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Let $X$ be a nodal maximally non-factorial Fano threefold. If there is $1$-node and no other singularities, they by the work of Kuznetsov-Shinder https://arxiv.org/pdf/2207.06477.pdf Lemma 6.18, $D^b(X)$ contains a categorical ordinary double point subcategory $\mathcal{P}\subset D^b(X)$, which is generated by a $\mathbb{P}^\infty$ object $P$, such that the category $\mathcal{P}^{\perp}$ is a smooth category.

I have a question

Let $X$ be a nodal maximally non-factorial Fano threefold with no other singularities, if $D^b(X)$ contains two categorical ordinary double point subcategories $\mathcal{P}$ and $\mathcal{P}'$ such that the $\mathbb{P}^{\infty}$ objects generating $\mathcal{P}$ and $\mathcal{P}'$ are $P$ and $P'$ respectively and $P\not\cong P'$, can we say that $X$ has at least two node?

Assume not, then $X$ has only one node, then by Lemma above, $D^b(X)=\langle\mathcal{P}^{\perp},\mathcal{P}\rangle$ such that $\mathcal{P}^{\perp}$ is a smooth category, if we can further show that $\mathcal{P}'\subset\mathcal{P}^{\perp}$, then this is a contradiction, then $X$ has at least two nodes.

But I am not sure if $\mathcal{P}'\subset\mathcal{P}^{\perp}$ is true.

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1 Answer 1

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If you assume that $\mathcal{P}$ and $\mathcal{P'}$ are semiorthogonal, this is true. The easiest way to see this is by looking at the singularity category. If $X$ has one node, (the idempotent completion of) the singularity category of $X$ is the category of $\mathbb{Z}/2$-graded vector spaces, and if $$ D^b(X) = \langle \mathcal{P}, \mathcal{P}', \dots \rangle, $$ the singularity category contains a direct sum of two copies of the category of $\mathbb{Z}/2$-graded vector spaces, a contradiction.

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  • $\begingroup$ I completely agree with you that if $\mathcal{P},\mathcal{P}'$ are semi-orthogonal, then it is true. In fact, what I want to show is a little bit weaker. Is it possible that $X$ is a 1-nodal maximally non-factorial Fano threefold and $D^b(X)$ contains two categorical ordinary double point $\mathcal{P},\mathcal{P}'$ but this two category are NOT semi-orthogonal to each other? I think it is not possible but since they are not semi-orthogonal, I do not know how to prove it. $\endgroup$
    – user41650
    Oct 3 at 21:16
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    $\begingroup$ Without the semiorthogonality condition this is of course possible --- just apply any autoequivalence (e.g. a line bundle twst) to $\mathcal{P}$, this will give you another copy. $\endgroup$
    – Sasha
    Oct 4 at 6:06
  • $\begingroup$ sorry about that, is this the only exceptional case that $\mathcal{P},\mathcal{P}'$ appear, i.e, one is created from the other by auto-equivalence of $D^b(X)$? $\endgroup$
    – user41650
    Oct 4 at 8:24
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    $\begingroup$ Another possibility is to replace $\mathcal{P} = \langle P \rangle$ by $\mathcal{P}' := \langle P^\vee \rangle$ or by the image of the latter under an autoequivalence. $\endgroup$
    – Sasha
    Oct 4 at 8:52
  • $\begingroup$ Thanks you very much, Sasha. $\endgroup$
    – user41650
    Oct 4 at 12:37

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