Suppose we are given a cocomplete closed symmetric monoidal stable $(\infty,1)$-category $\mathcal{C}$ with suspension $\Sigma$, and let $X \in \mathcal{C}$ be dualizable. I'd like to create a new stable cocomplete symmetric monoidal category $\mathcal{C}[\Sigma X^{-1}]$ together with a symmetric monoidal exact and continuous functor $\mathcal{C} \to \mathcal{C}[\Sigma X^{-1}]$ with the following property: For any symmetric monoidal stable category $\mathcal{D}$,the induced functor $Fun_{ex,cont}^{\otimes}(\mathcal{C}[\Sigma X^{-1}], \mathcal{D}) \to Fun_{ex,cont}^{\otimes}(\mathcal{C}, \mathcal{D})$ induces an equivalence of $Fun_{ex,cont}^{\otimes}(\mathcal{C}[\Sigma X^{-1}], \mathcal{D})$ with the full subcategory of $Fun_{ex,cont}^{\otimes}(\mathcal{C}, \mathcal{D})$ consisting of functors $\mathcal{C} \to \mathcal{D}$ that send $\Sigma X$ to an invertible object in $\mathcal{D}$. Here the notation $Fun_{ex,cont}^{\otimes}(\mathcal{C},\mathcal{D})$ denotes the category of monoidal functors that are both exact and continuous.

My idea was to localize at the set of maps $(\Sigma (\iota)) \otimes id_Y$, where $\iota$ is the map $X \otimes X^* \to \mathbb{1}$ coming from duality data for $X$. Here $Y$ ranges over all objects in $\mathcal{C}$. However, I have no idea if the resulting category would satisfy all the requirements listed above.

EDIT: Inverting $\Sigma X$ is clearly equivalent to inverting $X$.


1 Answer 1


In the case where $\mathcal{C}$ is presentable, this is constructed in proposition 2.9 of

Robalo, Marco, $K$-theory and the bridge from motives to noncommutative motives, Adv. Math. 269, 399-550 (2015). ZBL1315.14030.

Note that if $\mathcal{C}$ is stable, so is $\mathcal{C}[X^{-1}]$, since $\mathcal{C}$ contains an inverse for $S^1$ and $\mathcal{C}\to \mathcal{C}[X^{-1}]$ is symmetric monoidal.

Furthermore, corollary 2.22 shows that if $X$ is a symmetric object (i.e. the cyclic permutation acts trivially on $X^{\otimes n}$ for some $n>1$), then $\mathcal{C}[X^{-1}]$ can be obtained as a category of spectrum objects, i.e. as the $\infty$-category $$\operatorname{Stab}_X(\mathcal{C}):=\operatorname{colim}\left(\mathcal{C}\xrightarrow{X\otimes - }\mathcal{C}\xrightarrow{X\otimes-}\cdots\right)\cong \operatorname{lim}\left(\mathcal{C}\xleftarrow{\operatorname{hom}(X,-)}\mathcal{C}\xleftarrow{\operatorname{hom}(X,-)}\cdots\right)$$ where the first colimit is taken in $\mathrm{Pr}^L$.

To see an example where the latter identification does not hold, take $\mathcal{C}=\mathrm{Sp}$ and $X=\mathbb{S}\oplus \mathbb{S}$. Then it can be shown that $\mathcal{C}[X^{-1}]=0$ (indeed every additive $\infty$-category where the functor $y\mapsto y\oplus y$ is fully faithful is 0), but $\operatorname{Stab}_X\mathcal{C}$ is not 0, since its K-theory is $K(\mathbb{S})[1/2]$. For more discussion of this particular example see this paper.

  • $\begingroup$ If $X$ is not symmetric I assume the categories will not coincide. Will $Stab_X(\mathcal{C})$ be symmetric monoidal and stable when $X$ is not cyclic? $\endgroup$
    – Exit path
    Feb 25, 2019 at 7:58
  • $\begingroup$ @leibnewtz No, indeed. That's exactly the problem in proving the identification in general. This is discussed carefully in that section of Marco's paper $\endgroup$ Feb 25, 2019 at 7:59

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