Is a spectrum with trivial homology groups trivial?
If $X$ is a spectrum with trivial (integer-valued) homology groups, does it have to be weakly-equivalent to a point?
This is easy to prove for connective spectrum, as a Hurewitz-type argument is then possible, but what about the general case?
Furthermore, if this is not the case, how should I think of the functor $L_{Hmathbb{Z}}$ (Bousfield-localization at the spectrum EM spectrum $Hmathbb{Z}$)?
at.algebraic-topology stable-homotopy bousfield-localization
edited 23 mins ago
YCor
27.1k380132
asked 1 hour ago
user09127
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add a comment |
If $X$ is a spectrum with trivial (integer-valued) homology groups, does it have to be weakly-equivalent to a point?
This is easy to prove for connective spectrum, as a Hurewitz-type argument is then possible, but what about the general case?
Furthermore, if this is not the case, how should I think of the functor $L_{Hmathbb{Z}}$ (Bousfield-localization at the spectrum EM spectrum $Hmathbb{Z}$)?
at.algebraic-topology stable-homotopy bousfield-localization
edited 23 mins ago
YCor
27.1k380132
asked 1 hour ago
user09127
3066
add a comment |
If $X$ is a spectrum with trivial (integer-valued) homology groups, does it have to be weakly-equivalent to a point?
This is easy to prove for connective spectrum, as a Hurewitz-type argument is then possible, but what about the general case?
Furthermore, if this is not the case, how should I think of the functor $L_{Hmathbb{Z}}$ (Bousfield-localization at the spectrum EM spectrum $Hmathbb{Z}$)?
at.algebraic-topology stable-homotopy bousfield-localization
edited 23 mins ago
YCor
27.1k380132
asked 1 hour ago
user09127
3066
If $X$ is a spectrum with trivial (integer-valued) homology groups, does it have to be weakly-equivalent to a point?
This is easy to prove for connective spectrum, as a Hurewitz-type argument is then possible, but what about the general case?
Furthermore, if this is not the case, how should I think of the functor $L_{Hmathbb{Z}}$ (Bousfield-localization at the spectrum EM spectrum $Hmathbb{Z}$)?
at.algebraic-topology stable-homotopy bousfield-localization
at.algebraic-topology stable-homotopy bousfield-localization
edited 23 mins ago
YCor
27.1k380132
asked 1 hour ago
user09127
3066
edited 23 mins ago
YCor
27.1k380132
asked 1 hour ago
user09127
3066
edited 23 mins ago
YCor
27.1k380132
edited 23 mins ago
YCor
27.1k380132
edited 23 mins ago
YCor
27.1k380132
27.1k380132
asked 1 hour ago
user09127
3066
asked 1 hour ago
user09127
3066
asked 1 hour ago
user09127
3066
3066
add a comment |
add a comment |
1 Answer
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If $K(n)$ is the $n$-th Morava K-theory for $n>0$, then $K(n)otimes Hmathbb{Z}=0$ because, via the 2 complex orientations of $K(n)otimes Hmathbb{Z}$, there are two formal groups over the ring $pi_*(K(n)otimes Hmathbb{Z})$ and an isomorphism between them. But one has height 0 (additive formal group from $Hmathbb{Z}$) and the other has height $n>0$ (coming from $K(n)$). This is impossible unless $pi_*(K(n)otimes Hmathbb{Z})=0$. But of course $K(n)neq0$.
answered 1 hour ago
Christian Carrick
511
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That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
1 hour ago
add a comment |
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1 Answer
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If $K(n)$ is the $n$-th Morava K-theory for $n>0$, then $K(n)otimes Hmathbb{Z}=0$ because, via the 2 complex orientations of $K(n)otimes Hmathbb{Z}$, there are two formal groups over the ring $pi_*(K(n)otimes Hmathbb{Z})$ and an isomorphism between them. But one has height 0 (additive formal group from $Hmathbb{Z}$) and the other has height $n>0$ (coming from $K(n)$). This is impossible unless $pi_*(K(n)otimes Hmathbb{Z})=0$. But of course $K(n)neq0$.
answered 1 hour ago
Christian Carrick
511
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Christian Carrick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
1 hour ago
add a comment |
If $K(n)$ is the $n$-th Morava K-theory for $n>0$, then $K(n)otimes Hmathbb{Z}=0$ because, via the 2 complex orientations of $K(n)otimes Hmathbb{Z}$, there are two formal groups over the ring $pi_*(K(n)otimes Hmathbb{Z})$ and an isomorphism between them. But one has height 0 (additive formal group from $Hmathbb{Z}$) and the other has height $n>0$ (coming from $K(n)$). This is impossible unless $pi_*(K(n)otimes Hmathbb{Z})=0$. But of course $K(n)neq0$.
answered 1 hour ago
Christian Carrick
511
New contributor
Christian Carrick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
1 hour ago
add a comment |
If $K(n)$ is the $n$-th Morava K-theory for $n>0$, then $K(n)otimes Hmathbb{Z}=0$ because, via the 2 complex orientations of $K(n)otimes Hmathbb{Z}$, there are two formal groups over the ring $pi_*(K(n)otimes Hmathbb{Z})$ and an isomorphism between them. But one has height 0 (additive formal group from $Hmathbb{Z}$) and the other has height $n>0$ (coming from $K(n)$). This is impossible unless $pi_*(K(n)otimes Hmathbb{Z})=0$. But of course $K(n)neq0$.
answered 1 hour ago
Christian Carrick
511
New contributor
Christian Carrick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If $K(n)$ is the $n$-th Morava K-theory for $n>0$, then $K(n)otimes Hmathbb{Z}=0$ because, via the 2 complex orientations of $K(n)otimes Hmathbb{Z}$, there are two formal groups over the ring $pi_*(K(n)otimes Hmathbb{Z})$ and an isomorphism between them. But one has height 0 (additive formal group from $Hmathbb{Z}$) and the other has height $n>0$ (coming from $K(n)$). This is impossible unless $pi_*(K(n)otimes Hmathbb{Z})=0$. But of course $K(n)neq0$.
answered 1 hour ago
Christian Carrick
511
New contributor
Christian Carrick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Christian Carrick
511
New contributor
Christian Carrick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Christian Carrick
511
answered 1 hour ago
Christian Carrick
511
511
New contributor
Christian Carrick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Christian Carrick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Christian Carrick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
1 hour ago
add a comment |
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
1 hour ago
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
1 hour ago
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
1 hour ago
add a comment |
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