Collineations of Projective Planes on Points

Like with transformations in the plane - we can define transformations on projective planes over a field $F$ which we define below.

Definition: Let $F$ be any field and let $\mathbb{P}^2(F)$ be the projective plane over $F$. Let $M$ be a $3 \times 3$ invertible matrix whose entries are from $F$. A Collineation of $\mathbb{P}^2(F)$ is a bijective function $\phi_M : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ where for each point $\mathbf{x} = [x_1, x_2, x_3] \in \mathbb{P}^2(F)$ we have that $\phi_M(\mathbf{x}) = \mathbf{x}M$.

From the definition above, it is important to note that since $M$ is an invertible matrix that then $M^{-1}$ exists, so the equation $\mathbf{x}M = \mathbf{0}$ implies that $\mathbf{x} = \mathbf{0}M^{-1} = \mathbf{0}$, however, $[0, 0, 0]$ is not a point in the projective plane, and so from $M$ being invertible we have that no point in $\mathbb{P}^2(F)$ is mapped to $[0, 0, 0]$.

Secondly, it is important to note that indeed $\phi_M : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ defined for all $\mathbf{x} \in \mathbb{P}^2(F)$ by $\phi_M(\mathbf{x}) = \mathbf{x}M$ is indeed a bijective function. To prove this, we first show that $\phi_M$ is injective. Let $\mathbf{x}, \mathbf{y} \in \mathbb{P}^2(F)$ and suppose that $\phi_M(\mathbf{x}) = \phi_M(\mathbf{y})$. Then:

(1)

\begin{align} \quad \phi_M(\mathbf{x}) = \phi_M(\mathbf{y}) \\ \quad \mathbf{x}M = \mathbf{y}M \\ \quad (\mathbf{x} - \mathbf{y})M = \mathbf{0} \\ \quad (\mathbf{x} - \mathbf{y})MM^{-1} = \mathbf{0}M^{-1} \\ \quad \mathbf{x} = \mathbf{y} \end{align}

We now show that $\phi_M$ is surjective. Let $\mathbf{y} \in \mathbb{P}^2(F)$. Then for $\mathbf{x} = \mathbf{y}M^{-1}$ we have that:

(2)

\begin{align} \quad \phi_M(\mathbf{x}) = \mathbf{x}M = \mathbf{y}M^{-1}M = \mathbf{y} \end{align}

So for all $\mathbf{y} \in \mathbb{P}^2(F)$ there exists a $\mathbf{x} \in \mathbb{P}^2(F)$ such that $\phi_x(\mathbf{x}) = \mathbf{y}$, so $\phi_M$ is indeed surjective.

Since $\phi_M$ is both injective and surjective, we conclude that $\phi_M$ is bijective.

For example, consider the field of real numbers $\mathbb{R}$ and the collineation $\phi_M : \mathbb{P}^2(\mathbb{R}) \to \mathbb{P}^2(\mathbb{R})$ where $M = \begin{bmatrix} 1 & 2 & 3\\ 2 & 0 & 2\\ 1 & 3 & 1 \end{bmatrix}$. You should verify that $\det (M) = 12 \neq 0$ so $M$ is invertible. Suppose that we want to determine the image of the point $\mathbf{x} = [2, 3, 1] \in \mathbb{P}^2(\mathbb{R})$. Then:

(3)

\begin{align} \quad \phi_M(\mathbf{x}) = \mathbf{x}M = \begin{bmatrix} 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3\\ 2 & 0 & 2\\ 1 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 7 & 13 \end{bmatrix} \end{align}

Therefore the point $\mathbf{x} = [2, 3, 1]$ is mapped to $\phi_M(\mathbf{x}) = [9, 7, 13]$.