2
$\begingroup$

Let $k$ be a field of characteristic $0$, and let $\mathcal{C}= \mathbf{Vect}_k^{\leq 0}$ be the $\infty$-category of vector spaces concentrated in degrees $\leq 0$. Consider the category $\mathbf{Pr}(\mathcal{C}):= \operatorname{Fun}(\mathbf{CAlg}(\mathcal{C}), \mathcal{S})$ of prestacks over $k$, where $\mathcal{S}$ is the $\infty$-category of spaces or $\infty$-groupoids.

Suppose we have a grouplike prestack $G \in \mathbf{Pr}(\mathcal{C})$. That is, a functor $G: \mathbf{CAlg}(\mathcal{C}) \to \mathbf{Sp}^{\text{cn}}$, where $\mathbf{Sp}^{\text{cn}}$ is the $\infty$-category of connective spectra, thought of as a functor to spaces by composing with the forgetful functor $\mathbf{Sp}^{\text{cn}} \to \mathcal{S}$. We can then form the iterated classifying spaces $B^nG$.

Suppose we have a nice enough stack $X \in \mathbf{Pr}(\mathcal{C})$ (e.g. a perfect stack). When will the category $\mathbf{QCoh}(\text{Map}(X,B^nG))$ of quasicoherent sheaves on the mapping stack be compactly generated? Is the assumption that $X$ be perfect enough? Do we have to make any assumptions on $G$?

$\endgroup$
  • 3
    $\begingroup$ I don't see any reason for it to be compactly generated, even in a case as simple as $X$ a smooth projective curve, $n=1$, and $G=SL_2.$ There $\operatorname{Map}(X,BG)\equiv\operatorname{Bun}_G(X)$ is not quasi-compact, which you can maybe leverage to show that its category of quasicoherent sheaves is not compactly generated, maybe using techniques along the lines of section $12$ of arxiv.org/abs/1112.2402. $\endgroup$ – dhy Nov 28 '18 at 6:40

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.