the projective symmetry group of a conic

 A projective conic C has a matrix equation xTMx = 0, where M is a symmetric 3x3 matrix. The standard conic C0 : x2+y2-z2 = 0 has matrix equation xTDx = 0, where D = diag(1,1,-1). We know that any conic C is projective congruent to C0, so the symmetry group of C is conjugate to that of C0. If P,Q and R are p-points on the projective conic C, then there is a unique projective transformation which maps C to C0 and maps the points P,Q,R to X,Y,Z respectively. We can apply this to any three p-points on C0 to obtain a corresponding element of the projective symmetry group of C0. This shows that the group is very large. It also allows us to describe the elements to some extent. Given three p-points P,Q,R on C0, then the inverse p of the map guaranteed by the theorem maps X to P, Y to Q and Z to R. If we think of the matrix description of t, we see that the matrix has columns consisting of homogeneous coordinates for P, Q and R. The theorem states that the map is uniquely defined by the p-points, the actual multiples involved are constrained - we may use any verion of P, but then the multiples of Q and R are determined. Our main result needs the following fact which is not entirely obvious. Lemma If the symmetric matrices M and N define the same conic C, then N = kM for some non-zero k. The Projective Symmetry Theorem If the projective conic C has matrix M, then the projective symmetry group of C, S(C,P(2)), consists of the maps t([x]) = [Ax], with ATMA = M, det(A) = 1. Each such A determines a different element of the group.