5
$\begingroup$

In quantum field theory, we use the universal cover of the Lorentz group: $SL(2, \mathbb C)$, instead of $SO(3,1)$. The reason for this is, of course, that $SO(3,1)$ representations aren't able to describe spin $\frac12$ particles.

Therefore, I was wondering how the invariant speed of light is encoded in $SL(2, \mathbb C)$?

This curious fact of nature, is encoded in $SO(3,1)$, because this is exactly the group that leaves the Minkowski metric invariant. In contrast, $SL(2, \mathbb C)$ is just the group of complex $(2 \times 2)$ matrices with unit determinant.

$\endgroup$
6
$\begingroup$

In the spinorial representation of a Lorentz transformation, we represent an event in spacetime as an element of the 4D vector space of Hermitian $2\times2$ matrices:

$$X = t\,\mathrm{id} + x\,\sigma_x+y\,\sigma_y+z\,\sigma_z\tag{1}$$

where the $\sigma_j$ are, as always, the Pauli matrices and $(t,\,x,\,y,\,z)$ the four spacetime co-ordinates.

A member $\zeta\in SL(2,\,\mathbb{C})$ of the double cover of $SO(1,\,3)$ acts on such an event by the so-called spinor map:

$$X\mapsto \zeta^\dagger\,X\,\zeta\tag{2}$$

so we now seek to encode the line interval invariance into (2); in terms of the spacetime event $X$.

Exercise: Check that the Minkowskian length of $X$ is, indeed encoded as $\det X$.

Thus, from (2), the invariance of the spacetime interval length under the action of $\zeta$ is

$$\det(\zeta^\dagger\,X\,\zeta) = |\det\zeta|^2 \det X = \det X\tag{3}$$

and the fulfilling of (3) for all Hermitian $X$ is a necessary and sufficient condition for $\zeta$ to leave the spacetime interval invariant. In particular, unimodularity of $\zeta$ is sufficient (but not necessary) for this conservation.

This answers your question, but take heed that the whole set of matrices that conserve the interval is precisely $U(1)\times SL(2\,\mathbb{C})$, i.e. the set of matrices of the form $e^{i\,\phi}\,\zeta;\,\phi\in\mathbb{R}$, where $\zeta$ is unimodular. But note that the phase factor makes no difference to the spinor map (2); therefore, the group of spacetime interval conserving matrices $U(1)\times SL(2\,\mathbb{C})$ breaks up into equivalence classes of spinor maps, with precisely one member of $PSL(2,\,\mathbb{C})\cong SO^+(1,\,3)$ for each spinor map. The double cover $SL(2,\,\mathbb{C})$ comprises two members for each distinct spinor map, i.e. pairs of the form $\pm\zeta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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