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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).

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    $\begingroup$ 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 (mathoverflow.net/questions/7490/…). $\endgroup$
    – Qiaochu Yuan
    Feb 6, 2014 at 23:58
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    $\begingroup$ 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 $E^{\vee}_*(X):=\pi_*L_{K(n)}(E \wedge X)$ finitely generated. But I don't think $E^{\vee}_*(E) = \text{Hom}^c(\mathbb{G}_n,E_*)$ is finitely generated $\endgroup$
    – Drew Heard
    Feb 7, 2014 at 2:37
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    $\begingroup$ @Peter: for projective modules the two are equivalent. $\endgroup$
    – Qiaochu Yuan
    Feb 7, 2014 at 2:56
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    $\begingroup$ @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. $\endgroup$
    – Lennart Meier
    Feb 10, 2014 at 4:21
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    $\begingroup$ @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.) $\endgroup$
    – Cary
    Feb 10, 2014 at 21:21

2 Answers 2

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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.

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  • $\begingroup$ Thanks! I'm trying to understand if this is an example of $R$-module spectra for some ring $R$ - Schwede and Shipley's paper (homepages.math.uic.edu/~bshipley/classTopFinal.pdf, p.10) seems to say yes. $\endgroup$
    – Cary
    Feb 10, 2014 at 21:42
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    $\begingroup$ 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. $\endgroup$
    – Lennart Meier
    Feb 10, 2014 at 22:48
  • $\begingroup$ 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 $R$. $\endgroup$
    – Cary
    Feb 18, 2014 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.)

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