the projective symmetry group of the standard conic

Let C be the standard projective conic with equation xy + yz + zx = 0.
This passes through the standard p-points X[1,0,0], Y[0,1,0], Z[0,0,1].
To simplify the algebra, we scale the equation to 2xy + 2yz + 2zx = 0.
Then the equation can be written as xTKx = 0, where K is the matrix

K =
 | 0 1 1 | | 1 0 1 | | 1 1 0 |

Now suppose that t is the projective transformation t([x]) = [Ax], where
A is a non-singular 3x3 matrix.

Here, we will determine explicitly the matrices A for which t(C) = C.

Note that t(X) = [a1], where a1 is the first column of A. Similarly, we have
t(Y) = [a2] and t(Z) = [a3], where a2, a3 are the remaining columns of A.

If t maps C to C, then, as X,Y,Z lie on C, the columns of A must correspond
to three p-points X',Y',Z' of C. Since A is non-singular, these will be distinct.
Suppose that, if X' = [u1], Y' = [a2], Z' = [a3]. To say that the column a1
corresponds to X' means that a1 = λ1u1 for some non-zero λ1, and similarly
a2 = λ2u2 and a3 = λ3u3.

Suppose that X',Y',Z' are distinct p-points on C. The three point theorem
shows that there is a unique t mapping C to C, and (X,Y,Z) to (X',Y',Z'). The
uniqueness part implies that the multipliers are determined up to an overall
multiplier. We now proceed to show how these multipliers can be found.

From earlier work, we know that t(C) = C if and only if ATKA = κK with κ ≠ 0.
There, we showed that we can scale A to achieve κ = 1. Here, it is as well to
postpone this scaling.

If A has the columns as above, then AT has rows λ1u1T2u2T, λ3u3T, so
that the (i,j) entry of ATKA is λiλj (uiTKuj).
Provided that X',Y',Z' are on C, uiTKui = 0 for i = 1,2,3. Thus the diagonal
entries of ATKA are zero, so agree with those of κA for any κ.
As ATKA is symmetric (since K is symmetric), we need only choose the λi
such that the elements of ATKA above the diagonal agree with those of κA,
i.e. such that

(*)       λ1λ2(u1TKu2) = λ1λ3 (u1TKu3) = λ2λ3 (u2TKu3) = κ (≠ 0).

We must check that, for i ≠ j, uiTKuj ≠ 0. By earlier work, the equation of
the polar of V = [v] is vTKx = 0. For V on C, the polar is the tangent at V.
This meets C only in V. Thus, if [w] is another p-point on C, vTKw ≠ 0.
Provided that the [ui] are distinct p-points on C, we have the result.

Now we can see that the in the equations (*), we can choose any non-zero
value for λ1. Then the values of λ2 and λ3 and that of κ are determined.
Putting λ1 = α.u2TKu3, we get λ2 = α.u1TKu3 and λ3 = α.u1TKu2, and finally
κ = α2.u2TKu3. u2TKu3. u2TKu3. Changing α simply scales the matrix A and
so produces the same projective transformation t.

(**)      Note that, as ATKA = κK, det(A)2 = κ3.

The above calculation shows how to find the matrix A from any three distinct
p-points on C. To allow us to construct explicitly the elements of S(C,P(2)), we
need to find all triples of p-points on C.

This is achieved by the parametrization theorem. Each p-point on C other than
Z[0,0,1] has the form [r] = [r,1-r,r(r-1)], with r real. Note that, if r ≠ s, then
these produce different p-points since we cannot have s = kr and 1-s = k(1-r)
unless k = 1 and r = s.

We thus have two kinds of triple, those including Z, and those not including Z.
The calculation of the uiTKuj is fairly straight-forward - we leave the details
as an exercise. Suppose that z = (0,0,1), r = (r,1-r,r(r-1)), s = (s,1-s,s(s-1)).
(***)         rTKz = 1 and rTKs = (r-s)2.

If we choose the p-points [r], [s],[z] (in order) then we obtain a matrix A.
By (*),(**),(***), det(A)2 = (r-s)6. Again by direct calculation, we find that
(1/(r-s))A has determinant 1. It has columns (r/(r-s),(1-r)/(r-s),r(r-1)/(r-s)),
(s/(r-s),(1-s)/(r-s),s(s-1)/(r-s)) and (0,0,r-s).

Other cases involving Z are similar - the columns are permuted, and r-s may
have to be replaced by s-r to get determinant +1.

If u = (u,1-u,u(u-1)), the p-points [r],[s],[u] lead to a matrix A which has
det(A)2 = {(r-s)(s-u)(u-r)}6. If we scale by 1/{(r-s)(s-u)(u-r)}, then we can
check that the scaled matrix has determinant 1.

Thus, we can write down the elements of S(C,P(2)) explicitly, choosing in each
case the matrix scaled to have determinant 1.

The terminally curious can see typical elements.