Let $X:=X_{18}$ be an index one smooth prime Fano threefold of degree 18.

Consider its semi-orthogonal decomposition: $D^b(X)=\langle\mathcal{O}_X,\mathcal{E}^{\vee},\mathcal{A}_X\rangle=\langle\mathcal{Q}^{\vee},\mathcal{O}_X,\mathcal{A}_X\rangle$, where $\mathcal{E},\mathcal{Q}$ are tautological sub and quotient bundle on $X$ coming from Grassmannian $\mathrm{Gr}(2, 7)$. Note that $\mathcal{A}_X\cong ^{\perp}\langle Q^{\vee},\mathcal{O}_X\rangle$ or $\mathcal{A}_X\cong ^{\perp}\langle \mathcal{O}_X, \mathcal{E}^{\vee}\rangle$. It is known that $\mathcal{A}_X\cong D^b(C_2)$ where $C_2$ is a smooth genus 2 curve(hyperelliptic curve). It is also known that the group of auto-equivalences of $D^b(C_2)$ is generated by $Aut(C_2), [1], -\otimes\mathcal{L}$(automorphism of the curve, shift functor and tensoring with line bundles).

There is a natural involution $\tau\in Aut(C_2)$(hyperelliptic involution) inducing an auto-equivalence on $D^b(C_2)$(still denoted by $\tau$). My question: Is there any way to write the auto-equivalence $\tau:\mathcal{A}_X\rightarrow\mathcal{A}_X$ purely in terms of composition of functors associated to the objects in $D^b(X)$?

For example, if $X:=X_{10}$ is a special Gushel-Mukai threefold, its semi-orthogonal decomposition is given as $D^b(X)=\langle \mathcal{B}_X, \mathcal{E},\mathcal{O}_X\rangle$, where $\mathcal{B}_X\cong\langle\mathcal{E},\mathcal{O}_X\rangle^{\perp}$, the geometric involution $\tau$ on $X$ gives an auto-equivalence of $\mathcal{B}_X$ and one can write $\tau^{-1}$ as $\mathrm{L}_{\mathcal{E}}\circ\mathrm{L}_{\mathcal{O}_X}\circ(-\otimes\mathcal{O}_X(H))[-1]$.


Let me answer the question by myself. After a intensively literature research, I found that the habilitation of Faenzi,Daniele contains everything I need, here is the link http://dfaenzi.perso.math.cnrs.fr/publis/faenzi.hdr.pdf Section 3.2

  • $\begingroup$ Does this help also for $X_{18}$? It seems just set a correspondence between certain types of Fano 3folds. $\endgroup$
    – IMeasy
    Jul 14 at 14:07
  • $\begingroup$ @IMeasy, yes it does have everything for X18, he construct the first type auto-equivalence and the second type, which is exactly the one giving the involution on the genus two curve. $\endgroup$
    – user41650
    Jul 16 at 5:17
  • $\begingroup$ Do you mean that the same argument for $X_{10}$ works for $X_{18}$? $\endgroup$
    – IMeasy
    Jul 20 at 13:18
  • $\begingroup$ @IMeasy, I think in section 3.2, what he talks about is a genus 10 Fano threefold, which is degree 2\times 10-2=18. $\endgroup$
    – user41650
    Jul 20 at 23:13
  • $\begingroup$ sorry I have been terribly silly, I made a mistake in maling 10*2 - 2 .... $\endgroup$
    – IMeasy
    Jul 21 at 6:59

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