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Apr 28, 2023 at 13:06 comment added Z. M @MaximeRamzi Another reason for it not being Dold–Kan is that, the base category should be stable for this equivalence, while for Dold–Kan, the (weakly idempotent-complete) additivity suffices.
Apr 28, 2023 at 11:32 comment added Tim Campion 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.
Apr 28, 2023 at 11:22 comment added Tim Campion Wow, that's obvious in retrospect -- thanks! I wonder now about a "geometric interpretation"... $Spec(\Lambda)$ is somehow related to tangent spaces...
Apr 28, 2023 at 11:13 vote accept Tim Campion
Apr 28, 2023 at 8:09 comment added Maxime Ramzi 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 !
Apr 28, 2023 at 8:07 comment added Achim Krause 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)
Apr 28, 2023 at 8:03 comment added Maxime Ramzi 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 $-1$)
Apr 28, 2023 at 6:33 history answered Achim Krause CC BY-SA 4.0