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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 $\omega^\alpha, \tilde{\omega}^\dot{\alpha}$?

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  • $\begingroup$ I am slightly confused: If I am not mistaken the determinant of $X^{\alpha\dot\alpha}$ is $-x_3^2-x_1^2-x_2^2$. This is zero if and only if the vector $(x_1,x_2,x_3)$ is zero. $\endgroup$
    – Kurt G.
    Commented Oct 8, 2022 at 14:52
  • $\begingroup$ You are right Kurt. The determinants are zero if and only if $\vec{x}\dot\vec{x}=0$. $\endgroup$ Commented Oct 8, 2022 at 15:07
  • $\begingroup$ Reading the Wikipedia link in Eli's answer a bit. Apparently the determinant can be zero when $(x_1,x_2,x_3)\in\mathbb C^3$ and that's required for the $\otimes$-product representation of $X$. At the moment it does not look like to me that the cross product of two purely real and non zero three-vectors can be represented by means of spinors. $\endgroup$
    – Kurt G.
    Commented Oct 8, 2022 at 15:21

1 Answer 1

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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*}

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