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Feb 27, 2013 at 20:57 answer added Renato G. Bettiol timeline score: 5
Feb 26, 2013 at 16:21 comment added Venkataramana I think that the OP meant infinitely many orbits, not that the orbits were infinite sets(which, as you say, is obvious).
Feb 26, 2013 at 16:18 comment added YCor A subgroup of $PGL_n$ with only finite orbits on $P^{n-1}$ is finite. This is because $P^{n-1}$ has a finite subset $F$ such that every element of $PGL_n$ acting on $F$ as the identity is the identity everywhere (take $F$ to $\{e_1,e_2,\dots,e_n,e_1+e_2+\dots+e_n\}$ for some basis $(e_i)$).
Feb 26, 2013 at 15:53 answer added Venkataramana timeline score: 5
Feb 26, 2013 at 15:07 history asked Huangjun Zhu CC BY-SA 3.0