The Stacks project

20.52 Projection formula

In this section we collect variants of the projection formula. The most basic version is Lemma 20.52.2. After we state and prove it, we discuss a more general version involving perfect complexes.

Lemma 20.52.1. Let $X$ be a ringed space. Let $\mathcal{I}$ be an injective $\mathcal{O}_ X$-module. Let $\mathcal{E}$ be an $\mathcal{O}_ X$-module. Assume $\mathcal{E}$ is finite locally free on $X$, see Modules, Definition 17.14.1. Then $\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{I}$ is an injective $\mathcal{O}_ X$-module.

Proof. This is true because under the assumptions of the lemma we have

\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{I}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}( \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{E}^\vee , \mathcal{I}) \]

where $\mathcal{E}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}, \mathcal{O}_ X)$ is the dual of $\mathcal{E}$ which is finite locally free also. Since tensoring with a finite locally free sheaf is an exact functor we win by Homology, Lemma 12.27.2. $\square$

Lemma 20.52.2. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Let $\mathcal{E}$ be an $\mathcal{O}_ Y$-module. Assume $\mathcal{E}$ is finite locally free on $Y$, see Modules, Definition 17.14.1. Then there exist isomorphisms

\[ \mathcal{E} \otimes _{\mathcal{O}_ Y} R^ qf_*\mathcal{F} \longrightarrow R^ qf_*(f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F}) \]

for all $q \geq 0$. In fact there exists an isomorphism

\[ \mathcal{E} \otimes _{\mathcal{O}_ Y} Rf_*\mathcal{F} \longrightarrow Rf_*(f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F}) \]

in $D^{+}(Y)$ functorial in $\mathcal{F}$.

Proof. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $ on $X$. Note that $f^*\mathcal{E}$ is finite locally free also, hence we get a resolution

\[ f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F} \longrightarrow f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{I}^\bullet \]

which is an injective resolution by Lemma 20.52.1. Apply $f_*$ to see that

\[ Rf_*(f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F}) = f_*(f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{I}^\bullet ). \]

Hence the lemma follows if we can show that $f_*(f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F}) = \mathcal{E} \otimes _{\mathcal{O}_ Y} f_*(\mathcal{F})$ functorially in the $\mathcal{O}_ X$-module $\mathcal{F}$. This is clear when $\mathcal{E} = \mathcal{O}_ Y^{\oplus n}$, and follows in general by working locally on $Y$. Details omitted. $\square$

Let $f : X \to Y$ be a morphism of ringed spaces. Let $E \in D(\mathcal{O}_ X)$ and $K \in D(\mathcal{O}_ Y)$. Without any further assumptions there is a map
\begin{equation} \label{cohomology-equation-projection-formula-map} Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_ Y} K \longrightarrow Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} Lf^*K) \end{equation}

Namely, it is the adjoint to the canonical map

\[ Lf^*(Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_ Y} K) = Lf^*Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_ X} Lf^*K \longrightarrow E \otimes ^\mathbf {L}_{\mathcal{O}_ X} Lf^*K \]

coming from the map $Lf^*Rf_*E \to E$ and Lemmas 20.27.3 and 20.28.1. A reasonably general version of the projection formula is the following.

Lemma 20.52.3. Let $f : X \to Y$ be a morphism of ringed spaces. Let $E \in D(\mathcal{O}_ X)$ and $K \in D(\mathcal{O}_ Y)$. If $K$ is perfect, then

\[ Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_ Y} K = Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} Lf^*K) \]

in $D(\mathcal{O}_ Y)$.

Proof. To check ( is an isomorphism we may work locally on $Y$, i.e., we have to find a covering $\{ V_ j \to Y\} $ such that the map restricts to an isomorphism on $V_ j$. By definition of perfect objects, this means we may assume $K$ is represented by a strictly perfect complex of $\mathcal{O}_ Y$-modules. Note that, completely generally, the statement is true for $K = K_1 \oplus K_2$, if and only if the statement is true for $K_1$ and $K_2$. Hence we may assume $K$ is a finite complex of finite free $\mathcal{O}_ Y$-modules. In this case a simple argument involving stupid truncations reduces the statement to the case where $K$ is represented by a finite free $\mathcal{O}_ Y$-module. Since the statement is invariant under finite direct summands in the $K$ variable, we conclude it suffices to prove it for $K = \mathcal{O}_ Y[n]$ in which case it is trivial. $\square$

Here is a case where the projection formula is true in complete generality.

Lemma 20.52.4. Let $f : X \to Y$ be a morphism of ringed spaces such that $f$ is a homeomorphism onto a closed subset. Then ( is an isomorphism always.

Proof. Since $f$ is a homeomorphism onto a closed subset, the functor $f_*$ is exact (Modules, Lemma 17.6.1). Hence $Rf_*$ is computed by applying $f_*$ to any representative complex. Choose a K-flat complex $\mathcal{K}^\bullet $ of $\mathcal{O}_ Y$-modules representing $K$ and choose any complex $\mathcal{E}^\bullet $ of $\mathcal{O}_ X$-modules representing $E$. Then $Lf^*K$ is represented by $f^*\mathcal{K}^\bullet $ which is a K-flat complex of $\mathcal{O}_ X$-modules (Lemma 20.26.8). Thus the right hand side of ( is represented by

\[ f_*\text{Tot}(\mathcal{E}^\bullet \otimes _{\mathcal{O}_ X} f^*\mathcal{K}^\bullet ) \]

By the same reasoning we see that the left hand side is represented by

\[ \text{Tot}(f_*\mathcal{E}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{K}^\bullet ) \]

Since $f_*$ commutes with direct sums (Modules, Lemma 17.6.3) it suffices to show that

\[ f_*(\mathcal{E} \otimes _{\mathcal{O}_ X} f^*\mathcal{K}) = f_*\mathcal{E} \otimes _{\mathcal{O}_ Y} \mathcal{K} \]

for any $\mathcal{O}_ X$-module $\mathcal{E}$ and $\mathcal{O}_ Y$-module $\mathcal{K}$. We will check this by checking on stalks. Let $y \in Y$. If $y \not\in f(X)$, then the stalks of both sides are zero. If $y = f(x)$, then we see that we have to show

\[ \mathcal{E}_ x \otimes _{\mathcal{O}_{X, x}} (\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{F}_ y) = \mathcal{E}_ x \otimes _{\mathcal{O}_{Y, y}} \mathcal{F}_ y \]

(using Sheaves, Lemma 6.32.1 and Lemma 6.26.4). This equality holds and therefore the lemma has been proved. $\square$

Remark 20.52.5. The map ( is compatible with the base change map of Remark 20.28.3 in the following sense. Namely, suppose that

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]

is a commutative diagram of ringed spaces. Let $E \in D(\mathcal{O}_ X)$ and $K \in D(\mathcal{O}_ Y)$. Then the diagram

\[ \xymatrix{ Lg^*(Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_ Y} K) \ar[r]_ p \ar[d]_ t & Lg^*Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} Lf^*K) \ar[d]_ b \\ Lg^*Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_{Y'}} Lg^*K \ar[d]_ b & Rf'_*L(g')^*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} Lf^*K) \ar[d]_ t \\ Rf'_*L(g')^*E \otimes ^\mathbf {L}_{\mathcal{O}_{Y'}} Lg^*K \ar[rd]_ p & Rf'_*(L(g')^*E \otimes ^\mathbf {L}_{\mathcal{O}_{Y'}} L(g')^*Lf^*K) \ar[d]_ c \\ & Rf'_*(L(g')^*E \otimes ^\mathbf {L}_{\mathcal{O}_{Y'}} L(f')^*Lg^*K) } \]

is commutative. Here arrows labeled $t$ are gotten by an application of Lemma 20.27.3, arrows labeled $b$ by an application of Remark 20.28.3, arrows labeled $p$ by an application of (, and $c$ comes from $L(g')^* \circ Lf^* = L(f')^* \circ Lg^*$. We omit the verification.

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01E6. Beware of the difference between the letter 'O' and the digit '0'.