The Generalized Fundamental Theorem of Projective Planes

Recall from The Fundamental Theorem of Projective Planes page that if $F$ is a field, $\mathbb{P}^2(F)$ is the projective plane over $F$, and $\mathbf{p}, \mathbf{q}, \mathbf{r}, \mathbf{s} \in \mathbb{P}^2(F)$ are points such that no three of them are collinear, then for $\mathbf{z_1} = [1, 0, 0]$, $\mathbf{z_2} = [0, 1, 0]$, $\mathbf{z_3} = [0, 0, 1]$, and $\mathbf{z_4} = [1, 1, 1]$ there exists a unique $3 \times 3$ invertible matrix $M$ whose entries are from $F$ such that the collineation $\phi_M : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ defined for all $\mathbf{x} \in \mathbb{P}^2(F)$ by $\phi_M(\mathbf{x}) = \mathbf{x}M$ is such that $\phi_M(\mathbf{z_1}) = \mathbf{p}$, $\phi_M(\mathbf{z_2}) = \mathbf{q}$, $\phi_M(\mathbf{z_3}) = \mathbf{r}$, and $\phi_M(\mathbf{z_4}) = \mathbf{s}$.

We will now state a more generalized version of this theorem.

Theorem 1: Let $F$ be a field, $\mathbf{P}^2(F)$ be the projective plane over $F$, and let points $\mathbf{p}, \mathbf{q}, \mathbf{r}, \mathbf{s} \in \mathbb{P}^2(F)$ be such that no three of these points are collinear, and let $\mathbf{p'}, \mathbf{q'}, \mathbf{r'}, \mathbf{s'} \in \mathbb{P}^2(F)$ be such that no three of these points are collinear. Then there exists a $3 \times 3$ invertible matrix whose entries are from $F$ such that the collinear $\phi_M : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ defined for all $\mathbf{x} \in \mathbb{P}^2(F)$ is such that $\phi_M(\mathbf{p'}) = \mathbf{p}$, $\phi_M(\mathbf{q'}) = \mathbf{q}$, $\phi_M(\mathbf{r'}) = \mathbf{r}$, and $\phi_M(\mathbf{s'}) = \mathbf{s}$.

• Proof: By the Fundamental Theorem of Projective Planes, since $\mathbf{p}, \mathbf{q}, \mathbf{r}, \mathbf{s} \in \mathbb{P}^2(F)$ and such that no three of these points are collinear, there exists a unique matrix $M_1$ such that the collinear $\phi_{M_1} : \mathbb{P}^2 \to \mathbb{P}^2$ is such that:

(1)

\begin{align} \quad \phi_{M_1}([1, 0, 0]) = \mathbf{p} \\ \quad \phi_{M_1}([0, 1, 0]) = \mathbf{q} \\ \quad \phi_{M_1}([0, 0, 1]) = \mathbf{r} \\ \quad \phi_{M_1}([1, 1, 1]) = \mathbf{s} \end{align}

• Since $\phi_{M_1}$ is a bijection, $\phi_{M_1}^{-1}$ exists and is such that:

(2)

\begin{align} \quad \phi_{M_1}^{-1}(\mathbf{p}) = [1, 0, 0] \\ \quad \phi_{M_1}^{-1}(\mathbf{q}) = [0, 1, 0] \\ \quad \phi_{M_1}^{-1}(\mathbf{r}) = [0, 0, 1] \\ \quad \phi_{M_1}^{-1}(\mathbf{s}) = [1, 1, 1] \end{align}

• Now since $\mathbf{p'}, \mathbf{q'}, \mathbf{r'}, \mathbf{s'} \in \mathbb{P}^2(F)$ are such that no three of these points are collinear, then once again, there exists a unique matrix $M_2$ such that:

(3)

\begin{align} \quad \phi_{M_2}([1, 0, 0]) = \mathbf{p'} \\ \quad \phi_{M_2}([0, 1, 0]) = \mathbf{q'} \\ \quad \phi_{M_2}([0, 0, 1]) = \mathbf{r'} \\ \quad \phi_{M_2}([1, 1, 1]) = \mathbf{s'} \end{align}

• Consider the composition $\phi_{M_1}^{-1} \circ \phi_{M_2} : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$. Then:

(4)

\begin{align} \quad \phi_{M_1}^{-1} \circ \phi_{M_2}(\mathbf{p}) = \mathbf{p'} \\ \quad \phi_{M_1}^{-1} \circ \phi_{M_2}(\mathbf{q}) = \mathbf{q'} \\ \quad \phi_{M_1}^{-1} \circ \phi_{M_2}(\mathbf{r}) = \mathbf{r'} \\ \quad \phi_{M_1}^{-1} \circ \phi_{M_2}(\mathbf{s}) = \mathbf{s'} \\ \end{align}

• Therefore $M = M_1^{-1}M_2$ is the collineation satisfying the conditions above. $\blacksquare$