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I asked a question a year and 3 months ago on mathstackexchange but after 3 bounties and still no answer I've decided to try here. Here's the link: conformal compactification.

Construct a conformal compactification, $\overline G$ of $G:=\Bbb R^{1,1}_{\gt 0}$ and/or provide a diagram (could be hand drawn) of the conformal compactification of $G?$

conformal compactification

Let $G$ have the metric tensor (not necessarily positive definite): $$ds^2=\frac{dxdt}{xt}.$$

This link, Penrose diagram, (under the heading "basic properties") states the relation between Minkowski coordinates $(x,t)$ and Penrose coordinates $(u,v)$ via $$\tan(u \pm v)=x\pm t.$$

So I have been playing around with trying to relate all three coordinates. I should note that $G$ has "null coordinates:"

Light-cone coordinates

I think that the coordinates for $G$ should simply be $(e^x,e^t).$ And then I'd get $$\tan(e^u\pm e^v)=e^x\pm e^t.$$ But this doesn't seem quite right.

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  • $\begingroup$ What does the subscript "$>\!0$" signify? $\endgroup$
    – Qmechanic
    Dec 9, 2022 at 14:24
  • $\begingroup$ Related: physics.stackexchange.com/q/425371/2451 $\endgroup$
    – Qmechanic
    Dec 9, 2022 at 14:30
  • $\begingroup$ If you want to compactify $G$, why don’t you just use $x=\tan u$ and $y=\tan v$ like for a Penrose diagram? The $(u,v)$ manifold is a square (so compact) and the change of coordinates is conformal. $\endgroup$
    – LPZ
    Dec 9, 2022 at 15:09
  • $\begingroup$ @Qmechanic I'm using ">0" to mean that there are no negative values. For example $p=(-1,2)$ cannot be a point in the manifold because of the negative sign $\endgroup$
    – geocalc33
    Dec 9, 2022 at 22:03
  • $\begingroup$ Is "$>\!0$" spacelike or timelike in your convention? $\endgroup$
    – Qmechanic
    Dec 10, 2022 at 5:52

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