Example of curved spacetime with flat space layers

Question

curved spacetime is a famous geometrical and physical object. on the other hand there are debates on if space itself is curved or not. but is it really possible to a curved spacetime that all of its spatial layers be flat?
more precisely: every freely falling observer at every instance sees a 3-dimensional space. is it possible that all of this 3-dimensional spaces are flat while spacetime itself is curved?
is there a any explicit metric by this property?

Answer 1

The Friedmann metric with $k=0$ is spatially flat. It is consistent with current cosmological observations.

Answer 2

is it possible that all of this 3-dimensional spaces are flat while spacetime itself is curved? is there a any explicit metric by this property?

Sure, it's called the Friedman-Robertson-Walker metric and it describes the large-scale evolution of our Universe: $$ ds^2 = -dt^2 + a^2(t) (dx^2 + dy^2 + dz^2) $$ On any surface of constant $t$, the metric is that of flat Euclidean space. The Einstein equation gives an equation of motion for the so-called scale factor $a(t)$.

(Technically, this is only one of three possible forms of the FRW metric; there are also versions where the surfaces of constant $t$ are curved. But the version above is the simplest one, and also appears to be the one that describes our Universe.)

Answer 3

As @G. Smith said,

The Friedmann metric with $k=0$ is spatially flat,

but it is worth noting that this is an idealisation which depends on a perfectly uniform matter distribution. As such, we know it is an approximation to reality. It is only flat on a slice of constant cosmic time, which means it is only flat for observers with constant position in comoving coordinates (i.e. stationary in the CMB frame). For moving observers it would not be flat.

Also, when we consider observers in free fall, we are likely to be talking of observers in the vicinity of a gravitating body. Space in the vicinity of a gravitating body cannot be perfectly flat, although typically space curvature is so small that it is not directly measurable.

So when we debate whether space may be flat, we mean flat on cosmological scales. We know it is not perfectly flat.

Answer 4

A simple example of "spatial" slices not sharing the curvature of the background space is Euclidean space sliced into concentric spheres. This is similar to what happens in FLRW geometries. Like the spheres, the FLRW spatial slices are generally extrinsically curved (lines in them aren't geodesics).

I can't think of an example of a curved space with flat slices that is so easy to visualize, because it would need to be three-dimensional and to be easily visualizable it would need to be embedded in a flat background of four or more dimensions. But you could think of a conformally flat space with a metric of $a(z)^2(dx^2+dy^2+dz^2)$ or $a(t)^2(dx^2+dy^2-dt^2)$. It's obvious that slices of constant $z$ or $t$ are flat, and pretty easy to see that they are extrinsically curved iff $a'\ne0$ (because parallel paths on opposite sides of the surface have different lengths). Spatially flat FLRW cosmologies can be written in this form. The time coordinate is called conformal (cosmological) time, and is usually written with $η$ rather than $t$.

Answer 5

Schwarzschild metric describing an isolated non-rotating black hole (and also spacetime outside isolated spherically symmetric body) admits “slicing” of spacetime in flat spatial layers with the help of Gullstrand–Painlevé coordinates: $$ g = \left(1-\frac{2M}{r} \right)\, dt^2- 2\sqrt{\frac{2M}{r}} dt dr - dr^2-r^2 \, d\theta^2-r^2\sin^2\theta \, d\phi^2. $$ The geometry of a slice $t=\text{const}$ is a flat 3D space in spherical coordinates: $$ dl^2=dr^2+r^2 \, d\theta^2+r^2\sin^2\theta \, d\phi^2. $$

Metric of rotating black hole (Kerr metric), however, does not admit spatially flat slicing.

Note, that there seems to be some misconceptions that this question displays:

more precisely: every freely falling observer at every instance sees a 3-dimensional space. is it possible that all of this 3-dimensional spaces are flat while spacetime itself is curved?

1. Slicing of spacetime into spatial layers by itself does not have anything to do with observers. If we consider observers staying at the fixed values of spatial coordinates as metric evolves with time (so-called fiducial observers), then such observers generally would not be free-falling.

2. What an individual observer sees is not the geometry of spatial slice. Instead one should remember that light propagates with a finite velocity, so the light rays arriving to observer at a specific moment $t=t_0$ carry information from the past slices. Moreover, these light rays would generally be bent by gravitational field (even when spatial geometry is perfectly flat).