Is it true that the commutators of the gamma matrices form a representation of the Lie algebra of the Lorentz group?

Question

Wikipedia claims (http://en.wikipedia.org/wiki/Gamma_matrices):

The elements $\sigma^{\mu \nu} = \gamma^\mu \gamma^\nu - \gamma^\nu \gamma^\mu$ form a representation of the Lie algebra of the Lorentz group.

Using the convention

$$ \gamma^0 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix},\quad \gamma^1 = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} $$

$$ \gamma^2 = \begin{pmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \end{pmatrix},\quad \gamma^3 = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} $$

I evaluated the matrices $\sigma^{\mu\nu}$ in Mathematica. For example,

$$\sigma^{01} = \begin{pmatrix} 0 & 0 & 0 & 2 \\ 0 & 0 & 2 & 0 \\ 0 & 2 & 0 & 0 \\ 2 & 0 & 0 & 0 \end{pmatrix}$$

If this belongs to the Lie algebra for the Lorentz group then (correct me if I'm wrong here) I would expect that $M_{01}(t) = \exp(it\sigma^{01})$ would belong to the Lorentz group, so $M_{01}^T(t) \eta M_{01}(t) = \eta$ should hold, where $\eta = \operatorname{diag}(1, -1, -1, -1)$. So I evaluated this using Mathematica and got

$$M_{01}^T(t) \eta M_{01}(t) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -\cos 4t & -i \sin 4t & 0 \\ 0 & -i \sin 4t & -\cos 4t & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$$

which doesn't look like $\eta$. What did I do wrong in the above?

Answer 1

The products and linear combinations of gamma matrices form a clifford algebra; the algebraic properties of elements of this algebra do not depend in any way whatsoever on the actual matrices you use to represent the basis elements.

It helps, then, to think about the gamma matrices not as matrices but as basis vectors for Minkowski spacetime. In particular, they form an orthonormal basis. Thus, when I say "vector" in the following, I mean a linear combination of the gamma matrices.

A product of $k$ orthogonal vectors is called a $k$-blade. Linear combinations of $k$-blades might no longer be blades, but can still be called $k$-vectors. Both of these are said to have a "grade" of $k$.

The objects $\sigma^{\mu \nu}$ for $\mu \neq \nu$ are 2-blades, or "bivectors". Because the basis is orthonormal, the strict antisymmetrization should not be required: $\sigma^{\mu \nu} = 2 \gamma^\mu \gamma^\nu$ for $\mu \neq \nu$.

Note that bivectors of the form $\gamma^0 \gamma^i$, for $i = 1, 2, 3$, square as follows:

$$(\gamma^0 \gamma^i)(\gamma^0 \gamma^i) = -\gamma^0 \gamma^0 \gamma^i \gamma^i = -(1)(-1) = +1$$

(Again, interpret this as a matrix if you must. In vanilla clifford algebra, we would call this just the scalar $1$.)

As a result, exponentials of, say, $\gamma^0 \gamma^1$, yield hyperbolic functions:

$$\exp(\gamma^0 \gamma^1 \phi) = \cosh \phi + \gamma^0\gamma^1 \sinh \phi$$

In clifford algebra parlance, this might be called a "rotor". Note the similarity to quaternions: clifford algebra can generate the quaternion algebra in a similar way, using bivectors of the form $\gamma^i \gamma^j$ for $i \neq j$.

Let $R = \exp(-\gamma^0 \gamma^1 \phi/2)$ and let $R^\dagger = \exp(+\gamma^0 \gamma^1 \phi/2)$. Then a vector $a$ (read: linear combination of the gamma matrices) can be boosted in the following way:

$$a' = RaR^\dagger$$

This result is often proved geometrically, by considering reflection operations in the boost plane. The issue I perceive is that you computed, essentially, the rotor $R$, but not the whole boost map (which is double-sided, as you can see here). Thus, you tried to boost the metric using only half the pieces you should've used.

You should be able to verify that everything I have done here does not require an explicit form of the gamma matrices as matrices; all the properties follow from the basic clifford algebra properties.

A pure treatment of the clifford algebra of Minkowski spacetime should never use the complex imaginary $i$, either. It should only appear in the explicit matrix forms of the basis vectors.

Answer 2

Your calculation only shows that the $\sigma^{\mu\nu}$ do not satisfy the properties of the fundamental representation. If you worked in general dimension, this would already be obvious. The gamma matrices are $2^{\left \lfloor \frac{d}{2} \right \rfloor} \times 2^{\left \lfloor \frac{d}{2} \right \rfloor}$ while the metric (if you interpret it as a matrix) is of course $d \times d$. It is only a miracle that these coinside for $d = 4$.

Answer 3

The $M$s are in the bispinor representation of the Lorentz group. The indices are not spacetime indices, they are spinor indices. The matrices should satisfy something of the form $$M \gamma^\mu M^{-1} = \Lambda_\nu^{\;\mu} \gamma^\nu.$$