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
270110
$endgroup$
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
270110
$endgroup$
$begingroup$
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
270110
$endgroup$
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
270110
asked Nov 14 '18 at 19:06
user438991user438991
270110
edited Nov 14 '18 at 19:41
user438991
asked Nov 14 '18 at 19:06
user438991user438991
270110
asked Nov 14 '18 at 19:06
user438991user438991
270110
asked Nov 14 '18 at 19:06
user438991user438991
270110
270110
$begingroup$
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
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oldest
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$begingroup$
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
1,9061824
$endgroup$
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 |
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$begingroup$
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
1,9061824
$endgroup$
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
1,9061824
$endgroup$
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
1,9061824
$endgroup$
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
1,9061824
edited Nov 14 '18 at 21:12
answered Nov 14 '18 at 20:24
skdskd
1,9061824
answered Nov 14 '18 at 20:24
skdskd
1,9061824
answered Nov 14 '18 at 20:24
skdskd
1,9061824
1,9061824
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 |
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$begingroup$
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