For each scheme or algebraic stack their $\infty$-category of quasicoherent sheaves (resp., perfect complexes) on it is $k$-linear for a commutative ring $k$. That is (by a recent result of L.Cohn), it is a $\mathrm{DG}$-category over $k$.

But what about is the base ring of of $\mathrm{QC(X)}/\mathrm{Perf(X)}$ if $\mathrm{X}$ is a "derived" stack? That is, either

$(1)$ a derived Artin stack in the sense of "derived algebraic geometry" à la Toen and Vezzosi, based on simplicial commutative rings,

$(2)$ or a spectral Deligne-Mumford stack in the sense of "spectral algebraic geometry" à la Lurie, based on $\mathrm{E_{\infty}}$-rings.

There isn't much on the matter in existing papers and literature. The impression I got from browsing some papers on derived (resp., spectral) algebraic geometry, that Toen and Vezzosi consider $\mathrm{QC(X)}/\mathrm{Perf(X)}$ for a derived Artin stack as a $\mathrm{DG}$-category (that is, a $k$-linear $\infty$-category for a commutative ring $k$) while Lurie in his book "Spectral Algebraic Geometry" seems to treat more general $R$-linear $\infty$-categories for $R$ being an $\mathrm{E_{\infty}}$-ring (however, $\infty$-categories arising from spectral Deligne-Mumford stacks are still stable).

What I'm trying to ask is the following: Let $\mathrm{X}$ be a derived Artin stack or a spectral Deligne-Mumford stack. Then $\mathrm{QC(X)}$ and $\mathrm{Perf(X)}$ are $R$-linear (stable) $\infty$-category. What is $R$? An ordinary ring? A simplicial commutative ring? An $\mathrm{E_{\infty}}$-ring? Or what is the "base ring" of a derived stack?

Also, perhaps the better question would be the following: Given a derived (resp., spectral) stack $\mathrm{X}$, what $R$ can be "over"?

For example, derived Artin stacks are "built" from simplicial $k$-algebras for $k$ being a commutative ring. So even if there is a leap from commutative rings to simplicial commutative rings, ordinary commutative rings are still involved.

P.S. I assume that this question may seem a little naive for people who are familiar with derived algebraic geometry, but I don't know a lot on the matter, so forgive me if the question is of the low quality.

  • 1
    $\begingroup$ What is the question exactly? Whether QC(X) is linear over some E_infty-ring? A stupid answer is: of course, the sphere spectrum. $\endgroup$ – Dylan Wilson Jun 17 '17 at 18:55
  • $\begingroup$ @DylanWilson What type of rings QC(X) is linear over? Intuition may say that over E_infinity-rings, but if we X is a derived Artin stack, apparently, QC(X) is not linear over a simplicial commutative ring, but over an ordinary commutative ring. What is more, Lurie in his "Structures Spaces" define generalized schemes (a general theory that includes ordinary schemes, algebraic stacks, derived Artin stacks, derived schemes, spectral schemes, spectral Deligne-Mumford stacks et. al) over a commutative ring k. Though he doesn't develop sheaf theory for such generalized schemes. $\endgroup$ – jereckherr Jun 17 '17 at 19:56
  • 4
    $\begingroup$ If X is a derived stack over k, QC(X) is a k-linear stable infinity category. This holds whether k is E_infty or simplicial commutative or ordinary discrete commutative (in which case you may prefer to think of it as a k linear dg category). Same for Perf. geometrically this is just expressing the structure map from X to Spec k. $\endgroup$ – David Ben-Zvi Jun 17 '17 at 20:15
  • $\begingroup$ This question will likely attract more attention if it is clearer what you are really asking. "But what about is the base ring of of" comes at a key point and doesn't make grammatical sense, which isn't helping. $\endgroup$ – Hugh Thomas Jun 17 '17 at 20:17
  • 1
    $\begingroup$ @jereckherr What is $X$ a derived stack over? The answer to this question is also the answer to yours. $\endgroup$ – Denis Nardin Jun 17 '17 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.