Define projective plane $π½^2$ over a field $π½$. Define the notion of a projective line $L βπ½β^2$. Define what is meant by a projective transformation $Ο :π½β^2 βπ½β^2$. Show that a projective transformation $\tau$ takes projective lines to projective lines.
The projective plane $π½^2$ over a field $π½$ is the set of lines $[v]$ through the origin in a three-dimensional vector space $V β π½^3$ over the field $π½$.When are four points $P_0, P_1, P_2, P_3 βπ½β^2$ said to be in general position? Prove that if $\left\{P_0,β¦, P_3\right\}$, respectively $\left\{Q_0,β¦, Q_3\right\}$, are quadruples of points of $π½β^2$ in general position, then there is a unique projective transformation $Ο : π½^2 β π½ β^2$ with $Ο \left(P_i\right)=Q_i$ for all $i \in\{0,β¦, 3\}$.
In $π½β^2$, four points $P_0, P_1, P_2, P_3$ are said to be in general position, if each subset of 3 of these points can be represented by linearly independent representative vectors $p_i$ in $V$; equivalently if no three of them lie on a line.Show that it is possible to find two quadruples of pairwise distinct points $\left\{P_0,β¦, P_3\right\}$, respectively $\left\{Q_0,β¦, Q_3\right\}$, such that there is no projective transformation $Ο$ with $Ο \left(P_i\right)=Q_i$ for all $i β \{0,β¦, 3\}$.
Choose $P_0, P_1, P_2$ to be distinct points on a line, and $P_3$ to be any fourth point in the plane, distinct from the earlier ones. Choose $Q_0, Q_1, Q_2, Q_3$ to be in general position. Then for any projective transformation $Ο $, the three points $Ο \left(P_0\right), Ο \left(P_1\right), Ο \left(P_2\right)$ are collinear by (i). Hence they cannot be $Q_0, Q_1, Q_2$ since the latter are not collinear.Let $A, A', B, B'$ be pairwise distinct, non-collinear points of $π½β^2$. Prove that $A, A', B, B'$ are in general position if and only if there exists a projective transformation $Ο:π½β^2 βπ½β^2$ with $Ο (A)=B,Ο \left(A'\right)=B'$, and $Ο β Ο =\text{Id}$, the identity transformation on $π½^2$. [Hint: in one direction, you may wish to consider the lines $A B$ and $A' B'$ and their intersection. You may assume without proof that two distinct lines in the projective plane meet in a unique point.]
Assume first that there is such a $Ο$. Note that if $Ο (P)=Q$ then $Ο (Q)=P$ since $Ο β Ο =\text{Id}$. Consider the lines $L=A B$ and $L'=A' B'$. These lines are different, since otherwise the four points would be collinear. Let $O=L β© L'$ be their intersection. Then by what we proved above, $Ο (L)$ is a line so it is $L'$; similarly $Ο \left(L'\right)=L$. This implies that $Ο (O)=O$, so the point of intersection $O$ is a fixed point of $\tau$. But this means $O$ must be distinct from any of $A, A', B, B'$.Let $A, A', B, B'$ be given in projective coordinates by $A=[1,0,0], A'=[0,1,0]$, $B=[0,0,1]$ and $B'=[1,1,1]$. Find explicitly a projective transformation $Ο : π½β^2β π½ β^2$ with $Ο (A)=B, Ο \left(A'\right)=B'$, and $Ο β Ο =\text{Id}$
We need to write down the transformation $Ο$ that maps $A, A', B, B'$ to $B, B', A, A'$. For the first quadruple, for the representing vectors as given we already have $a+a'+b=b'$ so nothing to do. For the second quadruple, we have $a'=-b+b'-a$ so we must take $-b, b',-a$ as our representing triple. So we get that the matrix $$ A=\begin{pmatrix} 0 & 1 & -1 \\ 0 & 1 & 0 \\ -1 & 1 & 0 \end{pmatrix} $$ gives a projective transformation $Ο$ with the required properties.