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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.

Qmechanic
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geocalc33
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  • What does the subscript "$>!0$" signify? – Qmechanic Dec 09 '22 at 14:24
  • Related: https://physics.stackexchange.com/q/425371/2451 – Qmechanic Dec 09 '22 at 14:30
  • 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. – LPZ Dec 09 '22 at 15:09
  • @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 – geocalc33 Dec 09 '22 at 22:03
  • Is "$>!0$" spacelike or timelike in your convention? – Qmechanic Dec 10 '22 at 05:52
  • @Qmechanic I just mean that the coordinate axes are not negatively valued, they're strictly positive. So both space and time axes are not negatively valued in $\Bbb R^{1,1}_{\gt 0}$. This is in contrast to $\Bbb R^{1,1}$ where the coordinate axes can assume negative values – geocalc33 Dec 11 '22 at 17:10
  • @lpz not sure I understand completely – geocalc33 Dec 13 '22 at 14:11
  • From your metric $ds^2=\frac{dxdt}{xt}$ on $(x,t)\in (\mathbb R_+^)^2$, any transformation $x\to u=f(x),y\to v=g(y)$ is conformal, so you just need to send $\mathbb R_+^$ to a segment. What I was saying is that you can use $f=g=\tan$ like for Penrose diagrams (and even works on the entire $\mathbb R^2$). Your choice of $f=g=\exp$ does not work since the image is still $\mathbb R_+^*$, but fixable by a sign: $u=e^{-x},v=e^{-t}$ – LPZ Dec 13 '22 at 18:08

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