Cross product and spinor correspondence I wonder if there is a correspondence between a cross product of two vectors $\vec{x}, \vec{y} \in \mathbb{R}^3$ and their associated spinors $\lambda^\alpha, \tilde{\lambda}^\dot{\alpha}$ and $\omega^\alpha, \tilde{\omega}^\dot{\alpha}$.
Here is what I mean by that:
Given two vectors $\vec{x} = (x_1, x_2, x_3)$ and $\vec{y} = (y_1, y_2, y_3)$ one can associate the two complex matrices
\begin{equation}
    \vec{x} \mapsto X^{\alpha \dot{\alpha}} = 
        \begin{bmatrix}
            x_3 & x_1 - i x_2 \\
            x_1 + i x_2  & -x_3
        \end{bmatrix}
    \quad and \quad
    \vec{y} \mapsto Y^{\alpha \dot{\alpha}} = 
        \begin{bmatrix}
            y_3 & y_1 - i y_2 \\
            y_1 + i y_2  & -y_3
        \end{bmatrix}
,
\end{equation}
with
\begin{equation}
    det\left|X^{\alpha \dot{\alpha}}\right| = det\left|Y^{\alpha \dot{\alpha}}\right| = 0.
\end{equation}
Since the determinant of the matrices is zero these matrices may be written as an outer product of two complex 2-vectors:
\begin{equation}
    X^{\alpha \dot{\alpha}} = \lambda^\alpha \otimes \tilde{\lambda}^\dot{\alpha}
    \quad and \quad
    Y^{\alpha \dot{\alpha}} = \omega^\alpha \otimes \tilde{\omega}^\dot{\alpha}
\end{equation}
The cross product of $\vec{x}, \vec{y}$ can now be associated with these matrices like:
\begin{equation}
    \vec{x}\times\vec{y} = i\frac{1}{2}\left( XY-YX \right)
\end{equation}
My question now is, how can $i\frac{1}{2}\left( XY-YX \right)$ be expressed by means of the spinors $\lambda^\alpha, \tilde{\lambda}^\dot{\alpha}$ and $\omega^\alpha, \tilde{\omega}^\dot{\alpha}$?
 A: from the Wikipedia
$$\vec x\mapsto X\quad,\vec y\mapsto Y
\quad,\vec z=\vec x\times\vec y\mapsto Z$$
$$\frac 12\left(X\,Y-Y\,X\right)=i\,Z\quad,\rm det(Z)=0$$
with
\begin{align*}
&X=\begin{bmatrix}
  \xi_{x1} \\
   \xi_{x2}\\
\end{bmatrix}
\begin{bmatrix}
  -\xi_{x2} & \xi_{x1} \\
\end{bmatrix}\quad ,\vec x\cdot \vec x=0
\end{align*}
\begin{align*}
&Y=\begin{bmatrix}
  \xi_{y1} \\
   \xi_{y2}\\
\end{bmatrix}
\begin{bmatrix}
  -\xi_{y2} & \xi_{y1} \\
\end{bmatrix}\quad ,\vec y\cdot \vec y=0
\end{align*}
where
\begin{align*}
 &\xi_x=\begin{bmatrix}
  \xi_{x1} \\
   \xi_{x2}\\
\end{bmatrix}\quad,
\xi_y=\begin{bmatrix}
  \xi_{y1} \\
   \xi_{y2}\\
\end{bmatrix}
\end{align*}
are the spinors

Other solution
\begin{align*}
&\vec x=\begin{bmatrix}
          x_1 \\
          x_2 \\
          x_3 \\
        \end{bmatrix}=
\left[ \begin {array}{c} {\xi_{x1}}^{2}-{\xi_{x2}}^{2}
\\ i \left( {\xi_{x2}}^{2}+{\xi_{x1}}^{2}
 \right) \\  -2\,\xi_{x1}\xi_{x2}\end {array}
 \right]       \quad \text{with}~\vec{x}\cdot\vec{x}=0\\
  &\vec y=\begin{bmatrix}
          y_1 \\
          y_2 \\
          y_3 \\
        \end{bmatrix}=
\left[ \begin {array}{c} {\xi_{y1}}^{2}-{\xi_{y2}}^{2}
\\ i \left( {\xi_{y2}}^{2}+{\xi_{y1}}^{2}
 \right) \\  -2\,\xi_{y1}\xi_{y2}\end {array}
 \right]       \quad \text{with}~\vec{y}\cdot\vec{y}=0\\
\end{align*}
\begin{align*}
   \vec{z}&=\vec{x}\times\vec{y}\\
   &=\left[ \begin {array}{c} 0\\  -2\,\xi_{x1}\xi x_{
{2}}{\xi_{y1}}^{2}+2\,\xi_{x1}\xi_{x2}{\xi_{y2}}^{2}+2\,{
\xi_{x1}}^{2}\xi_{y1}\xi_{y2}-2\,{\xi_{x2}}^{2}\xi_{y1}
\xi_{y2}\\  0\end {array} \right]\\
&+i\,
\left[ \begin {array}{c} -2\,{\xi_{x2}}^{2}\xi_{y1}\xi_{y2}-
2\,{\xi_{x1}}^{2}\xi_{y1}\xi_{y2}+2\,\xi_{x1}\xi_{x2}{
\xi_{y2}}^{2}+2\,\xi_{x1}\xi_{x2}{\xi_{y1}}^{2}
\\  0\\  -2\,{\xi_{x2}}^{2}{\xi y
_{{1}}}^{2}+2\,{\xi_{x1}}^{2}{\xi_{y2}}^{2}\end {array} \right]
\end{align*}
