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Set up: Suppose that $\mathcal{X}$ is a derived stack (I'm using the conventions of Yaylali's "Notes on derived algebraic geometry," but I'd be happy to work with other conventions of the same flavor, e.g. Antieau-Gepner).

The $\infty$-category of quasi-coherent sheaves on $\mathcal{X}$ is defined to be limit $$\mathcal{D}_{\textrm{qc}}(\mathcal{X}) = \lim_{\operatorname{Spec}A\to\mathcal{X}}\mathcal{D}_{\textrm{qc}}(A),$$ where $\mathcal{D}_{\textrm{qc}}(A)$ denotes the $\infty$-category of $A$-modules for an animated ring $A.$

More General/Vague Question: Heuristically, it is often stated that an object of $\mathcal{D}_{\textrm{qc}}(\mathcal{X})$ is a "compatible family of modules $(\mathcal{F}_A)_A$ for every $\operatorname{Spec}A\to\mathcal{X},$" but in practice, how do we construct and check this compatibility?

Example: As an example, lemma 3.67 in Yaylali's notes states that if $f : X\to Y$ is a morphism of derived schemes, then $f$ admits a cotangent complex $\mathbf{L}_f.$ The proof constructs a candidate for $\mathbf{L}_f$ by first showing that $\mathbf{L}_X$ exists, and then checking that it satisfies the defining properties of a cotangent complex. In the construction, we consider a composition/triangle $\operatorname{Spec}B \to\operatorname{Spec}A \to X,$ where $\operatorname{Spec}A$ and $\operatorname{Spec}B$ are affine opens in $X$ (since in this case, the limit defining $\mathcal{D}_{\textrm{qc}}(X)$ may be taken over open immersions to $X$ from affines). Then we observe that $\operatorname{Spec}A$ admits a cotangent complex $\mathbf{L}_A,$ and that these are compatible with pullbacks in the sense that $\mathbf{L}_A\otimes_A B\simeq\mathbf{L}_B.$

At this point, Yaylali states that this defines an object $\mathbf{L}_X$ in the limit $\mathcal{D}_{\textrm{qc}}(X),$ observing that the relative cotangent complexes of monomorphisms were already shown to exist and be equivalent to $0.$

Specific Question 1: (Why) is it necessary and sufficient to check that the cotangent complex of a monomorphism is $0$ to verify compatibility?

Motivation: One construction I would like to make is a sheafy version of Hochschild homology for a morphism $\mathcal{X}\to\operatorname{Spec}k$ of derived schemes/stacks (which I will denote $\mathcal{HH}(\mathcal{X}/k)\in\mathcal{D}_{\textrm{qc}}(\mathcal{X})$). The "absolute variant," (denoted $\mathrm{HH}(\mathcal{X}/k)\in\mathcal{D}_{\textrm{qc}}(k)$), is defined as a Kan extension of $\mathrm{HH}(-/k),$ so there is no issue there.

However, if we want to view $\mathcal{HH}(\mathcal{X}/k)$ as living on $\mathcal{X}$ itself, this would be like taking a Kan extension of a "functor" where the target category depends on the input (if there's a way of doing this, I'd love to hear about it). However, I know several "compatibility" results which I suspect can be used to construct $\mathcal{HH}(\mathcal{X}/k).$ Namely:

  1. Given a morphism from $(k\to A)$ to $(k'\to A')$ in $\operatorname{Fun}(\Delta^1,\mathsf{aRing})$ (i.e., a "commutative square" of animated rings), there is a canonical base change morphism of animated rings $\mathcal{HH}(A/k)\otimes_A A'\to\mathcal{HH}(A'/k'),$ which is moreover an equivalence if the square is coCartesian.
  2. If $k\to A\to B$ is a composition of morphisms of animated rings with $A\to B$ etale, then the base change morphism $\mathcal{HH}(A/k)\otimes_A B\to\mathcal{HH}(B/k)$ is an equivalence.
  3. If $\operatorname{Spec}A\to\operatorname{Spec}k$ is a monomorphism of derived affine schemes, then $\mathcal{HH}(A/k)\simeq A.$

Specific Question 2: Are the above facts are enough to establish the desired compatibility to form $\mathcal{HH}(\mathcal{X}/k)$ as the object of the limit corresponding to the family "$(\operatorname{Spec}A\to\mathcal{X},\mathcal{HH}(A/k))$"? If so, how do we make this precise, and if not, what is missing (or what fails)?

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To check the compability you need to have $\mathbb{L}_{A/B} \simeq 0$, here the author invoke the argument of 3.63. and the fact that the relative cotangent of $\operatorname{Spec}(R) \rightarrow X$ are $0$ for $R = A$ and $B$, because they are monomorphisms. Hence you can show that $\mathbb{L}_{B}\otimes_B A \simeq \mathbb{L}_A$ because $$\mathbb{L}_B \otimes_B A \rightarrow \mathbb{L}_A \rightarrow \mathbb{L}_{A/B}$$ is exact (by 3.63. again).

By the way, to be fully rigorous about this construction, you have to know how to describe limits of $\infty$-categories. So if $X$ is your derived scheme, the limit of the diagram $\operatorname{Zar}_{/X}^{\operatorname{op}} \rightarrow \operatorname{Cat}_{\infty}$ sending an open affine $\operatorname{Spec}(R) \rightarrow X$ to $\operatorname{Mod}_A$ is $\operatorname{Qcoh}(X)$. To compute that limit you identify your functor with a cartesian fibration $E \rightarrow \operatorname{Zar}_{/X}$ and you take the $\infty$-category of cartesian sections of that fibration. This amounts to send to each $\operatorname{Spec}(R)\rightarrow X$ open, a module $M_R \in \operatorname{Mod}_R$ with equivalences $M_B \otimes_B A \simeq M_A$ + higher coherences data. For instance imagine if you have the following diagram : $$\operatorname{Spec}(A)\rightarrow\operatorname{Spec}(B)\rightarrow \operatorname{Spec}(C) \rightarrow X.$$ Over $\operatorname{Spec}(A)$ you have three equivalences : $$\mathbb{L}_B \otimes_B A \simeq \mathbb{L}_A,$$ $$\mathbb{L}_C \otimes_C A \simeq \mathbb{L}_A,\text{ and}$$ $$\left(\mathbb{L}_C \otimes_C B\right)\otimes_B A \simeq \mathbb{L}_B \otimes_B A.$$ You will have to provide a $2$-cell between those guarantying the correct compabilities. These coherences are satisfied for the cotangent complex but it is not clearly written in the notes and I believe it is hard to do it properly from the basic definitions of cotangent complex. To write it properly there are two ways of doing it, either you work in Lurie's framework and it is doable because in practice I think $E$ is given by the tangent category of $\operatorname{Zar}_{/X}$ (check section $7.3.$ of Higher Algebra for the affine theory of cotangent complex and 17.1.2. of SAG for the global theory). Or you can work in Toën-Vezzosi's framework (model categories) (check 1.4.1.11 of HAGII) I don't know the details but it seems more straighforward.

So if you want to properly define the Hochschild complex you will have to indeed check the $1$st level of coherences which seems true by what you say (an open immersion is always etale and etalness satisfies the two-out-of-three property, so in particular $\operatorname{Spec}(A)\rightarrow \operatorname{Spec}(B)$ is etale and then you have the first step of your cartesian section. But you will have to also check higher coherences data that isn't really doable by hand, so your data isn't enough.

For the Hochschild homology there is a cool trick with the loop space. So if $X$ is a derived stack you can define $\mathcal{L}X := \operatorname{Map}(S^1,X)$. If $X = \operatorname{Spec}(A)$ then $\mathcal{L} X = \operatorname{Spec}(A\otimes_{A\otimes A} A) = \operatorname{Spec}(\mathcal{HH}(A/k))$ (This is because $S^1 = *\sqcup_{*\sqcup *} *$). The underlying $A$-module structure is the pushforward along the evaluation at $1$ map $\mathcal{L} \operatorname{Spec}(A) \rightarrow \operatorname{Spec}(A)$. So in general you could define $\mathcal{HH}(X/k)$ as the pushforward of $\mathcal{O}_X$ along $\mathcal{L}X\rightarrow X$. See e.g. https://arxiv.org/abs/1002.3636 for cool stuff about loop spaces.

Using this definition I can at least check that if $X$ is a qcqs derived scheme (so that we can have base change) this definition lifts the affine definition. Let $p_X : \mathcal{L}X \rightarrow X$ be the projection and $i : \operatorname{Spec}(A) \rightarrow X$ an open immersion. Then $i^* \mathcal{HH}(X/k) \simeq \mathcal{HH}(A/k)$. This is an application of base change for the square :$\require{AMScd}$ $$ \begin{CD} \mathcal{L}\operatorname{Spec}(A) @>\mathcal{L}i>> \mathcal{L}X \\ @V{p_A}VV @VV{p_X}V\\ \operatorname{Spec}(A) @>i>> X. \end{CD} $$ So I claim this diagram is a pullback because $i$ is a monomorphism. So you can form a cube from : $$ \begin{CD} \mathcal{L}\operatorname{Spec}(A) @>>> \operatorname{Spec}(A) @>>> X @<<< \mathcal{L}X \\ @VVV @VVV @VVV @VVV\\ \operatorname{Spec}(A) @>>> \operatorname{Spec}(A)\times \operatorname{Spec}(A) @>>> X\times X @<<< X, \end{CD} $$ the middle square is a pullback because $i$ is a monomorphism, as well as the bottom square (given by all the bottom arrows) for the same reason. The other square are pullbacks because $\mathcal{L} X \simeq X\times_{X\times X} X$. So the upper one is a pullback that is what we want. You can apply base change because the loop space of a derived scheme is the shifted tangent bundle which is as well qcqs.

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  • $\begingroup$ Thanks for your response! The higher coherence data is exactly what I was worried about. As for the loop spaces, the remark in your linked paper about $U\mapsto\mathcal{L}U$ not forming an etale cosheaf in general seems a bit concerning, at least when it comes to ensuring that different definitions of Hochschild homology are compatible. For nice enough (e.g. qcqs) derived schemes over an affine base, the base change natural transformation for a Cartesian square will be an equivalence, and I believe this implies that the different definitions one wants to make will coincide. (continued) $\endgroup$ Commented Jan 27, 2025 at 15:41
  • $\begingroup$ But when the schemes become less controlled, or when we move to stacks, it is no longer clear (to me) that the different definitions one can make would coincide. For example, when defining $\mathrm{HH}(X/k)$ as a $k$-module, we have Kan extension/descent, pushforward from the loop space, or derived global sections of pushforward and pullback of the structure sheaf, all of which seem like they could differ if we don't have base change (or other information). $\endgroup$ Commented Jan 27, 2025 at 15:41
  • $\begingroup$ The stack is not a cosheaf but don't you believe the global functions form a sheaf for the etale topology ? And yes for general stacks I agree it's not clear the two definitions agree. But at least for DM stacks I think you can show the way you suggested it agrees. $\endgroup$ Commented Jan 27, 2025 at 16:09
  • $\begingroup$ Sure, I believe that global sections form an etale sheaf. I suppose my concern is actually the compatibility of the definitions, and in particular that an appropriate descent holds (as in the Kan extension definition). $\endgroup$ Commented Jan 27, 2025 at 17:06
  • $\begingroup$ I've added a proof of the fact that the global function definition works for qcqs derived schemes. I hope it is what you wanted. $\endgroup$ Commented Jan 27, 2025 at 18:37
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I believe I've managed to come up with a construction which is satisfying for my purposes. I'll record the idea below for anyone looking to do something similar.

In particular, first interpret Hochschild homology not as a functor valued in $\mathcal{D}(k),$ but rather valued in $\mathcal{QC}^{\textrm{op}}$, where $\mathcal{QC}\to\mathsf{d}\mathcal{S}\mathsf{tk}$ is the Cartesian fibration classifying the functor $\mathsf{d}\mathcal{S}\mathsf{tk}^{\textrm{op}}\to\mathcal{C}\mathsf{at}_{\infty}$ which sends $X\mapsto\mathcal{D}_{\textrm{qc}}(X).$

That is, we construct a functor $\left(\mathsf{d}\mathcal{A}\mathsf{ff}^{\Delta^1}\right)^{\textrm{op}}\to\mathcal{QC}^{\textrm{op}}$ sending $(\operatorname{Spec}A\to\operatorname{Spec}k)\mapsto (k,\mathrm{HH}(A/k)).$ Then we Kan extend this functor along the inclusion $\left(\mathsf{d}\mathcal{A}\mathsf{ff}^{\Delta^1}\right)^{\textrm{op}}\to\left(\mathsf{d}\mathcal{S}\mathsf{tk}^{\Delta^1}\right)^{\textrm{op}}$ to obtain Hochschild homology functorially in morphisms of derived stacks.

One can verify that this agrees with the standard definition by "pulling back" along $\{\operatorname{Spec} k\}\to\mathsf{d}\mathcal{A}\mathsf{ff}\to\mathsf{d}\mathcal{S}\mathsf{tk}$ (noting that $\left(\mathsf{d}\mathcal{A}\mathsf{ff}^{\Delta^1}\right)^{\textrm{op}}$ and $\left(\mathsf{d}\mathcal{S}\mathsf{tk}^{\Delta^1}\right)^{\textrm{op}}$ admit Cartesian fibrations to $\mathsf{d}\mathcal{A}\mathsf{ff}^{\textrm{op}}$ and $\mathsf{d}\mathcal{S}\mathsf{tk}^{\textrm{op}},$ respectively, and that $\mathsf{d}\mathcal{A}\mathsf{ff}\subseteq \mathsf{d}\mathcal{S}\mathsf{tk}$ is a replete sub-$\infty$-category) and appealing to lemma 4.3.3.9 in Higher Topos Theory.

The compatibility outlined above is essentially a verification that if $X\to S$ is a morphism of derived stacks, and if $f : \operatorname{Spec} B\to S$ is an arbitrary morphism, that $f^*\mathrm{HH}(X/S)\to\mathrm{HH}(X_B/B)$ is an equivalence. So, this construction provides $\mathrm{HH}(X/S)$ as a quasicoherent sheaf on $S$ "given by the family $(\operatorname{Spec}B\to S,\mathrm{HH}(X_B/B))$."

I also believe that with a bit more effort, this approach extends to give a way to define $\mathcal{HH}(X/S)\in\mathcal{D}_{\textrm{qc}}(X)$ functorially together with functorial equivalences $\mathrm{HH}(X/S)\simeq \pi_*\mathcal{HH}(X/S)$ if $\pi: X\to S$ denotes the morphism in question (as well as a similar description of $\mathcal{HH}(X/S)$ in terms of pullbacks to affines $\operatorname{Spec} B\to X$).

Hopefully I've managed to put the $\textrm{op}$'s in all the right places and get my co/Cartesian fibrations correct. Please let me know if you spot any errors.

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