3
$\begingroup$

Suppose we have a compactly generated triangulated category $\mathcal{T}$ such that the subcategory of compact objects $\mathcal{T}^c$ is essentially small. Let us take $\mathcal{A}, \mathcal{B}$ two thick subcategories of compact objects. I was wondering if the equality $\text{Loc}(\mathcal{A})\cap \text{Loc}(\mathcal{B})=\text{Loc}(\mathcal{A}\cap\mathcal{B})$ holds. Here $\text{Loc}(S)$ denotes the localizing subcategory generated by $S$ a class of objects of $\mathcal{T}$.

Since $\mathcal{T}$ is compactly generated Thomason's Localization Theorem states that if $S$ is a set of compact objects then $\text{Loc}(S)\cap \mathcal{T}^c=\text{Thick}(S)$, where the last term is the thick subcategory generated by $S$. Using this we easily deduce that both $\text{Loc}(\mathcal{A})\cap \text{Loc}(\mathcal{B})$ and $\text{Loc}(\mathcal{A}\cap\mathcal{B})$ have the same class of compact objects, namely $\mathcal{A}\cap \mathcal{B}$.

If also $\text{Loc}(\mathcal{A})\cap \text{Loc}(\mathcal{B})$ were generated by its compact objects, then the equality would follow. But I do not know if this is the case. I do not even understand if it is a reasonable expectation or counterexamples can be found easily. Any pointer or observation is welcome.

$\endgroup$
6
  • 1
    $\begingroup$ A small observation : if $T$ has a symmetric monoidal structure for which it is rigidly compactly generated, and $A,B$ are thick tensor idels, then the answer is yes, namely $Loc(A)\cap Loc(B) = Loc(A\cap B)$ $\endgroup$ Sep 7, 2022 at 7:00
  • $\begingroup$ Do you have a reference for this claim? $\endgroup$
    – N.B.
    Sep 7, 2022 at 7:47
  • 1
    $\begingroup$ I don't know one :) but the proof is relatively simple : if $X\in Loc(A)\cap Loc(B)$, then there is $a\in A$ with a nonzero map $a\to X$, so $a^\vee \otimes X$ is nonzero and in $Loc(B)$ still, so there is a $b\in B$ with a nonzero map $b\to a^\vee\otimes X$, i.e. a nonzero map $a\otimes b \to X$, and $a\otimes b\in A\cap B$. For general reasons, producing such a map is enough. As you can see, I really use the monoidal structure and the assumptions $\endgroup$ Sep 7, 2022 at 8:07
  • $\begingroup$ The fact that the existence of a non-zero map $a\rightarrow X$ is enough to guarantee $X \in \text{Loc}(\mathcal{A})$ seems really suspicious. Consider that since the ambient category is compactly generated we can construct a colocalization functor $C_{\mathcal{A}}$ such that for any object $X$ we have $C_{\mathcal{A}}X \in \text{Loc}(\mathcal{A})$ and $[a, C_{\mathcal{A}}X]\cong [a, X]$ for all $a \in \text{Loc}(\mathcal{A})$, with associated localization $L$. $\endgroup$
    – N.B.
    Sep 7, 2022 at 14:58
  • 1
    $\begingroup$ That's not my claim. My claim is global in $X$: if $C_0\subset C$ is a thick subcategory of compact objects, and for every $X$ in $C$ there is a nonzero map from some object of $C_0$, then $C= Loc(C_0)$ $\endgroup$ Sep 7, 2022 at 15:26

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.