Timeline for Witt vectors, the cotangent complex, and a solid construction of $B_{dR}^+$
Current License: CC BY-SA 4.0
11 events
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| Nov 3, 2021 at 18:41 | history | edited | Z. M | CC BY-SA 4.0 |
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| Nov 3, 2021 at 17:50 | comment | added | Dustin Clausen | This looks convincing. Thank you very much for these explanations! | |
| Nov 3, 2021 at 15:02 | comment | added | Z. M | @pupshaw For the acyclicity of LSym, it follows from Lazard's theorem by realizing flat modules as filtered colimits of free modules. See Lurie's Spectral Algebraic Geometry, Cor 25.2.3.3. In fact, I mistakenly referred to my thesis (the chapters of which are available at my homepage and on arXiv — click my profile if you are interested), which addresses a special version of a PD-variant of this which is sufficient to study the crystalline cohomology. Since the rest of that comment is informative, I did not remove it. | |
| Nov 3, 2021 at 14:44 | comment | added | pupshaw | it's certainly my intuition from char 0 situations where we have a less problematic Koszul duality that something like this ought to work, basically we need to know the Lie coalgebra structure on the shifted cotangent complex, and you might hope that a one dimensional such object has to be abelian. but we are on seemingly much more delicate ground with these sifted monads. I really would like to read your thesis at some point, but no hurry. | |
| Nov 3, 2021 at 14:40 | comment | added | pupshaw | I spent a bit of time with Raksit's article but evidently not enough, this will take me some time to digest but comes truly appreciated. Out of curiosity, is ordinary flatness a sufficient guarantee of acyclicity for the nonabelian derived functors LSym, or their funny Kan-extended versions? | |
| Nov 3, 2021 at 13:29 | history | edited | Z. M | CC BY-SA 4.0 |
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| Nov 3, 2021 at 13:25 | comment | added | Z. M | @Anonymous Thanks, that was a typo. | |
| Nov 3, 2021 at 13:24 | comment | added | Anonymous | Ah, thanks. But it seems you wrote (in the answer) that $gr^1$ is the suspension of the cotangent complex instead of the other way around ? | |
| Nov 3, 2021 at 13:18 | comment | added | Z. M | @Anonymous $\operatorname{gr}^n$ is the symmetric product of $\operatorname{gr}^1$, not the cotangent complex (by the way, a typo in your formula, the divided power of the cotangent complex should be shifted by $n$, not $1$), and the cotangent complex is the suspension of $\operatorname{gr}^1$. This is precisely the difference between $h_+$ and $h_-$ differential graded algebras — the $h_+$ ones have a PD-structure (roughly speaking) while the $h_-$-ones do not. | |
| Nov 3, 2021 at 12:20 | comment | added | Anonymous | I'm probably missing something, but paragraphs 4 and 5 seem incompatible outside characteristic $0$ to me. For instance, if $A = R/I$ with $I$ regular (say), then $L_{A/R} = I/I^2[1]$, so $Sym^n(L_{A/R}[1]) = \Gamma^n(I/I^2)[1]$, which suggests that the left adjoint produces the completed PD-envelope rather than the completion. (This affects the perfect ring assertion, but is of course not an issue in characteristic 0 where Dustin wanted to apply it.) | |
| Nov 3, 2021 at 12:11 | history | answered | Z. M | CC BY-SA 4.0 |