Stabilizers of pairs of ternary quadratic forms

Question

Let $A,B$ be two ternary quadratic forms with real coefficients, given by symmetric matrices

$$\displaystyle 2A = \begin{pmatrix} 2a_{11} & a_{12} & a_{13} \\ a_{12} & 2a_{22} & a_{23} \\ a_{13} & a_{23} & 2a_{33} \end{pmatrix}, 2B = \begin{pmatrix} 2b_{11} & b_{12} & b_{13} \\ b_{12} & 2b_{22} & b_{23} \\ b_{13} & b_{23} & 2b_{33} \end{pmatrix}.$$

Let $V_\mathbb{R}$ denote the 12 dimensional real vector space of $(A,B)$ over $\mathbb{R}$, and let $G(\mathbb{R}) = \operatorname{GL}_2(\mathbb{R}) \times \operatorname{SL}_3(\mathbb{R})$. Let $(r,g)$ be an element of $G(\mathbb{R})$, where

$$\displaystyle r = \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix}.$$

Then $(r,g)$ acts on $(A,B) \in V_\mathbb{R}$ by sending $(A,B)$ to $$(r_1 (gAg^T) + r_2 (gBg^T), r_3 (gAg^T) + r_4 (gBg^T)).$$

Bhargava, in his paper The density of discriminants of quartic rings and fields, stated that the stabilizer in $G(\mathbb{R})$ of an element $(A,B) \in V_\mathbb{R}$ has order 24 if $A,B$ have four common real zeroes over $\mathbb{P}^2$, order 8 if they have exactly one pair of real zeroes over $\mathbb{P}^2$, and order 4 if they have no common real zeroes. He simply said that "one easily checks" that this is the case. Can anyone give an explanation as to why this should be easy to see, and a proof of why it's true?

Answer 1

It's actually order 8 for no real zeroes and order 4 for two real zeroes, not the other way around. (See the bottom of page 1038 of the paper.)

The symmetry groups are taken modulo $\{ \pm 1 \}$, so we work projectively in ${\rm PGL}_2({\bf R}) \times {\rm SL}_3({\bf R})$. We must assume that we're in the generic case that $A,B$ span a two-dimensional space of conics that vanish on four distinct points of ${\bf P}^2$ in general linear position (i.e., no three on a line). Then over ${\bf C}$ the stabilizer is always $S_4$, because it is known (and easy) that ${\bf PGL}_3$ acts simply-transitively on ordered four-tuples in general linear position. But over ${\bf R}$ we must use permutations that commute with complex conjugation. This conjugation acts on the four points as the identity, a simple transposition, or a double transposition according as four, two, or none of them are real. The commutators have orders $24$, $4$, and $8$ respectively, as claimed.