# Local lorentz tranformations of fermion action

1. Aug 18, 2010

### jdstokes

The action for a fermion in curved spacetime is

$S = -\int d^4 x \sqrt{- \det(\eta^{ab} e_{a\mu}e_{b\nu})} \left[ i\overline{\psi} e^\mu_a \gamma^a D_\mu \psi + i m \overline{\psi}\psi \right]$

where $g_{\mu\nu} = \eta^{ab} e_{a\mu} e_{b\nu}$ and the derivative operator acting on fermions is given by

$D_\mu \psi = \partial_\mu \psi - \frac{i}{2} \omega_{ab\mu} S^{ab} \psi$

where $S^{ab} = -(i/4)[\gamma^a,\gamma^b]$.

I am failing to show that $D_\mu \psi$ transforms covariantly under local Lorentz transformations.

As far as I understand, the relation $\omega_{ab\mu} = e_a^\nu \nabla_\mu e_{b\nu}$ implies that the following transformation rules

$\psi(x) \mapsto \Lambda_{1/2}(x) \psi(x)$
$e_{a\mu}(x) \mapsto {\Lambda^a}_b(x) e_{b\mu}(x)$
$\omega_{ab\mu}(x) \mapsto {\Lambda^{\alpha}}_a(x) \omega_{\alpha\beta\mu}(x) {\Lambda^\beta}_b(x) + \eta_{\alpha\beta}{\Lambda^\alpha}_a(x)\partial_\mu {\Lambda^\beta}_b(x)$.

Plugging these transformations into $D_\mu \psi$ for an infinitesimal Lorentz transformation, for which
${\Lambda^a}_b = \delta^a_b + \theta^a_b$
$\Lambda_{1/2} = 1 + (i/2)\theta_{ab} S^{ab}$
I obtain

$D_\mu \psi \mapsto D_\mu' \psi' = (\partial_\mu \Lambda_{1/2})\psi + \Lambda_{1/2} \partial_\mu \psi - \frac{i}{2}\left[{\Lambda^{\alpha}}_a \omega_{\alpha\beta\mu} {\Lambda^\beta}_b + \eta_{\alpha\beta}{\Lambda^\alpha}_a \partial_\mu {\Lambda^\beta}_b \right]S^{ab}\Lambda_{1/2}\psi$

In infinitesimal form:

$D_\mu' \psi' = \frac{i}{2} \partial_\mu \theta_{ab} S^{ab}\psi + (1 + i/2\, \theta_{ab}S^{ab})\partial_\mu \psi -\frac{i}{2}\left[(\delta^\alpha_a + \theta^\alpha_a)\omega_{\alpha\beta\mu}(\delta^\beta_b + \theta^\beta_b) + \eta_{\alpha\beta} (\delta^\alpha_a + \theta^\alpha_a ) \partial_\mu \theta^\beta_b \right] S^{ab}(1 + i/2\, \theta_{cd}S^{cd})\psi$

Ignorning quadratic terms and using the anti-symmetry of $\theta_{ab}$:

$D_\mu' \psi' = (1 + i/2\, \theta_{ab}S^{ab})\partial_\mu \psi -\frac{i}{2} \left( \omega_{ab\mu} S^{ab} + \frac{i}{2} \omega_{ab\mu} \theta_{cd} S^{ab} S^{cd}\right)\psi - \frac{i}{2}\left( \omega_{a\beta\mu} \theta^\beta_b + \omega_{\alpha b \mu} \theta^\alpha_a\right) S^{ab} \psi$

The first two terms combine to give $(1+ i/2 \, \theta_{cd}S^{cd})(\partial_\mu - i/2\, \omega_{ab\mu}S^{ab})\psi = \Lambda_{1/2} D_\mu \psi$ as required, but the last term does not have anything to cancel with. Does anyone have any suggestions about where I might be going wrong?

Last edited: Aug 18, 2010
2. Aug 18, 2010

### jdstokes

I think the error lies in the fact that I should be considering the transformation of ${\omega^a}_{b}$ rather than $\omega_{ab}$. The transformation rule for ${\omega^a}_{b}$ is

${\omega^a}_{b\mu}(x) \mapsto {\Lambda^\beta}_b(x) {\omega^\alpha}_{\beta\mu}(x) {\Lambda_{\alpha}}^a(x) + [\partial_\mu {\Lambda^\alpha}_b(x)]{\Lambda_\alpha}^a(x)$

expanding this into infinitesimal form gives extra minus signs because of ${\Lambda^a}_b = \delta^a_b + \theta^a_b$ and ${\Lambda_a}^b = \delta^b_a - \theta^b_a$. These cause the last term in the expression for $D_\mu' \psi'$ to cancel.