shereen1 said:
no i mean gauging conformal group and studying the resulting action if invariant under diffeomorphism group. Do you have any idea about a refernce that could help me in proving this?
Thank you
By construction, the resulting action is invariant under diffeomorphsim, just like the Yang-Mills action on curved spacetime. Collect the 15 infinitesimal generators of SO(2,4) as J_{A} = \{ P_{a} , M_{ab}, K_{a}, D \} , where the index a = 0,1,2,3 is raised by the inverse Minkowski metric \eta^{ab}, then rewrite the Lie algebra so(2,4) in the standard form [J_{A},J_{B}] = C_{AB}{}^{C}J_{C} . The Cartan-Killing metric on so(2,4) is given in terms of the structure constants as G_{AB} = C_{AE}{}^{D} C_{BD}{}^{E} .
In the basis J_{A} , \ \ A = 1,2, \cdots , 15 , define an so(2,4)-valued connection \mathbb{A}_{\mu}(x) = A^{C}_{\mu}(x) J_{C} \equiv e^{a}_{\mu}(x) P_{a} + \omega^{ab}_{\mu}(x) M_{ab} + c^{a}_{\mu}(x) K_{a} + \alpha_{\mu}(x) D . The components of the field tensor \mathbb{F}_{\mu\nu} = F_{\mu\nu}^{C}J_{C} are given as usual by F^{C}_{\mu\nu} = \partial_{\mu}A_{\nu}^{C} - \partial_{\nu}A_{\mu}^{C} + C_{BD}{}^{C} A_{\mu}^{B}A_{\nu}^{D} . Now you can write down the following diffeomorphsim-invariant action
S = - \frac{1}{2 \alpha^{2}_{YM}} \int d^{4}x \ \sqrt{-g} \ g^{\mu\rho}g^{\nu\sigma} \ G_{AB} F^{A}_{\mu\nu}F^{B}_{\rho\sigma} .
