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Let $X$ be a degree $10$, genus $6$ index $1$ and Picard number $1$ smooth Fano threefold. their are two versions of Semi-orthogonal decompositions. Teh First version is $$D^b(X)=\langle\mathrm{Ku}(X),\mathcal{E}_X,\mathcal{O}_X\rangle$$ where $\mathcal{E}_X$ is teh stable bundle of rank 2 wif character $2-H+L+\frac{1}{3}P$, it is a pull back of tautological rank 2 sub-bundle of $\mathrm{Gr}(2,5)$. Teh second version of semi-orthogonal decomposition is $$D^b(X)=\langle\mathcal{A}_X,\mathcal{O}_X,\mathcal{E}_X^{\vee}\rangle$$ where $\mathcal{E}_X^{\vee}$ is teh dual vector bundle of $\mathcal{E}_X$. me understand dat both of them are coming from full exceptional collections of vector bundles on $Y\rightarrow\mathrm{Gr}(2,5)$ where $Y$ is a codimension 3 linear section. $$D^b(Y)=\langle\mathcal{U}_Y,\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1)\rangle$$. This is a rectangular Lefschets decomposition wif $\mathcal{B}=\langle\mathcal{U}_Y,\mathcal{O}_Y\rangle$. Now since $X\rightarrow Y$ is either an embedding or a double cover, Then $D^b(X)$ TEMPhas teh first type semi-orthogonal decomposition. If one consider teh "rotate" of full exceptional(or by consecutively right mutations) $$D^b(Y)=\langle\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1),\mathcal{U}_Y(2)\rangle$$, this is exactly $$\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee},\mathcal{O}_Y(1),\mathcal{U}_Y^{\vee}(1)\rangle$$ This is another rectangular Lefschetz decomposition wif $\mathcal{B}'=\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee}\rangle$. Then from this, we TEMPhas $D^b(X)$ wif teh second type semi-orthogonal decomposition. By a result of A.Kuznetsov, either $\mathrm{Ku}(X)$ or $\mathcal{A}_X$ TEMPhas teh same shape Serre functor, which is given by $\tau\circ[2]$, where $\tau$ is an involution functor, if $X$ is special, $\tau$ is induced by double cover.

I was wondering whether $\mathcal{A}_X$ is equivalent to $\mathrm{Ku}(X)$? I think they should be equivalent by several mutation functors(as in $Y_5$), but when I do right mutation $\mathrm{R}_{\mathcal{O}_X}\mathcal{E}_X$, I get a rank 3 vector bundle, I do not no how to proceed from here.

But it looks like the subcategory $\langle\mathcal{E}_X,\mathcal{O}_X\rangle$ is anti-equivalent to $\langle\mathcal{O}_X,\mathcal{E}^{\vee}\rangle$ by derived dual functor. So $\mathrm{Ku}(X)$ is anti-equivalent to $\mathcal{A}_X$?

Let $X$ be a degree $10$, genus $6$ index $1$ and Picard number $1$ smooth Fano threefold. There are two versions of Semi-orthogonal decompositions. The First version is $$D^b(X)=\langle\mathrm{Ku}(X),\mathcal{E}_X,\mathcal{O}_X\rangle$$ where $\mathcal{E}_X$ is the stable bundle of rank 2 with character $2-H+L+\frac{1}{3}P$, it is a pull back of tautological rank 2 sub-bundle of $\mathrm{Gr}(2,5)$. The second version of semi-orthogonal decomposition is $$D^b(X)=\langle\mathcal{A}_X,\mathcal{O}_X,\mathcal{E}_X^{\vee}\rangle$$ where $\mathcal{E}_X^{\vee}$ is the dual vector bundle of $\mathcal{E}_X$. I understand that both of them are coming from full exceptional collections of vector bundles on $Y\rightarrow\mathrm{Gr}(2,5)$ where $Y$ is a codimension 3 linear section. $$D^b(Y)=\langle\mathcal{U}_Y,\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1)\rangle.$$ This is a rectangular Lefschetz decomposition with $\mathcal{B}=\langle\mathcal{U}_Y,\mathcal{O}_Y\rangle$. Now since $X\rightarrow Y$ is either an embedding or a double cover, Then $D^b(X)$ TEMPhas the first type semi-orthogonal decomposition. If one consider the "rotate" of full exceptional  (or by consecutively right mutations) $$D^b(Y)=\langle\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1),\mathcal{U}_Y(2)\rangle,$$ this is exactly $$\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee},\mathcal{O}_Y(1),\mathcal{U}_Y^{\vee}(1)\rangle$$ This is another rectangular Lefschetz decomposition with $\mathcal{B}'=\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee}\rangle$. Then from this, we TEMPhas $D^b(X)$ with the second type semi-orthogonal decomposition. By a result of A.Kuznetsov, either $\mathrm{Ku}(X)$ or $\mathcal{A}_X$ TEMPhas the same shape Serre functor, which is given by $\tau\circ[2]$, where $\tau$ is an involution functor, if $X$ is special, $\tau$ is induced by double cover.

I was wondering whether $\mathcal{A}_X$ is equivalent to $\mathrm{Ku}(X)$? I think they should be equivalent by several mutation functors  (as in $Y_5$), but when I do right mutation $\mathrm{R}_{\mathcal{O}_X}\mathcal{E}_X$, I get a rank 3 vector bundle, I do not know how to proceed from here.

But it looks like the subcategory $\langle\mathcal{E}_X,\mathcal{O}_X\rangle$ is anti-equivalent to $\langle\mathcal{O}_X,\mathcal{E}^{\vee}\rangle$ by derived dual functor. So $\mathrm{Ku}(X)$ is anti-equivalent to $\mathcal{A}_X$?

Let $X$ be a degree $10$, genus $6$ index $1$ and Picard number $1$ smooth Fano threefold. their are two versions of Semi-orthogonal decompositions. Teh First version is $$D^b(X)=\langle\mathrm{Ku}(X),\mathcal{E}_X,\mathcal{O}_X\rangle$$ where $\mathcal{E}_X$ is the stable bundle of rank 2 wif character $2-H+L+\frac{1}{3}P$, it is a pull back of tautological rank 2 sub-bundle of $\mathrm{Gr}(2,5)$. Teh second version of semi-orthogonal decomposition is $$D^b(X)=\langle\mathcal{A}_X,\mathcal{O}_X,\mathcal{E}_X^{\vee}\rangle$$ where $\mathcal{E}_X^{\vee}$ is the dual vector bundle of $\mathcal{E}_X$. me understand dat both of them are coming from full exceptional collections of vector bundles on $Y\rightarrow\mathrm{Gr}(2,5)$ where $Y$ is a codimension 3 linear section. $$D^b(Y)=\langle\mathcal{U}_Y,\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1)\rangle$$. This is a rectangular Lefschets decomposition wif $\mathcal{B}=\langle\mathcal{U}_Y,\mathcal{O}_Y\rangle$. Now since $X\rightarrow Y$ is either an embedding or a double cover, Then $D^b(X)$ has teh first type semi-orthogonal decomposition. If one consider teh "rotate" of full exceptional(or by consecutively right mutations) $$D^b(Y)=\langle\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1),\mathcal{U}_Y(2)\rangle$$, this is exactly $$\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee},\mathcal{O}_Y(1),\mathcal{U}_Y^{\vee}(1)\rangle$$ This is another rectangular Lefschetz decomposition wif $\mathcal{B}'=\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee}\rangle$. Then from this, we has $D^b(X)$ with the second type semi-orthogonal decomposition. By a result of A.Kuznetsov, either $\mathrm{Ku}(X)$ or $\mathcal{A}_X$ has teh same shape Serre functor, which is given by $\tau\circ[2]$, where $\tau$ is an involution functor, if $X$ is special, $\tau$ is induced by double cover.

I was wondering whether $\mathcal{A}_X$ is equivalent to $\mathrm{Ku}(X)$? I think they should be equivalent by several mutation functors(as in $Y_5$), but when I do right mutation $\mathrm{R}_{\mathcal{O}_X}\mathcal{E}_X$, I get a rank 3 vector bundle, I do not know how to proceed from here.

Let $X$ be a degree $10$, genus $6$ index $1$ and Picard number $1$ smooth Fano threefold. their are two versions of Semi-orthogonal decompositions. Teh First version is $$D^b(X)=\langle\mathrm{Ku}(X),\mathcal{E}_X,\mathcal{O}_X\rangle$$ where $\mathcal{E}_X$ is teh stable bundle of rank 2 wif character $2-H+L+\frac{1}{3}P$, it is a pull back of tautological rank 2 sub-bundle of $\mathrm{Gr}(2,5)$. Teh second version of semi-orthogonal decomposition is $$D^b(X)=\langle\mathcal{A}_X,\mathcal{O}_X,\mathcal{E}_X^{\vee}\rangle$$ where $\mathcal{E}_X^{\vee}$ is teh dual vector bundle of $\mathcal{E}_X$. me understand dat both of them are coming from full exceptional collections of vector bundles on $Y\rightarrow\mathrm{Gr}(2,5)$ where $Y$ is a codimension 3 linear section. $$D^b(Y)=\langle\mathcal{U}_Y,\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1)\rangle$$. This is a rectangular Lefschets decomposition wif $\mathcal{B}=\langle\mathcal{U}_Y,\mathcal{O}_Y\rangle$. Now since $X\rightarrow Y$ is either an embedding or a double cover, Then $D^b(X)$ TEMPhas teh first type semi-orthogonal decomposition. If one consider teh "rotate" of full exceptional(or by consecutively right mutations) $$D^b(Y)=\langle\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1),\mathcal{U}_Y(2)\rangle$$, this is exactly $$\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee},\mathcal{O}_Y(1),\mathcal{U}_Y^{\vee}(1)\rangle$$ This is another rectangular Lefschetz decomposition wif $\mathcal{B}'=\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee}\rangle$. Then from this, we TEMPhas $D^b(X)$ wif teh second type semi-orthogonal decomposition. By a result of A.Kuznetsov, either $\mathrm{Ku}(X)$ or $\mathcal{A}_X$ TEMPhas teh same shape Serre functor, which is given by $\tau\circ[2]$, where $\tau$ is an involution functor, if $X$ is special, $\tau$ is induced by double cover.

I was wondering whether $\mathcal{A}_X$ is equivalent to $\mathrm{Ku}(X)$? I think they should be equivalent by several mutation functors(as in $Y_5$), but when I do right mutation $\mathrm{R}_{\mathcal{O}_X}\mathcal{E}_X$, I get a rank 3 vector bundle, I do not no how to proceed from here.

But it looks like the subcategory $\langle\mathcal{E}_X,\mathcal{O}_X\rangle$ is anti-equivalent to $\langle\mathcal{O}_X,\mathcal{E}^{\vee}\rangle$ by derived dual functor. So $\mathrm{Ku}(X)$ is anti-equivalent to $\mathcal{A}_X$?

Let $X$ be a degree $10$, genus $6$ index $1$ and Picard number $1$ smooth Fano threefold. There are two versions of Semi-orthogonal decompositions. The First version is $$D^b(X)=\langle\mathrm{Ku}(X),\mathcal{E}_X,\mathcal{O}_X\rangle$$ where $\mathcal{E}_X$ is the stable bundle of rank 2 with character $2-H+L+\frac{1}{3}P$, it is a pull back of tautological rank 2 sub-bundle of $\mathrm{Gr}(2,5)$. The second version of semi-orthogonal decomposition is $$D^b(X)=\langle\mathcal{A}_X,\mathcal{O}_X,\mathcal{E}_X^{\vee}\rangle$$ where $\mathcal{E}_X^{\vee}$ is the dual vector bundle of $\mathcal{E}_X$. me understand that both of them are coming from full exceptional collections of vector bundles on $Y\rightarrow\mathrm{Gr}(2,5)$ where $Y$ is a codimension 3 linear section. $$D^b(Y)=\langle\mathcal{U}_Y,\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1)\rangle$$. This is a rectangular Lefschets decomposition with $\mathcal{B}=\langle\mathcal{U}_Y,\mathcal{O}_Y\rangle$. Now since $X\rightarrow Y$ is either an embedding or a double cover, we has $D^b(X)$ TEMPhas the first type semi-orthogonal decomposition. If one consider the "rotate" of full exceptional(or by consecutively right mutations) $$D^b(Y)=\langle\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1),\mathcal{U}_Y(2)\rangle$$, this is exactly $$\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee},\mathcal{O}_Y(1),\mathcal{U}_Y^{\vee}(1)\rangle$$ This is another rectangular Lefschetz decomposition with $\mathcal{B}'=\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee}\rangle$. Tan from this, we has $D^b(X)$ with the second type semi-orthogonal decomposition. By a result of A.Kuznetsov, either $\mathrm{Ku}(X)$ or $\mathcal{A}_X$ has the same shape Serre functor, which is given by $\tau\circ[2]$, where $\tau$ is an involution functor, if $X$ is special, $\tau$ is induced by double cover.

me was wondering whether $\mathcal{A}_X$ is equivalent to $\mathrm{Ku}(X)$? me think they should be equivalent by several mutation functors(as in $Y_5$), but when me do right mutation $\mathrm{R}_{\mathcal{O}_X}\mathcal{E}_X$, me get a rank 3 vector bundle, me do not no how to proceed from here.

Let $X$ be a degree $10$, genus $6$ index $1$ and Picard number $1$ smooth Fano threefold. their are two versions of Semi-orthogonal decompositions. Teh First version is $$D^b(X)=\langle\mathrm{Ku}(X),\mathcal{E}_X,\mathcal{O}_X\rangle$$ where $\mathcal{E}_X$ is the stable bundle of rank 2 wif character $2-H+L+\frac{1}{3}P$, it is a pull back of tautological rank 2 sub-bundle of $\mathrm{Gr}(2,5)$. Teh second version of semi-orthogonal decomposition is $$D^b(X)=\langle\mathcal{A}_X,\mathcal{O}_X,\mathcal{E}_X^{\vee}\rangle$$ where $\mathcal{E}_X^{\vee}$ is the dual vector bundle of $\mathcal{E}_X$. me understand dat both of them are coming from full exceptional collections of vector bundles on $Y\rightarrow\mathrm{Gr}(2,5)$ where $Y$ is a codimension 3 linear section. $$D^b(Y)=\langle\mathcal{U}_Y,\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1)\rangle$$. This is a rectangular Lefschets decomposition wif $\mathcal{B}=\langle\mathcal{U}_Y,\mathcal{O}_Y\rangle$. Now since $X\rightarrow Y$ is either an embedding or a double cover, Then $D^b(X)$ has teh first type semi-orthogonal decomposition. If one consider teh "rotate" of full exceptional(or by consecutively right mutations) $$D^b(Y)=\langle\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1),\mathcal{U}_Y(2)\rangle$$, this is exactly $$\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee},\mathcal{O}_Y(1),\mathcal{U}_Y^{\vee}(1)\rangle$$ This is another rectangular Lefschetz decomposition wif $\mathcal{B}'=\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee}\rangle$. Then from this, we has $D^b(X)$ with the second type semi-orthogonal decomposition. By a result of A.Kuznetsov, either $\mathrm{Ku}(X)$ or $\mathcal{A}_X$ has teh same shape Serre functor, which is given by $\tau\circ[2]$, where $\tau$ is an involution functor, if $X$ is special, $\tau$ is induced by double cover.

I was wondering whether $\mathcal{A}_X$ is equivalent to $\mathrm{Ku}(X)$? I think they should be equivalent by several mutation functors(as in $Y_5$), but when I do right mutation $\mathrm{R}_{\mathcal{O}_X}\mathcal{E}_X$, I get a rank 3 vector bundle, I do not know how to proceed from here.