Suppose $X$ is a prime Fano threefold of index 1 such that $H = -K_X$ is ample. There is a full classification of the derived category of such threefolds depending on the genus of $X$; in the case that $g \geq 6$ we have $$ D^b(X) = \langle \operatorname{Ku}(X), \mathcal{E}, \mathscr{O}\rangle $$ where $\operatorname{Ku}(X) = \langle \mathcal{E}, \mathscr{O}\rangle^\perp$ and $\mathcal{E}$ is the pullback of the rank 2 tautological Grassmannian bundle..

Now suppose I take $\mathcal{D} = \langle \mathscr{O}\rangle^\perp = \langle \operatorname{Ku}(X), \mathcal{E}\rangle$ along with its respective inclusion functor $$ i : \mathcal{D} \hookrightarrow D^b(X) $$ Then $\mathcal{D}$ is an admissible subcategory, and in particular [1] shows that $$ S_\mathcal{D} = \mathbb{R}_{\mathscr{O}(-H)} \circ S_{D^b(X)} $$ (where $\mathbb{R}$ denotes right mutation). I'm currently using the fact that an admissible subcategory with a Serre functor relates left and right adjoints via $$ i^{!} = S_{\mathcal{D}} \circ i^{\ast} \circ S^{-1}_{D^b(X)} $$ to attempt to compute $i^! \mathscr{O}$, which should simplify as

$$ i^! \mathscr{O} = \mathbb{R}_{\langle \mathscr{O} \rangle} \circ S_{D^b(X)} \circ \mathbb{L}_{\langle \mathscr{O}\rangle } ( \mathscr{O}(H)[-3] ) $$

(since $\mathbb{L}_{\langle \mathscr{O} \rangle } = i^\ast$ and $S^{-1}_{D^b(X)} \mathscr{O} = \mathscr{O}(H)[-3]$).

However, its not exactly clear from the definition what $\mathbb{L}_{\langle \mathscr{O}\rangle} \mathscr{O}(H)[-3]$ should be — first off it doesnt necessarily commute with shift since mutations aren't always exact equivalences. And taking it as the cone in the triangle $$ \operatorname{RHom}(\mathscr{O}, \mathscr{O}(H)) \otimes \mathscr{O} \to \mathscr{O}(H) \to \mathbb{L}_{\langle \mathscr{O} \rangle} \mathscr{O}(H) $$ isn't any more enlightening. How should one reasonably compute $i^! \mathscr{O}$ in this scenario?

[1] Jacovskis, Liu, Zhang. Brill-Noether Theory for Kuznetsov Components and Refined Categorical Torelli Theorems for Index One Fano Threefolds, 2022


1 Answer 1


First, mutations are exact, hence commute with shifts. Second, $$ \mathrm{RHom}(\mathcal{O},\mathcal{O}(H)) = H^\bullet(X, \mathcal{O}(H)) = H^\bullet(\mathbb{P}^{g+1}, \mathcal{O}(H)), $$ hence the evaluation morphism on $X$ is the restriction of the analogous evaluation morphism on $\mathbb{P}^{g+1}$, where its cone is isomorphic to $\Omega(H)[1]$. Thus, $$ \mathbb{L}_{\mathcal{O}}(\mathcal{O}(H)) \cong \Omega(H)[1]\vert_X. $$

  • $\begingroup$ Many thanks @Sasha ! Are you using capital $\Omega$ to refer to the canonical bundle here? $\endgroup$
    – cdsb
    Commented Oct 19, 2022 at 22:15
  • 1
    $\begingroup$ No, $\Omega$ is the bundle of differental 1-forms on the projective space. $\endgroup$
    – Sasha
    Commented Oct 20, 2022 at 1:12

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