Past exam of the geometry course of the University of Oxford / 2007 / B / 5

Choose representative vectors $x_1,\dots ,x_{n+1}$ of the first $n+1$ points; they form a basis. Write a representative of the last point as ${\sum }_i{a}_i{x}_i$. General position forces each $a_i\ne 0$, since otherwise $n+1$ of the points would lie in a hyperplane. Rescale the basis vectors to $a_i{x}_i$. The tuple is then projectively equivalent to the standard frameApply the same normalization to the second tuple and compose the two transformations to obtain existence. An automorphism fixing the coordinate points is represented by a diagonal matrix. Fixing their sum forces all its diagonal entries to agree, so its projective action is the identity. This proves existence and uniqueness. At $n=1$ it says the Möbius group acts sharply triply transitively on the projective line: any ordered triple of distinct points can be sent to any other in exactly one way.

By the preceding frame result take the four points to be $[1:0:0],[0:1:0],[0:0:1],[1:1:1]$. The conics through them have equationsCoefficient vectors differing by a common nonzero scalar define the same conic. Their parameter space is the projectivization of a two-dimensional vector space, hence a projective line.

Each partition of the four points into two unordered pairs gives a reducible conic formed by the two joining lines. There are three pairings. Conversely a reducible conic is a union of two lines, possibly repeated. General position prevents any one line from carrying three of the points, so the lines must be distinct and carry exactly one of these pairings. Thus there are exactly three reducible members. In these coordinates they are the parameters $[0:1:-1]$, $[1:0:-1]$, $[1:-1:0]$, giving respectively $x(z-y)=0$, $y(z-x)=0$, and $z(y-x)=0$.

Write $P_5=[u:v:w]$ in the same normalized coordinates. Its coordinates are all nonzero, because the coordinate lines join pairs of the first three points. Evaluation at $P_5$ gives the additional linear equation $avw+bwu+cuv=0$ on the two-dimensional coefficient plane $a+b+c=0$. It is not identically zero there: that would imply $vw=wu=uv$, hence $u=v=w$ and $P_5=P_4$, contradicting the hypothesis. Its kernel is therefore one-dimensional. Thus there is exactly one projective conic through all five points. If it were reducible, $P_5$ would lie on a line joining two of the other points, contrary to no three being collinear. Hence the unique conic is irreducible.