Is a spectrum with trivial homology groups trivial?












2














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







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    2














    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







    share|cite|improve this question



























      2












      2








      2







      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







      share|cite|improve this question















      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






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      share|cite|improve this question













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      edited 23 mins ago









      YCor

      27.1k380132




      27.1k380132










      asked 1 hour ago









      user09127

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      3066






















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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 $K(n)neq0$.






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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











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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 $K(n)neq0$.






          share|cite|improve this answer








          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
















          5














          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 $K(n)neq0$.






          share|cite|improve this answer








          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














          5












          5








          5






          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 $K(n)neq0$.






          share|cite|improve this answer








          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 $K(n)neq0$.







          share|cite|improve this answer








          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.









          share|cite|improve this answer



          share|cite|improve this answer






          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

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          511




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          New contributor





          Christian Carrick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          Christian Carrick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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


















          • 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


















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