7
$\begingroup$

In Gaitsgory and Rozenblyum's Derived Algebraic Geometry book, they frequently use the following technique to transfer a $t$-structure from one category to another (for example, 1.5 in this paper or page 58 of this paper). The claim is (I think -- maybe some hidden assumptions?):

Let $F: C \rightarrow D$ be an exact functor which is a left adjoint, and suppose $D$ has a $t$-structure. Then, we define $C^{\leq 0}$ to be the full subcategory whose objects satisfy $F(X) \in D^{\leq 0}$. We define $C^{\geq 1}$ to be the right orthogonal to $C^{\leq 0}$. This defines a $t$-structure.

For example, in the first link, a proposition in Lurie's Higher Algebra is referenced. But this proposition assumes that the functor in question is a localization functor, which is exactly what I'm not sure about.

Here's an attempt. Let $Y \in C$. Then, there is an exact triangle $X \rightarrow FY \rightarrow Z$ where $X \in D^{\leq 0}$ and $Z \in D^{\geq 1}$. We have a map $Y \rightarrow GZ$ where $G$ is the right adjoint -- one can check that the right adjoint takes $D^{\geq 1}$ to $C^{\geq 1}$. The claim is that the fiber (i.e. cocone) of $Y \rightarrow GZ$ is in $C^{\leq 0}$.

Now, the only way to check this is to apply $F$. Since $F$ is exact, we have $$F(cocone) \rightarrow FX \rightarrow FGZ$$ However, it's not clear that the cocone of $FX \rightarrow FGZ$ is in $D^{\leq 0}$ to me.

$\endgroup$
2
$\begingroup$

A proof (of a dual statement for triangulated categories) is in Theorem 2.1.2 of Polishchuk, Constant families of t-structures on derived categories of coherent sheaves MR 2324559. Note that most of the assumption in the theorem can be ignored if your categories are cocomplete. In that case the result is essentially an immediate consequence of Theorem A.1 of Alonso Tarrío, Jeremías López, Souto Salorio, Construction of t-structures and equivalences of derived categories MR 1974001.

$\endgroup$
0

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.