Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Here's a question that recently came up for me that I feel sure must have a canonical answer.

Background. It's not so hard to prove that if you take a triple of points on a C^2 curve and take the limit of the inverse of the radius of the unique circle through the points as the points approach a single point x you get the curvature of the curve at x. Since the radius of this circumcircle can be written in terms of Cayley-Menger determinants, this gives curvature as a limit of (extrinsic) distance information.

Similarly, I think that for a C^1 curve, you can easily show that the curve is ``locally flat'' in the sense that the area of the triangle formed by a triple of points on the curve (divided by the square of the diameter of the triangle, say, to make it scale-invariant) approaches zero. This is probably even true for curves somewhat weirder than C^1, as long as they don't actually have corners.

Question. So what's the generalization for hypersurfaces (or just manifolds embedded in some Euclidean space)? It seems like exactly the same limits (inverse radius of circumsphere, quotient of volume of simplex by appropriate power of diameter) should be defined for n+2-tuples of points on an n-dimensional manifold. In particular, I'd like to know the following:

  • Is the limit of the inverse radius of circumsphere as points collapse to a single point the scalar curvature at the collapse point?

  • Is the ``locally flat'' condition (that is, "limit of (volume of simplex)/(diameter of simplex)^n+1 -> 0 as configuration collapses to a single point") implied by C^1? Does it imply C^1?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

The first question is cool, but unfortunately the answer's boring: The limit in general will not exist.

Counterexample: Consider the conic hypersurface $z=\lambda x^2+\mu y^2$. There are two symmetrical "1-parameter families of quadruples-of-points" for which the limit of the circumradius as $\epsilon\to 0$ is easy to calculate:

$\{(0,0,0), (\epsilon,0,\lambda\epsilon^2),(-\epsilon,0,\lambda\epsilon^2),(0,\epsilon,\mu\epsilon^2)\}$ and $\{(0,0,0), (\epsilon,0,\lambda\epsilon^2),(0,\epsilon,\mu\epsilon^2),(0,-\epsilon,\mu\epsilon^2)\}$.

For the first, the limit of the circumradius is $\frac{1}{2\lambda}$; for the second, it is $\frac{1}{2\mu}$.

However, here's a weaker statement that is true:

Fact: Let $\Sigma$ be a convex hypersurface of $\mathbb{R}^n$. Then $\frac{1}{\limsup}$ and $\frac{1}{\liminf}$ of the circumradius of an $(n+1)$-tuple of points, are the minimal and maximal [absolute-values-of-] principal curvatures respectively.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.