A question about covariant derivative on spinor field (Vierbein formalism)
Let's use Latin letters $a,b,c,\cdots$ for local <PERSON> frame indices, and Greek letters $\mu,\nu\,\lambda,\cdots$ for coordinate indices.
On one hand, we all know that the covariant derivatives for 1/2-spinor field can be written as: $$ D_\mu\psi=\partial_\mu\psi+\frac{1}{2}(\omega_{ab})\mu\Sigma^{ab}\psi $$ where $\Sigma^{ab}=\frac{1}{4}[\gamma^a,\gamma^b]$ is the <PERSON> group generator, while $(\omega{ab})\mu$ is the connection for local <PERSON> frame. In some references, $\Gamma\mu\equiv\frac{1}{2}(\omega_{ab})_\mu\Sigma^{ab}$.
On the other hand, from Riemannian geometry, covariant derivative for tangent vector can be written as: $$ \nabla_\mu V^\lambda=\partial_\mu V^\lambda+\Gamma^\lambda_{\nu\mu}V^\nu $$ where $\Gamma^\lambda_{\nu\mu}$ is <PERSON> symbol.
If one denotes the local frame transformation as $\theta^a_\mu$ and its inverse $e^\nu_b$, one can find the relationship between $(\omega_{ab})i$ and $\Gamma^\lambda{\nu\mu}$: $$ (\omega^a_{b})\mu=\theta^a\lambda\nabla_\mu e^\lambda_b $$
All the formula above can be found in any differential geometry textbooks or general relativity textbooks.(e.g. N.D.Birrell, P.C.W.Davies-Quantum Fields in Curved Space)
Here's the problem, we can prove that the Laplacian operator on a scalar field $\phi$ is $$ \Delta \phi=\nabla^\mu\nabla_\mu \phi=\frac{1}{\sqrt{|g|}} \partial^\mu\left(\sqrt{|g|}\partial_\mu \phi\right) $$ But for 1/2-spinor field $\psi$, is $$ \Delta \psi=D^\mu D_\mu \psi\tag 1 $$ true?
Or, is $$ \Delta \psi= D!!!/^2\psi\equiv(\gamma^\mu D_\mu)^2 \psi\tag 2 $$ true?
p.s. (1) and (2) seems not be the same since $[D,\gamma]\ne 0 $
**As an example, I have calculated the $\Gamma_\mu$ for a spherical surface:$S^2\times R$, I've got $\Gamma_\mu=(0,0,-\frac{i}{2}\gamma^0\cos\theta)$, when I have taken $\gamma^a=(\sigma_3,i\sigma_2,-i\sigma_1)$, but I find both (1) and (2) are wrong. 818