Metric of spacetime with zero speed of light Despite references to physical spaces, this question is purely mathematical on differential geometry.
While deriving coordinate transformations based on common assumptions of homogeneity and uniformity of space and time (see A Primer on Special Relativity or Nothing but Relativity), typically three logical options are considered:


*

*Finite speed of light - Minkowski spacetime.

*Infinite speed of light - Galilean spacetime, ruled out by observation.

*Imaginary speed of light (negative square) - Euclidean spacetime, ruled out by causality.
There exist however the forth logical option:


*Zero speed of light.


At first it appears unreal, but in fact there exists a conceptual case asymptotically close to this scenario.
Imagine a thin spherical shell approaching its Schwarzschild radius. Spacetime inside this empty shell is locally flat Minkowski with the same time dilation as at the shell. As the shell approaches an infinite time dilation at the Schwarzschild radius, the speed of light at and inside the shell approaches zero.
What is the metric structure of this space in the limit of the speed of light being exactly zero?
Metrics for other three cases are well known. The Euclidean and Minkowski metrics don't require an introduction. The Galilean structure is described here: What is a mathematical definition of the Maxwellian spacetime?
Would the $c=0$ spacetime collapse simply to a 3D Euclidean space with no time or would it have two separate metrics for space and time like the Galilean spacetime?
 A: From a mathematical perspective this is a question about signatures of quadratic forms. On $\mathbb{R}^4$ with space coordinates $(x, y, z)$ and time coordinate $t$ we can consider

*

*A Euclidean metric, corresponding to the quadratic form $x^2 + y^2 + z^2 + (ct)^2$, with signature $(+, +, +, +)$

*A Lorentzian / Minkowski metric, corresponding to the quadratic form $x^2 + y^2 + z^2 - (ct)^2$, with signature $(+, +, +, -)$ (or the  negative of this, depending on your conventions)

*A Galilean metric ($c \to \infty$), corresponding to the quadratic form $-t^2$, with signature $(0, 0, 0, -)$; we get this by dividing the previous expression by $c^2$ and just taking the limit in the most obvious sense

*A $c = 0$ metric, corresponding to the quadratic form $x^2 + y^2 + z^2$, with signature $(+, +, +, 0)$.

The physical significance of this is not entirely clear to me; I don't have much experience thinking about relativity, but here are some speculations off the top of my head.
If we think of these quadratic forms as describing spacelike vs. timelike vs. lightlike directions, then in Galilean spacetime every direction is timelike or lightlike, while in $c = 0$ spacetime every direction is spacelike or lightlike. I guess we can think of $c$ as describing the "slope of the light cone," in which case the Galilean limit $c \to \infty$ corresponds to everything being in your light cone, while the $c \to 0$ limit corresponds to nothing being in your light cone.
I guess your conception of the Galilean limit $c \to \infty$ is that it describes a universe where "time is infinitely more important than space," so first we have the quadratic form $-t^2$ dividing up time slices but then in each timeslice we have the usual 3d Euclidean metric, reflecting the fact that an infinite speed of light means we can in principle reach any point in space from any other point in space at a given time, but we still can't e.g. travel backwards in time. If so, then the corresponding $c \to 0$ limit describes a universe where "space is infinitely more important than time," so we can think of it as divided up into isolated points of 3d space, each of which has a timeline with a 1d time metric, reflecting the fact that a zero speed of light means nothing can move.
Perhaps we should call the $c = 0$ case Zenoan spacetime.
Edit, 9/4/20: The $c \to 0$ limit is briefly discussed in Freeman Dyson's Missed Opportunities; he calls the corresponding automorphism group $G$ the "Carroll group," after Lewis Carroll:

"A slow sort of country," said the Queen, "Now, here, you see, it takes
all the running you can do, to keep in the same place."

A: I don't know what would happen with the spherical shell; I'm not proficient with General Relativity. But I guess that nothing would move (or that everything "moves" in the same timelike direction), and I generally agree with Qiaochu Yuan's answer.

The standard Lorentzian metric $\eta$ applies to a pair of vectors $u,v$ as
$$\eta(u,v)=u^xv^x+u^yv^y+u^zv^z-c^2u^tv^t.$$
Sometimes it's better to think of $\eta$ as a function from vectors to covectors, in which case $\eta(u,v)$ should be written as $\eta(u)(v)$, with
$$\eta(u)=\eta(u^xe_x+u^ye_y+u^ze_z+u^te_t)=u^x\varepsilon^x+u^y\varepsilon^y+u^z\varepsilon^z-c^2u^t\varepsilon^t.$$
$e_\mu$ and $\varepsilon^\nu$ are the basis vectors and covectors, related by $\varepsilon^\nu(e_\mu)=\delta_\mu^\nu$. This form of $\eta$ has an inverse, a function from covectors to vectors:
$$\eta^{-1}(\omega)=\eta^{-1}(\omega_x\varepsilon^x+\omega_y\varepsilon^y+\omega_z\varepsilon^z+\omega_t\varepsilon^t)=\omega_xe_x+\omega_ye_y+\omega_ze_z-\frac{1}{c^2}\omega_te_t.$$
The Galilean metrics come from taking $c\to\infty$ :
$$\eta_\infty^+(u)=\lim_{c\to\infty}\frac{1}{c^2}\eta(u)=-u^t\varepsilon^t$$
$$\eta_\infty^-(\omega)=\lim_{c\to\infty}\eta^{-1}(\omega)=\omega_xe_x+\omega_ye_y+\omega_ze_z.$$
These have signature $(0,0,0,-)$ on vectors, and $(+,+,+,0)$ on covectors. (And the limit has broken their inverse relationship.) Now we look at zero lightspeed:
$$\eta_0^+(u)=\lim_{c\to0}\eta(u)=u^x\varepsilon^x+u^y\varepsilon^y+u^z\varepsilon^z$$
$$\eta_0^-(\omega)=\lim_{c\to0}c^2\eta^{-1}(\omega)=-\omega_te_t.$$
These have signature $(+,+,+,0)$ on vectors, and $(0,0,0,-)$ on covectors. In this sense, $c\to0$ and $c\to\infty$ are dual to each other.

Another way to express $\eta$ uses the tensor product:
$$\eta=\eta_{\mu\nu}\varepsilon^\mu\otimes\varepsilon^\nu=\varepsilon^x\otimes\varepsilon^x+\varepsilon^y\otimes\varepsilon^y+\varepsilon^z\otimes\varepsilon^z-c^2\varepsilon^t\otimes\varepsilon^t.$$
You asked in the comments (but not in the OP) about covariant derivatives. Let's require the metric to be "constant", differentiating with respect to $e_\xi$ :
$$0=\partial_\xi\eta=(\partial_\xi\eta_{\mu\nu})\varepsilon^\mu\otimes\varepsilon^\nu+\eta_{\mu\nu}(\partial_\xi\varepsilon^\mu)\otimes\varepsilon^\nu+\eta_{\mu\nu}\varepsilon^\mu\otimes(\partial_\xi\varepsilon^\nu)$$
(each $\eta_{\mu\nu}$ is just a number, $1$ or $0$ or $-c^2$)
$$=(0)+\eta_{\mu\nu}(-\Gamma^\mu_{\pi\xi}\varepsilon^\pi)\otimes\varepsilon^\nu+\eta_{\mu\nu}\varepsilon^\mu\otimes(-\Gamma^\nu_{\pi\xi}\varepsilon^\pi)$$
(renaming summation variables in the last term)
$$=-\eta_{\mu\nu}\Gamma^\mu_{\pi\xi}\varepsilon^\pi\otimes\varepsilon^\nu-\eta_{\pi\mu}\Gamma^\mu_{\nu\xi}\varepsilon^\pi\otimes\varepsilon^\nu$$
$$=-\Big(\eta_{\mu\nu}\Gamma^\mu_{\pi\xi}+\eta_{\mu\pi}\Gamma^\mu_{\nu\xi}\Big)\varepsilon^\pi\otimes\varepsilon^\nu.$$
All of the tensors $\varepsilon^x\otimes\varepsilon^x,\;\varepsilon^x\otimes\varepsilon^y,\;\varepsilon^y\otimes\varepsilon^x,\cdots$ are independent, so this sum vanishing means that each coefficient vanishes:
$$0=\eta_{\mu\nu}\Gamma^\mu_{\pi\xi}+\eta_{\mu\pi}\Gamma^\mu_{\nu\xi}$$
(and the sum over $\mu$ reduces to a single term, because most of $\eta_{\mu\nu}$ are $0$)
$$=\eta_{\nu\nu}\Gamma^\nu_{\pi\xi}+\eta_{\pi\pi}\Gamma^\pi_{\nu\xi}.$$
This is to be true for all $\xi,\pi,\nu$. Cycling the names,
$$0=\eta_{\pi\pi}\Gamma^\pi_{\xi\nu}+\eta_{\xi\xi}\Gamma^\xi_{\pi\nu}$$
$$0=\eta_{\xi\xi}\Gamma^\xi_{\nu\pi}+\eta_{\nu\nu}\Gamma^\nu_{\xi\pi}$$
and using $\Gamma^\nu_{\mu\xi}=\Gamma^\nu_{\xi\mu}$ (no torsion), these 3 equations have the form $a=-b=c=-a$ which implies $a=0$, that is,
$$\eta_{\nu\nu}\Gamma^\nu_{\pi\xi}=0.$$
Finally, with the Lorentzian metric (finite $c$), each $\eta_{\nu\nu}\neq0$, and thus all Christoffel symbols must vanish: $\Gamma^\nu_{\pi\xi}=0$. This means that the "straight lines" are the obvious ones.

Applying the same process to $\eta^{-1}$ instead, we would get
$$0=\eta^{\nu\nu}\Gamma^\pi_{\nu\xi}+\eta^{\pi\pi}\Gamma^\nu_{\pi\xi}$$
but we'd need to multiply by $\eta^{\xi\xi}$ (losing information in the non-Lorentzian cases) to combine the 3 cycled equations. The result is
$$\eta^{\nu\nu}\eta^{\xi\xi}\Gamma^\pi_{\nu\xi}=0$$
which, with the Lorentzian metric, again gives $\Gamma^\pi_{\nu\xi}=0$.

For the Galilean metrics, making $\eta_\infty^+$ constant gives only $\Gamma^t_{\pi\xi}=0$.
Making $\eta_\infty^-$ constant gives $\Gamma^\pi_{\nu\xi}=0$ for the spatial coordinates $\{\xi,\pi,\nu\}\subseteq\{x,y,z\}$, and $\Gamma^x_{xt}=\Gamma^y_{yt}=\Gamma^z_{zt}=\big(\Gamma^x_{yt}+\Gamma^y_{xt}\big)=\big(\Gamma^x_{zt}+\Gamma^z_{xt}\big)=\big(\Gamma^y_{zt}+\Gamma^z_{yt}\big)=0$.
There's still some freedom; for example $\Gamma^x_{tt}\neq0$, which says that a timelike geodesic may accelerate in the $x$ direction. Even if all other Christoffel symbols vanish, one component of the Riemann curvature is $R^x_{txt}=\partial_x\Gamma^x_{tt}$ which is not necessarily zero. (It is zero if $\Gamma^x_{tt}$ is uniform across space, which would happen if the Galilean spacetime is deformed in the sense of Cavalieri's principle, sliding the spacelike layers over each other while maintaining each layer's shape; this is equivalent to using an accelerating reference frame.) See also Schuller's lecture "Newtonian spacetime is curved".

For zero-lightspeed, making $\eta_0^+$ constant gives $\Gamma^x_{\pi\xi}=\Gamma^y_{\pi\xi}=\Gamma^z_{\pi\xi}=0$.
Making $\eta_0^-$ constant gives $\Gamma^t_{t\xi}=\Gamma^x_{t\xi}=\Gamma^y_{t\xi}=\Gamma^z_{t\xi}=0$ (but those last three are special cases of the previous line, with $\pi=t$).
Again there's some freedom; for example $\Gamma^t_{xx}\neq0$. Similar comments apply.
A: Your classification includes 4 cases. Paraphrasing Yoda: "no, there's another". But, first things first.
In the singular cases, the covariant and contravariant metrics are no longer inverses, but separate, as dual metrics.
In general, you may consider a 2-parameter family of geometries that have the following as their invariants:
$$
β dt^2 - α \left(dx^2 + dy^2 + dz^2\right), \\
β \left(\left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2\right) - α \left(\frac{∂}{∂t}\right)^2, \\
dt \frac{∂}{∂t} + dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z},
$$
where $(α,β) ≠ (0,0)$. The first invariant yields a line element for a metric, the second yields a differential operator that embodies the dual metric. The contraction of the metric and its dual yields the third invariant, up to a factor of $-αβ$. For singular metrics, this reduces to 0.
If $αβ > 0$, that yields a geometry possessing $c = \sqrt{β/α}$ as a finite non-zero invariant speed. If $αβ < 0$, then it's the 4-dimensional Euclidean geometry, where time is a spatial dimension - which (according to Hawking and Hartle) is the local geometry of the Universe before the Big Bang; i.e. a 4-dimensional time-less space, instead of a 3+1 dimensional space-time.
If $α = 0$ and $β ≠ 0$, that's the geometry that has Galilean symmetry, while for $α ≠ 0$ and $β = 0$, it is the geometry possessing Carrollean symmetry. The limit case $(α,β) = (0,0)$ is actually also meaningful; this describing the geometry that has the Static Group as its local symmetry group. In that case, however, additional geometric invariants arise.
So, for the Carrollean case, the metric and dual metric are embodied by the first two of these invariants, which - we normalize $α$ to $-1$ - become:
$$dx^2 + dy^2 + dz^2, \hspace 1em \left(\frac{∂}{∂t}\right)^2.$$
The symmetry group that preserves these two invariants and the third above-mentioned invariant is just the Carroll group.
You can embed these geometries in 5 dimensions, adding an extra coordinate $u$, by treating the line element an expression for $β ds^2$, with an invariant (or "proper") time $s$ decomposed into $s = t + α u$. Then, one has, instead, the following sets of invariants:
$$
dt + α du, \hspace 1em dx^2 + dy^2 + dz^2 + 2 β dt du + α β du^2, \\
\frac{∂}{∂u}, \hspace 1em β \left(\left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2\right) - α \left(\frac{∂}{∂t}\right)^2 + 2 \frac{∂}{∂t} \frac{∂}{∂u}, \\
dt \frac{∂}{∂t} + du \frac{∂}{∂u} + dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z},
$$
with the 4-dimensional geometry corresponding to the case where the line element for the metric is set to 0.
The symmetry groups that preserve these invariants are the respective central extensions of the symmetry groups for the first set of invariants. The central extensions are trivial if $α ≠ 0$, else non-trivial if $α = 0$. Yet, both Carroll and Static have isomorphic central extensions.
In each case, they are 5-dimensional geometries with a metric, a dual metric, an invariant vector field and invariant one-form. For $β = 0$, the metric and dual metric are respectively given by the invariants:
$$
dx^2 + dy^2 + dz^2, \hspace 1em - α \left(\frac{∂}{∂t}\right)^2 + 2 \frac{∂}{∂t} \frac{∂}{∂u}. \\
$$
The 2-parameter family of the symmetry groups (that are 5, up to isomorphism, 4 for their respective central extensions) is a subfamily of the 3-parameter family that consists of the 14 (13, after taking central extensions) that were laid out by Bacry and Lévy‐Leblond in 1968 in the Journal of Mathematical Physics:
Possible Kinematics
https://aip.scitation.org/doi/abs/10.1063/1.1664490
It is possible to expand the geometric representation to a form that provides a uniform 5-dimensional geometric representation for the entire family. The infinitesimal transforms that arise from the above invariants are given by:
$$
δ = × - βt + , \hspace 1em δt = -α· + τ, \hspace 1em δu = · + ψ, \\
δ(d) = ×d - βdt, \hspace 1em δ(dt) = -α·d, \hspace 1em δ(du) = ·d, \\
δ∇ = ×∇ + α \frac{∂}{∂t} - \frac{∂}{∂u}, \hspace 1em δ\left(\frac{∂}{∂t}\right) = β·∇, \hspace 1em δ\left(\frac{∂}{∂u}\right) = 0,
$$
where
$$ = (x, y, z), \hspace 1em d = (dx, dy, dz), \hspace 1em ∇ = \left(\frac{∂}{∂x}, \frac{∂}{∂y}, \frac{∂}{∂z}\right),$$
which includes the infinitesimal transforms $$ for rotations and $$ for boosts, which yield the homogeneous groups, and $$ for spatial translations, $τ$ for time translations and $ψ$ for the $u$-translations, which yield the respective inhomogeneous extensions of the homogeneous groups.
The representations for the other members of the Bacry / Lévy‐Leblond family are non-linear.
