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Sep 9, 2021 at 12:51 vote accept Martin Skilleter
Sep 9, 2021 at 12:49 comment added Johannes Hahn The same argument put more succinctly: The generating reflections are diagonalisable over $\mathbb{R}$ (because they're reflections) and commute by assumption. Therefore they are simultaneously diagonalisable, i.e. they are conjugated to a subgroup of $\{\pm 1\}^n$ which is finite. EDIT: Martin was a few seconds faster.
Sep 9, 2021 at 12:48 comment added Martin Skilleter Is the fact that $V$ is $G$-invariant simply because reflections are diagonalisable, and commuting diagonalisable operators can be simultaneously diagonalised? I can't see anywhere else you are using the abelian hypothesis.
Sep 9, 2021 at 12:30 history answered Geoff Robinson CC BY-SA 4.0