Suppose an $m$-dimensional manifold in an $n$-dimensional euclidean space, choose some point on this manifold and take an $n$-dimensional ball of certain radius $R$ centred in this point. If the volume of the manifold "enclosed" in this ball is $V$, what can we tell about the bounds for the different types of curvatures for the manifold inside this ball? Thanks very much in advance!
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Not a complete answer by any means, but a pretty and germane elementary theorem:
If $S$ is a sphere in $\mathbf{R}^3$ having radius $R_0$ and passing through the origin $O$, $R$ is a real number such that $0 < R \leq 2R_0$, and $\Sigma_R$ is the sphere of radius $R$ centered at $O$, then the area of the portion of $S$ inside $\Sigma_R$ is $\pi R^2$. (!)
Philosophically, the volume (here, area) of $S$ inside $\Sigma_R$ cannot distinguish $S$ (with constant positive curvature) from a flat plane until $R > 2R_0$.
Andrew D. Hwang
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$\Sigma_R$ is the sphere of radius $R$ in terms of geodesic distance, being $\Sigma$ a submanifold of $\mathbf{R}^3$, right? Is it extensible to other dimensions? – bourbaquez Aug 18 '13 at 16:32
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@bourbaquez: Here $R$ is the (extrinsic) Euclidean distance. I haven't checked in detail, but am pretty sure this theorem holds only in $\mathbf{R}^3$. (It's certainly false for circles in $\mathbf{R}^2$.) – Andrew D. Hwang Aug 18 '13 at 18:00
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Thanks! Where can i find this theorem? – bourbaquez Aug 18 '13 at 19:58
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It's a straightforward calculus exercise. :) I first came across it in one of the Marsden-Weinstein calculus books. – Andrew D. Hwang Aug 18 '13 at 21:00
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Yep, that's true. In any case, it would be nice to know how to extend it to higher dimensions/codimensions... – bourbaquez Aug 19 '13 at 11:28