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 [itex]SO(2,4)[/itex] as [tex]J_{A} = \{ P_{a} , M_{ab}, K_{a}, D \} ,[/tex] where the index [itex]a = 0,1,2,3[/itex] is raised by the inverse Minkowski metric [itex]\eta^{ab}[/itex], then rewrite the Lie algebra [itex]so(2,4)[/itex] in the standard form [tex][J_{A},J_{B}] = C_{AB}{}^{C}J_{C} .[/tex] The Cartan-Killing metric on [itex]so(2,4)[/itex] is given in terms of the structure constants as [tex]G_{AB} = C_{AE}{}^{D} C_{BD}{}^{E} .[/tex]
In the basis [itex]J_{A} , \ \ A = 1,2, \cdots , 15[/itex] , define an [itex]so(2,4)[/itex]-valued connection [tex]\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 .[/tex] The components of the field tensor [itex]\mathbb{F}_{\mu\nu} = F_{\mu\nu}^{C}J_{C}[/itex] are given as usual by [tex]F^{C}_{\mu\nu} = \partial_{\mu}A_{\nu}^{C} - \partial_{\nu}A_{\mu}^{C} + C_{BD}{}^{C} A_{\mu}^{B}A_{\nu}^{D} .[/tex] Now you can write down the following diffeomorphsim-invariant action
[tex]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} .[/tex]