The other day I managed to derive the generators of the Lorentz algebra from the assumption that the infinitesimal spacetime interval $ds^2$ is invariant under Lorentz transformations $x'=\Lambda x$ with $x$ the spacetime four vector and $\Lambda$ a general Lorentz transformation. I did this by requiring: \begin{equation} ds^2=dx^T\eta dx=dx^T \Lambda^T\eta\Lambda dx. \end{equation} I found that $\Lambda$ can be obtained by exponentiating a combination of the 6 generators generating the group SO(1,3), given by the following matrix: \begin{equation} \begin{pmatrix}0&a&b&c\\a&0&d&e\\b&-d&0&f\\c&-e&-f&0\end{pmatrix}. \end{equation} Now comes my question. I want to write the infinitesimal spacetime element using not the Minkowski metric, but by taking the determinant of the following 2x2 matrix: \begin{equation} ds^2 = det\begin{pmatrix}dt+dz&dx+idy\\dx-idy&dt-dz\end{pmatrix}\equiv det(dx_4). \end{equation} How should I now go about determining the 2x2 complex matrices $\Lambda$ corresponding to Lorentz transformations? Should I study the matrices $\Lambda$ satisfying $ds^2 = det(dx_4) = det(\Lambda dx_4)$? In case you are wondering, the above 2x2 matrix is obtained by combining the infinitesimal space and time intervals with either the identity matrix or one of the Pauli matrices in the following way: \begin{equation} dx_4 = dt \mathbb\sigma^0 + dx\sigma^1 + dy\sigma^2 + dz\sigma^3 \end{equation} and $\sigma^0 = \mathbb{I}_2$.

  • $\begingroup$ "How should I now go about determining the 2x2 complex matrices Λ corresponding to Lorentz transformations?" I may be wrong, but I don't think you can go this way because $\text{SL}(2,\mathbb{C})$ (the group acting on your quaternions) is the double covering of the Lorentz group. Since this is not an isomorphism, I think you are doomed. $\endgroup$ May 15, 2023 at 17:06
  • $\begingroup$ @JeanbaptisteRoux a double covering means it can be done... $\endgroup$ May 15, 2023 at 17:48
  • $\begingroup$ @naturallyInconsistent It would be then interesting to give the explicit formula inverting the formula (11.40, right) in the document you linked. Also, in this very document I can seem to find what OP is asking for. $\endgroup$ May 15, 2023 at 18:17
  • $\begingroup$ Related: physics.stackexchange.com/q/28505/2451 $\endgroup$
    – Qmechanic
    May 15, 2023 at 19:04

1 Answer 1


The direction you are going into, will not only get you a SU(2) double-cover representation of the Lorentz transformation group (actually two different copies), but leads naturally into Weyl spinors, and combining them, Dirac spinors.

The following link covers a bit about them. You might be able to find even better, even clearer expositions thereof, but it should cover at least this much of stuff.



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