Inverting homotopy groups of spectra












7












$begingroup$


Let $X$ be a spectrum. Is there a canonical construction/functor that would associate to this spectrum, an inverse spectrum $X'$, in the sense that $$pi_*(X)cong pi_{-*}(X')?$$



To be more precise, such a spectrum $X'$ can always be constructed by attaching cells to produce the right homotopy groups, but is there a more conceptual way of creating it?






at.algebraic-topology stable-homotopy







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$endgroup$

















    7












    $begingroup$


    Let $X$ be a spectrum. Is there a canonical construction/functor that would associate to this spectrum, an inverse spectrum $X'$, in the sense that $$pi_*(X)cong pi_{-*}(X')?$$



    To be more precise, such a spectrum $X'$ can always be constructed by attaching cells to produce the right homotopy groups, but is there a more conceptual way of creating it?






    at.algebraic-topology stable-homotopy







    share|cite|improve this question











    $endgroup$















      7












      7








      7


      2



      $begingroup$


      Let $X$ be a spectrum. Is there a canonical construction/functor that would associate to this spectrum, an inverse spectrum $X'$, in the sense that $$pi_*(X)cong pi_{-*}(X')?$$



      To be more precise, such a spectrum $X'$ can always be constructed by attaching cells to produce the right homotopy groups, but is there a more conceptual way of creating it?






      at.algebraic-topology stable-homotopy







      share|cite|improve this question











      $endgroup$




      Let $X$ be a spectrum. Is there a canonical construction/functor that would associate to this spectrum, an inverse spectrum $X'$, in the sense that $$pi_*(X)cong pi_{-*}(X')?$$



      To be more precise, such a spectrum $X'$ can always be constructed by attaching cells to produce the right homotopy groups, but is there a more conceptual way of creating it?






      at.algebraic-topology stable-homotopy




      at.algebraic-topology stable-homotopy






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 22 '18 at 18:24









      David White

      11.5k460101




      11.5k460101










      asked Dec 22 '18 at 16:56









      user09127user09127

      3888




      3888






















          1 Answer
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          15












          $begingroup$

          Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{mathbb{Q}/mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):



          $pi_{-ast}(I_{mathbb{Q}/mathbb{Z}}X) simeq Hom(pi_ast(X),mathbb{Q}/mathbb{Z})$



          So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.



          See here for more details






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
            $endgroup$
            – André Henriques
            Dec 23 '18 at 21:42








          • 1




            $begingroup$
            @AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
            $endgroup$
            – skd
            Dec 24 '18 at 5:47








          • 1




            $begingroup$
            @skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
            $endgroup$
            – André Henriques
            Dec 24 '18 at 16:18











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          1 Answer
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          1 Answer
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          active

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          active

          oldest

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          15












          $begingroup$

          Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{mathbb{Q}/mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):



          $pi_{-ast}(I_{mathbb{Q}/mathbb{Z}}X) simeq Hom(pi_ast(X),mathbb{Q}/mathbb{Z})$



          So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.



          See here for more details






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
            $endgroup$
            – André Henriques
            Dec 23 '18 at 21:42








          • 1




            $begingroup$
            @AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
            $endgroup$
            – skd
            Dec 24 '18 at 5:47








          • 1




            $begingroup$
            @skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
            $endgroup$
            – André Henriques
            Dec 24 '18 at 16:18
















          15












          $begingroup$

          Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{mathbb{Q}/mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):



          $pi_{-ast}(I_{mathbb{Q}/mathbb{Z}}X) simeq Hom(pi_ast(X),mathbb{Q}/mathbb{Z})$



          So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.



          See here for more details






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
            $endgroup$
            – André Henriques
            Dec 23 '18 at 21:42








          • 1




            $begingroup$
            @AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
            $endgroup$
            – skd
            Dec 24 '18 at 5:47








          • 1




            $begingroup$
            @skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
            $endgroup$
            – André Henriques
            Dec 24 '18 at 16:18














          15












          15








          15





          $begingroup$

          Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{mathbb{Q}/mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):



          $pi_{-ast}(I_{mathbb{Q}/mathbb{Z}}X) simeq Hom(pi_ast(X),mathbb{Q}/mathbb{Z})$



          So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.



          See here for more details






          share|cite|improve this answer











          $endgroup$



          Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{mathbb{Q}/mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):



          $pi_{-ast}(I_{mathbb{Q}/mathbb{Z}}X) simeq Hom(pi_ast(X),mathbb{Q}/mathbb{Z})$



          So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.



          See here for more details







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 23 '18 at 12:15

























          answered Dec 22 '18 at 17:37









          Sam GunninghamSam Gunningham

          4,3922029




          4,3922029








          • 1




            $begingroup$
            There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
            $endgroup$
            – André Henriques
            Dec 23 '18 at 21:42








          • 1




            $begingroup$
            @AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
            $endgroup$
            – skd
            Dec 24 '18 at 5:47








          • 1




            $begingroup$
            @skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
            $endgroup$
            – André Henriques
            Dec 24 '18 at 16:18














          • 1




            $begingroup$
            There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
            $endgroup$
            – André Henriques
            Dec 23 '18 at 21:42








          • 1




            $begingroup$
            @AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
            $endgroup$
            – skd
            Dec 24 '18 at 5:47








          • 1




            $begingroup$
            @skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
            $endgroup$
            – André Henriques
            Dec 24 '18 at 16:18








          1




          1




          $begingroup$
          There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
          $endgroup$
          – André Henriques
          Dec 23 '18 at 21:42






          $begingroup$
          There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
          $endgroup$
          – André Henriques
          Dec 23 '18 at 21:42






          1




          1




          $begingroup$
          @AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
          $endgroup$
          – skd
          Dec 24 '18 at 5:47






          $begingroup$
          @AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
          $endgroup$
          – skd
          Dec 24 '18 at 5:47






          1




          1




          $begingroup$
          @skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
          $endgroup$
          – André Henriques
          Dec 24 '18 at 16:18




          $begingroup$
          @skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
          $endgroup$
          – André Henriques
          Dec 24 '18 at 16:18


















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