I am wondering whether the region $H:=\{(t,x):x^2−t^2<1\}$ of $(1+1)$-dimensional Minkowski spacetime, equipped with the restriction $g\_H$ of the standard Minkowski metric $g=−\mathrm{d} t\otimes \mathrm{d}t + \mathrm{d}x \otimes \mathrm{d} x$, is globally conformally equivalent to the vertical strip $S:=\{(t,x):|x|<1\}$, again equipped with the restriction $g\_S$ of the Minkowski metric. To further clarify: by "globally conformally equivalent" I mean that there should be a diffeomorphism $\phi : H \to S$ with $\phi\_\* (g\_H)=\Omega^2 g\_S$ for some smooth, strictly positive function $\Omega$.


I posted this question on Mathematics Stack Exchange some time ago as I had the feeling there might be a straightforward argument involving null geodesics, but to this day I still haven't found the answer. Thanks for your help.



Global conformal equivalence of two regions of Minkowski spacetime
In $H$ you have two lightlike geodesics such that every lightlike geodesic intersects one of these two, namely the geodesics $\{x=y\}$ and $\{x=-y\}$. In $S$ you do not have such two geodesics. Hence, the regions are not conformally equivalent 


