I take a spacetime of the form $\mathcal{M}_{d+1}\times \mathbb{S}^n$, with $\mathcal{M}_{d+1}$ some generic non-compact $(d+1)$-dimensional spacetime and $\mathbb{S}^n$ an $n$-dimensional sphere, so that its metric has the following form:

$$\mathrm{d}s^2=\tilde{g}_{\mu\nu}(x)\mathrm{d}x^\mu\mathrm{d}x^\nu+\rho^2(x)\mathrm{d}\Omega_{(n)}^2\equiv G_{AB}\mathrm{d}X^A\mathrm{d}X^B. \tag{1}$$

Where $\rho^2(x)$ depends only on $x^\mu$ coordinates, that is to say it doesn't depend on the coordinates of the sphere. The Einstein-Hilbert action in $d+1+n$ dimensions is :

$$S\propto \int\mathrm{d}^{d+1+n}x\sqrt{-G}(\mathcal{R}^{(G)}-2\Lambda). \tag{2}$$

With $\mathcal{R}^{(G)}$ the scalar curvature of the $G_{AB}$ metric, and $\Lambda$ the cosmological constant. I want now to reduce the action on the sphere $\mathbb{S}^n$. I understand how to do the first steps, that is to say write, integrate over the sphere coordinates by separating $\mathrm{d}^{d+1+n}x=\mathrm{d}^{d+1}x\mathrm{d}^n\Omega$, and also simplify the determinant to $\sqrt{-G}=\sqrt{-\tilde{g}}\rho^n(x)$, so that I get :

$$S\propto \Omega_{(n)}\int\mathrm{d}^{d+1}x\,\sqrt{-\tilde{g}}\rho^2(x)\left(\mathcal{R}^{(G)}-2\Lambda\right). \tag{3}$$

But I do not see how to simplify this further. In particular, I think there should be a clever way to reduce $\mathcal{R}^{(G)}$ to $\mathcal{R}^{\tilde{g}}$ plus some terms proportional to $(\partial\rho)^2$ and $\rho$, but I don't see how to do this properly, and without redoing from scratch all the computations for the Christoffel symbols, Riemann tensor, Ricci tensor and scalar curvature.

Can someone please help me? I have found some derivations for the one-dimensional reduction on a circle, but usually it's done on some scalar field $\phi$ to show that some massive tower of states arise, and not about the Einstein-Hilbert action itself - and I have found close to nothing for the sphere case.

EDIT: After some more research, I've found a more detailed paper, and realized reading the section about the dimensional reduction of the $\mathcal{N}=4$ supergravity Lagrangian from D=7 to D=11 (around p.59 of the paper) that people were always doing KK reduction for very general ansatz of metrics, with off-diagonal terms which give rise to the gauge potentials in lower dimensions. But in my case, I really have a block-diagonal metric which should give only some dilaton $\varphi$ in $d+1$ dimensions, that I relate to $\rho$ via $\rho(x)=e^{\varphi(x)}$. That it why I decided to try to do the computations the hard way, i.e. by computing explicitly the Christoffel symbols, the Riemann tensor and the Ricci tensor components to explicitly get the $\rho$ dependence, and then contracted everything. The action I get in this case is then:

$$S\propto\Omega_{(n)}\int\mathrm{d}^{d+1}x\sqrt{-\tilde{g}}\left[\rho^n\mathcal{R}^{(\tilde{g})}+n(n-1)\rho^{n-2}(\partial\rho)^2+n(n-1)\rho^{n-2}-2\rho^n\Lambda \right] \tag{4}$$

Where $(\partial\rho)^2\equiv\tilde{g}^{\mu\nu}\partial_\mu\rho\partial_\nu\rho=G^{MN}\partial_M\rho\partial_B\rho$ since $\rho$ depends only on the coordinates of $\mathcal{M}_{d+1}$.

The action is in the string frame, so to go back in the usual Einstein Fram, I do a Weyl rescaling

$$\tilde{g}_{\mu\nu}\to g_{\mu\nu}=e^{2\omega(x)}\tilde{g}_{\mu\nu} \text{ with } \omega(x):=\frac{n}{d-1}\varphi(x) \tag{5}$$

To finally get:

$$S\propto \frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}\int\mathrm{d}^{d+1}x\sqrt{-g}\left[\mathcal{R}^{(g)}+\left(\frac{dn^2}{d-1}+n(n-1)\right)(\bar{\partial}\varphi)^2+\left(n(n-1)e^{-2\varphi}-2\Lambda\right)e^{-\frac{2n}{d-1}\varphi}\right] \tag{6}$$

Where I denote by $\bar{\partial}^\mu=e^{\frac{2n}{d-1}\varphi}\partial^\mu$ a contravariant vector with an index lifted with the rescaled metric $g_{\mu\nu}$. That seems to be exactly what we were searching for, since we get a kinetic term for $\varphi$ with $(\bar{\partial}\varphi)^2$ and an exponential potential $V(\varphi)$:

$$V(\varphi):=\left(n(n-1)e^{-2\varphi}-2\Lambda\right)e^{-\frac{2n}{d-1}\varphi} \tag{7}$$

I am not 100% sure that those are really the right coefficients, and I guess there can be some subtleties with the kinetic term to get the canonical normalization. Has someone already done this calculation, or maybe now that the problem is identifies more clearly, has someone seen some article where this computation was done to double check? Otherwise, I guess I mostly got my answer.

  • $\begingroup$ Interesting question. I also would like to see the calculations details. I expect to see some kind of Yang-Mills field lagrangian to be separated from this $\mathcal{R}^{(G)}$. Also, I don't think you could already integrate out the $n$ angular variables and extract that $\Omega_{(n)}$ in your last equation. Please, add a tag to each of your equations, using the command \tag{#}. $\endgroup$
    – Cham
    Jun 12, 2019 at 19:20
  • $\begingroup$ Theoretically, I know that I should find something like d+1 dimensional scalar curvature plus some kinetic and potential term for $\rho$, which would then play the role of the dilation after some Weyl rescaling - but that's just the idea, I don't see how to do this properly. As for the angular coordinates, nothing depends on them in the action, so theoretically I could integrate them out. Thank you for the edit proposition, it's corrected! $\endgroup$
    – Hakanaou
    Jun 12, 2019 at 20:44
  • $\begingroup$ How do you know that the scalar $\mathcal{R}^{(G)}$ doesn't depend on the $n$ angular coordinates? The metric $G$ depends on them (multiple $\sin{\vartheta} \ldots$). $\endgroup$
    – Cham
    Jun 13, 2019 at 0:15
  • $\begingroup$ Yes, I see your point. I supposed you could do some kind of change of variables to get the sphere in Euclidean coordinates, but that might alter the expression of $\mathcal{R}^{(G)}$ also, so maybe it's not so simple. $\endgroup$
    – Hakanaou
    Jun 13, 2019 at 10:59
  • $\begingroup$ can't you take the general KK ansatz and set vector fields to zero? $\endgroup$
    – Kosm
    Jun 14, 2019 at 3:42

1 Answer 1


This is a very interesting question, and though your answer is mostly correct, there are a few errors in addition to a missing conceptual simplification. I will pick up from Eqn$.~(2)$. We begin with the action for the product spacetime $\mathcal{M}_m = \mathcal{M}_{d+1}\times \tilde{\mathbb{S}}_n$ where the standard $n$-sphere is deformed by a smooth function $\rho(x)$ which depends only on the non-compact $d+1$ directions. The action in this case is

$$S\propto \int\mathrm{d}^{d+1+n}x\,\sqrt{-G}\left(\mathcal{R}^{(G)}-2\Lambda\right). \tag{2}$$

where I have not yet integrated out the angular dependence.

We may decompose the scalar curvature as $\mathcal{R}^{(G)} = G^{MN}\mathcal{R}_{MN} = G^{\mu\nu}\mathcal{R}_{\mu\nu} + G^{ij}\mathcal{R}_{ij}=\mathcal{R}^{(\tilde{g})}+\mathcal{R}^{(\tilde{\mathbb{S}}_n)}$ owing to the fact that the metric is block diagonal. To answer Cham's question, since the isometry group of the metric contains a subgroup which is invariant under spherical transformations, we have killing vectors along these spherical directions. In other words, "the intrinsic curvature along the $n$-sphere is the same everywhere" so there cannot be any special dependence on the angular variables. Now since the Lagrangian is independent of the spherical directions, this justifies expressing the action, correcting the typo on $\rho$, as

$$S\propto \Omega_{(n)}\int\mathrm{d}^{d+1}x\,\sqrt{-\tilde{g}}\rho^n(x)\left(\mathcal{R}^{(G)}-2\Lambda\right). \tag{3}$$

To illustrate the conceptual simplification, let us see what happens to the scalar curvature along the spherical directions. Suppose for the moment that were considering a standard $n$-sphere, with the relevant metric element being $\text{d}s^2_\Omega = r^2\text{d}\Omega_{(2)}^2$, in that case the scalar curvature is exactly known:

$$\mathcal{R}^{(\mathbb{S}_n)} = \frac{n(n-1)}{r^2}$$

For a deformed sphere $\text{d}s^2_\Omega = \rho(x)^2\text{d}\Omega_{(n)}^2$, the conceptual simplification is that the curvature must be the same since the spherical metrics are conformally equivalent up to a Laplacian term. I find the following expression

\begin{align}\mathcal{R}^{(\tilde{\mathbb{S}}_n)} &= \frac{n(n-1)}{\rho^2}-\frac{n(n-1)}{p-(d+1)}\left(\frac{1}{\sqrt{-G}}\partial_M\left[\sqrt{-G}\,{G}^{MN}\partial_N(\log(\rho))\right]\right)\tag{3.1}\\&=\frac{n(n-1)}{\rho^2}-{\beta_{n}}\nabla_M X^M\tag{3.2}\end{align}

where $p=n+d$ is the total number of spatial dimensions. The action then becomes


Depending on the fall off behavior of $\rho$ (along with the fact that the action integral considers dynamics in the bulk), the boundary piece may be subtracted off leaving a finite contribution. After the same Weyl rescaling,

$$S\propto \Omega_{(n)}\int_{\mathcal{M}_{d+1}}\mathrm{d}^{d+1}x\sqrt{-g}\left[\mathcal{R}^{(g)}+\left(\frac{dn^2}{d-1}\right)(\bar{\partial}\varphi)^2+\left(n(n-1)e^{-2\varphi}-2\Lambda\right)e^{-\frac{2n}{d-1}\varphi}\right] \tag{3.3}$$

where $d>1$. This has both the kinematics of the dilaton and a potential term as you have noted earlier.

A Side Note

For $d=1$, the Weyl rescaling breaks down, but it might be interesting to consider compactification of the $n$-sphere to a $1+1$ theory where pure gravity is already conformal and see what geometries are permissible when matter backreacts on the background. Again, great question!


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