Does deformation of spacetime imply deformation of space? In general relativity it is said that gravity is a deformation of spacetime. Does  this deformation take place only when I consider space and time as one entity, or is this a real deformation in space and in time individually?
For example, if I measure the internal angles of a gigantic triangle formed by the three stars, and there is a massive body in the center of this triangle, then, will the sum of the angles form 180° ?
Appreciate.
 A: One can have a non-flat spacetime, with time translation symmetry and a flat $t=$constant section. An example is
$$
d\tau^2 = dt^2- \delta_{ij}(dx^i - v^i(x)dt)(dx^j-v^j(x) dt).
$$
All  components of the curvature $R_{\lambda\mu\rho\sigma}$ that do not contain a "$t$" are zero. I think that
$$
R_{tkij}= \frac 12 \partial_k(\partial_i v_j-\partial_j v_i),
$$
and that $R_{titj}$ is also non-zero, but I have not checked these recently.
A: As Einstein showed, space can not exist independent of time and vice versa. Riemann tensor describes the curvature of spacetime where it's written as
$$R^{\lambda}_{\beta \nu \mu} = -\Gamma^{\lambda}_{\beta \nu , \mu} + \Gamma^{\lambda}_{\beta \mu , \nu} - \Gamma^{\sigma}_{\beta \nu} \Gamma^{\lambda}_{\sigma \mu} + \Gamma^{\sigma}_{\beta \mu} \Gamma^{\lambda}_{\sigma \nu}$$
Christoffel symbols:
$$\Gamma^{\alpha}_{\mu \rho} = \frac{1}{2}(g_{\mu \nu , \rho} + g_{\nu \rho , \mu} - g_{\rho \mu , \nu})$$
Where all Greek letters run $(0,1,2,3) \equiv (t, x, y, z)$
So you should consider spacetime as one entity since you can't seperate them. Speaking of sum of the angles in different geometries:
Euclidean: $180$ degrees
Non-Euclidean Elliptic: $> 180$ degrees
Non-Euclidean Hyperbolic: $< 180$ degrees
