Timeline for Commutative monoid gradings via group scheme actions
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2021 at 21:08 | comment | added | Emily | @skd Thanks, Sanath! This is exactly what I was looking for! :) | |
| Sep 4, 2021 at 13:11 | comment | added | Benjamin Steinberg | I would add it is a cocommutative bialgebra and not just a coalgebra | |
| Sep 4, 2021 at 12:47 | comment | added | skd | Yes, sorry - typo | |
| Sep 4, 2021 at 12:13 | comment | added | Benjamin Steinberg | Isn’t $Spec(Z[\Lambda])$ a monoid scheme | |
| Sep 4, 2021 at 7:45 | comment | added | Martin Brandenburg | I think that the second comment by skd is a full answer. (At least, this is what I would have answered :).) A reference is the book by Demazure-Gabriel. | |
| Sep 4, 2021 at 6:51 | comment | added | YCor | When the commutative semigroup $A$ has cancellation, it embeds into its group $B$ of fractions and an $A$-grading is a $B$-grading with a certain condition, which can be more or less described in terms of action. (The case of the semigroup $(\mathbf{N}_{>0},+)$, which is not a monoid, is quite important). I would have less intuition when, for instance, $A$ is non-cancelative, e.g. $A=(\{0,1\},\max)$. By the way $(\mathbf{N}_{>0},\times)\simeq \mathbf{N}^{(\mathbf{N})}$ is just a free commutative monoid on countably many generators. | |
| Sep 4, 2021 at 3:59 | comment | added | skd | Maybe to clarify: the general picture is that if $\Lambda$ is a commutative monoid in sets, then $\mathbf{Z}[\Lambda]$ is a commutative $\mathbf{Z}$-algebra, and the diagonal on $\Lambda$ makes $\mathbf{Z}[\Lambda]$ into a cocommutative coalgebra. So $\mathrm{Spec} \ \mathbf{Z}[\Lambda]$ is a commutative group scheme, and actions of this thing correspond to $\Lambda$-gradings (which you can think of as functors $\Lambda \to \mathrm{Mod}_\mathbf{Z}$, with $\Lambda$ viewed as a discrete category). This is a categorical version of Cartier duality. | |
| Sep 4, 2021 at 3:40 | comment | added | skd | Asking that a $\mathbf{Z}$-grading be a $\mathbf{N}$-grading just amounts to saying that the $\mathbf{G}_m = \mathrm{Spec} \ \mathbf{Z}[t^{\pm 1}]$-weights are nonnegative. One way you can say this is to consider $\mathbf{A}^1 = \mathrm{Spec} \ \mathbf{Z}[t]$ as a monoid with respect to multiplication. Then $\mathbf{A}^1$-actions correspond to nonnegative gradings. (If you consider $\mathrm{Spec}\ \mathbf{Z}[t^{-1}]$ instead, you'd get nonpositive gradings) | |
| Sep 4, 2021 at 3:30 | history | edited | Emily | CC BY-SA 4.0 |
added 85 characters in body
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| Sep 4, 2021 at 3:24 | history | asked | Emily | CC BY-SA 4.0 |