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Mar 14, 2013 at 22:02 comment added Geoff Robinson @Agol: A valid point; there are many ways to get there.
Mar 14, 2013 at 21:41 comment added Ian Agol Maybe one ought to remark here that any two non-commuting unipotent elements are conjugate to a pair of this type. A unipotent element fixes a unique point of $F\mathbb{P}^1$. The action of $SL(2,F)$ on $F\mathbb{P}^1$ is transitive on pairs of points in $F\mathbb{P}^1$, so one may assume the fixed point of one unipotent is $\infty$, and the other is $0$, giving the desired normalization (but where the lower left entry of the first matrix might not be $1$).
Mar 14, 2013 at 20:51 vote accept Jean Raimbault
Mar 14, 2013 at 20:46 history answered Geoff Robinson CC BY-SA 3.0