 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-48/1/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

M^{ρ σ} = \frac{1}{4} [γ^{ρ}, γ^{σ}] = \frac{1}{2} γ^{ρ} γ^{σ} - \frac{1}{2} η^{ρ σ} I_{4} .

They are antisymmetric in

ρ, σ

. Reordering a third gamma matrix with the Clifford algebra yields

[γ^{ρ} γ^{σ}, γ^{τ}] = 2 η^{σ τ} γ^{ρ} - 2 η^{ρ τ} γ^{σ}, [M^{ρ σ}, γ^{τ}] = η^{σ τ} γ^{ρ} - η^{ρ τ} γ^{σ} .

Use the commutator derivation identity on the two factors in

M^{τν} = \frac{1}{4} [γ^{τ}, γ^{ν}]

[M^{ρ σ}, M^{τν}] = \frac{1}{4} ([[M^{ρ σ}, γ^{τ}], γ^{ν}] + [γ^{τ}, [M^{ρ σ}, γ^{ν}]]) = η^{σ τ} M^{ρ ν} - η^{ρ τ} M^{σ ν} + η^{ρ ν} M^{σ τ} - η^{σ ν} M^{ρ τ} .

This derives every Lorentz algebra commutation relation, so the matrices $M^{ρ σ}$ form its spinor representation. Their finite exponentials describe the connected Spinor representation of the Lorentz group, more precisely the lift to its double cover.

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