Does somebody know how to show that the following equation is Weyl invariant?

$$\gamma^ae_a^\mu D_\mu \Psi=0$$

where: $D_\mu \Psi=\partial_\mu\Psi+A_\mu^{ab}\Sigma_{ab}\Psi$ is the spin-covariant derivative. Under a Weyl transformation the metric changes as $g^{'}_{\mu\nu}=\Omega^2g_{\mu\nu}$, with $\Omega$ positive function. In particular is to me not clear how spinors (and $D_\mu \Psi$) transform.

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    $\begingroup$ Discussion here beginning at page 81 might be useful to you. $\endgroup$
    – twistor59
    Jan 19 '13 at 21:40

The Weyl transformation (unlike diffeomorphisms) does not affect physical fields, only the geometry of space-time. Probably the best way to show that your equation is Weyl-invariant is to build an action which (when varied) yields the equation.

In your example it would be $$ S[\Psi] = \int d^4 x \: \sqrt{-g} \: \bar{\Psi} \gamma^a e^{\mu}_a D_{\mu} \Psi. $$

Then you can compute the stress-energy tensor: $$ T_{\mu \nu} = \bar{\Psi} \gamma^a e_{\nu a} D_{\mu} \Psi. $$

The action changes under Weyl transformation: $$ \delta S = \int d^4 x \: \sqrt{-g} \: \Omega(x) g^{\mu \nu} T_{\mu \nu}, $$

so the theory is Weyl-invariant if (and only if) the stress-energy tensor is traceless. You can see that the condition $T_{\mu}^{\mu} = 0$ in your case is equivalent to the original (Weyl) equation, so it is in fact Weyl-covariant.


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