86.9 Relative dualizing complexes for proper flat morphisms
Motivated by Duality for Schemes, Sections 48.12 and 48.28 and the material in Section 86.8 we make the following definition.
Definition 86.9.1. Let $S$ be a scheme. Let $f : X \to Y$ be a proper, flat morphism of algebraic spaces over $S$ which is of finite presentation. A relative dualizing complex for $X/Y$ is a pair $(\omega _{X/Y}^\bullet , \tau )$ consisting of a $Y$-perfect object $\omega _{X/Y}^\bullet $ of $D(\mathcal{O}_ X)$ and a map
\[ \tau : Rf_*\omega _{X/Y}^\bullet \longrightarrow \mathcal{O}_ Y \]
such that for any cartesian square
\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]
where $Y'$ is an affine scheme the pair $(L(g')^*\omega _{X/Y}^\bullet , Lg^*\tau )$ is isomorphic to the pair $(a'(\mathcal{O}_{Y'}), \text{Tr}_{f', \mathcal{O}_{Y'}})$ studied in Sections 86.3, 86.4, 86.5, 86.6, 86.7, and 86.8.
There are several remarks we should make here.
In Definition 86.9.1 one may drop the assumption that $\omega _{X/Y}^\bullet $ is $Y$-perfect. Namely, running $Y'$ through the members of an étale covering of $Y$ by affines, we see from Lemma 86.8.3 that the restrictions of $\omega _{X/Y}^\bullet $ to the members of an étale covering of $X$ are $Y$-perfect, which implies $\omega _{X/Y}^\bullet $ is $Y$-perfect, see More on Morphisms of Spaces, Section 76.52.
Consider a relative dualizing complex $(\omega _{X/Y}^\bullet , \tau )$ and a cartesian square as in Definition 86.9.1. We are going to think of the existence of the isomorphism $(L(g')^*\omega _{X/Y}^\bullet , Lg^*\tau ) \cong (a'(\mathcal{O}_{Y'}), \text{Tr}_{f', \mathcal{O}_{Y'}})$ as follows: it says that for any $M' \in D_\mathit{QCoh}(\mathcal{O}_{X'})$ the map
\[ \mathop{\mathrm{Hom}}\nolimits _{X'}(M', L(g')^*\omega _{X/Y}^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{Y'}(Rf'_*M', \mathcal{O}_{Y'}),\quad \varphi ' \longmapsto Lg^*\tau \circ Rf'_*\varphi ' \]
is an isomorphism. This follows from the definition of $a'$ and the discussion in Section 86.6. In particular, the Yoneda lemma guarantees that the isomorphism is unique.
If $Y$ is affine itself, then a relative dualizing complex $(\omega _{X/Y}^\bullet , \tau )$ exists and is canonically isomorphic to $(a(\mathcal{O}_ Y), \text{Tr}_{f, \mathcal{O}_ Y})$ where $a$ is the right adjoint for $Rf_*$ as in Lemma 86.3.1 and $\text{Tr}_ f$ is as in Section 86.6. Namely, given a diagram as in the definition we get an isomorphism $L(g')^*a(\mathcal{O}_ Y) \to a'(\mathcal{O}_{Y'})$ by Lemma 86.5.1 which is compatible with trace maps by Lemma 86.6.1.
This produces exactly enough information to glue the locally given relative dualizing complexes to global ones. We suggest the reader skip the proofs of the following lemmas.
Lemma 86.9.2. Let $S$ be a scheme. Let $X \to Y$ be a proper, flat morphism of algebraic spaces which is of finite presentation. If $(\omega _{X/Y}^\bullet , \tau )$ is a relative dualizing complex, then $\mathcal{O}_ X \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\omega _{X/Y}^\bullet , \omega _{X/Y}^\bullet )$ is an isomorphism and $Rf_*\omega _{X/Y}^\bullet $ has vanishing cohomology sheaves in positive degrees.
Proof.
It suffices to prove this after base change to an affine scheme étale over $Y$ in which case it follows from Lemma 86.8.3.
$\square$
Lemma 86.9.3. Let $S$ be a scheme. Let $X \to Y$ be a proper, flat morphism of algebraic spaces which is of finite presentation. If $(\omega _ j^\bullet , \tau _ j)$, $j = 1, 2$ are two relative dualizing complexes on $X/Y$, then there is a unique isomorphism $(\omega _1^\bullet , \tau _1) \to (\omega _2^\bullet , \tau _2)$.
Proof.
Consider $g : Y' \to Y$ étale with $Y'$ an affine scheme and denote $X' = Y' \times _ Y X$ the base change. By Definition 86.9.1 and the discussion following, there is a unique isomorphism $\iota : (\omega _1^\bullet |_{X'}, \tau _1|_{Y'}) \to (\omega _2^\bullet |_{X'}, \tau _2|_{Y'})$. If $Y'' \to Y'$ is a further étale morphism of affines and $X'' = Y'' \times _ Y X$, then $\iota |_{X''}$ is the unique isomorphism $(\omega _1^\bullet |_{X''}, \tau _1|_{Y''}) \to (\omega _2^\bullet |_{X''}, \tau _2|_{Y''})$ (by uniqueness). Also we have
\[ \text{Ext}^ p_{X'}(\omega _1^\bullet |_{X'}, \omega _2^\bullet |_{X'}) = 0, \quad p < 0 \]
because $\mathcal{O}_{X'} \cong R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{X'}}(\omega _1^\bullet |_{X'}, \omega _1^\bullet |_{X'}) \cong R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{X'}}(\omega _1^\bullet |_{X'}, \omega _2^\bullet |_{X'})$ by Lemma 86.9.2.
Choose a étale hypercovering $b : V \to Y$ such that each $V_ n = \coprod _{i \in I_ n} Y_{n, i}$ with $Y_{n, i}$ affine. This is possible by Hypercoverings, Lemma 25.12.2 and Remark 25.12.9 (to replace the hypercovering produced in the lemma by the one having disjoint unions in each degree). Denote $X_{n, i} = Y_{n, i} \times _ Y X$ and $U_ n = V_ n \times _ Y X$ so that we obtain an étale hypercovering $a : U \to X$ (Hypercoverings, Lemma 25.12.4) with $U_ n = \coprod X_{n, i}$. The assumptions of Simplicial Spaces, Lemma 85.35.1 are satisfied for $a : U \to X$ and the complexes $\omega _1^\bullet $ and $\omega _2^\bullet $. Hence we obtain a unique morphism $\iota : \omega _1^\bullet \to \omega _2^\bullet $ whose restriction to $X_{0, i}$ is the unique isomorphism $(\omega _1^\bullet |_{X_{0, i}}, \tau _1|_{Y_{0, i}}) \to (\omega _2^\bullet |_{X_{0, i}}, \tau _2|_{Y_{0, i}})$ We still have to see that the diagram
\[ \xymatrix{ Rf_*\omega _1^\bullet \ar[rd]_{\tau _1} \ar[rr]_{Rf_*\iota } & & Rf_*\omega _1^\bullet \ar[ld]^{\tau _2} \\ & \mathcal{O}_ Y } \]
is commutative. However, we know that $Rf_*\omega _1^\bullet $ and $Rf_*\omega _2^\bullet $ have vanishing cohomology sheaves in positive degrees (Lemma 86.9.2) thus this commutativity may be proved after restricting to the affines $Y_{0, i}$ where it holds by construction.
$\square$
Lemma 86.9.4. Let $S$ be a scheme. Let $X \to Y$ be a proper, flat morphism of algebraic spaces which is of finite presentation. Let $(\omega ^\bullet , \tau )$ be a pair consisting of a $Y$-perfect object of $D(\mathcal{O}_ X)$ and a map $\tau : Rf_*\omega ^\bullet \to \mathcal{O}_ Y$. Assume we have cartesian diagrams
\[ \xymatrix{ X_ i \ar[r]_{g_ i'} \ar[d]_{f_ i} & X \ar[d]^ f \\ Y_ i \ar[r]^{g_ i} & Y } \]
with $Y_ i$ affine such that $\{ g_ i : Y_ i \to Y\} $ is an étale covering and isomorphisms of pairs $(\omega ^\bullet |_{X_ i}, \tau |_{Y_ i}) \to (a_ i(\mathcal{O}_{Y_ i}), \text{Tr}_{f_ i, \mathcal{O}_{Y_ i}})$ as in Definition 86.9.1. Then $(\omega ^\bullet , \tau )$ is a relative dualizing complex for $X$ over $Y$.
Proof.
Let $g : Y' \to Y$ and $X', f', g', a'$ be as in Definition 86.9.1. Set $((\omega ')^\bullet , \tau ') = (L(g')^*\omega ^\bullet , Lg^*\tau )$. We can find a finite étale covering $\{ Y'_ j \to Y'\} $ by affines which refines $\{ Y_ i \times _ Y Y' \to Y'\} $ (Topologies, Lemma 34.4.4). Thus for each $j$ there is an $i_ j$ and a morphism $k_ j : Y'_ j \to Y_{i_ j}$ over $Y$. Consider the fibre products
\[ \xymatrix{ X'_ j \ar[r]_{h_ j'} \ar[d]_{f'_ j} & X' \ar[d]^{f'} \\ Y'_ j \ar[r]^{h_ j} & Y' } \]
Denote $k'_ j : X'_ j \to X_{i_ j}$ the induced morphism (base change of $k_ j$ by $f_{i_ j}$). Restricting the given isomorphisms to $Y'_ j$ via the morphism $k'_ j$ we get isomorphisms of pairs $((\omega ')^\bullet |_{X'_ j}, \tau '|_{Y'_ j}) \to (a_ j(\mathcal{O}_{Y'_ j}), \text{Tr}_{f'_ j, \mathcal{O}_{Y'_ j}})$. After replacing $f : X \to Y$ by $f' : X' \to Y'$ we reduce to the problem solved in the next paragraph.
Assume $Y$ is affine. Problem: show $(\omega ^\bullet , \tau )$ is isomorphic to $(\omega _{X/Y}^\bullet , \text{Tr}) = (a(\mathcal{O}_ Y), \text{Tr}_{f, \mathcal{O}_ Y})$. We may assume our covering $\{ Y_ i \to Y\} $ is given by a single surjective étale morphism $\{ g : Y' \to Y\} $ of affines. Namely, we can first replace $\{ g_ i: Y_ i \to Y\} $ by a finite subcovering, and then we can set $g = \coprod g_ i : Y' = \coprod Y_ i \to Y$; some details omitted. Set $X' = Y' \times _ Y X$ with maps $f', g'$ as in Definition 86.9.1. Then all we're given is that we have an isomorphism
\[ (\omega ^\bullet |_{X'}, \tau |_{Y'}) \to (a'(\mathcal{O}_{Y'}), \text{Tr}_{f', \mathcal{O}_{Y'}}) \]
Since $(\omega _{X/Y}^\bullet , \text{Tr})$ is a relative dualizing complex (see discussion following Definition 86.9.1) there is a unique isomorphism
\[ (\omega _{X/Y}^\bullet |_{X'}, \text{Tr}|_{Y'}) \to (a'(\mathcal{O}_{Y'}), \text{Tr}_{f', \mathcal{O}_{Y'}}) \]
Uniqueness by Lemma 86.9.3 for example. Combining the displayed isomorphisms we find an isomorphism
\[ \alpha : (\omega ^\bullet |_{X'}, \tau |_{Y'}) \to (\omega _{X/Y}^\bullet |_{X'}, \text{Tr}|_{Y'}) \]
Set $Y'' = Y' \times _ Y Y'$ and $X'' = Y'' \times _ Y X$ the two pullbacks of $\alpha $ to $X''$ have to be the same by uniqueness again. Since we have vanishing negative self exts for $\omega _{X'/Y'}^\bullet $ over $X'$ (Lemma 86.9.2) and since this remains true after pulling back by any projection $Y' \times _ Y \ldots \times _ Y Y' \to Y'$ (small detail omitted – compare with the proof of Lemma 86.9.3), we find that $\alpha $ descends to an isomorphism $\omega ^\bullet \to \omega _{X/Y}^\bullet $ over $X$ by Simplicial Spaces, Lemma 85.35.1.
$\square$
Lemma 86.9.5. Let $S$ be a scheme. Let $X \to Y$ be a proper, flat morphism of algebraic spaces which is of finite presentation. There exists a relative dualizing complex $(\omega _{X/Y}^\bullet , \tau )$.
Proof.
Choose a étale hypercovering $b : V \to Y$ such that each $V_ n = \coprod _{i \in I_ n} Y_{n, i}$ with $Y_{n, i}$ affine. This is possible by Hypercoverings, Lemma 25.12.2 and Remark 25.12.9 (to replace the hypercovering produced in the lemma by the one having disjoint unions in each degree). Denote $X_{n, i} = Y_{n, i} \times _ Y X$ and $U_ n = V_ n \times _ Y X$ so that we obtain an étale hypercovering $a : U \to X$ (Hypercoverings, Lemma 25.12.4) with $U_ n = \coprod X_{n, i}$. For each $n, i$ there exists a relative dualizing complex $(\omega _{n, i}^\bullet , \tau _{n, i})$ on $X_{n, i}/Y_{n, i}$. See discussion following Definition 86.9.1. For $\varphi : [m] \to [n]$ and $i \in I_ n$ consider the morphisms $g_{\varphi , i} : Y_{n, i} \to Y_{m, \alpha (\varphi )}$ and $g'_{\varphi , i} : X_{n, i} \to X_{m, \alpha (\varphi )}$ which are part of the structure of the given hypercoverings (Hypercoverings, Section 25.12). Then we have a unique isomorphisms
\[ \iota _{n, i, \varphi } : (L(g'_{n, i})^*\omega _{n, i}^\bullet , Lg_{n, i}^*\tau _{n, i}) \longrightarrow (\omega _{m, \alpha (\varphi )(i)}^\bullet , \tau _{m, \alpha (\varphi )(i)}) \]
of pairs, see discussion following Definition 86.9.1. Observe that $\omega _{n, i}^\bullet $ has vanishing negative self exts on $X_{n, i}$ by Lemma 86.9.2. Denote $(\omega _ n^\bullet , \tau _ n)$ the pair on $U_ n/V_ n$ constructed using the pairs $(\omega _{n, i}^\bullet , \tau _{n, i})$ for $i \in I_ n$. For $\varphi : [m] \to [n]$ and $i \in I_ n$ consider the morphisms $g_\varphi : V_ n \to V_ m$ and $g'_\varphi : U_ n \to U_ m$ which are part of the structure of the simplicial algebraic spaces $V$ and $U$. Then we have unique isomorphisms
\[ \iota _\varphi : (L(g'_\varphi )^*\omega _ n^\bullet , Lg_\varphi ^*\tau _ n) \longrightarrow (\omega _ m^\bullet , \tau _ m) \]
of pairs constructed from the isomorphisms on the pieces. The uniqueness guarantees that these isomorphisms satisfy the transitivity condition as formulated in Simplicial Spaces, Definition 85.14.1. The assumptions of Simplicial Spaces, Lemma 85.35.2 are satisfied for $a : U \to X$, the complexes $\omega _ n^\bullet $ and the isomorphisms $\iota _\varphi $1. Thus we obtain an object $\omega ^\bullet $ of $D_\mathit{QCoh}(\mathcal{O}_ X)$ together with an isomorphism $\iota _0 : \omega ^\bullet |_{U_0} \to \omega _0^\bullet $ compatible with the two isomorphisms $\iota _{\delta ^1_0}$ and $\iota _{\delta ^1_1}$. Finally, we apply Simplicial Spaces, Lemma 85.35.1 to find a unique morphism
\[ \tau : Rf_*\omega ^\bullet \longrightarrow \mathcal{O}_ Y \]
whose restriction to $V_0$ agrees with $\tau _0$; some details omitted – compare with the end of the proof of Lemma 86.9.3 for example to see why we have the required vanishing of negative exts. By Lemma 86.9.4 the pair $(\omega ^\bullet , \tau )$ is a relative dualizing complex and the proof is complete.
$\square$
Lemma 86.9.6. Let $S$ be a scheme. Consider a cartesian square
\[ \xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]
of algebraic spaces over $S$. Assume $X \to Y$ is proper, flat, and of finite presentation. Let $(\omega _{X/Y}^\bullet , \tau )$ be a relative dualizing complex for $f$. Then $(L(g')^*\omega _{X/Y}^\bullet , Lg^*\tau )$ is a relative dualizing complex for $f'$.
Proof.
Observe that $L(g')^*\omega _{X/Y}^\bullet $ is $Y'$-perfect by More on Morphisms of Spaces, Lemma 76.52.6. The other condition of Definition 86.9.1 holds by transitivity of fibre products.
$\square$
Comments (0)