Timeline for Find an action of $\mathbb{Z}/2$ on $\mathbb{C}P^1$ which is compatible with the fraction linear transform of $SL(2,\mathbb{R})$

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Jul 27, 2012 at 3:42	comment	added	YCor		more generally if $G$ is a group and $H$ a subgroup, the group of permutations of $G/H$ commuting with the $G$-action of $G/H$ is naturally identified with $N(H)/H$ where $N(H)$ is the normalizer of $H$ in $G$: if $k\in N(H)$ the corresponding permutation is $gH\mapsto k^{-1}H=gHk^{-1}$. In this special case, $G=SL_2$ and $H$ is the triangular group, so $N(H)/H$ is the trivial group (so you don't even use the order 2 condition, nor the continuity).
Jul 27, 2012 at 0:37	comment	added	Vitali Kapovitch		@Zhaoting Wei even if you allow fixed points you won't get much. it's easy to see that the only order 2 diffeomorphisms of $\mathbb CP^1$ that commute with the action of $SL(2,\mathbb R)$ are the identity and the complex conjugation.
Jul 26, 2012 at 18:43	comment	added	Zhaoting Wei		I get your point. Yes it is impossible to require that there is no fixed point.
Jul 26, 2012 at 18:42	vote	accept	Zhaoting Wei
Jul 26, 2012 at 18:30	history	answered	Misha	CC BY-SA 3.0