# Visualizing the conformal compactification diagram of $G$

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.

• What does the subscript "$>\!0$" signify? Dec 9, 2022 at 14:24
• Dec 9, 2022 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 9, 2022 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 Dec 9, 2022 at 22:03
• Is "$>\!0$" spacelike or timelike in your convention? Dec 10, 2022 at 5:52