Timeline for Are algebras of smooth functions formally smooth?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 22, 2022 at 9:53 | vote | accept | Tobias Fritz | ||
| Nov 22, 2022 at 9:27 | answer | added | Achim Krause | timeline score: 10 | |
| Nov 21, 2022 at 12:30 | comment | added | Tobias Fritz | @MartinBrandenburg: perhaps you can spell out the details in an answer? That would be helpful insofar as it would indicate that the question should be restricted to the compact case. | |
| Nov 21, 2022 at 12:02 | comment | added | Martin Brandenburg | I believe that $k^I$ is not formally smooth over $k$ when $k$ is a field and $I$ is an infinite set. | |
| Nov 21, 2022 at 10:16 | comment | added | Tobias Fritz | Am I understanding correctly that these comments are suggesting to employ Quillen's criterion that a ring map $A \to B$ is formally smooth iff $\Omega_{B/A}$ is projective and the cotangent complex is homotopy equivalent to it in degree zero? According to this other question, this also holds without finite presentation, right? | |
| Nov 21, 2022 at 9:05 | comment | added | Z. M | Is it formally smooth when $M=\mathbb N$, namely, what about the cotangent complex of ${\mathbb R}^{\mathbb N}$ over $\mathbb R$? | |
| Nov 20, 2022 at 22:17 | comment | added | Achim Krause | (one would hope before basechange it is given by sections of $T^*M$, but I'm currently not sure whether we can detect this at the residue fields) | |
| Nov 20, 2022 at 22:15 | comment | added | Achim Krause | This is only a partial answer, maybe someone sees how to finish: for a point $x\in M$, we have the evaluation map $C^\infty(M)\to \mathbb{R}$. Its cotangent complex is really just $T^*_xM[1]$. To see this, we may localise to assume $M=\mathbb{R}^n$ with $x=0$, and then the kernel of the evaluation map is generated by the regular sequence of coordinates $x_1,..,x_n$. So from the conormal sequence we at least get that the base-changes of the cotangent complex $L_{C^\infty(M)/\mathbb{R}} $ along the evaluation maps are projective, given by $T_x^*M$. | |
| Nov 20, 2022 at 21:14 | history | edited | YCor |
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| Nov 20, 2022 at 20:34 | history | asked | Tobias Fritz | CC BY-SA 4.0 |