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I've been thinking about compact spactimes lately. My understanding of the Yamabe problem is that one can always conformally transform a (compact) spacetime to one of constant scalar curvature, something like:

$$g_{\mu\nu}=\tilde{g}_{\mu\nu}e^{\phi(x)}$$

If we narrow the field a bit to conformally flat spacetimes, we can also say that we can write our metric as a conformal transformation of the Minkowski metric, something like:

$$g_{\mu\nu}=\eta_{\mu\nu}e^{\phi(x)}$$

Here's what I've been wondering about. Supposing the spacetime is globally conformally flat, we can write the above no problem. Now though we want to add localized distortions which preserve the 4-volume of the spacetime in question. I know that such transformations (volume preserving) are called the special affine group , which for 4 dimensions is $SL(4,R)$. In this case we have something like:

$$g_{\mu\nu}=\Lambda(x)\eta_{\mu\nu}e^{\phi(x)}$$

Where of course $\Lambda(x)\epsilon SL(4,R)$. Here's where I'm not sure....$SL(4,R)$ is isomorphic (I believe) to $SL(2,C)$ which I understand is the double cover of the (restricted) Poincare group (please correct me if I'm wrong).

So basically my thinking was that I could represent a conformally flat spacetime (say general FLRW universe for example) as a composition of the Minkowski metric with a conformal transformation (representing the specific FLRW shape and volume at a givent cosmological time and it's growing/contracing with time) along with an $SL(2,C)$ Poincare transformation representing all deviations from homogeneity.

Is this do-able? What I like is that all the matter and fields en-total contribute to a globalized conformal term, while localized metric distortions (via local matter/fields) are contained within the $SL(2,C)$ term. (FYI I'm using a spinor formalism, probably Newman-Penrose when all is said and done). When I go from $SL(4,R)$ to $SL(2,C)$ maybe I have to hit the metric two sided with the Lambdas? Sorry it's late and I was curious.

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  • $\begingroup$ FWIW, the Lie group $SL(4,\mathbb{R})$ has 15 real dimensions, while $SL(2,\mathbb{C})$ only has 6 real dimensions. $\endgroup$ – Qmechanic Mar 3 at 13:05
  • $\begingroup$ @Qmechanic Gotcha, so it would be an have to still be an $SL(4,R)$ transformation. I'll try and look up what geometric constaints correspond to going from one to the other to get a better grasp of such a condition $\endgroup$ – R. Rankin Mar 3 at 13:22
  • $\begingroup$ There are “Ashtekar variables” for gravity that have $SL(2,\mathbb{C})$ connection. But those are used for canonical formulation of general relativity. $\endgroup$ – A.V.S. Mar 3 at 14:07

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