Past exam of the short options course of the University of Oxford / 2014 / 3

The required general linear position condition is that every $n+1$ of the $n+2$ representing vectors are linearly independent. Such an ordered configuration is a projective frame.

Choose representatives $p_0,\dots ,p_n$ of the first $n+1$ projective points; they form a basis. Write a representative of the last point as $p_{n+1}={\sum }_i{a}_i{p}_i$. Every $a_i$ is nonzero: otherwise the last point together with the other $n$ basis points would fail the independence condition. Rescale the basis representatives to ${\tilde{p}}_i=a_i{p}_i$, so that $p_{n+1}={\sum }_i{\tilde{p}}_i$.

Do the same for the target configuration, obtaining representatives ${\tilde{q}}_i$ with $q_{n+1}={\sum }_i{\tilde{q}}_i$. The unique linear map sending the basis ${\tilde{p}}_i$ to ${\tilde{q}}_i$ is invertible and also sends $p_{n+1}$ to $q_{n+1}$. Its induced projective transformation sends every point to its prescribed partner. Thus any two ordered projective frames are projectively equivalent.

A projective quadric is the zero locus of a nonzero homogeneous quadratic polynomial $q(v)=v^{T}Bv$, with $B$ symmetric over a field of characteristic different from two. A projective point on it is singular when all first partial derivatives vanish. Since $\nabla q(v)=2Bv$, the quadric is nonsingular exactly when $B$ is invertible.

Over the complex numbers, a nonzero quadratic form in three variables can be diagonalized by an invertible change of basis. Its rank classifies a projective conic:

The zero polynomial is excluded from the definition of a conic. Thus these last two cases are precisely the singular conics.

Use part (a) to put the four projective points atA quadratic polynomial vanishing at the first three has no square terms, and vanishing at the fourth givesThese form a two-dimensional vector space. Choose the two membersTheir coefficient triples are independent and both satisfy the relation. Their symmetric bilinear form matrices are of the formFor $(A,B,C)=(1,1,-2)$ and $(1,-2,1)$, both determinants are $-1/2$, so both selected conics are nonsingular. Any coefficient triple with zero sum is uniquely a linear combination of these two. Hence every conic through the four points has matrixUndoing the projective transformation carries the entire family back to the original four points, preserving nonsingularity. This is a pencil of conics through four points.

For the explicit basis in part (c), $q_0+t{q}_1$ has coefficients $(1+t,1-2t,t-2)$. ThusAt these values the conics factor, up to nonzero constants, asGeometrically these are exactly the three pairings of the four base points into two pairs, with a projective line joining each pair. No double line can contain four points in general linear position.

The numerical parameter values depend on the selected basis, but the three distinct singular members do not. Another pair of independent nonsingular basis conics gives an invertible projective change of pencil parameter. Its parameter infinity is nonsingular because the second basis form is nonsingular. Therefore all three singular members again occur at three distinct finite values of $t$. This proves the claimed count for any such basis.