Tightness and the Convex Hull:

For an immersion to be tight, it must have minimal total absolute curvature [LINK]. In showing that the total absolute curvature is bounded below by 4 - X(M) [More], where X(M) is the Euler characteristic of the surface, the inequality is introduced when an integral is divided into two non-negative parts, one of which is simply dropped. To have minimal total absolute curvature, this dropped integral must actually equal zero; so a surface is tight if, and only if,
that is, the integral of the Gaussian curvature, K, over the region of the surface where the curvature is positive but not part of the convex envelope, H, must be zero.

Since the integrand is always positive, the only way this integral can be zero is for the region of integration to be empty. This means that all the positive curvature must be on the convex envelope of the surface, an observation that leads to the following important geometric description of tight surfaces:

Theorem: An immersion of a closed, compact, connected surface M is tight if, and only if, it can be decomposed into two components, M+ and M-, such that

  1. The curvature is non-negative on M+ and non-positive on M-,
  2. The image of M+ is an embedding and is equal to the convex envelope of the image of M minus a finite number of planar, convex disks,
  3. The boundary of each of these disks is the image of a non-trivial 1-cycle in M. These curves are called top cycles.

For example, the torus of revolution is tight (below left). Its decomposition into the M+ and M- regions is shown (right). Notice the two top cycles that form the common boundary of the M+ and M- regions.

[Right] Top sets and top cycles
[Right] Tightness and polar height functions
[Left] Tightness and its consequences
[Up] Kuiper's original question

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8/8/94 dpvc@geom.umn.edu -- The Geometry Center