Timeline for Distinguished triangle of dualizing complexes and/or determinants?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 14, 2022 at 11:01 | comment | added | Jason Starr | @MarcHoyois I thought your arrow was in the category of schemes (thus considering $k[\epsilon]$ as a $k$-algebra). Now I see that your morphism is in the category of rings. Sorry for the confusion. | |
| Apr 14, 2022 at 6:42 | vote | accept | Leo Herr | ||
| Apr 14, 2022 at 5:41 | comment | added | Marc Hoyois | @JasonStarr Yes, but my morphism is a retraction of yours (geometrically, a section), so by the distinguished triangle for cotangent complexes we have $\mathbb L_{k/k[\epsilon]}=\mathbb L_{k[\epsilon]/k}[1]$, which lies in cohomological degrees [-2,-1]. | |
| Apr 14, 2022 at 0:09 | comment | added | Jason Starr | @MarcHoyois. Isn't the $k$-algebra $k[x]/\langle x^2\rangle$ an LCI $k$-algebra? Maybe we have different definitions. I mean a morphism that locally factors as a smooth morphism precomposed with a "regular embedding". | |
| Apr 13, 2022 at 21:11 | comment | added | Leo Herr | Thanks, fellow Leo. I didn't realize this followed from functoriality. I knew there were "quasi-lci" morphisms with L truncated in degrees [-2, 0] but couldn't reconcile this with Avramov's result; thanks Marc. I don't know if one can make sense of det L = \omega generally; it was just for my intuition. | |
| Apr 13, 2022 at 19:19 | comment | added | Leo Alonso | I think there is a confusion between determinant in the categorical sense, as in Marc's quotation, and in a more informal sense, as the higher exterior power, in which case it wouldn't make sense for the unbounded case. I was referring to the latter. I added this precision to the answer. | |
| Apr 13, 2022 at 19:18 | history | edited | Leo Alonso | CC BY-SA 4.0 |
Add further explanation on how to derived the formula of OP from the facts presented
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| Apr 13, 2022 at 17:50 | comment | added | Marc Hoyois | Avramov's result is only for morphisms of finite Tor-dimension, e.g., $k[x]/(x^2)\to k$ is not lci but has a perfect cotangent complex (in this example the determinant does give the dualizing sheaf, but I don't know any general result like this beyond the lci case). | |
| Apr 13, 2022 at 10:34 | comment | added | Jason Starr | The formula for the dualizing sheaf as a determinant of the cotangent complex can only hold under the LCI hypothesis. Avramov proved that for non-LCI morphisms, the cotangent complex is unbounded, so there is no "determinant" (at least I have never heard of a theory of determinants for unbounded complexes). In the LCI case, it should be possible to get the formula even from the quite short exposition of dualizing complexes in the book by Koll\'ar -- Mori (but I am sure that it is also in Hartshorne's "Residues and Duality"). | |
| Apr 13, 2022 at 10:23 | history | answered | Leo Alonso | CC BY-SA 4.0 |