# Dual objects in the $\infty$-category of spectra

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 $\Sigma^{\infty} S^0$.

• The stable subcategory generated by suspension spectra of finite complexes. – Dylan Wilson Dec 29 '17 at 19:56
• 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... – Qiaochu Yuan Dec 29 '17 at 20:52
• ...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. – Qiaochu Yuan Dec 29 '17 at 20:52
• 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) – Dylan Wilson Dec 29 '17 at 21:08
• By the way, a keyword: ncatlab.org/nlab/show/Spanier-Whitehead+duality – Qiaochu Yuan Dec 29 '17 at 23:44

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.
• 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? – leibnewtz Dec 30 '17 at 10:03