General relativity tells us that the universe is bent by gravity, but this curvature is intrinsic to the universe (the universe bends, but not in a fourth spatial dimension, the universe having only three bent spatial dimensions).
How could we know the difference between the intrinsic curvature of a 3D spatial universe, and a the extrinsic curvature of a 3D shape in a 4D universe?
Obviously, there is a similar question about our curved 4D spacetime inside a 5D spacetime.
As I understand it, the question seems to be asking how one distinguishes the intrinsic curvature of a $d= 3$ manifold $\Sigma$ from the extrinsic curvature of instead some $\Sigma \subset M$ for a $d=4$ manifold $M$.
In differential geometry, there is a formalism due to Gauss and Codazzi that expresses the relation between the curvature of a sub-manifold and the manifold in which it is embedded. In particular, for some $\Sigma \subset M$ one has that,$^\dagger$
$$^{(\Sigma)}R^a_{bcd} = {}^{(M)}R^{a'}_{b'c'd'}h^a_{a'}h^{b'}_bh^{c'}_ch^{d'}_d \pm \sum_{i=1}^{\mathrm{codim}(\Sigma)}{K_i^a}_c K_{ibd} - {K_i^a}_d K_{ibc}$$
where $h$ is the fundamental form on $\Sigma$ and $K$ is the extrinsic curvature of $\Sigma$ for the $i$th normal, of which there are in number the same as the co-dimension of $\Sigma$.
Thus, the intrinsic curvature of a sub-manifold $\Sigma \subset M$ has a contribution from the intrinsic curvature of the manifold it is embedded in as well as from its own extrinsic curvature, which is itself a sort of consequence of the embedding.
So, the curvatures are all interlinked. If we had a three-dimensional universe which we thought of as not being embedded at all, then there would be no contributions from $K$ to the intrinsic curvature of the manifold.
Re-arranging the Gauss-Codazzi equation, one can then see that if we had a three-dimensional universe (or some hyper-surface) embedded in a four-dimensional one, then the extrinsic curvature picks up contributions from the intrinsic curvature of both manifolds in a sense.
$\dagger$ The sign ambiguity, $\pm$, is due to the fact that if for a particular $i$, the $i$th normal $n_i$ is time-like, that is, $n_in^i >0$ then one subtracts, and if $n_i n^i < 0$, then one adds the contribution instead. For a Riemannian manifold of totally positive signature, it is always a minus.
@Jamal showed a good way to calculate it, and that they can be somewhat intertwined. Still, for the cosmological solution one can easily see what causes the curvature, and what is curved. A good way to understand it is to solve the equations for general relativity. If you do it for d=4, with one of the dimensions timelike, you get the Robertson Walker metric, and for the modeled equations of state (which account for matter, radiation and if one introduces the cosmological constant, dark energy) one gets the full FLRW solution.
The metric solution can be written as, with k = 0, 1 or -1,
$$ds^2 = dt^2 - a(t)^2[dr^2/(1-kr^2) + d\Sigma^2]$$
with the item in square parenthesis the metric of a constant curvature k spatial section, and the last term is simply the angular components
$$d\theta^2 + sin^2(\theta)d\phi^2$$
You can calculate curvatures, but it is clear the spatial section has intrinsic curvature k, and the spatial slices are growing with a linear scale factor a(t) in each spatial dimension.
The Ricci scalar R, which is an invariant measure of the 4d curvature is dependent on k and a(t). See the reference to Wikipedia below for the exact dependence. This says there is a spacetime curvature, and it is true even if k = 0, which gives flat spatial slices.
So yes, space may be curved (cosmological data has k =0 to within a few percent, i.e. flat space) but spacetime definitely is curved if a(t) changes, which we know it does.
See wiki at https://en.m.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric
This is intrinsic curvature, for the spacetime, and for the spatial slices you can see where is the contribution, if k is not zero, and that otherwise they have zero curvature. The wiki article gives you the non-zero terms for the 4d Ricci components, and in fact $R_{tt}$ is proportional to the second derivative of a(t). Other Ricci components also depend on derivatives of a(t).
You can do all this without invoking any other dimensions.
A good way to understand it is to solve the equations for general relativity. If you do it for d=4, with one of the dimensions timelike, you get the Robertson Walker metric, and for the modeled equations of state (which account for matter, radiation and if one introduces the cosmological constant, dark energy) one gets the full FLRW solution.
The metric solution can be written as, when for simplicity we take the spatially flat case favored by the cosmological data (and does not
$$ds^2 = dt^2 - a(t)^2[dr^2/(1-kr^2) + d\Sigma^2]$$
with the item in square parenthesis the metric of a constant curvature k spatial section, and the last term is simply the angular components
$$d\theta^2 + sin^2(\theta)d\phi^2$$
You can calculate curvatures, but it is clear the spatial section has curvature k, and the spatial slices are growing with a linear scale factor a(t) in each spatial dimension.
The Ricci scalar R, which is an invariant measure of the 4d curvature is dependent on k and a(t). See the reference to Wikipedia below for the exact dependence. This says there is a spacetime curvature, and it is true even if k = 0, which gives flat spatial slices.
So yes, space may be curved (cosmological data has k =0 to within a few percent, i.e. flat space) but spacetime definitely is curved if a(t)changes, which we know it does.
All of this is intrinsic curvatures, for the spacetime, and any spatial slices.
The answer is in the "inverse square law". Gravity would follow inverse cube law if it was using the 4th dimension for curving. Gravity only knows three dimensions. In a 5 dimensional universe, gravity would follow inverse fourth power law and so on.
As inverse square law remains valid in "most realistic scenarios", it indicates that there are no more than 3 spatial dimensions in universe.
Therefore any curving has to be intrinsic.
