Lemma 48.17.1. In Situation 48.16.1 let $Y$ be an object of $\textit{FTS}_ S$ and let $j : X \to Y$ be an open immersion. Then there is a canonical isomorphism $j^! = j^*$ of functors.
48.17 Properties of upper shriek functors
Here are some properties of the upper shriek functors.
For an étale morphism $f : X \to Y$ of $\textit{FTS}_ S$ we also have $f^* \cong f^!$, see Lemma 48.18.2.
Proof. In this case we may choose $\overline{X} = Y$ as our compactification. Then the right adjoint of Lemma 48.3.1 for $\text{id} : Y \to Y$ is the identity functor and hence $j^! = j^*$ by definition. $\square$
Lemma 48.17.2. In Situation 48.16.1 let be a commutative diagram of $\textit{FTS}_ S$ where $j$ and $j'$ are open immersions. Then $j^* \circ f^! = g^! \circ (j')^*$ as functors $D^+_\mathit{QCoh}(\mathcal{O}_ Y) \to D^+(\mathcal{O}_ U)$.
\[ \xymatrix{ U \ar[r]_ j \ar[d]_ g & X \ar[d]^ f \\ V \ar[r]^{j'} & Y } \]
Proof. Let $h = f \circ j = j' \circ g$. By Lemma 48.16.3 we have $h^! = j^! \circ f^! = g^! \circ (j')^!$. By Lemma 48.17.1 we have $j^! = j^*$ and $(j')^! = (j')^*$. $\square$
Lemma 48.17.3. In Situation 48.16.1 let $Y$ be an object of $\textit{FTS}_ S$ and let $f : X = \mathbf{A}^1_ Y \to Y$ be the projection. Then there is a (noncanonical) isomorphism $f^!(-) \cong Lf^*(-) [1]$ of functors.
Proof. Since $X = \mathbf{A}^1_ Y \subset \mathbf{P}^1_ Y$ and since $\mathcal{O}_{\mathbf{P}^1_ Y}(-2)|_ X \cong \mathcal{O}_ X$ this follows from Lemmas 48.15.1 and 48.13.3. $\square$
Lemma 48.17.4. In Situation 48.16.1 let $Y$ be an object of $\textit{FTS}_ S$ and let $i : X \to Y$ be a closed immersion. Then there is a canonical isomorphism $i^!(-) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ X, -)$ of functors.
Proof. This is a restatement of Lemma 48.9.7. $\square$
Remark 48.17.5 (Local description upper shriek). In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Using the lemmas above we can compute $f^!$ locally as follows. Suppose that we are given affine opens Since $j^! \circ f^! = g^! \circ i^!$ (Lemma 48.16.3) and since $j^!$ and $i^!$ are given by restriction (Lemma 48.17.1) we see that for any $E \in D^+_\mathit{QCoh}(\mathcal{O}_ X)$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(R)$ and let $\varphi : R \to A$ be the finite type ring map corresponding to $g$. Choose a presentation $A = P/I$ where $P = R[x_1, \ldots , x_ n]$ is a polynomial algebra in $n$ variables over $R$. Choose an object $K \in D^+(R)$ corresponding to $E|_ V$ (Derived Categories of Schemes, Lemma 36.3.5). Then we claim that $f^!E|_ U$ corresponds to where $R\mathop{\mathrm{Hom}}\nolimits (A, -) : D(P) \to D(A)$ is the functor of Dualizing Complexes, Section 47.13 and where $\varphi ^! : D(R) \to D(A)$ is the functor of Dualizing Complexes, Section 47.24. Namely, the choice of presentation gives a factorization Applying Lemma 48.17.3 exactly $n$ times we see that $(\mathbf{A}^ n_ V \to V)^!(E|_ V)$ corresponds to $K \otimes _ R^\mathbf {L} P[n]$. By Lemmas 48.9.5 and 48.17.4 the last step corresponds to applying $R\mathop{\mathrm{Hom}}\nolimits (A, -)$.
\[ \xymatrix{ U \ar[r]_ j \ar[d]_ g & X \ar[d]^ f \\ V \ar[r]^ i & Y } \]
\[ (f^!E)|_ U = g^!(E|_ V) \]
\[ \varphi ^!(K) = R\mathop{\mathrm{Hom}}\nolimits (A, K \otimes _ R^\mathbf {L} P)[n] \]
\[ U \rightarrow \mathbf{A}^ n_ V \to \mathbf{A}^{n - 1}_ V \to \ldots \to \mathbf{A}^1_ V \to V \]
Lemma 48.17.6. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Then $f^!$ maps $D_{\textit{Coh}}^+(\mathcal{O}_ Y)$ into $D_{\textit{Coh}}^+(\mathcal{O}_ X)$.
Proof. The question is local on $X$ hence we may assume that $X$ and $Y$ are affine schemes. In this case we can factor $f : X \to Y$ as
\[ X \xrightarrow {i} \mathbf{A}^ n_ Y \to \mathbf{A}^{n - 1}_ Y \to \ldots \to \mathbf{A}^1_ Y \to Y \]
where $i$ is a closed immersion. The lemma follows from By Lemmas 48.17.3 and 48.9.6 and Dualizing Complexes, Lemma 47.15.10 and induction. $\square$
Lemma 48.17.7. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. If $K$ is a dualizing complex for $Y$, then $f^!K$ is a dualizing complex for $X$.
Proof. The question is local on $X$ hence we may assume that $X$ and $Y$ are affine schemes. In this case we can factor $f : X \to Y$ as
\[ X \xrightarrow {i} \mathbf{A}^ n_ Y \to \mathbf{A}^{n - 1}_ Y \to \ldots \to \mathbf{A}^1_ Y \to Y \]
where $i$ is a closed immersion. By Lemma 48.17.3 and Dualizing Complexes, Lemma 47.15.10 and induction we see that the $p^!K$ is a dualizing complex on $\mathbf{A}^ n_ Y$ where $p : \mathbf{A}^ n_ Y \to Y$ is the projection. Similarly, by Dualizing Complexes, Lemma 47.15.9 and Lemmas 48.9.5 and 48.17.4 we see that $i^!$ transforms dualizing complexes into dualizing complexes. $\square$
Lemma 48.17.8. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Let $K$ be a dualizing complex on $Y$. Set $D_ Y(M) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(M, K)$ for $M \in D_{\textit{Coh}}(\mathcal{O}_ Y)$ and $D_ X(E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E, f^!K)$ for $E \in D_{\textit{Coh}}(\mathcal{O}_ X)$. Then there is a canonical isomorphism for $M \in D_{\textit{Coh}}^+(\mathcal{O}_ Y)$.
\[ f^!M \longrightarrow D_ X(Lf^*D_ Y(M)) \]
Proof. Choose compactification $j : X \subset \overline{X}$ of $X$ over $Y$ (More on Flatness, Theorem 38.33.8 and Lemma 38.32.2). Let $a$ be the right adjoint of Lemma 48.3.1 for $\overline{X} \to Y$. Set $D_{\overline{X}}(E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{\overline{X}}}(E, a(K))$ for $E \in D_{\textit{Coh}}(\mathcal{O}_{\overline{X}})$. Since formation of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits $ commutes with restriction to opens and since $f^! = j^* \circ a$ we see that it suffices to prove that there is a canonical isomorphism
\[ a(M) \longrightarrow D_{\overline{X}}(L\overline{f}^*D_ Y(M)) \]
for $M \in D_{\textit{Coh}}(\mathcal{O}_ Y)$. For $F \in D_\mathit{QCoh}(\mathcal{O}_ X)$ we have
\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{\overline{X}}( F, D_{\overline{X}}(L\overline{f}^*D_ Y(M))) & = \mathop{\mathrm{Hom}}\nolimits _{\overline{X}}( F \otimes _{\mathcal{O}_ X}^\mathbf {L} L\overline{f}^*D_ Y(M), a(K)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y( R\overline{f}_*(F \otimes _{\mathcal{O}_ X}^\mathbf {L} L\overline{f}^*D_ Y(M)), K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y( R\overline{f}_*(F) \otimes _{\mathcal{O}_ Y}^\mathbf {L} D_ Y(M), K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y( R\overline{f}_*(F), D_ Y(D_ Y(M))) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(R\overline{f}_*(F), M) \\ & = \mathop{\mathrm{Hom}}\nolimits _{\overline{X}}(F, a(M)) \end{align*}
The first equality by Cohomology, Lemma 20.42.2. The second by definition of $a$. The third by Derived Categories of Schemes, Lemma 36.22.1. The fourth equality by Cohomology, Lemma 20.42.2 and the definition of $D_ Y$. The fifth equality by Lemma 48.2.5. The final equality by definition of $a$. Hence we see that $a(M) = D_{\overline{X}}(L\overline{f}^*D_ Y(M))$ by Yoneda's lemma. $\square$
Lemma 48.17.9. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Assume $f$ is perfect (e.g., flat). Then
$f^!$ maps $D_{\textit{Coh}}^ b(\mathcal{O}_ Y)$ into $D_{\textit{Coh}}^ b(\mathcal{O}_ X)$,
the map $\mu _{f, K} : Lf^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} f^!\mathcal{O}_ Y \to f^!K$ of Lemma 48.16.5 is an isomorphism for all $K \in D_\mathit{QCoh}^+(\mathcal{O}_ Y)$.
Proof. (A flat morphism of finite presentation is perfect, see More on Morphisms, Lemma 37.61.5.) We begin with a series of preliminary remarks.
We already know that $f^!$ sends $D_{\textit{Coh}}^+(\mathcal{O}_ Y)$ into $D_{\textit{Coh}}^+(\mathcal{O}_ X)$, see Lemma 48.17.6.
If $f$ is an open immersion, then (a) and (b) are true because we can take $\overline{X} = Y$ in the construction of $f^!$ and $\mu _ f$. See also Lemma 48.17.1.
If $f$ is a perfect proper morphism, then (b) is true by Lemma 48.13.3.
If there exists an open covering $X = \bigcup U_ i$ and (a) is true for $U_ i \to Y$, then (a) is true for $X \to Y$. Same for (b). This holds because the construction of $f^!$ and $\mu _ f$ commutes with passing to open subschemes.
If $g : Y \to Z$ is a second perfect morphism in $\textit{FTS}_ S$ and (b) holds for $f$ and $g$, then $f^!g^!\mathcal{O}_ Z = Lf^*g^!\mathcal{O}_ Z \otimes _{\mathcal{O}_ X}^\mathbf {L} f^!\mathcal{O}_ Y$ and (b) holds for $g \circ f$ by the commutative diagram of Lemma 48.16.5.
If (a) and (b) hold for both $f$ and $g$, then (a) and (b) hold for $g \circ f$. Namely, then $f^!g^!\mathcal{O}_ Z$ is bounded above (by the previous point) and $L(g \circ f)^*$ has finite cohomological dimension and (a) follows from (b) which we saw above.
From these points we see it suffices to prove the result in case $X$ is affine. Choose an immersion $X \to \mathbf{A}^ n_ Y$ (Morphisms, Lemma 29.39.2) which we factor as $X \to U \to \mathbf{A}^ n_ Y \to Y$ where $X \to U$ is a closed immersion and $U \subset \mathbf{A}^ n_ Y$ is open. Note that $X \to U$ is a perfect closed immersion by More on Morphisms, Lemma 37.61.8. Thus it suffices to prove the lemma for a perfect closed immersion and for the projection $\mathbf{A}^ n_ Y \to Y$.
Let $f : X \to Y$ be a perfect closed immersion. We already know (b) holds. Let $K \in D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$. Then $f^!K = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ X, K)$ (Lemma 48.17.4) and $f_*f^!K = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(f_*\mathcal{O}_ X, K)$. Since $f$ is perfect, the complex $f_*\mathcal{O}_ X$ is perfect and hence $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(f_*\mathcal{O}_ X, K)$ is bounded above. This proves that (a) holds. Some details omitted.
Let $f : \mathbf{A}^ n_ Y \to Y$ be the projection. Then (a) holds by repeated application of Lemma 48.17.3. Finally, (b) is true because it holds for $\mathbf{P}^ n_ Y \to Y$ (flat and proper) and because $\mathbf{A}^ n_ Y \subset \mathbf{P}^ n_ Y$ is an open. $\square$
Lemma 48.17.10. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. If $f$ is flat, then $f^!\mathcal{O}_ Y$ is a $Y$-perfect object of $D(\mathcal{O}_ X)$ and $\mathcal{O}_ X \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(f^!\mathcal{O}_ Y, f^!\mathcal{O}_ Y)$ is an isomorphism.
Proof. Both assertions are local on $X$. Thus we may assume $X$ and $Y$ are affine. Then Remark 48.17.5 turns the lemma into an algebra lemma, namely Dualizing Complexes, Lemma 47.25.2. (Use Derived Categories of Schemes, Lemma 36.35.3 to match the languages.) $\square$
Lemma 48.17.11. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Assume $f : X \to Y$ is a local complete intersection morphism. Then
$f^!\mathcal{O}_ Y$ is an invertible object of $D(\mathcal{O}_ X)$, and
$f^!$ maps perfect complexes to perfect complexes.
Proof. Recall that a local complete intersection morphism is perfect, see More on Morphisms, Lemma 37.62.4. By Lemma 48.17.9 it suffices to show that $f^!\mathcal{O}_ Y$ is an invertible object in $D(\mathcal{O}_ X)$. This question is local on $X$ and $Y$. Hence we may assume that $X \to Y$ factors as $X \to \mathbf{A}^ n_ Y \to Y$ where the first arrow is a Koszul regular immersion. See More on Morphisms, Section 37.62. The result holds for $\mathbf{A}^ n_ Y \to Y$ by Lemma 48.17.3. Thus it suffices to prove the lemma when $f$ is a Koszul regular immersion. Working locally once again we reduce to the case $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$, where $A = B/(f_1, \ldots , f_ r)$ for some regular sequence $f_1, \ldots , f_ r \in B$ (use that for Noetherian local rings the notion of Koszul regular and regular are the same, see More on Algebra, Lemma 15.30.7). Thus $X \to Y$ is a composition
\[ X = X_ r \to X_{r - 1} \to \ldots \to X_1 \to X_0 = Y \]
where each arrow is the inclusion of an effective Cartier divisor. In this way we reduce to the case of an inclusion of an effective Cartier divisor $i : D \to X$. In this case $i^!\mathcal{O}_ X = \mathcal{N}[1]$ by Lemma 48.14.1 and the proof is complete. $\square$
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