Timeline for Simplicial resolution for commutative group scheme
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2023 at 0:02 | comment | added | Sam | Sorry, this is my fault. $S^n(G)$ is not an algebraic group and one has only the map of schemes $S^n(G)\to G$. Nevertheless, $S^\bullet(G)$ is a commutative monoidal scheme and the question is still valid | |
| Oct 7, 2023 at 0:03 | comment | added | Denis T | @S.Carnahan +1. For example, symmetric power of an elliptic curve cannot ever be an algebraic group by cohomological reasons; its cohomology is not an exterior algebra. | |
| Oct 6, 2023 at 23:26 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 242 characters in body
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| Oct 6, 2023 at 8:40 | comment | added | Jon Pridham | Regarding the extra degeneracy: in any adjunction of that form, it's very rare for that morphism in $\mathcal{C}$ to lift to a morphism in $\mathcal{D}$; that's effectively asking for the generators of a free algebra to be a subalgebra. | |
| Oct 6, 2023 at 0:15 | comment | added | S. Carnahan♦ | Why is $S^d(G)$ a group? For example, if $d=2$, how do you multiply the unordered pair $(x,y)$ by itself when $x \neq y$? | |
| Oct 5, 2023 at 19:25 | history | asked | Sam | CC BY-SA 4.0 |