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On $\mathbb{CP}^1$ with standard complex structure, how to prove that there are only two types of antiholomorphic involution, given by $$ \tau :[z:w]\mapsto [\bar w, \bar z] \qquad \eta :[z:w]\mapsto [\bar w, -\bar z]\ ?$$

For $\mathbb {CP}^n$, there should be one involution for even $n$, two for odd $n$.

As suggested: in homogeneous coordinates, let us write our involution as $z \to \overline {Az}$ for some $A \in GL_{n+1}(C)$. Then we have $\overline{A} A = \lambda id$ for some real $\lambda$. By rescaling $\lambda$ and taking determinant, we see that for even $n$ only $\overline{A}A=1$ is possible (class $\tau$ above), while for odd $n$ also $\overline{A}A=-1$ is possible (so also $\eta$ above).

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    $\begingroup$ In even dimension you have only one, but in odd dimension you have two such. I suggest you to write your antiholomorphic involution as the product of the standard one with a biholomorphic map, and then to write down what is such a map. $\endgroup$ Jun 1, 2014 at 16:24
  • $\begingroup$ Cross posted on MSE. $\endgroup$
    – MvG
    Jun 2, 2014 at 6:28
  • $\begingroup$ It's a question about involutive elements in $PGL\setminus PSL$, hence just something about eigenvalues (simple linear algebra). I presume that you are speaking about the standard complex structure; it has nothing to do with the metric. $\endgroup$ Jun 2, 2014 at 6:58
  • $\begingroup$ @JérémyBlanc Do you know a reference to the result? Just to avoid repeatition of this well-known linear algebra computation in a research paper. $\endgroup$ Nov 13, 2019 at 5:37
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    $\begingroup$ @AlexDegtyarev: no, $PGL_{n+1}(C)=PSL_{n+1}(C)$. It's a question about an overgroup of index 2 of $PGL_{n+1}(C)$. $\endgroup$
    – YCor
    Nov 26, 2019 at 9:54

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Coda (Lemma 11 from arXiv:2002.01355):

Projective transformations of $\mathbb{C}P^{n}$ act on the solutions of the equation $L\bar L=I$, respectively $L\bar L=-I$, by transformations $L\mapsto U^{-1}L\bar U$, where $U$ is an invertible complex $(n+1)\times (n+1)$ matrix. It remains to show that each solution can be transformed to $I$ and $J=\left(\begin{smallmatrix} 0 & -I\\ I & 0\end{smallmatrix}\right)$ respectively, i.e., each matrix $L$ satisfying $L\bar L=I$ (respectively, $L\bar L=-I$) has form $L=U^{-1}\bar U$ (respectively, $L=U^{-1}J\bar U$ ) for some invertible matrix $U$.

If $L\bar L=I$, then take $a\in\mathbb{R}$ such that $-e^{2ia}$ is not an eigenvalue of $\bar L$. Then $U:=e^{ia}I+e^{-ia}\bar L$ is the required invertible matrix because $UL=(e^{ia}I+e^{-ia}\bar L)L=e^{ia}L+e^{-ia}I=\bar U$ by the equation $\bar LL=L\bar L=I$.

If $L\bar L=-I$, then take $a\in\mathbb{R}$ such that $e^{2ia}$ is not an eigenvalue of $J\bar L$. Then $U:=e^{ia}J+e^{-ia}\bar L$ is the required matrix: $UL=(e^{ia}J+e^{-ia}\bar L)L=e^{ia}JL+e^{-ia}J\bar J=J\bar U$ because $\bar LL=L\bar L=-I=J\bar J$.

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