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

\begin{equation} 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. \end{equation} with $F=d A$ being the field strength. Take for example the gauge equations:

\begin{equation} d(\tau\star F + \theta F) =0. \end{equation}

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 $\tau\star F$?
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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 $\tau(\phi)$.

If you plan to work in an area using these things, you need to learn the calculus of differential forms. Indeed, even if you do not plan such work, you should learn this calculus as it opens many doors.

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