Let's use Latin letters $a,b,c,\cdots$ for local Minkowski 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 Lorentz group generator, while $(\omega_{ab})_\mu$ is the connection for local Minkowski 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 Christoffel 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.


1 Answer 1


The spinor Laplacian is (1). The square (2) of the Dirac operator minus the spinor Laplacian (1) is a quarter of the scalar Levi–Civita curvature, cf. Weitzenböck-Lichnerowicz identity.

  • $\begingroup$ Thanks, I am able to repeat the result $D\!\!\!\!\!/^2-D^aD_a=R_{sc}/4$. But neither $D\!\!\!\!\!/^2$ nor $D^aD_a$ is $\partial^2_t-\vec{\nabla}^2$, there are additional terms. $\endgroup$
    – bacon
    Dec 19, 2016 at 18:03
  • $\begingroup$ I make sense now, the reasonable Laplacian must be $D^aD_a$. Thanks a lot. $\endgroup$
    – bacon
    Dec 19, 2016 at 18:24

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