Conformal compactification and Penrose diagram

Remarks

Since $\tilde g$ and $g$ differ by a positive factor on $\psi(M)$, they have the same light cones there, so the compactification keeps the causal structure of $M$.

Example: for Minkowski space with $ds^2 = -dt^2 + dr^2 + r^2 d\Omega^2$, set $U = \arctan(t - r)$ and $V = \arctan(t + r)$, so $-\pi/2 < U \le V < \pi/2$. A direct computation gives $$ds^2 = \frac{1}{4\cos^2 U \cos^2 V} \left( -4\, dU\, dV + \sin^2(V - U)\, d\Omega^2 \right).$$ With $\Omega = 2 \cos U \cos V$, the metric $\tilde g = \Omega^2 g$ is, with $T = V + U$ and $R = V - U$, the metric $-dT^2 + dR^2 + \sin^2 R \, d\Omega^2$ of the Einstein static universe $\R \times S^3$. Minkowski space becomes the region $\abs{T} + R < \pi$, $R \ge 0$, and its Penrose diagram is a triangle. Its boundary consists of future and past null infinity $\mathscr{I}^\pm$ (where $V = \pi/2$, resp. $U = -\pi/2$), spatial infinity $i^0$ and future and past timelike infinity $i^\pm$.

Non-example: the identity map of Minkowski space onto itself, with $\Omega = 1$, satisfies clause 1, but $\overline{\psi(M)} = M$ is not compact, so it is not a conformal compactification.