Product Collineations of Projective Planes

# Product Collineations of Projective Planes

Let $F$ be a field and $\mathbb{P}^2(F)$ be the projective plane over $F$. Consider two $3 \times 3$ invertible matrices $M$ and $N$. As we have seen, we can define collineations on $\mathbb{P}^2(F)$ as functions $\phi_M, \phi_N : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ for all $\mathbf{x} \in \mathbb{P}^2(F)$ by $\phi_M(\mathbf{x}) = \mathbf{x}M$ and $\phi_N(\mathbf{x}) = \mathbf{x}N$.

Now from linear algebra, recall that if $M$ and $N$ are invertible matrices, then their products $MN$ and $NM$ are invertible matrices. In particular, we have that:

(1)
\begin{align} \quad (MN)^{-1} = N^{-1}M^{-1} \end{align}

Since $(MN)(MN)^{-1} = MNN^{-1}M^{-1} = MIM^{-1} = MM^{-1} = I$ and $(MN)^{-1}(MN) = N^{-1}M^{-1}MN = N^{-1}IN = N^{-1}N = I$.

So we have verified that $MN$ is an invertible matrix… so what can be said about the collineation $\phi_{MN} : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ defined for all $\mathbf{x} \in \mathbb{P}^2(F)$ by $\phi_{MN} (\mathbf{x}) = \mathbf{x}MN$?

 Theorem 1: Let $F$ be a field, $\mathbb{P}^2(F)$ be the projective plane over $F$, and $M$ and $N$ be $3 \times 3$ invertible matrices where $\phi_M, \phi_N : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ for all $\mathbf{x} \in \mathbb{P}^2(F)$ define the collineations $\phi_M(\mathbf{x}) = \mathbf{x}M$ and $\phi_N(\mathbf{x}) = \mathbf{x}M$. Then $\phi_{MN} : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ is given by $\phi_{MN} = \phi_N \circ \phi_M$ for all $\mathbf{x} \in \mathbb{P}^2(F)$ and $\phi_{MN}^{-1} = \phi_M^{-1} \circ \phi_N^{-1}$.
• Proof: Let $\mathbf{x} \in F$. Then:
(2)
\begin{align} \quad (\phi_N \circ \phi_M)(\mathbf{x}) = \phi_N(\phi_M(\mathbf{x})) = \phi_M(\mathbf{x}M) = (\mathbf{x}M)N = \mathbf{x}MN = \phi_{MN} \end{align}
• Furthermore we have that:
(3)
\begin{align} \quad (\phi_M^{-1} \circ \phi_N^{-1})(\mathbf{x}) = \phi_M^{-1}(\phi_N^{-1}(\mathbf{x})) = \phi_M^{-1}(\mathbf{x}N^{-1}) = \mathbf{x}N^{-1}M^{-1} \end{align}
• So then:
(4)
\begin{align} \quad (\phi_M^{-1} \circ \phi_N^{-1})(\phi_{MN}(\mathbf{x})) = (\mathbf{x}MN)N^{-1}M^{-1} = \mathbf{x} \end{align}
• And also:
(5)
\begin{align} \quad \phi_{MN}((\phi_M^{-1} \circ \phi_N^{-1})(\mathbf{x})) = \phi_MN(\mathbf{x}N^{-1}M^{-1}) = (\mathbf{x}N^{-1}M^{-1})MN = \mathbf{x} \end{align}
• Therefore $\phi_{MN}^{-1} = \phi_M^{-1} \circ \phi_N^{-1}$. $\blacksquare$