# Confusion on summand of Hochschild homology of D-modules

I've encountered the following confusion.

Start with the category of perfect D-modules on $$\mathbb{A} ^{1}$$, which I'll denote $$D$$. We have the object $$\mathcal{O}$$, which is exceptional in the sense of having derived endomorphism ring $$k$$.

I believe that this implies that the subcategory generated by $$\mathcal{O}$$ is an admissible subcategory and so its Hochschild homology is a summand of that of $$D$$. On the other hand the Hochschild homology of the category $$D$$ ought to be the Hochschild homology of the algebra of differential operators on $$\mathbb{A} ^{1}$$, which has vanishing zero degree component, and thus the confusion.

Evidently I'm missing an assumption somewhere, but I've been unable to locate exactly where and help would be appreciated.

Edit: I should point out that Morita theory implies that the subcategory generated by $$\mathcal{O}$$ is isomorphic to perfect complexes over $$\mathbb{R}End(\mathcal{O})=k$$.

I think that if we consider the category perfect modules(which we have to in order to get a non-trivial answer for Hochschild homology) then the inclusion of the subcategory generated by $$E$$ doesn't have any adjoints. For example, the usual construction of the left adjoint is $$M\mapsto RHom(M, E)^{*}\otimes_k E$$ but for $$M=\mathcal{D}$$(free module of rank $$1$$) we have $$RHom(\mathcal{D},\mathcal{O})=Hom(\mathcal{D},{\mathcal{O}})=\mathcal{O}$$ but $$\mathcal{O}$$ is not a finite-dimensional vector space. One could probably turn this observation into a rigorous argument by showing that the adjoint, if exists, should be compatible with that on the category of all modules.

Edit: As noticed by EBz and Sam Gunningham in the comments, what follows is incorrect and either of the orthogonal complements to $$\mathcal{O}$$ in the category of perfect $$\mathcal{D}$$-modules is non-zero.

[Another way to see that there is no orthogonal decomposition of the form $$D=\langle \langle \mathcal{O}\rangle, A\rangle$$ is that the right(and also left) orthogonal $$\mathcal{O}^{\perp}$$ is zero in the category of perfect D-modules. Indeed, if $$M$$ is an object of the derived category of D-modules with finitely generated cohomology in a bounded range of degrees which is right orthogonal to $$\mathcal{O}$$, the spectral sequence $$E^{pq}_2=Ext^q(\mathcal{O},H^p(M))\Rightarrow H^{p+q}(RHom(\mathcal{O},M))$$ converges to $$0$$. Let $$p$$ be the minimal degree where $$H^p(M)\neq 0$$. There are no differentials coming in or out of the object $$E^{p,0}_r$$ for any $$r$$ so $$E^{p,0}_2$$ has to vanish for the spectral sequence to be converging to $$0$$. Since $$\mathcal{O}$$ is the cokernel of the right multiplication by $$\partial_x$$ on $$\mathcal{D}$$, it implies that $$\partial_x$$ is an automorphism of $$H^p(M)$$. But since $$H^p(M)$$ is finitely generated, it has to be $$0$$ which contradicts with the definition of $$p$$.]

• Hi, am I right in saying that you mixed up E and M in the first bit of the answer? Nonetheless it still seems to work. I'm starting to think that the missing assumption is homological compactness of the category, ie the thing that implies finitude of hom spaces. Does this seem correct? This would at least explain all the literature dealing with these decompositions being about proper varieties.
– EBz
Feb 7, 2019 at 9:46
• Sure, I should have swapped them, edited. Yes, it seems that under the assumption that all $H^d(RHom(A,B))$ are finite-dimensional it seems that the semi-orthogonal decomposition on the category of all modules descends to that of the category of perfect modules. Feb 8, 2019 at 17:34
• I accepted the answer bc I'm now sure compactness of the dg cat is the required condition but I'm still a bit confused by something. It seems \partial can act as an automorphism on a finite type D module. Indeed functions on the punctured line are finite type as a D module and just Fourier transform this?
– EBz
Feb 9, 2019 at 11:30
• I agree with @EBz, there are many perfect D-modules in the right orthogonal to $\mathcal O$. For another example, take the (clean) extension of a flat connection on $\mathbb A^1- \{0\}$ with non-unipotent monodromy. Feb 11, 2019 at 17:48
• @SamGunningham EBz Yes, you're completely right, I shouldn't have fallen for the commutative intuition. Feb 17, 2019 at 16:54