2
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ Discussion here beginning at page 81 might be useful to you. $\endgroup$
    – twistor59
    Jan 19 '13 at 21:40
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.