6
$\begingroup$

I have a series of vague questions, related to localization of symmetric monoidal categories.

Here is the context. Say we are working over a field of characteristic zero. Then the "one category level higher" version of (DG) commutative ring is a (DG) symmetric monoidal category. It is well-known that for $X$ a scheme (or even, IIUC, a Noetherian stack with affine diagonal) $X$ can be recovered from the DG symmetric monoidal category of its quasicoherent sheaves. (Moreover, the functor $\text{Schemes}\to\text{SymMonCat}$ is fully faithful, including in an $\infty$-categorical context.

Now if $R$ is a commutative ring, we say that $S$ is a localization of $R$ if it can be obtained from $R$ by inverting some set of elements. If $R, S$ are both Noetherian, then there is a very nice alternative way of characterizing localizations:

(*) A map $R\to S$ is a localization if and only if the product map $S\otimes_R S\to S$ (derived tensor product) is an equivalence.

Now there are (at least) three interesting notions of localization for a DG symmetric monoidal category $\mathcal{C}$ (note all make sense also for just monoidal categories, and that when taking universal objects in the category of categories I'm going to be vague about what I'm requiring from categories: I'm willing to assume compactly generated, idempotent complete, etc.).

  1. Localization along a morphism $f:X\to Y.$
  2. Localization along an object $X$ (defined as the universal symmetric monoidal DG category admitting a functor from $\mathcal{C}$ where $X$ is $\otimes$-invertible, maybe satisfying some additional conditions).
  3. "Parallel" localization along a morphism: if $f:X\to Y$ is a morphism, I'm defining this to be the initial category in which $X,Y$ are invertible and there exists a map $f':X^{-1}\to Y^{-1}$ with $f\otimes f' = \text{id}:\mathbb{I}\to \mathbb{I},$ where the equality is understood via an appropriate system of coherences.

(Of course 2. is a special case of 3.)

Here are some questions.

  • Defining $\otimes$ in terms of colimit in the category of symmetric monoidal categories, is there a context where (*) holds for symmetric monoidal categories (i.e., localizations can be characterized by a tensor-idempotence condition)?
  • Are there "interesting" localizations of this type of the category $\mathcal{C}$ of DG quasicoherent coherent sheaves on a scheme $X$ which do not come from geometric localizations? If no, are there any examples in more general contexts that have been studied or computed in some sense? (For example, what happens if you $\otimes$-localize the category of vector spaces along a two-dimensional vector space?)
  • By functoriality and universality, localizations of any of the types 1., 2., 3. can be "combined" (in a commutative way), and (by formal nonsense), the result of applying two localizations $\mathcal{C}\to \mathcal{C}_1$ and $\mathcal{C}\to \mathcal{C}_2$ is the colimit $\mathcal{C}_1\otimes_{\mathcal{C}}\mathcal{C}_2$. **Definition**. For $\mathcal{C}_1\leftarrow\mathcal{C}\to \mathcal{C}_2$ a pair of localizations of $\mathcal{C}$ as above, set $$\mathcal{C}_{12}: = \mathcal{C}_1\otimes_{\mathcal{C}}\mathcal{C}_2$$ for the "combined" localization. Say that $\mathcal{C}_1$ and $\mathcal{C}_2$ *cover* $\mathcal{C}$ if $\mathcal{C}$ is the limit of the diagram $\mathcal{C}_1\to \mathcal{C}_{12}\leftarrow \mathcal{C}_2.$ My question now is: are there interesting examples of covers of DG symmetric monoidal categories in this sense other than Zariski covers in geometry?
$\endgroup$
3
$\begingroup$

Given a symmetric monoidal functor $\mathcal{C} \rightarrow \mathcal{C}'$, the property that $\mathcal{C}' \otimes_{\mathcal{C}} \mathcal{C}' \rightarrow \mathcal{C}'$ be an isomorphism is equivalent to $\mathcal{C} \rightarrow \mathcal{C}'$ being an epimorphism in the category of symmetric monoidal dg categories. This is a purely formal fact: in any cocartesian symmetric monoidal category $\mathcal{E}$ a map out of the initial object $0 \rightarrow Z$ is an epimorphism if and only if the map $Z \coprod Z \rightarrow Z$ is an isomorphism. Our case follows from this by taking $\mathcal{E}$ to be the category of symmetric monoidal categories over $\mathcal{C}$.

The three notions of localization mentioned in the question are epimorphisms, since a map out of the localization is by definition a map out of $\mathcal{C}$ satisfying a property (namely, that a certain arrow becomes invertible, that a certain object becomes invertible, etc). Hence all three notions satisfy the tensor idempotence condition.

Inverting an arrow $f: X \rightarrow Y$ amounts to passing to the quotient by the ideal generated by the cofiber of $f$. A symmetric monoidal functor $F: \mathcal{C} \rightarrow \mathcal{D}$ maps $\operatorname{cofib}(f)$ to zero if and only if it inverts $1_{\mathcal{C}} \oplus \operatorname{cofib}(f)$. Therefore your first notion of localization is a particular case of the second one.

Your third notion of localization is in fact equivalent to the second one. Given a map $f: X \rightarrow Y$ between invertible objects, any such map $f': X^{-1} \rightarrow Y^{-1}$ is necessarily dual to an inverse to $f$. Hence localizing in your third way along an arbitrary map $f:X \rightarrow Y$ is equivalent to first inverting $X, Y$ and then inverting $f$, which we already observed can be reduced to the second notion.


There are epimorphisms that do not arise as quotients by ideals: Consider for instance the category $\operatorname{Sh}(M)$ of sheaves of (complexes of) vector spaces on a manifold $M$. Let $x$ be a point in $M$ and $U$ its complement. The star pullback functor $\operatorname{Sh}(M) \rightarrow \operatorname{Sh}(\lbrace x \rbrace)$ exhibits $\operatorname{Sh}(\lbrace x \rbrace)$ as the quotient of $\operatorname{Sh}(M)$ by the ideal of sheaves with vanishing stalk at $x$. Similarly, $\operatorname{Sh}(U)$ is the quotient of $\operatorname{Sh}(M)$ by the ideal of sheaves supported at $x$. The ideal generated by the union of these two ideals is the whole $\operatorname{Sh}(M)$, so we see that $\operatorname{Sh}(\lbrace x \rbrace) \otimes_{\operatorname{Sh}(M)} \operatorname{Sh}(U) = 0$. It follows that the functor $\operatorname{Sh}(M) \rightarrow \operatorname{Sh}(\lbrace x \rbrace) \times \operatorname{Sh}(U)$ satisfies the tensor-idempotence condition, but it doesn't arise as the quotient by an ideal since its right adjoint $\operatorname{Sh}(\lbrace x \rbrace) \times \operatorname{Sh}(U) \rightarrow \operatorname{Sh}(M)$ (given by star-pushforward in each coordinate) is not fully faithful.


Under tameness conditions all notions of localization agree: I don't know if every epimorphism arises by inverting an object in general, but under certain tameness conditions one can show that this is the case:

Claim: Let $\mathcal{C}$ be a symmetric monoidal dg category compactly generated by its dualizable objects and $\mathcal{C}' $ be a compactly generated symmetric monoidal dg category equipped with a symmetric monoidal functor $\mathcal{C} \rightarrow \mathcal{C}'$ that preserves compact objects, and such that the map $\mathcal{C}' \otimes_{\mathcal{C}} \mathcal{C}' \rightarrow \mathcal{C}'$ is an isomorphism. Then the functor $\mathcal{C} \rightarrow \mathcal{C}'$ arises by passing to the quotient by an ideal of $\mathcal{C}$.

Sketch of proof: Let $\mathcal{K}$ be the full subcategory of $\mathcal{C}'$ generated under colimits by the image of the functor $\mathcal{C}\rightarrow \mathcal{C}'$. Our conditions guarantee that the right adjoint to the inclusion $\mathcal{C} \rightarrow \mathcal{K}$ is colimit preserving and monadic. Note moreover that $\mathcal{K}$ is a $\mathcal{C}$-module and the functor $\mathcal{C} \rightarrow \mathcal{K}$ is a map of $\mathcal{C}$-modules. Its right adjoint in principle only commutes with the $\mathcal{C}$-action up to natural transformations, but the fact that $\mathcal{C}$ is compactly generated by its dualizable objects guarantees that the natural transformations are isomorphisms, and so the functor $\mathcal{K}\rightarrow \mathcal{C}$ is also a morphism of $\mathcal{C}$-modules. It follows that $\mathcal{K}$ is the category of algebras for a $\mathcal{C}$-linear monad on $\mathcal{C}$, and so we have an identification $\mathcal{K} = A\operatorname{-mod}$ for some algebra $A$ in $\mathcal{C}$. The fact that $\mathcal{C}' \otimes_{\mathcal{C}} \mathcal{C}' = \mathcal{C}'$ implies that the multiplication map $A \otimes A \rightarrow A$ is an isomorphism. This means that $\mathcal{K}$ is in fact the category of algebras for an idempotent $\mathcal{C}$-linear monad, and so it arises as the quotient of $\mathcal{C}$ by an ideal. This whole thing reduces you to understanding the case when $\mathcal{C} \rightarrow \mathcal{C}'$ is fully faithful. Since the canonical map $\mathcal{C}\otimes_{\mathcal{C}}\mathcal{C}' \rightarrow \mathcal{C}'\otimes_{\mathcal{C}} \mathcal{C}'$ is an isomorphism, we have that $\mathcal{C}'/\mathcal{C} \otimes_{\mathcal{C}} \mathcal{C}'$ vanishes. This contains $\mathcal{C}'/\mathcal{C} \otimes_{\mathcal{C}} \mathcal{C} = \mathcal{C}'/\mathcal{C}$ as a full subcategory, so we see that $\mathcal{C}' = \mathcal{C}$ is the trivial localization.


The algebro-geometric case: The above includes for instance the case of $\mathcal{C} = \operatorname{QCoh}(X)$ for $X$ a separated scheme. Moreover, from the proof we see that the resulting localizations are categories of modules for quasicoherent sheaves of algebras $A$ over $X$ such that the multiplication map $A \otimes A \rightarrow A$ is an isomorphism. In the Noetherian case you can use the result stated in the question to deduce that $A$ is locally a localization of the structure sheaf, so you see that all localizations in this sense are classified by collections of points of $X$ closed under specialization.

If you drop the condition that the functor $\mathcal{C} \rightarrow \mathcal{C}'$ preserves compact objects there are more examples, even in the geometric case. Indeed, any ideal of $\operatorname{QCoh}(X)$ provides an example, and these are classified (in the Noetherian case) by arbitrary collections of (non necessarily closed) points of $X$. This classification goes back to Hopkins, Neeman, and by now there is a whole industry about it - key words being tensor triangular geometry and classification of localizing subcategories.


Beyond algebraic geometry: If you are looking for interesting covers beyond Zariski covers in algebraic geometry one source could be topology. If you have a manifold $M$ and $U$ is an open set of $M$, the category $\operatorname{Sh}(U)$ is the category of comodules for an idempotent coalgebra in $\operatorname{Sh}(M)$ with underlying sheaf $k_U$, and so it can be obtained as the colimit $$\operatorname{Sh}(M) \xrightarrow{\otimes k_U}\operatorname{Sh}(M) \xrightarrow{\otimes k_U} \operatorname{Sh}(M)\xrightarrow{\otimes k_U} \ldots .$$ This is the same diagram you would use to invert $k_U$, and so $\operatorname{Sh}(U)$ in fact results from inverting $k_U$. From this it follows that $\operatorname{Sh}(U)\otimes_{\operatorname{Sh}(M)} \operatorname{Sh}(V) = \operatorname{Sh}(U \cup V)$ for every pair of opens, and so you see that any open cover of $M$ provides a cover of symmetric monoidal categories in your sense.

You can build many more examples by variations of this theme. You could take $M$ to be the union of two manifolds $U, V$ along a closed submanifold (for instance $M$ could be the union of the $x$ and $y$ axes in $\mathbb{R}^2$) and you still have that $\operatorname{Sh}(M)$ is covered by $\operatorname{Sh}(U)$ and $\operatorname{Sh}(V)$. You could even require your sheaves to be constructible with respect to a stratification and $U, V$ to respect the stratification to get covers of categories of modules over quivers.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you German for this insightful and thorough answer! I thought the third localization may follow from the first two but couldn't see how. Does your argument assume rigidity, or does inverting $X$ automatically imply that $X, X^{-1}$ are dual and thus maps $X\to Y$ have dual maps $Y^{-1}\to X^{-1}$? Your implication (2) => (1) is very nice, and thank you for the reference to Ballmer, Hopkins, Neeman and co. $\endgroup$ – Dmitry Vaintrob Apr 6 at 4:18
  • 1
    $\begingroup$ @DmitryVaintrob No rigidity needed. I believe it is a general fact for any symmetric monoidal category that if an object $X$ is invertible then it is dualizable with dual $X^{-1}$. The idea is that for any object $Y$ one has that $X^{-1}\otimes Y$ is an internal Hom between $X$ and $Y$. Therefore the map $[X, 1] \otimes X \rightarrow [X, X]$ is an isomorphism (where brackets denote internal hom), which is equivalent to $X$ being dualizable. Full details seem to be supplied in pages.uoregon.edu/ddugger/invcoh.pdf proposition 4.11. As you say, this provides the dual maps that you need. $\endgroup$ – G. Stefanich Apr 6 at 23:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.