This question is motivated by the construction of the Kuznetsov component on a prime Fano threefold $X$ of index 1 (say genus $g \geq 6$, $g \neq 7, 9$): $$ D^b(X) = \langle Ku(X), E, \mathcal{O}_X \rangle $$ where $E$ is the pullback of the rank 2 tautological subbundle on $Gr(2, g/2 + 2)$.

In [1], the authors construct explicit formulas for the gluing data of $E$ to $Ku(X)$ by considering $D := \langle \mathcal{O}_X \rangle^\perp = \langle Ku(X), E \rangle$ along with inclusion functor $i : Ku(X) \hookrightarrow D$ and go on to state the left adjoint is $i^{\ast} = \mathbf{L}_E$.

Now consider the inclusion $j : D \hookrightarrow D^b(X)$. Clearly $j \circ i$ is the inclusion of $Ku(X)$ into $D^b(X)$, so that the left adjoint $( j \circ i )^\ast$ is the usual projection functor defined for semi-orthogonal decompositions. What would the left adjoint $j^\ast$ be in this case?

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

  • $\begingroup$ the last author is Zhang $\endgroup$
    – user41650
    Sep 19, 2022 at 8:24

1 Answer 1


If $\mathcal{A} \subset \mathcal{T}$ is a right admissible subcategory and $$ \mathcal{T} = \langle \mathcal{A}^\perp, \mathcal{A} \rangle $$ is the corresponding semiorthogonal decomposition, the left mutation functor $\mathbf{L}_{\mathcal{A}}$ is defined as the left adjoint functor of the embedding $\mathcal{A}^\perp \hookrightarrow \mathcal{T}$.

In particular, in your case $j^* = \mathbf{L}_{\mathcal{O}_X}$.


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