Dirac Equation in General Relativity Dirac equation for the massless fermions in curved spase time is $γ^ae^μ_aD_μΨ=0$, where $e^μ_a$ are the tetrads. I have to show that Dirac spinors obey the following equation:
$$(−D_μD^μ+\frac{1}{4}R)Ψ=0\qquad(1)$$
where $R$ is the Ricci scalar.
I already know that $[D_\mu,D_\nu]A^\rho={{R_{\mu\nu}}^\rho}_\sigma A^\sigma$, but a key point is to know what $[D_\mu,D_\nu]\Psi$ is.
($D_μΨ=∂_μΨ+A^{ab}_μΣ_{ab}$ is the covariant derivative of the spinor field and $Σ_{ab}$ the Lorentz generators involving gamma matrices).
 A: Denoting by $\gamma^a$ the Minkowski space gamma matrices with respect to the Lorentz tetrad $\{e^a\}$, and covariant derivative $D_a$, then the gammas are covariantly constant.
Start with the massless Dirac equation $$ \gamma^{b}D_{b}\Psi = 0$$
Act again with the Dirac operator $$\gamma^{a}D_{a}\gamma^{b}D_{b}\Psi=0 $$  So, since $D$ annihilates $\gamma$ $$\gamma^{a}\gamma^{b}D_{a}D_{b}\Psi = 0 $$ so $$\frac{1}{2}\{\gamma^{a},\gamma^{b}\}D_{a}D_{b}\Psi + \frac{1}{2}\gamma^{a}\gamma^{b}[D_a,D_b]\Psi = 0 \ \ (1) $$  But $$\{\gamma^{a},\gamma^{b}\}=2\eta^{ab} $$ and $$ [D_a,D_b]\Psi = {\mathcal{R}_{ab}}\Psi $$ Where ${\mathcal{R}}_{ab}$  is the spin-curvature (antisymmetric in a and b).   ${\mathcal{R}}_{ab}$ satisfies the identity $$ -\gamma^b{\mathcal{R}}_{ab} = {\mathcal{R}}_{ab}\gamma^b = \frac{1}{2}\gamma^b R_{ab}$$ where $R_{ab}$ is the Ricci tensor (in the Lorentz tetrad). so  (1) becomes $$ [D^aD_a+\frac{1}{4}\gamma^a\gamma^bR_{ab}]\Psi = 0 $$ i.e. $$ [D^aD_a-\frac{1}{4}R]\Psi = 0 $$
