Base change

Recall in commutative algebras given a diagram
$$\begin{array}{ccc}B& \to & R\\ \downarrow & & \\ S& & \end{array}$$ we can form a pushout (or fiber coproduct) $$\begin{array}{ccc}B& \to & R\\ \downarrow & & \downarrow \\ S& \to & S{\otimes }_{B}R\end{array}$$ using the tensor product $S{\otimes }_{B}R$.

Consequently, in the category of affine schemes, we have fiber products (or pullbacks) $$\begin{array}{ccc}X{\times }_{Z}Y& \to & Y\\ \downarrow & & \downarrow \\ X& \to & Z\end{array}$$ where, if $X=\mathrm{Spec}R$, $Y=\mathrm{Spec}S$, and $Z=\mathrm{Spec}B$, we have $$X{\times }_{Z}Y=\mathrm{Spec}(S{\otimes }_{B}R).$$

For schemes generally, we can use this to build fiber products. Given $$\begin{array}{ccc}& & Y\\ & & \downarrow g\\ X& \underset{f}{\to }& Z\end{array}$$ $X{\times }_{Z}Y$ exists. Given affine $U\subset X,V\subset Y$, and $W\subset Z$ with $f(U),g(V)\subset W$, we have an affine chart $$U{\times }_{W}V$$ and these can be glued together appropriately to give the fiber product as a scheme.

A square $$\begin{array}{ccc}W& \to & Y\\ \downarrow & & \downarrow \\ X& \to & Z\end{array}$$ isomorphic to a fiber product square is commonly called Cartesian.

Given a Cartesian square $$\begin{array}{ccc}W& \underset{\tilde{f}}{\to }& Y\\ \downarrow \tilde{g}& & \downarrow g\\ X& \underset{f}{\to }& Z\end{array}$$ and a $E\in D(\mathrm{Qcoh}X)$ can we compare $$\mathbf{L}{g}^{\ast }\mathbf{R}{f}_{\ast }E\text{and }\mathbf{R}{\tilde{f}}_{\ast }\mathbf{L}{\tilde{g}}^{\ast }E$$ In general, they will not be quasi-isomorphic. However, under some assumptions, this is the case.

Theorem. Assume that $$\mathcal{T}o{r}_i^{{\mathcal{O}}_Z}({\mathcal{O}}_X,{\mathcal{O}}_Y)=0$$ for $i\ne 0$. Then, there is a natural isomorphism $$\mathbf{L}{g}^{\ast }\mathbf{R}{f}_{\ast }E\cong \mathbf{R}{\tilde{f}}_{\ast }\mathbf{L}{\tilde{g}}^{\ast }E$$

In particular, if either ${\mathcal{O}}_X$ or ${\mathcal{O}}_Y$ is flat over ${\mathcal{O}}_Z$, then we can exchange pushing and pulling.

The previous result is often called (flat) base change. We won't provide more details on this result. Lipman has some very thorough notes on Grothendieck duality which include exposition on Tor-independent base change.

Along with the projection formula, base change will serve as regular tool in our future investigations of derived categories.

Kernels and integral transforms

We saw last semester that for two rings $S$ and $R$ the category of $(S,R)$-bimodules was an imporant source of functors between $D(\mathrm{Mod}S)$ and $D(\mathrm{Mod}R)$.

Here we globalize this idea. Note that an $(S,R)$-bimodule is the same data as a $S{\otimes }_{\mathbb{Z}}R$-module. Geometrically, $S\otimes R$ corresponds to the fiber product. Therefore, it is natural to look to objects on the (fiber) product for a source of functors.

For $$X\stackrel{{\pi }_X}{\leftarrow }X{\times }_{Z}Y\stackrel{{\pi }_Y}{\to }Y$$ and $Q\in D(\mathrm{Qcoh}X{\times }_{Z}Y)$, we get a functor $$\begin{array}{rl}{\Phi }_Q:D(\mathrm{Qcoh}X)& \to D(\mathrm{Qcoh}Y)\\ E& \mapsto \mathbf{R}{\pi }_{Y\ast }\left(\mathbf{L}{\pi }_X^{\ast }E\stackrel{\mathbf{L}}{\otimes }Q\right)\end{array}$$

We call ${\Phi }_Q$ an integral transform and $Q$ its kernel. If ${\Phi }_Q$ is an equivalence, we say ${\Phi }_Q$ and $Q$ are Fourier-Mukai and $X$ and $Y$ are Fourier-Mukai partners.

The language is inspired by the analogy with integral transformations in analysis. $$f(x)\mapsto {\int }_{X}Q(x,y)f(x)dx.$$

Each integral transform comes with a right adjoint thanks to Grothendieck duality.

Lemma. We have $${\Phi }_Q⊣{\Psi }_Q:=\mathbf{R}{\pi }_{X\ast }\mathbf{R}\mathcal{H}o{m}_{X\times Y}(Q,{\pi }_Y^{!}(-))$$