Ernst potential from Kaluza-Klein reduction of axisymmetric space-time

Question

Following appendix A of "Ergoregions in Magnetised Black Hole Spacetimes" by G. W. Gibbons, A. H. Mujtaba and C. N. Pope, starting from the Lagrangian

$$\mathcal{L} = \hat{R} - \hat{F}_{\mu\nu}\hat{F}^{\mu\nu},$$ metric and field

$$\mathrm{d}\hat{s}^2 = e^{2\phi} \mathrm{d}s^2 + e^{-2\phi}(\mathrm{d}z + 2\mathcal{A})^2,\\ \hat{A} = A + \chi(\mathrm{d}z +2 \mathcal{A}),$$

after Kaluza-Klein reduction, with Killing vector $K\equiv\partial_z$ corresponding to a spatial dimension, we obtain the reduced lagrangian. After using Lagrange multipliers and dualizing the fields, we obtain $$\hat{F} = -e^{2\phi} \star \mathrm{d}\psi + d \chi \wedge (\mathrm{d}z + 2\mathcal{A}),\quad e^{-2\phi} \star F = \mathrm{d} \psi, \quad F \equiv \mathrm{d} A + 2\chi \mathrm{d}\mathcal{A},$$

where the hatted quantities are 4-dimensional, none of the fields depends on $z$ and $\star$ is the Hodge dual with respect to $\mathrm{d}s^2$. At the end, they define the complex Ernst potential by $d \Phi = i_K(\hat{\star}\hat{F} + \mathrm{i} \hat{F})$, $\Phi = \psi + \mathrm{i}\chi$, where $i_K$ is the interior product by $K$. $i_K \hat{F} = -\mathrm{d} \chi$ is easy to derive, but $\mathrm{d}\psi = i_K \hat{\star} \hat{F}$ has proved to be not so easy.

My question is about this last equation. Considering that $\det \hat{g}_{MN} = e^{4\phi} \det g_{\mu\nu}$, I obtain $$i_K \hat{\star} \hat{F} = - e^{2\phi} \star(F + 2 \mathrm{d} \chi \wedge \mathcal{A}),$$ which is obviously wrong. Could somebody provide some hint about this last equation? It must be easy to derive, but I cannot see the way at the moment.

Answer 1

We have $i_K\hat{\star}d \psi = \star d\psi$. Indeed, $$i_K \hat{\star} d \psi = (d \psi)^a \hat{\epsilon}_{abcd} K^b = (d\psi)_e \hat{g}^{ea} e^{2\phi} \epsilon_{acd} = (d\psi)_e g^{ea} \epsilon_{acd} = \star d \psi,$$ since $\hat{\epsilon}_{abcd} K^b = \hat{e}^0\wedge \hat{e}^1 \wedge \hat{e}^2 \wedge i_K\hat{e}^z = e^{2\phi} e^0\wedge e^1 \wedge e^2 = e^{2\phi}\epsilon_{acd}$. Also, the identity $$i_K \hat{\star} d\psi = \hat{\star} (d\psi\wedge K)$$ holds, and if we use it in the definition of $\hat{F}$, we obtain $\hat{F} = e^{2\phi}[- \hat{\star}(d \psi\wedge K) + d \chi\wedge K]$, from which the equation $i_K\hat{\star}\hat{F} = - d\psi$ follows. We observe that there is a sign error in the article. Actually, the definition of $\Phi$ is wrong, but this only becomes apparent if one attempts to relate it to the Ernst potentials.