Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Let $\Lambda$ be the Lorentz transformation parameterized by the asymmetric real matrix $w_{\mu \nu}$. That is, let $\Lambda = \exp(\frac{w_{\mu \nu}}{2}J^{\mu \nu})$, where $(J^{\mu \nu})_{\alpha \beta} = \delta^\mu_\alpha \delta^\nu_\beta\ - \delta^\mu_\beta \delta^\nu_\alpha$. All indices run from $0$ to $3$, and I am using the metric signature $+---$.

The spin 1/2 representation of the Lorentz group maps $\Lambda$ to $R[\Lambda]\stackrel{\mathrm{def}}{=}\exp(\frac{w_{\mu \nu}}{2}\gamma^\mu \gamma^\nu)$, where $\{ \gamma^\mu \}_{\mu = 0,1,2,3}$ are 4-by-4 complex matrices satisfying $\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu}$.


If we interpret the $\gamma^\mu$ not as matrices, but as the basis vectors of the geometric algebra $Cl(1,3)$, then we have

$R[\Lambda] x^\mu \gamma_\mu R^{-1}[\Lambda] = \Lambda^{\mu}_\nu x^\nu \gamma_\mu$.

Why is this true? I have no trouble doing the calculation - I am looking for a deeper understanding.

It seems a total coincidence to me that the matrix representation $R[\Lambda]$ happens to be also be the rotor for the Lorentz transformation $\Lambda$. Maybe this indicates that we can dispense with the matrix representation entirely, and somehow formulate the equivalent of a spin 1/2 rep. using $Cl(1,3)$ alone?

Thanks in advance for any help.

share|cite|improve this question

I think the problem here is the tail wagging the dog: you've defined the spin 1/2 representation of a rotation and then found that this is indeed also the rotor that performs rotations in a bilinear fashion. I think it's better to look at it the other way around: one can construct a rotor-based transformation that happens to be a rotation, and then prove that those rotors are the spinors of spin 1/2 that you know about.

Here's how that logic goes: if $n$ is a vector normal to a hyperplane, then write $n = n^\mu \gamma_\mu$ and see that a reflection across this hyperplane is $\underline N(a) = -nan^{-1}$.

Then, observe that two such reflections performs a Lorentz rotation. Thus, a rotation can be written $\underline R(a) = mnan^{-1} m^{-1}$ for two vectors $m,n$.

$mn$ is then a Lorentz spinor. The quantity $w_{\mu \nu} \gamma^\mu \gamma^\nu$ then has the orientation of $m \wedge n$ and the magnitude of $\theta/2 = \cos^{-1} (m \cdot n/|m||n|)/2$. The spinor $mn$ then has a scalar part $m \cdot n$ and a bivector part $m \wedge n$.

In this sense, I thinnk the answer to your ultimate question is yes, it's possible to dispense with the matrices, as they almost always simply represent some kind of object. $w_{\mu \nu}$ for instance is just a collection of components of basis bivectors, describing the plane in which the transformation is being done. The components of a transformation as a whole just tell us how basis vectors change. As a matter of course, all rotations can be performed using the rotors and geometric products without interpreting any of the basis vectors or other quantities as matrices.

share|cite|improve this answer
I'm just curious. Where does $\cos$ in your definition of $\theta$ come from? The metric is not Euclidean. – Andrey Sokolov Nov 27 '13 at 4:52
You're right; it's impossible to say which trig function goes there without knowing whether $m \wedge n$ squares to $+1$ or $-1$. – Muphrid Nov 27 '13 at 5:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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