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192    I. Ivrissimtzis and H.-P. Seidel










                           Fig. 10.7. From left to right: (a) The unit cube. (b) P 110 =(0.2, 0.0, 0.0).(c) P 110 =
                           (0.2, 0.2, 0.0). (d) P 110 =(0.0, 0.0, 0.2). (e) P 111 =(0.2, 0.2, 0.2).



                           algorithm. The row of the matrix giving the new position of the vertex corresponding
                           to g is
                                                            3 n−σ(g)
                                                       a g =                            (10.17)
                                                              4 n
                           For example, if n =2 we get the matrix
                                                      ⎛       ⎞
                                                        9331
                                                      ⎜ 3913  ⎟
                                                      ⎜       ⎟  /16                    (10.18)
                                                      ⎝ 3193 ⎠
                                                        1339
                              We have

                           Proposition 4. The eigenvalue corresponding to the eigenvector v g of the subdivi-
                           sion matrix of the n-dimensional quadratic spline is  2 σ(g) .
                                                                       1
                              For a sketch of the proof, we notice by Eq. 10.10, 10.17 the eigenvalue corre-
                           sponding to the character χ g is

                                                                3 n−σ(h)
                                                      =    χ g (h)                      (10.19)
                                                  λ χ g             n
                                                                   4
                                                        h∈G
                           giving,
                                                     (3 + 1) n−σ(g) (3 − 1) σ(g)
                                                   =                                    (10.20)
                                               λ χ g           n
                                                              4
                           To see this, we expand the product (3+1) n−σ(g) (3−1) σ(g)  and rearrange the factors
                           so that the terms (3-1) are placed at the positions where δ(g)=0. Finally, from
                           Eq. 10.20 we get
                                                               1
                                                           =                            (10.21)
                                                       λ χ g
                                                             2 σ(g)
                             # $
                              In the limit, the cell converges to a single point, which is its barycenter. Assuming
                           that the barycenter is the origin, the limit shape is given by the eigencomponents of
                           the next eigenvalues, that is by the n components with eigenvalue 1/2. After scaling
                           the cell to counter the shrinkage effect we get
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