# nLab Bondal-Orlov reconstruction theorem

Bondal-Orlov reconstruction theorem

# Bondal-Orlov reconstruction theorem

## Idea

Thomason and Balmer showed that the derived category of coherent sheaves on a smooth variety, when considered as a monoidal category (i.e. with the tensor product) in addition to its triangulated category structure (i.e. as a tensor triangulated category), completely determines the variety uniquely; see spectrum of a triangulated category. However, the derived category still turns out to be an interesting invariant when considered without the monoidal category structure; moreover triangulated equivalences and autoequivalences are also important in relation to the homological mirror symmetry and similar phenomena. In fact, Bondal and Orlov show how to reconstruct a smooth variety from its derived category of coherent sheaves when its canonical sheaf is ample or anti-ample, using only the graded structure (i.e. the translation functor).

## Statement

Let $X$ be a smooth projective variety and suppose that its canonical sheaf $\omega_X$ is ample or anti-ample. If $Y$ is another smooth projective variety such that there is an equivalence $F : D^b(X) \stackrel{\sim}{\to} D^b(Y)$ between the bounded derived categories of coherent sheaves on $X$ and $Y$, then there is an isomorphism $X \simeq Y$.

## Sketch of proof

The idea of the proof is to characterize “cohomologically” the closed points, invertible sheaves, and Zariski topology.

The main tool we use is the Serre functor. Due to Grothendieck-Serre duality?, both $D^b(X)$ and $D^b(Y)$ have Serre functors $S_X = (\cdot \otimes \omega_X)[n]$ and $S_Y = (\cdot \otimes \omega_Y)[m]$, $n$ and $m$ being the dimensions of $X$ and $Y$ respectively.

For simplicity assume $\omega_X$ is ample; the anti-ample case is analogous.

Step 1. First we characterize the closed points of $X$ and $Y$.

###### Definition

Let $\mathcal{C}$ be a k-linear graded category?. A point object of $\mathcal{C}$ is an object $P$ satisfying

• (PO-1) $S(P) \simeq P[n]$,
• (PO-2) $\Hom^i(P, P) = 0$ for $i \lt 0$,
• (PO-3) $k(P) := \Hom(P, P)$ is a field.

It is clear that the skyscraper sheaf $\mathcal{O}_x = \mathrm{Sky}_x(k(x))$ of the residue field at any closed point $x \in X$ is a point object of $D^b(X)$, as is any translation $\mathcal{O}_x[r]$. In fact, when we impose the ampleness condition on $\omega_X$, it turns out that all point objects are of this form.

###### Proposition

If $\omega_X$ is ample, then a complex $\mathcal{F}^\bullet$ in $D^b(X)$ is a point object iff $\mathcal{F}^\bullet \simeq \mathcal{O}_x[r]$ for some closed point $x \in X$, $r \in \mathbb{Z}$.

In fact, even though we don’t know that $\omega_Y$ is also ample, we get the same result for $D^b(Y)$. Let $G$ be the inverse equivalence to $F$ and note that $F$ and $G$ preserve point objects. Suppose there was a point object $P$ in $D^b(Y)$ not isomorphic to any $\mathcal{O}_y[s]$, and note $G(P) \simeq \mathcal{O}_{x_0}[r]$ for some closed $x_0$. Now for any $y \in Y$, $\Hom(P, \mathcal{O}_y) = \Hom(G(P), G(\mathcal{O}_y)) = \Hom(\mathcal{O}_{x_0}[r], \mathcal{O}_x[r']) = 0$ since $x_0 \ne x$. But $\{\mathcal{O}_y : y \in Y\}$ form a spanning class? of $D^b(Y)$ which implies that $P = 0$.

Step 2. Next we characterize the invertible sheaves on both varieties.

###### Definition

Let $\mathcal{C}$ be a $k$-linear graded category. An invertible object of $\mathcal{C}$ is an object $L$ satisfying for some $s \in \mathbb{Z}$

• (IO-1) $\Hom^s(L, P) = k(P)$ for all point objects $P$,
• (IO-2) $\Hom^i(L, P) = 0$ for $i \ne s$.
###### Proposition

If all point objects of $D^b(X)$, for a smooth projective variety $X$, are translations of skyscraper sheaves of closed points, then all invertible objects of $D^b(X)$ are translations of invertible sheaves on $X$.

This means that the invertible objects of both $D^b(X)$ and $D^b(Y)$ are translations of invertible sheaves, by step 1.

Step 3. Now we establish a bijection between the sets of points of the varieties.

Since $\mathcal{O}_X$ is an invertible object, we can assume without loss of generality that $F(\mathcal{O}_X)$ is isomorphic to some invertible sheaf $\mathcal{L}_0$ on $Y$. For any $x \in X$,

$k(x) = \Hom(\mathcal{O}_X, \mathcal{O}_x) = \Hom(\mathcal{L}_0, \mathcal{O}_y[s])$

for some $y \in Y$, which implies $s=0$ and $F(\mathcal{O}_x) \simeq \mathcal{O}_y$, $k(x) \simeq k(y)$. We get a bijection of sets $\phi : X \to Y$ by mapping $x \mapsto y$.

Step 4. Now we can get an isomorphism of the canonical ring?s of the varieties.

Assume WLOG $F(\mathcal{O}_X) \simeq \mathcal{O}_Y$. Note that $S_X^k(\mathcal{O}_X)[-nk] \simeq \omega_X^{nk}$ for any $k$. By commutativity of $F$ with the Serre functor, $F(\omega_X^{nk}) = S_Y^k(\mathcal{O}_Y)[-nk] = \omega_Y^{nk}$. Therefore

$H^0(X, \omega_X^{nk}) = \Hom(\mathcal{O}_X, \omega_X^{nk}) = \Hom(\mathcal{O}_Y, \otimes_Y^{nk}) = H^0(Y, \omega_Y^{nk})$

and $\bigoplus_k H^0(X, \omega_X^{nk}) \simeq \bigoplus_k H^0(Y, \omega_Y^{nk})$, that is we have an isomorphism of the canonical rings $A(X)$ and $A(Y)$.

Step 5. Since $X = \mathrm{Proj}(A(X))$ iff $\omega_X$ is ample, it remains therefore only to show that $\omega_Y$ is ample. To do this we characterize the Zariski topology “cohomologically”.

For any invertible sheaf $\mathcal{L}$ on $X$, $\alpha \in \Hom(\mathcal{O}_X, \mathcal{L})$, and $x \in X$, let $\alpha_X^* : \Hom(\mathcal{L}, \mathcal{O}_x) \to \Hom(\mathcal{O}_X, \mathcal{O}_x)$ be the induced map that takes $f$ to $f \circ \alpha$, and define the subset $U_\alpha = \{ x \in X : \alpha_x^* \ne 0 \} \subset X$. Now $\omega_X$ being ample is equivalent to the subcollection $\{ U_\alpha : \alpha \in \Hom(\mathcal{O}_X, \omega_X^m), m \in \mathbb{Z} \}$ forming a basis of the Zariski topology on $X$. Now the equivalence $F$ maps any $\mathcal{O}_X \stackrel{\alpha}{\to} \mathcal{L} \to \mathcal{O}_x$ to

$\mathcal{O}_Y \stackrel{F(\alpha)}{\to} F(\mathcal{L}) \to \mathcal{O}_{\phi(x)}[s].$

$F(\alpha)$ corresponds bijectively to some $\beta \in \Hom(\mathcal{O}_Y, \omega_Y^m)$ and in fact our bijection $f : X \to Y$ becomes a homeomorphism mapping $U_\alpha$ to $V_{F(\alpha)}$. It follows that $\{ V_\beta : \beta \in \Hom(\mathcal{O}_Y, \omega_Y^m), m \in \mathbb{Z} \}$ forms a basis of the Zariski topology on $Y$, which implies $\omega_Y$ is ample.