# Results relying on higher derived algebraic geometry

Are there any results in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures but whose most natural (known) proof uses derived $$n$$-Artin stacks for $$n>1$$? We are using Toen--Vezzosi terminology.

EDIT: the $$n$$-Artin stack in question should not to be obtained as the quotient of the constant groupoid associated to a $$(n-1)$$-Artin stack. I did not specify this initially, my fault, but I think the requirement is pretty natural. Neither of the two answers I can see at the time of the edit address this point.

• Derived categories (of quasicoherent sheaves, of sheaves of $D$-modules, of etale sheaves ...) are simultanously higher and derived. Do they count? – Charles Rezk May 19 '19 at 14:15
• @CharlesRezk do they prove any result in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures? – user138661 May 19 '19 at 14:26
• If derived categories are going to count, then this question suggests an answer (and I'm sure there are others in the same vein): mathoverflow.net/questions/321852/… . However, I would have thought this is not what you had in mind. It is certainly not what I would call derived geometry. Also, people usually study derived categories of coherent sheaves as a triangulated category, i.e. forgetting the higher structure to some extent. – Sam Gunningham May 19 '19 at 15:28
• To a certain extent, all derived geometry is necessarily "higher'", for example in the sense that the functor of points of a derived scheme must take values in the $\infty$-category of spaces. Perhaps you want "derived" to mean only using derived schemes as opposed to derived stacks, and "higher" to refer to (not necessarily derived) $\infty$-stacks? – Sam Gunningham May 19 '19 at 15:36
• I am very confused by this question. Would the solution to the Weibel conjecture count? – Denis Nardin May 19 '19 at 17:42

Here is an example from Bhargav Bhatt's talk "Using DAG" at MSRI last week. Needless to say, any mistakes are mine.

Theorem. Let $$X$$ be a coherent (quasi-compact and quasi-separated) scheme, let $$A$$ be a ring complete with respect to an ideal $$I\subseteq A$$. Then $$X(A) \to \varprojlim_n X(A/I^{n+1})$$ is bijective.

Before going into the proof, let us consider the case $$X$$ is affine. Then $$X(A) = {\rm Hom}(\Gamma(X, \mathcal{O}_X), A) = \varprojlim_n {\rm Hom}(\Gamma(X, \mathcal{O}_X), A/I^{n+1}) = \varprojlim_n X(A/I^{n+1}) .$$

The idea for the general (coherent) case is to replace $$\Gamma(X, \mathcal{O}_X)$$ with $${\rm Perf}(X)$$, the category of perfect complexes on $$X$$.

Slogan. Affine schemes have "enough functions". Coherent schemes have "enough vector bundles (perfect complexes)".

The second idea may be due to Thomason.

More precisely, we have:

Proposition. Let $$X$$ and $$Y$$ be schemes.

(a) If $$X$$ is affine, then $${\rm Hom}(Y, X) \to {\rm Hom}(\Gamma(X, \mathcal{O}_X), \Gamma(Y, \mathcal{O}_Y))$$ is bijective.

(b) If $$X$$ is coherent, then $${\rm Hom}(Y, X) \to {\rm Hom}({\rm Perf}(X), {\rm Perf}(Y))$$ is an equivalence.

We must specify what (b) means (here is where DAG enters the picture). We consider $${\rm Perf}(X)$$ as the symmetric monoidal $$\infty$$-category of perfect complexes on $$X$$ (complexes locally quasi-isomorphic to a bounded complex of locally free sheaves of finite rank). The $${\rm Hom}$$ on the right means the $$\infty$$-groupoid (space) exact $$\otimes$$-functors. So in particular (b) implies that this space is discrete. The map in (b) sends $$f$$ to $$f^*$$, the pull-back functor.

"Proof" of Theorem. We repeat the proof of the affine case, replacing rings with categories of perfect complexes: $$X(A) = {\rm Hom}({\rm Perf}(X), {\rm Perf}(A)) = \varprojlim_n {\rm Hom}({\rm Perf}(X), {\rm Perf}(A/I^{n+1})) = \varprojlim_n X(A/I^{n+1}) .$$ Unlike in the affine case, the middle equality needs some justification, which I am not ready to give.

End remarks.

(1) I think Bhargav mentioned that an idea due to Gabber allows one to get rid of the assumption that $$X$$ is coherent in the Theorem.

(2) He also said that the above proof is the only one he is aware of.

(3) Reference for the above (thanks to the user crystalline):

Bhargav Bhatt Algebraization and Tannaka duality arxiv.org/abs/1404.7483.

• This doesn't actually use DAG, only higher categories. Key word is Tannaka duality, reference for the above: arxiv.org/abs/1404.7483. – crystalline May 19 '19 at 18:06
• Thank you! I thought looking at $X$ through the symmetric monoidal $\infty$-category ${\rm Perf}(X)$ counts as a use of DAG, though indeed, derived schemes do not make an appearance. – Piotr Achinger May 19 '19 at 20:44
• Certainly very derived techniques, just not DAG in the sense of the question. – crystalline May 21 '19 at 17:40

The derived moduli stack of perfect complexes $$RPerf$$ is a derived Artin stack which admits a filtration by open sub stacks $$RPerf^{[a,b]}$$. The latter is a derived $$(b-a+1)$$-Artin stack. See:

Moduli of objects in dg-categories. Annales scientifiques de l'École Normale Supérieure, Serie 4, Volume 40 (2007) no. 3, pp. 387-444. doi : 10.1016/j.ansens.2007.05.001. http://www.numdam.org/item/ASENS_2007_4_40_3_387_0/

This derived stack plays a critical role in the recently hot topic of shifted symplectic structures and shifted deformation quantization, see:

Pantev, T., Toën, B., Vaquié, M. et al. Publ.math.IHES (2013) 117: 271. https://doi.org/10.1007/s10240-013-0054-1

• Wrt applications to classical stuff, existence of symmetric obstruction theories in certain classical (obstructed!) moduli problems (typically DT type moduli of sheaves stuff) – EBz May 19 '19 at 18:57
• @EBz you mean that is needed to define the virtual fundamental class to define counting invariants? – user138661 May 19 '19 at 19:05
• Is it a theorem that $RPerf^{[a, b]}$ is a strict $(b-a+1)$-Artin stack, i.e. not obtained as the quotient of the constant groupoid associated to a $(b-a)$-Artin stack? – user138661 May 19 '19 at 19:21
• @schematic_boi Not sure I get the question; $RPerf^{[a,b]}$ is definitely not $(b-a)$-Artin if that's what you're asking. Even think about $RPerf^{[0,0]}$, it's 1-Artin but obviously not 0-Artin. And $RPerf$ itself is not even close to being 1-Artin. – crystalline May 21 '19 at 17:52
• @schematic_boi Being $n$-Artin places strong bounds on your stack. If I have the indexing right, say $X$ is $n$-Artin, then the functor $X : dRing \to sSet$, restricted to ordinary rings, takes values in $n-1$-trunc. spaces. Obviously that's not the case for $RPerf^{[0,1]}$ as perfect complexes of amplitude $[0,1]$ form an $\infty$-category that is not a $1$-category. – crystalline May 21 '19 at 18:43