higher geometry / derived geometry
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derived smooth geometry
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).
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 $X \simeq Y$.
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 $Y$.
Let $\mathcal{C}$ be a k-linear graded category?. A point object of $\mathcal{C}$ is an object $P$ satisfying
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.
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.
Let $\mathcal{C}$ be a $k$-linear graded category. An invertible object of $\mathcal{C}$ is an object $L$ satisfying for some $s \in \mathbb{Z}$
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 $x \in X$,
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
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
$F(\alpha)$ 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.
The original paper:
A nice exposition of the proof with more details can be found in section 3.1 of the course notes of Igor Dolgachev on derived categories:
The theorem is also discussed in:
Raphaël Rouquier, Catégories dérivées et géométrie birationnelle, Séminaire Bourbaki, no 947, March 2005.
Daniel Huybrechts, Fourier-Mukai transforms in algebraic geometry, 2006.
Last revised on January 1, 2015 at 15:21:17. See the history of this page for a list of all contributions to it.