In 3-dimensional projective geometry I have a point-point map (collineation) $c$ with matrix $A$. Then $A^{-1t}$ is the matrix for the plane-plane map for the same $c$. These matrices are 4x4 and determined up to a factor $\rho \neq 0$.

In the same way a point-plane mapping (correlation) $d$ with matrix $B$ has matrix $B^{-1t}$ for the plane-point mapping for the same $d$.

The image of a line can be calculated by choosing two points on the line, or two planes through the line and then join or meet the two images that can be calculated with one of the matrices given above.

I am looking for 6x6 matrices that map line-coordinates to new line-coordinates. Line-coordinates or Plücker coordinates have 6 coordinates and can be pointwise (contravariant) or planewise (covariant). In case of a collineation there must be a 6x6 matrix $C$ that maps contravariant line-coordinates to contravariant line-coordinates. Then $C^{-1t}$ will be good for covariant line-coordinates. In case of a correlation there will be a 6x6 matrix $D$ that maps contravariant line-coordinates to covariant line-coordinates, and $D^{-1t}$ the other way around.

My question is: is there a formula how to calculate $C$ c.q. $D$, when $A$ c.q. $B$ is given?


The coordinates of the Plücker coordinate vector of a line are the $2\times2$ sub-determinants of the coordinates of the points which span that line. So if $a$ and $b$ span a line $g$, you have

$$ g_{ij}=\begin{vmatrix}a_i&a_j\\b_i&b_j\end{vmatrix} $$

Now you transform your points to $a'=M\cdot a$ and $b'=M\cdot b$. This is your matrix $A$ but I want to avoid confusion with the point $a$. You obtain a new line $g'$ whose coordinates are

\begin{align*} g_{ij}'&=\begin{vmatrix}a_i'&a_j'\\b_i'&b_j'\end{vmatrix} \\&=\begin{vmatrix} M_{i1}a_1+M_{i2}a_2+M_{i3}a_3+M_{i4}a_4 & M_{j1}a_1+M_{j2}a_2+M_{j3}a_3+M_{j4}a_4 \\ M_{i1}b_1+M_{i2}b_2+M_{i3}b_3+M_{i4}b_4 & M_{j1}b_1+M_{j2}b_2+M_{j3}b_3+M_{j4}b_4 \end{vmatrix} \\&= \begin{vmatrix}M_{i1}&M_{i2}\\M_{j1}&M_{j2}\end{vmatrix}\cdot \begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix} + \begin{vmatrix}M_{i1}&M_{i3}\\M_{j1}&M_{j3}\end{vmatrix}\cdot \begin{vmatrix}a_1&a_3\\b_1&b_3\end{vmatrix} + \cdots \\&= \begin{vmatrix}M_{i1}&M_{i2}\\M_{j1}&M_{j2}\end{vmatrix}\cdot g_{12} + \begin{vmatrix}M_{i1}&M_{i3}\\M_{j1}&M_{j3}\end{vmatrix}\cdot g_{13} + \cdots \end{align*}

So the elements of the matrix $C$ you are looking for are just the $2\times2$ determinants you obtain from the input matrix which you called $A$ but which I called $M$.

The step from $B$ to $D$ can be handled by similar considerations, but I'll leave that as an excercise. In part because I'm used to a different convention which handles signs in a different way, so I'd have to concentrate in order to not mix things up.

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