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?



Six functor formalism for quasi-coherent $D$-modules
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.


