Let's take a look at the equation for the Riemann tensor in terms of an arbitrary 1-form:


Where we are using the Levi-Cevita connection (we might consider other connections later). The Killing equation reads as:

$$\nabla_{\mu}K_{\nu}+\nabla_{\nu}K_{\mu}=0$$ Where $K$ is a killing 1-form.

Let us suppose for the moment that our spacetime is perfectly isotropic and homogenous (we will address the fact that this isn't the case in a bit). We have then that the killing vectors span the space. Using the Jacobi identity, and symmetries of the Riemann tensor we can, for each Killing vector field (i), write the relation:


Where $\nabla^{\alpha}\nabla_{\alpha}$ is the Laplace-Beltrami operator, $R$ is the Ricci scalar curvature, and $n$ is the dimension of our spacetime (taken from now on to be $n=4$). (This is borrowed from a paper [here]1 Now the magnitude of the ith Killing field corresponding to translation $|K_{i}|^{2}=1$ and there are four of them. We may write the above equation then as:


In which case we can simply write The Einstein Hilbert action (for such a trivial case) as:


The Laplace Beltrami Operator however Enjoys the quite beautiful feature that, due to it's Self-Adjointness, it may be written as (for arbitrary $A,B$):

$$\intop_{M}\left(A^{\mu}\nabla^{\alpha}\nabla_{\alpha}B_{\mu}\right)\sqrt{|g|}d^{4}x=-\intop_{M}\langle dA,dB\rangle\sqrt{|g|}d^{4}x=\intop_{M}\left(B^{\mu}\nabla^{\alpha}\nabla_{\alpha}A_{\mu}\right)\sqrt{|g|}d^{4}x$$

Thus we Can write the Einstein-Hilbert action for our maximally symmetric spacetime in terms of Killing forms $K$:

$$S_{EH}=\intop_{M}-\frac{1}{4}\langle dK_{(i)},dK_{(j)}\rangle\sqrt{|g|}d^{4}x$$

For the Levi-Cevita connection the Covariant exterior derivative has the form:

$$dK_{(i)}=\partial_{\mu}K_{(i)\nu}-\partial_{\nu}K_{(i)\mu} $$

At this point it just makes sense to define a field strength tensor $dK_{(i)}=F_{\mu\nu}^{i}$ such that we can write:


Even though we're in a nonrealistic perfectly homogenous isotopic universe, the familiar form of the above ought to strike the reader's interest. Note that Our Killing vectors form a Lie algebra and are thus invariant under it's action (that is the Lie algebra to the group of isometries for our spacetime). We have Killing vectors as gauge fields!

This seems to be saying that the Ricci curvature of a space is equal to the inner product of the curvature of the group of isometries of that space...

I have only come across a Field Strength tensor as a sort of “square root” of the Ricci tensor in one place: Wheeler's Geometrodynamics (wherein he called the Electromagnetic Field strength tensor the “Maxwell square root” of the Ricci tensor). Misner and him called this their "already unified theory" of electromagnetism and general relativity.

It should also be noted that each of the three different FLRW cosmologies (that are perfectly homogenous and isotropic) would lead to different gauge symmetries corresponding to the isometry groups of those spaces.

Considering the different possible FLRW universes for example we have closed ($S^3$), Open ($H^3$), and flat ($R^3$) corresponding to $SO(4), SO(3,1)$, and $ISO(3)$ . Of which the closed universe appears the most interesting.

Note, were missing the nonabelian terms in our field tensor which go like the latter term in Cartans curvature two-form:

$$\Omega=dA+\frac{1}{2}[A\wedge A]$$

My initial guess is they're missing because we didn't use all of the Killing vector fields (we actually have 10 in all for our 4d spacetime and we only used 4!), Also we could have used a different connection Like in Qmechanic's answer here

The Pessimistic reader might say “So what? We don't have a perfectly Homogenous isotropic universe!!” To that I answer: “Have you ever Heard of the Yamabe problem?”.

For those unfamiliar, the Yamabe problem basically boils down to this: Given a (pseudo)Riemannian manifold does there always exist a conformal transformation $\lambda$ of the metric $g$ such that:


Where $\tilde{g}$ is a space of constant curvature (ie our homogenous isotropic space).

The answer it turns out is a yes for compact manifolds. Of course we are going the other way, we want to conformally map the above action “away from” a perfectly homogenous isotropic universe. Note that while we're at it we can conformally map the spatial portion to be expanding (in line with observation).

The scale parameter (not arbitrary in the closed case), the rate of expansion of space, these would have to show up in terms of the lie algebra structure constants or as a parameter in the breaking of those symmetries (the latter for the terms describing expansion of the universe).

In such a case, we would want to vary the action with respect to a local conformal transformation, and we'd get to see how the symmetries of our gauge fields are broken by different families of such transformations. Note there's not a lot of room to "fool around" in that the possible Lie groups here are "predetermined" by observation of our universe (though the non-simply connected possibility expands this quite a bit). For the closed universe for instance we might choose the universal cover for $SO(4)$ being $SU(2)_{L}\times SU(2)_{R}$ and see how that breaks for an expanding universe.

I'm not going to write out the conformally transformed version here, but I think the attentive reader gets the idea (or maybe I'm off my rocker haha!)

Anyway, I thought It was kind of interesting, especially after Reading Wheeler's book, which only discusses U(1) gauge fields in this context.

Any questions or comments on this kind of approach to doing general relativity? I really like how it explicitly demonstrates global topological choices affecting families of field symmetries locally. There is much more of a standard model type of feel to it. (:

Or did I make a silly mistake somewhere.

  • $\begingroup$ Your new action is a functional of $g$ and $K$, but there is nothing in the action that would ensure that the fields $K$ have the desired properties. Also, constant scalar curvature of the Yamabe problem still does not guarantee the existence of Killing vectors. $\endgroup$ – A.V.S. Nov 29 '18 at 9:46
  • $\begingroup$ @A.V.S. I was restricting myself to homogenous isotropic spaces, which definitely do admit the full set of Killing vectors, and are constant curvature spaces. Since conformal transformations are diffeomorphisms, they would have to keep all topological properties of the initial manifold. As for your first sentence, yeah, there's definitely some fun to be had playing with that, remember, after a conformal transformation (which this action is NOT invariant under), the transformation itself will be one of the variables of the functional. So we'll have g,K,$\phi$ $\endgroup$ – R. Rankin Nov 29 '18 at 10:14
  • $\begingroup$ There's a really good treatment of the killing vectors in such spaces Here: iopscience.iop.org/article/10.3367/UFNe.2016.05.037808/pdf $\endgroup$ – R. Rankin Nov 29 '18 at 10:30

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