More concretely, I understand that an expression like $dx_1/dx_0$ is not tensorial in open spacetime coordinates $x_0, x_1, \ldots$. And using covariant derivatives, say $\nabla_{\partial_{x_0}} \partial_{x_1}$ to represent a "velocity" is no better, since the $finite$ values of a tensor are not invariant w.r.t. diffeomorphisms, but only the zeros and poles $\pm \infty$ of the tensor $\nabla_{\partial_{x_0}} \partial_{x_1}$ are invariant. (C.f. Einstein's Point-Coincidence Argument).

@BenCrowell. Yes, that distinction seems ad hoc (e.g., MTW, pp. 37) and convenient only when the metric is diagonal, in which case the local Lorentz frame appears canonical. Could you please recommend a reference to further elaborate on the distinction? It seems to preclude even the possibility of a spacetime metric with variable luminal velocity, e.g. what is a metric where speed of light increases perfectly linearly with time?

Minor comment. Your metric represents a perfectly fine model of the spacetime we live in IF we could observe the speed of light to be perfectly linearly increasing with "time".

pp.9 of the article mural.maynoothuniversity.ie/10079/1/DW-Fundamental-1998.pdf says Yes, and that a result of Gromov proves the group is generated by at most $(2.5)^{\dim(X)}$ generators.

@Venkataramana Yes, i was mistaken. Representing the semidirect product as $\begin{pmatrix} A & v \\ 0 & 1 \end{pmatrix}$ for $A\in PGL_n$, $v\in \mathbb{R}^n$, we see the $\mathbb{Z}$-points are defined by stabilizing the lattice $\mathbb{Z}^{n+1}$.

@skupers i think you make mistake considering YCor's above examples as arithmetic. Yes the free group $F_2$ is commensurable to $PGL(\mathbb{Z}^2)$, but not in the case of thin groups (YCor's second example). Likewise the extension (first example) is not commensurable to $\textbf{G}(\mathbb{Z})$ for any linear algebraic group scheme $\textbf{G}$.

Very interesting question.

Compare Section 4 in web.math.princeton.edu/sarnak/InternalApollonianPackings09.pdf, where the divergence of $\sum r_i$ is credited to O. Wesler, “An infinite packing theorem for spheres,” PAMS Vol. 11, pp. 324-326, (1960).

The article (or at least as much as i can read it here books.google.ca/…) does not have any conclusive answer... apparently Levi-Civita, Schrodinger, Bauer were all critical of the pseudotensor $t_{ij}$.

The recent article of math.toronto.edu/mccann/papers/LimMcCann20.pdf contains results on such variational questions.

By the phrase "the infinite branches do not have a common intersection" i mean the "graph-edges" containing the corners do not share a common "vertex".

@GabeK You're correct that any infinite-sided polygon inscribed in unit disk has cut loci and medial axes with infinitely many branches. Indeed $M(A)$ intersects the boundary at points with divergent curvature (the corners). I'm claiming that the infinite branches do not have a common intersection except when they are located on a spherical arc and focus to a common point. You're correct this deserves to be clarified.

@Gabe K. Maybe the correct argument is simply this: $C$ is compact when $A$ is open and bounded. QED. There are no accumulation points.

The pathological smooth shapes (relative to cut loci) are those with constant curvature, which is everywhere destroyed by generic perturbations. But the pathological continuous curves are much wilder, e.g. sin(1/x) graphs, and i avoid them.