It would be nice to have a cute method that uses Lorentz transformations of basis vectors by exponential transformation using gamma matrices. To avoid confusion, let's assume -+++ signature. Given $\gamma_\mu$ as gamma matrices that satisfy ${\rm tr}(\gamma_\mu\gamma_\nu) = 4\eta_{\mu\nu}$, then we have ${\rm tr}(\gamma'_\mu\gamma'_\nu) = 4\eta_{\mu\nu}$ if we put:
$\gamma'_\mu = \exp(-A)\gamma_\mu\exp(+A)$
where $A$ is any matrix.

Boosts and rotations use $A$ as a bivector. For example, with $\alpha$ a real number, $A=\alpha\gamma_0\gamma_3$ boosts in the z direction while $A=\alpha\gamma_1\gamma_2$ gives a rotation around the z axis.

Solving for the value of $\alpha$ that gives a boost with velocity $\beta = v/c$ appears to be straightforward. But how to do the rotations?

And by the way, what happens when you generalize $A$ to be something other than bivectors?

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    $\begingroup$ I am tempted to write something up on Geometric Algebra for this. $\endgroup$ Mar 13, 2011 at 13:02
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    $\begingroup$ I suppose you realize that bivectors actually form a representation of the Lorentz Lie algebra in the Clifford algebra and that's why these matrices (and none others) are important for transformation of gamma vectors (which in turn form a representation of the Lorentz group). $\endgroup$
    – Marek
    Mar 13, 2011 at 13:05
  • $\begingroup$ Poincare Group, you are interested in the lie algebra of this group, which can be found in most standard references. See for example, QFT by Ryder. $\endgroup$
    – user442
    Mar 13, 2011 at 16:05

1 Answer 1


The most general Lorentz transformation that is connected to the identity is given by the conjugation by $\exp(-A)$ where $$ A = \frac 12 \omega_{\mu\nu} \gamma^\mu \gamma^\nu $$ and $\omega_{\mu\nu}$ is an antisymmetric tensor containing $D(D-1)/2$ parameters. The group of all such transformations is isomorphic to $Spin(D-1,1)$. If $\omega$ only contains one component $0\mu$, then it is a boost, and the nonzero numerical value of $\omega$ is the rapidity - the "hyperbolic angle" $\eta$ such that $v/c=\tanh\eta$.

If only one doubly spatial component of $\omega$ is nonzero, then this component $\omega_{\mu\nu} = -\omega_{\nu\mu}$ is obviously the angle itself. Note that the spatial-spatial terms in $A$ are anti-Hermitean, producing unitary transformations; the mixed temporal-spatial terms in $A$ are Hermitean and they don't product unitary transformations on the 4-component space of spinors (but they become unitary if they're promoted to transformations of the full Hilbert space of quantum field theory).

In 4 dimensions, a general antisymmetric matrix $4\times 4$ contains 6 independent parameters and has eigenvalues $\pm i a, \pm i b$, so in 3+1 dimensions, one can always represent a general Lorentz transformation as a rotation around an axis in the 4-dimensional space followed by a boost in the complementary transverse 2-plane. This is the counterpart of the statement that any $SU(2)$ rotations in 3 dimensions is a rotation around a particular axis by an angle.

If you allowed $A$ to contain something else than $\gamma^{\mu\nu}$ matrices which generate the Lorentz group, you could get other groups. Only for a properly chosen subset of allowed values of $\omega$, you would get a closed group from the resulting exponentials (under multiplication). In particular, if you allowed $A$ to be an arbitrary complex combination of any products of gamma matrices, well, then you would allow $A$ to be any complex $4\times 4$ matrix, and its exponentials would produce the full group $GL(4,C)$ - surprising, Carl? ;-) It's not a terribly useful groups in physics because actions are usually not invariant under this "full group", are they?

Also, there are not too many groups in between $Spin(3,1)$ and $GL(4,C)$ - I guess that there's no proper group of $GL(4,C)$ that has a proper $Spin(3,1)$ subgroup. Obviously, there are many subgroups of $Spin(3,1)$ - such as $Spin(3)$, $Spin(1,1)\times Spin(2)$, and others.


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