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We say (according to https://ncatlab.org/nlab/show/%28infinity%2Cn%29-category+with+duals) that a symmetric monoidal $(\infty,1)$ category $\mathcal{C}$ has duals if its homotopy category $h\mathcal{C}$ is rigid monoidal.

I'm interested in the $\infty$-category $Sp$ of spectra. What is the largest stable $\infty$-subcategory $\mathcal{C}$ of $Sp$ such that $\mathcal{C}$ has duals? Is such a category nontrivial? If so does it have a nice description? In this case the dual object of $X \in Sp$ is $Map(X, \mathbb{S})$, where $\mathbb{S}$ is the sphere spectrum $\Sigma^{\infty} S^0$.

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    $\begingroup$ The stable subcategory generated by suspension spectra of finite complexes. $\endgroup$ – Dylan Wilson Dec 29 '17 at 19:56
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    $\begingroup$ Any dualizable object $X$ satisfies $[X, Y] \cong [S, X^{\ast} \otimes Y]$ where $[-, -]$ is the mapping space and $S$ is the sphere spectrum; it follows that every dualizable object is compact in the sense that $[X, -]$ preserves filtered colimits. Now, I believe it's true that every spectrum is a filtered colimit of finite spectra (suspension spectra of finite complexes), so writing $X$ as such a filtered colimit, the identity map $X \to X$ necessarily factors through a finite spectrum, so $X$ is a retract of a finite spectrum. This is essentially the same as the proof that... $\endgroup$ – Qiaochu Yuan Dec 29 '17 at 20:52
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    $\begingroup$ ...the dualizable objects in $k$-modules, $k$ a commutative ring, must be finitely presented projective. More generally, in a symmetric monoidal ($\infty$)-category, if the unit object is compact then every dualizable object is compact. $\endgroup$ – Qiaochu Yuan Dec 29 '17 at 20:52
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    $\begingroup$ Actually maybe that only tells you every dualizable thing is a retract of a finite spectrum? Takes a tiny bit more to show those are closed under retracts (after all, the analogous statement for spaces is false) $\endgroup$ – Dylan Wilson Dec 29 '17 at 21:08
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    $\begingroup$ By the way, a keyword: ncatlab.org/nlab/show/Spanier-Whitehead+duality $\endgroup$ – Qiaochu Yuan Dec 29 '17 at 23:44
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As requested, the comments turned into answers:

  1. The dualizable objects in spectra are precisely the finite spectra (i.e. spectra of the form $\Sigma^{-k}\Sigma^{\infty}X$ where $X$ is a finite complex.)
  2. If you only want the statement 'dualizable objects are finite spectra and their retracts' there is a very formal proof that works in great generality: (a) the unit in $\mathsf{Sp}$ is a compact and $\mathsf{Sp}$ is closed symmetric monoidal, (b) every object in $\mathsf{Sp}$ is a filtered colimit of finite spectra; it follows that every dualizable object $X$ is compact and that the identity map $X \to X$ factors through a finite spectrum, hence $X$ is a retract of a finite spectrum.
  3. If you want to know that retracts of finite spectra are themselves finite, this is a bit less formal. One may argue directly or use (2) to show that, for $X$ dualizable, $H_*(X, \mathbb{Z})$ is finitely generated and projective, hence free and concentrated in finitely many degrees. Now inductively build cell complexes $Y_j$ and maps $Y_j \to X$ which are equivalences on $H_m(-,\mathbb{Z})$ for $m\le j$. This stops at some finite step and you get a map $Y \to X$ from a finite complex to $X$ which is an equivalence on $H_*(-,\mathbb{Z})$. But $X$ is connective (lots of ways to see this... for example, by (2) it's a retract of something connective), so this map is an equivalence by Hurewicz. (Note: for spaces, this argument works as soon as $X$ is simply connected, but can fail because of Wall's finiteness obstruction when $X$ is not simply connected).

Obligatory: none of this had anything to do with $\infty$-categories.

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  • $\begingroup$ Small nitpick: the general argument just tells you that all dualizable spectra are retracts of finite spectra, not the viceversa (of course it is easy to prove that finite spectra are dualizable by induction on the number of cells) $\endgroup$ – Denis Nardin Dec 30 '17 at 9:23
  • $\begingroup$ Perhaps this question is too general for the comments, but why do we define dualizability on the level of the homotopy category? I assume you said your answer has nothing to do with $\infty$-categories since we only care about the homotopy category. This just seems to contradict a general trend in higher category theory, where we require diagrams to commute up to coherent homotopy, not just homotopy. Does this make sense? $\endgroup$ – leibnewtz Dec 30 '17 at 10:03
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    $\begingroup$ @leibnewtz: what is special about duals is that, like inverses, when they exist they exist uniquely up to a contractible space of choices. So there is no need to keep track of coherences. $\endgroup$ – Qiaochu Yuan Dec 30 '17 at 11:17

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