## 48.15 Right adjoint of pushforward in examples

In this section we compute the right adjoint to pushforward in some examples. The isomorphisms are canonical but only in the weakest possible sense, i.e., we do not prove or claim that these isomorphisms are compatible with various operations such as base change and compositions of morphisms. There is a huge literature on these types of issues; the reader can start with the material in [RD], (these citations use a different starting point for duality but address the issue of constructing canonical representatives for relative dualizing complexes) and then continue looking at works by Joseph Lipman and collaborators.

Lemma 48.15.1. Let $Y$ be a quasi-compact and quasi-separated scheme. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ Y$-module of rank $n + 1$ with determinant $\mathcal{L} = \wedge ^{n + 1}(\mathcal{E})$. Let $f : X = \mathbf{P}(\mathcal{E}) \to Y$ be the projection. Let $a$ be the right adjoint for $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ of Lemma 48.3.1. Then there is an isomorphism

$c : f^*\mathcal{L}(-n - 1)[n] \longrightarrow a(\mathcal{O}_ Y)$

In particular, if $\mathcal{E} = \mathcal{O}_ Y^{\oplus n + 1}$, then $X = \mathbf{P}^ n_ Y$ and we obtain $a(\mathcal{O}_ Y) = \mathcal{O}_ X(-n - 1)[n]$.

Proof. In (the proof of) Cohomology of Schemes, Lemma 30.8.4 we constructed a canonical isomorphism

$R^ nf_*(f^*\mathcal{L}(-n - 1)) \longrightarrow \mathcal{O}_ Y$

Moreover, $Rf_*(f^*\mathcal{L}(-n - 1))[n] = R^ nf_*(f^*\mathcal{L}(-n - 1))$, i.e., the other higher direct images are zero. Thus we find an isomorphism

$Rf_*(f^*\mathcal{L}(-n - 1)[n]) \longrightarrow \mathcal{O}_ Y$

This isomorphism determines $c$ as in the statement of the lemma because $a$ is the right adjoint of $Rf_*$. By Lemma 48.4.4 construction of the $a$ is local on the base. In particular, to check that $c$ is an isomorphism, we may work locally on $Y$. In other words, we may assume $Y$ is affine and $\mathcal{E} = \mathcal{O}_ Y^{\oplus n + 1}$. In this case the sheaves $\mathcal{O}_ X, \mathcal{O}_ X(-1), \ldots , \mathcal{O}_ X(-n)$ generate $D_\mathit{QCoh}(X)$, see Derived Categories of Schemes, Lemma 36.16.3. Hence it suffices to show that $c : \mathcal{O}_ X(-n - 1)[n] \to a(\mathcal{O}_ Y)$ is transformed into an isomorphism under the functors

$F_{i, p}(-) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\mathcal{O}_ X(i), (-)[p])$

for $i \in \{ -n, \ldots , 0\}$ and $p \in \mathbf{Z}$. For $F_{0, p}$ this holds by construction of the arrow $c$! For $i \in \{ -n, \ldots , -1\}$ we have

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\mathcal{O}_ X(i), \mathcal{O}_ X(-n - 1)[n + p]) = H^ p(X, \mathcal{O}_ X(-n - 1 - i)) = 0$

by the computation of cohomology of projective space (Cohomology of Schemes, Lemma 30.8.1) and we have

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\mathcal{O}_ X(i), a(\mathcal{O}_ Y)[p]) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(Rf_*\mathcal{O}_ X(i), \mathcal{O}_ Y[p]) = 0$

because $Rf_*\mathcal{O}_ X(i) = 0$ by the same lemma. Hence the source and the target of $F_{i, p}(c)$ vanish and $F_{i, p}(c)$ is necessarily an isomorphism. This finishes the proof. $\square$

Example 48.15.2. The base change map (48.5.0.1) is not an isomorphism if $f$ is perfect proper and $g$ is perfect. Let $k$ be a field. Let $Y = \mathbf{A}^2_ k$ and let $f : X \to Y$ be the blowup of $Y$ in the origin. Denote $E \subset X$ the exceptional divisor. Then we can factor $f$ as

$X \xrightarrow {i} \mathbf{P}^1_ Y \xrightarrow {p} Y$

This gives a factorization $a = c \circ b$ where $a$, $b$, and $c$ are the right adjoints of Lemma 48.3.1 of $Rf_*$, $Rp_*$, and $Ri_*$. Denote $\mathcal{O}(n)$ the Serre twist of the structure sheaf on $\mathbf{P}^1_ Y$ and denote $\mathcal{O}_ X(n)$ its restriction to $X$. Note that $X \subset \mathbf{P}^1_ Y$ is cut out by a degree one equation, hence $\mathcal{O}(X) = \mathcal{O}(1)$. By Lemma 48.15.1 we have $b(\mathcal{O}_ Y) = \mathcal{O}(-2)[1]$. By Lemma 48.9.7 we have

$a(\mathcal{O}_ Y) = c(b(\mathcal{O}_ Y)) = c(\mathcal{O}(-2)[1]) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ X, \mathcal{O}(-2)[1]) = \mathcal{O}_ X(-1)$

Last equality by Lemma 48.14.2. Let $Y' = \mathop{\mathrm{Spec}}(k)$ be the origin in $Y$. The restriction of $a(\mathcal{O}_ Y)$ to $X' = E = \mathbf{P}^1_ k$ is an invertible sheaf of degree $-1$ placed in cohomological degree $0$. But on the other hand, $a'(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}) = \mathcal{O}_ E(-2)[1]$ which is an invertible sheaf of degree $-2$ placed in cohomological degree $-1$, so different. In this example the hypothesis of Tor indepence in Lemma 48.6.2 is violated.

Lemma 48.15.3. Let $Y$ be a ringed space. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be a sheaf of ideals. Set $\mathcal{O}_ X = \mathcal{O}_ Y/\mathcal{I}$ and $\mathcal{N} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{I}/\mathcal{I}^2, \mathcal{O}_ X)$. There is a canonical isomorphism $c : \mathcal{N} \to \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^1_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)$.

Proof. Consider the canonical short exact sequence

48.15.3.1
$$\label{duality-equation-second-order-thickening} 0 \to \mathcal{I}/\mathcal{I}^2 \to \mathcal{O}_ Y/\mathcal{I}^2 \to \mathcal{O}_ X \to 0$$

Let $U \subset X$ be open and let $s \in \mathcal{N}(U)$. Then we can pushout (48.15.3.1) via $s$ to get an extension $E_ s$ of $\mathcal{O}_ X|_ U$ by $\mathcal{O}_ X|_ U$. This in turn defines a section $c(s)$ of $\mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^1_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)$ over $U$. See Cohomology, Lemma 20.42.1 and Derived Categories, Lemma 13.27.6. Conversely, given an extension

$0 \to \mathcal{O}_ X|_ U \to \mathcal{E} \to \mathcal{O}_ X|_ U \to 0$

of $\mathcal{O}_ U$-modules, we can find an open covering $U = \bigcup U_ i$ and sections $e_ i \in \mathcal{E}(U_ i)$ mapping to $1 \in \mathcal{O}_ X(U_ i)$. Then $e_ i$ defines a map $\mathcal{O}_ Y|_{U_ i} \to \mathcal{E}|_{U_ i}$ whose kernel contains $\mathcal{I}^2$. In this way we see that $\mathcal{E}|_{U_ i}$ comes from a pushout as above. This shows that $c$ is surjective. We omit the proof of injectivity. $\square$

Lemma 48.15.4. Let $Y$ be a ringed space. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be a sheaf of ideals. Set $\mathcal{O}_ X = \mathcal{O}_ Y/\mathcal{I}$. If $\mathcal{I}$ is Koszul-regular (Divisors, Definition 31.20.2) then composition on $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)$ defines isomorphisms

$\wedge ^ i(\mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^1_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)) \longrightarrow \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ i_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)$

for all $i$.

Proof. By composition we mean the map

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \otimes _{\mathcal{O}_ Y}^\mathbf {L} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)$

of Cohomology, Lemma 20.42.5. This induces multiplication maps

$\mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ a_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \otimes _{\mathcal{O}_ Y} \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ b_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \longrightarrow \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^{a + b}_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)$

Please compare with More on Algebra, Equation (15.63.0.1). The statement of the lemma means that the induced map

$\mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^1_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \otimes \ldots \otimes \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^1_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \longrightarrow \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ i_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)$

factors through the wedge product and then induces an isomorphism. To see this is true we may work locally on $Y$. Hence we may assume that we have global sections $f_1, \ldots , f_ r$ of $\mathcal{O}_ Y$ which generate $\mathcal{I}$ and which form a Koszul regular sequence. Denote

$\mathcal{A} = \mathcal{O}_ Y\langle \xi _1, \ldots , \xi _ r\rangle$

the sheaf of strictly commutative differential graded $\mathcal{O}_ Y$-algebras which is a (divided power) polynomial algebra on $\xi _1, \ldots , \xi _ r$ in degree $-1$ over $\mathcal{O}_ Y$ with differential $\text{d}$ given by the rule $\text{d}\xi _ i = f_ i$. Let us denote $\mathcal{A}^\bullet$ the underlying complex of $\mathcal{O}_ Y$-modules which is the Koszul complex mentioned above. Thus the canonical map $\mathcal{A}^\bullet \to \mathcal{O}_ X$ is a quasi-isomorphism. We obtain quasi-isomorphisms

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{A}^\bullet ) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ X)$

by Cohomology, Lemma 20.46.9. The differentials of the latter complex are zero, and hence

$\mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ i_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \cong \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{A}^{-i}, \mathcal{O}_ X)$

For $j \in \{ 1, \ldots , r\}$ let $\delta _ j : \mathcal{A} \to \mathcal{A}$ be the derivation of degree $1$ with $\delta _ j(\xi _ i) = \delta _{ij}$ (Kronecker delta). A computation shows that $\delta _ j \circ \text{d} = - \text{d} \circ \delta _ j$ which shows that we get a morphism of complexes

$\delta _ j : \mathcal{A}^\bullet \to \mathcal{A}^\bullet [1].$

Whence $\delta _ j$ defines a section of the corresponding $\mathop{\mathcal{E}\! \mathit{xt}}\nolimits$-sheaf. Another computation shows that $\delta _1, \ldots , \delta _ r$ map to a basis for $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{A}^{-1}, \mathcal{O}_ X)$ over $\mathcal{O}_ X$. Since it is clear that $\delta _ j \circ \delta _ j = 0$ and $\delta _ j \circ \delta _{j'} = - \delta _{j'} \circ \delta _ j$ as endomorphisms of $\mathcal{A}$ and hence in the $\mathop{\mathcal{E}\! \mathit{xt}}\nolimits$-sheaves we obtain the statement that our map above factors through the exterior power. To see we get the desired isomorphism the reader checks that the elements

$\delta _{j_1} \circ \ldots \circ \delta _{j_ i}$

for $j_1 < \ldots < j_ i$ map to a basis of the sheaf $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{A}^{-i}, \mathcal{O}_ X)$ over $\mathcal{O}_ X$. $\square$

Lemma 48.15.5. Let $Y$ be a ringed space. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be a sheaf of ideals. Set $\mathcal{O}_ X = \mathcal{O}_ Y/\mathcal{I}$ and $\mathcal{N} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{I}/\mathcal{I}^2, \mathcal{O}_ X)$. If $\mathcal{I}$ is Koszul-regular (Divisors, Definition 31.20.2) then

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ Y) = \wedge ^ r \mathcal{N}[-r]$

where $r : Y \to \{ 1, 2, 3, \ldots \}$ sends $y$ to the minimal number of generators of $\mathcal{I}$ needed in a neighbourhood of $y$.

Proof. We can use Lemmas 48.15.3 and 48.15.4 to see that we have isomorphisms $\wedge ^ i\mathcal{N} \to \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ i_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)$ for $i \geq 0$. Thus it suffices to show that the map $\mathcal{O}_ Y \to \mathcal{O}_ X$ induces an isomorphism

$\mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ r_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ Y) \longrightarrow \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ r_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)$

and that $\mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ i_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ Y)$ is zero for $i \not= r$. These statements are local on $Y$. Thus we may assume that we have global sections $f_1, \ldots , f_ r$ of $\mathcal{O}_ Y$ which generate $\mathcal{I}$ and which form a Koszul regular sequence. Let $\mathcal{A}^\bullet$ be the Koszul complex on $f_1, \ldots , f_ r$ as introduced in the proof of Lemma 48.15.4. Then

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ Y) = \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ Y)$

by Cohomology, Lemma 20.46.9. Denote $1 : \mathcal{A}^\bullet \to \mathcal{O}_ Y$ the map of differential graded $\mathcal{O}_ Y$-algebras given by the identity map of $\mathcal{A}^0 = \mathcal{O}_ Y \to \mathcal{O}_ Y$ in degree $0$. With $\delta _ j$ as in the proof of Lemma 48.15.4 we get an isomorphism of graded $\mathcal{O}_ Y$-modules

$\mathcal{O}_ Y\langle \delta _1, \ldots , \delta _ r\rangle \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ Y)$

by mapping $\delta _{j_1} \ldots \delta _{j_ i}$ to $1 \circ \delta _{j_1} \circ \ldots \circ \delta _{j_ i}$ in degree $i$. Via this isomorphism the differential on the right hand side induces a differential $\text{d}$ on the left hand side. By our sign rules we have $\text{d}(1) = - \sum f_ j \delta _ j$. Since $\delta _ j : \mathcal{A}^\bullet \to \mathcal{A}^\bullet [1]$ is a morphism of complexes, it follows that

$\text{d}(\delta _{j_1} \ldots \delta _{j_ i}) = (- \sum f_ j \delta _ j )\delta _{j_1} \ldots \delta _{j_ i}$

Observe that we have $\text{d} = \sum f_ j \delta _ j$ on the differential graded algebra $\mathcal{A}$. Therefore the map defined by the rule

$1 \circ \delta _{j_1} \ldots \delta _{j_ i} \longmapsto (\delta _{j_1} \circ \ldots \circ \delta _{j_ i})(\xi _1 \ldots \xi _ r)$

will define an isomorphism of complexes

$\mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ Y) \longrightarrow \mathcal{A}^\bullet [-r]$

if $r$ is odd and commuting with differentials up to sign if $r$ is even. In any case these complexes have isomorphic cohomology, which shows the desired vanishing. The isomorphism on cohomology in degree $r$ under the map

$\mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ Y) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ X)$

also follows in a straightforward manner from this. (We observe that our choice of conventions regarding Koszul complexes does intervene in the definition of the isomorphism $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ X, \mathcal{O}_ Y) = \wedge ^ r \mathcal{N}[-r]$.) $\square$

Lemma 48.15.6. Let $Y$ be a quasi-compact and quasi-separated scheme. Let $i : X \to Y$ be a Koszul-regular closed immersion. Let $a$ be the right adjoint of $Ri_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ of Lemma 48.3.1. Then there is an isomorphism

$\wedge ^ r\mathcal{N}[-r] \longrightarrow a(\mathcal{O}_ Y)$

where $\mathcal{N} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{C}_{X/Y}, \mathcal{O}_ X)$ is the normal sheaf of $i$ (Morphisms, Section 29.31) and $r$ is its rank viewed as a locally constant function on $X$.

Proof. Recall, from Lemmas 48.9.7 and 48.9.3, that $a(\mathcal{O}_ Y)$ is an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$ whose pushforward to $Y$ is $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(i_*\mathcal{O}_ X, \mathcal{O}_ Y)$. Thus the result follows from Lemma 48.15.5. $\square$

Lemma 48.15.7. Let $S$ be a Noetherian scheme. Let $f : X \to S$ be a smooth proper morphism of relative dimension $d$. Let $a$ be the right adjoint of $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ S)$ as in Lemma 48.3.1. Then there is an isomorphism

$\wedge ^ d \Omega _{X/S}[d] \longrightarrow a(\mathcal{O}_ S)$

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

Proof. Set $\omega _{X/S}^\bullet = a(\mathcal{O}_ S)$ as in Remark 48.12.5. Let $c$ be the right adjoint of Lemma 48.3.1 for $\Delta : X \to X \times _ S X$. Because $\Delta$ is the diagonal of a smooth morphism it is a Koszul-regular immersion, see Divisors, Lemma 31.22.11. In particular, $\Delta$ is a perfect proper morphism (More on Morphisms, Lemma 37.61.7) and we obtain

\begin{align*} \mathcal{O}_ X & = c(L\text{pr}_1^*\omega _{X/S}^\bullet ) \\ & = L\Delta ^*(L\text{pr}_1^*\omega _{X/S}^\bullet ) \otimes _{\mathcal{O}_ X}^\mathbf {L} c(\mathcal{O}_{X \times _ S X}) \\ & = \omega _{X/S}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} c(\mathcal{O}_{X \times _ S X}) \\ & = \omega _{X/S}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} \wedge ^ d(\mathcal{N}_\Delta )[-d] \end{align*}

The first equality is (48.12.8.1) because $\omega _{X/S}^\bullet = a(\mathcal{O}_ S)$. The second equality by Lemma 48.13.3. The third equality because $\text{pr}_1 \circ \Delta = \text{id}_ X$. The fourth equality by Lemma 48.15.6. Observe that $\wedge ^ d(\mathcal{N}_\Delta )$ is an invertible $\mathcal{O}_ X$-module. Hence $\wedge ^ d(\mathcal{N}_\Delta )[-d]$ is an invertible object of $D(\mathcal{O}_ X)$ and we conclude that $a(\mathcal{O}_ S) = \omega _{X/S}^\bullet = \wedge ^ d(\mathcal{C}_\Delta )[d]$. Since the conormal sheaf $\mathcal{C}_\Delta$ of $\Delta$ is $\Omega _{X/S}$ by Morphisms, Lemma 29.32.7 the proof is complete. $\square$

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