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

  • $\begingroup$ 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


from the Wikipedia

$$\vec x\mapsto X\quad,\vec y\mapsto Y \quad,\vec z=\vec x\times\vec y\mapsto Z$$

$$\frac 12\left(X\,Y-Y\,X\right)=i\,Z\quad,\rm det(Z)=0$$


\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*}


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.