Weyl rescaling of massless Dirac equation in curved spacetime

Question

Let us consider a $D$-dimensional spacetime $M$ with metric tensor $g_{\mu\nu}$ whose spin connection $D_\mu$ is defined by $\omega_\mu^{\ ab}$. I would like to prove that under a Weyl rescaling $$ g_{\mu\nu}(x)\longmapsto \tilde g_{\mu\nu}(x)=\Omega(x)g_{\mu\nu}(x), $$ where $\Omega(x)$ is an everywhere-smooth and positive function how are the relevant, the massless Dirac equation $$ i \gamma^\mu D_\mu\psi =0 $$ is invariant, where $\psi$ is a spin-$1/2$ field, namely $$ D_\mu \psi = \partial_\mu \psi+\frac{1}{4}\omega_\mu^{\ ab}\gamma_{ab}\psi,\\ \gamma_{ab}=\frac{1}{2}(\gamma_a\gamma_b-\gamma_b\gamma_a), $$ with suitable conformal weight under conformal rescaling.

Answer 1

Let us recall that the spin connection is given in terms of the following quantity $$ \omega_{\mu\nu\rho}=\frac{1}{2}\big[ e_{\rho a}(\partial_\mu e_\nu^a - \partial_\nu e_\mu^a)-e_{\mu a}(\partial_\nu e_\rho^a-\partial_\rho e_\nu^a)+e_{\nu a}(\partial_\rho e_\mu^a-\partial_\mu e_\rho^a), \big] $$ where $e_\mu^a$ is a set of $D$ one-forms labelled by an index $a=0,1,\ldots,D-1$, satisfying $$ e_\mu^a\eta_{ab}e^b_\nu=g_{\mu\nu} $$ at each point, $\eta_{ab}$ being a diag$(\pm1,\ldots, \pm1)$ with the same signature as $g_{\mu\nu}$. Then, under Weyl rescalings $$ \tilde e_\mu^a=\Omega^{1/2}e_\mu^a $$ and correspondingly $$ \tilde \omega_{\mu\nu\rho}=\Omega\left(\omega_{\mu\nu\rho}+\frac{1}{2}g_{\mu[\nu}Y_{\rho]}\right), $$ where $Y_\mu=\partial_\mu \log\Omega$ and square brackets denote antisymmetrization without weight. Recalling $\omega_{\mu}^{\ ab}=e^{\nu a} e^{\rho b}\omega_{\mu\nu\rho}$, we get $$ \tilde \omega_{\mu}^{\ ab}=\omega_\mu^{\ ab}+\frac{1}{2}e_{\mu}^{[a}Y^{b]}. $$ Assuming $\tilde \psi=\Omega^\sigma \psi$ we have $$ i\tilde \gamma^{\mu}\tilde D_\mu \tilde \psi=i \Omega^{\sigma-1/2}\left( \gamma^\mu D_\mu \psi+ \sigma\, \gamma\cdot Y \psi+\frac{1}{8}\gamma^{[b}Y^{c]}\gamma_{bc}\psi\right). $$ Using the Clifford algebra $\{\gamma_a,\gamma_b\}=2\eta_{ab}$ we obtain $\gamma^a\gamma_a=D$ and hence $\gamma^a\gamma_{ab}=(D-1)\gamma_b$, which in turn yields $$ \gamma^{[b}Y^{c]}\gamma_{bc}=2(D-1)\gamma\cdot Y. $$ Thus, $$ i\tilde \gamma^{\mu}\tilde D_\mu \tilde \psi = i\Omega^{\sigma-1/2}\left[ \gamma^\mu D_\mu \psi + \left(\sigma + \frac{D-1}{4}\right) \gamma\cdot Y\psi\right]. $$ Invariance is therefore achieved for $$ \sigma=\frac{1-D}{4}. $$