Mixing
The complex topological $K$-theory spectrum is not an $Hmathbb{Z}$-module.
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Why is it true that the complex topological $K$-theory spectrum is not an $Hmathbb{Z}$-module? I mean, why the nontriviality of the first $k$-invariant implies the claim above?
algebraic-topology k-theory topological-k-theory
asked Jan 30 at 19:28
user438991user438991
16016
$endgroup$
add a comment |
$begingroup$
Why is it true that the complex topological $K$-theory spectrum is not an $Hmathbb{Z}$-module? I mean, why the nontriviality of the first $k$-invariant implies the claim above?
algebraic-topology k-theory topological-k-theory
asked Jan 30 at 19:28
user438991user438991
16016
$endgroup$
add a comment |
$begingroup$
Why is it true that the complex topological $K$-theory spectrum is not an $Hmathbb{Z}$-module? I mean, why the nontriviality of the first $k$-invariant implies the claim above?
algebraic-topology k-theory topological-k-theory
asked Jan 30 at 19:28
user438991user438991
16016
$endgroup$
Why is it true that the complex topological $K$-theory spectrum is not an $Hmathbb{Z}$-module? I mean, why the nontriviality of the first $k$-invariant implies the claim above?
algebraic-topology k-theory topological-k-theory
algebraic-topology k-theory topological-k-theory
asked Jan 30 at 19:28
user438991user438991
16016
asked Jan 30 at 19:28
user438991user438991
16016
asked Jan 30 at 19:28
user438991user438991
16016
asked Jan 30 at 19:28
user438991user438991
16016
asked Jan 30 at 19:28
user438991user438991
16016
16016
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $E$ be a $Hmathbb{Z}$-module spectrum and consider the smash $Hmathbb{Z}wedge E$. I claim in this case that it is the wedge
$$Hmathbb{Z}wedge Esimeq bigvee_{kinmathbb{Z}} Sigma^kHG_k$$
where $G_k=H_k(E)$. To see this, for each $kinmathbb{Z}$ take the Moore spectrum $Sigma^kMG_k$ and a map $alpha_k:Sigma^kMG_krightarrow Hmathbb{Z}wedge E$ which induces an isomorphism
$$alpha_k:pi_k(HG_k)cong H_k(E)xrightarrowcongpi_k(Hmathbb{Z}wedge E)=H_k(E).$$
It is a standard result that such a map should exist, and it is indeed easy to construct one explicitly starting with a given presentation of $MG_k$ as a cofiber of a map between wedges of spheres.
Now fix $k$ and consider the map
$$beta_k:Hmathbb{Z}wedgeSigma^kMG_kxrightarrow{1wedgealpha_k}Hmathbb{Z}wedge Hmathbb{Z}wedge Exrightarrow{muwedge 1}Hmathbb{Z}wedge E$$
where $mu$ is the ring-spectrum product on $Hmathbb{Z}$. It is clear from the definitions that this map induces an isomorphism on $pi_k$, and the trivial homomorphism on $pi_l$ for $lneq k$. Therefore, when we define
$$bigvee_{kinmathbb{Z}} Sigma^kHG_kxrightarrow{bigveebeta_k}Hmathbb{Z}wedge E$$
we easily see that it is a weak equivalence, and we get our claim.
The point is that an $Hmathbb{Z}$-module spectrum is a GES (generalised Eilenberg-Mac Lane Spectrum, i.e. a wedge of suspensions of EM-spectra) and in particular has no non-trivial Postnikov invariants. On the other hand, the K-theory spectrum $KU$ indeed does have non-trivial Postnikov invariants. Hence it is not a GES, and so not an $Hmathbb{Z}$-module.
answered Jan 31 at 11:14
TyroneTyrone
5,31711226
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1 Answer
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1 Answer
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$begingroup$
Let $E$ be a $Hmathbb{Z}$-module spectrum and consider the smash $Hmathbb{Z}wedge E$. I claim in this case that it is the wedge
$$Hmathbb{Z}wedge Esimeq bigvee_{kinmathbb{Z}} Sigma^kHG_k$$
where $G_k=H_k(E)$. To see this, for each $kinmathbb{Z}$ take the Moore spectrum $Sigma^kMG_k$ and a map $alpha_k:Sigma^kMG_krightarrow Hmathbb{Z}wedge E$ which induces an isomorphism
$$alpha_k:pi_k(HG_k)cong H_k(E)xrightarrowcongpi_k(Hmathbb{Z}wedge E)=H_k(E).$$
It is a standard result that such a map should exist, and it is indeed easy to construct one explicitly starting with a given presentation of $MG_k$ as a cofiber of a map between wedges of spheres.
Now fix $k$ and consider the map
$$beta_k:Hmathbb{Z}wedgeSigma^kMG_kxrightarrow{1wedgealpha_k}Hmathbb{Z}wedge Hmathbb{Z}wedge Exrightarrow{muwedge 1}Hmathbb{Z}wedge E$$
where $mu$ is the ring-spectrum product on $Hmathbb{Z}$. It is clear from the definitions that this map induces an isomorphism on $pi_k$, and the trivial homomorphism on $pi_l$ for $lneq k$. Therefore, when we define
$$bigvee_{kinmathbb{Z}} Sigma^kHG_kxrightarrow{bigveebeta_k}Hmathbb{Z}wedge E$$
we easily see that it is a weak equivalence, and we get our claim.
The point is that an $Hmathbb{Z}$-module spectrum is a GES (generalised Eilenberg-Mac Lane Spectrum, i.e. a wedge of suspensions of EM-spectra) and in particular has no non-trivial Postnikov invariants. On the other hand, the K-theory spectrum $KU$ indeed does have non-trivial Postnikov invariants. Hence it is not a GES, and so not an $Hmathbb{Z}$-module.
answered Jan 31 at 11:14
TyroneTyrone
5,31711226
$endgroup$
add a comment |
$begingroup$
Let $E$ be a $Hmathbb{Z}$-module spectrum and consider the smash $Hmathbb{Z}wedge E$. I claim in this case that it is the wedge
$$Hmathbb{Z}wedge Esimeq bigvee_{kinmathbb{Z}} Sigma^kHG_k$$
where $G_k=H_k(E)$. To see this, for each $kinmathbb{Z}$ take the Moore spectrum $Sigma^kMG_k$ and a map $alpha_k:Sigma^kMG_krightarrow Hmathbb{Z}wedge E$ which induces an isomorphism
$$alpha_k:pi_k(HG_k)cong H_k(E)xrightarrowcongpi_k(Hmathbb{Z}wedge E)=H_k(E).$$
It is a standard result that such a map should exist, and it is indeed easy to construct one explicitly starting with a given presentation of $MG_k$ as a cofiber of a map between wedges of spheres.
Now fix $k$ and consider the map
$$beta_k:Hmathbb{Z}wedgeSigma^kMG_kxrightarrow{1wedgealpha_k}Hmathbb{Z}wedge Hmathbb{Z}wedge Exrightarrow{muwedge 1}Hmathbb{Z}wedge E$$
where $mu$ is the ring-spectrum product on $Hmathbb{Z}$. It is clear from the definitions that this map induces an isomorphism on $pi_k$, and the trivial homomorphism on $pi_l$ for $lneq k$. Therefore, when we define
$$bigvee_{kinmathbb{Z}} Sigma^kHG_kxrightarrow{bigveebeta_k}Hmathbb{Z}wedge E$$
we easily see that it is a weak equivalence, and we get our claim.
The point is that an $Hmathbb{Z}$-module spectrum is a GES (generalised Eilenberg-Mac Lane Spectrum, i.e. a wedge of suspensions of EM-spectra) and in particular has no non-trivial Postnikov invariants. On the other hand, the K-theory spectrum $KU$ indeed does have non-trivial Postnikov invariants. Hence it is not a GES, and so not an $Hmathbb{Z}$-module.
answered Jan 31 at 11:14
TyroneTyrone
5,31711226
$endgroup$
add a comment |
$begingroup$
Let $E$ be a $Hmathbb{Z}$-module spectrum and consider the smash $Hmathbb{Z}wedge E$. I claim in this case that it is the wedge
$$Hmathbb{Z}wedge Esimeq bigvee_{kinmathbb{Z}} Sigma^kHG_k$$
where $G_k=H_k(E)$. To see this, for each $kinmathbb{Z}$ take the Moore spectrum $Sigma^kMG_k$ and a map $alpha_k:Sigma^kMG_krightarrow Hmathbb{Z}wedge E$ which induces an isomorphism
$$alpha_k:pi_k(HG_k)cong H_k(E)xrightarrowcongpi_k(Hmathbb{Z}wedge E)=H_k(E).$$
It is a standard result that such a map should exist, and it is indeed easy to construct one explicitly starting with a given presentation of $MG_k$ as a cofiber of a map between wedges of spheres.
Now fix $k$ and consider the map
$$beta_k:Hmathbb{Z}wedgeSigma^kMG_kxrightarrow{1wedgealpha_k}Hmathbb{Z}wedge Hmathbb{Z}wedge Exrightarrow{muwedge 1}Hmathbb{Z}wedge E$$
where $mu$ is the ring-spectrum product on $Hmathbb{Z}$. It is clear from the definitions that this map induces an isomorphism on $pi_k$, and the trivial homomorphism on $pi_l$ for $lneq k$. Therefore, when we define
$$bigvee_{kinmathbb{Z}} Sigma^kHG_kxrightarrow{bigveebeta_k}Hmathbb{Z}wedge E$$
we easily see that it is a weak equivalence, and we get our claim.
The point is that an $Hmathbb{Z}$-module spectrum is a GES (generalised Eilenberg-Mac Lane Spectrum, i.e. a wedge of suspensions of EM-spectra) and in particular has no non-trivial Postnikov invariants. On the other hand, the K-theory spectrum $KU$ indeed does have non-trivial Postnikov invariants. Hence it is not a GES, and so not an $Hmathbb{Z}$-module.
answered Jan 31 at 11:14
TyroneTyrone
5,31711226
$endgroup$
Let $E$ be a $Hmathbb{Z}$-module spectrum and consider the smash $Hmathbb{Z}wedge E$. I claim in this case that it is the wedge
$$Hmathbb{Z}wedge Esimeq bigvee_{kinmathbb{Z}} Sigma^kHG_k$$
where $G_k=H_k(E)$. To see this, for each $kinmathbb{Z}$ take the Moore spectrum $Sigma^kMG_k$ and a map $alpha_k:Sigma^kMG_krightarrow Hmathbb{Z}wedge E$ which induces an isomorphism
$$alpha_k:pi_k(HG_k)cong H_k(E)xrightarrowcongpi_k(Hmathbb{Z}wedge E)=H_k(E).$$
It is a standard result that such a map should exist, and it is indeed easy to construct one explicitly starting with a given presentation of $MG_k$ as a cofiber of a map between wedges of spheres.
Now fix $k$ and consider the map
$$beta_k:Hmathbb{Z}wedgeSigma^kMG_kxrightarrow{1wedgealpha_k}Hmathbb{Z}wedge Hmathbb{Z}wedge Exrightarrow{muwedge 1}Hmathbb{Z}wedge E$$
where $mu$ is the ring-spectrum product on $Hmathbb{Z}$. It is clear from the definitions that this map induces an isomorphism on $pi_k$, and the trivial homomorphism on $pi_l$ for $lneq k$. Therefore, when we define
$$bigvee_{kinmathbb{Z}} Sigma^kHG_kxrightarrow{bigveebeta_k}Hmathbb{Z}wedge E$$
we easily see that it is a weak equivalence, and we get our claim.
The point is that an $Hmathbb{Z}$-module spectrum is a GES (generalised Eilenberg-Mac Lane Spectrum, i.e. a wedge of suspensions of EM-spectra) and in particular has no non-trivial Postnikov invariants. On the other hand, the K-theory spectrum $KU$ indeed does have non-trivial Postnikov invariants. Hence it is not a GES, and so not an $Hmathbb{Z}$-module.
answered Jan 31 at 11:14
TyroneTyrone
5,31711226
answered Jan 31 at 11:14
TyroneTyrone
5,31711226
answered Jan 31 at 11:14
TyroneTyrone
5,31711226
answered Jan 31 at 11:14
TyroneTyrone
5,31711226
5,31711226
add a comment |
add a comment |
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