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There is a natural fraction linear transform of $SL(2,\mathbb{R})$ on $\mathbb{C}P^1$ given by: $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot[z,w]=[az+bw,cz+dw]. $$

Let $\mathbb{Z}/2=\{ 1,s \}$ be the group with 2 elements.

My question is: is there an action of $\mathbb{Z}/2$ on $\mathbb{C}P^1$ such that it is compatible with the $SL(2,\mathbb{R})$ action and the action of the nontrivial element $s$ has no fixed point.

By compatible with the $SL(2,\mathbb{R})$ action I mean $\forall g\in SL(2,\mathbb{R}), x\in \mathbb{C}P^1$, we have $$ g\cdot(s\cdot x)=s\cdot(g\cdot x). $$

The action of $s$ need to be a diffeomorphism of $\mathbb{C}P^1=S^2$ but not required to be holomorphic.

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Such $s$ does not exist since it would have to fix fixed points of all parabolic elements of $SL(2,R)$ and, hence, a circle. Note that your compatibility notion is usually called commutation.

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  • $\begingroup$ I get your point. Yes it is impossible to require that there is no fixed point. $\endgroup$
    – Zhaoting Wei
    Commented Jul 26, 2012 at 18:43
  • $\begingroup$ @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. $\endgroup$
    – Vitali Kapovitch
    Commented Jul 27, 2012 at 0:37
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    $\begingroup$ 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). $\endgroup$
    – YCor
    Commented Jul 27, 2012 at 3:42

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