Let $Y$ be a smooth cubic 4-fold in $\mathbf{P}^5$. The derived category of $Y$ admits a semiorthogonal decomposition

$$D^b(Y) = \langle \mathcal{A}_Y, \mathcal{O}_Y, \mathcal{O}_{Y}(1), \mathcal{O}_{Y}(2) \rangle,$$

where $\mathcal{A}_Y$ is the right-orthogonal to the triangulated subcategory generated by $\mathcal{O}_Y, \mathcal{O}_{Y}(1), \mathcal{O}_{Y}(2)$. Namely, $\mathcal{A}_Y$ is the full subcategory with objects those $F \in D^b(Y)$ such that $Hom_{D^b(Y)}(\mathcal{O}_Y(j), F[i]) = 0$ for $j=0,1,2$ and all $i$.

Kuznetsov studies $\mathcal{A}_Y$ in his paper "Derived categories of cubic fourfolds" http://arxiv.org/abs/0808.3351. He parenthetically remarks that the Serre functor of $\mathcal A_Y$ is the shift by two functor $[2]$.

The question is: How do you show this?

The only reference I could find for this result is Corollary 4.4 of the paper http://arxiv.org/abs/math/0303037. However, I don't understand the proof given there. Specifically, I don't know how to justify the following points in Lemma 4.2 preceding the Corollary (notation as in the paper):

1) How do you get the expression Kuznetsov gives for the kernel of $O^d$ from the kernel of $O$?

2) Why is the restriction to $Y \times Y$ of the Beilinson resolution of the diagonal in $\mathbf{P}^{n+1} \times \mathbf{P}^{n+1}$ quasi-isomorphic to ${\Delta_{Y}}_*(\mathcal O_{Y})$? It seems the paper is being sloppy here with what degrees the complexes live in: Let's call $A$ the restriction of the Beilinson resolution (including the final term ${\Delta_{Y}}_*(\mathcal O_{Y}(d))$ as in the paper). Let's place A in nonpositive degrees so that ${\Delta_{Y}}_*(\mathcal O_{Y}(d))$ is in degree $0$ (I assume this is what is intended in the paper). Then I think the precise statement should be that $A$ is quasi-isomorphic to ${\Delta_{Y}}_*(\mathcal O_{Y})[2]$. Moreover, since $A$ is the restriction of a resolution of $\Delta_*(\mathcal O_{\mathbf{P}^{n+1}}(d))$ by locally free sheaves, this can be rephrased as the following computation of left-derived functors: if $i : Y \times Y \hookrightarrow \mathbf{P}^{n+1} \times \mathbf{P}^{n+1}$ is the inclusion, then $$L_1 i^*(\Delta_*(\mathcal O_{\mathbf{P}^{n+1}}(d))) = {\Delta_{Y}}_*(\mathcal O_{Y}) $$ and $$L_k i^*(\Delta_*(\mathcal O_{\mathbf{P}^{n+1}}(d))) = 0$$ for $k > 1$. How do you show this?


1 Answer 1


The first is a straightforward computation. For example, to compute $O^2$ you note that the convolution of kernels preserves exactness of the triangles, hence one has distinguished triangles $$ \Delta_{Y*}O_Y(1) \circ K_1 \to O_Y(2)\boxtimes O_Y \to \Delta_*O_Y(2), $$ $$ (O_Y(1)\boxtimes O_Y)\circ K_1 \to V^*\otimes O_Y(1)\boxtimes O \to O_Y(1)\boxtimes O_Y(1), $$ (where $V^* = H^0(O_Y(1))$), and $$ K_1\circ K_1 \to (O_Y(1)\boxtimes O_Y)\circ K_1 \to \Delta_{Y*}O_Y(1) \circ K_1. $$ The second triangle implies that $(O_Y(1)\boxtimes O_Y)\circ K_1 \cong O_Y(1)\boxtimes \Omega(1)_{|Y}$, so substituting this and the first triangle into the third triangle you get an expression for $O^2$. Then you repeat this computation.

For the second you need to compute $L(\alpha\times\alpha)^*\Delta_*O$ (in the notation of the paper). Let us first compute $(\alpha\times\alpha)_*L(\alpha\times\alpha)^*\Delta_*O$ instead. By projection formula it is isomorphic to $$ (\alpha\times\alpha)_*L(\alpha\times\alpha)^*\Delta_*O \cong (\alpha\times\alpha)_*O_{Y\times Y} \otimes^L \Delta_*O_P. $$ To compute this we use the resolution $$ 0 \to O(-d)\boxtimes O(-d) \to O(-d)\boxtimes O \oplus O \boxtimes O(-d) \to O \boxtimes O $$ of $(\alpha\times\alpha)_*O_{Y\times Y}$. Tensoring it with $\Delta_*O_P$ one gets the complex $$ 0 \to \Delta_*O_P(-2d) \to \Delta_*O_P(-d) \oplus \Delta_*O_P(-d) \to \Delta_*O_P $$ which is a direct sum od the resolution of $\Delta_*O_Y$ and of $\Delta_*O_Y(-d)[1]$. Thus $$ (\alpha\times\alpha)_*L(\alpha\times\alpha)^*\Delta_*O \cong \Delta_*O_Y \oplus \Delta_*O_Y(-d)[1]. $$ Finally, since $\alpha\times\alpha$ is a closed embedding, the functor $(\alpha\times\alpha)_*$ is exact and conservative, hence the claim.


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