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?
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.