Is it necessary to embed a 4D surface in 5D space? Lets consider the line element:
$$ds^2=dr^2+r^2[d\theta^2+\sin^2\theta d\phi^2]$$
There are three variables r,theta and phi.
If we use a surface constraint like r=constant the number of
independent variables is reduced by one--now we have two independent
variables.These surfaces[corresponding to r=const] may be embedded in
a three dimensional space.
Now lets consider Schwarzschild's metric:
$$ds^2=(1-2m/r)dt^2 - (1-2m/r)^{-1} dr^2  - r^2[d \theta^2+\sin^2 \theta d\phi^2]$$
If we use a surface constraint[for example: t=constant] we have three
independent variables.The resulting time slices are three dimensional surfaces which are naturally embedded in a 4D space.
The General Relativity metric  has four variables: one relating to
time and three relating to the spatial coordinates.Any surface
constraint would reduce the number of variables to three.
In fact any arbitrary spacetime curve[world-line] may be made to lie
on a 3D- Surface obtained by applying some suitable constraint on 4D
space.The constraint may not be a simple one like t= constant or r= constant. It may be of a complicated nature.
For the purpose of embedding in GR a 4D space seems to be sufficient.
Queries:


*

*Is it essential the we should consider a 4D surface embedded in 5D space to understand or interpret GR?

*It appears that the curve[4D path] is more important than the surface on which it is lying since the same curve may be made to lie on several distinct surfaces at the same time[you may extend this to the case of  4D surf embedded in 5D space].Is this interpretation correct?

 A: Any enclosing space is outside of the problem domain of GR: All results can be obtained from within the space-time. Physically, it makes no sense to talk about an enclosing space which has no impact whatsoever on measurements. In particular, even though we say space-time is curved, the question Where does it curve to? makes no sense in the framework of GR.
Also, 5 dimensions are not enough to contain arbitrary 4-dimensional pseudo-Riemannian manifolds of index 1 if you want to preserve the metric. Quoting C. J. S. Clarke, On the Global Isometric Embedding of Pseudo-Riemannian Manifolds:

The space-time of general relativity can be embedded isometrically in $E^{2,q+2}$ (pseudo-Euclidean space of signature $q-2$) where $q=46$ or $q=87$ for compact or non-compact space-time, respectively.

However, the result is only valid for finite $k$ and not $C^\infty$.
A: No, it is not essential to consider a surface to be embedded in a higher-dimensional space. In fact the embedding is pretty much irrelevant. GR is based on differential geometry, and part of the reason differential geometry was developed the way it was is that it allows you to analyze surfaces without having to consider their embeddings in higher-dimensional spaces. You can calculate the geodesics or whatever else you want using only properties that can be measured from "within" the space.
