Does some type of curvature require the space be an embedded manifold in a higher-dimensional space? [closed]

Assuming an appropriate definition of 'curvature', is there a theorem that says: "At least n+1 dimensions are necessary for a particular curvature to exist in an n-dimensional space, and the n-space must be an embedded manifold in the (n+1)-space."?

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closed as not a real question by Misha, Gerald Edgar, Lee Mosher, Deane Yang, Anton PetruninApr 16 '13 at 16:52

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

What metric do you use on the ambient space? Do you want embedding to be isometric? If answer is "yes" to both, then the answer to your question is: If metric on n-manifold is not flat, then the manifold does not embed isometrically in Euclidean n-space. – Misha Apr 15 '13 at 22:18
Dear Paul: Honestly, I do not think there is a real question here. My suggestion for you is to pick up a textbook on differential geometry of curves and surfaces (like do Carmo) and read about definition of Gaussian curvature, 2nd fundamental form, etc. Voting to close. – Misha Apr 16 '13 at 4:30

Here is an attempted answer-cum-interpretation, modulo adjusting some dimensions: A closed surface of negative Gaussian curvature cannot be isometrically embedded in $\mathbb{R}^3$. So in this case, at least $n+2$ dimensions are needed (where $n=2$).