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Jun 14, 2023 at 14:00 history edited Kenta Suzuki CC BY-SA 4.0
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Jun 14, 2023 at 8:37 comment added Dror Speiser I'm sure you know this, the third example isn't that much different, as changing the variety to GLn/B and the multiplicative group to $\mathbb{F}^\times _q {}^n$ gives a case of the first example
Jun 14, 2023 at 2:43 history edited Kenta Suzuki CC BY-SA 4.0
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Jun 14, 2023 at 2:42 comment added Kenta Suzuki Sorry, there was a typo: $G\times H$ acts on $G$ by $(g,h)\cdot x:=gxh^{-1}$. This naturally extends to a $G\times H$-action on $\mathbb C[G]$.
Jun 14, 2023 at 2:12 comment added semisimpleton Sorry, in the first example, could you explicitly describe the action by $G\times H$ on $\mathbb{C}[G]$? Did you perhaps mean to say: for $x\in\mathbb{C}[G]$, the action is given by $x\cdot (g,h) = (x\cdot g)\cdot h$? But this doesn't define an action...
Jun 14, 2023 at 0:30 history answered Kenta Suzuki CC BY-SA 4.0