# Values of a pair of determinants

Let $$\mathbf{x} = (x_0, x_1, x_2), \mathbf{y} = (y_0, y_1, y_2)$$ be vectors over a field $$\mathbb{F}$$ of characteristic zero. Define the function

$$\displaystyle S(\mathbf{x}, \mathbf{y}) = x_2 (y_0^2 - 2 y_1 y_2) + x_1 (2 y_2^2 - y_0 y_1) + x_0 (y_1^2 - y_0 y_2) = \begin{vmatrix} x_2 & x_1 & x_0 \\ y_2 & y_1 & y_0 \\ y_1 & y_0 & 2 y_2 \end{vmatrix}$$

and $$T(\mathbf{x}, \mathbf{y}) = S(\mathbf{y}, \mathbf{x})$$.

Curiously, I found that for fixed $$(s,t) \in \mathbb{F}^2$$ the set of solutions to $$s = S(\mathbf{x}, \mathbf{y}), t = T(\mathbf{x}, \mathbf{y})$$ is stable under the map

$$\mathbf{x} \mapsto \begin{bmatrix} x_0 + 2 x_1 + 2 x_2 \\ x_0 + x_1 + 2 x_2 \\ x_0 + x_1 + x_2 \end{bmatrix}, \mathbf{y} \mapsto \begin{bmatrix} y_0 + 2 y_1 + 2 y_2 \\ y_0 + y_1 + 2 y_2 \\ y_0 + y_1 + y_2 \end{bmatrix}.$$

Moreover, the matrix defining this linear map which is

$$M = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 1 \end{bmatrix}$$

has determinant one.

Thus, if we define the group $$\mathcal{G} \subset \text{GL}_3(\mathbb{F})$$ to be the set of $$3 \times 3$$ matrices $$A$$ over $$\mathbb{F}$$ such that $$S(\mathbf{x}, \mathbf{y}) = S(A \mathbf{x}, A \mathbf{y})$$ for all $$\mathbf{x}, \mathbf{y} \in \mathbb{F}^3$$, then we have shown that $$M \in \mathcal{G}$$. Further, $$M$$ has infinite order so $$\mathcal{G}$$ contains infinitely many elements.

Is it possible to determine $$\mathcal{G}$$ in a reasonable way?

First, by transposing and interchanging two rows $$S({\bf x},{\bf y}) = -\left|\begin{matrix} x_0 & y_0 & 2y_2 \\ x_1 & y_1 & y_0 \\ x_2 & y_2 & y_1 \end{matrix} \right|$$. Notice also that if $$T = \left[\begin{matrix}0&0&2 \\ 1&0 &0 \\ 0& 1& 0\end{matrix} \right]$$ then $$T{\bf y} = \left[\begin{matrix} 2y_2 \\ y_0 \\ y_1\end{matrix}\right]$$. Hence, $$S({\bf x},{\bf y}) = -\left| {\bf x}\ \ {\bf y}\ \ T{\bf y} \right|$$
Suppose $$A\in \mathcal G$$ then for all $${\bf x},{\bf y} \in \mathbb F^3$$ $$S({\bf x},{\bf y}) = S(A{\bf x}, A{\bf y}) = -\left|{\bf x}\ \ {\bf y} \ \ |A|A^{-1}TA{\bf y}\right|.$$ Now when $${\bf x} = e_3$$ then this gives $$y_0^2 - 2y_1y_2 = y_0\langle|A|A^{-1}TA{\bf y}, e_2\rangle - y_1\langle|A|A^{-1}TA{\bf y}, e_1\rangle.$$ Subsequently, when $${\bf y} = e_1$$ this gives $$1 = \langle|A|A^{-1}TAe_1, e_2\rangle$$, and when $${\bf y} = e_2$$ this gives $$0 = \langle|A|A^{-1}Tae_2, e_1\rangle$$. Moreover, when $${\bf y} = e_1+e_3$$ then after some computation $$\langle|A|A^{-1}TAe_1, e_1\rangle = \langle|A|A^{-1}TAe_3, e_3\rangle$$
When $${\bf x} = e_2$$ this gives $$2y_2^2 - y_0y_1 = y_2\langle|A|A^{-1}TA{\bf y}, e_1\rangle - y_0\langle|A|A^{-1}TA{\bf y}, e_3\rangle,$$ $${\bf y} = e_1$$ gives $$0 = \langle|A|A^{-1}TAe_1, e_3\rangle$$, and $${\bf y} = e_3$$ gives $$2 = \langle|A|A^{-1}Tae_3, e_1\rangle$$. Moreover, when $${\bf y} = e_1+e_2$$ then $$\langle|A|A^{-1}TAe_1, e_1\rangle = \langle|A|A^{-1}TAe_2, e_2\rangle$$
When $${\bf x} = e_1$$ this gives $$y_1^2 - y_0y_2 = y_1\langle|A|A^{-1}TA{\bf y}, e_3\rangle - y_2\langle|A|A^{-1}TA{\bf y}, e_2\rangle$$ $${\bf y}=e_2$$ gives $$1 = \langle|A|A^{-1}TAe_2, e_3\rangle$$ and $${\bf y} = e_3$$ gives $$0 = \langle|A|A^{-1}TAe_3, e_2\rangle$$
If we define $$\lambda := \langle|A|A^{-1}TAe_1, e_1\rangle$$ then the above argument concludes $$|A|A^{-1}TA = \left[\begin{matrix}\lambda&0&2 \\ 1&\lambda &0 \\ 0& 1& \lambda\end{matrix} \right] = \lambda I + T$$ and so $$TA = \frac{1}{|A|}\lambda A + \frac{1}{|A|}AT.$$ One won't get anymore out of these equations. Furthermore, any $$A$$ satisfying this equation gives $$S(A{\bf x}, A{\bf y}) = -|{\bf x} \ \ {\bf y} \ \ \lambda{\bf y} + T{\bf y}| = S({\bf x}, {\bf y})$$ by column replacement. Your $$M$$ corresponds to the case where $$\lambda = 0$$, that is commutes with $$T$$, and $$|A| = 1$$. However, there may be more matrices in $$\mathcal G$$.
Therefore, $$\mathcal G = \left\{ A\in {\rm GL}_3(\mathbb F) : TA = \frac{1}{|A|}\lambda A + \frac{1}{|A|}AT, \lambda\in \mathbb F \right\}$$