6
$\begingroup$

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?

$\endgroup$

1 Answer 1

3
$\begingroup$

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, $$ \mathcal G = \left\{ A\in {\rm GL}_3(\mathbb F) : TA = \frac{1}{|A|}\lambda A + \frac{1}{|A|}AT, \lambda\in \mathbb F \right\} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.