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

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

## 1 Answer

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

• Does this help also for $X_{18}$? It seems just set a correspondence between certain types of Fano 3folds. Jul 14 at 14:07
• @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. Jul 16 at 5:17
• Do you mean that the same argument for $X_{10}$ works for $X_{18}$? Jul 20 at 13:18
• @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. Jul 20 at 23:13
• sorry I have been terribly silly, I made a mistake in maling 10*2 - 2 .... Jul 21 at 6:59