4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.