# 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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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 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 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 $n=2$).