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Set Up

In the original classical Kaluza Klein theory, you have a $d+1$ dimensional manifold where one space dimension is a circle $S^1$. In the "low energy limit," none of the metric components depend on this compact dimension. This independence of the compact direction is called the "cylinder condition." If one breaks up the metric as $$ g_{AB} = \begin{pmatrix} g_{\mu \nu} + \phi^2 A_\mu A_\nu & \phi^2 A_\mu \\ \phi^2 A_\nu & \phi^2 \end{pmatrix} $$ then, after a rather tedious calculation, one can find that the Ricci tensor, scalar curvature, and Einstein tensor, are \begin{align} \newcommand{\oh}{\tfrac{1}{2}} \newcommand{\of}{\tfrac{1}{4}} \newcommand{\tf}{\tfrac{3}{4}} R_{yy} &= - \phi \nabla^2 \phi + \of \phi^4 F^2 \\ R_{\mu y} &= - \oh \phi^{-1} \nabla_\nu( \phi^3 F^\nu_{\;\; \mu} ) + A_\mu R_{yy} \\ R_{\mu \nu} &= R_{\mu \nu}^{(d)} - \phi^{-1} \nabla_\nu \nabla_\mu \phi - \oh \phi^2 F_\mu^{\; \rho} F_{\nu \rho} - A_\mu A_\nu R_{yy} + A_\nu R_{\mu y} + A_\mu R_{\nu y} \\ R &= R^{(d)} - \of \phi^2 F^2 - 2 \phi^{-1} \nabla^2 \phi \\ G_{yy} &= \tfrac{3}{8} \phi^4 F^2 - \oh \phi^2 R^{(d)} \\ G_{\mu y} &= - \oh \phi^{-1} \nabla_\nu ( \phi^3 F^\nu_{\; \mu} ) + A_\mu G_{yy} \\ G_{\mu \nu} &= G_{\mu \nu}^{(d)} - \phi^{-1} ( \nabla_\nu \nabla_\mu \phi - \nabla^2 \phi) - \oh \phi^2 ( F_\mu^{\; \rho} F_{\nu \rho} - \of g_{\mu \nu} F^2 ) \\ & \hspace{0.4 cm} - A_\mu A_\nu G_{yy} + A_\nu G_{y \mu} + A_\mu G_{y \nu}. \end{align} (We have denoted the full $d+1$ dimensional indices with $A,B$, the $d$ dimensional base space indices with $\mu, \nu$, and the $S^1$ coordinate with $y$. All tensors on the right hand side of the above equations (covariant derivates, etc.) are defined using the $d$ dimensional metric $g_{\mu \nu}$.)

Setting the above expressions equal to $0$ can be regarded as vaccuum equations of motion. Note that the gauge-dependent terms which depend explicitly on $A_\mu$ are $0$ on shell.

Famously, the $d$-dimensional theory has three fields: a metric, a photon, and the scalar dilaton $\phi$. The VEV of $\phi$, which scales the radius of the compact direction, also famously scales the effective coupling constants of the theory. The fact that $\phi$ can change from place to place means that $\epsilon_0$ and Newton's constant $G$ can vary from place to place in the $d$ dimensional base space theory.

Question

Because the dilaton is coupled in strange ways to the other fields, it is not entirely clear to me if this theory is "stable" or not. An unstable theory, in my mind, would allow for "generic" solutions to have things like $\phi$ grow or shrink without bound. Note that I am only interested in the classical stability, not the quantum stability. I have seen it claimed, for instance in the introduction of Ed Witten's 1982 paper on the quantum instability, "Instability of the Kaluza Klein Vacuum" that classical Kaluza Klein theory is stable. How do we know that, given the strange dependence on $\phi$? I know we can do certain field redefinitions to make the equations look more familiar, but the fact that the scalar field will multiply the other fields in one way or another will not go away. I would be interested in an answer that discusses stability properties using the field equations I wrote above, or something similar. For instance, using the $R_{yy} = 0$ equation, it is clear that if $\phi$ is a constant then that is equivalent to constraining $F^2 = 0$, so any runaway of $\phi$ would depend on there being a nonzero $F^2$.

Furthermore, I also read sections 1.10 and 1.11 of Roger Penrose's 2016 book "Fashion, Faith, and Fantasy" (which I do not fully understand) where Penrose claims about generic manifolds with compactifications

But matters are far more serious than this, because practically all such perturbations will lead to an evolution... that is singular... This means, in effect, that the extra dimensions must be expected to crumple up to something where curvatures diverge to infinity and further evolution of the classical equations becomes impossible. This conclusion follows from mathematical singularity theorems that were proved in the late 1960s – most particularly one that Stephen Hawking and I had established a little before 1970 [Hawking and Penrose 1970] – which showed, among other things, that almost any $n$-dimensional space-time $(n\geq 3)$ containing a compact spacelike $(n − 1)$-surface... but which does not contain closed timelike loops, must evolve to a space-time singularity if its Einstein tensor $G^(n)$ satisfies an energy (non-negativity) condition called the strong energy condition.

Penrose here seems to be citing some sort of theorem (?) which he claims implies that for "almost any" case, compact dimensions are unstable. Does the original Kaluza Klein theory evade these (unclear to me) instability criterion for some reason?

It occurs to me that my question can really be broken up into four cases. For instance, we can consider the evolution of the "vacuum" Kaluza Klein theory, but we can also include our own scalar matter field in the $d+1$ spacetime. This has the potential to alter the "stability" of the theory greatly. Furthermore, we can also consider what happens when we remove the "cylinder condition."

So, my admittedly very huge question is, "Is the original Kaluza Klein theory stable in the cases where

  1. You have the vacuum equations and the cylinder condition
  2. You have the vacuum equations and no cylinder condition
  3. You have a matter field and the cylinder condition
  4. You have a matter field and no cylinder condition

and if so, how can we see that from the equations of motion? Furthermore, do Penrose's claims about the generic instability of compactifications apply or not? Finally, is there an intuitive way to understand how stable/unstable solutions behave?"

Partial answers are welcome. I also have a poor understanding of the literature and would be happy to receive any references which address some aspects of these questions.

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    $\begingroup$ well, the singularity theorem do not say how fast the spacetime evolves toward a singularity. If it is in the distant future it is not an observational constraint $\endgroup$ – lurscher May 28 at 3:15

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