**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,*

*every local extreme vertex of**M*is a global extreme vertex of*M*,*every edge of the convex hull**M*is an edge of*M*,*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.

* 9/29/94 dpvc@geom.umn.edu -- *

*The Geometry Center*