The Stacks project

21.50 Projection formula

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

Namely, it is the adjoint to the canonical map

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

coming from the map $Lf^*Rf_*E \to E$ and Lemmas 21.18.4 and 21.19.1. A reasonably general version of the projection formula is the following.

Lemma 21.50.1. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D})$ be a morphism of ringed topoi. Let $E \in D(\mathcal{O}_\mathcal {C})$ and $K \in D(\mathcal{O}_\mathcal {D})$. If $K$ is perfect, then

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

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

Proof. To check ( is an isomorphism we may work locally on $\mathcal{D}$, i.e., for any object $V$ of $\mathcal{D}$ we have to find a covering $\{ V_ j \to V\} $ 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 $\mathcal{O}_\mathcal {D}$-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 $\mathcal{O}_\mathcal {D}$-modules. In this case a simple argument involving stupid truncations reduces the statement to the case where $K$ is represented by a finite free $\mathcal{O}_\mathcal {D}$-module. Since the statement is invariant under finite direct summands in the $K$ variable, we conclude it suffices to prove it for $K = \mathcal{O}_\mathcal {D}[n]$ in which case it is trivial. $\square$

Remark 21.50.2. The map ( is compatible with the base change map of Remark 21.19.3 in the following sense. Namely, suppose that

\[ \xymatrix{ (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}_{\mathcal{C}'}) \ar[r]_{g'} \ar[d]_{f'} & (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \ar[d]^ f \\ (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}'), \mathcal{O}_{\mathcal{D}'}) \ar[r]^ g & (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D}) } \]

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

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

is commutative. Here arrows labeled $t$ are gotten by an application of Lemma 21.18.4, arrows labeled $b$ by an application of Remark 21.19.3, 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 0943. Beware of the difference between the letter 'O' and the digit '0'.