MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let us consider the group $PGL(2,\mathbb{R})$ as the group of automorphisms of real projective line and $H\subset PGL(2,\mathbb{R})$ is a subgroup of prime order $> 2$. Is it true that there always exists a fixed point of $H$ action on $P^1$?

share|cite|improve this question
I don´t think so. View $P_1$ as the boundary of the upper half-plane. Take a circle $SO_2$ in your group. It contains cyclic subgroups of any order, whose elements are all elliptic. They have a fix point in the upper half plane, but not in the boundary. – Claudio Gorodski Mar 26 '13 at 20:54
up vote 3 down vote accepted

The answer is no for every prime $p$. Set $\alpha=\pi/p$. Then the image of $\begin{pmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{pmatrix}$ in $PGL(2,\mathbb R)$ has order $p$, but no fixed points.

share|cite|improve this answer
I'm verry sorry, of course I mean subgroups of odd prime order. – Ewan Spencer Mar 26 '13 at 21:01
There is no relation between the order of the group and the fact that you have fixed points. The important thing is that the extension $\mathbb{C}:\mathbb{R}$ is of degree $2$. Hence, the two fixed points on $\mathbb{P}^1$ may be defined over $\mathbb{C}$ but not over $\mathbb{R}$. In contrast, acting on $\mathbb{P}^2$ with a cyclic group you always have a fixed point. – Jérémy Blanc Mar 26 '13 at 21:34
Jeremy, real fixed point? How to see that? – Ewan Spencer Mar 27 '13 at 5:51
On $\mathbb{P}^2_{\mathbb{C}}$, the action of a cyclic group is linearizable. Looking at the possible "eigenvalues" (up to multiple) you find either three points fixed or one line and one isolated fixed point. In both cases, the action of the anti-holomorphic involution has to fix one point, so you find a fixed real point. This does not work in dimension odd, as pointed by Peter Mueller. By the way, you should accept his answer to your question by clicking on the mark at the left. – Jérémy Blanc Mar 27 '13 at 7:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.