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 ndimensional space, and the nspace 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 Petrunin Apr 16 '13 at 16:52It'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. 


Here is an attempted answercuminterpretation, 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 $n=2$). 

