As part of a hw problem for a class, we're supposed to be deriving the equivalence given in equation 2.3 of this paper http://arxiv.org/abs/1107.5563. I was wondering if there is some special relation involving the Ricci Curvature in 5d's relationship to one in 4d. Since with a general metric like the one given in 2.1, calculating the Christoffel symbols would seem to be an enormous and not particularly smart idea.

  • $\begingroup$ I do think you need to do the explicit calculation, but it's less work than you think. The full metric is given by $G_{MN}$, but only $G_{yy}$ and $G_{y \mu} \sim A_\mu$ are interesting here. The same thing holds for the Ricci tensor, since ${}^4R$ leaves the $R_{\mu \nu}$ untouched. In any case, the relevant Christoffels will be expressed in terms of $\phi, A_\mu$ only so they cannot be that difficult. $\endgroup$
    – Vibert
    May 15 '13 at 22:47
  • $\begingroup$ Thanks for the reply! Just to clarify, would this be akin to making "corrections" to the curvature to deal with our new dimension(s)? $\endgroup$
    – user15961
    May 16 '13 at 2:11
  • $\begingroup$ @Vibert I think that could actually be an answer. $\endgroup$
    – David Z
    May 16 '13 at 4:39

The Cartan formalism is ideal for working in a great deal of generality, one does need even need to fix the dimension of spacetime. I will start with the Kaluza-Klein metric ansatz, namely,

$$\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu \mathrm{d}x^\nu - e^{2\sigma}\left[\mathrm{d}\psi + A_{\mu}\mathrm{d}x^\mu \right]$$

where $A$ is a potential $1$-form, $\sigma$ is a scalar field (the dilaton) and $g_{\mu\nu}$ is the metric of the purely four-dimensional part of the metric. We begin by defining an orthonormal basis,

$$\omega^\psi = e^\sigma \left[ \mathrm{d}\psi + A\right]$$

and we denote the basis for the $g$ metric as $\omega^a$. Taking exterior derivatives yields,

$$\mathrm{d}\omega^\psi = e^{\sigma} \sigma_{,a} \omega^a \wedge \omega^\psi + e^\sigma F$$

where $F=\mathrm{d}A$ is the $2$-form field-strength. We can read off the components of the spin connection from Cartan's first equation (with the torsionless condition),

$$\mathrm{d}\omega^a = \theta^a_b \wedge \omega^b$$

$$\therefore \theta^\psi_a=\sigma_a \omega^\psi + \frac{1}{2}e^\sigma F_{ab} \omega^b \quad \theta^a_b = \theta^a_{0b} +\frac{1}{2}e^\sigma F^a_b \omega^\psi$$

where $\theta_0$ refers to the pure 4D spin connection. The Ricci tensor is given by Cartan's equation,

$$R^a_b = \mathrm{d}\theta^a_b + \theta^a_c \wedge \theta^c_b$$

After incredibly tedious manipulations, we obtain,

$$R^a_b=\mathrm{d}\theta^a_{0b} +\frac{1}{2}e^\sigma \left[ \sigma_{,c}\omega ^c\wedge \omega^\psi F^a_b + F^a_{b,c}\omega^c \wedge \omega^\psi + F^a_b \left(\sigma_{,c} \omega^c \wedge \omega^\psi + e^\sigma F \right)\right] \\ + \theta^a_{0c} \wedge \theta^c_{0b} + \frac{1}{2}e^\sigma \left[ \theta^a_{0c}\wedge F^{c}_{b} \omega^\psi +F^a_c \omega^\psi \wedge \theta^c_{0b} \right]\\ + \frac{1}{2}\sigma^{,a}\omega^\psi \wedge e^\sigma F_{bc}\omega^c + \frac{1}{2}F^a_c \omega^c \wedge \sigma_{,b}\omega^\psi + \frac{1}{4}e^{2\sigma}F^a_c \omega^c \wedge F_{bd}\omega^d$$

The $R^\psi_a$ may be found similarly, and the Riemann components are given by,

$$R^a_b \sim R^a_{bcd}\omega^c \wedge \omega^d$$Taking the usual contraction over indices, we eventually arrive at the Ricci tensor,

$$R_{ab}=R_{0ab}-\nabla_a \nabla_b \sigma -\nabla_a \sigma \nabla_b \sigma + \frac{1}{2}e^{2\sigma}F_{ac}F^c_b$$

and contracting with the metric yields the Ricci scalar,

$$R_5=R_0 +\frac{1}{4}e^{2\sigma}F^2-2\square \sigma -2(\nabla \sigma)^2$$

Plugging into the Einstein-Hilbert action, and assuming the fifth dimension is periodic with period $L$:

$$S=-\frac{L}{16\pi G}\int \mathrm{d}^4 x \, \sqrt{g} \, e^\sigma\left[ R_{\mathrm{4D}} +\frac{1}{4}e^{2\sigma}F^2\right]$$

If we set the dilaton to a constant, the action reduces to pure Einstein-Maxwell.


While @JamalS's answer is probably a more rigourous way to do this, a lot can be said by looking simply at the symmetries of the KK reduction. In the standard KK picture, the metric in $D=5$ is $$ ds^2 = g_{\mu\nu} dx^\mu dx^\nu - e^{2\sigma} \left( d t + A_\mu dx^\mu \right)^2 $$ To reduce $R_5 \to R_4$, we note that $\sigma$ is a scalar of 4 dimensions and $A_\mu$ is a gauge field. The gauge symmetry corresponds to local redefinition of the origin on the compactified $S^1$, $t \to t+\lambda$, $A_\mu \to A_\mu - \partial_\mu \lambda$. Now, the mass dimensions of the quantities at hand are $[\sigma]=[A_\mu]=[g_{\mu\nu}]=0$, while $[R_4]=2$. Simple dimensional analysis and gauge invariance implies $$ \int dt d^4 x \sqrt{g_5} R_5 = 2\pi R \int d^4 x \sqrt{g} \left[ a R_4 + \frac{b}{4} F_{\mu\nu} F^{\mu\nu} + c (\nabla \sigma)^2 \right] $$ where appropriate integration by parts have been done to reduce the form. The remaining dimensionless quantities $a$, $b$, and $c$ can in general be functions of $\sigma$, but not $A_\mu$ (due to gauge invariance)

More can be said about these quantities by looking at some scaling arguments. For example, scaling $t \to \lambda t$, $A_\mu \to \lambda A_\mu$ and $e^{2\sigma} \to \lambda^{-2} e^{2\sigma}$ leaves $ds^2$ unchanged. Under the same change $R \to \lambda R$, $F_{\mu\nu} F^{\mu\nu} \to \lambda^2 F_{\mu\nu} F^{\mu\nu}$ and $\nabla_\mu \sigma \to \nabla_\mu \sigma$. However, since the entire action must remain the same, we must have $a \to \lambda^{-1} a$, $b \to \lambda^{-3} b$ and $c \to \lambda^{-1} c$. This fixes the $\sigma$ dependence of these quantities as $$ a \propto e^\sigma,~b \propto e^{3\sigma},~c \propto e^\sigma $$ Thus, we have $$ \int dt d^4 x \sqrt{g_5} R_5 = 2\pi R \int d^4 x \sqrt{g} e^\sigma \left[ \alpha R_4 + \frac{\beta }{4} e^{2\sigma} F_{\mu\nu} F^{\mu\nu} + \gamma (\nabla \sigma)^2 \right] $$ where now $\alpha$, $\beta$, and $\gamma$ are just numbers. These can now be fixed by taking special cases for $ds^2$. For example, if $A_\mu = \sigma = 0$, we find $R_5 = R_4$ and hence $\alpha = 1$. Similar methods can be used to fix $\beta$ and $\gamma$.

  • $\begingroup$ A nice approach too :) $\endgroup$
    – JamalS
    May 9 '14 at 14:28
  • $\begingroup$ I believe there are a couple of inaccuracies in @Prahar 's answer. First there should also be a term $d \nabla^2 \sigma$. In the end that term combines with the $(\nabla \sigma)^2$ term to give the $e^{-\sigma}\nabla^2 e^\sigma$ in $R$. Also $ R$ does not scale as $R\rightarrow \lambda R$, but is invariant under this scaling. However $\sqrt{g_5} = e^{\sigma} \sqrt{g}$ so all the rest remains the same. $\endgroup$ Jun 5 '20 at 10:14

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