Outlines
Elimination Theory, Commutative Algebra and Applications
Laurent Bus´e
INRIA Sophia-Antipolis, France
March 2, 2005
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Part I: Residual resultants in 2 Outlines P Part II: Implicitization Problem Part I: Solving polynomial systems with resultants
1 Overview of the classical approach The Macaulay resultant Solving zero-dimensional polynomial system Limitations
2 The residual resultant approach The residual resultant Solving polynomial systems An example: cylinders passing through 5 points
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Part I: Residual resultants in 2 Outlines P Part II: Implicitization Problem Part II: Implicitization problem for curves and surfaces
3 Implicitization of rational plane curves Using resultants Using moving lines Using syzygies
4 Implicitization of rational surfaces Using resultants Using syzygies
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Classical approach Residual approach
Part I
Solving polynomial systems with resultants
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations The Macaulay resultant
Consider a system of 3 homogeneous polynomials in X0, X1, X2: P α f0 = c0,α.X |α|=d0 f = P c .X α 1 |α|=d1 1,α f = P c .X α 2 |α|=d2 2,α
α α0 α1 α2 X denotes a monomial X0 X1 X2 (αi ≥ 0). The ci,α’s are the parameters of the system with value in an algebraically closed field k.
Theorem (Macaulay - 1902) There exists an irreducible and homogeneous polynomial in k[ci,α], denoted Res(f0, f1, f2), satisfying:
2 Res(f0, f1, f2) = 0 ⇔ ∃x ∈ Pk such that f0(x) = f1(x) = f2(x) = 0.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations The Macaulay resultant
Consider a system of 3 homogeneous polynomials in X0, X1, X2: P α f0 = c0,α.X |α|=d0 f = P c .X α 1 |α|=d1 1,α f = P c .X α 2 |α|=d2 2,α
α α0 α1 α2 X denotes a monomial X0 X1 X2 (αi ≥ 0). The ci,α’s are the parameters of the system with value in an algebraically closed field k.
Theorem (Macaulay - 1902) There exists an irreducible and homogeneous polynomial in k[ci,α], denoted Res(f0, f1, f2), satisfying:
2 Res(f0, f1, f2) = 0 ⇔ ∃x ∈ Pk such that f0(x) = f1(x) = f2(x) = 0.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Computing the Macaulay resultant
Degrees: Res(f0, f1, f2) is homogeneous w.r.t. each fi : d d d deg = 0 1 2 with i ∈ {0, 1, 2} fixed. ci,α di
Computation: Let ν := d0 + d1 + d2 − 2, set A := k[ci,α] and consider the map (the first map of a Koszul complex)
3 M A[X0, X1, X2]ν−di → A[X0, X1, X2]ν i=1 (g0, g1, g2) 7→ g0f0 + g1f1 + g2f2.
It depends on the ci,α’s and has maximal rank Some maximal minors are the Macaulay matrices. Their gcd (3 are sufficients) gives Res(f0, f1, f2).
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Computing the Macaulay resultant
Degrees: Res(f0, f1, f2) is homogeneous w.r.t. each fi : d d d deg = 0 1 2 with i ∈ {0, 1, 2} fixed. ci,α di
Computation: Let ν := d0 + d1 + d2 − 2, set A := k[ci,α] and consider the map (the first map of a Koszul complex)
3 M A[X0, X1, X2]ν−di → A[X0, X1, X2]ν i=1 (g0, g1, g2) 7→ g0f0 + g1f1 + g2f2.
It depends on the ci,α’s and has maximal rank Some maximal minors are the Macaulay matrices. Their gcd (3 are sufficients) gives Res(f0, f1, f2).
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Computing the Macaulay resultant
Degrees: Res(f0, f1, f2) is homogeneous w.r.t. each fi : d d d deg = 0 1 2 with i ∈ {0, 1, 2} fixed. ci,α di
Computation: Let ν := d0 + d1 + d2 − 2, set A := k[ci,α] and consider the map (the first map of a Koszul complex)
3 M A[X0, X1, X2]ν−di → A[X0, X1, X2]ν i=1 (g0, g1, g2) 7→ g0f0 + g1f1 + g2f2.
It depends on the ci,α’s and has maximal rank Some maximal minors are the Macaulay matrices. Their gcd (3 are sufficients) gives Res(f0, f1, f2).
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Recovering the Chow form
V (f1, f2) Let f1, f2 ∈ R = k[X0, X1, X2] be homo- geneous polynomials defining isolated 2 points (a complete intersection) in P .
Introduce L(X) = c0X0 + c1X1 + c2X2. L(X )
Proposition (Chow form)
Y µ(ξ) Res(L, f1, f2) = L(ξ) ,
ξ∈V (f1,f2) where µ(ξ) denotes the multiplicity (or degree) of ξ.
⇒ An Absolute factorisation problem
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Recovering the Chow form
V (f1, f2) Let f1, f2 ∈ R = k[X0, X1, X2] be homo- geneous polynomials defining isolated 2 points (a complete intersection) in P .
Introduce L(X) = c0X0 + c1X1 + c2X2. L(X )
Proposition (Chow form)
Y µ(ξ) Res(L, f1, f2) = L(ξ) ,
ξ∈V (f1,f2) where µ(ξ) denotes the multiplicity (or degree) of ξ.
⇒ An Absolute factorisation problem
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Recovering multiplication maps And one may assume that there is The Macaulay matrix with no solutions on X = 0: right degree in L : 0 EF EF . . A B . . X0E L f1, f2 Rν − − − . . Rν \ X0E . . C D
Proposition
The matrix of the multiplication by L in R/(f1, f2)|X0=1 is given by:
−1 ML = (A − BD C)|X0=1.
⇒ The roots ξ are obtained through eigen-computations.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Recovering multiplication maps And one may assume that there is The Macaulay matrix with no solutions on X = 0: right degree in L : 0 EF EF . . A B . . X0E L f1, f2 Rν − − − . . Rν \ X0E . . C D
Proposition
The matrix of the multiplication by L in R/(f1, f2)|X0=1 is given by:
−1 ML = (A − BD C)|X0=1.
⇒ The roots ξ are obtained through eigen-computations.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Recovering multiplication maps And one may assume that there is The Macaulay matrix with no solutions on X = 0: right degree in L : 0 EF EF . . A B . . X0E L f1, f2 Rν − − − . . Rν \ X0E . . C D
Proposition
The matrix of the multiplication by L in R/(f1, f2)|X0=1 is given by:
−1 ML = (A − BD C)|X0=1.
⇒ The roots ξ are obtained through eigen-computations.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations The limitations
Problems:
If V (f1, f2) is not finite then Res(L, f1, f2) is identically zero.
If V (f1, f2) has base points (roots which are independant of the parameters ci,α’s) then Res(L, f1, f2) is also identically zero. Questions:
How to remove the non finite part of V (f1, f2)? More generally, how to compute “a part” (even zero-dimensional) of V (f1, f2)?
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations The limitations
Problems:
If V (f1, f2) is not finite then Res(L, f1, f2) is identically zero.
If V (f1, f2) has base points (roots which are independant of the parameters ci,α’s) then Res(L, f1, f2) is also identically zero. Questions:
How to remove the non finite part of V (f1, f2)? More generally, how to compute “a part” (even zero-dimensional) of V (f1, f2)?
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Three circles in the projective plane
Consider the following system:
2 2 2 f0 = c0,1X0 + c0,2X0X1 + c0,3X0X2 + c0,4(X1 + X2 ) 2 2 2 f1 = c1,1X0 + c1,2X0X1 + c1,3X0X2 + c1,4(X1 + X2 ) 2 2 2 f2 = c2,1X0 + c2,2X0X1 + c2,3X0X2 + c2,4(X1 + X2 ) Problem: When do they have a common point ?
f1 Two “base points”: (0 : 1 : ±i).
f0 This implies Res(f0, f1, f2) ≡ 0. Refined question: When do they have a common root which is not f2 (0 : 1 : ±i)? 2 P
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Three circles in the projective plane
Consider the following system:
2 2 2 f0 = c0,1X0 + c0,2X0X1 + c0,3X0X2 + c0,4(X1 + X2 ) 2 2 2 f1 = c1,1X0 + c1,2X0X1 + c1,3X0X2 + c1,4(X1 + X2 ) 2 2 2 f2 = c2,1X0 + c2,2X0X1 + c2,3X0X2 + c2,4(X1 + X2 ) Problem: When do they have a common point ?
f1 Two “base points”: (0 : 1 : ±i).
f0 This implies Res(f0, f1, f2) ≡ 0. Refined question: When do they have a common root which is not f2 (0 : 1 : ±i)? 2 P
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The Macaulay resultant Classical approach Solving zero-dimensional polynomial system Residual approach Limitations Three circles in the projective plane
Consider the following system:
2 2 2 f0 = c0,1X0 + c0,2X0X1 + c0,3X0X2 + c0,4(X1 + X2 ) 2 2 2 f1 = c1,1X0 + c1,2X0X1 + c1,3X0X2 + c1,4(X1 + X2 ) 2 2 2 f2 = c2,1X0 + c2,2X0X1 + c2,3X0X2 + c2,4(X1 + X2 ) Problem: When do they have a common point ?
f1 Two “base points”: (0 : 1 : ±i).
f0 This implies Res(f0, f1, f2) ≡ 0. Refined question: When do they have a common root which is not f2 (0 : 1 : ±i)? 2 P
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points The setting
We keep the same notation; R = k[X0, X1, X2]. Let g1(X),..., gm(X) be homogeneous polynomials in R of positive degree k1,..., km, respectively. Let d0, d1, d2 be integers such that di ≥ kj for all (i, j). One considers the system : Pm f0 = i=1 hi,0(X) gi(X) total degree d0 Pm f1 = i=1 hi,1(X) gi(X) total degree d1 Pm f2 = i=1 hi,2(X) gi(X) total degree d2 where h (X ) = P ci,j X α. i,j |α|=dj −ki α Main fact
The Polynomials fi are generic of degree di in the ideal G = (g1,..., gm). Rk : The case G = (1) corresponds to the previous setting.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points The setting
We keep the same notation; R = k[X0, X1, X2]. Let g1(X),..., gm(X) be homogeneous polynomials in R of positive degree k1,..., km, respectively. Let d0, d1, d2 be integers such that di ≥ kj for all (i, j). One considers the system : Pm f0 = i=1 hi,0(X) gi(X) total degree d0 Pm f1 = i=1 hi,1(X) gi(X) total degree d1 Pm f2 = i=1 hi,2(X) gi(X) total degree d2 where h (X ) = P ci,j X α. i,j |α|=dj −ki α Main fact
The Polynomials fi are generic of degree di in the ideal G = (g1,..., gm). Rk : The case G = (1) corresponds to the previous setting.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points The setting
We keep the same notation; R = k[X0, X1, X2]. Let g1(X),..., gm(X) be homogeneous polynomials in R of positive degree k1,..., km, respectively. Let d0, d1, d2 be integers such that di ≥ kj for all (i, j). One considers the system : Pm f0 = i=1 hi,0(X) gi(X) total degree d0 Pm f1 = i=1 hi,1(X) gi(X) total degree d1 Pm f2 = i=1 hi,2(X) gi(X) total degree d2 where h (X ) = P ci,j X α. i,j |α|=dj −ki α Main fact
The Polynomials fi are generic of degree di in the ideal G = (g1,..., gm). Rk : The case G = (1) corresponds to the previous setting.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points The setting
We keep the same notation; R = k[X0, X1, X2]. Let g1(X),..., gm(X) be homogeneous polynomials in R of positive degree k1,..., km, respectively. Let d0, d1, d2 be integers such that di ≥ kj for all (i, j). One considers the system : Pm f0 = i=1 hi,0(X) gi(X) total degree d0 Pm f1 = i=1 hi,1(X) gi(X) total degree d1 Pm f2 = i=1 hi,2(X) gi(X) total degree d2 where h (X ) = P ci,j X α. i,j |α|=dj −ki α Main fact
The Polynomials fi are generic of degree di in the ideal G = (g1,..., gm). Rk : The case G = (1) corresponds to the previous setting.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Existence
Notations : i,j F := (f0, f1, f2) ⊂ A[X0, X1, X2] with A = k[cα ], (F : G) = {f ∈ A[X0, X1, X2] such that fG ⊂ F }. Theorem
If G is a local complete intersection and max(di ) ≥ min(kj ) + 1 then there exists an irreducible and homogeneous polynomial in A, i,j denoted Res(f0, f1, f2), which satisfies: for any cα → k
Res(f0, f1, f2) = 0 ⇔ V (F : G) 6= ∅.
n ⇔ the fi vanish on a point of P “outside” the locus V (G). Rk: Macaulay resultants ⇔ G = R: n V (F : R) = V (F ) 6= ∅ ⇔ ∃x ∈ P : f0(x) = f1(x) = f2(x) = 0.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Existence
Notations : i,j F := (f0, f1, f2) ⊂ A[X0, X1, X2] with A = k[cα ], (F : G) = {f ∈ A[X0, X1, X2] such that fG ⊂ F }. Theorem
If G is a local complete intersection and max(di ) ≥ min(kj ) + 1 then there exists an irreducible and homogeneous polynomial in A, i,j denoted Res(f0, f1, f2), which satisfies: for any cα → k
Res(f0, f1, f2) = 0 ⇔ V (F : G) 6= ∅.
n ⇔ the fi vanish on a point of P “outside” the locus V (G). Rk: Macaulay resultants ⇔ G = R: n V (F : R) = V (F ) 6= ∅ ⇔ ∃x ∈ P : f0(x) = f1(x) = f2(x) = 0.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Existence
Notations : i,j F := (f0, f1, f2) ⊂ A[X0, X1, X2] with A = k[cα ], (F : G) = {f ∈ A[X0, X1, X2] such that fG ⊂ F }. Theorem
If G is a local complete intersection and max(di ) ≥ min(kj ) + 1 then there exists an irreducible and homogeneous polynomial in A, i,j denoted Res(f0, f1, f2), which satisfies: for any cα → k
Res(f0, f1, f2) = 0 ⇔ V (F : G) 6= ∅.
n ⇔ the fi vanish on a point of P “outside” the locus V (G). Rk: Macaulay resultants ⇔ G = R: n V (F : R) = V (F ) 6= ∅ ⇔ ∃x ∈ P : f0(x) = f1(x) = f2(x) = 0.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Computing residual resultants (1)
Assume that G is l.c.i. and moreover a codim 2 ACM ideal. This implies that G is completely determined by its first syzygies : m−1 m M φ M (g1,...,gm) 0 → R(−li ) −→ R(−ki ) −−−−−−→ G → 0. i=1 i=1 p1,1 ... p1,m−1 . . L := Mat(φ) = . . . pm,1 ... pm,m−1
Proposition (multi-degree)
The degree is completely explicit in the di ’s, kj ’s and lt ’s:
n−1 n d d d P l2 − P k2 deg (Res(f , f , f )) = 0 1 2 − i=1 j i=1 j . fi 0 1 2 di 2
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Computing residual resultants (1)
Assume that G is l.c.i. and moreover a codim 2 ACM ideal. This implies that G is completely determined by its first syzygies : m−1 m M φ M (g1,...,gm) 0 → R(−li ) −→ R(−ki ) −−−−−−→ G → 0. i=1 i=1 p1,1 ... p1,m−1 . . L := Mat(φ) = . . . pm,1 ... pm,m−1
Proposition (multi-degree)
The degree is completely explicit in the di ’s, kj ’s and lt ’s:
n−1 n d d d P l2 − P k2 deg (Res(f , f , f )) = 0 1 2 − i=1 j i=1 j . fi 0 1 2 di 2
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Computing residual resultants (2)
Construct the m × (m + 2) matrix gluing L and the hi,j ’s: p1,1 ... p1,m−1 h1,0 h1,1 h1,2 . . . . . L|H := . . . . . , pm,1 ... pm,m−1 hm,0 hm,1 hm,2
m Set ν := d0 + d1 + d2 − 2 − 2 min kj and let I ⊂ N be the set indexing the m × m minors of L|H. Construct the matrix M of the map (first map of an Eagon-Northcott complex): M A[X0, X1, X2]ν−deg(∆s) → A[X0, X1, X2]ν s∈I X (..., bs ,...) 7→ bs ∆s s∈I Proposition
This matrix “represents” Res(f0, f1, f2).
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Computing residual resultants (2)
Construct the m × (m + 2) matrix gluing L and the hi,j ’s: p1,1 ... p1,m−1 h1,0 h1,1 h1,2 . . . . . L|H := . . . . . , pm,1 ... pm,m−1 hm,0 hm,1 hm,2
m Set ν := d0 + d1 + d2 − 2 − 2 min kj and let I ⊂ N be the set indexing the m × m minors of L|H. Construct the matrix M of the map (first map of an Eagon-Northcott complex): M A[X0, X1, X2]ν−deg(∆s) → A[X0, X1, X2]ν s∈I X (..., bs ,...) 7→ bs ∆s s∈I Proposition
This matrix “represents” Res(f0, f1, f2).
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Computing residual resultants (2)
Construct the m × (m + 2) matrix gluing L and the hi,j ’s: p1,1 ... p1,m−1 h1,0 h1,1 h1,2 . . . . . L|H := . . . . . , pm,1 ... pm,m−1 hm,0 hm,1 hm,2
m Set ν := d0 + d1 + d2 − 2 − 2 min kj and let I ⊂ N be the set indexing the m × m minors of L|H. Construct the matrix M of the map (first map of an Eagon-Northcott complex): M A[X0, X1, X2]ν−deg(∆s) → A[X0, X1, X2]ν s∈I X (..., bs ,...) 7→ bs ∆s s∈I Proposition
This matrix “represents” Res(f0, f1, f2).
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Recovering the Chow form
Let f1, f2 ∈ R = k[X0, X1, X2] be homo- V (f1, f2) geneous polynomials defining some iso- lated points Z and V (G) (l.c.i. codim V (G) 2 2 ACM) in P .
Introduce L(X) = c0X0 + c1X1 + c2X2 and choose f vanishing on V (G) and not on Z. L(X ) Proposition (Chow form)
Y µ(ξ) Res(Lf , f1, f2) = L(ξ) , ξ∈V (F :G) where µ(ξ) denotes the multiplicity (or degree) of ξ.
⇒ An absolute factorisation problem
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Recovering the Chow form
Let f1, f2 ∈ R = k[X0, X1, X2] be homo- V (f1, f2) geneous polynomials defining some iso- lated points Z and V (G) (l.c.i. codim V (G) 2 2 ACM) in P .
Introduce L(X) = c0X0 + c1X1 + c2X2 and choose f vanishing on V (G) and not on Z. L(X ) Proposition (Chow form)
Y µ(ξ) Res(Lf , f1, f2) = L(ξ) , ξ∈V (F :G) where µ(ξ) denotes the multiplicity (or degree) of ξ.
⇒ An absolute factorisation problem
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Recovering multiplication maps And one may assume that there is A Macaulay type matrix no residual root on X = 0: with right degree in L : 0 EF EF . . A B . . X0E L f1, f2 Rν − − − . . Rν \ X0E . . C D
Proposition
The matrix of the multiplication by L in R/(F : G)|X0=1 is given by:
−1 ML = (A − BD C)|X0=1.
⇒ The roots ξ are obtained through eigen-computations.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Recovering multiplication maps And one may assume that there is A Macaulay type matrix no residual root on X = 0: with right degree in L : 0 EF EF . . A B . . X0E L f1, f2 Rν − − − . . Rν \ X0E . . C D
Proposition
The matrix of the multiplication by L in R/(F : G)|X0=1 is given by:
−1 ML = (A − BD C)|X0=1.
⇒ The roots ξ are obtained through eigen-computations.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Recovering multiplication maps And one may assume that there is A Macaulay type matrix no residual root on X = 0: with right degree in L : 0 EF EF . . A B . . X0E L f1, f2 Rν − − − . . Rν \ X0E . . C D
Proposition
The matrix of the multiplication by L in R/(F : G)|X0=1 is given by:
−1 ML = (A − BD C)|X0=1.
⇒ The roots ξ are obtained through eigen-computations.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Cylinders passing through 5 points (1)
Suppose given five points p1,..., p5 in the space. We are looking for the cylinders passing through them (reconstruction problem in CAD). A cylinder ⇔ a unitary direction −→ t = (l, m, n) ⇔ p1 a point (l:m:n) p2 in 2. P p −→ 3 Passing through 4 t
points: a p4 homogeneous p polynomial of 5 −→ degree 3. orthogonal plane to t
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Cylinders passing through 5 points (2)
We are looking for the roots of the polynomial system
C1234(l, m, n) = C1235(l, m, n) = 0,
Using the classical approach: 9 points, including unwanted directions p1p2, p1p3, p2p3. Let G defining these 3 points (on a line and a cubic) The residual resultant gives the 6 points; and it is a geometric method: −→ −→ t t
p5 p1, p2 p1, p2 p5
p4 p3 p3 p4
remove p1p2 keep p1p2
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Cylinders passing through 5 points (2)
We are looking for the roots of the polynomial system
C1234(l, m, n) = C1235(l, m, n) = 0,
Using the classical approach: 9 points, including unwanted directions p1p2, p1p3, p2p3. Let G defining these 3 points (on a line and a cubic) The residual resultant gives the 6 points; and it is a geometric method: −→ −→ t t
p5 p1, p2 p1, p2 p5
p4 p3 p3 p4
remove p1p2 keep p1p2
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Cylinders passing through 5 points (2)
We are looking for the roots of the polynomial system
C1234(l, m, n) = C1235(l, m, n) = 0,
Using the classical approach: 9 points, including unwanted directions p1p2, p1p3, p2p3. Let G defining these 3 points (on a line and a cubic) The residual resultant gives the 6 points; and it is a geometric method: −→ −→ t t
p5 p1, p2 p1, p2 p5
p4 p3 p3 p4
remove p1p2 keep p1p2
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications The residual resultant Classical approach Solving polynomial systems Residual approach An example: cylinders passing through 5 points Cylinders passing through 5 points (2)
We are looking for the roots of the polynomial system
C1234(l, m, n) = C1235(l, m, n) = 0,
Using the classical approach: 9 points, including unwanted directions p1p2, p1p3, p2p3. Let G defining these 3 points (on a line and a cubic) The residual resultant gives the 6 points; and it is a geometric method: −→ −→ t t
p5 p1, p2 p1, p2 p5
p4 p3 p3 p4
remove p1p2 keep p1p2
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Implicitization of surfaces
Part II
Implicitization problem
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies The implicitization problem for curves
Suppose given a generically finite rational map
1 φ 2 P −→ P x := (x0 : x1) 7→ (f1(x): f2(x): f3(x)),
where f1(x), f2(x), f3(x) are homogeneous of the same degree d. 2 Its closed image is an irreducible curve C in P . We would like to compute its implicit equation that we will denote C(T1, T2, T3). Proposition (degree of C)
0 if φ not gen. finite, d − deg(gcd(f , f , f )) = 1 2 3 deg(φ)deg(C) otherwise.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies The implicitization problem for curves
Suppose given a generically finite rational map
1 φ 2 P −→ P x := (x0 : x1) 7→ (f1(x): f2(x): f3(x)),
where f1(x), f2(x), f3(x) are homogeneous of the same degree d. 2 Its closed image is an irreducible curve C in P . We would like to compute its implicit equation that we will denote C(T1, T2, T3). Proposition (degree of C)
0 if φ not gen. finite, d − deg(gcd(f , f , f )) = 1 2 3 deg(φ)deg(C) otherwise.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Implicitizing with resultants
Assume that there is no base point, which means here that the gcd(f1, f2, f3) is a constant. Then:
Proposition
deg(φ) Res(f1(x) − T1f3(x), f2(x) − T2f3(x)) = C(T1, T2, 1) .
Here Res denotes the Sylvester resultant in x = (x0, x1). It can be computed as the determinant of a square matrix. Observe that we get a universal formula
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Implicitizing with “moving lines”
Definition (Sederberg-Chen ’95) A moving line of degree ν following C is
H(x; T1, T2, T3) := g1(x)T1 + g2(x)T2 + g3(x)T3
P3 such that degx(H) = ν and i=1 gi (x)fi (x) = 0.
Assume that there is no base point, then there exists d linearly independant moving lines of degree d − 1, say H(1),..., H(d). (i) Pν (i) j ν−j rewrite H = j=0 Lj (T1, T2, T3)x0x1 , (i) Construct the square matrix M := (Lj ) deg(φ) det(M) = C(T1, T2, T3)
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Implicitizing with “moving lines”
Definition (Sederberg-Chen ’95) A moving line of degree ν following C is
H(x; T1, T2, T3) := g1(x)T1 + g2(x)T2 + g3(x)T3
P3 such that degx(H) = ν and i=1 gi (x)fi (x) = 0.
Assume that there is no base point, then there exists d linearly independant moving lines of degree d − 1, say H(1),..., H(d). (i) Pν (i) j ν−j rewrite H = j=0 Lj (T1, T2, T3)x0x1 , (i) Construct the square matrix M := (Lj ) deg(φ) det(M) = C(T1, T2, T3)
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Implicitizing with “moving lines”
Definition (Sederberg-Chen ’95) A moving line of degree ν following C is
H(x; T1, T2, T3) := g1(x)T1 + g2(x)T2 + g3(x)T3
P3 such that degx(H) = ν and i=1 gi (x)fi (x) = 0.
Assume that there is no base point, then there exists d linearly independant moving lines of degree d − 1, say H(1),..., H(d). (i) Pν (i) j ν−j rewrite H = j=0 Lj (T1, T2, T3)x0x1 , (i) Construct the square matrix M := (Lj ) deg(φ) det(M) = C(T1, T2, T3)
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Implicitizing with “moving lines”
Definition (Sederberg-Chen ’95) A moving line of degree ν following C is
H(x; T1, T2, T3) := g1(x)T1 + g2(x)T2 + g3(x)T3
P3 such that degx(H) = ν and i=1 gi (x)fi (x) = 0.
Assume that there is no base point, then there exists d linearly independant moving lines of degree d − 1, say H(1),..., H(d). (i) Pν (i) j ν−j rewrite H = j=0 Lj (T1, T2, T3)x0x1 , (i) Construct the square matrix M := (Lj ) deg(φ) det(M) = C(T1, T2, T3)
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Implicitizing with “moving lines”
Definition (Sederberg-Chen ’95) A moving line of degree ν following C is
H(x; T1, T2, T3) := g1(x)T1 + g2(x)T2 + g3(x)T3
P3 such that degx(H) = ν and i=1 gi (x)fi (x) = 0.
Assume that there is no base point, then there exists d linearly independant moving lines of degree d − 1, say H(1),..., H(d). (i) Pν (i) j ν−j rewrite H = j=0 Lj (T1, T2, T3)x0x1 , (i) Construct the square matrix M := (Lj ) deg(φ) det(M) = C(T1, T2, T3)
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Commutative algebra revisiting moving lines
Set A := K[x0, x1][T1, T2, T3] bigraded: A(n; m). Consider both Koszul complexes
d d d1=(f1 f2 f3) 0 → A(−3d; 0) −→3 A(−2d; 0)3 −→2 A(−d; 0)3 −−−−−−−−→ A(0; 0),
v v v1=(T1 T2 T3) 0 → A(0; −3) −→3 A(0; −2)3 −→2 A(0; −1)3 −−−−−−−−−→ A(0; 0), they induce the following bigraded complex of A-modules:
v3 v2 v1 Ker(d3) −→ Ker(d2)(0; −2) −→ Ker(d1)(0; −1) −→ A(0; 0) P3 = 0 (g1, g2, g3) 7→ i=1 gi (x)Ti . called the Z-approximation complex (Simis-Vasconcelos ’81).
Moving lines following C are exactly v1(Ker(d1)).
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Commutative algebra revisiting moving lines
Set A := K[x0, x1][T1, T2, T3] bigraded: A(n; m). Consider both Koszul complexes
d d d1=(f1 f2 f3) 0 → A(−3d; 0) −→3 A(−2d; 0)3 −→2 A(−d; 0)3 −−−−−−−−→ A(0; 0),
v v v1=(T1 T2 T3) 0 → A(0; −3) −→3 A(0; −2)3 −→2 A(0; −1)3 −−−−−−−−−→ A(0; 0), they induce the following bigraded complex of A-modules:
v3 v2 v1 Ker(d3) −→ Ker(d2)(0; −2) −→ Ker(d1)(0; −1) −→ A(0; 0) P3 = 0 (g1, g2, g3) 7→ i=1 gi (x)Ti . called the Z-approximation complex (Simis-Vasconcelos ’81).
Moving lines following C are exactly v1(Ker(d1)).
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Commutative algebra revisiting moving lines
Set A := K[x0, x1][T1, T2, T3] bigraded: A(n; m). Consider both Koszul complexes
d d d1=(f1 f2 f3) 0 → A(−3d; 0) −→3 A(−2d; 0)3 −→2 A(−d; 0)3 −−−−−−−−→ A(0; 0),
v v v1=(T1 T2 T3) 0 → A(0; −3) −→3 A(0; −2)3 −→2 A(0; −1)3 −−−−−−−−−→ A(0; 0), they induce the following bigraded complex of A-modules:
v3 v2 v1 Ker(d3) −→ Ker(d2)(0; −2) −→ Ker(d1)(0; −1) −→ A(0; 0) P3 = 0 (g1, g2, g3) 7→ i=1 gi (x)Ti . called the Z-approximation complex (Simis-Vasconcelos ’81).
Moving lines following C are exactly v1(Ker(d1)).
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Implicitizing with syzygies
Theorem The determinant of the complex of free K[T1, T2, T3]-modules
v2 v1 0 → Ker(d2)[ν](−2) −→ Ker(d1)[ν](−1) −→ A[ν](0),
deg(φ) equals C(T1, T2, T3) for all ν ≥ d − 1.
In general C deg(φ) is represented either as • a quotient D1(T1,T2,T3) . D2(T1,T2,T3) • the gcd of the maximal minors of the first map.
If there is no base point then Ker(d2)(0; 0) ' A(−d; 0), and
deg(φ) v1 C(T1, T2, T3) = det Ker(d1)[d−1](−1) −→ A[d−1](0) , which is nothing but the moving lines matrix.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Implicitizing with syzygies
Theorem The determinant of the complex of free K[T1, T2, T3]-modules
v2 v1 0 → Ker(d2)[ν](−2) −→ Ker(d1)[ν](−1) −→ A[ν](0),
deg(φ) equals C(T1, T2, T3) for all ν ≥ d − 1.
In general C deg(φ) is represented either as • a quotient D1(T1,T2,T3) . D2(T1,T2,T3) • the gcd of the maximal minors of the first map.
If there is no base point then Ker(d2)(0; 0) ' A(−d; 0), and
deg(φ) v1 C(T1, T2, T3) = det Ker(d1)[d−1](−1) −→ A[d−1](0) , which is nothing but the moving lines matrix.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Using resultants Implicitization of curves Using moving lines Implicitization of surfaces Using syzygies Implicitizing with syzygies
Theorem The determinant of the complex of free K[T1, T2, T3]-modules
v2 v1 0 → Ker(d2)[ν](−2) −→ Ker(d1)[ν](−1) −→ A[ν](0),
deg(φ) equals C(T1, T2, T3) for all ν ≥ d − 1.
In general C deg(φ) is represented either as • a quotient D1(T1,T2,T3) . D2(T1,T2,T3) • the gcd of the maximal minors of the first map.
If there is no base point then Ker(d2)(0; 0) ' A(−d; 0), and
deg(φ) v1 C(T1, T2, T3) = det Ker(d1)[d−1](−1) −→ A[d−1](0) , which is nothing but the moving lines matrix.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies The implicitization problem for surfaces
Suppose given a generically finite rational map
2 φ 3 P −→ P x := (x0 : x1 : x2) 7→ (f1(x): f2(x): f3(x): f4(x))
where f1(x),..., f4(x) are homogeneous of the same degree d.
Its closed image is an irreducible surface S: S(T1, T2, T3, T4).
Proposition (degree of S)
Assume that V (f1, f2, f3, f4) is finite (base points are isolated):
X 0 if φ not generically finite, d2− e = p deg(φ)deg(S) otherwise. p∈V (f1,f2,f3,f4)
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies The implicitization problem for surfaces
Suppose given a generically finite rational map
2 φ 3 P −→ P x := (x0 : x1 : x2) 7→ (f1(x): f2(x): f3(x): f4(x))
where f1(x),..., f4(x) are homogeneous of the same degree d.
Its closed image is an irreducible surface S: S(T1, T2, T3, T4).
Proposition (degree of S)
Assume that V (f1, f2, f3, f4) is finite (base points are isolated):
X 0 if φ not generically finite, d2− e = p deg(φ)deg(S) otherwise. p∈V (f1,f2,f3,f4)
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies Implicitization formulas with resultants
If there is no base point then
deg(φ) Res(f1(x)−T1f4(x), f2(x)−T2f4(x), f3(x)−T3f4(x)) = S(T1, T2, T3, 1)
If there are base points then we can do similarly with A sparse resultant (no base points in a certain toric variety) A residual resultant (l.c.i. base points in P2) . . . your own ad hoc resultant.
Remark: These formulas are universal for certain classes of parameterizations.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies Implicitization formulas with resultants
If there is no base point then
deg(φ) Res(f1(x)−T1f4(x), f2(x)−T2f4(x), f3(x)−T3f4(x)) = S(T1, T2, T3, 1)
If there are base points then we can do similarly with A sparse resultant (no base points in a certain toric variety) A residual resultant (l.c.i. base points in P2) . . . your own ad hoc resultant.
Remark: These formulas are universal for certain classes of parameterizations.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies Implicitization formulas with resultants
If there is no base point then
deg(φ) Res(f1(x)−T1f4(x), f2(x)−T2f4(x), f3(x)−T3f4(x)) = S(T1, T2, T3, 1)
If there are base points then we can do similarly with A sparse resultant (no base points in a certain toric variety) A residual resultant (l.c.i. base points in P2) . . . your own ad hoc resultant.
Remark: These formulas are universal for certain classes of parameterizations.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies Implicitizing with syzygies (1)
Set A := K[x0, x1, x2][T1, T2, T3, T4] bigraded: A(n; m). As for curve implicitization, consider both Koszul complexes
(f ,...,f ) 0 → A(−4d; 0) −→d4 A(−3d; 0)4 −→d3 A(−2d; 0)6 −→d2 A(−d; 0)4 −−−−−→1 4 A(0; 0),
v =(T ,...,T ) 0 → A(0; −4) −→v4 A(0; −3)4 −→v3 A(0; −2)6 −→v2 A(0; −1)4 −−−−−−−−→1 1 4 A(0; 0).
They induce the following bigraded complex of A-modules:
v4 v3 v2 Ker(d4) = 0 −→ Ker(d3)(0; −3) −→ Ker(d2)(0; −2) −→
v1 Ker(d1)(0; −1) −→ A(0; 0) P4 (g1, g2, g3, g4) 7→ i=1 gi (x)Ti (x).
Moving planes following S are exactly v1(Ker(d1))
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies Implicitizing with syzygies (2)
Set I := (f1, f2, f3, f4) ⊂ K[x0, x1, x2]. Theorem 2 Assume V (I ) ⊂ P is finite and an almost l.c.i. (locally at most 3 generators). Then, for all integers
sat ν ≥ ν0 := 2(d − 1) − indeg(I ),
the determinant of the complex of free K[T1,..., T4]-modules
v3 v2 v1 0 → Ker(d3)[ν](−3) −→ Ker(d2)[ν](−2) −→ Ker(d1)[ν](−1) −→ A[ν](0),
2 P is homogeneous of degree d − p∈V (I ) dp, deg(φ) is a multiple of S(T1, T2, T3, T4) , is exactly Sdeg(φ) if and only if I is a l.c.i..
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies Implicitizing with syzygies (3)
⇒ In general Sdeg(φ) is represented either as • a quotient D1(T1,T2,T3,T4).D3(T1,T2,T3,T4) . D2(T1,T2,T3,T4) • the gcd of the maximal minors of the first map v1.
sat ⇒ The bound ν0 = 2(d − 1) − indeg(I ): sat sat indeg(I ) := min{t ∈ N : It 6= 0}. It is a geometric invariant of the scheme V (I ): the smallest degree of a (non-zero) hypersurface passing through all the points defined by I . We have: sat • ν0 = 2d − 2 if I has no base points (I = K[x0, x1, x2]), • 2d − 3 ≥ ν0 ≥ d − 2 if I has base points,
• ν0 = d − 2 if I is saturated (and we also have D3 = 1). Remark: The presence of base points simplifies the problem!
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies Implicitizing with syzygies (3)
⇒ In general Sdeg(φ) is represented either as • a quotient D1(T1,T2,T3,T4).D3(T1,T2,T3,T4) . D2(T1,T2,T3,T4) • the gcd of the maximal minors of the first map v1.
sat ⇒ The bound ν0 = 2(d − 1) − indeg(I ): sat sat indeg(I ) := min{t ∈ N : It 6= 0}. It is a geometric invariant of the scheme V (I ): the smallest degree of a (non-zero) hypersurface passing through all the points defined by I . We have: sat • ν0 = 2d − 2 if I has no base points (I = K[x0, x1, x2]), • 2d − 3 ≥ ν0 ≥ d − 2 if I has base points,
• ν0 = d − 2 if I is saturated (and we also have D3 = 1). Remark: The presence of base points simplifies the problem!
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies Implicitizing with syzygies (3)
⇒ In general Sdeg(φ) is represented either as • a quotient D1(T1,T2,T3,T4).D3(T1,T2,T3,T4) . D2(T1,T2,T3,T4) • the gcd of the maximal minors of the first map v1.
sat ⇒ The bound ν0 = 2(d − 1) − indeg(I ): sat sat indeg(I ) := min{t ∈ N : It 6= 0}. It is a geometric invariant of the scheme V (I ): the smallest degree of a (non-zero) hypersurface passing through all the points defined by I . We have: sat • ν0 = 2d − 2 if I has no base points (I = K[x0, x1, x2]), • 2d − 3 ≥ ν0 ≥ d − 2 if I has base points,
• ν0 = d − 2 if I is saturated (and we also have D3 = 1). Remark: The presence of base points simplifies the problem!
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies Examples
• A cubic with 6 base points (Sederberg-Chen):
2 3 2 2 2 2 3 f1 = s t + 2t + s u + 4stu + 4t u + 3su + 2tu + 2u , 3 2 2 2 2 3 f2 = −s − 2st − 2s u − stu + su − 2tu + 2u , 3 2 2 2 2 2 2 f3 = −s − 2s t − 3st − 3s u − 3stu + 2t u − 2su − 2tu , 3 2 3 2 2 2 2 3 f4 = s + s t + t + s u + t u − su − tu − u .
I is saturated ⇒ ν0 = d − 2 = 1. The surface is represented by the matrix: x −z − w y + w y x − 2y + z − 2w 2y − z . z −x − 2w y + 2w
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications Implicitization of curves Using resultants Implicitization of surfaces Using syzygies Examples
• An example with a fat base point:
3 2 2 2 2 f1 = s − 6s t − 5st − 4s u + 4stu − 3t u, 3 2 2 2 2 f2 = −s − 2s t − st − 5s u − 3stu − 6t u, 3 2 2 3 2 2 f3 = −4s − 2s t + 4st − 6t + 6s u − 6stu − 2t u, 3 2 2 3 2 2 f4 = 2s − 6s t + 3st − 6t − 3s u − 4stu + 2t u.
2 I defines exactly the fat point p = (s, t) (dp = 3 and ep = 4). 2 The method gives a multiple of degree d − dp = 9 − 3 = 6 (and not 5). Here ν0 = 2(3 − 1) − 2 = 2, we obtain a single 6 × 6 matrix.
Laurent Bus´e Elimination Theory, Commutative Algebra and Applications