Implicit matrix representation of curves via quadratic relations - PowerPoint PPT Presentation

Implicit matrix representation of curves via quadratic relations Fatmanur Yıldırım Aromath Team Inria Sophia Antipolis M´ editerran´ ee joint work with Laurent Bus´ e and Cl´ ement Laroche

What are parametric curves? R n φ := R → � � f 1 ( s ) f 0 ( s ) , · · · , f n ( s ) s �→ , f 0 ( s ) image of φ defines a curve in R n .

Implicitization Example Unit circle in R 2 .

What are implicit equations used for ? Example: (plane curves) ◮ Is a given point on a given plane curve C ? p = ( x , y ) : point in R 2 , F ( T 1 , T 2 ) = 0 : implicit equation of the C . Question Is F ( x , y ) = 0 ?

What are implicit equations used for ? Example: (plane curves) ◮ Is a given point on a given plane curve C ? p = ( x , y ) : point in R 2 , F ( T 1 , T 2 ) = 0 : implicit equation of the C . Question Is F ( x , y ) = 0 ? Example: (space curves) ◮ Is a given point p = ( x 1 , · · · , x n ) on a given space curve C in R n ? F 1 , · · · , F r with F 1 � = · · · � = F r define the C . Question Are F 1 ( x 1 , · · · , x n ) = 0 , · · · , F r ( x 1 , · · · , x n ) = 0 ?

What are implicit equations used for ? ◮ Intersection of curves C 1 and C 2 , both in R 2 : C 1 is given by the parameterization R 2 → R � � f 1 ( s ) f 0 ( s ) , f 2 ( s ) s �→ , f 0 ( s ) C 2 is given by the implicit equation F ( T 1 , T 2 ) = 0. Question � � f 1 ( s ) f 0 ( s ) , f 2 ( s ) ◮ Is F = 0 ? f 0 ( s ) ◮ If yes, for which s values ?

What are implicit equations used for ? ◮ Intersection of the curves C 1 and C 2 , both in R n , n ≥ 2 C 1 is given by the parameterization R n → R � � f 1 ( s ) f 0 ( s ) , · · · , f n ( s ) �→ s , f 0 ( s ) C 2 is given by the implicit equations F 1 ( T 1 , · · · , T n ) = 0 , · · · F r ( T 1 , · · · , T n ) = 0. Question � � � � f 1 ( s ) f 0 ( s ) , · · · , f n ( s ) f 1 ( s ) f 0 ( s ) , · · · , f n ( s ) ◮ Are F 1 = 0 , · · · , F r = 0 ? f 0 ( s ) f 0 ( s ) ◮ If yes, for which s values ? Difficulty ◮ Several substitutions, ◮ high degree of polynomials to manipulate.

Plane curves Let K be a field. Algebraic parameterization φ is defined as follows P 2 φ := P → ( s : t ) �→ ( f 0 ( s , t ) : f 1 ( s , t ) : f 2 ( s , t )) , and its image defines the curve C . We assume that the f i ’s are of degree d for all i = 0 , 1 , 2 .

Plane curves Implicitization via Sylvester matrix, notation : Syl T 0 , T 1 , T 2 : new indeterminates. I := ( f 0 T 1 − f 1 T 0 , f 0 T 2 − f 2 T 0 ) ⊂ K [ s , t , T 0 , T 1 , T 2 ] ideal. I contains implicit equation of the curve C . Example f 0 = s 3 − 1 9 st 2 − t 3 , f 1 = 14 s 3 + 2 3 s 2 t − 3 st 2 + t 3 , 2 s 2 t + 5 4 s 3 − 12 s 2 t − 4 3 st 2 − 97 t 3 . Then, f 2 = − 1 Syl ( f 0 T 1 − f 1 T 0 , f 0 T 2 − f 2 T 0 ) = − 1 2 T 1 − 2 5  T 1 − 14 T 0 3 T 0 9 T 1 + 3 T 0 − T 1 − T 0 0 0  − 1 2 T 1 − 2 5 0 T 1 − 14 T 0 3 T 0 9 T 1 + 3 T 0 − T 1 − T 0 0     − 1 2 T 1 − 2 5 0 0 T 1 − 14 T 0 9 T 1 + 3 T 0 − T 1 − T 0 3 T 0     T 2 + 1 − 1 5 9 T 2 + 4 2 T 2 + 12 T 0 − T 2 + 97 T 0 0 0  4 T 0 3 T 0    T 2 + 1 − 1 5 9 T 2 + 4   0 4 T 0 2 T 2 + 12 T 0 3 T 0 − T 2 + 97 T 0 0   T 2 + 1 − 1 9 T 2 + 4 5 0 0 4 T 0 2 T 2 + 12 T 0 3 T 0 − T 2 + 97 T 0 is a 6 × 6 matrix with linear entries in T 0 , T 1 , T 2 , and its determinant yields a polynomial of degree 6 in T 0 , T 1 , T 2 .

Plane curves Definition Syzygy module of the parameterization φ , denoted by Syz , is ∈ K [ s , t ] 3 : p 0 ( s , t ) f 0 ( s , t )+ Syz ( f 0 , f 1 , f 2 ) := { ( p 0 , p 1 , p 2 ) p 1 ( s , t ) f 1 ( s , t ) + p 2 ( s , t ) f 2 ( s , t ) = 0 } . p 0 , p 1 , p 2 are called syzygies of f 0 , f 1 , f 2 . Moreover, Syz ( f 0 , f 1 , f 2 ) is a free module of K [ s , t ] with 2 generators p and q in K [ s , t ] 3 : p := ( p 0 , p 1 , p 2 ) and q := ( q 0 , q 1 , q 2 ) .

Plane curves Definition Syzygy module of the parameterization φ , denoted by Syz , is ∈ K [ s , t ] 3 : p 0 ( s , t ) f 0 ( s , t )+ Syz ( f 0 , f 1 , f 2 ) := { ( p 0 , p 1 , p 2 ) p 1 ( s , t ) f 1 ( s , t ) + p 2 ( s , t ) f 2 ( s , t ) = 0 } . p 0 , p 1 , p 2 are called syzygies of f 0 , f 1 , f 2 . Moreover, Syz ( f 0 , f 1 , f 2 ) is a free module of K [ s , t ] with 2 generators p and q in K [ s , t ] 3 : p := ( p 0 , p 1 , p 2 ) and q := ( q 0 , q 1 , q 2 ) . Definition { p , q } are called µ -basis of the parametric curve, if ◮ { p , q } is a basis of Syz ( f 0 , f 1 , f 2 ) and ◮ p , q have the lowest degree among all the basis of Syz ( f 0 , f 1 , f 2 ) . Moreover, deg ( p ) = µ 1 , deg ( q ) = µ 2 and d = µ 1 + µ 2 . We assume that µ 2 ≥ µ 1 .

Implicitization by resultant matrices with respect to p , q Notation p = p 0 ( s , t ) T 0 + p 1 ( s , t ) T 1 + p 2 ( s , t ) T 2 and q = q 0 ( s , t ) T 0 + q 1 ( s , t ) T 1 + q 2 ( s , t ) T 2 . Example f 0 = s 3 − 1 9 st 2 − t 3 , f 1 = 14 s 3 + 2 3 s 2 t − 3 st 2 + t 3 , 2 s 2 t + 5 4 s 3 − 12 s 2 t − 4 3 st 2 − 97 t 3 . f 2 = − 1 1007 / 84 s 2 + 233 / 168 st + 5431 / 56 t 2    2072314393 / 993502048 s + 491833577 / 124187756 t  p 0 q 0 − 97 / 112 s 2 − 43 / 504 st − 389 / 56 t 2  = p 1 q 1 − 147910417 / 993502048 s − 293063387 / 1490253072 t    − 23 / 42 s 2 + 97 / 126 st − 15 / 14 t 2 p 2 q 2 1568555 / 248375512 s − 9123809 / 2128932960 t Then, µ 1 = 1 and µ 2 = 2 . Syl ( p , q ) is a matrix of 3 × 3 size, with linear entries in T 0 , T 1 , T 2 , and its determinant yields a polynomial of degree 3 in T 0 , T 1 , T 2 .

Implicitization by resultant matrices with respect to p , q Notation p = p 0 ( s , t ) T 0 + p 1 ( s , t ) T 1 + p 2 ( s , t ) T 2 and q = q 0 ( s , t ) T 0 + q 1 ( s , t ) T 1 + q 2 ( s , t ) T 2 . 1007 / 84 s 2 + 233 / 168 st + 5431 / 56 t 2    2072314393 / 993502048 s + 491833577 / 124187756 t  p 0 q 0 − 97 / 112 s 2 − 43 / 504 st − 389 / 56 t 2  = p 1 q 1 − 147910417 / 993502048 s − 293063387 / 1490253072 t    − 23 / 42 s 2 + 97 / 126 st − 15 / 14 t 2 p 2 q 2 1568555 / 248375512 s − 9123809 / 2128932960 t We have µ 1 = 1 , µ 2 = 2 . Definition B´ ezout matrix, denoted by Bez ( p , q ) = ( b ij ) 1 ≤ i , j ≤ µ 2 , is defined to be p ( τ, σ ) q ( s , t ) − p ( s , t ) q ( τ, σ ) � b ij t i − 1 s µ 2 − i +1 τ j − 1 σ µ 2 − j +1 . = s τ − t σ i , j =1 ◮ Bez ( p , q ) is a matrix of 2 × 2 size, with only quadratic entries in T 0 , T 1 , T 2 , and its determinant yields to a polynomial of degree 2 µ 2 in T 0 , T 1 , T 2 .

Implicitization by resultant matrices with respect to p , q Notation p = p 0 ( s , t ) T 0 + p 1 ( s , t ) T 1 + p 2 ( s , t ) T 2 and q = q 0 ( s , t ) T 0 + q 1 ( s , t ) T 1 + q 2 ( s , t ) T 2 . ◮ Hybird B´ ezout matrix, HBez ( p , q ) is composed of the last µ 2 − µ 1 rows of Syl ( p , q ) in coefficients of q and the first µ 1 rows of Bez ( p , q ). Hence, again for the same example HBez ( p , q ) is a matrix of 2 × 2 size, with linear and quadratic entries in T 0 , T 1 , T 2 , and its determinant yields a polynomial of degree d = µ 2 + µ 1 in T 0 , T 1 , T 2 . � � last row of Syl s ( p , q ) ∗ ∗ HBez ( p , q ) T = 1st row of Bez s ( p , q ) ∗ ∗

Hybrid B´ ezout p = a 0 ( T 1 , T 2 ) t µ 1 + a 1 ( T 1 , T 2 ) st µ 1 − 1 + · · · + a µ 1 ( T 1 , T 2 ) s µ 1 , q = b 0 ( T 1 , T 2 ) t µ 2 + b 1 ( T 1 , T 2 ) st µ 2 − 1 + · · · + b µ 2 ( T 1 , T 2 ) s µ 2 . b µ 2 b µ 2 − 1 · · · b 0   ...           ...       · · ·  b µ 2 b µ 2 − 1 b 0    Syl ( p , q ) =  a µ 1 a µ 1 − 1 · · · a 0       ...          ...     a µ 1 a µ 1 − 1 a µ 1 − 2   a µ 1 a µ 1 − 1 · · · a 0 last row of Syl ( p , q ) ∗ · · · ∗   d-1th row of Syl ( p , q ) ∗ · · · ∗   .   .     . ∗ · · · ∗   HBez ( p , q ) T =   d − µ 2 + µ 1 + 1th row of Syl ( p , q ) ∗ · · · ∗   1st row of Bez ( p , q )  ∗ · · · ∗      .   .   . ∗ · · · ∗   µ 1 th row of Bez ( p , q ) ∗ · · · ∗ Remark If µ 2 = µ 1 , then HBez ( p , q ) does not have any rows of Syl ( p , q ) , i.e. any rows with linear entries in T 0 , T 1 , T 2 .