# Maxwell-Chern-Simons equation: Translating from differential form to component form

I am trying to solve the scalar-coupled Maxwell-CS equations (which is one of the equation of motions in $$N=2$$ supergravity coupled to 3 vector multiplets), which is written in this form in the differential form language $$\begin{eqnarray} 0=d(Q_{IJ}\star F^J)+\frac{1}{4}C_{IJK}F^J\wedge F^K \end{eqnarray}$$ where $$F^I = dA^I$$ with $$I=1,2,3$$ are the 3 Abelian field strengths, $$Q_{IJ}$$ is a matrix of 3 scalars that are coupled to the gauge field strengths, and $$C_{IJK}=\epsilon_{IJK}$$ where $$\epsilon_{IJK}$$ is a totally antisymmetric tensor. First, I'd like to translate this equation back to component form, as I do below.

There are two terms in this equation, which we want to compute in component forms. The first term is $$\begin{eqnarray} d(Q_{IJ}\star F^J) &=& d Q_{IJ}\,\star F^J + Q_{IJ}\, d\star F^J \nonumber\\ % &=& (\partial_\mu Q_{IJ}) \left[\frac{\sqrt{-g}}{24} F^{J\alpha\beta}\epsilon_{\alpha\beta\rho\sigma\tau}\right]dx^\mu \wedge dx^\rho \wedge dx^\sigma \wedge dx^\tau \nonumber\\&&+ Q_{IJ}\left[\frac{1}{24}\left[(\partial_\mu\sqrt{-g}) F^{J\alpha\beta}+ \sqrt{-g}(\partial_\mu F^{J\alpha\beta}) \right]\epsilon_{\alpha\beta\rho\sigma\tau} \,dx^\mu\wedge dx^\rho \wedge dx^\sigma \wedge dx^\tau\right]\nonumber\\ % &=& \frac{1}{24}\left\{ \sqrt{-g}(\partial_\mu Q_{IJ})F^{J\alpha\beta}+ Q_{IJ}(\partial_\mu\sqrt{-g}) F^{J\alpha\beta}+ Q_{IJ}\sqrt{-g}(\partial_\mu F^{J\alpha\beta})\right\}\epsilon_{\alpha\beta\rho\sigma\tau}dx^\mu\wedge dx^\rho \wedge dx^\sigma \wedge dx^\tau\nonumber\\ \end{eqnarray}$$ and the second term is $$\begin{eqnarray} \frac{1}{4}C_{IJK}F^J\wedge F^K &=& \frac{1}{4\,4!}\epsilon^{\mu\nu\rho\sigma\tau}C_{IJK} F^{J}_{\nu\rho}F^{K}_{\sigma\tau} dx^\nu\wedge dx^\rho \wedge dx^{\sigma}\wedge dx^\tau \end{eqnarray}$$ Recall that for a $$r$$-form $$\omega$$

$$\begin{eqnarray} \omega = \frac{1}{r!}\omega_{\mu_1\,...\,\mu_r}dx^{\mu_1}\wedge\,...\,\wedge dx^{\mu_r} \in \Omega^r(M) &\Rightarrow &\star: \Omega^r(M) \Rightarrow \Omega^{d-r}(M) \\&& \star \omega = \frac{\sqrt{|g|}}{r!(d-r)!}\omega_{\mu_1\,...\,\mu_r} \epsilon^{\mu_1\,...\,\mu_r}_{\mu_{r+1}\,...\,\mu_d} dx^{\mu_{r+1}}\wedge \,...\,\wedge dx^{\mu_d} \end{eqnarray}$$

so we have $$\begin{eqnarray}F^J = \frac{1}{2} F^J_{\mu\nu} dx^\mu \wedge dx^\nu &\Rightarrow &\star F^J = \star F^J_{\rho\sigma\tau} dx^\rho\wedge dx^\sigma\wedge dx^\tau =\frac{\sqrt{-g}}{2! 3!}F^J_{\mu\nu}\epsilon^{\mu\nu}_{\rho\sigma\tau} dx^\rho \wedge dx^\sigma \wedge dx^\tau\nonumber\\ % &\Rightarrow & d\star F^J = d\left(\frac{\sqrt{-g}}{2! 3!}F^J_{\mu\nu}g^{\mu\alpha} g^{\nu\beta}\epsilon_{\alpha\beta\rho\sigma\tau}dx^\rho \wedge dx^\sigma \wedge dx^\tau \right)\nonumber\\ &\Rightarrow & d\star F^J = \frac{1}{24}\left[(\partial_\mu\sqrt{-g}) F^{J\alpha\beta}+ \sqrt{-g}(\partial_\mu F^{J\alpha\beta}) \right]\epsilon_{\alpha\beta\rho\sigma\tau} \,dx^\mu\wedge dx^\rho \wedge dx^\sigma \wedge dx^\tau \nonumber \end{eqnarray}$$ Putting the 2 terms in the Maxwell-CS equations together gives the following 15 equations for the 15 components ($$I,\mu$$) where $$(I = 1,2,3; \mu = 0,\,...\,4)$$ $$\begin{eqnarray} 0&=&d(Q_{IJ}\star F^J) +\frac{1}{4}C_{IJK}F^J\wedge F^K \nonumber\\ &=& \frac{1}{24}\left\{ \sqrt{-g}(\partial_\mu Q_{IJ})F^{J\alpha\beta}+ Q_{IJ}(\partial_\mu\sqrt{-g}) F^{J\alpha\beta}+ Q_{IJ}\sqrt{-g}(\partial_\mu F^{J\alpha\beta})\right\}\epsilon_{\alpha\beta\rho\sigma\tau}dx^\mu\wedge dx^\rho \wedge dx^\sigma \wedge dx^\tau\nonumber\\ && +\frac{1}{96}\epsilon^{\mu\nu\rho\sigma\tau}C_{IJK} F^{J}_{\nu\rho}F^{K}_{\sigma\tau} dx^\nu\wedge dx^\rho \wedge dx^{\sigma}\wedge dx^\tau \nonumber\\ \nonumber\\ % \end{eqnarray}$$ At this step, I'm stuck because if I get rid of the basis, the $$dx^\nu\wedge dx^\rho \wedge dx^{\sigma}\wedge dx^\tau$$ part in both terms, the remaining indices still contain the $$\rho\sigma\tau$$ part, which I don't know how to get rid of. There should be only $$I$$ and $$\mu$$ left in both terms. Any help or suggestion is appreciated. Thanks !