Timeline for spectrum of an induced algebra
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 7, 2013 at 9:01 | comment | added | Jan Weidner | The affinization of a scheme $X$ is $spec(\Gamma(X,\mathcal O)$. In the situation of the question this means $G\times^B X=spec(S)$ if and only if $G\times^B X$ is affine. For example if $X=u$ then $G\times^B u$ is the nilpotent cone, which is affine hence the desired equality holds. | |
| Nov 6, 2013 at 10:48 | comment | added | NN guest | I'm sorry for not being clear in my last question. I just didn't understand what "affinization" means. I was considering the example when $X=\mathfrak{u}$, the Lie algebra of the unipotent radical subgroup of $B$, and $B$ acts on it as conjugation, denoted by a dot "$\cdot$". Then we have $G\cdot\mathfrak{u}$ is the nilpotent cone of Lie$(G)$ which is the spectrum of $(k[G]\otimes k[\mathfrak{u}])^B$. | |
| Nov 6, 2013 at 9:54 | comment | added | Jan Weidner | What do you mean by $G \cdot X$? | |
| Nov 5, 2013 at 20:37 | comment | added | NN guest | Thanks a lot Jan! Is it possible to determine $spec(S)$? In some cases, it looks like $G\cdot X$ but I'm not sure if it's true in general. | |
| Nov 5, 2013 at 20:33 | vote | accept | NN guest | ||
| Nov 5, 2013 at 17:44 | comment | added | Jan Weidner | I had to type slightly less :) | |
| Nov 5, 2013 at 17:42 | comment | added | Sam Gunningham | Ah, you got there first... | |
| Nov 5, 2013 at 17:41 | history | answered | Jan Weidner | CC BY-SA 3.0 |