Light ray in Friedmann–Lemaître–Robertson–Walker metric What are the main differences in the path of light ray for different values of $k$ in the FLRW metric?
Is it possible to build an experiment (even a mental one) with a single light ray to distinguish which variant of the metric we live in?
 A: Interesting that you should ask, since I have been working on trying to refine a thought experiment to deal with this issue.
If $\Omega_k$ is sufficiently larger or smaller than zero, it should be theoretically possible to observe that the intensity (watts per square meter) of CMB radiation is correspondingly larger or smaller than what it should be for $\Omega_k = 0$.
If you would like more details, please let me know.
ADDED
I found an error in my idea. I will not bother showing the math unless someone wants to see it. My idea was that in a curved universe the watts per sq meter ($W/M^2$) of CMB radiation reaching an observer's measuring instrument would vary with the curvature. For a hyper-spherical universe the $W/M^2$ emitted would be less than for a flat universe, and for a hyper-hyperbolic universe $W/M^2$ would be more than for a flat universe. What I has overlooked was that the fraction of the emitted radiation that would reach an observer's device also varied with curvature. For a hyper-spherical universe the $W/M^2$ emitted would be more than for a flat universe, and for a hyper-hyperbolic universe the fraction would be less than a flat universe. Furthermore, these two effects exactly cancelled each other out.
ANOTHER ADDED
Math for showing no change due to curvature regarding the CMB photon power received by an observer’s measurement device
I have organized this presentation as two steps. Step 1 will show the development of equations comparing the watts of photon energy produced from two spherical surfaces with the same radius, but with respect to two different geometries. Step 2 will show the development of equations comparing the fractions of the emitted photons from these two spherical surfaces that will hit the observer’s measurement device. The two geometries are (1) infinite flat Euclidean and (2) finite curved hyper-spherical.
Step 1. Sphere $S_1$ is in a 3D Euclidean flat space. Sphere $S_2$, is in a 3D spherically curved space which can be imagined as the 3D boundary of a 4D hyper-sphere. Both of these 2D spheres represents a spherical surface centered on the Earth with a radius equal to the distance between (a) the current location of the Earth, and (b) the surface where it is now in a much expanded universe, which long ago emitted the photons of the CMB. I call this radius distance “R”. To simplify the presentation, I will first develop the math in terms of 2D circles with 1D circumference boundaries rather than in terms of 3D spheres with 2D boundaries.
The following are notations for the two above described spheres.
$O$: center of the sphere where the Earth observer has the measurement device
$R$: the current radial distance between $O$ and the photon radiating sphere.
$A_1$: is the surface area of sphere $S_1$ (in flat space)
$A_2$: is the surface area of sphere $S_2$ (in hyper-spherical space)
$C_1$: is the circumference of a circle of radius $R$ on a flat plane
$C_2$: is the circumference of a circle of radius $R$ on a spherical surface
$\rho$: is the ratio $A_2\ /\ A_1\ = (C_2\ /\ C_1)^2.$
The following is a simplified analysis of the curved space in which the 3D hyper-sphere space is represented as the 2D surface of a 3D globe-like sphere $G$, and the $S_2$ photon radiating sphere is represented as a 1D latitude circle $L$.  Imagine a globe-sphere (sort of like the Earth) with (1) a North Pole $N$ and a South Pole, (2) many circles of longitude each with radius $r$, and each passing through both the North and South Poles, (3) an equator of radius $r$ equidistant between North and South poles, and (4) many circles of latitude each with its plane parallel to the equator, and having a radius less than $r$.
The following are additional notations for the points, angles, lines and distances related to this simplified version of $S_2$.
$G$: The 2D sphere on which an observer measures photon radiation from a latitude circle $L$
$o$: center of the sphere $G$
$N$: north pole on $G$
$L$: a particular latitude circle on $G$
$r$: radius of the 2D spherical “globe” $G$
$s$: radius of $L$
$l$: a particular point on circle $L$
$n$: the particular point on the line between $O$ and $N$ which is closest to the point $l$
$\theta$: the angle at point $n$ in the right triangle $O$-$l$-$n$
(If you have trouble visualizing this geometry, I suggest you draw a circle with the points $N$ and $l$ on its circumference, and then draw the right triangle $O$-$l$-$n$.)
The first fact is that the length $C_2$ of the circumference of a circle of radius $R$ in curved 2D spherical space $S_2$ is not the same as it is in $S_1$ in the 2D flat Euclidean space. In Euclidean space $S_1$
$$C_1 = \pi R.$$
The radius $s$ of the latitude circle $L$ is
$$s = r\ sin\ \theta.$$
Therefore the circumference of the circle with the curved distance $R$ is
$$C_2\ = \pi\ s\ =\ \pi\ r\ sin\ \theta.$$
Note that
$$\theta\ =\ R/r,$$
so the ratio $\rho$ of the curved space $S_1$ circumference to the flat space $S_2$ circumference is
$$\rho\ =\ (r\ sin\ (R/r)\ /\ R)^2.$$
To get a sense of the value of this ratio I expand the square root to two terms as follows.
$$\rho\ ~=\ 1 - (R/r)^2/3.$$
Now for step 2. Here I will calculate the fraction of produced photons that will hit the observer’s measurement devise for both $S_1$ and $S_2$. For both geometries, think of each point $p$ of the photon emitting surface as the center of a sphere of radius $R$ with surface areas $A_1$ and $A_2$. Imagine the measuring device as a small sphere of radius $m$ with the same center as a sphere $S_1$ or  $S_2$. A sphere of radius $R$ includes every possible point that an emitted photon can hit at that distance. The measurement device, has a target surface area of $\pi/ m^2$ facing the direction of the point $p$. Therefore the probability $P_1$ that the emitted photon will hit the measuring device in space $S_1$ is
$$P_1\ =\ \pi\ m^2\ /\ A_1,$$
and for space $S_2$ the probability is
$$P_2\ =\ \pi\ m^2\ /\ A_2.$$
Note that the ratio
$$P_2\ /\ P_1\ =\ A_1\ /  A_2\ =\ 1\ /\ \rho.$$
Thus the total photon energy hitting the target measurement device is identically the same for the two geometries.
