Find all projective transformations that fix two conics

Question

Given two distinct nonsingular conics $C_1,C_2$ in a complex projective plane, I try to find all projective transformations that fix $C_1$ and fix $C_2$.

Then I consider the intersections of $C_1,C_2$. There are 5 cases:

1. four single points;
2. two single points, one double point;
3. two double points;
4. one single point, one triple point;
5. one quadruple point.

Case 1. Obtain three pairs of lines by joining disjoint pairs of these four points, each such pair of lines meets in a point $v_i$ where $\{v_1, v_2, v_3\}$ is a basis relative to which the matrices of $C_1,C_2$ are simultaneously diagonalizable. So there is a basis in which the two conics are defined by ($\lambda_i$ are distinct) $$x_1^2+x_2^2+x_3^2=0, \quad \lambda_1 x_1^2+\lambda_2 x_2^2+\lambda_3 x_3^2=0$$ so there are four projective transformations that fix $C_1$ and fix $C_2$ $$\begin{array}l [x_1,x_2,x_3]\mapsto[x_1,x_2,x_3]\cr [x_1,x_2,x_3]\mapsto[-x_1,x_2,x_3]\cr [x_1,x_2,x_3]\mapsto[x_1,-x_2,x_3]\cr [x_1,x_2,x_3]\mapsto[x_1,x_2,-x_3] \end{array}$$ The projective transformation induces a permutation on the four points, the permutations in $S_4$ that preserve cross ratio form a subgroup $V_4$, and image of the four points determine the projective transformation, so there are only 4 projective transformations that fix $C_1$ and fix $C_2$.

Case 3. Similar to case 1, there is a basis in which the two conics are defined by ($\lambda\ne1$) $$x_1^2+x_2^2+x_3^2=0, \quad \lambda x_1^2+x_2^2+x_3^2=0$$
There are infinitely many projective transformations that fix $C_1$ and fix $C_2$. $$ [x_1,x_2,x_3]\mapsto[x_1,\cos(t)x_2-\sin(t)x_3,\sin(t)x_2+\cos(t)x_3]\\ [x_1,x_2,x_3]\mapsto[x_1,\cos(t)x_2+\sin(t)x_3,\sin(t)x_2-\cos(t)x_3] $$

The other cases: the number of distinct intersections is odd, so the matrices of $C_1,C_2$ are not simultaneously diagonalizable, so I don't know how to write down the equations.

I think, in case $2$ there are only two projective transformations that fix $C_1$ and fix $C_2$: one is identity transformation, the other swaps two single points. I think, the image of $3$ points determine the projective tranformation uniquely on $C_1$ and $C_2$, which determine the projective transformation on the whole plane uniquely.

In the cases $4, 5$ I guess there are infinitely many projective transformations that fix $C_1$ and fix $C_2$, but I cannot prove it.

Answer 1

Let's use $G \subset PGL(n, \mathbb{C})$ to denote the group of projective transformations preserving $C_1$ and $C_2$.

Case 2 Yes, your reasoning is correct. $G \cong \mathbb{Z} / 2\mathbb{Z}$.

Case 4 In fact, only the identity transformation preserves $C_1$ and $C_2$. $G \cong \{ e \}$.

Proof: Let $p_1, p_2$ denote the intersection points of degree 1 and 3 respectively. Let $\ell_1$ denote the common tangent line through $p_1$, and let $\ell_2$ denote the other common tangent line. The line $\ell_2$ intersects $C_1$ and $C_2$ at points $q_1$ and $q_2$ respectively. The points $p_1, p_2, q_1, q_2$ are all pairwise different, and no three of them are colinear. A projective transformation preserving $C_1$ and $C_2$ would have to preserve each of these four points, so it can only be the identity.

Case 5 Yes, there are infinitely many such transformations. We have $G \cong (\mathbb{Z} / 2 \mathbb{Z}) \times (\mathbb{C}, +)$.

Let $p$ denote the intersection point of $C_1$ and $C_2$, and let $\ell$ denote the common tangent line. We may assume $\ell$ to be the line at infinity, and that $C_1$ and $C_2$ are defined by the affine equations $y = x^2$ and $y = x^2 + 1$ respectively. Then $G$ is the group of affine linear transformations preserving $C_1$ and $C_2$. It can be seen by explicit computations that $g \in G$ if and only if it is on form $g : (x, y) \mapsto (b x + c, y + 2bcx + c^2)$ where $b = \pm 1$. The group of such automorphisms is exactly $(\mathbb{Z} / 2 \mathbb{Z}) \times (\mathbb{C}, +)$.