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 $D^b(X)$.



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> Is $\mathcal{E}^{\star}=D^b(X)$?
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I know it is true in the case of $X=\mathbb{P}^n$.



triangulated hull and thick hull of $\mathcal{O}_X,\dots,\mathcal{O}_X(n)$
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.


