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I am struggling hours with this problem, which probably has a fast solution, but it hasn't come up in my mind yet:

Let $P_i$ and $Q_i$, $i \in \{1,2,3\}$, be two projective frames of $\mathbb{CP^1}$, determine a projectivity $\tau: \mathbb{CP^1} \to \mathbb{CP^1}$ such that $\tau(P_i)=Q_i$.

$P_1: \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} $, $P_2: \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} $ $P_3: \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} $

$Q_1: \begin{bmatrix} 1 \\ -i \\ \end{bmatrix} $ $Q_2: \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} $, $Q_2: \begin{bmatrix} 1 \\ i \\ \end{bmatrix} $

Basically, I must solve this system:

$\tau= \begin{bmatrix} a&b \\ c&d \\ \end{bmatrix} $, so that: $\begin{cases} b=k \\ d=-ik \end{cases}$, $\begin{cases} a=s \\ c=s\end{cases}$, $\begin{cases} a+b =t\\ c+d=it\end{cases}$

Any suggestion? Thank you

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Denote $z:=[z, 1]$, $\infty := [1, 0]$.

Hint: Do an Euclidean division first to write the projectivity in the form $a+bz$ or $a + \frac b{(c-1)z+1}$. This simplifies the system, you'll be able to find the parameters one by one.

We're not interested in the first one because it fixes $\infty$. The second sends sends $0\mapsto a+b$, sends $\infty \mapsto a$ and $1 \mapsto a+b/c$.

Thus we can solve for $a, b, c$ in that order! We have $a = \tau(\infty) = 1$. Next, $b = \tau(0) - a = i - 1$. Finally $c = b/(\tau(1)-a) = \frac{i-1}{-i-1} = -i$. We find $$\tau(z) = 1+ \frac{i-1}{(-i-1)z+1} = \frac{(-i-1)z+i}{(-i-1)z+1}$$ That is, $$\tau = \begin{bmatrix}-i-1&i \\-i-1&1 \\\end{bmatrix}$$

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    $\begingroup$ Yes, exactly. And keep in mind that $[a,b] = [1, b/a]$. $\endgroup$ – punctured dusk May 24 '18 at 9:39
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    $\begingroup$ I have a question too, are you used to view $\lambda \in \mathbb C$ as $[1, \lambda]$ or as $[\lambda, 1]$? $\endgroup$ – punctured dusk May 24 '18 at 9:40
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    $\begingroup$ Ok! In my answer I used $[\lambda, 1]$ so I hope that isn't confusing... $\endgroup$ – punctured dusk May 24 '18 at 9:43
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    $\begingroup$ @Arcticmonkey See edit; we can 'read' them when we write $\tau(z)$ as a fraction. $\endgroup$ – punctured dusk May 24 '18 at 16:03
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    $\begingroup$ @Arcticmonkey $\tau(0) := \tau([0,1]) = [1,-i] = [i,1] =: i$ and similarly for the others $\endgroup$ – punctured dusk May 24 '18 at 17:52

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