7
$\begingroup$

An old chestnut is that filtered objects are the same as sheaves over $\mathbb A^1 / \mathbb G_m$.

Question: Is there a similar description of chain complexes?

More precisely, if $\mathcal C$ is a category, then define the category of filtered objects in $\mathcal C$ to be the functor category $Fil(\mathcal C) = Fun(\mathbb Z, \mathcal C)$, where $\mathbb Z$ is the integers regarded as a poset. Then if $X$ is a scheme, there is a canonical equivalence $Fil(QCoh(X)) \simeq QCoh(X \times \mathbb A^1 / \mathbb G_m)$, where $\mathbb G_m$ acts on $\mathbb A^1$ in the usual way. What this says is that $\mathbb G_m$-equivariant sheaves on $\mathbb A^1 \times X$ are the same as filtered sheaves on $X$. As $\mathbb G_m$-actions are the same as gradings, this says in other words that a graded object equipped with an an endomorphism of degree 1 is the same as a filtered object.

I'd like a similar description of the category of chain complexes $Ch(QCoh(X)) \simeq QCoh(X \times S)$, where $S$ is some fixed stack, probably a quotient $S = T / G$ for some scheme $T$ and some action by a group scheme $G$.

Note: I believe that if $\mathcal C$ is stable, then $Fil(\mathcal C) \simeq Ch(\mathcal C)$ via some sort of $\infty$-categorical Dold-Kan correspondence (at any rate, I'm quite sure this is true if we talk about nonnegatively-graded chain complexes and nonnegative filtrations). So the stack $S$ will have to be derived-equivalent to $\mathbb A^1 / \mathbb G_m$, but perhaps not equivalent in an underived sense.

$\endgroup$
7
  • 1
    $\begingroup$ Perhaps you are interested in the work of Balmer et al on Tensor Triangulated Categories. Briefly, a space is associated with such a category. Note that filtered objects carry a natural tensor structure. Without a tensor structure, it is not obvious where the product/co-product for your Hopf-like structure will come from. $\endgroup$
    – Kapil
    Apr 28, 2023 at 3:20
  • $\begingroup$ Unless I’m severely misunderstanding your question, it seems that one candidate for your S is BG_a^#, where G_a^# is the divided power hull of the origin in G_a. (This’ll recover cochain complexes, corresponding to complete filtered objects.) This is coming from the fact that G_a^# is Cartier dual to the formal completion of G_a at the origin. $\endgroup$
    – skd
    Apr 28, 2023 at 3:47
  • $\begingroup$ Sorry, I meant the classifying stack of the semidirect product G_a^# x| G_m, where G_m is acting with weight 1. $\endgroup$
    – skd
    Apr 28, 2023 at 3:54
  • $\begingroup$ @skd I'd love to hear more about this! I don't understand the role played by Cartier duality here, for one thing. $\endgroup$
    – Tim Campion
    Apr 28, 2023 at 4:19
  • 1
    $\begingroup$ @TimCampion am I missing something that all the answers so far give geometric objects over $\mathbb Z$, so they'd work for something like $\mathbb Z$-linear stable $\infty$ categories, rather than arbitrary infinity categories? For example, in spectra, filtered objects are sheaves over $\mathbb A^1/ \mathbb G_m$, where $\mathbb A^1 = Spec(\mathbb S[\mathbb N]), \mathbb G_, = Spec(\mathbb S[\mathbb Z])$. Over the sphere it's not so clear to me what Achim's $\Lambda$ should be, even less clear what Sanath's $\mathbb G_a^\#$ should be.... $\endgroup$
    – Bbb
    Oct 3, 2023 at 13:26

2 Answers 2

8
$\begingroup$

Shouldn't it just be $\operatorname{Spec}(\Lambda)/\mathbb{G}_m$, where $\Lambda = \mathbb{Z}[d]/d^2$ and the $\mathbb{G}_m$ action encodes the grading with $d$ in degree $-1$? This is just the observation that chain complexes are the same as graded modules over an exterior algebra in a degree $-1$ generator, similar to how the $\mathbb{A}^1/\mathbb{G}_m$ description of filtered objects relates to their description as graded modules over a polynomial algebra $\mathbb{Z} [\tau]$.

$\endgroup$
6
  • 2
    $\begingroup$ Can one recover Dold-Kan from this via some form of Koszul duality ? (I'm thinking that $\Lambda$ looks a lot like $\mathbb Z\otimes_{\mathbb Z[\tau]}\mathbb Z$, up to $d$ being in degree $1$ rather than $-1$) $\endgroup$ Apr 28, 2023 at 8:03
  • $\begingroup$ Probably! I'm not super convinced that the equivalence between filtered objects and cochain complexes should be called Dold-Kan, shouldn't that refer to something with simplicial objects? In any case, It's been written up by Stefano Ariotta in arxiv.org/abs/2109.01017. He doesn't express it in terms of graded modules and instead makes a statement about functor categories, but it looks like the type of argument you're suggesting. (The $+1$ rather than $-1$ corresponds to the fact that cochain complexes appear) $\endgroup$ Apr 28, 2023 at 8:07
  • $\begingroup$ Oh yeah you're absolutely right that I should not call this Dold-Kan :D I hadn't read Stefano's paper, I was just aware of it, but thanks ! $\endgroup$ Apr 28, 2023 at 8:09
  • 2
    $\begingroup$ Wow, that's obvious in retrospect -- thanks! I wonder now about a "geometric interpretation"... $Spec(\Lambda)$ is somehow related to tangent spaces... $\endgroup$
    – Tim Campion
    Apr 28, 2023 at 11:22
  • 2
    $\begingroup$ It looks like I understood more about this a year ago than I do now! There was a discussion on the algebraic topology discord about this. See here for an invite link. $\endgroup$
    – Tim Campion
    Apr 28, 2023 at 11:32
5
$\begingroup$

$\newcommand{\AA}{\mathbf{A}}\newcommand{\GG}{\mathbf{G}}\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\fil}{\mathrm{fil}}\newcommand{\QCoh}{\mathrm{QCoh}} \newcommand{\spec}{\mathrm{Spec}}$Let $R$ be a commutative ring, and consider $\hat{\GG}_a/\GG_m$ over $R$, where the $\GG_m$-action on $\GG_a = \spec(R[t])$ has weight $1$. Then the ($\infty$-)category of complete filtered $R$-modules is equivalent to $\QCoh(\hat{\GG}_a/\GG_m)$, i.e., the category of quasicoherent sheaves over $\GG_a/\GG_m$ whose pullback to $\GG_a$ is $t$-complete. Let $\GG_a^\sharp$ denote the PD-hull $\spec(R[\frac{x^n}{n!}])$. Then the pairing $\GG_a^\sharp \times \hat{\GG}_a \to \GG_m$ sending $(x, t) \mapsto e^{xt}$ exhibits $\GG_a^\sharp$ as the Cartier dual of $\hat{\GG}_a$. This pairing is $\GG_m$-equivariant as long as the coordinate $x$ on $\GG_a^\sharp$ has weight $-1$. General principles of Cartier duality now imply that $\QCoh(\hat{\GG}_a/\GG_m)$ is equivalent to $\QCoh((B\GG_a^\sharp)/\GG_m)$, where this equivalence swaps the usual tensor product with convolution. (This is from my comment.) Now, $B\GG_a^\sharp$ is affine (as a derived stack), and $R\Gamma(B\GG_a^\sharp; \mathcal{O}) \cong R[d]/d^2$ where $d$ is in degree $-1$. So $(B\GG_a^\sharp)/\GG_m = \mathrm{Spec}(R[d]/d^2)/\GG_m$, and one recovers Achim Krause's answer.

As for the relation to tangent spaces: indeed, the appearance of $B\GG_a^\sharp$ in both places is not a coincidence. Namely, if $X$ is a (smooth, say) $R$-scheme, the mapping stack $\mathrm{Map}(B\GG_a^\sharp, X)$ can be identified with $\spec_{\mathcal{O}_X}\left(\bigoplus_{j\geq 0} \Omega^j_{X/R}[j]\right) = BT_X^\sharp$, where $T_X^\sharp$ is the PD-hull of the zero section in the tangent bundle of $X$. See, e.g., Section 2.5 of https://www.math.ias.edu/~bhatt/teaching/mat549f22/lectures.pdf . This mapping stack has a natural $B(\GG_a^\sharp\rtimes \GG_m)$-action, placing $\Omega^j_{X/R}[j]$ in weight $j$; writing $B\GG_a^\sharp = \mathrm{Spec}(R[d]/d^2)$ with $d$ in weight $-1$ and homological degree $-1$, the resulting coaction of $R[d]/d^2$ corresponds to the de Rham differential. Under these Koszul/Cartier duality identifications, this object corresponds to the Hodge filtration $\Omega^{\geq \star}_{X/R}$ on the algebraic de Rham complex. (I just want to mention that this perspective is useful in that it can be generalized to other situations, where one replaces $\hat{\GG}_a$ by a general $1$-dimensional formal group over $R$. The resulting category of filtered spectra with its symmetric monoidal structure encodes the Leibniz rule on the $q$-de Rham complex in the case of the rescaled multiplicative group law $x + y + (q-1) xy$, and more generally complexes studied in my joint work https://arxiv.org/abs/2304.04739 .)

$\endgroup$
4
  • $\begingroup$ Thanks, that's very interesting! I should learn more about Cartier duality. One idea I have in mind is that the functor $R \mapsto D(R)$, might be the "height zero case" of some more general family of functors $CRing \to Pr^L_{st}$. It sounds like this thought might be related to the paper you link? Correspondingly, the functor $R \mapsto HR$ might be the "height zero case" of some more general family of functors $CRing \to CAlg(Spt)$... $\endgroup$
    – Tim Campion
    Apr 28, 2023 at 14:10
  • $\begingroup$ I guess what I'm calling Cartier duality between G_a-hat and BG_a^# is just Koszul duality between R[[t]] and R[d]/d^2, so it might not really be necessary to think about it in terms of Cartier duality. I'm not sure what the family of functors CRing --> CAlg(Sp) you have in mind is supposed to be, can you elaborate? $\endgroup$
    – skd
    Apr 28, 2023 at 14:36
  • $\begingroup$ Start with $R \mapsto QCoh(Spec R \times S)$ for one of these variant stacks $S$ coming from FGLs like you say. Then take "the stable module category / singularity category construction" (or some version thereof)-- localize the 1-category $QCoh(Spec R \times S)$ to kill the dualizable objects or something like that to obtain a stable symmetric monoidal $\infty$-category. (Does this track so far with what you have in mind?) I'm then wildly speculating that the resulting sym mon $\infty$-category might be of the form $Mod_E$ for some $E \in CAlg(Spt)$ and maybe $E$ is even functorial in $R$. $\endgroup$
    – Tim Campion
    Apr 28, 2023 at 15:52
  • $\begingroup$ I suppose I don't want to kill all dualizable objects -- just the projectives, maybe. $\endgroup$
    – Tim Campion
    Apr 29, 2023 at 18:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.