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If $f$ is a homography of the real projective line, $f^2=id$ (is an involution), and $f$ has exactly two fixed points, how can I construct (geometrically) the image of an arbitrary point?

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I assume you mean knowing where the fixed points are.

Assume for simplicity that the fixed points of $f$ are $0$ and $\infty$. Then it's not too hard to see (for example by the classification of Möbius transformations) that $f$ has to be $x \mapsto -x$. You know how to geometrically construct the image of an arbitrary point by that transformation.

In the general case, you can show that $f$ is the inversion with respect to the circle that contains the two points and whose center is on the real projective line.

Edit: I'm adding some details following OP's questions in the comments below.

Here I'm using the usual coordinate $x$ on the real projective line (defined by $[x_1, x_2] \mapsto x = \frac{x_1}{x_2}$) so that $\mathbb{R}\mathbf{P}^1 \approx \mathbb{R} \cup \infty$ and a homography acts like $x \mapsto \frac{ax+b}{cx+d}$.

In particular we can see homographies of the real projective line as special examples of homographies of the complex projective line $\mathbb{C}\mathbf{P}^1 \approx \mathbb{C} \cup \infty$ (aka Möbius transformations), they are those with real coefficients, or equivalently those which preserve $\mathbb{R}\mathbf{P}^1 \subset \mathbb{C}\mathbf{P}^1$.

Circles (or should I say "circle-or-line"s) are well defined in $\mathbb{C}\mathbf{P}^1$ and there is a geometric transformation called inversion with respect to a circle (see link above). I am saying that in restriction to the real projective line, a "real" homography that fixes two points on the real projective line acts like the inversion with respect to the circle (in $\mathbb{C}\mathbf{P}^1$) whose center is on the real projective line and contains the two points.

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  • $\begingroup$ Okay I understand some of what you're saying but not completely. First what do you mean by "know where they are"? What I meant is something like: let $P,Q\in\mathbb P^1$ (the projective line) be the two fixed points. Next, is $x$ a coordinate in a coordinate system? If it is then which? If not then what is $x$? A point in the line? In that case $-x$ doesn't make sense seeing as there's no addition of points, or at least none which can be considered an operation. Lastly how is a circle defined in a 1-dim. real projective space? All honest questions, no offence meant (in case it was taken). $\endgroup$ – GPerez Jan 6 '14 at 15:08
  • $\begingroup$ No offense taken of course. I've tried to clear things up for you. Is it better now? $\endgroup$ – Seub Jan 6 '14 at 15:31
  • $\begingroup$ Hmm yes now I see completely what you meant, although I was hoping for an expression of $f$ as a composition of perspectivities (by including the line inside the projective plane). It's ok though, I should have been more specific $\endgroup$ – GPerez Jan 6 '14 at 16:31

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