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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.

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    $\begingroup$ Derived categories (of quasicoherent sheaves, of sheaves of $D$-modules, of etale sheaves ...) are simultanously higher and derived. Do they count? $\endgroup$ – Charles Rezk May 19 at 14:15
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    $\begingroup$ @CharlesRezk do they prove any result in number theory or algebraic geometry whose statement does not involve either higher categories or any derived structures? $\endgroup$ – user138661 May 19 at 14:26
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    $\begingroup$ 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. $\endgroup$ – Sam Gunningham May 19 at 15:28
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    $\begingroup$ 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? $\endgroup$ – Sam Gunningham May 19 at 15:36
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    $\begingroup$ I am very confused by this question. Would the solution to the Weibel conjecture count? $\endgroup$ – Denis Nardin May 19 at 17:42
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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.

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    $\begingroup$ This doesn't actually use DAG, only higher categories. Key word is Tannaka duality, reference for the above: arxiv.org/abs/1404.7483. $\endgroup$ – crystalline May 19 at 18:06
  • $\begingroup$ 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. $\endgroup$ – Piotr Achinger May 19 at 20:44
  • $\begingroup$ Certainly very derived techniques, just not DAG in the sense of the question. $\endgroup$ – crystalline May 21 at 17:40
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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

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  • $\begingroup$ Wrt applications to classical stuff, existence of symmetric obstruction theories in certain classical (obstructed!) moduli problems (typically DT type moduli of sheaves stuff) $\endgroup$ – EBz May 19 at 18:57
  • $\begingroup$ @EBz you mean that is needed to define the virtual fundamental class to define counting invariants? $\endgroup$ – user138661 May 19 at 19:05
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    $\begingroup$ 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? $\endgroup$ – user138661 May 19 at 19:21
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    $\begingroup$ @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. $\endgroup$ – crystalline May 21 at 17:52
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    $\begingroup$ @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. $\endgroup$ – crystalline May 21 at 18:43

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