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In the book by Beltrametti, he states that if $\omega$ is a projectivity of a line $r$, it has two fixed points, $u$ and $v$. If $u$ and $v$ are distinct, then the cross ratio $(u,v,\lambda, \omega(\lambda))$ does not depend on $\lambda$.

I tried to prove it.

$$(u,v,\lambda, \omega(\lambda))=\frac{(\lambda-u)(\omega(\lambda)-v)}{(\lambda-v)(\omega(\lambda)-u)}=\frac{\lambda \omega(\lambda) -\lambda v -u \omega(\lambda)+uv}{\lambda\omega(\lambda)-\lambda u-v\omega(\lambda)+uv}=\frac{\lambda \frac{a_{11}\lambda+a_{10}}{a_{01}\lambda+a_{00}} -\lambda v -u \frac{a_{11}\lambda+a_{10}}{a_{01}\lambda+a_{00}}+uv}{\lambda\frac{a_{11}\lambda+a_{10}}{a_{01}\lambda+a_{00}}-\lambda u-v\frac{a_{11}\lambda+a_{10}}{a_{01}\lambda+a_{00}}+uv}$$

Then I simplified it to

$$\frac{\lambda^2(a_{11}-a_{01}v)+\lambda (a_{10}-a_{00}v-a_{11}u+uva_{01})-a_{10}u+uv a_{00}}{\lambda^2(a_{11}-a_{01}u)+\lambda (a_{10}-a_{00}u-a_{11}v+uva_{01})-a_{10}v+uv a_{00}}$$

I was trying to show that the coefficients of $\lambda$ and $\lambda^2$ are zeros using the fact that $u=\frac{a_{11}u+a_{10}}{a_{01}u+a_{00}}$ and $v=\frac{a_{11}v+a_{10}}{a_{01}v+a_{00}}$. But it seems they are not. So how can I show that it does not depend on $\lambda$? Thank you for the help!

KittyL
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1 Answers1

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I got my answer here: http://en.wikipedia.org/wiki/M%C3%B6bius_transformation

I'd like to post it to share.

Consider a projectivity $g$ that maps $u,v$ to $0,\infty$: $$g(\lambda)=\frac{\lambda-u}{\lambda-v}$$ Then the projectivity $g\circ\omega\circ g^{-1}$ maps $0,\infty$ to $0,\infty$. So it is a dilation\rotaion: $$g\circ\omega\circ g^{-1}(\lambda)=k(\lambda)$$ We can then write $$\frac{\omega(\lambda)-u}{\omega(\lambda)-v}=k \frac{\lambda-u}{\lambda-v}$$ i.e., $$\frac{(\lambda-u)(\omega(\lambda)-v)}{(\lambda-v)(\omega(\lambda)-u)}=k.$$

KittyL
  • 16,965