Famous Fourier-Mukai Results II (Orlov’s Result and the Beilinson Resolution)

This a continuation of this post, and this post follows the paper of Orlov.  I’m going to give a rough outline to the following result

(Orlov’s Result) Any functor $F \colon D^b(A) \to D^b(B)$ which is full, faithful and exact is represented by an object on the product.

The proof is long and complicated so I’ll only attempt to give a flavor of the ideas used in the proof.  One central part of the proof is the Beilinson resolution of the diagonal of projective space:

$0 \to G^{-N} \to G^{-N + 1} \to ... \to G^{-1} \to O \boxtimes O \to O_\Delta \to 0$ (0)

where $O$ is the structure sheaf on $\mathbb{P}^N$ and $G^{-i} = O(-i)\boxtimes \Omega^i(i)$.

Constructing the resolution

Start with what is sometimes called the Euler sequence on $\mathbb{P}^N$ (here $\mathscr{T}$ is the tangent bundle):

$0 \to O(-1) \to O^{n+1} \to \mathscr{T}(-1) \to 0$

locally:

$k \cdot \vec v \subset k^{n+1} \twoheadrightarrow k^{n+1}/k\cdot \vec v$ (1)

Note $H^o(O(1)) = V^*$ for $V = k^{n+1}$ and $H^0(\mathscr{T}(-1) = V$ from the les in cohomology.  Now from the Kunneth formula in algebraic geometry:

$H^0(\mathbb{P}^N\times \mathbb{P}^N, O(1)\boxtimes \mathscr{T}(-1)) = V^* \otimes V = \hom(V,V)$

I’ll make use of the global section corresponding to the identity.  Now $O(1)\boxtimes \mathscr{T}(-1) = \mathcal{H}om(O_{\mathbb{P}^N\times \mathbb{P}^N}, O(1)\boxtimes \mathscr{T}(-1)) = \mathcal{H}om(p_1^*O(-1), p_2^*\mathscr{T}(-1))$.  Locally the identity corresponds to a map

$s \colon p_1^*O(-1) \to p_2^*\mathscr{T}(-1)$

so at $(\bar v, \bar w ) \in \mathbb{P}^N \times \mathbb{P}^N$ this is

$s \colon O(-1)|_{\bar v} \to \mathscr{T}(-1)|_{\bar w}$

and in view of (1) its locally the inclusion followed by the projection:

$k\cdot v \subset k^{n+1} \to k^{n+1}/k \cdot w$

the only point in this local description is that $s = 0$ iff $v \in k\cdot w$, i.e. $\bar v = \bar w$, in other words $s$ vanishes exactly on the diagonal.

If $v \in V$ is a vector then there is a contraction map $\wedge^i V^* \to \wedge^{i-1} V^*$ via

$w_1 \wedge ... \wedge w_i \mapsto \sum_j (-1)^j w_j(v) w_1 \wedge ... \widehat{w_j} ...\wedge w_i$.

Now contraction with $s$ gives a map $\wedge^i \bigl(O(-1)\boxtimes \Omega(1)\bigr) \to \wedge^{i-1} \bigl(O(-1)\boxtimes \Omega(1)\bigr)$ and these are exactly the maps that appear in the Beilinson resolution.

A rough outline

Now to Orlov’s result.  From the data $F \colon D^b(A) \to D^b(B)$ I need to produce an object $E \in D^b(A\times B)$.

The first step is to use that $A$ is projective to get an embedding $j \colon A \to \mathbb{P}$ and consider the functor

$F' \colon D^b(\mathbb{P}^N) \xrightarrow{j^*}D^b(A) \to D^b(B)$

Now that $\mathbb{P}^N$ has entered the picture we can utilize the Beilinson resolution (0), and using $F'$ obtain a complex

$0 \to H^{-N} \to H^{-N + 1} \to ... \to G^{-1} \to H^0$ (2)

where $H^{-i} = O(-i)\boxtimes F'(\Omega^i(i))$.  I’m brushing a lot under the rug, it takes a bit of work to actually come up with this complex.

Convolution

Now I need a powerful tool which I don’t have a firm grasp of and I can’t really explain here.

Let $X^{-i} \to X^{-i+1} \to ... \to X^{-1} \to X^0$ be a bounded complex.  A left Postnikov system of $X^\bullet$ is a diagram:

where the stared triangles are distinguished and the triangles with circles are commutative.  An object $E$ is a left convolution of $X^\bullet$ if there is a left Postnikov system such that $E = Y_0$.  Denote by $Tot(X^\bullet)$ the class of all convolutions of $X^\bullet$.

Prop. If $\hom(X^a,X^b)[i] = 0$ for $i < 0$ and $a< b$ then $X^\bullet$ has a convolution $Y^0$.  Further, if $\hom(X^a,Y^0)[i] = 0 \forall a$ and $i<0$ then all convolutions are canonically isomorphic.

Remark: I don’t have a great way of motivating this convolution business but its not a very geometric tool and in this post I’m trying to focus on the geometry that goes into this proof.

Continuing, the idea is now to the use the proposition with the complex (2) to obtain an object $E' \in D^b(\mathbb{P}^n \times B)$.  Next one can show that $F'(O(k)) \cong \Phi_{E'}(O(k))$ basically by showing that both are convolutions of the same complex.  With this result one can ultimately show $F' \cong \Phi_{E'}$.  I’m not including the details because I want to focus on the rough idea and I want to avoid making an overly long post.

What’s Left

The functor $F' = F \circ j^*$ has been represented by an object on the product.  Using general Fourier-Mukai properties see e.g. this post, it remains to find an object $E \in D^b(A\times B)$ such that $Rj_*E' = E$.

The object $E$ is produced in much the same way as $E'$ is produced.  Using and ample line bundle on $A$ obtain a resolution for the diagonal on $A\times A$ and using $F$ obtain a complex on $A \times B$ much like (2), then use the proposition to get a convolution $G \in D^b(A\times B)$.  The details are a little different and more complicated because $\mathbb{P}^N$ is understandably more explicit than a general smooth projective variety $A$.

The object $G$ does not represent $F$.  Instead, using cohomological properties, one decomposes $G = C \oplus E$ and $E$ and then a another argument is needed go show $Rj_*E = E'$.

I’m tempted to say that this was much less then even a rough outline of the proof.  But I really only wanted to discuss the Beilinson resolution and even though I was brief with Orlov’s result I think its clear that the Beilinson resolution is one of the key ideas behind it.