Here I am dealing with the difference of triangulated hull and thick hull. Let $\mathcal{D}$ be a triangulated category and $\mathcal{E}\subset\mathcal{D}$ be a collection of objects. The triangulated hull $\mathcal{E}^{\star}$ in $\mathcal{D}$ is the smallest triangulated subcategory of $\mathcal{D}$ containing $\mathcal{E}$, the thick hull $\langle\mathcal{E}\rangle$ in $\mathcal{D}$ is the smallest thick subcategory of $\mathcal{D}$ containing $\mathcal{E}$.

Consider $X$ a complex smooth projective variety of dimension $n$ with a polarization $\mathcal{O}_X(1)$ or say $\mathcal{O}_X(1)$ is very ample, then we know that the thick hull of $\mathcal{E}=\{\mathcal{O}_X,\dots,\mathcal{O}_X(n)\}$ in $D^b(X)$ is $D^b(X)$.

Is $\mathcal{E}^{\star}=D^b(X)$?

I know it is true in the case of $X=\mathbb{P}^n$.


1 Answer 1


No. The simplest counterexample is a smooth quadric surface $X = \mathbb{P}^1 \times \mathbb{P}^1 \subset \mathbb{P}^3$. If $F$ belongs to the triangulated hull of $\mathcal{O}_X$, $\mathcal{O}_X(1)$, $\mathcal{O}_X(2)$, then its first Chen class is proportional to $$ c_1(\mathcal{O}_X(1)) = h_1 + h_2, $$ where $h_i$ are the classes of the rulings. In particular, the line bundle $\mathcal{O}_X(h_1)$ is not in this triangulated hull.

  • $\begingroup$ Thank you for the example. Could you please explain your statement about the first Chern class? $\endgroup$
    – user485941
    Jul 18, 2022 at 21:02
  • $\begingroup$ All three line bundles have this property, and the cone of a morphism of two objects with this property also has this property. So, it follows that any object in the triangulated hull has this property. $\endgroup$
    – Sasha
    Jul 18, 2022 at 21:06

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