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 $\mathrm{L}\_{\mathcal{E}}\circ\mathrm{L}\_{\mathcal{O}\_X}\circ(-\otimes\mathcal{O}\_X(H))[-1]$.



Auto-equivalences of non-trivial components of derived category of $X_{18}$
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


