There is a quantity that we can integrate over a 4-manifold $$ \frac{1}{8 \pi^2}\int_{M} Tr(R \wedge R ) $$ such that the $R$ is a Riemann curvature two form or tensor. The trace $Tr$ is over the Lie algebra of the tangent space of manifold $M$. Typically $so(4)$ or $so(3,1)$ in physics.
Question:
- How do we write this expression $ \frac{1}{8 \pi^2}\int_{M} Tr(R \wedge R )$ expicitly in terms of the Riemann tensor with 4 indices? Namely by employing this: $$ R^{\rho}{}_{\sigma\mu\nu} = \partial_{\mu}\Gamma^{\rho}{}_{\nu\sigma} - \partial_{\nu}\Gamma^{\rho}{}_{\mu\sigma} + \Gamma^{\rho}{}_{\mu\lambda}\Gamma^{\lambda}{}_{\nu\sigma} - \Gamma^{\rho}{}_{\nu\lambda}\Gamma^{\lambda}{}_{\mu\sigma}$$
Is that $$ \frac{1}{8 \pi^2}\int_{M} Tr(R \wedge R ) =\frac{1}{8 \pi^2}\int_{M} \epsilon^{\mu\nu \mu'\nu'} R^{\rho}{}_{\sigma\mu\nu} R^{\rho'}{}_{\sigma'\mu'\nu'} d^4 x =\frac{1}{8 \pi^2}\int_{M} \epsilon^{\mu\nu \mu'\nu'} R^{\rho}{}_{\sigma\mu\nu} R^{\sigma}{}_{\rho\mu'\nu'} d^4 x? $$ into the formula.
Is that the effect of the Lie algebra trace $Tr$ is the contraction of indices $..{}^{\rho}{}_{\sigma\mu\nu} ..{}^{\sigma}{}_{\rho\mu'\nu'}$?
Here $\epsilon^{\mu\nu \mu'\nu'} d^4 x= d x^{\mu} dx^{\nu} dx^{\mu'} dx^{\nu'}$.
- It is know that this quantity gives rise to the signature of four-manifold. Does $\frac{1}{8 \pi^2}\int_{M} Tr(R \wedge R )$ apply to other dimensions, other than four-manifold $M_4$? Yes or no? Why or why not?