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A spectrum $X$ is dualizable if the natural map $$Map(X,\mathbb S) \wedge X \rightarrow Map(X,X)$$ is an equivalence of spectra. This is equivalent to having evaluation and coevaluation maps in the stable homotopy category $$ X \wedge DX \rightarrow \mathbb S $$ $$ \mathbb S \rightarrow DX \wedge X $$ for which the usual composites $$ X \rightarrow X \wedge DX \wedge X \rightarrow X $$ $$ DX \rightarrow DX \wedge X \wedge DX \rightarrow DX $$ give the identity in the homotopy category (cf. Lewis-May-Steinberger III.1.2). It is also well-known that this implies that $X \rightarrow D(DX)$ is an equivalence of spectra (LMS III.1.3(i)), but my question is

Is the converse true? Does $X \overset\sim\rightarrow D(DX)$ imply that $X$ is dualizable?

I understand that everything I have said holds in an arbitrary closed symmetric monoidal category, but I am willing to consider arguments that only work for spectra (or $R$-module spectra).

Cary
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    IIRC this is already false for, say, $R$-modules: an $R$-module is dualizable iff it is projective and finitely presented but there are examples of reflexive modules not having this property (http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules). – Qiaochu Yuan Feb 06 '14 at 23:58
  • Qiaochu, don't you mean finitely generated, rather than finitely presented? – Peter May Feb 07 '14 at 01:54
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    I don't know an example in spectra, but if you are willing to work in the $K(n)$-local category then I think the Morava $E$-theory spectrum is an example. Strickland has shown that $D(E_n) = F(E_n,L_{K(n)}S^0) = \Sigma^{-n^2}E_n$, which in turn can be used to show that the natural map $E_n \to D^2E_n$ is an equivalence. In the $K(n)$-local category $X$ dualisable is equivalent to $E^{\vee}*(X):=\piL_{K(n)}(E \wedge X)$ finitely generated. But I don't think $E^{\vee}_(E) = \text{Hom}^c(\mathbb{G}n,E*)$ is finitely generated – Drew Heard Feb 07 '14 at 02:37
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    @Peter: for projective modules the two are equivalent. – Qiaochu Yuan Feb 07 '14 at 02:56
  • @Drew: Thanks, that at least reassures me that these two notions are probably not the same. Still wondering about $\mathbb S$-modules though. – Cary Feb 09 '14 at 22:38
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    @Qiaochu: You have to be a little bit careful with the 'already'. If we view $R$ as a ring spectrum $HR$, then Hom becomes derived Hom (which does not agree with Hom if $X$ is not projective). Thus, a reflexive, but not projective $R$-module does not directly give a counterexample. – Lennart Meier Feb 10 '14 at 04:21
  • @Lennart: I wasn't claiming that it did; just giving evidence that this seemed unlikely because the analogous statement in the underived setting is false. – Qiaochu Yuan Feb 10 '14 at 04:40
  • You don't say this explicitly, but the dualizable spectra are precisely the finite spectra, right? – Tom Goodwillie Feb 10 '14 at 13:05
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    @Tom: Yes, X is dualizable iff it is finite. From finite to dualizable is easy: you prove $DX \wedge Y \rightarrow F(X,Y)$ is an equivalence for all $Y$ by induction on the cells of $X$. For the other direction, you show the coevaluation map $\mathbb S \rightarrow DX \wedge X$ must factor through $DX \wedge X'$ for some finite spectrum X', and then you use the above "composite is the identity" condition to show that $X$ is a retract of $X'$, therefore finite. (I took that second argument from EKMM.) – Cary Feb 10 '14 at 21:21

2 Answers2

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K(n) is weakly dualizable in the K(n)-local category, but it is not strongly dualizable. See Hahn and Mitchell, Section 8, Iwasawa theory of K(1)-local spectra, for the case when n=1.

Mark Hovey
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  • Thanks! I'm trying to understand if this is an example of $R$-module spectra for some ring $R$ - Schwede and Shipley's paper (http://homepages.math.uic.edu/~bshipley/classTopFinal.pdf, p.10) seems to say yes. – Cary Feb 10 '14 at 21:42
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    You have to be careful as the $K(n)$-local sphere is not small in the $K(n)$-local homotopy category (Section 8 of Hovey, Strickland - Morava K-theories and localisation). But the unit of the smash product is small in the category of $R$-modules for a ring spectrum $R$. Thus, there cannot be a multiplicative equivalence. – Lennart Meier Feb 10 '14 at 22:48
  • I see! They give an "additive" Quillen equivalence but that's not enough to preserve the notion of duality since duality is defined using smash products. So we still don't have a counterexample here of the form "$R$-module spectra" for some ring spectrum $R$. – Cary Feb 18 '14 at 19:40
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Let $R$ be bounded below, bounded above and nontrivial, and work in $R$-modules. Let $X = \bigvee_{n\in\mathbb{Z}} \Sigma^n R$. Then $X \overset{\simeq}\to D(DX)$ is an equivalence, but $X$ is not dualizable. (Edit: Increased generality, based on a comment by Nardin.)

John Rognes
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