Timeline for Equivariant sheafs and $G$ actions on modules
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 8, 2019 at 13:19 | comment | added | Louis Jaburi | I think I am starting to understand! So I assume instead of $\mathbb{C}$ we could take any algebraically closed field $k$. Then I get an equivalence {Comodule map $X\to \mathcal{O}(G) \otimes X$} <-> {an action $\mathcal{O}(G) (k) \times X\to X$ such that the above condition is fulfilled}. Is that right? | |
| Jun 8, 2019 at 12:40 | comment | added | Sam Gunningham | ...The fact that this action is algebraic means that for each $v\in X$ there is a formula of the form $g \cdot v = \sum f_i(g) v_i$ for some $f_i \in \mathcal O(G)$. But this is just the expression for the coaction $X \to \mathcal O(G) \otimes X$. Namely, $v \mapsto \sum f_i \otimes v_i$. | |
| Jun 8, 2019 at 12:38 | comment | added | Sam Gunningham | More generally, suppose you have an algebraic representation $X$ of an affine algebraic group $G$. As you say, $X$ can be formalized of as a comodule for $\mathcal O(G)$. But sometimes it is convenient to express this data in terms of an action map $G\times X \to X$, $(g,x) \mapsto g\cdot x$ (I guess technically I have in mind that $g$ denotes a $\mathbb C$-point of $G$)... | |
| Jun 8, 2019 at 12:16 | comment | added | Louis Jaburi | I think the author is giving me an automorphism $\lambda^*:\mathbb{C}[z,z^{-1}]\otimes V\to \mathbb{C}[z,z^-1]\otimes V$ for each $\lambda\in \mathbb{C}^*$ (which he denotes by $\mathbb{G}_m$, alright). And intuition wise it is somehow natural that this corresponds to the grading you gave. Nevertheless I can't formalize it... | |
| Jun 8, 2019 at 12:00 | comment | added | Sam Gunningham | I think it is a fairly standard abuse of notation to equate a scheme over $\mathbb C$ (e.g. $\mathbb G_m$) with its $\mathbb C$-points ($\mathbb G_m(\mathbb C) = \mathbb C^\times$). In my opinion it is quite natural to express the data of a $\mathbb G_m$ representation in the way the author does above. Such an expression can easily be translated in to your preferred form: the $i$th-graded piece of the module $\mathbb C[z,z^{-1}] \otimes V$ above consists of expressions of the form $z^{-i} \otimes v$. Perhaps this was not the cause of confusion here though? | |
| Jun 7, 2019 at 20:00 | review | First posts | |||
| Jun 7, 2019 at 21:17 | |||||
| Jun 7, 2019 at 19:55 | history | asked | Louis Jaburi | CC BY-SA 4.0 |