The Stacks project

20.46 Duals

In this section we characterize the dualizable objects of the category of complexes and of the derived category. In particular, we will see that an object of $D(\mathcal{O}_ X)$ has a dual if and only if it is perfect (this follows from Example 20.46.6 and Lemma 20.46.7).

Lemma 20.46.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. The category of complexes of $\mathcal{O}_ X$-modules with tensor product defined by $\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet = \text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{G}^\bullet )$ is a symmetric monoidal category (for sign rules, see More on Algebra, Section 15.68).

Proof. Omitted. Hints: as unit $\mathbf{1}$ we take the complex having $\mathcal{O}_ X$ in degree $0$ and zero in other degrees with obvious isomorphisms $\text{Tot}(\mathbf{1} \otimes _{\mathcal{O}_ X} \mathcal{G}^\bullet ) = \mathcal{G}^\bullet $ and $\text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathbf{1}) = \mathcal{F}^\bullet $. to prove the lemma you have to check the commutativity of various diagrams, see Categories, Definitions 4.42.1 and 4.42.9. The verifications are straightforward in each case. $\square$

Example 20.46.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}^\bullet $ be a locally bounded complex of $\mathcal{O}_ X$-modules such that each $\mathcal{F}^ n$ is locally a direct summand of a finite free $\mathcal{O}_ X$-module. In other words, there is an open covering $X = \bigcup U_ i$ such that $\mathcal{F}^\bullet |_{U_ i}$ is a strictly perfect complex. Consider the complex

\[ \mathcal{G}^\bullet = \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{F}^\bullet , \mathcal{O}_ X) \]

as in Section 20.37. Let

\[ \eta : \mathcal{O}_ X \to \text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{G}^\bullet ) \quad \text{and}\quad \epsilon : \text{Tot}(\mathcal{G}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{F}^\bullet ) \to \mathcal{O}_ X \]

be $\eta = \sum \eta _ n$ and $\epsilon = \sum \epsilon _ n$ where $\eta _ n : \mathcal{O}_ X \to \mathcal{F}^ n \otimes _{\mathcal{O}_ X} \mathcal{G}^{-n}$ and $\epsilon _ n : \mathcal{G}^{-n} \otimes _{\mathcal{O}_ X} \mathcal{F}^ n \to \mathcal{O}_ X$ are as in Modules, Example 17.17.1. Then $\mathcal{G}^\bullet , \eta , \epsilon $ is a left dual for $\mathcal{F}^\bullet $ as in Categories, Definition 4.42.5. We omit the verification that $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_{\mathcal{F}^\bullet }$ and $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}_{\mathcal{G}^\bullet }$. Please compare with More on Algebra, Lemma 15.68.3.

Lemma 20.46.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}^\bullet $ be a complex of $\mathcal{O}_ X$-modules. If $\mathcal{F}^\bullet $ has a left dual in the monoidal category of complexes of $\mathcal{O}_ X$-modules (Categories, Definition 4.42.5) then $\mathcal{F}^\bullet $ is a locally bounded complex whose terms are locally direct summands of finite free $\mathcal{O}_ X$-modules and the left dual is as constructed in Example 20.46.2.

Proof. By uniqueness of left duals (Categories, Remark 4.42.7) we get the final statement provided we show that $\mathcal{F}^\bullet $ is as stated. Let $\mathcal{G}^\bullet , \eta , \epsilon $ be a left dual. Write $\eta = \sum \eta _ n$ and $\epsilon = \sum \epsilon _ n$ where $\eta _ n : \mathcal{O}_ X \to \mathcal{F}^ n \otimes _{\mathcal{O}_ X} \mathcal{G}^{-n}$ and $\epsilon _ n : \mathcal{G}^{-n} \otimes _{\mathcal{O}_ X} \mathcal{F}^ n \to \mathcal{O}_ X$. Since $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_{\mathcal{F}^\bullet }$ and $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}_{\mathcal{G}^\bullet }$ by Categories, Definition 4.42.5 we see immediately that we have $(1 \otimes \epsilon _ n) \circ (\eta _ n \otimes 1) = \text{id}_{\mathcal{F}^ n}$ and $(\epsilon _ n \otimes 1) \circ (1 \otimes \eta _ n) = \text{id}_{\mathcal{G}^{-n}}$. Hence we see that $\mathcal{F}^ n$ is locally a direct summand of a finite free $\mathcal{O}_ X$-module by Modules, Lemma 17.17.2. Since the sum $\eta = \sum \eta _ n$ is locally finite, we conclude that $\mathcal{F}^\bullet $ is locally bounded. $\square$

Lemma 20.46.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $K$ be a perfect object of $D(\mathcal{O}_ X)$. Then $K^\vee = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \mathcal{O}_ X)$ is a perfect object too and $(K^\vee )^\vee \cong K$. There are functorial isomorphisms

\[ M \otimes ^\mathbf {L}_{\mathcal{O}_ X} K^\vee = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M) \]

and

\[ H^0(X, M \otimes ^\mathbf {L}_{\mathcal{O}_ X} K^\vee ) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K, M) \]

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

Proof. We will use without further mention that formation of internal hom commutes with restriction to opens (Lemma 20.38.3). We may check $K^\vee $ is perfect locally on $X$. There is a canonical map

\[ K = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ X, \mathcal{O}_ X) \otimes _{\mathcal{O}_ X}^\mathbf {L} K \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \mathcal{O}_ X), \mathcal{O}_ X) = (K^\vee )^\vee \]

see Lemma 20.38.9. It suffices to prove this map is an isomorphism locally. By Lemma 20.38.8 to see the final statement it suffices to check that the map (20.38.8.1)

\[ M \otimes ^\mathbf {L}_{\mathcal{O}_ X} K^\vee \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M) \]

is an isomorphism. This is local on $X$ as well. Hence it suffices to prove the lemma when $K$ is represented by a strictly perfect complex.

Assume $K$ is represented by the strictly perfect complex $\mathcal{E}^\bullet $. Then it follows from Lemma 20.42.9 that $K^\vee $ is represented by the complex whose terms are $(\mathcal{E}^{-n})^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}^{-n}, \mathcal{O}_ X)$ in degree $n$. Since $\mathcal{E}^{-n}$ is a direct summand of a finite free $\mathcal{O}_ X$-module, so is $(\mathcal{E}^{-n})^\vee $. Hence $K^\vee $ is represented by a strictly perfect complex too and we see that $K^\vee $ is perfect. The map $K \to (K^\vee )^\vee $ is an isomorphism as it is given up to sign by the evaluation maps $\mathcal{E}^ n \to ((\mathcal{E}^{-n})^\vee )^\vee $ which are isomorphisms. To see that (20.38.8.1) is an isomorphism, represent $M$ by a complex $\mathcal{F}^\bullet $. By Lemma 20.42.9 the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$ is represented by the complex with terms

\[ \bigoplus \nolimits _{n = p + q} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}^{-q}, \mathcal{F}^ p) \]

On the other hand, the object $M \otimes ^\mathbf {L}_{\mathcal{O}_ X} K^\vee $ is represented by the complex with terms

\[ \bigoplus \nolimits _{n = p + q} \mathcal{F}^ p \otimes _{\mathcal{O}_ X} (\mathcal{E}^{-q})^\vee \]

Thus the assertion that (20.38.8.1) is an isomorphism reduces to the assertion that the canonical map

\[ \mathcal{F} \otimes _{\mathcal{O}_ X} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}, \mathcal{O}_ X) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}, \mathcal{F}) \]

is an isomorphism when $\mathcal{E}$ is a direct summand of a finite free $\mathcal{O}_ X$-module and $\mathcal{F}$ is any $\mathcal{O}_ X$-module. This follows immediately from the corresponding statement when $\mathcal{E}$ is finite free. $\square$

Lemma 20.46.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. The derived category $D(\mathcal{O}_ X)$ is a symmetric monoidal category with tensor product given by derived tensor product with usual associativity and commutativity constraints (for sign rules, see More on Algebra, Section 15.68).

Proof. Omitted. Compare with Lemma 20.46.1. $\square$

Example 20.46.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $K$ be a perfect object of $D(\mathcal{O}_ X)$. Set $K^\vee = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \mathcal{O}_ X)$ as in Lemma 20.46.4. Then the map

\[ K \otimes _{\mathcal{O}_ X}^\mathbf {L} K^\vee \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K) \]

is an isomorphism (by the lemma). Denote

\[ \eta : \mathcal{O}_ X \longrightarrow K \otimes _{\mathcal{O}_ X}^\mathbf {L} K^\vee \]

the map sending $1$ to the section corresponding to $\text{id}_ K$ under the isomorphism above. Denote

\[ \epsilon : K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} K \longrightarrow \mathcal{O}_ X \]

the evaluation map (to construct it you can use Lemma 20.38.5 for example). Then $K^\vee , \eta , \epsilon $ is a left dual for $K$ as in Categories, Definition 4.42.5. We omit the verification that $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_ K$ and $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}_{K^\vee }$.

Lemma 20.46.7. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $M$ be an object of $D(\mathcal{O}_ X)$. If $M$ has a left dual in the monoidal category $D(\mathcal{O}_ X)$ (Categories, Definition 4.42.5) then $M$ is perfect and the left dual is as constructed in Example 20.46.6.

Proof. Let $x \in X$. It suffices to find an open neighbourhood $U$ of $x$ such that $M$ restricts to a perfect complex over $U$. Hence during the proof we can (finitely often) replace $X$ by an open neighbourhood of $x$. Let $N, \eta , \epsilon $ be a left dual.

We are going to use the following argument several times. Choose any complex $\mathcal{M}^\bullet $ of $\mathcal{O}_ X$-modules representing $M$. Choose a K-flat complex $\mathcal{N}^\bullet $ representing $N$ whose terms are flat $\mathcal{O}_ X$-modules, see Lemma 20.26.11. Consider the map

\[ \eta : \mathcal{O}_ X \to \text{Tot}(\mathcal{M}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{N}^\bullet ) \]

After shrinking $X$ we can find an integer $N$ and for $i = 1, \ldots , N$ integers $n_ i \in \mathbf{Z}$ and sections $f_ i$ and $g_ i$ of $\mathcal{M}^{n_ i}$ and $\mathcal{N}^{-n_ i}$ such that

\[ \eta (1) = \sum \nolimits _ i f_ i \otimes g_ i \]

Let $\mathcal{K}^\bullet \subset \mathcal{M}^\bullet $ be any subcomplex of $\mathcal{O}_ X$-modules containing the sections $f_ i$ for $i = 1, \ldots , N$. Since $\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{N}^\bullet ) \subset \text{Tot}(\mathcal{M}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{N}^\bullet )$ by flatness of the modules $\mathcal{N}^ n$, we see that $\eta $ factors through

\[ \tilde\eta : \mathcal{O}_ X \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{N}^\bullet ) \]

Denoting $K$ the object of $D(\mathcal{O}_ X)$ represented by $\mathcal{K}^\bullet $ we find a commutative diagram

\[ \xymatrix{ M \ar[rr]_-{\eta \otimes 1} \ar[rrd]_{\tilde\eta \otimes 1} & & M \otimes ^\mathbf {L} N \otimes ^\mathbf {L} M \ar[r]_-{1 \otimes \epsilon } & M \\ & & K \otimes ^\mathbf {L} N \otimes ^\mathbf {L} M \ar[u] \ar[r]^-{1 \otimes \epsilon } & K \ar[u] } \]

Since the composition of the upper row is the identity on $M$ we conclude that $M$ is a direct summand of $K$ in $D(\mathcal{O}_ X)$.

As a first use of the argument above, we can choose the subcomplex $\mathcal{K}^\bullet = \sigma _{\geq a} \tau _{\leq b}\mathcal{M}^\bullet $ with $a < n_ i < b$ for $i = 1, \ldots , N$. Thus $M$ is a direct summand in $D(\mathcal{O}_ X)$ of a bounded complex and we conclude we may assume $M$ is in $D^ b(\mathcal{O}_ X)$. (Recall that the process above involves shrinking $X$.)

Since $M$ is in $D^ b(\mathcal{O}_ X)$ we may choose $\mathcal{M}^\bullet $ to be a bounded above complex of flat modules (by Modules, Lemma 17.16.6 and Derived Categories, Lemma 13.15.4). Then we can choose $\mathcal{K}^\bullet = \sigma _{\geq a}\mathcal{M}^\bullet $ with $a < n_ i$ for $i = 1, \ldots , N$ in the argument above. Thus we find that we may assume $M$ is a direct summand in $D(\mathcal{O}_ X)$ of a bounded complex of flat modules. In particular, $M$ has finite tor amplitude.

Say $M$ has tor amplitude in $[a, b]$. Assuming $M$ is $m$-pseudo-coherent we are going to show that (after shrinking $X$) we may assume $M$ is $(m - 1)$-pseudo-coherent. This will finish the proof by Lemma 20.45.4 and the fact that $M$ is $(b + 1)$-pseudo-coherent in any case. After shrinking $X$ we may assume there exists a strictly perfect complex $\mathcal{E}^\bullet $ and a map $\alpha : \mathcal{E}^\bullet \to M$ in $D(\mathcal{O}_ X)$ such that $H^ i(\alpha )$ is an isomorphism for $i > m$ and surjective for $i = m$. We may and do assume that $\mathcal{E}^ i = 0$ for $i < m$. Choose a distinguished triangle

\[ \mathcal{E}^\bullet \to M \to L \to \mathcal{E}^\bullet [1] \]

Observe that $H^ i(L) = 0$ for $i \geq m$. Thus we may represent $L$ by a complex $\mathcal{L}^\bullet $ with $\mathcal{L}^ i = 0$ for $i \geq m$. The map $L \to \mathcal{E}^\bullet [1]$ is given by a map of complexes $\mathcal{L}^\bullet \to \mathcal{E}^\bullet [1]$ which is zero in all degrees except in degree $m - 1$ where we obtain a map $\mathcal{L}^{m - 1} \to \mathcal{E}^ m$, see Derived Categories, Lemma 13.27.3. Then $M$ is represented by the complex

\[ \mathcal{M}^\bullet : \ldots \to \mathcal{L}^{m - 2} \to \mathcal{L}^{m - 1} \to \mathcal{E}^ m \to \mathcal{E}^{m + 1} \to \ldots \]

Apply the discussion in the second paragraph to this complex to get sections $f_ i$ of $\mathcal{M}^{n_ i}$ for $i = 1, \ldots , N$. For $n < m$ let $\mathcal{K}^ n \subset \mathcal{L}^ n$ be the $\mathcal{O}_ X$-submodule generated by the sections $f_ i$ for $n_ i = n$ and $d(f_ i)$ for $n_ i = n - 1$. For $n \geq m$ set $\mathcal{K}^ n = \mathcal{E}^ n$. Clearly, we have a morphism of distinguished triangles

\[ \xymatrix{ \mathcal{E}^\bullet \ar[r] & \mathcal{M}^\bullet \ar[r] & \mathcal{L}^\bullet \ar[r] & \mathcal{E}^\bullet [1] \\ \mathcal{E}^\bullet \ar[r] \ar[u] & \mathcal{K}^\bullet \ar[r] \ar[u] & \sigma _{\leq m - 1}\mathcal{K}^\bullet \ar[r] \ar[u] & \mathcal{E}^\bullet [1] \ar[u] } \]

where all the morphisms are as indicated above. Denote $K$ the object of $D(\mathcal{O}_ X)$ corresponding to the complex $\mathcal{K}^\bullet $. By the arguments in the second paragraph of the proof we obtain a morphism $s : M \to K$ in $D(\mathcal{O}_ X)$ such that the composition $M \to K \to M$ is the identity on $M$. We don't know that the diagram

\[ \xymatrix{ \mathcal{E}^\bullet \ar[r] & \mathcal{K}^\bullet \ar@{=}[r] & K \\ \mathcal{E}^\bullet \ar[u]^{\text{id}} \ar[r]^ i & \mathcal{M}^\bullet \ar@{=}[r] & M \ar[u]_ s } \]

commutes, but we do know it commutes after composing with the map $K \to M$. By Lemma 20.42.8 after shrinking $X$ we may assume that $s \circ i$ is given by a map of complexes $\sigma : \mathcal{E}^\bullet \to \mathcal{K}^\bullet $. By the same lemma we may assume the composition of $\sigma $ with the inclusion $\mathcal{K}^\bullet \subset \mathcal{M}^\bullet $ is homotopic to zero by some homotopy $\{ h^ i : \mathcal{E}^ i \to \mathcal{M}^{i - 1}\} $. Thus, after replacing $\mathcal{K}^{m - 1}$ by $\mathcal{K}^{m - 1} + \mathop{\mathrm{Im}}(h^ m)$ (note that after doing this it is still the case that $\mathcal{K}^{m - 1}$ is generated by finitely many global sections), we see that $\sigma $ itself is homotopic to zero! This means that we have a commutative solid diagram

\[ \xymatrix{ \mathcal{E}^\bullet \ar[r] & M \ar[r] & \mathcal{L}^\bullet \ar[r] & \mathcal{E}^\bullet [1] \\ \mathcal{E}^\bullet \ar[r] \ar[u] & K \ar[r] \ar[u] & \sigma _{\leq m - 1}\mathcal{K}^\bullet \ar[r] \ar[u] & \mathcal{E}^\bullet [1] \ar[u] \\ \mathcal{E}^\bullet \ar[r] \ar[u] & M \ar[r] \ar[u]^ s & \mathcal{L}^\bullet \ar[r] \ar@{..>}[u] & \mathcal{E}^\bullet [1] \ar[u] } \]

By the axioms of triangulated categories we obtain a dotted arrow fitting into the diagram. Looking at cohomology sheaves in degree $m - 1$ we see that we obtain

\[ \xymatrix{ H^{m - 1}(M) \ar[r] & H^{m - 1}(\mathcal{L}^\bullet ) \ar[r] & H^ m(\mathcal{E}^\bullet ) \\ H^{m - 1}(K) \ar[r] \ar[u] & H^{m - 1}(\sigma _{\leq m - 1}\mathcal{K}^\bullet ) \ar[r] \ar[u] & H^ m(\mathcal{E}^\bullet ) \ar[u] \\ H^{m - 1}(M) \ar[r] \ar[u] & H^{m - 1}(\mathcal{L}^\bullet ) \ar[r] \ar[u] & H^ m(\mathcal{E}^\bullet ) \ar[u] } \]

Since the vertical compositions are the identity in both the left and right column, we conclude the vertical composition $H^{m - 1}(\mathcal{L}^\bullet ) \to H^{m - 1}(\sigma _{\leq m - 1}\mathcal{K}^\bullet ) \to H^{m - 1}(\mathcal{L}^\bullet )$ in the middle is surjective! In particular $H^{m - 1}(\sigma _{\leq m - 1}\mathcal{K}^\bullet ) \to H^{m - 1}(\mathcal{L}^\bullet )$ is surjective. Using the induced map of long exact sequences of cohomology sheaves from the morphism of triangles above, a diagram chase shows this implies $H^ i(K) \to H^ i(M)$ is an isomorphism for $i \geq m$ and surjective for $i = m - 1$. By construction we can choose an $r \geq 0$ and a surjection $\mathcal{O}_ X^{\oplus r} \to \mathcal{K}^{m - 1}$. Then the composition

\[ (\mathcal{O}_ X^{\oplus r} \to \mathcal{E}^ m \to \mathcal{E}^{m + 1} \to \ldots ) \longrightarrow K \longrightarrow M \]

induces an isomorphism on cohomology sheaves in degrees $\geq m$ and a surjection in degree $m - 1$ and the proof is complete. $\square$

The following two lemmas belong somewhere else.

Lemma 20.46.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(K_ n)_{n \in \mathbf{N}}$ be a system of perfect objects of $D(\mathcal{O}_ X)$. Let $K = \text{hocolim} K_ n$ be the derived colimit (Derived Categories, Definition 13.33.1). Then for any object $E$ of $D(\mathcal{O}_ X)$ we have

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, E) = R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_{\mathcal{O}_ X} K_ n^\vee \]

where $(K_ n^\vee )$ is the inverse system of dual perfect complexes.

Proof. By Lemma 20.46.4 we have $R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_{\mathcal{O}_ X} K_ n^\vee = R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$ which fits into the distinguished triangle

\[ R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \to \prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \to \prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \]

Because $K$ similarly fits into the distinguished triangle $\bigoplus K_ n \to \bigoplus K_ n \to K$ it suffices to show that $\prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\bigoplus K_ n, E)$. This is a formal consequence of (20.38.0.1) and the fact that derived tensor product commutes with direct sums. $\square$

Lemma 20.46.9. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $K$ and $E$ be objects of $D(\mathcal{O}_ X)$ with $E$ perfect. The diagram

\[ \xymatrix{ H^0(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} E^\vee ) \times H^0(X, E) \ar[r] \ar[d] & H^0(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} E^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E) \ar[d] \\ \mathop{\mathrm{Hom}}\nolimits _ X(E, K) \times H^0(X, E) \ar[r] & H^0(X, K) } \]

commutes where the top horizontal arrow is the cup product, the right vertical arrow uses $\epsilon : E^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E \to \mathcal{O}_ X$ (Example 20.46.6), the left vertical arrow uses Lemma 20.46.4, and the bottom horizontal arrow is the obvious one.

Proof. We will abbreviate $\otimes = \otimes _{\mathcal{O}_ X}^\mathbf {L}$ and $\mathcal{O} = \mathcal{O}_ X$. We will identify $E$ and $K$ with $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E)$ and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, K)$ and we will identify $E^\vee $ with $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{O})$.

Let $\xi \in H^0(X, K \otimes E^\vee )$ and $\eta \in H^0(X, E)$. Denote $\tilde\xi : \mathcal{O} \to K \otimes E^\vee $ and $\tilde\eta : \mathcal{O} \to E$ the corresponding maps in $D(\mathcal{O})$. By Lemma 20.31.1 the cup product $\xi \cup \eta $ corresponds to $\tilde\xi \otimes \tilde\eta : \mathcal{O} \to K \otimes E^\vee \otimes E$.

We claim the map $\xi ' : E \to K$ corresponding to $\xi $ by Lemma 20.46.4 is the composition

\[ E = \mathcal{O} \otimes E \xrightarrow {\tilde\xi \otimes 1_ E} K \otimes E^\vee \otimes E \xrightarrow {1_ K \otimes \epsilon } K \]

The construction in Lemma 20.46.4 uses the evaluation map (20.38.8.1) which in turn is constructed using the identification of $E$ with $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E)$ and the composition $\underline{\circ }$ constructed in Lemma 20.38.5. Hence $\xi '$ is the composition

\begin{align*} E = \mathcal{O} \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E) & \xrightarrow {\tilde\xi \otimes 1} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, K) \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{O}) \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E) \\ & \xrightarrow {\underline{\circ } \otimes 1} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E) \\ & \xrightarrow {\underline{\circ }} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, K) = K \end{align*}

The claim follows immediately from this and the fact that the composition $\underline{\circ }$ constructed in Lemma 20.38.5 is associative (insert future reference here) and the fact that $\epsilon $ is defined as the composition $\underline{\circ } : E^\vee \otimes E \to \mathcal{O}$ in Example 20.46.6.

Using the results from the previous two paragraphs, we find the statement of the lemma is that $(1_ K \otimes \epsilon ) \circ (\tilde\xi \otimes \tilde\eta )$ is equal to $(1_ K \otimes \epsilon ) \circ (\tilde\xi \otimes 1_ E) \circ (1_\mathcal {O} \otimes \tilde\eta )$ which is immediate. $\square$


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