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Let $X$ be a smooth scheme over a field $k$ and let $\mathsf{D}_{\text{qc}}(\mathcal{D}_X)$ be the full subcategory of $\mathsf{D}(\mathcal{D}_X\mathsf{-Mod})$ composed of the complexes of $\mathcal{D}_X$-modules with quasi-coherent cohomology.

If $f:X\to Y$ is a morphism between such schemes, we have natural functors $$f_*:\mathsf{D}_{\text{qc}}(\mathcal{D}_X)\leftrightarrows \mathsf{D}_{\text{qc}}(\mathcal{D}_Y):f^!.$$

If we restrict to the subcategory of complexes with holonomic cohomology, then the Verdier duality functor allows us to find left adjoints $f^*$ and $f_!$ of $f_*$ and $f^!$, respectively.

I wonder if those adjoints already exist in $\mathsf{D}_{\text{qc}}(\mathcal{D}_X)$.

Perhaps we can use Brown representability for left adjoints (as in Neeman's book about triangulated categories) or some other adjoint functor theorem...

In Neeman's paper The Grothendieck Duality Theorem via Bousfield's Techniques, it is proven that $f_*$ has a right adjoint (even though in the text it is said that it's a left adjoint). Perhaps we can use this functor to construct our left adjoints?

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  • $\begingroup$ Here we see a difference between \mathsf{} and \textsf{}: $\mathsf{D}(\mathcal{D}_X\mathsf{-Mod})$ versus $\mathsf{D}(\mathcal{D}_X\textsf{-Mod}).$ The latter gives you a hyphen rather than a minus sign. A hyphen would be standard here, if I understand you correctly. $\endgroup$ Oct 5, 2021 at 18:19
  • $\begingroup$ @MichaelHardy Indeed, I usually have a macro for categories so I forgot how to do it right. Thank you $\endgroup$
    – Gabriel
    Oct 5, 2021 at 18:28

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They do not exist in general. The simplest example is maybe to take $X\rightarrow Y$ to be the closed embedding of the origin inside $\mathbb{A}^1.$ Then $f_*$ sends a vector space $V$ to the $\mathcal{D}_{\mathbb{A}^1}$-module $V\otimes\delta_0$, where $\delta_0$ is the irreducible $\mathcal{D}_{\mathbb{A}^1}$-module set-theoretically supported at $0$. The point is that because $\delta_0$ is infinite-dimensional as a vector space, this functor does not commute with products and hence cannot have a left adjoint.

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  • $\begingroup$ this seems like the right direction but I'd think you need an example with nonholonomic D-modules, and not a closed embedding where you have Kashiwara's equivalence $\endgroup$ Oct 6, 2021 at 3:49
  • $\begingroup$ @DavidBen-Zvi I don't think Kashiwara's is a problem - the category $D(\mathbb{A}^1)_0$ of D-modules supported at $0$ is not closed under limits, no? (I.e. it admits limits but they don't agree with the limits in $D(\mathbb{A}^1).$) $\endgroup$
    – dhy
    Oct 6, 2021 at 3:51
  • $\begingroup$ ah thanks, I guess the point is the inclusion of (ind)holonomics into all D modules doesn’t preserve limits, so the holonomic adjunction doesn’t help $\endgroup$ Oct 6, 2021 at 4:11

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