= Solution
Start with the four-dimensional <Minkowski metric> $ds^2=-dt^2+dr^2+r^2d\Omega^2$, where $r\geq0$. Choose an arbitrary length $L>0$, and use the <retarded and advanced null coordinates> $u=t-r$, $v=t+r$. For the <Minkowski conformal compactification>, set
$$
p=\arctan(u/L),\quad q=\arctan(v/L),\qquad T=p+q,\quad R=q-p.
$$
Both $p,q$ lie between $-\pi/2$ and $\pi/2$. Since $v\geq u$, we have $R\geq0$; the remaining inequalities are $|T|+R<\pi$. Moreover,
$$
r=\frac{L\sin R}{2\cos p\cos q},\qquad ds^2=\frac{L^2}{4\cos^2p\cos^2q}\left(-dT^2+dR^2+\sin^2R\,d\Omega^2\right).
$$
Multiply by the square of the <conformal factor> $\Omega_c=2\cos p\cos q/L$. The resulting metric is regular on the appropriate boundary pieces and preserves the <null directions>. Suppressing the angular two-spheres gives a triangular <Penrose diagram> with radial <null geodesics> at $45$ degrees.
The line $R=0$ is the ordinary timelike centre $r=0$. The upper sloping edge $T+R=\pi$ is <future null infinity>, reached with $v\to+\infty$ and finite $u$; the lower sloping edge $T-R=-\pi$ is <past null infinity>, reached with $u\to-\infty$ and finite $v$. The vertices $(T,R)=(\pi,0)$ and $(-\pi,0)$ are future and past <timelike infinity>, denoted $i^+$ and $i^-$. The vertex $(0,\pi)$ is <spacelike infinity>, $i^0$. These are limiting endpoints in the <conformal completion>, rather than ordinary physical events. In particular, finite diagram coordinates at <null infinity> do not imply finite physical <affine parameter>.
\b[The four-dimensional radial diagram is the triangle $R\geq0$, $|T|+R\leq\pi$.] If one instead draws two-dimensional <Minkowski spacetime> with a signed Cartesian spatial coordinate, the diagram is the full diamond. The centre is a boundary of the radial quotient, not a boundary of the physical four-dimensional <Minkowski spacetime>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-52-conformal-diagrams.png]
{title=Kruskal extension and the radial Minkowski Penrose diagram}
{height=650}
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