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 yesterday
YCor
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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 yesterday
YCor
27.1k380132
asked yesterday
user09127
3618
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 yesterday
YCor
27.1k380132
asked yesterday
user09127
3618
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 yesterday
YCor
27.1k380132
asked yesterday
user09127
3618
edited yesterday
YCor
27.1k380132
asked yesterday
user09127
3618
edited yesterday
YCor
27.1k380132
edited yesterday
YCor
27.1k380132
edited yesterday
YCor
27.1k380132
27.1k380132
asked yesterday
user09127
3618
asked yesterday
user09127
3618
asked yesterday
user09127
3618
3618
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 yesterday
Christian Carrick
1812
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That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
yesterday
add a comment |
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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 yesterday
Christian Carrick
1812
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
yesterday
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 yesterday
Christian Carrick
1812
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
yesterday
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 yesterday
Christian Carrick
1812
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 yesterday
Christian Carrick
1812
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 yesterday
Christian Carrick
1812
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 yesterday
Christian Carrick
1812
answered yesterday
Christian Carrick
1812
1812
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
yesterday
add a comment |
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
yesterday
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
yesterday
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
– user09127
yesterday
add a comment |
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