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Mar 24, 2014 at 22:37 comment added Nilay Kumar Thanks a bunch for the references - glad to see there's a nice way to do this in general.
Mar 24, 2014 at 22:35 vote accept Nilay Kumar
Mar 24, 2014 at 13:56 comment added Faisal Yes, $O(n)$ here is really $O(\mathbb C^n, B)$, as stated in the OP.
Mar 24, 2014 at 7:51 comment added Ben McKay @AlexDegtyarev: I think that the question is being asked in the algebraic category, so that $G$ is a complex algebraic group. The notation $O(n)$ is unclear, but means $O(n,\mathbb{C})$.
Mar 24, 2014 at 6:15 comment added Alex Degtyarev There seems to be a misunderstanding here. Do you actually mean $O(n)$? How does it act? Should it not be $U(n)$ instead?
Mar 24, 2014 at 4:09 comment added Faisal @Mariano: No, it's not: if $g$ is in $O(2)$ then $g \cdot i = (\det g) i = \pm i$.
Mar 24, 2014 at 4:03 comment added Mariano Suárez-Álvarez In the case of P1 the action is transitive, no?
Mar 24, 2014 at 3:59 history edited Faisal CC BY-SA 3.0
deleted 24 characters in body
Mar 24, 2014 at 3:54 history answered Faisal CC BY-SA 3.0