Obstructions to realisation of dual finite spectra as suspension spectra












5












$begingroup$


Suppose $X$ is a finite dimensional CW-complex with top cell at dimmension $n$ and consider its S-dual denoted by $DX$. I wonder if there are any obstructions to find a space $Y$ and an interger $kgeqslant n$ so that
$$Sigma^kD(X)simeqSigma^infty Y_+ ?$$
For example, in the case of $X=S^m$ the answer is positive.



EDIT In the case of finite dimensional projective spaces, one may start from spaces $mathbb{R}P_m^n$ and choosing $m$ and $n$ in accordance with James periodicity, we can actually compute such a $k$.



The question is that is there any such $k$ for a given CW-complex of finite type (which of course after Neil's answer below I realise the answer is positive if one is to choose $k$ enough large, and obstructions are for specific $k$'s)



I would be very grateful for any references.






at.algebraic-topology







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$endgroup$

















    5












    $begingroup$


    Suppose $X$ is a finite dimensional CW-complex with top cell at dimmension $n$ and consider its S-dual denoted by $DX$. I wonder if there are any obstructions to find a space $Y$ and an interger $kgeqslant n$ so that
    $$Sigma^kD(X)simeqSigma^infty Y_+ ?$$
    For example, in the case of $X=S^m$ the answer is positive.



    EDIT In the case of finite dimensional projective spaces, one may start from spaces $mathbb{R}P_m^n$ and choosing $m$ and $n$ in accordance with James periodicity, we can actually compute such a $k$.



    The question is that is there any such $k$ for a given CW-complex of finite type (which of course after Neil's answer below I realise the answer is positive if one is to choose $k$ enough large, and obstructions are for specific $k$'s)



    I would be very grateful for any references.






    at.algebraic-topology







    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      Suppose $X$ is a finite dimensional CW-complex with top cell at dimmension $n$ and consider its S-dual denoted by $DX$. I wonder if there are any obstructions to find a space $Y$ and an interger $kgeqslant n$ so that
      $$Sigma^kD(X)simeqSigma^infty Y_+ ?$$
      For example, in the case of $X=S^m$ the answer is positive.



      EDIT In the case of finite dimensional projective spaces, one may start from spaces $mathbb{R}P_m^n$ and choosing $m$ and $n$ in accordance with James periodicity, we can actually compute such a $k$.



      The question is that is there any such $k$ for a given CW-complex of finite type (which of course after Neil's answer below I realise the answer is positive if one is to choose $k$ enough large, and obstructions are for specific $k$'s)



      I would be very grateful for any references.






      at.algebraic-topology







      share|cite|improve this question











      $endgroup$




      Suppose $X$ is a finite dimensional CW-complex with top cell at dimmension $n$ and consider its S-dual denoted by $DX$. I wonder if there are any obstructions to find a space $Y$ and an interger $kgeqslant n$ so that
      $$Sigma^kD(X)simeqSigma^infty Y_+ ?$$
      For example, in the case of $X=S^m$ the answer is positive.



      EDIT In the case of finite dimensional projective spaces, one may start from spaces $mathbb{R}P_m^n$ and choosing $m$ and $n$ in accordance with James periodicity, we can actually compute such a $k$.



      The question is that is there any such $k$ for a given CW-complex of finite type (which of course after Neil's answer below I realise the answer is positive if one is to choose $k$ enough large, and obstructions are for specific $k$'s)



      I would be very grateful for any references.






      at.algebraic-topology




      at.algebraic-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 26 '18 at 19:05







      user51223

















      asked Nov 26 '18 at 9:31









      user51223user51223

      1,341617




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          8












          $begingroup$

          Firstly, you say that the answer is negative for finite-dimensional projective spaces. However, in this case, and for any finite complex $X$, the answer will be positive if we take $k$ sufficiently large. Perhaps you are just thinking of the case $k=n$? Anyway, I will assume that we have fixed some particular $kgeq n$ and we want to know whether $Sigma^kDX$ is a suspension spectrum.



          The most obvious point is that $H^*(Sigma^kD(X);mathbb{F}_p)$ needs to be an unstable module over the Steenrod algebra. This is easy to check if you have a good understanding of $H^*(X;mathbb{F}_p)$ as a Steenrod module. Similarly, the module $M=K^0(Sigma^kDX)$ has naturally defined Adams operations $psi^qcolon Mto M[q^{-1}]$, and if $Sigma^kDX$ is a suspension spectrum then these will lift in a compatible way to give operations $psi^qcolon M to M$. If this does not settle the question then one can consider additive unstable operations in $BP$ theory or $MU$ theory. These are harder to handle explicitly but the basic slogan is as follows: $MU^*(Sigma^kDX)$ is functorial for isomorphisms of formal groups, and if $Sigma^kDX$ is a suspension spectrum then this extends to give functoriality for all homomorphisms of formal groups.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for correcting me on my statement about projective spaces for your comments on the general case. I will edit the statement about the projective spaces.
            $endgroup$
            – user51223
            Nov 26 '18 at 19:08











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          $begingroup$

          Firstly, you say that the answer is negative for finite-dimensional projective spaces. However, in this case, and for any finite complex $X$, the answer will be positive if we take $k$ sufficiently large. Perhaps you are just thinking of the case $k=n$? Anyway, I will assume that we have fixed some particular $kgeq n$ and we want to know whether $Sigma^kDX$ is a suspension spectrum.



          The most obvious point is that $H^*(Sigma^kD(X);mathbb{F}_p)$ needs to be an unstable module over the Steenrod algebra. This is easy to check if you have a good understanding of $H^*(X;mathbb{F}_p)$ as a Steenrod module. Similarly, the module $M=K^0(Sigma^kDX)$ has naturally defined Adams operations $psi^qcolon Mto M[q^{-1}]$, and if $Sigma^kDX$ is a suspension spectrum then these will lift in a compatible way to give operations $psi^qcolon M to M$. If this does not settle the question then one can consider additive unstable operations in $BP$ theory or $MU$ theory. These are harder to handle explicitly but the basic slogan is as follows: $MU^*(Sigma^kDX)$ is functorial for isomorphisms of formal groups, and if $Sigma^kDX$ is a suspension spectrum then this extends to give functoriality for all homomorphisms of formal groups.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for correcting me on my statement about projective spaces for your comments on the general case. I will edit the statement about the projective spaces.
            $endgroup$
            – user51223
            Nov 26 '18 at 19:08
















          8












          $begingroup$

          Firstly, you say that the answer is negative for finite-dimensional projective spaces. However, in this case, and for any finite complex $X$, the answer will be positive if we take $k$ sufficiently large. Perhaps you are just thinking of the case $k=n$? Anyway, I will assume that we have fixed some particular $kgeq n$ and we want to know whether $Sigma^kDX$ is a suspension spectrum.



          The most obvious point is that $H^*(Sigma^kD(X);mathbb{F}_p)$ needs to be an unstable module over the Steenrod algebra. This is easy to check if you have a good understanding of $H^*(X;mathbb{F}_p)$ as a Steenrod module. Similarly, the module $M=K^0(Sigma^kDX)$ has naturally defined Adams operations $psi^qcolon Mto M[q^{-1}]$, and if $Sigma^kDX$ is a suspension spectrum then these will lift in a compatible way to give operations $psi^qcolon M to M$. If this does not settle the question then one can consider additive unstable operations in $BP$ theory or $MU$ theory. These are harder to handle explicitly but the basic slogan is as follows: $MU^*(Sigma^kDX)$ is functorial for isomorphisms of formal groups, and if $Sigma^kDX$ is a suspension spectrum then this extends to give functoriality for all homomorphisms of formal groups.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for correcting me on my statement about projective spaces for your comments on the general case. I will edit the statement about the projective spaces.
            $endgroup$
            – user51223
            Nov 26 '18 at 19:08














          8












          8








          8





          $begingroup$

          Firstly, you say that the answer is negative for finite-dimensional projective spaces. However, in this case, and for any finite complex $X$, the answer will be positive if we take $k$ sufficiently large. Perhaps you are just thinking of the case $k=n$? Anyway, I will assume that we have fixed some particular $kgeq n$ and we want to know whether $Sigma^kDX$ is a suspension spectrum.



          The most obvious point is that $H^*(Sigma^kD(X);mathbb{F}_p)$ needs to be an unstable module over the Steenrod algebra. This is easy to check if you have a good understanding of $H^*(X;mathbb{F}_p)$ as a Steenrod module. Similarly, the module $M=K^0(Sigma^kDX)$ has naturally defined Adams operations $psi^qcolon Mto M[q^{-1}]$, and if $Sigma^kDX$ is a suspension spectrum then these will lift in a compatible way to give operations $psi^qcolon M to M$. If this does not settle the question then one can consider additive unstable operations in $BP$ theory or $MU$ theory. These are harder to handle explicitly but the basic slogan is as follows: $MU^*(Sigma^kDX)$ is functorial for isomorphisms of formal groups, and if $Sigma^kDX$ is a suspension spectrum then this extends to give functoriality for all homomorphisms of formal groups.






          share|cite|improve this answer









          $endgroup$



          Firstly, you say that the answer is negative for finite-dimensional projective spaces. However, in this case, and for any finite complex $X$, the answer will be positive if we take $k$ sufficiently large. Perhaps you are just thinking of the case $k=n$? Anyway, I will assume that we have fixed some particular $kgeq n$ and we want to know whether $Sigma^kDX$ is a suspension spectrum.



          The most obvious point is that $H^*(Sigma^kD(X);mathbb{F}_p)$ needs to be an unstable module over the Steenrod algebra. This is easy to check if you have a good understanding of $H^*(X;mathbb{F}_p)$ as a Steenrod module. Similarly, the module $M=K^0(Sigma^kDX)$ has naturally defined Adams operations $psi^qcolon Mto M[q^{-1}]$, and if $Sigma^kDX$ is a suspension spectrum then these will lift in a compatible way to give operations $psi^qcolon M to M$. If this does not settle the question then one can consider additive unstable operations in $BP$ theory or $MU$ theory. These are harder to handle explicitly but the basic slogan is as follows: $MU^*(Sigma^kDX)$ is functorial for isomorphisms of formal groups, and if $Sigma^kDX$ is a suspension spectrum then this extends to give functoriality for all homomorphisms of formal groups.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 9:47









          Neil StricklandNeil Strickland

          36.4k695188




          36.4k695188












          • $begingroup$
            Thanks for correcting me on my statement about projective spaces for your comments on the general case. I will edit the statement about the projective spaces.
            $endgroup$
            – user51223
            Nov 26 '18 at 19:08


















          • $begingroup$
            Thanks for correcting me on my statement about projective spaces for your comments on the general case. I will edit the statement about the projective spaces.
            $endgroup$
            – user51223
            Nov 26 '18 at 19:08
















          $begingroup$
          Thanks for correcting me on my statement about projective spaces for your comments on the general case. I will edit the statement about the projective spaces.
          $endgroup$
          – user51223
          Nov 26 '18 at 19:08




          $begingroup$
          Thanks for correcting me on my statement about projective spaces for your comments on the general case. I will edit the statement about the projective spaces.
          $endgroup$
          – user51223
          Nov 26 '18 at 19:08


















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