The action of the Kaluza–Klein reduction (Chapter 4 of "D-branes (Clifford Johnson)")

Question

In the first part of Section 4, the author gives \begin{equation} S = \frac1{16\pi G^N_{(5)}}\int(-G_{(5)})^{1/2}R^{(5)}d^5x = \frac1{16\pi G^N_{(4)}}\int(-G_{(4)})^{1/2}\Big(R^{(4)} - \frac32\partial_\mu\phi\partial^\mu\phi - \frac14e^{3\phi}F_{\mu\nu}F^{\mu\nu}\Big)d^4x. \end{equation} How is this reduction obtained?

I show the calculation I did below but it does not match: \begin{align} S &= \frac1{16\pi G^N_{(5)}}\int(-G_{(5)})^{1/2}R^{(5)}d^5x = \frac1{16\pi G^N_{(5)}}\int(-G_{(5)})^{1/2} \Big(R^{(4)} - 2e^{-\phi}\nabla^2e^\phi - \frac14e^{2\phi}F_{\mu\nu}F^{\mu\nu}\Big)d^5x \end{align} The five dimensional metric is decomposed as \begin{equation} G_{(5)} = \det\begin{bmatrix} G^{(4)}_{\mu\nu} + G_{44}A_\mu A_\nu& G_{44}A_\mu\\ G_{44}A_\nu& G_{44} \end{bmatrix} = \det\begin{bmatrix} G^{(4)}_{\mu\nu}& 0\\ G_{44}A_\nu& G_{44} \end{bmatrix} = G_{44}\det(G^{(4)}_{\mu\nu}) =: G_{44}G_{(4)},\;\; G_{44} = e^{2\phi}. \end{equation}

$\nabla^2f = g^{\mu\nu}\nabla_\mu\partial_\nu f = g^{\mu\nu}\partial_\mu\partial_\nu f - g^{\mu\nu}\Gamma^\rho_{\mu\nu}\partial_\rho f$, where the connection is changed as \begin{equation} \tilde\Gamma^\rho_{\mu\nu} = \frac12e^{-2\Omega}g^{\rho\lambda} (\partial_\mu(e^{2\Omega}g_{\lambda\nu}) + \partial_\nu(e^{2\Omega}g_{\lambda\mu}) - \partial_\lambda(e^{2\Omega}g_{\mu\nu})) = \Gamma^\rho_{\mu\nu} + 2\delta^\rho_{(\nu}\partial_{\mu)}\Omega - g_{\mu\nu}\partial^\rho\Omega \end{equation} Then the Laplacian changes as \begin{equation} \tilde\nabla^2f = e^{-2\Omega}(g^{\mu\nu}\nabla_\mu\partial_\nu f + (2-4)\partial^\rho\Omega\partial_\rho f) = e^{-2\Omega}(\nabla^2f - 2\partial^\rho\Omega\partial_\rho f) \end{equation}

\begin{align} S &= \frac1{16\pi G^N_{(5)}}\int(-G_{(5)})^{1/2}R^{(5)}d^5x = \frac1{16\pi G^N_{(5)}}\int(-G_{(5)})^{1/2} \Big(R^{(4)} - 2e^{-\phi}\nabla^2e^\phi - \frac14e^{2\phi}F_{\mu\nu}F^{\mu\nu}\Big)d^5x \end{align} The five dimensional metric is decomposed as \begin{equation} G_{(5)} = \det\begin{bmatrix} G^{(4)}_{\mu\nu} + G_{44}A_\mu A_\nu& G_{44}A_\mu\\ G_{44}A_\nu& G_{44} \end{bmatrix} = \det\begin{bmatrix} G^{(4)}_{\mu\nu}& 0\\ G_{44}A_\nu& G_{44} \end{bmatrix} = G_{44}\det(G^{(4)}_{\mu\nu}) =: G_{44}G_{(4)},\;\; G_{44} = e^{2\phi}. \end{equation}

The transformation of the curvature under the conformal transformation is for $\tilde g_{\mu\nu} = e^{2\Omega}g_{\mu\nu}$, $\tilde R = e^{-2\Omega}[R - 2(D-1)\nabla^2\Omega - (D-2)(D-1)\partial_\mu\Omega\partial^\mu\Omega]$. Now applying it for $g_{\mu\nu} = G^{(4)}_{\mu\nu}$ and $D=4$ gives \begin{equation} R^{(4)} = e^{2\Omega}(\tilde R^{(4)} + 6\nabla^2\Omega - 6\partial_\mu\Omega\partial^\mu\Omega). \end{equation} and the determinant factor changes as $(-G_{44}G_{(4)})^{1/2} = (-e^{2\phi}e^{-8\Omega}\tilde G_{(4)})^{1/2}$.

\begin{align} S &= \frac{2\pi R}{16\pi G^N_{(5)}}\int(-G_{(4)})^{1/2}e^{\phi-4\Omega} \Big(e^{2\Omega}\tilde R^{(4)} + 6e^{2\Omega}\nabla^2\Omega - 6e^{2\Omega}\partial_\mu\Omega\partial^\mu\Omega\nonumber\\ &\hspace{5cm} - 2e^{-\phi}e^{2\Omega}(\nabla^2e^\phi + 2\partial^\rho\Omega\partial_\rho e^\phi) - \frac14e^{2\phi+4\Omega}F_{\mu\nu}F^{\mu\nu}\Big)d^4x. \end{align} By setting $\Omega = \phi/2$, the terms including $\Omega$ and $\phi$ are \begin{equation} 3\nabla^2\phi - \frac32\partial_\mu\phi\partial_\mu\phi - 2e^{-\phi}(\nabla^2e^\phi + \partial^\mu\phi\partial_\mu e^\phi) \rightarrow - \frac72\partial_\mu\phi\partial_\mu\phi, \end{equation} where we removed the total derivative terms.

Answer 1

Unfortunately I don't have access to the text mentioned, so I'll show a similar calculation I've done in the past for the standard Kaluza-Klein compactification from $D=d+1$ to $d$ dimensions. This might be slightly different to the calculation in Johnson's book, but it's the best I can do without the text. It gives the answer required and also matches the answer given in many popular works, e.g. here, Eq (582). Note that usually one rescales the fields so that the kinetic terms are normalised. I assume Johnson does this later at some point, and it is a simple step.

Let me know if you have any confusion with my slightly different notation. (I use $R_D$ and $G$ for the $D$-dimensional Ricci scalar and metric respectively, etc).

Starting with the action \begin{equation} \tag{1} S = \frac{1}{\kappa_{D}^2} \int d^{D}x \sqrt{-G}R_{D} \ , \end{equation} and the metric \begin{equation} \tag{2} ds_{0}^2 = G_{MN}(x)dx^M dx^N = g_{\mu \nu}(x)dx^{\mu}dx^{\nu}+ \textrm{e}^{2\Phi(x)}\Big(A_{\mu}(x)dx^{\mu}+dy \Big)^2 \ , \end{equation} let us compactify onto a circle of radius $L$. First we decompose the Ricci scalar, \begin{equation} \tag{3} R_{D} = R_{d} - 2\textrm{e}^{-\Phi} \nabla^{2}\textrm{e}^{\Phi} - \frac{1}{4} \textrm{e}^{2 \Phi} F_{\mu \nu} F^{\mu \nu} \ , \end{equation} where $R_d$ now denotes the curvature in $d$ dimensions, and $F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$ is the field strength in $d$ dimensions. The covariant derivative squared is defined by \begin{equation} (\nabla^2 \Phi) = \nabla_{\mu} \nabla^{\mu} \Phi \ , \end{equation} which we can discarded as a total derivative in the transformation above: i.e. in the action (1) this term appears as \begin{equation} -2 \int d^dx \sqrt{-g} \nabla^2 \textrm{e}^{\Phi} = -2 \int d^dx ~ \partial_{\mu} (\sqrt{-g} \nabla^{\mu}\textrm{e}^{\phi}) = 0 \ . \end{equation}

Now let us go back and deal with the metric determinant which is simply \begin{equation} \tag{4} \sqrt{-G} = \sqrt{-g \textrm{e}^{2 \Phi}} = \textrm{e}^{\Phi} \sqrt{-g} \ . \end{equation}

Putting (3) and (4) back into the action and integrating out the $y$ coordinate (the $D^{th}$ dimension) gives the compactified $d$-dimensional action, \begin{equation} \tag{5} S_{(0)} = \frac{2 \pi L}{\kappa_{D}^2} \int d^d x \sqrt{-g} \textrm{e}^{\Phi} \Big[R_{d} - \frac{1}{4} \textrm{e}^{2 \Phi} F_{\mu \nu} F^{\mu \nu} \Big] \ , \end{equation} the $2 \pi L$ coming from the circumference of the compactified dimensions (I think this may be absorbed into the $G^{N}_{4}$ in your question but I'm not sure). The action is in the Jordan frame, so we apply the conformal/Weyl rescaling of the metric to write the action in the Einstein frame (i.e. with the usual minimal coupling of the EH-action), \begin{equation} \tag{6} g_{\mu \nu} = \Omega^2 \tilde{g}_{\mu \nu} \ \ \textrm{where} \ \ \Omega^2 = \textrm{e}^{2 \omega} \ . \end{equation} We easily find $\omega$ by seeing that \begin{equation} \sqrt{-g} \cdot \textrm{e}^{\Phi} \cdot R_d \rightarrow \sqrt{-\tilde{g}} \textrm{e}^{d \omega} \cdot \textrm{e}^{\Phi} \cdot \tilde{R}_d \textrm{e}^{-2 \omega} = \sqrt{-\tilde{g}} \tilde{R}_d \ , \end{equation} \begin{equation} \Rightarrow \omega = \frac{\Phi}{2-d} \ . \end{equation} The Ricci scalar and metric determinant then become \begin{equation} \tag{7} R_d = \textrm{e}^{\frac{-2 \Phi}{2-d}} \Big[\tilde{R}_d - \frac{2(d-1)}{2-d} \tilde{\nabla}^2 \Phi - \frac{d-1}{d-2} \tilde{g}^{\mu \nu}\tilde{\nabla}_{\mu}\Phi \tilde{\nabla}_{\nu} \Phi \Big] \ \ ; \ \ \ \sqrt{-g}=\sqrt{-\tilde{g}} \textrm{e}^{\frac{d \Phi}{2-d}} \ \ , \end{equation} where we've used the standard conformal transformation formula for the Ricci scalar. The field strength terms $F_{\mu \nu}$ transform under the conformal rescaling in the action like \begin{equation} \tag{8} %F^{\mu \nu}F_{\mu \nu} = g^{\mu \rho} g^{\nu \sigma} F_{\rho \sigma} g_{\mu \rho} g_{\nu \sigma} F^{\rho \sigma} = \textrm{e}^{-4 \omega} \tilde{g}^{\mu \rho} \tilde{g}^{\nu \sigma} \tilde{F}_{\rho \sigma} \textrm{e}^{4 \omega} \tilde{g}_{\mu \rho} \tilde{g}_{\nu \sigma} \tilde{F}^{\rho \sigma} = \tilde{F}_{\rho \sigma} \tilde{F}^{\rho \sigma} \ . \sqrt{-g} g^{\mu \rho} g^{\nu \sigma} F_{\rho \sigma} F_{\rho \sigma} = \sqrt{-\tilde{g}} \textrm{e}^{d \omega} \textrm{e}^{-2 \omega} \textrm{e}^{-2 \omega} \tilde{g}^{\mu \rho} \tilde{g}^{\nu \sigma} \tilde{F}_{\mu \nu} \tilde{F}_{\rho \sigma} = \sqrt{-\tilde{g}} \textrm{e}^{\frac{\phi (d-4)}{2-d}} \tilde{g}^{\mu \rho} \tilde{g}^{\nu \sigma} \tilde{F}_{\mu \nu} \tilde{F}_{\rho \sigma} \ , \end{equation} which is invariant in $d=4$ dimensions.

Substituting (7) and (8) into the compactified action (5), and dropping the total derivative again, gives the $d$-dimensional action in the Einstein frame, \begin{equation} \tag{9} S_{(0)}^E = \frac{2 \pi L}{\kappa_{D}^2} \int d^d x \sqrt{-\tilde{g}} \Big[\tilde{R}_d - \Big(\frac{d-1}{d-2} \Big) (\tilde{\nabla} \Phi)^2 - \frac{1}{4} \textrm{e}^{\frac{\phi(4d-10)}{(d-2)}} \tilde{F}^2 \Big] \ \ , \end{equation} where $F^2=F_{\mu \nu}F^{\mu \nu}$ and $(\nabla \Phi)^2 = g^{\mu \nu} \nabla_{\mu} \Phi \nabla_{\nu} \Phi $. Now simply put in $d=4$ and we arrive at \begin{equation} \tag{10} S_{(0)}^E = \frac{2 \pi L}{\kappa_{D}^2} \int d^d x \sqrt{-\tilde{g}} \Big[\tilde{R}_d - \frac{3}{2}\partial_{\mu} \Phi \partial^{\mu} \Phi - \frac{1}{4} \textrm{e}^{3\Phi} \tilde{F}^2 \Big] \ \ . \end{equation} where we've noted that $\Phi$ is a scalar so $\tilde{\nabla}_{\mu} \Phi = \partial_{\mu} \Phi$. You can then go onto to rescale the field $\Phi$ to give the correct kinetic term, but this answer is already very long and the details are in the reference at the end. You might have to work out how the gravitational coupling constants align with the text you're working from.