The Stacks project

22.35 Variant of derived tensor product

Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Then we have the functors

\[ \text{Comp}(\mathcal{O}) \to K(\mathcal{O}) \to D(\mathcal{O}) \]

and as we've seen above we have differential graded enhancement $\text{Comp}^{dg}(\mathcal{O})$. Namely, this is the differential graded category of Example 22.26.6 associated to the abelian category $\textit{Mod}(\mathcal{O})$. Let $K^\bullet $ be a complex of $\mathcal{O}$-modules in other words, an object of $\text{Comp}^{dg}(\mathcal{O})$. Set

\[ (E, \text{d}) = \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{O})}(K^\bullet , K^\bullet ) \]

This is a differential graded $\mathbf{Z}$-algebra. We claim there is an analogue of the derived base change in this situation.

Lemma 22.35.1. In the situation above there is a functor

\[ - \otimes _ E K^\bullet : \text{Mod}^{dg}_{(E, \text{d})} \longrightarrow \text{Comp}^{dg}(\mathcal{O}) \]

of differential graded categories. This functor sends $E$ to $K^\bullet $ and commutes with direct sums.

Proof. Let $M$ be a differential graded $E$-module. For every object $U$ of $\mathcal{C}$ the complex $K^\bullet (U)$ is a left differential graded $E$-module as well as a right $\mathcal{O}(U)$-module. The actions commute, so we have a bimodule. Thus, by the constructions in Sections 22.12 and 22.28 we can form the tensor product

\[ M \otimes _ E K^\bullet (U) \]

which is a differential graded $\mathcal{O}(U)$-module, i.e., a complex of $\mathcal{O}(U)$-modules. This construction is functorial with respect to $U$, hence we can sheafify to get a complex of $\mathcal{O}$-modules which we denote

\[ M \otimes _ E K^\bullet \]

Moreover, for each $U$ the construction determines a functor $\text{Mod}^{dg}_{(E, \text{d})} \to \text{Comp}^{dg}(\mathcal{O}(U))$ of differential graded categories by Lemma 22.29.1. It is therefore clear that we obtain a functor as stated in the lemma. $\square$

Lemma 22.35.2. The functor of Lemma 22.35.1 defines an exact functor of triangulated categories $K(\text{Mod}_{(E, \text{d})}) \to K(\mathcal{O})$.

Proof. The functor induces a functor between homotopy categories by Lemma 22.26.5. We have to show that $- \otimes _ E K^\bullet $ transforms distinguished triangles into distinguished triangles. Suppose that $0 \to K \to L \to M \to 0$ is an admissible short exact sequence of differential graded $E$-modules. Let $s : M \to L$ be a graded $E$-module homomorphism which is left inverse to $L \to M$. Then $s$ defines a map $M \otimes _ E K^\bullet \to L \otimes _ E K^\bullet $ of graded $\mathcal{O}$-modules (i.e., respecting $\mathcal{O}$-module structure and grading, but not differentials) which is left inverse to $L \otimes _ E K^\bullet \to M \otimes _ E K^\bullet $. Thus we see that

\[ 0 \to K \otimes _ E K^\bullet \to L \otimes _ E K^\bullet \to M \otimes _ E K^\bullet \to 0 \]

is a termwise split short exact sequences of complexes, i.e., a defines a distinguished triangle in $K(\mathcal{O})$. $\square$

Lemma 22.35.3. The functor $K(\text{Mod}_{(E, \text{d})}) \to K(\mathcal{O})$ of Lemma 22.35.2 has a left derived version defined on all of $D(E, \text{d})$. We denote it $- \otimes _ E^\mathbf {L} K^\bullet : D(E, \text{d}) \to D(\mathcal{O})$.

Proof. We will use Derived Categories, Lemma 13.14.15 to prove this. As our collection $\mathcal{P}$ of objects we will use the objects with property (P). Property (1) was shown in Lemma 22.20.4. Property (2) holds because if $s : P \to P'$ is a quasi-isomorphism of modules with property (P), then $s$ is a homotopy equivalence by Lemma 22.22.3. $\square$

Lemma 22.35.4. Let $R$ be a ring. Let $\mathcal{C}$ be a site. Let $\mathcal{O}$ be a sheaf of commutative $R$-algebras. Let $K^\bullet $ be a complex of $\mathcal{O}$-modules. The functor of Lemma 22.35.3 has the following property: For every $M$, $N$ in $D(E, \text{d})$ there is a canonical map

\[ R\mathop{\mathrm{Hom}}\nolimits (M, N) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(M \otimes _ E^\mathbf {L} K^\bullet , N \otimes _ E^\mathbf {L} K^\bullet ) \]

in $D(R)$ which on cohomology modules gives the maps

\[ \mathop{\mathrm{Ext}}\nolimits ^ n_{D(E, \text{d})}(M, N) \to \mathop{\mathrm{Ext}}\nolimits ^ n_{D(\mathcal{O})} (M \otimes _ E^\mathbf {L} K^\bullet , N \otimes _ E^\mathbf {L} K^\bullet ) \]

induced by the functor $- \otimes _ E^\mathbf {L} K^\bullet $.

Proof. The right hand side of the arrow is the global derived hom introduced in Cohomology on Sites, Section 21.36 which has the correct cohomology modules. For the left hand side we think of $M$ as a $(R, A)$-bimodule and we have the derived $\mathop{\mathrm{Hom}}\nolimits $ introduced in Section 22.31 which also has the correct cohomology modules. To prove the lemma we may assume $M$ and $N$ are differential graded $E$-modules with property (P); this does not change the left hand side of the arrow by Lemma 22.31.3. By Lemma 22.31.5 this means that the left hand side of the arrow becomes $\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(M, N)$. In Lemmas 22.35.1, 22.35.2, and 22.35.3 we have constructed a functor

\[ - \otimes _ E K^\bullet : \text{Mod}^{dg}_{(E, \text{d})} \longrightarrow \text{Comp}^{dg}(\mathcal{O}) \]

of differential graded categories and we have shown that $- \otimes _ E^\mathbf {L} K^\bullet $ is computed by evaluating this functor on differential graded $E$-modules with property (P). Hence we obtain a map of complexes of $R$-modules

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(M, N) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{O})} (M \otimes _ E K^\bullet , N \otimes _ E K^\bullet ) \]

For any complexes of $\mathcal{O}$-modules $\mathcal{F}^\bullet $, $\mathcal{G}^\bullet $ there is a canonical map

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{O})} (\mathcal{F}^\bullet , \mathcal{G}^\bullet ) = \Gamma (\mathcal{C}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{F}^\bullet , \mathcal{G}^\bullet )) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(\mathcal{F}^\bullet , \mathcal{G}^\bullet ). \]

Combining these maps we obtain the desired map of the lemma. $\square$

Lemma 22.35.5. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $K^\bullet $ be a complex of $\mathcal{O}$-modules. Then the functor

\[ - \otimes _ E^\mathbf {L} K^\bullet : D(E, \text{d}) \longrightarrow D(\mathcal{O}) \]

of Lemma 22.35.3 is a left adjoint of the functor

\[ R\mathop{\mathrm{Hom}}\nolimits (K^\bullet , -) : D(\mathcal{O}) \longrightarrow D(E, \text{d}) \]

of Lemma 22.32.1.

Proof. The statement means that we have

\[ \mathop{\mathrm{Hom}}\nolimits _{D(E, \text{d})}(M, R\mathop{\mathrm{Hom}}\nolimits (K^\bullet , L^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O})}(M \otimes ^\mathbf {L}_ E K^\bullet , L^\bullet ) \]

bifunctorially in $M$ and $L^\bullet $. To see this we may replace $M$ by a differential graded $E$-module $P$ with property (P). We also may replace $L^\bullet $ by a K-injective complex of $\mathcal{O}$-modules $I^\bullet $. The computation of the derived functors given in the lemmas referenced in the statement combined with Lemma 22.22.3 translates the above into

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(E, \text{d})})} (P, \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(K^\bullet , I^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{O})}(P \otimes _ E K^\bullet , I^\bullet ) \]

where $\mathcal{B} = \text{Comp}^{dg}(\mathcal{O})$. There is an evaluation map from right to left functorial in $P$ and $I^\bullet $ (details omitted). Choose a filtration $F_\bullet $ on $P$ as in the definition of property (P). By Lemma 22.20.1 and the fact that both sides of the equation are homological functors in $P$ on $K(\text{Mod}_{(E, \text{d})})$ we reduce to the case where $P$ is replaced by the differential graded $E$-module $\bigoplus F_ iP$. Since both sides turn direct sums in the variable $P$ into direct products we reduce to the case where $P$ is one of the differential graded $E$-modules $F_ iP$. Since each $F_ iP$ has a finite filtration (given by admissible monomorphisms) whose graded pieces are graded projective $E$-modules we reduce to the case where $P$ is a graded projective $E$-module. In this case we clearly have

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(E, \text{d})}} (P, \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(K^\bullet , I^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{O})}(P \otimes _ E K^\bullet , I^\bullet ) \]

as graded $\mathbf{Z}$-modules (because this statement reduces to the case $P = E[k]$ where it is obvious). As the isomorphism is compatible with differentials we conclude. $\square$

Lemma 22.35.6. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $K^\bullet $ be a complex of $\mathcal{O}$-modules. Assume

  1. $K^\bullet $ represents a compact object of $D(\mathcal{O})$, and

  2. $E = \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{O})}(K^\bullet , K^\bullet )$ computes the ext groups of $K^\bullet $ in $D(\mathcal{O})$.

Then the functor

\[ - \otimes _ E^\mathbf {L} K^\bullet : D(E, \text{d}) \longrightarrow D(\mathcal{O}) \]

of Lemma 22.35.3 is fully faithful.

Proof. Because our functor has a left adjoint given by $R\mathop{\mathrm{Hom}}\nolimits (K^\bullet , -)$ by Lemma 22.35.5 it suffices to show for a differential graded $E$-module $M$ that the map

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

is an isomorphism. We may assume that $M = P$ is a differential graded $E$-module which has property (P). Since $K^\bullet $ defines a compact object, we reduce using Lemma 22.20.1 to the case where $P$ has a finite filtration whose graded pieces are direct sums of $E[k]$. Again using compactness we reduce to the case $P = E[k]$. The assumption on $K^\bullet $ is that the result holds for these. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09LU. Beware of the difference between the letter 'O' and the digit '0'.