Reducibility of Lorentz group generators, contrasted with irreducible gamma-matrix representation

After introducing the gamma matrices as $$\gamma^0=-i \pmatrix{\begin{matrix} 0 & \Bbb I_{2x2} \\ \Bbb I_{2x2} & 0 \\ \end{matrix}} , \qquad \gamma^i=-i\pmatrix{\begin{matrix} 0 & \sigma^i \\ -\sigma^i & 0 \\ \end{matrix}}$$ we can find the matrix representation of the generators of the homogeneous Lorentz group in the spinor representation (i.e., $$J^{\mu\nu}=-\frac{i}{4}[\gamma^{\mu},\gamma^{\nu}])$$ so: $$J^{ij}=\frac{1}{2}\varepsilon_{ijk}\pmatrix{\begin{matrix} \sigma^k & 0 \\ 0 & \sigma^k \\ \end{matrix}}\quad\text{ and }\quad J^{i0}=\frac{i}{2}\pmatrix{\begin{matrix} \sigma^i & 0 \\ 0 & -\sigma^i \\ \end{matrix}}.$$ We see that the matrices of the generators are block diagonal in this form so the representation is reducible. On the other hand we know that 4x4 gamma matrices in the form above furnish an irreducible representation of Lorentz group, because the maximal number of antisymmetric independent tensors created using gamma matrices is 16 in 4D spacetime, so the minimal dimensionality to represent the gamma matrices are at least form an 4x4 matrix representation — which is why the form we see above furnishes an irreducible representation.

How can we reconcile the fact that the generators $$J$$ are reducible, while the 4x4 gamma matrices constitute an irreducible representation?

• – Qmechanic Apr 3 at 15:53
• The gamma matrices generate Clifford. They do not generate Lorentz. – AccidentalFourierTransform Apr 3 at 19:49