Intrinsic and extrinsic curvature Does having intrinsic curvature always mean that there is extrinsic curvature? Can you have one without the other? Is it possible to have one without the other all the time?
And is it hypothetically possible for the intrinsic curvature of spacetime to change in different regions such that one regions curvature is modeled by Einstein's equations whereas another region might be Euclidean and not curve at all?
 A: You can trivially have intrinsic curvature without extrinsic if your manifold is not embedded in anything.
Extrinsic curvature depends on what you embed the manifold in. If you have a circular cylindrical manifold (zero intrinsic curvature) embedded in $R^3$ it can have positive extrinsic curvature around its circumference. But it can also be embedded in the very different space $R^2 \times [0,2\pi)$ where you identify the opposite sides of the interval (an infinite slab that repeats itself) so that there is no extrinsic curvature. This embedding-dependence is why intrinsic curvature usually is "better". Especially since inside a manifold you do not get any information about what space you are embedded in, if any.
The Einstein equations tell you how the intrinsic curvature changes in different locations, and are entirely fine with flat space if there is no matter around.
Addendum: As answered here, it is possible to have non-zero intrinsic curvature and zero extrinsic curvature if you use the right curvature measure. If one demands that all the extrinsic curvatures vanish the only intrinsic curvature that can remain is the one induced by the embedding space (which could be itself curved). But if the mean curvature is zero then minimal surfaces in $R^3$ like the catenoid have non-zero intrinsic and yet zero extrinsic (mean) curvature.
(The other surface curvature measure used for 2D surfaces embedded in 3D space, Gaussian curvature, does not allow this: zero Gaussian curvature surfaces are developable surfaces and have zero intrinsic curvature.)
A: It depends on the properties of the manifolds in question. Let $S$ be the manifold and $M$ be the manifold in which $S$ is embedded, both with dimension greater than 1. This makes $S$ a submanifold of $M$. The question can be divided into two cases depending on whether $M$ is curved.
If $M$ has intrinsic curvature, then the flattest submanifold one can get is a totally geodesic submanifold, which means that any geodesic in $S$ is also a geodesic in $M$. In this case, $S$ has zero extrinsic curvature. Its intrinsic curvature simply matches that of $M$. In other words, $S$ is "flat" relative to $M$.
If $M$ is flat, then the answer also depends on how you define extrinsic curvature. If the extrinsic curvature is taken to be the second fundamental form, then any submanifold $S$ with zero extrinsic curvature must also have zero intrinsic curvature. The converse is not true: Surfaces with zero intrinsic curvature can be given non-zero extrinsic curvature (such as the commonly-cited cylinder example). However, if you take extrinsic curvature to be the mean curvature, then it is still possible to have intrinsically-curved submanifolds with zero mean curvature. Such submanifolds are known as minimal surfaces.
