The structure of complex cobordism cohomology of the Eilenberg-Maclane spectrum
$begingroup$
Let $MU$ be the complex bordism spectrum and let $HmathbbZ$ be the Eilenberg-Maclane spectrum.
Is it know what the structure of the complex cobordism cohomology $MU^*(HmathbbZ)$ is?
EDIT: What if instead $HmathbbZ$, one consider $HmathbbZ/(p)$ for a prime $p$?
at.algebraic-topology homotopy-theory cobordism
asked Nov 14 '18 at 19:06
user438991user438991
1378
$endgroup$
add a comment |
$begingroup$
Let $MU$ be the complex bordism spectrum and let $HmathbbZ$ be the Eilenberg-Maclane spectrum.
Is it know what the structure of the complex cobordism cohomology $MU^*(HmathbbZ)$ is?
EDIT: What if instead $HmathbbZ$, one consider $HmathbbZ/(p)$ for a prime $p$?
at.algebraic-topology homotopy-theory cobordism
asked Nov 14 '18 at 19:06
user438991user438991
1378
$endgroup$
$begingroup$
Hrmm.. we know $MU_*HmathbbZ=HmathbbZ_*MU=mathbbZ[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
$endgroup$
– Denis Nardin
Nov 14 '18 at 19:23
2
$begingroup$
If I remember correctly, it is $0$, but I don't remember a reference off the head.
$endgroup$
– user43326
Nov 14 '18 at 19:24
add a comment |
$begingroup$
Let $MU$ be the complex bordism spectrum and let $HmathbbZ$ be the Eilenberg-Maclane spectrum.
Is it know what the structure of the complex cobordism cohomology $MU^*(HmathbbZ)$ is?
EDIT: What if instead $HmathbbZ$, one consider $HmathbbZ/(p)$ for a prime $p$?
at.algebraic-topology homotopy-theory cobordism
asked Nov 14 '18 at 19:06
user438991user438991
1378
$endgroup$
Let $MU$ be the complex bordism spectrum and let $HmathbbZ$ be the Eilenberg-Maclane spectrum.
Is it know what the structure of the complex cobordism cohomology $MU^*(HmathbbZ)$ is?
EDIT: What if instead $HmathbbZ$, one consider $HmathbbZ/(p)$ for a prime $p$?
at.algebraic-topology homotopy-theory cobordism
at.algebraic-topology homotopy-theory cobordism
asked Nov 14 '18 at 19:06
user438991user438991
1378
asked Nov 14 '18 at 19:06
user438991user438991
1378
edited Nov 14 '18 at 19:41
user438991
asked Nov 14 '18 at 19:06
user438991user438991
1378
asked Nov 14 '18 at 19:06
user438991user438991
1378
asked Nov 14 '18 at 19:06
user438991user438991
1378
1378
$begingroup$
Hrmm.. we know $MU_*HmathbbZ=HmathbbZ_*MU=mathbbZ[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
$endgroup$
– Denis Nardin
Nov 14 '18 at 19:23
2
$begingroup$
If I remember correctly, it is $0$, but I don't remember a reference off the head.
$endgroup$
– user43326
Nov 14 '18 at 19:24
add a comment |
$begingroup$
Hrmm.. we know $MU_*HmathbbZ=HmathbbZ_*MU=mathbbZ[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
$endgroup$
– Denis Nardin
Nov 14 '18 at 19:23
2
$begingroup$
If I remember correctly, it is $0$, but I don't remember a reference off the head.
$endgroup$
– user43326
Nov 14 '18 at 19:24
$begingroup$
Hrmm.. we know $MU_*HmathbbZ=HmathbbZ_*MU=mathbbZ[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
$endgroup$
– Denis Nardin
Nov 14 '18 at 19:23
$begingroup$
Hrmm.. we know $MU_*HmathbbZ=HmathbbZ_*MU=mathbbZ[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
$endgroup$
– Denis Nardin
Nov 14 '18 at 19:23
2
2
$begingroup$
If I remember correctly, it is $0$, but I don't remember a reference off the head.
$endgroup$
– user43326
Nov 14 '18 at 19:24
$begingroup$
If I remember correctly, it is $0$, but I don't remember a reference off the head.
$endgroup$
– user43326
Nov 14 '18 at 19:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One can prove that $mathrmMap(HmathbfF_p,MU)$ is contractible. We know that $HmathbfF_p$ 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 $mathrmMap(HmathbfZ,MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $HmathbfZ$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrmMap(HmathbfZ,MU)$ and $mathrmMap(L_E HmathbfZ,MU)$. We therefore need to understand $L_E HmathbfZ$. By the discussion at this question, we can conclude that $L_E_p HmathbfZ simeq HmathbfQ_p$. It therefore suffices to understand $MU^ast(HmathbfQ)$. But $HmathbfQ$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(HmathbfQ)$ 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 $mathrmExt^1_mathbfZ(mathbfQ,Z) cong widehatmathbfZ/mathbfZ$.
answered Nov 14 '18 at 20:24
skdskd
1,8361824
$endgroup$
1
$begingroup$
I don't think that the last bit of this is right. $MU^0(HmathbbQ_p)$ is not a ring. We can write $HmathbbQ=SmathbbQ$ 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 $F(H,MU)=F(HmathbbQ,MU)=F(SmathbbQ,MU)$. The second description relates $[SmathbbQ,MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $textExt(mathbbQ,mathbbZ)=widehatmathbbZ/mathbbZ$.
$endgroup$
– Neil Strickland
Nov 14 '18 at 20:59
$begingroup$
@NeilStrickland you're right, thanks! I'll edit my answer.
$endgroup$
– skd
Nov 14 '18 at 21:04
5
$begingroup$
Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $HmathbbF_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcalA^*$; this is already enough to show that $MU^*(HmathbbF_p) = 0$. Then the sequence $mathbbZ to mathbbQ to mathbbQ/mathbbZ$ tells you that $Y^*mathrmHmathbbZ$ will always be a humongous sum of $mathrmExt(mathbbQ, ?)$'s if $Y^*(HmathbbF_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
$endgroup$
– Dylan Wilson
Nov 14 '18 at 23:21
add a comment |
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',
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%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
$begingroup$
One can prove that $mathrmMap(HmathbfF_p,MU)$ is contractible. We know that $HmathbfF_p$ 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 $mathrmMap(HmathbfZ,MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $HmathbfZ$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrmMap(HmathbfZ,MU)$ and $mathrmMap(L_E HmathbfZ,MU)$. We therefore need to understand $L_E HmathbfZ$. By the discussion at this question, we can conclude that $L_E_p HmathbfZ simeq HmathbfQ_p$. It therefore suffices to understand $MU^ast(HmathbfQ)$. But $HmathbfQ$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(HmathbfQ)$ 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 $mathrmExt^1_mathbfZ(mathbfQ,Z) cong widehatmathbfZ/mathbfZ$.
answered Nov 14 '18 at 20:24
skdskd
1,8361824
$endgroup$
1
$begingroup$
I don't think that the last bit of this is right. $MU^0(HmathbbQ_p)$ is not a ring. We can write $HmathbbQ=SmathbbQ$ 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 $F(H,MU)=F(HmathbbQ,MU)=F(SmathbbQ,MU)$. The second description relates $[SmathbbQ,MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $textExt(mathbbQ,mathbbZ)=widehatmathbbZ/mathbbZ$.
$endgroup$
– Neil Strickland
Nov 14 '18 at 20:59
$begingroup$
@NeilStrickland you're right, thanks! I'll edit my answer.
$endgroup$
– skd
Nov 14 '18 at 21:04
5
$begingroup$
Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $HmathbbF_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcalA^*$; this is already enough to show that $MU^*(HmathbbF_p) = 0$. Then the sequence $mathbbZ to mathbbQ to mathbbQ/mathbbZ$ tells you that $Y^*mathrmHmathbbZ$ will always be a humongous sum of $mathrmExt(mathbbQ, ?)$'s if $Y^*(HmathbbF_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
$endgroup$
– Dylan Wilson
Nov 14 '18 at 23:21
add a comment |
$begingroup$
One can prove that $mathrmMap(HmathbfF_p,MU)$ is contractible. We know that $HmathbfF_p$ 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 $mathrmMap(HmathbfZ,MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $HmathbfZ$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrmMap(HmathbfZ,MU)$ and $mathrmMap(L_E HmathbfZ,MU)$. We therefore need to understand $L_E HmathbfZ$. By the discussion at this question, we can conclude that $L_E_p HmathbfZ simeq HmathbfQ_p$. It therefore suffices to understand $MU^ast(HmathbfQ)$. But $HmathbfQ$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(HmathbfQ)$ 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 $mathrmExt^1_mathbfZ(mathbfQ,Z) cong widehatmathbfZ/mathbfZ$.
answered Nov 14 '18 at 20:24
skdskd
1,8361824
$endgroup$
1
$begingroup$
I don't think that the last bit of this is right. $MU^0(HmathbbQ_p)$ is not a ring. We can write $HmathbbQ=SmathbbQ$ 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 $F(H,MU)=F(HmathbbQ,MU)=F(SmathbbQ,MU)$. The second description relates $[SmathbbQ,MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $textExt(mathbbQ,mathbbZ)=widehatmathbbZ/mathbbZ$.
$endgroup$
– Neil Strickland
Nov 14 '18 at 20:59
$begingroup$
@NeilStrickland you're right, thanks! I'll edit my answer.
$endgroup$
– skd
Nov 14 '18 at 21:04
5
$begingroup$
Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $HmathbbF_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcalA^*$; this is already enough to show that $MU^*(HmathbbF_p) = 0$. Then the sequence $mathbbZ to mathbbQ to mathbbQ/mathbbZ$ tells you that $Y^*mathrmHmathbbZ$ will always be a humongous sum of $mathrmExt(mathbbQ, ?)$'s if $Y^*(HmathbbF_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
$endgroup$
– Dylan Wilson
Nov 14 '18 at 23:21
add a comment |
$begingroup$
One can prove that $mathrmMap(HmathbfF_p,MU)$ is contractible. We know that $HmathbfF_p$ 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 $mathrmMap(HmathbfZ,MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $HmathbfZ$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrmMap(HmathbfZ,MU)$ and $mathrmMap(L_E HmathbfZ,MU)$. We therefore need to understand $L_E HmathbfZ$. By the discussion at this question, we can conclude that $L_E_p HmathbfZ simeq HmathbfQ_p$. It therefore suffices to understand $MU^ast(HmathbfQ)$. But $HmathbfQ$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(HmathbfQ)$ 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 $mathrmExt^1_mathbfZ(mathbfQ,Z) cong widehatmathbfZ/mathbfZ$.
answered Nov 14 '18 at 20:24
skdskd
1,8361824
$endgroup$
One can prove that $mathrmMap(HmathbfF_p,MU)$ is contractible. We know that $HmathbfF_p$ 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 $mathrmMap(HmathbfZ,MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $HmathbfZ$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrmMap(HmathbfZ,MU)$ and $mathrmMap(L_E HmathbfZ,MU)$. We therefore need to understand $L_E HmathbfZ$. By the discussion at this question, we can conclude that $L_E_p HmathbfZ simeq HmathbfQ_p$. It therefore suffices to understand $MU^ast(HmathbfQ)$. But $HmathbfQ$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(HmathbfQ)$ 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 $mathrmExt^1_mathbfZ(mathbfQ,Z) cong widehatmathbfZ/mathbfZ$.
answered Nov 14 '18 at 20:24
skdskd
1,8361824
edited Nov 14 '18 at 21:12
answered Nov 14 '18 at 20:24
skdskd
1,8361824
answered Nov 14 '18 at 20:24
skdskd
1,8361824
answered Nov 14 '18 at 20:24
skdskd
1,8361824
1,8361824
1
$begingroup$
I don't think that the last bit of this is right. $MU^0(HmathbbQ_p)$ is not a ring. We can write $HmathbbQ=SmathbbQ$ 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 $F(H,MU)=F(HmathbbQ,MU)=F(SmathbbQ,MU)$. The second description relates $[SmathbbQ,MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $textExt(mathbbQ,mathbbZ)=widehatmathbbZ/mathbbZ$.
$endgroup$
– Neil Strickland
Nov 14 '18 at 20:59
$begingroup$
@NeilStrickland you're right, thanks! I'll edit my answer.
$endgroup$
– skd
Nov 14 '18 at 21:04
5
$begingroup$
Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $HmathbbF_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcalA^*$; this is already enough to show that $MU^*(HmathbbF_p) = 0$. Then the sequence $mathbbZ to mathbbQ to mathbbQ/mathbbZ$ tells you that $Y^*mathrmHmathbbZ$ will always be a humongous sum of $mathrmExt(mathbbQ, ?)$'s if $Y^*(HmathbbF_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
$endgroup$
– Dylan Wilson
Nov 14 '18 at 23:21
add a comment |
1
$begingroup$
I don't think that the last bit of this is right. $MU^0(HmathbbQ_p)$ is not a ring. We can write $HmathbbQ=SmathbbQ$ 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 $F(H,MU)=F(HmathbbQ,MU)=F(SmathbbQ,MU)$. The second description relates $[SmathbbQ,MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $textExt(mathbbQ,mathbbZ)=widehatmathbbZ/mathbbZ$.
$endgroup$
– Neil Strickland
Nov 14 '18 at 20:59
$begingroup$
@NeilStrickland you're right, thanks! I'll edit my answer.
$endgroup$
– skd
Nov 14 '18 at 21:04
5
$begingroup$
Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $HmathbbF_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcalA^*$; this is already enough to show that $MU^*(HmathbbF_p) = 0$. Then the sequence $mathbbZ to mathbbQ to mathbbQ/mathbbZ$ tells you that $Y^*mathrmHmathbbZ$ will always be a humongous sum of $mathrmExt(mathbbQ, ?)$'s if $Y^*(HmathbbF_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
$endgroup$
– Dylan Wilson
Nov 14 '18 at 23:21
1
1
$begingroup$
I don't think that the last bit of this is right. $MU^0(HmathbbQ_p)$ is not a ring. We can write $HmathbbQ=SmathbbQ$ 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 $F(H,MU)=F(HmathbbQ,MU)=F(SmathbbQ,MU)$. The second description relates $[SmathbbQ,MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $textExt(mathbbQ,mathbbZ)=widehatmathbbZ/mathbbZ$.
$endgroup$
– Neil Strickland
Nov 14 '18 at 20:59
$begingroup$
I don't think that the last bit of this is right. $MU^0(HmathbbQ_p)$ is not a ring. We can write $HmathbbQ=SmathbbQ$ 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 $F(H,MU)=F(HmathbbQ,MU)=F(SmathbbQ,MU)$. The second description relates $[SmathbbQ,MU]_*$ to $lim^0$ and $lim^1$ of multiplication by $2,3,4,dotsc$ on $pi_*(MU)$. Here $lim^0=0$ but $lim^1$ involves $textExt(mathbbQ,mathbbZ)=widehatmathbbZ/mathbbZ$.
$endgroup$
– Neil Strickland
Nov 14 '18 at 20:59
$begingroup$
@NeilStrickland you're right, thanks! I'll edit my answer.
$endgroup$
– skd
Nov 14 '18 at 21:04
$begingroup$
@NeilStrickland you're right, thanks! I'll edit my answer.
$endgroup$
– skd
Nov 14 '18 at 21:04
5
5
$begingroup$
Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $HmathbbF_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcalA^*$; this is already enough to show that $MU^*(HmathbbF_p) = 0$. Then the sequence $mathbbZ to mathbbQ to mathbbQ/mathbbZ$ tells you that $Y^*mathrmHmathbbZ$ will always be a humongous sum of $mathrmExt(mathbbQ, ?)$'s if $Y^*(HmathbbF_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
$endgroup$
– Dylan Wilson
Nov 14 '18 at 23:21
$begingroup$
Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $HmathbbF_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $mathcalA^*$; this is already enough to show that $MU^*(HmathbbF_p) = 0$. Then the sequence $mathbbZ to mathbbQ to mathbbQ/mathbbZ$ tells you that $Y^*mathrmHmathbbZ$ will always be a humongous sum of $mathrmExt(mathbbQ, ?)$'s if $Y^*(HmathbbF_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this.
$endgroup$
– Dylan Wilson
Nov 14 '18 at 23:21
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%2f315320%2fthe-structure-of-complex-cobordism-cohomology-of-the-eilenberg-maclane-spectrum%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
$begingroup$
Hrmm.. we know $MU_*HmathbbZ=HmathbbZ_*MU=mathbbZ[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated.
$endgroup$
– Denis Nardin
Nov 14 '18 at 19:23
2
$begingroup$
If I remember correctly, it is $0$, but I don't remember a reference off the head.
$endgroup$
– user43326
Nov 14 '18 at 19:24