Background
I am interested in the computation of the four-dimensional Einstein-Hilbert action seen as the "inverse" of the Kaluza-Klein procedure. That is, I want to write something like: \begin{equation} \int d^4\hat{x}\sqrt{-\hat{g}} \hat{R} \sim \int d^2 x\sqrt{-g}R+S^{\text{2D}}_\text{other} \tag{1} \end{equation} Where $S^{\text{2D}}_{\text{other}}$ contains all the remaining degrees of freedom and is a two-dimensional action. To do so I supposed the following ansatz. \begin{equation} \hat{g}_{AB}=\left( \begin{matrix} g_{\mu \nu}+\tilde{g}_{ij} L^i_\mu L^j_\nu & \tilde{g}_{ij}L^i_\nu \\ \tilde{g}_{ij}L^j_\mu & \tilde{g}_{ij}\end{matrix} \right),\,\,\,L^i_\alpha = K^i_b A^b_\alpha \tag{2} \end{equation} With $K^i_b$ the Killing vectors of the manifold associated with $\tilde{g}_{ij}$. The Greek indices run from $0$ to $1$, and the lower case Latin indices run from $2$ to $3$. The signature of $\hat{g}_{AB}$ is $(-,+,+,+)$. The document Kaluza-Klein theories treats a similar question but in the case explicated there, there are special dependencies in the position variables for the quantities inside the four-dimensional metric. While here I want the general case which should, as said right after $(2.8)$, display scalars (before compactification), and then massless modes and a tower of massive modes after compactification.
In accordance with the section $\text{III}$ of Elementary features of Kaluza-Klein theories, there should be a term of the form: \begin{equation} \mathcal{L}_{F^2}\sim\beta \tilde{g}_{ij}K^i_a K^j_b F^a_{\mu \nu} F^{a,\mu\nu}, \tag{3} \end{equation} where $F$ is the curvature of the gauge field $A$, and $\beta$ is a simple constant. As seen in the first document I linked, one can easily impose some scaling of the Killing vectors such that there is only $F_{\mu \nu}^a F^{a,\mu\nu}$ that remains in the action.
Calculation
For this calculation, I've been inspired by this PSE post, where one imposes successively some conditions on the form of the answer's counterpart of $\hat{g}_{AB}$. I don't have to impose what I should in order to find $(3)$ since we are assured it will pop out from the maths. The first term of the action is the first term of the RHS of $(1)$ as one would expect. Next I imposed $g_{\mu \nu}=\eta_{\mu \nu}$ the 2D Minkowski metric, and $L^i_\alpha =0$. I've found that the Christoffel symbols that don't vanish are: \begin{align} \Gamma^k_{ij} &= \frac{1}{2}\tilde{g}^{kl}\left( \partial_j \tilde{g}_{il}+ \partial_i \tilde{g}_{jl}-\partial_l \tilde{g}_{ij}\right) \tag{4} \\ \Gamma^k_{i\nu} &= \Gamma^k_{\nu i} = \frac{1}{2} \tilde{g}^{kl} \partial_\nu \tilde{g}_{li} \tag{5} \end{align} The Ricci tensor is then: \begin{align} \hat{R}_{AB} =& \left( \begin{matrix} 0 & \partial_k \Gamma^{k}_{j\nu} \\ \partial_k \Gamma^{k}_{\mu i} & \partial_k \Gamma^{k}_{ij} \end{matrix} \right) - \left( \begin{matrix} \partial_\mu \Gamma^k_{k\nu} & \partial_j \Gamma^k_{k\nu} \\ \partial_\mu \Gamma^k_{ki} & \partial_j \Gamma^k_{ki} \end{matrix} \right) \\ &+\left( \begin{matrix} 0 & \Gamma^l_{lk}\Gamma^k_{j\nu} \\ \Gamma^l_{lk}\Gamma^k_{\mu i} & \Gamma^l_{lk}\Gamma^k_{ij} \end{matrix} \right)-\left( \begin{matrix} \Gamma^l_{\mu k} \Gamma^k_{l \nu} & \Gamma^l_{j k} \Gamma^k_{l \nu} \\ \Gamma^l_{\mu k} \Gamma^k_{l i} & \Gamma^l_{j k} \Gamma^k_{l i}\end{matrix} \right) \tag{6} \end{align} From this it is straightforward to calculate the Ricci scalar in the frame of the imposed metric, $\text{diag}(\eta_{\mu \nu}, \tilde{g}_{ij})$: \begin{equation} \hat{R} = \tilde{R}-\frac{1}{2}\partial^\mu \tilde{g}^{kl}\partial_\mu \tilde{g}_{kl}-\frac{1}{4}\tilde{g}^{il}\tilde{g}^{km}\partial^\mu \tilde{g}_{lk} \partial_\mu \tilde{g}_{mi} \tag{7} \end{equation} Then, the action is for this part the Einstein-Hilbert action of the metric $\tilde{g}$ plus some weird gravitons with quartic potential. So it seems that my action is then of the form: \begin{align} S\sim& \int d^2 x \sqrt{-g} R-\frac{1}{4}\int d^2 x \sqrt{g}F^a_{\mu \nu} F^{a,\mu\nu} \\ &+\int d^4 x \sqrt{\hat{g}}\tilde{R}-\int d^4 x \sqrt{\hat{g}} \left( \frac{1}{2}\partial^\mu \tilde{g}^{kl}\partial_\mu \tilde{g}_{kl}+\frac{1}{4}\tilde{g}^{il}\tilde{g}^{km}\partial^\mu \tilde{g}_{lk} \partial_\mu \tilde{g}_{mi} \right) \tag{8} \end{align} I see nowhere scalar fields, and there is the $\tilde{R}$ that is troublesome. But I suppose they arise from the compactification.
Question
Did I make any mistake in my derivation of this action? How should I treat the compactification in order to get rid of the four-dimensional integrals and of the $\tilde{R}$ term?
EDIT
Ok so, I tried a compactification over a 2-torus to see what will happen, and it seems that $\tilde{R}$ reduces to: \begin{equation} \tilde{R}\leadsto \tilde{g}^{jl}\left[ \Gamma^i_{ik}\Gamma^k_{jl}-\Gamma^i_{lk}\Gamma^k_{ij} \right]\leadsto-\frac{1}{2}\sum_{\vec{n}}M^2_{\vec{n}}\tilde{g}^{(\vec{n})}_{ij}\tilde{g}^{(\vec{n}),ij}, \end{equation} with $M^2_{\vec{n}} \equiv \frac{1}{2}\frac{n_k n^k}{R_k R^k}$ where $R_k$ are the radii of the torus. This means that the $\tilde{R}$ contribution transforms into masses of the massive modes.
However, I don't see how to deal with the quartic interaction term. I tried to express the regular derivatives in the kinetic term as covariant derivatives, but I don't get the correct factor... Does anyone have an idea?
