4
$\begingroup$

An object $T$ in a triangulated category $\mathcal{D}$ is called a generator if $T^\perp=0$, which means that for any nonzero $X$ in $\mathcal{D}$, there are $i\in\mathbb{Z}$ and a nonzero morphism $T[-i]\to X$.

For example, it is well known that $T=\mathcal{O}\oplus\mathcal{O}(1)\oplus\cdots\oplus\mathcal{O}(n)$ is a generator of $D^b(\mathbb{P}^n)$, the bounded derived category of coherent sheaves on $\mathbb{P}^n$. However, it is claimed in the paper Tilting generators via ample line bundles (link) by Toda and Uehara that $T$ is also a generator of the bounded above derived category $D^-(\mathbb{P}^n)$ (see Lemma 2.3 and Example 3.2). Why?

In general, suppose to have a generator $T$ (say, a vector bundle) of the bounded derived category $D^b(X)$ of coherent sheaves on an algebraic variety $X$. Is it true that $T$ is also a generator of the unbounded derived categories of coherent sheaves, and of the (bounded and unbounded) derived categories of quasi-coherent sheaves?

For example, the question has a positive answer for the bounded below derived category $D^+(X)$: for any nonzero $F^\bullet\in D^+(X)$ we have, for some integers $m$ and $i$, a nonzero morphism $T[-i]\to \tau_{\leq m}F^\bullet$, and this remains nonzero after composition with $\tau_{\leq m}F^\bullet\to F^\bullet$, as it induces a nonzero map in cohomology, in some degree $\ell\leq m$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.