How can contracting dimensions lead to cosmological inflation?

Question

Using the Kasner metric, given by

$$ ds^2 = -dt^2 + \sum_{j=1}^D t^{2p_j}(dx^j) $$

it is possible to not only describe the cosmological expansion of some space directions (the ones with positive Kasner exponents $p_j$, but this metric allows for some dimensions to contract too, those have negative $p_j$. The two Kasner conditions

$$ \sum_{j=1}^{D-1} p_j = 1 $$

and

$$ \sum_{j=1}^{D-1} (p_j)^2 = 1 $$

say that there have to be contracting and expanding dimensions at the same time, as the $p_j$ can not all have the same sign.

In a comment I have read, that in models with for example 3 expanding and $n>1$ contracting dimennsions, the contracting dimensions drive the inflation in the other directions by leading their expansion to accelerate without a cosmological constant. This is interesting and about this I'd like to learn some more.

So can somebody a bit more explicitely explain how such inflation models work? For example what exactly would the vacuum energy from a physics point of view be in this case? Up to now I only heard about inflation models where the vacuum energy density is the potential energy of some inflaton field(s) in a little bit more detail.

Answer 1

You seem to be talking about inflation and expansion as if they were the same thing; they aren't. A Kasner metric has expansion and contraction, but it doesn't have anything like inflation. Inflation is exponential and is driven by a scalar field; the Kasner metrics are vacuum solutions and their behavior isn't exponential.

[...]what exactly would the vacuum energy from a physics point of view be in this case?

There is no vacuum energy in a Kasner metric; the Kasner metrics are vacuum solutions, i.e., solutions of the Einstein field equations with zero stress-energy tensor and zero cosmological constant.

[...] the contracting dimensions drive the inflation in the other directions by leading their expansion to accelerate without a cosmological constant.

This is something that has to happen because it's a vacuum solution. In a vacuum solution, the Ricci tensor vanishes. The interpretation of a vanishing Ricci tensor is that the only gravitational forces are tidal in character, as opposed to the kind of gravitational forces you get from a source that's present in that the region of space where you're measuring the curvature. One way to distinguish a tidal from a non-tidal force is that if you release a cloud of test particles in a purely tidal field, the volume of the cloud is conserved. If you want to conserve volume, you can't have expansion along all axes.