1
$\begingroup$

In the paper "Locally finite triangulated categories" Xiao and Zhu define locally finite triangulated categories. These are triangulated categories $\mathcal{A}$ such that $$\sum_{X \in \mathrm{ind}\mathcal{A}} \mathrm{dim}_k \mathrm{Hom}(X,Y) < \infty$$ for all indecomposable object $Y \in \mathcal{A}$.

Is the (bounded) derived category $D^b(X)$ of a smooth projective scheme $X$ locally finite?

EDIT: In another paper (Han: A note on generic objects and triangulated categories) a triangulated category is defined to be locally finite if for each indecomposable object $X$, there are only finitely many isoclasses of indecomposable objects $Y$ such that $\mathrm{Hom}(X,Y)\neq 0$.

$\endgroup$
2
  • 1
    $\begingroup$ I believe that the countably many distinct objects $\mathcal{O}_X(d)[0]$, $d\in \mathbb{Z}$, are all indecomposable. For every $d\leq 0$, $\text{Hom}(\mathcal{O}_X(d)[0],\mathcal{O}_X(0)[0])$ is nonzero. Thus, the sum of the dimensions of these countably many nonzero vector spaces is infinite. $\endgroup$ Mar 9, 2017 at 17:47
  • 1
    $\begingroup$ Thank you. How if we assume further that there are no isomorphic indecomposable objects? For example, according to Beilinson's resolution of the diagonal, $D^b(\mathbb{P}^n)$ is quasi-equivalent to a semisimple category with $n+1$ indecomposable objects. $\endgroup$ Mar 9, 2017 at 18:49

0

You must log in to answer this question.

Browse other questions tagged .