Lemma 57.7.1. Let $f : X' \to X$ be a proper birational morphism of integral Noetherian schemes with $X$ regular. The map $\mathcal{O}_ X \to Rf_*\mathcal{O}_{X'}$ canonically splits in $D(\mathcal{O}_ X)$.
57.7 Representability in the regular proper case
Theorem 57.6.3 also holds for regular (for example smooth) proper varieties. This is proven in [BvdB] using their general characterization of “right saturated” $k$-linear triangulated categories. In this section we give a quick and dirty proof of this result using a little bit of duality.
Proof. Set $E = Rf_*\mathcal{O}_{X'}$ in $D(\mathcal{O}_ X)$. Observe that $E$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.11.3. By Derived Categories of Schemes, Lemma 36.11.8 we find that $E$ is a perfect object of $D(\mathcal{O}_ X)$. Since $\mathcal{O}_{X'}$ is a sheaf of algebras, we have the relative cup product $\mu : E \otimes _{\mathcal{O}_ X}^\mathbf {L} E \to E$ by Cohomology, Remark 20.28.7. Let $\sigma : E \otimes E^\vee \to E^\vee \otimes E$ be the commutativity constraint on the symmetric monoidal category $D(\mathcal{O}_ X)$ (Cohomology, Lemma 20.47.6). Denote $\eta : \mathcal{O}_ X \to E \otimes E^\vee $ and $\epsilon : E^\vee \otimes E \to \mathcal{O}_ X$ the maps constructed in Cohomology, Example 20.47.7. Then we can consider the map
We claim that this map is a one sided inverse to the map in the statement of the lemma. To see this it suffices to show that the composition $\mathcal{O}_ X \to \mathcal{O}_ X$ is the identity map. This we may do in the generic point of $X$ (or on an open subscheme of $X$ over which $f$ is an isomorphism). In this case $E = \mathcal{O}_ X$ and $\mu $ is the usual multiplication map and the result is clear. $\square$
Lemma 57.7.2. Let $X$ be a proper scheme over a field $k$ which is regular. Let $K \in \mathop{\mathrm{Ob}}\nolimits (D_\mathit{QCoh}(\mathcal{O}_ X))$. The following are equivalent
$K \in D^ b_{\textit{Coh}}(\mathcal{O}_ X) = D_{perf}(\mathcal{O}_ X)$, and
$\sum _{i \in \mathbf{Z}} \dim _ k \mathop{\mathrm{Ext}}\nolimits ^ i_ X(E, K) < \infty $ for all perfect $E$ in $D(\mathcal{O}_ X)$.
Proof. The equality in (1) holds by Derived Categories of Schemes, Lemma 36.11.8. The implication (1) $\Rightarrow $ (2) follows from Lemma 57.5.3. The implication (2) $\Rightarrow $ (1) follows from More on Morphisms, Lemma 37.66.6. $\square$
Lemma 57.7.3. Let $X$ be a proper scheme over a field $k$ which is regular.
Let $F : D_{perf}(\mathcal{O}_ X)^{opp} \to \text{Vect}_ k$ be a $k$-linear cohomological functor such that
for all $E \in D_{perf}(\mathcal{O}_ X)$. Then $F$ is isomorphic to a functor of the form $E \mapsto \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$ for some $K \in D_{perf}(\mathcal{O}_ X)$.
Let $G : D_{perf}(\mathcal{O}_ X) \to \text{Vect}_ k$ be a $k$-linear homological functor such that
for all $E \in D_{perf}(\mathcal{O}_ X)$. Then $G$ is isomorphic to a functor of the form $E \mapsto \mathop{\mathrm{Hom}}\nolimits _ X(K, E)$ for some $K \in D_{perf}(\mathcal{O}_ X)$.
Proof. Proof of (1). The derived category $D_\mathit{QCoh}(\mathcal{O}_ X)$ has direct sums, is compactly generated, and $D_{perf}(\mathcal{O}_ X)$ is the full subcategory of compact objects, see Derived Categories of Schemes, Lemma 36.3.1, Theorem 36.15.3, and Proposition 36.17.1. By Lemma 57.6.2 we may assume $F(E) = \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$ for some $K \in \mathop{\mathrm{Ob}}\nolimits (D_\mathit{QCoh}(\mathcal{O}_ X))$. Then it follows that $K$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Lemma 57.7.2.
Proof of (2). Consider the contravariant functor $E \mapsto E^\vee $ on $D_{perf}(\mathcal{O}_ X)$, see Cohomology, Lemma 20.47.5. This functor is an exact anti-self-equivalence of $D_{perf}(\mathcal{O}_ X)$. Hence we may apply part (1) to the functor $F(E) = G(E^\vee )$ to find $K \in D_{perf}(\mathcal{O}_ X)$ such that $G(E^\vee ) = \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$. It follows that $G(E) = \mathop{\mathrm{Hom}}\nolimits _ X(E^\vee , K) = \mathop{\mathrm{Hom}}\nolimits _ X(K^\vee , E)$ and we conclude that taking $K^\vee $ works. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)