Kaluza-Klein Reduction of Einstein Gravity

Kaluza–Klein reduction reorganizes a higher-dimensional metric into a lower-dimensional metric, a gauge potential, and a scalar measuring the size of the extra direction. The gauge symmetry is inherited from higher-dimensional diffeomorphisms, while the scalar cannot generally be frozen without imposing an additional field equation. Pure geometry therefore appears as interacting lower-dimensional fields without modifying the parent Einstein theory.

Required background. Use Killing fields and isometry groups for the compact isometry and the Einstein equation for the higher-dimensional parent system.

Helpful background. Hypersurface normals and tangential projectors provides an analogous normal/tangential decomposition.

A five-dimensional metric with one Killing direction

Let $z$ parametrize a spacelike circle of coordinate period $L$, and assume the cylinder condition

$$\partial_z\widehat g_{MN}=0.$$

For four-dimensional signature $(-,+,+,+)$, take the five-dimensional line element

$$d\widehat s^2 =e^{-\varphi/\sqrt3}g_{\mu\nu}dx^\mu dx^\nu +e^{2\varphi/\sqrt3} \bigl(dz+2A_\mu dx^\mu\bigr)^2.$$

The exponents choose the four-dimensional Einstein frame, in which the Ricci scalar has canonical normalization and $\varphi$ has a canonical kinetic term. Other parametrizations are legitimate, but their displayed scalar and Maxwell couplings differ. Overduin and Wesson show the unrescaled metric, cylinder condition, and conformal-frame transition in Overduin and Wesson 1997, §§3.2–3.7, pp. 13–18, PDF.

The block determinant gives

$$\sqrt{-\widehat g}=e^{-\varphi/\sqrt3}\sqrt{-g}.$$

A coordinate change along the circle,

$$z'=z-2\chi(x),$$

preserves the form of the ansatz provided

$$A'_\mu=A_\mu+\partial_\mu\chi.$$

Thus the lower-dimensional $U(1)$ gauge transformation is a surviving five-dimensional diffeomorphism. The invariant field strength is $F=dA$.

Reduction to Einstein–Maxwell–dilaton theory

Start with the five-dimensional Einstein–Hilbert action

$$\widehat S =\frac{1}{16\pi G_5} \int d^4x\,dz\,\sqrt{-\widehat g}\,\widehat R.$$

Substituting the ansatz, using $z$ independence, and discarding the total derivative generated by the conformal transformation gives

$$S =\frac{1}{16\pi G_4} \int d^4x\,\sqrt{-g} \left[ R-\frac12(\nabla\varphi)^2 -e^{\sqrt3\varphi}F_{\mu\nu}F^{\mu\nu} \right],$$

where $G_4=G_5/L$ for the chosen coordinate period. The factor multiplying $F^2$ reflects the normalization $dz+2A$; rescaling $A$ moves a constant between this coefficient and the gauge field. The positive exponent is fixed by the metric ansatz above. A convention using $e^{-\sqrt3\varphi}F^2$ must also replace $\varphi\mapsto-\varphi$ in both metric exponents.

To see the sign without trusting a memorized reduction formula, temporarily hold $\varphi$ constant and write $\widetilde g_{\mu\nu}=e^{-\varphi/\sqrt3}g_{\mu\nu}$ and $b=e^{\varphi/\sqrt3}$. For a circle of radius factor $b$, the curvature contains $-b^2(2F)^2_{\widetilde g}/4$. The volume factor is $\sqrt{-\widehat g}=b\sqrt{-\widetilde g}$, while raising the two indices on $F$ supplies $e^{2\varphi/\sqrt3}$. Therefore the coefficient in the Einstein-frame action is

$$e^{-\varphi/\sqrt3}\,e^{2\varphi/\sqrt3}\, e^{2\varphi/\sqrt3}=e^{\sqrt3\varphi}.$$

Allowing the scalar to vary also produces its kinetic term and a total derivative. The full calculation is given in Pope n.d., §7.1, pp. 81–84, Eqs. (7.7)–(7.16), PDF: use $D=4$, the source’s metric exponent $\alpha=-1/(2\sqrt3)$, and its gauge field $\mathcal A=2A$ to obtain the normalization here. Dropping the total derivative is appropriate for the bulk equations; a variational problem with boundaries must reduce its boundary terms as well.

The three sets of lower-dimensional equations are

$$G_{\mu\nu} =\frac12\nabla_\mu\varphi\nabla_\nu\varphi -\frac14g_{\mu\nu}(\nabla\varphi)^2 +2e^{\sqrt3\varphi} \left(F_{\mu\rho}F_\nu{}^\rho-\frac14g_{\mu\nu}F^2\right),$$ $$\nabla_\mu\left(e^{\sqrt3\varphi}F^{\mu\nu}\right)=0, \qquad \Box\varphi-\sqrt3e^{\sqrt3\varphi}F^2=0.$$

These equations display how the $15$ independent components of a five-metric become $10$ metric components, $4$ vector components, and $1$ scalar. Four-dimensional diffeomorphisms and the inherited $U(1)$ redundancy remove coordinate descriptions, leaving the same five physical polarizations as a massless graviton in five dimensions: two graviton, two photon, and one scalar modes.

Frames, consistency, and checks

In the unrescaled or Jordan-frame metric, the orbit-radius scalar multiplies the lower-dimensional Ricci scalar. A Weyl rescaling moves this nonminimal factor into the scalar kinetic and gauge couplings. A “frame” is therefore more than notation once other matter is added: one must state which metric couples to that matter. Within the pure reduced gravity sector, invertible local field redefinitions preserve the solution content where the conformal factor is finite and nonzero.

Setting $A_\mu=0$ is a consistent sector because the Maxwell equation is then satisfied. Setting $\varphi$ to a constant while retaining a generic Maxwell field is not: the scalar equation demands $F_{\mu\nu}F^{\mu\nu}=0$ in this simple model. This is the classic check emphasized after the reduced equations by Overduin and Wesson 1997, §3.4, pp. 14–15, PDF. A charged Coulomb field has $F^2\ne0$, so ordinary Einstein–Maxwell solutions do not all uplift with a constant circle radius.

The zero-mode cylinder condition is also a truncation, not the statement that the full compactified theory has no higher Fourier modes. Fields on a circle admit modes $e^{2\pi inz/L}$ with masses proportional to $|n|/L$. Keeping only $n=0$ is justified as an exact invariant sector under the stated ansatz; using it as a low-energy approximation requires energies well below the first Kaluza–Klein scale and control of sources for discarded modes.

Exercise

Verify the gauge transformation above directly. Show that the one-form $dz+2A$ is unchanged by $z'=z-2\chi(x)$ and $A'=A+d\chi$.

Solution

Differentiating the coordinate transformation gives $dz'=dz-2d\chi$, or $dz=dz'+2d\chi$. Therefore

$$dz+2A=dz'+2d\chi+2A=dz'+2(A+d\chi)=dz'+2A'.$$

The five-dimensional line element retains exactly the same form. Applying $d$ gives $F'=dA'=dA+d^2\chi=F$, so the reduced action depends on the gauge-invariant field strength. This also identifies the gauge symmetry as a restricted parent diffeomorphism rather than a new arbitrary postulate.

References

• Overduin, J. M., and P. S. Wesson. “Kaluza–Klein Gravity.” Physics Reports 283 (1997): 303–378. DOI. Open PDF.
• Pope, C. N. Geometry and Topology in Physics II: Applications. Undated lecture notes, §7.1, pp. 80–89. Open PDF.