Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 56 / 1 / Solution

Work in three-dimensional projective space, with four homogeneous coordinates indexed $0,1,2,3$. Let $A$ and $B$ be independent representatives of two points. The Plücker coordinates of their line areThis antisymmetric matrix has rank two and is defined only up to a nonzero scalar. Replacing the pair by another basis of the same two-dimensional subspace multiplies $L$ by the determinant of that basis change. Its six independent entries satisfy $L^{01}{L}^{23}-L^{02}{L}^{13}+L^{03}{L}^{12}=0$, since the exterior product $(A\wedge B)\wedge (A\wedge B)$ vanishes.

To describe the same line as an intersection of planes, choose a volume orientation with ${\epsilon }_{0123}=1$ and define the dual Plücker coordinates byThe independent components pair asConversely, $L^{ij}=\frac{1}{2}{\epsilon }^{ijkl}{L}_{kl}$ with the matching volume convention. The lower-index form annihilates both defining points: $L_{ij}{A}^j=L_{ij}{B}^j=0$. Its columns span the pencil of planes containing their line. This dualization uses the volume form, not a Euclidean metric for raising and lowering indices. Two-index point and plane representations are dual in projective dimension three; in higher dimensions the corresponding dual exterior form has a different degree.

For a general plane covector $F_i$, contract the upper-index form:This is a linear combination of $A,B$, so it is on the line, and $F_i{P}^i=0$ by antisymmetry, so it is on the plane. Provided the resulting vector is nonzero, its homogeneous coordinates give the unique intersection, including intersections at infinity. If $F(A)=F(B)=0$, the whole line lies in the plane and the zero result correctly indicates that there is no unique intersection. These are the line-plane incidence in Plücker coordinates operations used in the subparts.