Is a spectrum with trivial homology groups trivial?
$begingroup$
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 Jan 4 at 10:31
YCor
28.2k483136
asked Jan 4 at 9:03
user09127user09127
3958
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 10:31
YCor
28.2k483136
asked Jan 4 at 9:03
user09127user09127
3958
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 10:31
YCor
28.2k483136
asked Jan 4 at 9:03
user09127user09127
3958
$endgroup$
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 Jan 4 at 10:31
YCor
28.2k483136
asked Jan 4 at 9:03
user09127user09127
3958
edited Jan 4 at 10:31
YCor
28.2k483136
asked Jan 4 at 9:03
user09127user09127
3958
edited Jan 4 at 10:31
YCor
28.2k483136
edited Jan 4 at 10:31
YCor
28.2k483136
edited Jan 4 at 10:31
YCor
28.2k483136
28.2k483136
asked Jan 4 at 9:03
user09127user09127
3958
asked Jan 4 at 9:03
user09127user09127
3958
asked Jan 4 at 9:03
user09127user09127
3958
3958
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 Jan 4 at 9:38
Christian CarrickChristian Carrick
32612
$endgroup$
$begingroup$
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
$endgroup$
– user09127
Jan 4 at 9:45
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 Jan 4 at 9:38
Christian CarrickChristian Carrick
32612
$endgroup$
$begingroup$
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
$endgroup$
– user09127
Jan 4 at 9:45
add a comment |
$begingroup$
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 Jan 4 at 9:38
Christian CarrickChristian Carrick
32612
$endgroup$
$begingroup$
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
$endgroup$
– user09127
Jan 4 at 9:45
add a comment |
$begingroup$
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 Jan 4 at 9:38
Christian CarrickChristian Carrick
32612
$endgroup$
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 Jan 4 at 9:38
Christian CarrickChristian Carrick
32612
answered Jan 4 at 9:38
Christian CarrickChristian Carrick
32612
answered Jan 4 at 9:38
Christian CarrickChristian Carrick
32612
answered Jan 4 at 9:38
Christian CarrickChristian Carrick
32612
32612
$begingroup$
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
$endgroup$
– user09127
Jan 4 at 9:45
add a comment |
$begingroup$
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
$endgroup$
– user09127
Jan 4 at 9:45
$begingroup$
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
$endgroup$
– user09127
Jan 4 at 9:45
$begingroup$
That's what I thought, thanks for the answer. Can you say something about the functor $L_{Hmathbb{Z}}$?
$endgroup$
– user09127
Jan 4 at 9:45
add a comment |
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