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I'm trying to prove the following statement in $\mathbb{R}P^2$: If I have $4$ distinct points $P_1,P_2,P_3,P_4$ on the unit circle, there is always a projective transformation that maps the unit circle onto the unit circle and the four points to four points of the form $(x,y), (-x,y), (x,-y), (-x,-y)$.

I have proved that by choosing $x$ and $y$ we can get any value in $\mathbb{R}\setminus\{0,1\}$ as the cross-ratio of those four points (and for $0$, $1$ and $\infty$ we would have some points coinciding). So now I just need to be able to argue that we can always find a projective transformation that maps the unit circle to itself and any $3$ points to any other $3$ of our choice. Then, by choosing the target points such that they have the same cross-ratio as our original $4$ points and using the fact that the cross-ratio is preserved under projective transformations, the fourth point will automatically be mapped to our fourth desired target.

I also know that in $\mathbb{R}P^1$ we can map any $3$ points to any other $3$ points and the unit circle is homeomorphic to $\mathbb{R}P^1$ (through stereographic projection). But how do I know such a projective transformation (that maps the unit circle to itself and lets me freely choose where 3 points on the unit circle are mapped to) exists?

Alex Ravsky
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yagod
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2 Answers2

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I'd recommend a look at Richter-Gebert's book Perspectives on Projective Geometry. Section 10.4 studies projective transformations that leave a given fixed conic $C$ invariant, and also concludes (as you noted) that the group of all projective transformations that leaves a nondegenerate conic invariant is isomorphic to the group of projective transformations of $\mathbb{RP}^1$. Ideally you can find a copy of the book, but I'll summarize the important points you need. The main theorem is as follows:

Theorem 10.4. Let $a, b, c, d$, and $a′, b′, c′, d′$ be two quadruples of distinct points on a nondegenerate conic $C$ such that $(a,b; c,d)_C = (a′,b′; c′,d′)_C$ . Then there exists a unique projective transformation $\tau: \mathbb{RP}^2 \to \mathbb{RP}^2$ with $\tau(a) = a′,\tau(b) = b′, \tau(c) = c′, \tau(d) = d′$ that furthermore leaves $C$ invariant.

In your case $C$ is the unit circle, and you want points $a, b, c$ to map to $a′, b′, c′$. To apply the theorem let $d$ be the harmonic conjugate of $c$ w.r.t. $a,b$. (to construct $d$, let $e$ be the intersection of the tangents at $a,b$, then $d$ is the intersection of $ce$ with $C$.). Similarly let $d'$ be the harmonic conjugate of $c'$ w.r.t. $a',b'$. This way the two quadruples have the same cross ratio $(a,b; c,d)_C = (a′,b′; c′,d′)_C=-1$ so you can use Theorem 10.4.

In case you can't find the book, the proof of Theorem 10.4 is as follows:

The transformation $\tau$ is uniquely determined by the preimage points $a,b,c,d$ and the image points $a′,b′,c′,d′$. Thus we have only to show that $\tau$ indeed leaves $C$ invariant. Since a nondegenerate conic is uniquely determined by five points on it, it suffices to prove that there exists one more point $p$ on $C$ whose image $\tau(p)$ is also on $C$. For this, let $p$ be an arbitrary point distinct from the points $a,b,c,d$. Thus we have

\begin{align} (a′,b′; c′,d′)_C &= (a,b; c,d)_C \\ &= (a,b; c,d)_p \\ &= (\tau(a),\tau(b); \tau(c),\tau(d))_{\tau(p)}\;\text{(since $\tau$ is a projective transformation)}\\ & = (a′,b′; c′,d′)_{\tau(p)}. \end{align}

The fact that $(a′,b′; c′,d′)_C = (a′,b′; c′,d′)_{τ(p)}$ shows that $τ(p)$ also must lie on the conic $C$.

brainjam
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There is a very useful fact: for any point $Q$ inside the unit circle $\omega$ there exists a projective transformation mapping $\omega$ onto itself and sending $Q$ to the center. Although the are some computational proofs (e.g. this one), there exists also a purely geometric argument. Consider the affine part of $\mathbb{RP}^2$ as a plane $\pi$ in $\mathbb R^3$ and let $q$ be the polar of $Q$ w.r.t. $\omega$. Let $\alpha$ be the plane containing the line $q$ and orthogonal to $\pi$. Since $q$ does not intersect $\omega$, there exists a sphere $S$ such that $S\cap\pi=\omega$ and $S$ is tangent to the plane $\pi$ at some point $N$. Finally, let $\pi'$ be any plane parallel to $\alpha$. Then the central projection $f\colon\mathbb R^3\to\pi'$ from $N$ restricts to $S$ as a stereographic projection and therefore maps $\omega$ to a circle $\omega'\subset\pi'$. Moreover, $f$ sends $q$ to the line at infinity and hence maps the point $Q$ to the center of $\omega'$. Since the restriction $f|_{\pi}$ is a projective transformation, the result follows.

Returning to your question, if $Q=(P_1 P_3)\cap (P_2 P_4)$, then the above mapping sends $P_i$ to the vertices of some rectangle. Also, for any three distinct points $P_1,P_2,P_3\in\omega$ you can take $P_4\in\omega$ to be harmonic conjugate to $P_2$ w.r.t. $P_1$ and $P_3$ and again put $Q=(P_1 P_3)\cap (P_2 P_4)$. Then the above mapping sends $P_i$ to the vertices of a square. After composing with an isometry this gives you a projective transformation of $\mathbb{RP}^2$ mapping three given points of the unit circle to some three fixed points. Using this argument twice provides with a projective transformation mapping three points on the unit circle to the given ones.

Gorg
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