Kevin McHugh said:
In the absence of mass, G=0. Does this imply R=0, and that space is flat in between massive objects? If space is always curved is that intrinsic or extrinsic curvature? Thanks in advance for your insight.
If the Einstein tensor ##G_{ab}## is zero, as other posters have noted the Ricci tensor ##R_{ab}## is zero, but this doesn't imply that "space is flat", because the fulll Riemann tensor ##R_{abcd}## is not necessarily zero.
When we talk about curvature in GR, we're pretty much always talking about intrinsic curvature, just because we are imagining that we live within the universe, not outside it. To have extrinsic curvature, we'd need a concept of what lay outside the universe. This isn't a part of most theories, mainly due to experimental difficulties in stepping outside the universe to "take a look".
The curvature of space-time can be reduced to the intrinsic (aka Gaussian) curvature of all possible space-time slices. See for instance
https://en.wikipedia.org/wiki/Gaussian_curvature. So we can regard the Riemann tensor as giving us the Gaussian curvature of any two dimensional subspace. And if we know the Gaussian curvature of all two dimensional subspaces, we can find the Riemann tensor.
To make a precise statement about "space being curved" one needs a definition of space. A good candidate for this definition would be the Bel decomposition of the Riemann tensor,
https://en.wikipedia.org/wiki/Bel_decomposition. One could regard the "topgravitic" part of this tensor as representing the idea of "space curvature".
GIven this precise definition of what "spatial curvature" is, one can additionally say that in a vacuum region of space-time, the topogravitic part of the Riemann is equal to the electrogravitic part. So the presence of any tidal forces (which are represented by the electrogravitic part of the Riemann tensor) in a vacuum space-time implies that "space is curved", in the sense that the topogravitic part of the tensor must be non-zero as well.
The tidal for components would be, for instance ##R_{txtx}, R_{tyty}, R_{tztz}##. If we orient or coordinate axis correctly, we can make all the non-diagonal terms of the electrogravitic tensor zero, leaving only these three diagonal terms. The corresponding numerically equal topogravitic components of the tensor would be given by the Hodges dual, where (tx)->(yz), (ty)->(xy), (tz)->(yz), so that the numerically equal components would be ##R_{txtx}->R_{yzyz}##, and similarly for the other two.
Thus if (tx) plane of space-time has a non-zero Gaussian curvature, we can say that ##R_{txtx}## is nonzerom which implies that ##R_{yzyz}## is nonzero (assuming a vacuum space-time), and thus we can conclude that the purely spatial (yz) plane has a non-zero Gaussian curvature.