# classify antiholomorphic involutions of projective space

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).

• 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. Jun 1, 2014 at 16:24
• Cross posted on MSE.
– MvG
Jun 2, 2014 at 6:28
• 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. Jun 2, 2014 at 6:58
• @JérémyBlanc Do you know a reference to the result? Just to avoid repeatition of this well-known linear algebra computation in a research paper. Nov 13, 2019 at 5:37
• @AlexDegtyarev: no, $PGL_{n+1}(C)=PSL_{n+1}(C)$. It's a question about an overgroup of index 2 of $PGL_{n+1}(C)$.
– YCor
Nov 26, 2019 at 9:54

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 $$\bar LL=L\bar L=-I=J\bar J$$.