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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?

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  • $\begingroup$ Just a clarification, my second question has been misunderstood. I want to know if say the geometric properties of space can arbitrarily change from region to region, “imagine our solar system passing through an area of space described by Einstein equations to an area described by euclidian geometry. This is purely hypothetical but i want to know if it entails any mathematical contradictions (violating axioms of GR isn’t a contradiction as axioms describing spacetime would change) $\endgroup$ Aug 18, 2021 at 17:00

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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.)

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    $\begingroup$ Aside from the trivial case in the first sentence, this answer is backward to the question. The cylinder has zero intrinsic curvature and (potentially) non-zero extrinsic. I thought of that first too. But the question is for non-zero intrinsic and zero extrinsic. I couldn't think of an example of that immediately. $\endgroup$
    – Brick
    Aug 18, 2021 at 12:53
  • $\begingroup$ @Brick - Well, if you assume the manifold has non-zero intrinsic curvature you should state that in the question. Rather hard to respond to unstated assumptions. $\endgroup$ Aug 18, 2021 at 12:55
  • $\begingroup$ It's not my question, but I think that part was clearly stated in the first line: "Does having intrinsic curvature always mean that there is extrinsic curvature?" $\endgroup$
    – Brick
    Aug 18, 2021 at 12:57
  • $\begingroup$ @Brick - Thanks for prodding me, I found a positive answer. $\endgroup$ Aug 19, 2021 at 7:53
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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.

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  • $\begingroup$ "If $M$ has intrinsic curvature, then it is not possible for $S$ to have zero intrinsic curvature." If $S$ is one dimensional, this is trivially false. $\endgroup$
    – TimRias
    Aug 18, 2021 at 13:46
  • $\begingroup$ It is not necessarily true otherwise either. (E.g. consider $R^2\to R^2 \times S^2$) $\endgroup$
    – TimRias
    Aug 18, 2021 at 13:51
  • $\begingroup$ @mmeent Maybe I phrased it wrongly. I mean that the curvature of $S$ will follow that of $M$. In this case $R^2$ does have zero sectional curvature. $\endgroup$ Aug 18, 2021 at 13:54

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