Non-Tight Polyhedral Surfaces:

Lemma: An immersed polyhedral surface M is tight if, and only if,
1. every local extreme vertex of M is a global extreme vertex of M,
2. every edge of the convex hull M is an edge of M,
3. no extreme vertex of M is in the double set of M.

All three conditions are required, as indicated by the diagram below which contains three surfaces, each satisfying two of the conditions but not the third.

The first surface is a square-based pyramid whose base has been pushed up into its interior by the addition of a new vertex. The convex hull is the original pyramid and all the edges are still in place so condition (2) is satisfied, and since there is no self-intersection, (3) also holds, but the surface is not tight; a horizontal plane can cut off both the top of the pyramid and the interior vertex, so the surface does not have the two-piece property, so it is not tight.

The second surface is the double cone over a triangle where the cone points are skewed vertically. All the local extreme vertices are on the convex hull satisfying (1), but an edge of the convex hull is missing (the one between the left- and right-most vertices of the surface). A horizontal plane can cut off these vertices so the surface does not have the two-piece property and is not tight.

Finally, the third surface has all its local extreme vertices on the convex hull, satisfying (1), and all the edges of the convex hull are part of the surface, satisfying (2), but the apex of the surface is doubly covered, once from the left and once from the right (this is a sphere with two points touching). Again, a horizontal plane can slice off the top, which will cut off two pieces from the rest of the surface, so it is not tight.

Tightness for Polyhedral Surfaces
The polyhedral solution

` 8/10/94 dpvc@geom.umn.edu -- ` `The Geometry Center`