Page 188 - Geometric Modeling and Algebraic Geometry

190 I. Ivrissimtzis and H.-P. Seidel
000 1 1 1 1 1 1 1 1
100 1 -1 1 1 -1 -1 1 -1
00 1 1 1 1 010 1 1 -1 1 -1 1 -1 -1
10 1 -1 1 -1 001 1 1 1 -1 1 -1 -1 -1
/4 /8
01 1 1 -1 -1 110 1 -1 -1 1 1 -1 -1 1
11 1 -1 -1 1 101 1 -1 1 -1 -1 1 -1 1
011 1 1 -1 -1 -1 -1 1 1
111 1 -1 -1 -1 1 1 1 -1

Table 10.1. The eigenvectors of Z 2 and Z 2 are the rows of the tables.
3
2
Similarly to the polygonal case, the point P is the barycenter of the cube
(0,...,0)
and we may assume it to be the origin. Then, there are n eigenvectors with σ(g)=1,
e.g. 10 and 01 for n =2 and 100, 010 and 001 for n =3. The components of these
n
n
eigenvectors are equal to 1/2 at all vertices of one face, and −1/2 at all vertices

of the opposite face. Thus, a point P with σ(g)=1 is at the barycenter of a face,
g

which means that the barycenter of the opposite face is at −P . Generally, there are
g

n
k elements of G with σ(g)= k.
Below we study in more detail the cases n =2, 3.
The quadrilateral (n=2): Using the geometrically intuitive cyclic ordering of the
vertices instead of ordering them by the value of σ as above, let the initial quadri-

lateral be (P ,P ,P ,P . The component (P ,P ,P ,P is its barycenter

00 10 11 01 00 00 00 00

and we may assume it is the origin. The next components, (P ,P , −P , −P )

10 10 10 10

and (P , −P , −P ,P ) join the middles of opposite edges of the quad. If the
01 01 01 01

fourth component (P , −P ,P , −P ) is zero, then the quad is a parallelogram,

11 11 11 11

with (P , −P ) and (−P ,P ) giving its two directions. The magnitude of the

10 10 01 01
fourth component can be thought as a measure of how far is the quad from being

a parallelogram. Notice that (P , −P ) joins the middles of the diagonals of the

11 11
quad, while a quad is a parallelogram if and only if its diagonals bisect. A quad is
planar if and only if P , P and P are linearly dependent.

01
11
It is interesting to compare the decomposition of the quad given by G = Z 2
10
2
with the one obtained with G = Z 4 , cf. Fig. 10.5. The difference in the two decom-
positions is the use of the eigenvectors (1,i, −1, −i) and (1, −i, −1,i) instead of
(1, 1, −1, −1) and (1, −1, −1, 1). The former, represent two squares with opposite
orientation and a linear combination of them is the affine image of a square, i.e., a
parallelogram [2].
The hexahedron (n=3): Similarly to the case n =2, P 000 is the barycenter of the
hexahedron while the three components with σ(g)=1 give the three directions
of a parallelepiped. The hexahedron is a parallelepiped if and only if the four other
components are zero. The geometric interpretation of the first three of these four con-
ditions is recursively related to the two dimensional case. Indeed, a hexahedron has
twelve edges, which can be separated into three subsets of four edges with the same