Gdtyjr

The Segal conjecture describes the Spanier-Whitehead dual $D Sigma^infty_+ BG$ for certain $G$. Is there a similar description of $DSigma^infty_+ K(G,n)$ when $n geq 2$ when $G$ is finite (and abelian)?

Notes:

• I'd be happy to understand the case of cyclic groups $G = C_p$.




• $K(G,n)$ can be modeled by an abelian topological group, but I'm not sure it falls under the umbrella of other known generalizations of the Segal conjecture, although when $G = mathbb Z$ and $n=2$ there is a known decomposition (see Ravenel). For $G = mathbb Z^n$ and $n=2$ there is also this.



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  Let me recall that the Segal conjecture (proved by Carlsson) says that when $G$ is finite, the Spanier-Whitehead dual $DSigma^infty_+ BG$ is a certain completion of $vee_(H) subseteq G Sigma^infty_+ BW_G(H)$ where $(H) subseteq G$ ranges over conjugacy classes of subgroups and $W_G(H) = N_G(H) / H$ is the Weyl group of $H$ in $G$. In particular, when $G = C_p$ it says that

  
  
  

  $$DSigma^infty_+ BC_p = mathbb S vee(Sigma^infty_+ BC_p )^wedge_p$$

  
  
  

  where $mathbb S$ is the sphere spectrum (corresponding to the subgroup $C_p subseteq C_p$; the other term corresponds to the trivial subgroup $0 subseteq C_p$) and $(-)^wedge_p$ is $p$-completion.

  
  



• Lin showed that $D H G = 0$ when $G$ is a finite abelian group, where $H$ indicates taking Eilenberg-MacLane spectra. Since $HG = varinjlim_n Sigma^infty-n K(G,n)$, we have $0 = DHG = varprojlim_n Sigma^n DSigma^infty K(G,n)$, and from the Milnor exact sequence we conclude that $varprojlim_n pi_ast-n DSigma^infty K(G,n) = varprojlim^1_n pi_ast-n D Sigma^infty K(G,n) = 0$. But I'm not sure how much information that is, really.




• If we work in the $K(h)$-local or the $T(h)$-local category then by ambidexterity we have $F(Sigma^infty_+ K(G,n), Lmathbb S) = L Sigma^infty_+ K(G,n)$ where $L$ is the relevant localization. But it seems that the relevant limit does not commute with localization here.

In the 1980's, Chun Nip Lee showed that the Spanier Whitehead dual of (the suspension spectrum of) $K(mathbb Z/p, n)$ is contractible for $n >1$. (The key case is $n=2$. The idea" view $K(A,n+1)$ as the bar construction on $K(A,n)$ and

(No time right now to write more ... but maybe this is enough.)