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This question might be really easy (or stupid), but I have only vague (heard-about) knowledge of DG categories, so I don't know where to look for an answer.

Let $X$ be a smooth projective variety over a field $k$ (I am mostly interested in $k = \mathbb{C}$). Assume that I have a class $\alpha \in H^k(X,\mathcal{O}_X)$, for some $0 \leq k \leq \dim X$. Then $\alpha$ gives a natural tansformation of triangulated functors:

$$ \alpha \otimes id : id_X \rightarrow id_X[k],$$

where $id_X$ is the identity functor in the derived category $D^b(X)$ (seen as a triangulated category).

Let's now endow $D^b(X)$ with its DG enhancement. Is it possible to lift $\alpha \otimes id_X$ to a natural transformation of DG functors:

$$\tilde{\alpha} : id_X \rightarrow id_X,$$

such that $\mathcal{H}^k(\tilde{\alpha}) = \alpha \otimes id$ and $\mathcal{H}^p(\tilde{\alpha}) = 0$ for all $p \neq k$?

Perhpas it is not possible in the general case, but are there conditions on $X$ which would guarantee that it is possible in some cases?

Thanks a lot in advance!

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  • $\begingroup$ The derived tensor product with a representing cycle does the job. $\endgroup$ Commented Feb 5, 2016 at 21:40
  • $\begingroup$ @FernandoMuro Dear Fernando, thanks for the answer. But could you elaborate a bit since I really know very little about DG categories? I sometimes heard that tensor products and DG engancement of $D^b(X)$ are not compatible. Is it the case here? Or not? Or Have I misunderstood what this "non-compatibility" means? $\endgroup$
    – Libli
    Commented Feb 5, 2016 at 21:46
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    $\begingroup$ Where'd you hear that? its not true, the derived tensor product will give you a symmetric monoidal structure on the dg category (interpreted in the appropriate homotopical sense). $\endgroup$ Commented Feb 5, 2016 at 23:11
  • $\begingroup$ @crystalline Thanks for the answer. Well I can't really remember, probably a seminar talk... I thought somethink like if $P$ is a (co)-fibrant resolution of $A$ and $Q$ is a (co)-fibrant resolution of $B$, then $P \otimes Q$ is not necessarily a (co)-fibrant resolution of $A \otimes B$. But perhaps this does not cause any problem for my question... As already mentionned above, I really know very little about DG categories. $\endgroup$
    – Libli
    Commented Feb 5, 2016 at 23:17
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    $\begingroup$ i think what you're thinking of is the dg category of dg categories. there you have that problem, that the symmetric monoidal structure isnt compatible with the model structure. but you definitely have a symmetric monoidal model structure on complexes in a (Grothendieck?) abelian category. $\endgroup$ Commented Feb 5, 2016 at 23:48

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It is possible to lift $\alpha \otimes \mathrm{id}$ to a dg-enhancement, one way is the following. Just to fix notation, this natural transformation is induced from $$ \alpha \colon \mathcal{O}_X \to \mathcal{O}_X[k] \in \mathrm{Hom}(\mathcal{O}_X,\mathcal{O}_X[k]) = \mathrm{H}^k(X,\mathcal{O}_X). $$ by tensoring with $\mathrm{id} \colon A \to A$ for $A \in \mathcal{D}^b(X)$.

As a dg-enhancement, I choose $\mathrm{Inj}(X)$, the bounded complex of injective sheaves with coherent cohomology. Let $I^\bullet \in \mathrm{Inj}(X)$ be an injective resolution of $\mathcal{O}_X$. Then $\alpha$ can be lifted to a morphism of complexes $$ \tilde\alpha \colon I^\bullet \to I^\bullet[k]. $$ The next step should be again to tensor with $\mathrm{id}\colon J^\bullet \to J^\bullet$ for $J^\bullet \in \mathrm{Inj}(X)$ and obtain the desired natural transformation of $\mathrm{Inj}(X)$.

This is not possible directly, but can be done using enriched model category theory. As Olaf Schnürer worked out in details in Six operations on dg enhancements of derived categories of sheaves, the whole six-functor-formalism can be set up for $\mathrm{Inj}(X)$, in particular there is a tensor product.

Actually, he establishes all these functors for the unbounded derived category of quasi-coherent sheaves $\mathcal{D}(X)$ using unbounded h-injective complexes as a dg-enhancement. So by passing to the unbounded categories, the variety $X$ does not even need to be smooth.

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