Question

(1) let ( t : mathbb{rp}^2 \to mathbb{rp}^2 ) be a projective transformation. let ( l cong mathbb{rp}^1 subset mathbb{rp}^2 ) be a line. show that ( t(l) ) is also a line, and the induced map ( t|_l : mathbb{rp}^1 \to mathbb{rp}^1 ) is projective.

Explanation:

Step1: Recall properties of projective transformations

Projective transformations \(T:\mathbb{R}P^{2}\to\mathbb{R}P^{2}\) are linear transformations on homogeneous coordinates (up to scalar multiples). A line \(L\subseteq\mathbb{R}P^{2}\) can be represented in homogeneous coordinates. Let \([x_0:x_1:x_2]\) be homogeneous coordinates of a point in \(\mathbb{R}P^{2}\). A line \(L\) has the equation \(ax_0 + bx_1+cx_2 = 0\), where \([a:b:c]\in\mathbb{R}P^{2}\) (the dual space).

A projective transformation \(T\) is given by \(T([x_0:x_1:x_2])=[(A\mathbf{x})_0:(A\mathbf{x})_1:(A\mathbf{x})_2]\), where \(A\) is a non - singular \(3\times3\) matrix and \(\mathbf{x}=(x_0,x_1,x_2)^T\).

Step2: Show \(T(L)\) is a line

Let \(p = [x_0:x_1:x_2]\in L\), so \(ax_0+bx_1 + cx_2=0\). After applying \(T\), \(T(p)=[y_0:y_1:y_2]\) where \(\mathbf{y}=A\mathbf{x}\). Since \(A\) is invertible, \(\mathbf{x}=A^{-1}\mathbf{y}\). Substituting \(\mathbf{x}\) into the equation of \(L\): \(a(A^{-1}\mathbf{y})_0 + b(A^{-1}\mathbf{y})_1 + c(A^{-1}\mathbf{y})_2=0\). Let \(d_i=\sum_{j = 0}^{2}(A^{-1})_{ij}a_j\) (\(i = 0,1,2\), where \(a_0=a,a_1 = b,a_2=c\)). Then \(d_0y_0 + d_1y_1 + d_2y_2=0\), which is the equation of a line in \(\mathbb{R}P^{2}\). So \(T(L)\) is a line.

Step3: Show \(T|_L\) is projective

In homogeneous coordinates, a projective transformation on \(\mathbb{R}P^{1}\) is of the form \(S([u:v])=[\alpha u+\beta v:\gamma u+\delta v]\) with \(

$$\begin{vmatrix}\alpha&\beta\\\gamma&\delta\end{vmatrix}$$

eq0\).

We can use a coordinate system. Let \(L\) be parameterized as \([t:1:0]\) (using a suitable choice of homogeneous coordinates). Then \(T([t:1:0])=[(A

$$\begin{pmatrix}t\\1\\0\end{pmatrix}$$

)_0:(A

$$\begin{pmatrix}t\\1\\0\end{pmatrix}$$

)_1:(A

$$\begin{pmatrix}t\\1\\0\end{pmatrix}$$

)_2]\). Since \(T(L)\) is a line, we can consider the non - zero components (say \(y_0\) and \(y_1\) if \(y_2 = 0\) for points on \(T(L)\)).

Let \(A=

$$\begin{pmatrix}a_{00}&a_{01}&a_{02}\\a_{10}&a_{11}&a_{12}\\a_{20}&a_{21}&a_{22}\end{pmatrix}$$

\). Then \(A

$$\begin{pmatrix}t\\1\\0\end{pmatrix}$$

=

$$\begin{pmatrix}a_{00}t + a_{01}\\a_{10}t+a_{11}\\a_{20}t + a_{21}\end{pmatrix}$$

\). If \(T(L)\) has the equation \(d_0y_0 + d_1y_1 + d_2y_2=0\), and we can write \(y_2 = 0\) (by re - scaling if necessary) for the parameterization of \(T(L)\). Then \(T|_L([t:1])=[(a_{00}t + a_{01}):(a_{10}t + a_{11})]\) (after re - scaling to remove the non - zero third component, since \(T(L)\) is a line). This is a projective transformation on \(\mathbb{R}P^{1}\) because it is a linear (up to scalar) transformation of the parameter \(t\).

Answer:

We have shown that \(T(L)\) is a line (by expressing the equation of \(T(L)\) in homogeneous coordinates) and that \(T|_L:\mathbb{R}P^{1}\to\mathbb{R}P^{1}\) is projective (by parameterizing the line \(L\) and \(T(L)\) and showing the induced map is a linear (up to scalar) transformation on the parameter space).