thanks! I'll edit my answer.
    – skd
    yesterday






  • 4




    Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $Hmathbb{F}_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcal{A}^*$; this is already enough to show that $MU^*(Hmathbb{F}_p) = 0$. Then the sequence $mathbb{Z} to mathbb{Q} to mathbb{Q}/mathbb{Z}$ tells you that $Y^*mathrm{H}mathbb{Z}$ will always be a humongous sum of $mathrm{Ext}(mathbb{Q}, ?)$'s if $Y^*(Hmathbb{F}_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
    – Dylan Wilson
    yesterday










  • Good to know, thanks @DylanWilson!
    – skd
    yesterday















up vote
11
down vote













One can prove that $mathrm{Map}(Hmathbf{F}_p,MU)$ is contractible. We know that $Hmathbf{F}_p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories"), but $MU$ is harmonic (Theorem 4.2 of that paper). Since dissonant spectra (resp. harmonic spectra) are by definition $E$-acyclic (resp. local) for the spectrum $E = bigvee_p E_p$, where $E_p = bigvee_{0leq n<infty} K(n)$, the claim follows. (I just realized that this is Corollary 4.10 of Ravenel's paper.)



It is, however, not the case that $mathrm{Map}(Hmathbf{Z},MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $Hmathbf{Z}$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrm{Map}(Hmathbf{Z},MU)$ and $mathrm{Map}(L_E Hmathbf{Z},MU)$. We therefore need to understand $L_E Hmathbf{Z}$. By the discussion at this question, we can conclude that $L_{E_p} Hmathbf{Z} simeq Hmathbf{Q}_p$. It therefore suffices to understand $MU^ast(Hmathbf{Q})$. But $Hmathbf{Q}$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(Hmathbf{Q})$ admits a description in terms of $lim^0$ and $lim^1$ of multiplication by $2,3,5,7,cdots$ on $pi_ast MU$. In particular, the $lim^1$ term is $mathrm{Ext}^1_mathbf{Z}(mathbf{Q,Z}) cong widehat{mathbf{Z}}/mathbf{Z}$.






share|cite|improve this answer



















  • 1




    I don't think that the last bit of this is right. $MU^0(Hmathbb{Q}_p)$ is not a ring. We can write $Hmathbb{Q}=Smathbb{Q}$ as the telescope of multiplication by $2,3,4,5,dotsc$ on $H$ or on $S$. From the first description together with $F(H/n,MU)=0$ we get $F(H,MU)=F(Hmathbb{Q},MU)=F(Smathbb{Q},MU)$. The second description relates $[Smathbb{Q},MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $text{Ext}(mathbb{Q},mathbb{Z})=widehat{mathbb{Z}}/mathbb{Z}$.
    – Neil Strickland
    yesterday










  • @NeilStrickland you're right, thanks! I'll edit my answer.
    – skd
    yesterday






  • 4




    Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $Hmathbb{F}_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcal{A}^*$; this is already enough to show that $MU^*(Hmathbb{F}_p) = 0$. Then the sequence $mathbb{Z} to mathbb{Q} to mathbb{Q}/mathbb{Z}$ tells you that $Y^*mathrm{H}mathbb{Z}$ will always be a humongous sum of $mathrm{Ext}(mathbb{Q}, ?)$'s if $Y^*(Hmathbb{F}_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
    – Dylan Wilson
    yesterday










  • Good to know, thanks @DylanWilson!
    – skd
    yesterday













up vote
11
down vote










up vote
11
down vote









One can prove that $mathrm{Map}(Hmathbf{F}_p,MU)$ is contractible. We know that $Hmathbf{F}_p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories"), but $MU$ is harmonic (Theorem 4.2 of that paper). Since dissonant spectra (resp. harmonic spectra) are by definition $E$-acyclic (resp. local) for the spectrum $E = bigvee_p E_p$, where $E_p = bigvee_{0leq n<infty} K(n)$, the claim follows. (I just realized that this is Corollary 4.10 of Ravenel's paper.)



It is, however, not the case that $mathrm{Map}(Hmathbf{Z},MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $Hmathbf{Z}$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrm{Map}(Hmathbf{Z},MU)$ and $mathrm{Map}(L_E Hmathbf{Z},MU)$. We therefore need to understand $L_E Hmathbf{Z}$. By the discussion at this question, we can conclude that $L_{E_p} Hmathbf{Z} simeq Hmathbf{Q}_p$. It therefore suffices to understand $MU^ast(Hmathbf{Q})$. But $Hmathbf{Q}$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(Hmathbf{Q})$ admits a description in terms of $lim^0$ and $lim^1$ of multiplication by $2,3,5,7,cdots$ on $pi_ast MU$. In particular, the $lim^1$ term is $mathrm{Ext}^1_mathbf{Z}(mathbf{Q,Z}) cong widehat{mathbf{Z}}/mathbf{Z}$.






share|cite|improve this answer














One can prove that $mathrm{Map}(Hmathbf{F}_p,MU)$ is contractible. We know that $Hmathbf{F}_p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories"), but $MU$ is harmonic (Theorem 4.2 of that paper). Since dissonant spectra (resp. harmonic spectra) are by definition $E$-acyclic (resp. local) for the spectrum $E = bigvee_p E_p$, where $E_p = bigvee_{0leq n<infty} K(n)$, the claim follows. (I just realized that this is Corollary 4.10 of Ravenel's paper.)



It is, however, not the case that $mathrm{Map}(Hmathbf{Z},MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $Hmathbf{Z}$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrm{Map}(Hmathbf{Z},MU)$ and $mathrm{Map}(L_E Hmathbf{Z},MU)$. We therefore need to understand $L_E Hmathbf{Z}$. By the discussion at this question, we can conclude that $L_{E_p} Hmathbf{Z} simeq Hmathbf{Q}_p$. It therefore suffices to understand $MU^ast(Hmathbf{Q})$. But $Hmathbf{Q}$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(Hmathbf{Q})$ admits a description in terms of $lim^0$ and $lim^1$ of multiplication by $2,3,5,7,cdots$ on $pi_ast MU$. In particular, the $lim^1$ term is $mathrm{Ext}^1_mathbf{Z}(mathbf{Q,Z}) cong widehat{mathbf{Z}}/mathbf{Z}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









skd

1,6861623




1,6861623








  • 1




    I don't think that the last bit of this is right. $MU^0(Hmathbb{Q}_p)$ is not a ring. We can write $Hmathbb{Q}=Smathbb{Q}$ as the telescope of multiplication by $2,3,4,5,dotsc$ on $H$ or on $S$. From the first description together with $F(H/n,MU)=0$ we get $F(H,MU)=F(Hmathbb{Q},MU)=F(Smathbb{Q},MU)$. The second description relates $[Smathbb{Q},MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $text{Ext}(mathbb{Q},mathbb{Z})=widehat{mathbb{Z}}/mathbb{Z}$.
    – Neil Strickland
    yesterday










  • @NeilStrickland you're right, thanks! I'll edit my answer.
    – skd
    yesterday






  • 4




    Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $Hmathbb{F}_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcal{A}^*$; this is already enough to show that $MU^*(Hmathbb{F}_p) = 0$. Then the sequence $mathbb{Z} to mathbb{Q} to mathbb{Q}/mathbb{Z}$ tells you that $Y^*mathrm{H}mathbb{Z}$ will always be a humongous sum of $mathrm{Ext}(mathbb{Q}, ?)$'s if $Y^*(Hmathbb{F}_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
    – Dylan Wilson
    yesterday










  • Good to know, thanks @DylanWilson!
    – skd
    yesterday














  • 1




    I don't think that the last bit of this is right. $MU^0(Hmathbb{Q}_p)$ is not a ring. We can write $Hmathbb{Q}=Smathbb{Q}$ as the telescope of multiplication by $2,3,4,5,dotsc$ on $H$ or on $S$. From the first description together with $F(H/n,MU)=0$ we get $F(H,MU)=F(Hmathbb{Q},MU)=F(Smathbb{Q},MU)$. The second description relates $[Smathbb{Q},MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $text{Ext}(mathbb{Q},mathbb{Z})=widehat{mathbb{Z}}/mathbb{Z}$.
    – Neil Strickland
    yesterday










  • @NeilStrickland you're right, thanks! I'll edit my answer.
    – skd
    yesterday






  • 4




    Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $Hmathbb{F}_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcal{A}^*$; this is already enough to show that $MU^*(Hmathbb{F}_p) = 0$. Then the sequence $mathbb{Z} to mathbb{Q} to mathbb{Q}/mathbb{Z}$ tells you that $Y^*mathrm{H}mathbb{Z}$ will always be a humongous sum of $mathrm{Ext}(mathbb{Q}, ?)$'s if $Y^*(Hmathbb{F}_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
    – Dylan Wilson
    yesterday










  • Good to know, thanks @DylanWilson!
    – skd
    yesterday








1




1




I don't think that the last bit of this is right. $MU^0(Hmathbb{Q}_p)$ is not a ring. We can write $Hmathbb{Q}=Smathbb{Q}$ as the telescope of multiplication by $2,3,4,5,dotsc$ on $H$ or on $S$. From the first description together with $F(H/n,MU)=0$ we get $F(H,MU)=F(Hmathbb{Q},MU)=F(Smathbb{Q},MU)$. The second description relates $[Smathbb{Q},MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $text{Ext}(mathbb{Q},mathbb{Z})=widehat{mathbb{Z}}/mathbb{Z}$.
– Neil Strickland
yesterday




I don't think that the last bit of this is right. $MU^0(Hmathbb{Q}_p)$ is not a ring. We can write $Hmathbb{Q}=Smathbb{Q}$ as the telescope of multiplication by $2,3,4,5,dotsc$ on $H$ or on $S$. From the first description together with $F(H/n,MU)=0$ we get $F(H,MU)=F(Hmathbb{Q},MU)=F(Smathbb{Q},MU)$. The second description relates $[Smathbb{Q},MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $text{Ext}(mathbb{Q},mathbb{Z})=widehat{mathbb{Z}}/mathbb{Z}$.
– Neil Strickland
yesterday












@NeilStrickland you're right, thanks! I'll edit my answer.
– skd
yesterday




@NeilStrickland you're right, thanks! I'll edit my answer.
– skd
yesterday




4




4




Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $Hmathbb{F}_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcal{A}^*$; this is already enough to show that $MU^*(Hmathbb{F}_p) = 0$. Then the sequence $mathbb{Z} to mathbb{Q} to mathbb{Q}/mathbb{Z}$ tells you that $Y^*mathrm{H}mathbb{Z}$ will always be a humongous sum of $mathrm{Ext}(mathbb{Q}, ?)$'s if $Y^*(Hmathbb{F}_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
– Dylan Wilson
yesterday




Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $Hmathbb{F}_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcal{A}^*$; this is already enough to show that $MU^*(Hmathbb{F}_p) = 0$. Then the sequence $mathbb{Z} to mathbb{Q} to mathbb{Q}/mathbb{Z}$ tells you that $Y^*mathrm{H}mathbb{Z}$ will always be a humongous sum of $mathrm{Ext}(mathbb{Q}, ?)$'s if $Y^*(Hmathbb{F}_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
– Dylan Wilson
yesterday












Good to know, thanks @DylanWilson!
– skd
yesterday




Good to know, thanks @DylanWilson!
– skd
yesterday


















 

draft saved


draft discarded



















































 


draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f315320%2fthe-structure-of-complex-cobordism-cohomology-of-the-eilenberg-maclane-spectrum%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

QoS: MAC-Priority for clients behind a repeater

Ивакино (Тотемский район)

Can't locate Autom4te/ChannelDefs.pm in @INC (when it definitely is there)