# Chern-Simons, General relativity, and notations

Consider the Einstein-Maxwell-dilaton theory with an additional Chern-Simons term as in this paper

$$$$S = \int d^4 \sqrt{-g} \left[ \frac{1}{2} R - \frac{1}{2} (\partial\varphi)^2 - \frac{\tau(\varphi)}{4} F^2 - V(\varphi) \right] -\frac{1}{2} \int \theta(\varphi) F\wedge F.$$$$ with $$F=d A$$ being the field strength. Take for example the gauge equations:

$$$$d(\tau\star F + \theta F) =0.$$$$

I have two questions coming from my ignorance on the topic:

1. What does the operation $$\wedge$$ mean and what is the explicit expression for $$F \wedge F$$?
2. What does the operation $$\star$$ mean and what is the explicit expression for $$\tau\star F$$?

The $$\wedge$$ in $$F\wedge F$$ is the wedge product of two copies of the curvature 2-form $$F= \frac 12 F_{\mu\nu} dx^\mu \wedge dx^\nu=dA.$$ where $$A= A_\mu dx^\mu$$ and $$dA= d(A_\nu dx^\nu ) = (\partial_\mu A_\nu) dx^\mu\wedge dx^\nu= \frac 12 (\partial_\mu A_\nu-\partial_\nu A_\mu) dx^\mu\wedge dx^\nu.$$ Thus $$F\wedge F= \frac 14 F_{\mu\nu}F_{\rho\sigma}dx^\mu \wedge dx^\nu \wedge dx^\rho\wedge dx^\sigma = \frac 14 \epsilon^{\mu\nu\rho\sigma} F_{\mu\nu}F_{\rho\sigma}d^4x$$
The $$\star$$ is the Hodge star dual and $$\tau$$ is just multiplication by the function $$\tau(\phi)$$.