## 48.17 Properties of upper shriek functors

Here are some properties of the upper shriek functors.

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.

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

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

be a commutative diagram of $\textit{FTS}_ S$ where $j$ and $j'$ are open immersions. Then $j^* \circ f^! = g^! \circ (j')^*$ as functors $D^+_\mathit{QCoh}(\mathcal{O}_ Y) \to D^+(\mathcal{O}_ U)$.

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

$\xymatrix{ U \ar[r]_ j \ar[d]_ g & X \ar[d]^ f \\ V \ar[r]^ i & Y }$

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

$(f^!E)|_ U = g^!(E|_ V)$

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

$\varphi ^!(K) = R\mathop{\mathrm{Hom}}\nolimits (A, K \otimes _ R^\mathbf {L} P)[n]$

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

$U \rightarrow \mathbf{A}^ n_ V \to \mathbf{A}^{n - 1}_ V \to \ldots \to \mathbf{A}^1_ V \to V$

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 $R\mathop{\mathrm{Hom}}\nolimits (A, -)$.

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

$f^!M \longrightarrow D_ X(Lf^*D_ Y(M))$

for $M \in D_{\textit{Coh}}^+(\mathcal{O}_ Y)$.

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

1. $f^!\mathcal{O}_ Y$ is in $D_{\textit{Coh}}^ b(\mathcal{O}_ X)$,

2. $f^!\mathcal{O}_ Y$ has finite tor dimension in $D(f^{-1}\mathcal{O}_ Y)$,

3. $\mathcal{O}_ X \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(f^!\mathcal{O}_ Y, f^!\mathcal{O}_ Y)$ is an isomorphism,

4. $f^!$ maps $D_{\textit{Coh}}^ b(\mathcal{O}_ Y)$ into $D_{\textit{Coh}}^ b(\mathcal{O}_ X)$,

5. 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.) Assertions (a), (b), and (c) are local on $X$. Thus we may assume $X$ and $Y$ are affine. Then Remark 48.17.5 turns (a), (b), and (c) into (1)(a), (1)(b), and (1)(c) of Dualizing Complexes, Lemma 47.25.2. (Use Derived Categories of Schemes, Lemmas 36.10.5, 36.10.1, 36.10.3 36.10.8 to match the assertions.)

To prove (d) and (e) we begin with a series of preliminary remarks.

1. We already know that $f^!$ sends $D_{\textit{Coh}}^+(\mathcal{O}_ Y)$ into $D_{\textit{Coh}}^+(\mathcal{O}_ X)$, see Lemma 48.17.6.

2. If $f$ is an open immersion, then $f^! = f^*$ and (d) and (e) are true because we can take $\overline{X} = Y$ in the construction of $f^!$ and $\mu _ f$, see Lemma 48.17.1.

3. If $f$ is a perfect proper morphism, then (e) is true by Lemma 48.13.3.

4. If there exists an open covering $X = \bigcup U_ i$ and (d) is true for $U_ i \to Y$, then (d) is true for $X \to Y$. Same for (e). This holds because the construction of $f^!$ and $\mu _ f$ commutes with passing to open subschemes.

5. If $g : Y \to Z$ is a second perfect morphism in $\textit{FTS}_ S$ and (e) 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 (e) holds for $g \circ f$ by the commutative diagram of Lemma 48.16.5.

6. If (d) and (e) hold for both $f$ and $g$, then (d) and (e) 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 (d) follows from (e) which we saw above.

From these points we see it suffices to prove (d) and (e) 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 (d) 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 (d) holds. Some details omitted.

Let $f : \mathbf{A}^ n_ Y \to Y$ be the projection. Then (d) holds by repeated application of Lemma 48.17.3. Finally, (e) 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

1. $f^!\mathcal{O}_ Y$ is an invertible object of $D(\mathcal{O}_ X)$, and

2. $f^!$ 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. $\square$

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).