Page 105 - Geometric Modeling and Algebraic Geometry
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6 Classification of Surfaces  103
                           b) Two conjugate couples: type III

                           Lemma 11. We assume that (t 1 ,u 1 )= (t 2 ,u 2 ) and (t 3 ,u 3 )= (t 4 ,u 4 ). It exists two

                           real homographies η 1 ,η 2 : P (R)→P (R) and two values θ, θ ∈ [0,π] such that:
                                                  1
                                                         1
                                                                    iθ
                                       η 1 (t 1 )= i, η 1 (t 2 )= −i, η 1 (t 3 )= e ,η 1 (t 4 )= e −iθ
                                      η 2 (u 1 )= i, η 2 (u 2 )= −i, η 2 (u 3 )= e iθ   ,η 2 (u 4 )= e −iθ
                              Therefore, by choosing the four special tangent planes as (X =0), (Y =
                           0), (Z =0), (T =0) and by a similar demonstration as in the generic complex
                           case, we have the parametric complex representation of the surface in the affine chart
                           s = v =1:
                                               ⎧
                                               ⎪ X =(t − i)(u − i) 2
                                               ⎪
                                                 Y =(t + i)(u + i) 2
                                               ⎨
                                                          iθ
                                               ⎪ Z =(t − e )(u − e iθ   ) 2
                                               ⎪
                                               ⎩          −iθ      −iθ
                                                 T =(t − e   )(u − e  ) 2
                              By similar transformation as in the case (a), we obtain the following proposition
                           (with two moduli θ and θ ):

                           Proposition 12. A surface of type III has a real parameterization as follows:
                                ⎧
                                  X = tu − t − 2u
                                        2
                                ⎪
                                ⎪
                                ⎪ Y =2tu + u − 1
                                ⎪
                                            2
                                ⎨
                           (S):   Z =(  t  − cotan θ)((  u  − cotan θ ) − 1) − 2(  u  − cotan θ )

                                                                     2
                                ⎪      sin θ          sin θ                   sin θ
                                ⎪
                                ⎪        t              u                 u
                                ⎪
                                  T =2(     − cotan θ)(    − cotan θ )+(     − cotan θ ) − 1
                                ⎩                                                      2
                                        sin θ         sin θ             sin θ
                           6.5 Non generic cases
                           We now list the following particular cases arising in the intersection of two curves of
                           bidegree (1,2) whose equations are ϕ 1 (t, u) and ϕ 2 (t, u).
                           6.5.1 Their intersection is finite
                           Set ϕ 1 (t, u) ∩ ϕ 2 (t, u)= {(t 1 ,u 1 ), (t 2 ,u 2 ), (t 3 ,u 3 ), (t 4 ,u 4 )}. We distinguish the
                           following cases:
                              a) 4 distinct points.
                                We have two cases: either (t 1 = t 2 and t 3  = t 4 )or(t 1 = t 2 and t 3 = t 4 ).
                              b) 2 distinct points and 1 double point (t 3 ,u 3 )=(t 4 ,u 4 ).
                                 We have 3 cases: either (t 1 − t 2 )(t 2 − t 3 )(t 1 − t 3 )  =0,or t 1 = t 2 or
                           t 1 = t 3 (= t 4 ).
                              c) 2 double points.
                              d) 1 triple point and 1 simple point.
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