Proof of the Tightness Lemma:

First, recall that tightness is equivalent to the two-piece property for surfaces [More].

Remark: Since the number of components of a polyhedron (or its intersection with a half-space) depends only on its 1-skeleton, a polyhedral surface has the two-piece property if, and only if, its 1-skeleton does.

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.

Proof: Suppose first that M is tight (i.e., has the two-piece property), and let v be a local extreme vertex of M. Then there is a height function for which v is an isolated local maximum, and a plane perpendicular to the direction of the height function and just below v cuts off a small neighborhood of v. Since M has the two-piece property, the remainder of the surface must be below the plane (there can be only one piece on each side of the plane), so v is the (unique) global maximum for this height function, hence it lies on the convex hull of M and so is an extreme vertex. This verifies property (1).

Now suppose e = uv is an edge of the convex hull of M, and let N be the average of the (outward) normals of the two faces of the convex hull containing e. Consider the height function on M in the direction of N: a plane perpendicular to N slightly below e cuts off u and v from M (and no other vertices of M). Since M has the two-piece property, the region cut off by the plane must be connected, and by the remark above, it's 1-skeleton is connected. The only possible edge in this region is e = uv, so e must be in M. This verifies (2).

Finally, suppose a and b are two vertices of M that map to a common extreme vertex v. Then there is a height function for which v is an isolated global maximum, and a plane perpendicular to this direction, but slightly below v, cuts off a neighborhood of v; that is, a neighborhood of a and a neighborhood of b. If the slice is close enough to v, the neighborhoods will contain only one vertex each. Now since M has the two-piece property, the two neighborhoods must be connected, but since each contains only one vertex, it must be that the two neighborhoods are identical; that is, a equals b. This verifies (3).

Conversely, suppose M satisfies the three properties of the lemma, and let P be a plane intersecting M. Let N be the normal to this plane, and consider the height function in this direction. Each component cut off by P must contain a local maximum for this function, so each component contains a local extreme vertex of M. By (1), these must be global extreme vertices of M, and by (3) they must be distinct. These extreme vertices all belong to the convex envelope of M, and since convex surfaces have the two-piece property, these components lie within the same component of the convex envelope when it is cut by P. By the remark above they must be connected within the 1-skeleton of the convex envelope. By (2) all these edges are edges of M, so they connect the extreme vertices not only on the convex envelope, but in M itself. Thus there is only one component of M above P. A similar argument applies to the opposite side of P, and we see that P cuts M into exactly two pieces. This is true for any plane dividing M, so M has the two-piece property, hence is tight.

[Left] Tightness for polyhedral surfaces
[Left] The polyhedral solution
[Up] Introduction

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9/29/94 -- The Geometry Center