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S May 4, 2023 at 8:24 history bounty ended Gabriel
S May 4, 2023 at 8:24 history notice removed Gabriel
May 4, 2023 at 8:24 vote accept Gabriel
May 2, 2023 at 12:17 answer added Peter Scholze timeline score: 18
May 2, 2023 at 9:11 comment added Pavel Safronov Just like the Betti stack, you can extend it to be constant in the "algebraic" direction to obtain a condensed stack. For constructible sheaves: you have to use the exit-path category in the Betti setting or the Galois category in the $\ell$-adic setting (see Chapter 13 of arxiv.org/abs/1807.03281). Note that these are not (condensed) $\infty$-groupoids, but rather (condensed) $\infty$-categories.
May 2, 2023 at 8:30 comment added Gabriel Dear @PavelSafronov, do you know if it's possible to see this condensed shape as some sort of stack and/or if this also works for constructible sheaves?
May 2, 2023 at 7:58 comment added Pavel Safronov For $\ell$-adic cohomology there is the condensed/pyknotic shape which is a condensed space whose category of local systems gives the category of lisse $\ell$-adic sheaves, see e.g. Appendix A in arxiv.org/abs/2012.02853.
S May 2, 2023 at 7:35 history bounty started Gabriel
S May 2, 2023 at 7:35 history notice added Gabriel Draw attention
Apr 27, 2023 at 18:03 comment added Gabriel Dear @PiotrAchinger, if I understand correctly, given a scheme $X$ of finite-type over $\mathbb{C}$, its "Betti" space $X_B$ is just the constant sheaf (of anima) equal to the topological space $X_\text{an}$. (There's more about this in [Sim] and in the end of the first chapter in Scholze's notes about six-functor formalisms.) Do you have any idea for an étale counterpart?
Apr 27, 2023 at 17:27 comment added Piotr Achinger I don’t understand the definition of $K_B$ in Porta–Sala but at first glance it looks like something which should have an etale homotopy counterpart.
Apr 27, 2023 at 15:54 comment added Gabriel Since this post is already too big, let me explain the term ring stack in the title for those curious: in many cases it was remarked that it suffices to construct the stack $X_\text{stk}$ for $X=\mathbb{A}^1$. The resulting object is a "ring stack" and we have $X_\text{stk}=X\circ \mathbb{A}^1_\text{stk}$ as a functor.
Apr 27, 2023 at 15:52 history asked Gabriel CC BY-SA 4.0