Marc Dohm

Severinas Zube

The implicit equation of a canal surface *

Keywords: canal surface, implicit equation, resultant, µ-basis, offset

The implicit equation of a canal surface

Introduction

In surface design, the user often needs to perform rounding or filleting between two intersecting surfaces. Mathematically, the surface used in making the rounding is defined as the envelope of a family of spheres which are tangent to both surfaces. This envelope of spheres centered at c(t) ∈ R 3 with radius r(t), where c(t) and r(t) are rational functions, is called a canal surface with spine curve E = {(c(t), r(t)) ∈ R 4 |t ∈ R}. If the radius r(t) is constant the surface is called a pipe surface. Moreover, if additionally we reduce the dimension (take c(t) in a plane and consider circles instead of spheres) we obtain the offset to the curve. Canal surfaces are very popular in Geometric Modelling, as they can be used as a blending surface between two surfaces. For example, any two circular cones with a common inscribed sphere can be blended by a part of a Dupin cyclide bounded by two circles as it was shown by [Pratt(1990), Pratt(1995)] (see Figure 1). Cyclides are envelopes of special quadratic families of spheres. For other examples of blending with canal surfaces we refer to [Kazakeviciute(2005)].

Here we study the implicit equation of a canal surface C and its implicit degree. The implicit equation of a canal surface can be obtained after elimination of the family variable t from the system of two equations g 1 (y, t) = g 2 (y, t) = 0 (here g 1 , g 2 are quadratic in the variables y = (y 1 , y 2 , y 3 , y 4 )), i.e. by taking the resultant with respect to t. However, this resultant can have extraneous factors. In the paper we explain how these factors appear and how we can eliminate Figure 1: A Dupin cyclide used for blending circular cones. them. By using Lie and Laguerre geometry, we see that the above system of equations is related to a system h 1 (ŷ, t) = h 2 (ŷ, t) = Q(ŷ) = 0, where h 1 , h 2 are linear in the variables ŷ = (u, y 0 , y 1 , y 2 , y 3 , y 4 ) and Q(ŷ) is the Lie quadric (for the exact definition see formula ( 6)). It turns out that the variety defined by the system of equations h 1 (ŷ, t) = h 2 (ŷ, t) = 0 is a dual variety to the curve Ê ∈ P 5 , where Ê is a curve on the Lie quadric determined by the spine curve E (for the explicit definition see formula ( 11)). For the dual variety V( Ê) we define the µ-basis, which consists of two polynomials p 1 (ŷ, t), p 2 (ŷ, t) which are linear in ŷ, and of degree d 1 , d 2 in t are such that d 1 + d 2 is minimal. It turns out that the resultant of p 1 and p 2 with respect to t gives the implicit equation of the variety V( Ê). There is a simple substitution formula (see the algorithm at the end of section 5) to compute the implicit equation of the canal surface from the implicit equation of the variety V( Ê).

Partial solutions to the problem of finding the implicit equation (and degree) for canal surfaces have been given in other papers. For instance, the degree of offsets to curves is studied in [Segundo, Sendra(2005)]. In [Xu et al.(2006)], there is a degree formula for the implicit equation of a polynomial canal surface. Quadratic canal surfaces (parametric and implicit representation) have been studied in [Krasauskas, Zube(2007)].

We close the introduction by noting that the implicit degree of a canal surface is important for the parametric degree. Our observation is that if the canal surface has the minimal parametrization of bi-degree (2, d) then its implicit degree is close to 2d. On the minimal bi-degree (2, d) parametrizations of the canal surface we refer to [Krasauskas(2007)].

The paper is organized as follows. In the next section we develop some algebraic formalism about modules with two quasi-generators. We define the µ-basis for these modules and present an algorithm for its computation. In the following section, we recall some needed facts about Lie and Laguerre sphere geometry. Then using Lie and Laguerre geometry we describe the canal surface explicitly. Also, we introduce the Γ-hypersurface which contains all d-offsets to the canal surface. Using the µ-basis algorithm we compute the implicit equations of the dual variety V( Ê), the Γ-hypersurface and the canal surface C. Next we apply the results of the previous section to the dual variety V( Ê) of the curve and explain how to compute the implicit degree of the Γ-hypersurface (without computation of the implicit equation). Finally, we give some computational examples.

2 Modules with two quasi-generators and the µbasis.

Let R[t] be polynomial ring over the field of real numbers, and denote R[t] d the R-module of d-dimensional row vectors with entries in R[t]. Let R(t) be the field of rational functions in t. For a pair of vectors

A = (A 1 , A 2 , . . . , A d ),B = (B 1 , B 2 , ..., B d ) ∈ R[t] d the set M = A, B = {aA + bB ∈ R[t] d | a, b ∈ R(t), A, B ∈ R[t] d } ⊂ R[t] d (1)
A ∧ B = ([1, 2], [1, 3], ..., [d -1, d]) ∈ R[t] d(d-1)/2 , where [i, j] = A i B j -A j B i .
In other words, A ∧ B is the vector of 2-minors of the matrix

W A,B = A 1 A 2 • • • A d B 1 B 2 • • • B d
and we denote by deg(A ∧ B) = max i,j {deg(A i B j -A j B i )} the degree of the Plücker coordinate vector, i.e. the maximal degree of a 2-minor of W A,B . Let a polynomial vector A ∈ R[t] d be presented as

A = n i=0 α i t i , α i ∈ R d , i = 0, ..., n; α n = 0.
We denote the leading vector α n by LV (A) and the degree of A by deg A = n. Note that if LV (A) and LV (B) are linearly independent over R then deg A∧

B = deg A + deg B and LV (A ∧ B) = LV (A) ∧ LV (B). We define deg M = min{deg( Ã ∧ B) | Ã, B ∈ R[t] d such that Ã, B = M }
to be the degree of the module M with two quasi-generators. Remark 2. By abuse of notation, we will continue to denote parameters t, however in the geometric definitions that follow, they should be understood as parameters (t : s) ∈ P 1 and polynomials in R[t] should be thought of as homogenized with respect to a new variable s.

We define the following subspace of R d for the module M = A, B .

L(M, t 0 ) = {x ∈ R d | C(t 0 ) • x = 0 for all C ∈ M } where C(t) = (C 1 (t), C 2 (t), . . . , C d (t)) ∈ R[t] d , x = (x 1 , x 2 , . . . , x d ) T and C(t) • x = x 1 C 1 (t) + x 2 C 2 (t) + . . . + x d C d (t).
We have the inequality dim L(M, t 0 ) ≥ d -2, because the module M has only two quasi-generators. In fact, we have dim L(M, t 0 ) = d -2 for all t 0 , as we will see in Proposition 8.2. Whenever two vectors A(t 0 ) and B(t

0 ) are linearly independent in R d then L(M, t 0 ) is the intersection of two hyperspaces {x ∈ R d | A(t 0 ) • x = 0} and {x ∈ R d | B(t 0 ) • x = 0}.
Using those subspaces, we can associate a hypersurface S M in the real projective space P d-1 = P(R d ) with the module M

S M := t P(L(M, t)) ⊂ P d-1 .
(2)

Note that this definition and the definition of L(M, t 0 ) depend only on the module M and not on the choice of quasi-generators. It is useful to compare the hypersurface S M with the hypersurface S A,B defined as

S A,B := t ({A(t) • x} ∩ {B(t) • x}) ⊂ P d-1
(3)

where In fact, we will see in Proposition 6 that this phenomenon does not occur for µ-bases, i.e. if Ã, B is a µ-basis of the module M then S M = S Ã, B and there are no extraneous factors as before.

Remark 3. We should explain why we use the term µ-basis. The above definition is a generalization of the usual definition for the µ-basis of a rational ruled surface (as in [Cox, Sederberg, Chen (1998)], [Chen et al.(2001)] or [START_REF] Dohm | [END_REF]]). They coincide in the special case d = 4. M is the analogue of the syzygy module (i.e. the module of moving planes following the parametrization of the ruled surface) and the subspaces L(M, t), which in this case are 2-dimensional and hence define projective lines, are exactly the family of lines which constitute the ruled surface. Similarly, the case d = 3 corresponds at the theory of µ-bases for rational curves and our definition is equivalent to the usual definition as in [Chen, Wang(2003), Theorem 3, Condition 3].

However, the approach used here is actually inverse to the approach in the cited papers. In the latter the ruled surface is defined by a parametrization and then the module of moving planes is studied, whereas here we fix a module that "looks like" such a moving plane module and then study the (generalized) ruled surface that corresponds to it. Note that by definition of the subspaces L(M, t) any element C of M can be considered a moving plane following S M , in the sense that for all x ∈ S M there is a parameter t such that C(t) • x = 0.

Note that A ∧ B defines the so-called Plücker curve P in P d(d-1)/2-1 by

ϕ P : P 1 P d(d-1)/2-1 t → ([1, 2] : [1, 3] : ... : [d -1, d])
where [i, j] = A i B j -A j B i . We will denote k = deg ϕ P the degree of the parametrization, which is the cardinality of the fiber of a generic point in the image of ϕ P . Note that ϕ P and k are the same for any choice of quasi-generators of M . Proposition 4. For any pair of quasi-generators A, B of M we have the degree formula k

• deg S M = deg(A ∧ B) -deg q A,B ,
where q A,B = gcd(A ∧ B) and k = deg ϕ P . Moreover, we have deg P = deg S M .

Proof. The proposition and the proof are similar to Lemma 1 in [Chen et al.(2001)] and to Theorem 5.3 in [Pottmann et al.(1998)].

The implicit degree of the hypersurface S A,B is the number of intersections between a generic line and the hypersurface. The generic line L(s) is defined by two points in the space L(s) = H 0 + sH 1 , where

H i = (h i1 , h i2 , ..., h id ), i = 0, 1. The line L(s) intersects the hyperplane {A(t) • x} if and only if H 0 • A(t) + sH 1 • A(t) = 0.
Since the line L(s) should intersect the hyperplane {B(t) • x} too, we see that the implicit degree is the number of intersections of two curves in the (t, s) plane:

H 0 • A(t) + sH 1 • A(t) = 0, H 0 • B(t) + sH 1 • B(t) = 0.
Eliminating s from the above equation we have

H 0 • A(t) H 1 • A(t) H 0 • B(t) H 1 • B(t) = (H 0 ∧ H 1 ) • (A(t) ∧ B(t)) = 0, (4)
where C • D means a standard scalar product of two vectors C, D ∈ R d(d-1)/2 . The number of solutions of ( 4) is the number of intersection points of the Plücker curve with a generic hyperplane in P d(d-1)/2-1 , so deg P = deg S M . Now it is known (see for example [START_REF] Dohm | [END_REF], Theorem 1]) that

k • deg P = deg(A ∧ B) -deg q A,B
and the proposition follows.

We have yet to show the existence of the µ-basis. To this end, we propose an algorithm for its computation, the basic idea of which is to reduce q A,B = gcd(A ∧ B) to a constant using the so-called Smith form of the 2 × d matrix

W A,B = A 1 A 2 • • • A d B 1 B 2 • • • B d
and then render the leading vectors linearly independent by a simple degree reduction. The Smith form is a decomposition

W A,B = U •S•V , with unimodular U ∈ R[t] 2×2 , V ∈ R[t]
d×d , and

S = 1 0 0 • • • 0 0 q A,B 0 • • • 0 ∈ R[t] 2×d
It always exists and can be computed efficiently by standard computer algebra systems.

Algorithm (µ-basis)

1. INPUT: Quasi-generators A = (A 1 , A 2 , . . . , A d ),B = (B 1 , B 2 , ..., B d ) ∈ R[t] d of the module M 2. Set W A,B = A 1 A 2 • • • A d B 1 B 2 • • • B d .

Compute a Smith form

W A,B = U • 1 0 0 • • • 0 0 q A,B 0 • • • 0 • V with unimodular U ∈ R[t] 2×2 ,V ∈ R[t] d×d .
4. Set W ′ to be the 2 × d-submatrix consisting of the first two rows of V .

5. If the vector of leading terms (with respect to the variable t) of the first row is h times the one of the second row,

h ∈ R[t], set W ′ := 1 -h 0 1 • W ′ .
6. If the vector of leading terms (with respect to the variable t) of the second row is h times the one of the first row,

h ∈ R[t], set W ′ := 1 0 -h 1 • W ′ .
7. If the preceding two steps changed W ′ go back to Step 5.

8. Set Ã, B to be the rows of W ′ .

OUTPUT: A µ-basis Ã, B of the module M

As we shall see in Section 5, the case we are interested in is the case d = 6, so we are dealing with very small matrices and the computations are extremely fast. Note that we actually only need the first two rows of V , so we could optimize the algorithm by modifying the Smith form algorithm used as not to compute the unnecessary entries of the matrices U and V . Generally, the number of elementary matrix operations in Step 5 and 6 is very low. In the worst case, it is bounded by the maximal degree of the entries of the matrix W ′ in Step 4 of the algorithm, since each step reduces the maximal degree in one of the rows of W ′ . Next, we will show that the output of the above algorithm is a µ-basis and that the resultant of a µ-basis Ã, B of the module M = A, B is an implicit equation of S M . In Section 5, we will use these results for a special choice of A and B to compute the implicit equation of a canal surface.

k • deg S M = deg( Ã ∧ B) -deg(q) = deg( Ã ∧ B) = deg( Ã) + deg( B) Moreover, by definition we have deg M ≤ deg( Ã ∧ B) and if A, B are quasi- generators such that deg(A ∧ B) is minimal, the degree formula gives deg( Ã ∧ B) = deg(A ∧ B) -deg(q) ≤ deg M , which shows that k • deg S M = deg M , and as a consequence that Ã(t), B(t) is indeed a µ-basis. Proposition 6. Let Ã(t), B(t) be a µ-basis of M and x = (x 1 , x 2 , . . . , x d ) T variables. Then Res t ( Ã(t) • x, B(t) • x) = F k SM
where F SM is the implicit equation of the hypersurface S M .

Proof. First, we will show in the same way as in [START_REF] Dohm | [END_REF]

W = {(t, x) ∈ P 1 × P d-1 | Ã(t) • x = B(t) • x = 0} ⊂ P 1 × P d-1
is a vector bundle over P 1 and hence irreducible. So the projection on P d-1 is irreducible as well and its equation, which is by definition the hypersurface defined by Res t ( Ã(t) • x, B(t) • x), is a power of an irreducible polynomial.

As we have remarked earlier, the resultant of two quasi-generators is always a multiple of the implicit equation of S M , so Res

t ( Ã(t) • x, B(t) • x) is a power of F SM .
But using the degree property above we see

deg(Res t ( Ã(t) • x, B(t) • x)) = deg( Ã) + deg( B) = k • deg S M which implies that Res t ( Ã(t) • x, B(t) • x) equals F k SM . Remark 7.
It is known that the Plücker curve P can be properly reparametrized, i.e. there exists a rational function h of degree k such that A ∧ B = C • h, where C is a proper parametrization of P. It is tempting to use this proper reparametrization in order to represent the implicit equation F SM of S M directly as a resultant as in the proof of [START_REF] Dohm | [END_REF], Theorem 3]. However, h does not necessarily factorize A and B, i.e. it is not sure that there exist A ′ and B ′ with A = A ′ • h and B = B ′ • h, which would be needed to do this.

In the following we present some properties of µ-bases. Note that the properties in Propositions 8, 9 are similar to [Chen, Wang(2003)] Theorems 1,3. However, we give different proofs by deducing them from the degree formula and Lemma 5.

Proposition 8. Let M = A, B and let Ã, B be a µ-basis of the module M . Then the following properties hold: 1. The vectors LV ( Ã), LV ( B) are linearly independent. 2. Ã(t 0 ), B(t 0 ) are linearly independent over C for any parameter value t 0 ∈ C.

Proof. 1. If LV ( Ã), LV ( B) were linearly dependent, this would imply that k • deg S M = deg( Ã ∧ B) -deg(q Ã, B ) < deg( Ã) + deg( B) = deg M which is a contradiction to Lemma 5.
2. Suppose that Ã(t 0 ), B(t 0 ) are linearly dependent for some t 0 ∈ C. This is equivalent to saying that the matrix W Ã, B is not of full rank, which means that all 2-minors vanish. So t 0 is a root of q Ã, B and as above we deduce k

• deg S M = deg( Ã ∧ B) -deg(q Ã, B ) < deg( Ã) + deg( B) = deg M which is again a contradiction to Lemma 5.
Proposition 9. Let M = Ã, B and assume that Ã, B satisfy conditions 1,2 from Proposition 8. Then any element D ∈ M has the following expression: Suppose that deg à ≤ deg B and let M = P 1 , P 2 . Then we proved that 3 Elements of Lie and Laguerre Sphere Geometry

D = h 1 Ã + h 2 B for some h 1 , h 2 ∈ R[t], i.e. Ã,
P i = h i1 Ã+h i2 B, i = 1, 2 for some polynomials h ij ∈ R[t]. Since LV ( Ã), LV ( B) are linearly independent LV (h i1 Ã) and LV (h i2 B), i = 1, 2 do not cancel each other. Therefore, deg P 1 ≥ deg B (if h 12 = 0) or deg P 2 ≥ deg B (if h 22 = 0). Also deg P 1 ≥ deg Ã
Here we shortly recall the elements of Lie and Laguerre Sphere Geometry (cf. [Cecil(1992), Pottmann, Peternell(1998), Krasauskas, Mäurer(2000)]). We start from the construction of Lie's geometry of oriented spheres and planes in R 3 . Let p ∈ R 3 , r ∈ R. The oriented sphere S p,r in R 3 is the set

S p,r = {v ∈ R 3 |(v -p) • (v -p) = r 2 },
where by v•w we denote the standard positive definite scalar product in R 3 . The orientation is determined by the sign of r: the normals are pointing outwards if r > 0. If r = 0 then S p,0 = {p} is a point. Let n ∈ R 3 with n • n = 1 and h ∈ R. The oriented plane P n,h in R 3 is the set

P n,h = {v ∈ R 3 |v • n = h}.
The Lie scalar product with signature (4, 2) in R 6 is defined by the formula

[x, z] = -x 1 z 2 -x 2 z 1 2 + x 3 z 3 + x 4 z 4 + x 5 z 5 -x 6 z 6 .
for x = (x 1 , . . . , x 6 ) and z = (z 1 , . . . , z 6 ). In matrix notation we have

[x, z] = xCz T , where xC = (-x 2 /2, -x 1 /2, x 3 , x 4 , x 5 , -x 6 ). (5)
Denote ŷ = (u : y 0 : y 1 : y 2 : y 3 : y 4 ) ∈ P(R 6 ) = P 5 and define the quadric

Q = {ŷ ∈ P 5 | [ŷ, ŷ] = -uy 0 + y 2 1 + y 2 2 + y 2 3 -y 2 4 = 0} (6)
where [., .] is the obvious extension of the Lie scalar product to P 5 . Q is called Lie quadric.

We represent an oriented sphere S p,r (or an oriented plane P n,h ) as a point Lie(S p,r ) (resp. Lie(P n,h )) on the Lie quadric:

Lie(S p,r ) = (2(p • p -r 2 ), 2, 2p, 2r) ∈ Q, p ∈ R 3 , r ∈ R, Lie(P n,h ) = (2h, 0, n, 1) ∈ Q, n ∈ R 3 , h ∈ R.
It is easy to see that we have determined a bijective correspondence between the set of points on the Lie quadric Q and the set of all oriented spheres/planes in R 3 . Here we assume that a point q = (1 : 0 : 0 : 0 : 0 : 0) ∈ Q on the Lie quadric Q corresponds to an infinity, i.e. to a point in the compactification of R 3 . We say that q = (1 : 0 : 0 : 0 : 0 : 0) is the improper point on the Lie quadric. Notice that oriented planes in R 3 correspond to points Q ∩ T q , where T q = {ŷ = (u : y 0 : y 1 : y 2 : y 3 : y 4 ) ∈ P 5 |y 0 = 0} is a tangent hyperplane to the Lie quadric at the improper point q.

Two oriented spheres S p 1 ,r1 , S p 2 ,r2 are in oriented contact if they are tangent and have the same orientation at the point of contact. The analytic condition for oriented contact is

p 1 -p 2 = |r 1 -r 2 |,
where p 1 -p 2 denotes the usual distance between two points in the Euclidean space R 3 . One can check directly that the analytical condition of oriented contact on the Lie quadric is equivalent to the equation

[Lie(S p 1 ,r1 ), Lie(S p 2 ,r2 )] = 0.
It is known that the Lie quadric contains projective lines but no linear subspaces of higher dimension (Chapter 1, Corollary 5.2 in [Cecil(1992)]). Moreover, the line in P 5 determined by two points k 1 , k 2 of Q lies on Q if and only [k 1 , k 2 ] = 0, i.e. the corresponding spheres to k 1 , k 2 are in an oriented contact (Chapter 1, Theorem 1.5.4 in [Cecil(1992)]). The points on a line on Q form so called parabolic pencil of spheres. All spheres which correspond to a line on Q are precisely the set of all spheres in an oriented contact.

Remark 10. Here we use a slightly different coordinate system in Lie Geometry than in the book [Cecil(1992)]. The scalar product as in [Cecil(1992)] may be obtained applying the following transformation:

x ′ 1 = (x 1 + x 2 )/2, x ′ 2 = (x 2 -x 1 )/2, x ′ 3 = x 3 , x ′ 4 = x 4 , x ′ 5 = x 5 , x ′ 6 = x 6 .
We show now that the set of points ŷ in Q with y 0 = 0 is naturally diffeomorphic to the affine space R 4 . This diffeomorphism is defined by the map

φ : Q \ T q → R 4 , (u : y 0 : y 1 : y 2 : y 3 : y 4 ) → y1 y0 , y2 y0 , y3 y0 , y4 y0 ,
where T q = {ŷ = (u : y 0 : y 1 : y 2 : y 3 : y 4 ) ∈ P 5 | y 0 = 0} as before, i.e. the tangent hyperplane to the Lie quadric Q at the improper point q = (1 : 0 : 0 :

0 : 0 : 0). Let v = (v 1 , v 2 , v 3 , v 4 ), w = (w 1 , w 2 , w 3 , w 4 ) ∈ R 4 and denote by v, w = v 1 w 1 + v 2 w 2 + v 3 w 3 -v 4 w 4
the Lorentz scalar product on R 4 , which can be seen as the restriction of the Lie scalar product [., .] to R 4 . The affine space R 4 with the Lorentz scalar product is called the Lorentz space and denoted by R 4 1 . Let y = (y 1 , y 2 , y 3 , y 4 ) ∈ R 4 . One can check that inverse map of φ is given by the formula:

φ -1 (y) = ( y, y , 1, y) ∈ Q \ T q
Notice, that φ(Lie(S p,r )) = (p, r), i.e. the sphere S p,r ∈ R 3 corresponds to a point (p, r) ∈ R 4 1 . The map φ can be extended to a linear projection Φ from Q \ {q} to P 4 defined as Φ :

Q \ {q} → P 4 (u : y 0 : y 1 : y 2 : y 3 : y 4 ) → (y 0 : y 1 : y 2 : y 3 : y 4 )

The points of Q ∩ T q can be represented as Lie(P n,h ) = (2h, 0, n, 1) and these points correspond to planes in R 3 . Note that

Φ(Lie(P n,h )) = (0, n, 1) ∈ Ω = {y 0 = 0, y 2 1 + y 2 2 + y 2 3 -y 2 4 = 0}
are infinite points to the natural extension of R 4 to P 4 which correspond to a pencil of parallel planes in R 3 . The quadric Ω is called absolute quadric. The preimage of the map Φ has the following form Φ -1 (y) = ( y, y : y 2 0 : y 0 y 1 : y 0 y 2 : y 0 y 3 :

y 0 y 4 ) ∈ Q \ {q} (7)
where y = (y 0 : y 1 : y 2 : y 3 : y 4 ) ∈ P 4 and y = (y 1 , y 2 , y 3 , y 4 ) as before.

A direct computation shows that for v, w ∈ R 4

-2[φ -1 (v), φ -1 (w)] = v -w, v -w (8)
The formula shows that two oriented spheres defined by v, w (i.e. spheres S (v1,v2,v3),v4 and S (w1,w2,w3),w4 ) are in oriented contact if and only if v -w, vw = 0. Let us define two maps: an embedding i d : R 3 → R 4 , i d (p) = (p, d), d ∈ R and a projection π : R 4 → R 3 , π(p, r) = p, where r ∈ R. We will treat points i 0 (R 3 ) as spheres with zero radius and identify them with R 3 . All interrelations between the spaces introduced above can be described in the following diagram

Q \ T q ⊂ Q \ {q} ⊂ P 5 ↓ φ ↓ Φ R 3 i d → R 4 ⊂ P 4 ↓ π R 3 = R 3 (
) 2 + (a 2 -a 0 y 2 ) 2 + (a 3 -a 0 y 3 ) 2 = (a 4 -a 0 d) 2 , i.e. π(G d ((a 0 : a))| R 4 ) = S a 1 a 0 , a 2 a 0 , a 3 a 0 , a 4 -a 0 d a 0 = Env -d S a 1 a 0 , a 2 a 0 , a 3 a 0 , a 4 a 0 and G d ((a 0 : a))| R 4 = i d S a 1 a 0 , a 2 a 0 , a 3 a 0 , a 4 -a 0 d a 0
Therefore, in this case, the isotropic hypersurface G((a 0 : a)) may be treated as a union all envelopes to the sphere S a 1 a 0

,

a 2 a 0 , a 3 a 0 , a 4 a 0
. In the next section we generalize the definition of the isotropic hypersurface G(M) for a curve M in R 4 (or P 4 ). All lines in R 4 1 with directional vectors v can be classified into three types depending on the sign of v, v : (+)-lines, (0)-lines (also called isotropic lines), and (-)-lines.

The isotropic hypersurface and d-envelopes

In this section, we will see that the definition of the canal surface is not obvious and and we will introduce some geometrical object related to it. A canal surface is given by a so-called spine curve E, which is the closed image (with respect to the Zariski topology) of a rational map

R R 4 t → e1(t) e0(t) , e2(t) e0(t) , e3(t) e0(t) , e4(t) e0(t)
with polynomials e 0 (t), . . . , e 4 (t) ∈ R[t] such that n = max i=0,.,4 {deg(e i (t))}.

For abbreviation, we usually skip the variable t in the notations. The spine curve describes a family of spheres {S e 1 (t) e 0 (t) , e 2 (t) e 0 (t) , e 3 (t) e 0 (t) , e 4 (t) e 0 (t)

| t ∈ R} whose centers are given by the first three coordinates e1(t) e0(t) , e2(t) e0(t) , e3(t) e0(t) and whose radii are given by the last coordinate e4(t) e0(t) . Intuitively, the canal surface is the envelope of this family of spheres, but there are some subtleties to consider before we can make a precise definition.

We can also consider the spine curve as a projective curve E given as the closed image of a parametrization

P 1 P 4 t
→ (e 0 (t) : e 1 (t) : e 2 (t) : e 3 (t) : e 4 (t))

with the non-restrictive condition gcd(e 0 , . . . , e 4 ) = 1, which means that there are no base-points (i.e. parameters for which the map is not well-defined). Note that in this case the polynomials e i are actually to be considered as homogenized to the same degree n with respect to a new variable s. As there is a one-to-one correspondence between the univariate polynomials of a certain degree and their homogeneous counterparts, we will keep the notation from above and distinguish between the affine and projective case only where it is necessary to avoid confusion.

In the following we use the notations e = (e 1 , e 2 , e 3 , e 4 ), y = (y 1 , y 2 , y 3 , y 4 ), e = (e 0 : e 1 : e 2 : e 3 : e 4 ), y = (y 0 : y 1 : y 2 : y 3 : y 4 ).

We first proceed to define a hypersurface in P 4 which is closely related to the canal surface.

Definition 12. The isotropic hypersurface G(E) = {y | G(y) = 0} ⊂ P 4 associated with the (projective) spine curve E is the variety in P 4 defined by the polynomial G(y) = Res t (g 1 , g 2 ) where g 1 (y, t) = g(e, y) = (e 0 y 1 -e 1 y 0 ) 2 + (e 0 y 2 -e 2 y 0 ) 2 + (e 0 y 3 -e 3 y 0 ) 2 -(e 0 y 4 -e 4 y 0 ) 2 = e 0 y -y 0 e, e 0 y -y 0 e = e 2 0 y, y -2 e 0 e, y 0 y + y 2 0 e, e , g 2 (y, t) = ∂g 1 (y, t) ∂t = 2(e 0 e ′ 0 y, y -(e 0 e) ′ , y 0 y + y 2 0 e ′ , e ).

So, we define G(E) as the envelope of the family of isotropic hypersurfaces G(e) = G((e 0 (t) : e(t)).

In the previous section we showed that .

G d (e)| R 4 = Env -d S e 1 e 0 , e 2
This interpretation leads to the following definition.

Definition 13. The d-envelope associated with the (projective) spine curve E is defined as the hypersurface Env d (E) ⊂ P 3 given by the implicit equation is a hyperplane section of G(E), which can be interpreted as a parametrization of all offsets (with respect to the parameter y 4 ). The special case d = 0 is particularly important. For the real part of Env 0 (E) to be non-empty, one has to suppose that E has tangent (+)-lines almost everywhere, or equivalently that e, e > 0 almost everywhere. Env 0 (E) is the envelope of the family of spheres in R 3 given by the spine curve E and Env d (E) is the envelope of the same family of spheres with radii augmented by d. For instance, circular cylinders or circular cones (call them just cones) are envelopes Env 0 (L) of (+)-lines L and vice versa. In the literature, the canal surface C is usually defined as this envelope Env 0 (E). However, we will show in an example that these envelopes can contain "unwanted" extraneous factors, which are geometrically counterintuitive.

G d (y 0 , y 1 , y 2 , y 3 ) = Res t (g 1 | y4=-dy0 , g 2 | y4=-dy0 ) = Res t (g 1 , g 2 )| y4=-
(t) = 1 -t 2 1 + t 2 , 2t 1 + t 2 , 0, 1 2 .
The first three coordinates describe a circle in the plane and moving spheres of constant radius along this curve, so intuitively the envelope should be a torus T . But it turns out that the implicit equation of Env 0 (E) is up to a constant computed as

G 0 = Res t (g 1 , g 2 )| y4=0,y0=1 = (y 2 1 + y 2 2
) 2 (4y 2 1 + 4y 2 2 + 4y 2 3 + 8y 1 + 3)F T where F T is indeed the equation of the torus. To understand where the other factors come from, consider the following: For a given parameter t, the equations g 1 and g 2 define spheres S 1 (t) and S 2 (t) in R 3 and

Env 0 (E) = t S 1 (t) ∩ S 2 (t)
of the intersections of these spheres (actually this is nothing else than the geometric definition of the resultant). Now, while for almost all t this intersection is a transversal circle on the torus (often called characteristic circle in the literature), it can happen that the spheres degenerate either to planes or to the whole space. In our example, for the parameters t = i and t = -i we have g 1 (i) = g 1 (-i) = 0, g 2 (i) = -iy 1 + y 2 and g 2 (-i) = iy 1 + y 2 , so the intersection in those parameters actually degenerates to (complex) planes which correspond to the factor (-iy 1 + y 2 )(iy 1 + y 2 ) = y 2 1 + y 2 2 . In the parameter t = ∞, both g 1 and g 2 define the same sphere whose equation 4y 2 1 + 4y 2 2 + 4y 2 3 + 8y 1 + 3 is the other extraneous factor. This kind of phenomenon can also happen for real parameter values, but it is interesting to remark that even though we consider a real parametrization, non-real parameters can interfere with the envelope, because the resultant "knows" about them.

This example shows that Env 0 (E) is not a suitable definition for the canal surface C and we will later develop one that avoids the kind of extraneous components we have observed.

Remark 15. Sometimes, in the literature, Env 0 (E) is defined in affine space as the resultant Ǧ0 (y 1 , y 2 , y 3 ) = Res t (ǧ 1 (y 1 , y 2 , y 3 , t), ǧ2 (y 1 , y 2 , y 3 , t)), where ǧ1 = e 2 0 f1 , ǧ2 = e 3 0 f2 ,

f1 = y 1 - e 1 e 0 2 + y 2 - e 2 e 0 2 + y 3 - e 3 e 0 2 - e 4 e 0 2 , f2 = ∂ f1
∂t or in other words by deriving the affine equation of the sphere after the substitutions y 4 = 0, y 0 = 1 and homogenizing afterwards. Note that in this case f2 and ǧ2 are linear in y 1 , y 2 , y 3 . Let f1 = g 1 | y4=0,y0=1 and f2 = g 2 | y4=0,y0=1 . An easy computation shows that we have the following equalities ǧ1 = f1 , ǧ2 = e 0 f2 -2e ′ 0 f1 . Therefore, by properties of the resultant (29),(30), we have

Res t (ǧ 1 , ǧ2 ) = Res t ( f1 , e 0 f2 -2e ′ 0 f1 ) = Res t ( f1 , e 0 f2 ) = Res t ( f1 , e 0 ) • Res t ( f1 , f2 ).
Hence, we have Ǧ0 = Res t ( f1 , e 0 )•G 0 , so there are even more extraneous factors than before due to the roots of e 0 .

Linearizing the problem

The main idea to understand and eliminate the extraneous components that appeared in the example is to linearize the equations g 1 and g 2 by replacing the quadratic term y, y by a new variable u (or more precisely uy 0 to keep the equations homogeneous). This will make the results developed in Section 2 applicable. Geometrically, this means that we will pull back the spine curve to Q via the correspondence Φ.

For a spine curve E ∈ R 4 1 we define a proper pre-image Ê in the Lie quadric Q as the closure of the set Ê = Φ -1 (E) in Q. It is immediate by ( 7) that the parametrization of Ê is

P 1 Q ⊂ P 5 t
→ ( e(t), e(t) : e 2 0 (t) : e 0 (t)e 1 (t) : e 0 (t)e 2 (t) : e 0 (t)e 3 (t) : e 0 (t)e 4 (t)) (11) We can now define the envelopes associated with this new spine curve as follows.

Definition 16. The variety H( Ê) ⊂ P 5 associated with Ê is the hypersurface in P 5 defined by the implicit equation H(ŷ) = Res t (h 1 , h 2 ) where ŷ = (u : y 0 : y 1 : y 2 : y 3 : y 4 ) and

h 1 (ŷ, t) = -2[ŷ, Ê(t)] = ue 2 0 + y 0 e, e -2 e 0 e, y , h 2 (ŷ, t) = ∂h 1 (ŷ, t) ∂t = -2[ŷ, Ê′ (t)] = 2(ue 0 e ′ 0 + y 0 e ′ , e -(e 0 e) ′ , y ).
Similarly, the variety H d ( Ê) ⊂ P 4 is defined by the implicit equation

H d (u, y 0 , y 1 , y 2 , y 3 ) = Res t (h 1 | y4=-dy0 , h 2 | y4=-dy0 ) = Res t (h 1 , h 2 )| y4=-dy0 ,
i.e. the equation obtained by replacing y 4 in H(y) by -dy 0 , where d ∈ R.

Of course this is nothing else than substituting y, y in g 1 and g 2 by uy 0 and dividing by y 0 , so g i (y) = h i ( y, y , y 2 0 , y 0 y), i = 1, 2, i.e. g i = h i • Φ -1 , i = 1, 2. Now as an immediate corollary we obtain Proposition 17. With the notations as above we have

G(y) = H( y, y , y 2 0 , y 0 y), i.e. G = H • Φ -1 , (12)
and

G d (y 0 , y 1 , y 2 , y 3 ) = H d (y 2 1 + y 2 2 + y 2 3 -d 2 y 2 0 , y 2 0 , y 0 y 1 , y 0 y 2 , y 0 y 3 ). ( 13
)
To sum up, we have defined two hypersurfaces as resultants of two quadratic forms: Env d (E) ⊂ P 3 , which are the offsets to the spine curve E, and G(E) ⊂ P 4 , which can be interpreted as a parametrization of those offsets. As seen in an example, these definitions can lead to additional components which are against the geometric intuition, so it is desirable to give another definition which avoids those extra factors. To this end, we have linearized the problem by replacing the quadratic polynomials g 1 and g 2 by linear forms h 1 and h 2 by substituting the quadratic term by a new variable and have seen how to reverse this substitution. Geometrically, this means that we replace the hypersurfaces Env d (E) and G(E) by hypersurfaces H d ( Ê) and H( Ê) in one dimension higher.

This has the advantage that we can now apply the technique of µ-bases developed earlier to understand and eliminate the extraneous factors of H d ( Ê) and H( Ê) and then come back to P 3 (resp. P 4 ) with the substitution formulae of Proposition 17.

5 The dual variety, offsets, and the canal surface.

In this section, we will finally be able to define the canal surface C (and more general offsets to it) and the so-called dual variety Γ(E), which can be seen as a parametrization of the offsets to C. Up to the constant -2 the system h 1 = h 2 = 0 is equal to

   ŷ, Ê(t) = Ê(t)C ŷT = 0, ŷ, Ê′ (t) = Ê′ (t)C ŷT = 0, ( 14
)
where the matrix C is defined by the formula (5).

We can interpret the variety H( Ê) defined by ( 14) as a dual variety to the curve Ê with respect to the Lie quadric Q, i.e. the dual variety to the curve Ê(t)C. Indeed, this dual variety consists of the hyperplanes which touch the curve Ê(t)C. The first equation in ( 14) means that the hyperplane contains the point Ê(t)C, the second equation means that the hyperplane contains the tangent vector Ê′ (t)C to the curve Ê(t)C.

In order to simplify notation we denote

E = Ê(t)C = - e 2 0 2
, -e, e 2 , e 0 e 1 , e 0 e 2 , e 0 e 3 , -e 0 e 4

E ′ = Ê′ (t)C = (-e 0 e ′ 0 , -e ′ , e , e ′ 0 e 1 + e 0 e ′ 1 , e ′ 0 e 2 + e 0 e ′ 2 , e ′ 0 e 3 + e 0 e ′ 3 , -e ′ 0 e 4 -e 0 e ′ 4 ) ( 16)

and we have that H( Ê) = S E,E ′ by (3). As we have seen in Section 2, this surface contains extraneous factors which correspond to the roots of the 2-minors of the matrix W E,E ′ , but which can be eliminated by replacing E, E ′ by a µ-basis of the module E, E ′ . It is thus natural to make the following definition.

Definition 18. We define the dual variety V( Ê) ⊂ P 5 to the curve Ê as the hypersurface

V( Ê) = S E,E ′ (17)
where E, E ′ is the module quasi-generated by E and E ′ .

By the results of Section 2, it is immediate that V( Ê) ⊂ H( Ê) does not contain the components of H( Ê) caused by parameters t where W E,E ′ (t) is not of full rank or equivalently, where the intersection of the hyperplanes defined by h 1 and h 2 is of codimension 1, i.e. the hyperplanes coincide. So we can deduce Proposition 19. Let E 1 , E 2 be a µ-basis of the module quasi-generated by E and E ′ and let k be the degree of the parametrization

E ∧ E ′ as in Section 2. Then k • deg V( Ê) = deg E 1 + deg E 2 = deg(E ∧ E ′ ) -deg q E,E ′ ,
where q E,E ′ = gcd(E ∧ E ′ ) and

Res t (E 1 • ŷT , E 2 • ŷT ) = F k V( Ê) ,
where F V( Ê) is the implicit equation of V( Ê).

Proof. It follows directly from Propositions 4 and 6.

Of course, the same considerations can be applied to the hypersurfaces H d ( Ê) and we make the analogous definitions. Substituting y 4 = -dy 0 in h 1 and h 2 corresponds to replacing E and E ′ by two linear forms Definition 20. We define the hypersurface V d ( Ê) as

V d ( Ê) = S D,D ′ ⊂ P 4 (19)
where D, D ′ is the module quasi-generated by D and D ′ .

In this case also, V d ( Ê) ⊂ H d ( Ê) does not contain extraneous factors due to the parameters t where the rank of W D,D ′ (t) drops. At this point, it should be remarked that while we clearly always have

V d ( Ê) ⊂ V( Ê) ∩ {y 4 = -dy 0 }
this inclusion is not necessarily an equality (note that we had Env d (E) = G(E) ∩ {y 4 = -dy 0 } for the corresponding varieties). Analogously to Proposition 19 the following holds.

Res t (D 1 • (u, y 0 , y 1 , y 2 , y 3 ) T , D 2 • (u, y 0 , y 1 , y 2 , y 3 ) T ) = F k V d ( Ê)
where

F V d ( Ê) is the implicit equation of V d ( Ê).
Proof. It follows directly from Propositions 4 and 6.

Finally, we can use the correspondance of Proposition 17 to define the canal surface.

Definition 22. The Γ-hypersurface is defined as

Γ(E) = Φ(V( Ê) ∩ Q),
and the offset Off d (E) at distance d to the canal surface C is

Off d (E) = Φ 0 (V d ( Ê) ∩ Q d ),
where Q d = {(u : y 0 : y 1 : y 2 : y 3 ) ∈ P 4 | -uy 0 + y 2 1 + y 2 2 + y 2 3 -d 2 y 2 0 = 0} and Φ 0 (u, y 0 , y 1 , y 2 , y 3 ) = (y 0 , y 1 , y 2 , y 3 ). The canal surface itself is the special case d = 0 or in other words C = Off 0 (E).

Note that the extraneous factors of H d ( Ê) and H( Ê) are in one-to-one correspondence with the extraneous factors of the corresponding hypersurfaces Env d (E) and G(E) since they are caused by parameter values where the intersection of h 1 and h 2 (resp. g 1 and g 2 ) is of codimension one. So Γ(E) and C d (E) contain no such factors.

In this section and the previous one, many different geometric objects have been defined. We illustrate in the following diagram how they are related in order to make the situation clearer.

P 4 P 5 P 5 P 4 ∪ ∪ ∪ ∪ G(E) Φ ←-H( Ê) ∩ Q ⊇ V( Ê) ∩ Q Φ -→ Γ(E) ∪ ∪ ∪ ∪ Env d (E) Φ d ←-H d ( Ê) ∩ Q d ⊇ V d ( Ê) ∩ Q d Φ d -→ Off d (E) ∩ ∩ ∩ ∩ P 3 P 4 P 4 P 3 (20)
Note that the hypersurfaces in the third row are included in the corresponding hypersurfaces in the second row. The first column is the naive definition of the objects to be studied: Env d (E) is more or less a d-offset to the canal offsets and G(E) a hypersurface in one dimension higher containing all those offsets. However, they contain extraneous factors. So by passing to the second column, we linearize the hypersurfaces (i.e. we express them as resultants of linear forms) and can apply µ-bases to eliminate the extraneous factor, which gives the third column and finally go back down in dimension (by intersecting with Q and applying Φ to obtain the objects we are interested in: the offsets Off d (E) (in particular the canal surface C = Off 0 (E)) and the Γ-hypersurface.

The implicit equation.

We can now describe how to compute powers of the implicit equations of the dual varieties V( Ê) and V d ( Ê), the hypersurface Γ(E) and the offsets surface C d (E). We should remark that these powers (which are the degrees of the parametrizations of the corresponding Plücker curves) are in a way inherent to the geometry of the problem, as we shall illustrate in Example 27. They can be interpreted as the number of times the surface is traced by the spine curve. Note also that this not necessarily due to the non-properness of the spine curve: Even for a proper spine curve it can happen that the canal surface (or its offsets) is multiply traced, as in Example 27.

Algorithm (implicit equations)

1. INPUT: A rational vector e(t) ∈ R(t) 4 as in formula (4).

Since C(0) = q we can compute the multiplicity m = i(q, {Res(h 1 , h 2 , t)} ∩ C) as follows. Lets denote by ch

1 = h 1 | C , ch 2 = h 2 | C the restriction of polynomials h 1 , h 2 to the conic C: ch 1 = -e 2 0 (ku -1)/2 -kf u 2 /2 + e 0 e, ku 2 a + u(b -a) , (33)
ch 2 = u((2f e ′ 0 -f ′ e 0 )ku/2 + e 0 w, kua + b -a ), where f = e, e . (34)
The computation of the resultant gives Res t (ch 1 , ch 2 ) = u m (A + Bu + ...), here m = i(q, {Res t (h 1 , h 2 )} ∩ C). On the other side we can compute the number of common points (0, t 0 ) on curves {ch 1 (u, t)} and {ch 2 (u, t)} counted with multiplicities. This number coincides with m (see [Buse et al.(2005)], Proposition 5).

The second curve ch 2 (u, t) = u • ch 3 (u, t) is reducible. Therefore, the resultant with respect to t is

Res(ch 1 , ch 2 ) = Res(ch 1 , u) Res(ch 1 , ch 3 ) = u deg(h1,t) Res(ch 1 , ch 3 ) = u 2n Res(ch 1 , ch 3 )
The second factor has a representation ch 3 = uD(t) -N (t), where D(t) = (2f e ′ 0 -f ′ e 0 )k/2 + e 0 w, ka and N (t) = e 0 e ′ e 0 -ee ′ 0 , a -b . It is easy to see that gcd(N (t), e 2 0 ) = e 0 . If ch 1 (0, t 0 ) = ch 3 (0, t 0 ) = 0 then e 0 (t 0 ) = 0. The first curve {ch 1 (u, t)} is hyper-elliptic, i.e. the projection to t axes pr : {ch 1 } → t is a map two-to-one. The curve {ch 1 (u, t)} has the following discriminant with respect to u:

disc(ch 1 , u) = e 2 0 ( (ke 0 /2 -e, b -a ) 2 -2e 0 e, ka + kf ). ( 35
)
It is ease to see that point (0, t 0 ) is a singular point on the curve {ch 1 (u, t)} if and only if e 0 (t 0 ) = 0. Therefore the point (0, t 0 ) has multiplicity at least two as a point of the intersection of two curves {ch 1 } ∩ {ch 3 }. We will prove that the multiplicity of the intersection of two curves {ch 1 } ∩ {ch 3 } at the point (0, t 0 ) equals to two if e 0 (t 0 ) = 0. For simplicity we assume that t 0 = 0. The first equation ( 33 An easy computation with MAPLE shows that Res t (ch 1 , ch 3 ) = u 2 (K 4 u 4 + K 3 u 3 + K 2 u 2 + K 1 u + K 0 ).

Therefore the point (0, 0) has multiplicity two if and only if K 0 = 0, i.e. K 0 is a unit in the local ring R. A straightforward computation shows that Since f (0) = 0 and e ′ 0 (0) = 0 by the condition (27) we conclude that K 0 = 0 for a general conic. Hence, the multiplicity i(q, {Res t (h 1 , h 2 )} ∩ Q) is equal to deg(h 1 , t) + 2 deg(e 0 , t) = 4n. Therefore, we have d w (Res t (h 1 , h 2 )) = 2 deg(Res t (h 1 , h 2 )) -i(q, {Res t (h 1 , h 2 )} ∩ Q) = 2(5n -2) -4n = 6n -4.

Remark 26. We conjecture that the degree of the hypersurface Γ(E) is deg V( Ê)+ deg(w) -deg(gcd(w)), where w is defined by the formula (26).

7 Examples and special cases.

Let n be the degree of the spine curve E = {e(t) ∈ R 4 }. And let c n (E) be the degree of the hypersurface Γ(E) with the spine curve E.

Polynomial case. Assume that the spine curve is polynomial, i.e. e 0 = 1. By the theorem in [Xu et al.(2006)] the degree of hypersurface Γ(E) with the polynomial spine is at most 4n -2.

For the general spine curve we have c n (E) = 6n -4, i.e. c n (E) ≤ 6n -4. The lower bound is not clear. There are examples of spine curves with the following degrees: c 2 (E) = 3, 4, 5, 6, 8; c 3 (E) = 6, 7, 8, 9, 10, 11, 12, 14; c 4 (E) = 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20. It seems that there does not exist a spine curve such that c n (E) = 6n -5.

We consider three examples.

Example 27. Let us consider the following spine curve: e(t) = 0, 0, 8t 1+t 2 , 3-3t 2 1+t 2

, n = 2. This is a proper parametrization of an ellipse in R 4 . We find the Plücker coordinate vector P = Ê(t)C ∧ Ê′ (t)C and q = gcd(P ) = 1. Also, we see that deg V( Ê) = deg P -deg(q, t) = 8 and γ = deg(w) = 2. If we run the µ-basis algorithm with two input vectors E(t)C, E ′ (t)C we get the output two vectors E 1 and E 2 :

E 1 • ŷT = 4 t 3 y 3 + (-u -41 y 0 ) t 2 + 12 ty 3 + 9 y 0 -u -6 y 4 , E 2 • ŷT = (u -9 y 0 -6 y 4 ) t 3 -12 t 2 y 3 + (41 y 0 + u) t -4 y 3 .

Now we can find the implicit equation G = Res(E 1 • ŷT , E 2 • ŷT , t) of the dual variety V( Ê). The polynomial G contains 26 monomials and has degree 6 (as in Proposition 23). The equation of the hypersurface Γ(E) is defined by the Example 29. In the next example we take the following spine curve:

e(t) = (1-t 2 ) 2 (1+t 2 ) 2 , 2
t(1-t 2 ) (1+t 2 ) 2 , 2 t 1+t 2 , 1 , n = 4. The first three coordinates define the Viviani curve, i.e. it is intersection curve of the sphere and the tangent cylinder. We find the Plücker coordinate vector P = Ê(t)C ∧ Ê′ (t)C and q = gcd(P ) = 1. Also, we see that deg V( Ê) = deg P -deg q = 6. If we run the µ-basis algorithm with two input vectors Ê(t)C, Ê′ (t)C we get output of two vectors E 1 = 0, 4 + 4 t 2 , 4 -4 t 2 , 6 t -2 t 3 , 6 t + 2 t 3 , 4 + 4 t 2 , E 2 = 0, 4 t + 4 t 3 , 4 t -1 + t 2 , 2 -6 t 2 , 2 + 6 t 2 , 4 t + 4 t 3 , both of degree 3 and find the implicit equation G of the dual variety V( Ê) (it contains 58 monomials). Finally, we find that c 4 (E) = d w (G) = 10. For this example, we have deg C = deg Γ(E), i.e. the implicit degree of the canal surface is 10.

Definition 1 .

 1 Two quasi-generators Ã, B of the module M = A, B are called a µ-basis of the module M if deg M = deg à + deg B. As we always have the inequality deg(A ∧ B) ≤ deg A + deg B, this means in particular that the sum deg à + deg B is minimal. A µ-basis always exists, as we shall see at the end of the section. Let us explain the geometric motivation behind this definition.

Lemma 5 .

 5 The output of the above algorithm is a µ-basis and we have k • deg S M = deg M , where k = deg ϕ P . Proof. Let Ã(t), B(t) be the output of the above algorithm. By construction it is clear that Ã(t), B(t) are quasi-generators of M and that qA,B = gcd( Ã ∧ B) = 1. Furthermore, we have deg( Ã ∧ B) = deg( Ã) + deg( B), because the vectors of leading terms of Ã(t) and B(t) are linearly independent. So by Proposition 4 we deduce

  B are generators of the module M over the polynomial ring R[t]. Moreover, the pair Ã, B is a µ-basis of the module M . Proof. Let D ∈ M , it can be expressed as a, b, c, d ∈ R[t] and co-prime numerators and denominators in the rational functions a b and c d . Furthermore, we may assume that gcd(a, c) = 1, because if D gcd(a,c) is a linear combination of Ã, B, then so is D. Multiplying both sides of the above equation with bd we obtain bdD = ad à + bc B or equivalently b(dD-c B) = ad à and since b divides neither a nor à (if it divided Ã, for any root t 0 of b and any constant α we would deduce the relation 0 = α Ã(t 0 ) + 0 • B(t 0 ), which contradicts property 2 in Proposition 8), one concludes that b divides d and by a symmetric argument that d divides b, so we may assume b = d. So we have bD = a à + c B and plugging a root t 0 of b into the equation, we would obtain a non-trivial linear relation between à and B, again a contradiction to Proposition 8. This implies that b and d are constant, which shows that any D ∈ M can be expressed as linear combination of à and B over R[t]. In other words: à and B are not only quasi-generators of M , but actually generators in the usual sense, i.e. over R[t].

  and deg P 2 ≥ deg Ã. So, we see that deg P 1 + deg P 2 ≥ deg à + deg B, i.e. a pair Ã, B is a µ-basis of the module M .

.

  9)Definition 11. For an oriented surface (curve or point) M ⊂ R 3 define an isotropic hypersurface G(M) ⊂ P 4 as the union of all points in R 4 which correspond to oriented tangent spheres of M. Let G d (M) = G(M) ∩ {y 4 = dy 0 } be a variety which corresponds to tangent spheres with radiusd of M. The set Env d (M) = π(G d (M)| R 4 ) ⊂ R 3 are centers of spheres with radius d tangent to M. The set Env d (M) is called d-envelope of the variety M. Since G(M) = d G d (M)we can treat the isotropic hypersurface G(M) as the union of all d-envelope to the variety M.If y, a ∈ R 4 , a 0 , y 0 ∈ R, we define a function g((a 0 : a), (y 0 : y)) = ay 0 -a 0 y, ay 0 -a 0 y = y 2 a -2a 0 y 0 a, y + a 2 0 y, y . Let (y 0 : y) be such that g((a 0 : a), (y 0 : y)) = 0. By the formula (8) we see that spheres S contact. Therefore, in the same manner as previously, we define the isotropic hypersurface G((a 0 : a)) as follows G((a 0 : a)) = {(y 0 , y) ∈ P 4 | g((a 0 : a), (y 0 : y)) = 0} ⊂ P 4 .In fact, G((a 0 : a)) is a quadratic cone with a singular point at a vertex (a 0 : a) ∈ P 4 and may be viewed as the set of all spheres which touches the fixed sphere S After the restriction to the linear subspace y 4 = dy 0 this hypersurface consists of all spheres with radius d which are in oriented contact with the sphere S a 1 as G d ((a 0 : a)) = G((a 0 : a)) ∩ {y 4 = dy 0 }. We notice that G d ((a 0 : a))| y0=1 is defined by the equation (a 1 -a 0 y 1

  dy0 , i.e. the equation obtained by replacing y 4 in G(y) by -dy 0 , where d ∈ R. The affine envelope Env d (E) at distance d is the restriction of Env d (E) to the affine space R 3 , defined by the equation G d | y0=1 = Res t (g 1 , g 2 )| y4=-dy0,y0=1 , i.e. by setting y 0 = 1. So G(E) contains all offsets associated with the spine curve E. Indeed, the surface Env d (E) = G(E) ∩ {y 4 = -dy 0 }

  0 e 4 , e 0 e 1 , e 0 e 2 , e 0 e 3 (18) D ′ = (-e 0 e ′ 0 , -e ′ , e + d(e ′ 0 e 4 -e 0 e ′ 4 ), e ′ 0 e 1 + e 0 e ′ 1 , e ′ 0 e 2 + e 0 e ′ 2 , e ′ 0 e 3 + e 0 e ′ 3 ) with D, D ′ ∈ R 5 . Now H d ( Ê) = S D,D ′ and one makes an analogous definition:

Proposition 21 .

 21 Let D 1 , D 2 be a µ-basis of the module quasi-generated by D and D ′ and let k be the degree of the parametrization D ∧ D ′ as in Section 2. Then deg V d ( Ê) = deg D 1 + deg D 2 = deg(D ∧ D ′ ) -deg q D,D ′ , where q D,D ′ = gcd(D ∧ D ′ ) and

  2. Define E, E ′ ∈ R[t] 6 as in formula (15) and D, D ′ ∈ R[t] 5 as in formula (18). 3. Compute a µ-basis E 1 , E 2 of the module E, E ′ and a µ-basis D 1 , D 2 of the module D, D ′ using the algorithm in Section 2. such that a, a = b, b = 0. With the notations as above the general conic C := P ∩ Q has the following parametrization: C(u) := (ku -1, ku 2 , ku 2 a + u(b -a)), where k = 2 a, b .

  ) in the local ring R = R[u, t] u,t is ch 1 = a 21 u 2 t + a 20 u 2 + a 12 ut 2 -a 11 ut + a 02 t 2 , where [a 21 , a 20 , a 12 , a 11 , a 02 ] = [ẽ 0 k e, a , -f k/2, -ẽ 2 0 k/2, ẽ0 e, a -b , 1/2 ẽ2 0 ](36) and tẽ 0 = e 0 . The second equation (34) in the local ring R is ch 2 = u(ch 3 ), ch 3 = b 12 ut 2 + b 11 ut + b 10 u + b 02 t 2 + b 01 t, where [b 12 , b 11 , b 10 , b 02 , b 01 ] = [ẽ 2 0 k e ′ , a , -f ′ ẽ0 -ẽ0 e ′ 0 k e, a , kf (e ′ 0 + e ′ 0 ) , -ẽ 2 0 k e ′ , a -b , ẽ0 e ′ 0 e, a -b ].

K 0 =

 0 a 02 b 10 2 a 02 + b 01 2 a 20 + b 01 b 10 a 11 = f kẽ 4

  is the R[t]-module with two polynomial quasi-generators A, B. Here, we assume that A, B are R[t]-linearly independent, i.e. aA + bB = 0 with a, b ∈ R[t] if and only if a = b = 0.

Remark: Note that the vectors A, B may not be generators of the module M over R[t] because a and b in the definition (1) are from the field R(t) of rational functions. For example, if A = pD with p ∈ R[t], D ∈ R[t] d and deg p > 0 then A, B are not generators of the module M . For A = (A 1 , A 2 , . . . , A d ),B = (B 1 , B 2 , ..., B d ) ∈ R[t] d we define the Plücker coordinate vector A ∧ B as follows:

  Example 14. Consider the spine curve E given by

	e 1 (t) e 0 (t)	,	e 2 (t) e 0 (t)	,	e 3 (t) e 0 (t)	,	e 4 (t) e 0

4. Set F V( Ê) = Res t (E 1 •ŷ T , E 2 •ŷ T ) and F V d ( Ê) = Res t (D 1 •(u, y 0 , y 1 , y 2 , y 3 ) T , D 2 • (u, y 0 , y 1 , y 2 , y 3 ) T ) = 0.

5. Let F Γ(E) (y 0 , y 1 , y 2 , y 3 , y 4 ) = y k 0 F V( Ê) ((y 2 1 +y 2 2 +y 2 3 -y 2 4 )/y 0 , y 0 , y 1 , y 2 , y 3 , y 4 ), where k is a minimal integer such that F Γ(E) is a polynomial. Similarly, set F C d (E) (y 0 , y 1 , y 2 , y 3 ) = y k 0 F V d ( Ê) ((y 2 1 + y 2 2 + y 2 3 -d 2 y 2 0 )/y 0 , y 0 , y 1 , y 2 , y 3 ).

6. OUTPUT:

, and F C d (E) , which are powers of the implicit equation of the varieties V( Ê), V d ( Ê), Γ(E) and C d (E)

Note that the affine parts of these equations can be obtained by replacing y 0 = 1 before the resultant computation.

The parametrization of the dual variety.

We can describe the parametrization of V( Ê). The hyperplane defined by the equation det(ŷ, Ê(t)C, Ê′ (t)C, a 1 , a 2 , a 3 ) = A 1 u + A 2 y 0 + A 3 y 1 + ... + A 6 y 4 = 0 ( 21) is tangent to the curve Ê(t)C , a i ∈ R 6 , i = 1, 2, 3 are three arbitrary points.

By the definition a point on the dual variety V( Ê) is (A 1 , ..., A 6 ). Define D = ( Ê(t)C, Ê′ (t)C, a 1 , a 2 , a 3 ) to be the 5×6 matrix with five rows Ê(t)C, Ê′ (t)C, a 1 , a 2 , a 3 . And let D i , i = 1, ..., 6 be 5 × 5 matrices obtained from D by removing the i-th column. Then using the Laplacian expansion by minors for the first row of the determinant (21) we obtain the parametrization of V( Ê) as follows:

where m = gcd(D 1 , ..., D 6 ). Here t, a 1 , a 2 , a 3 are arbitrary parameters.

6 The implicit degree of the hypersurface Γ(E).

The aim of this section is to get some formula for the implicit degree of the hypersurface Γ(E) in terms of the rational spine curve E = {e(t) ∈ P 4 }. Notice that the implicit degree of the canal surface C is less or equal than deg Γ(E) because we always have the inclusion

So this formula gives upper bound for the degree of the canal surface. In the case of a polynomial spine curve the upper bound was obtained in the paper [Xu et al.(2006)]. Note that for the computation of the implicit degree we do not need the implicit equation of the hypersurface. We believe that this formula is useful for higher degree spine curves because the computation of the implicit equation may be very difficult in practice.

Let us remind that the pre-image Φ -1 (Γ(E)) ⊂ Q is defined by the intersection of two varieties V( Ê) ∩ Q. Let denote by G(u, y 0 , y 1 , y 2 , y 3 , y 4 ) = 0 the equation of V( Ê). The Lie quadric has the equation uy 0 = y, y (recall that y, y = y 2 1 +y 2 2 +y 2 3 -y 2 4 ). By the definition (22) the equation of Γ(E) is obtained after the elimination of the variable u from the equations of Q and V( Ê), i.e.

Γ(E) : F (y 0 , y 1 , y 2 , y 3 , y 4 ) = y k 0 G y, y y 0 , y 0 , y 1 , y 2 , y 3 , y 4 = 0 ,

where k is a minimal integer such that the left side of the equation ( 23) is polynomial. We introduce the following weighted degree

Let us assume that the curve E = {e(t, s) ∈ R 4 | (t, s) ∈ P 1 } has a homogeneous parametrization. We introduce the following notations w = (w 1 , w 2 , w 3 , w 4 ), w j = e ′ j e 0 -e j e ′ 0 , j = 1, 2, 3, 4, ( 25)

where e ′ i means derivative with respect to t. We will say that the curve E ∈ P 4 is of general type if : gcd(w 1 , w 2 , w 3 , w 4 ) = gcd(e 0 , e ′ 0 ) = gcd(e 0 , e, e ) = 1, deg E = deg(e 0 , t) and the parametrization degree of the Plücker curve ϕ

(i.e. deg ϕ P = 1), where e = (e 1 , e 2 , e 3 , e 4 ) and E as in formula (15). The first equation in the system (14) has the following form:

The polynomial h 1 is linear in the variables u, y 0 , y 1 , y 2 , y 3 , y 4 and has degree 2n in the variable t. The elimination of the variable t from the system

, where LC(h 1 ) is the leading coefficient of the polynomial h 1 with respect to the variable t and

Proof. For the curve of general type by the Proposition 19 we have deg

We can compute components of the Plücker vector E ∧ E ′ = ([1, 2] : [1, 3] : ... : [5, 6]) ∈ P 14 . For example, [1, 2 + j] = e 2 0 w j /2, j = 1, 2, 3, 4 and [2, 3] = (-e, e (e 0 e 1 ) ′ + e 0 e 1 e, e ′ )/2. By assumption (27) we see gcd ([1, 3], [1,4], [1,5], [1,6]) = e 2 0 and gcd(e 0 , [2, 3]) = 1. Therefore gcd(E∧E ′ ) = 1. So we have deg

20

Thereinafter, we will show that for the curve of general type, we have d w (G) = 6n -4, where n = deg E. For this we consider another resultant Res t (h 1 , h 2 ) and show that d w (Res t (h 1 , h 2 )) = d w (G). We define h 2 in the following way. Let g = e 2 0 then we have the following equality h ′ 1 g -h 1 g ′ = e 0 h 2 , where h 2 = (2 e, e e ′ 0 -e, e ′ e 0 )y 0 /2+e 0 e ′ e 0 -ee ′ 0 , y = (2 e, e e ′ 0 -e, e ′ e 0 )y 0 /2+e 0 w, y .

In other words, h 2 is the numerator of a rational function h1 g ′ . We often use the following properties of the resultant [Cox et al.(1998)], p. 73), ( 29) [Cox et al.(1998)], p. 70), ( 30)

We need an explicit formula for factors of the resultant

where LC(h 1 ) is a leading coefficient with respect to variable t of the polynomial h 1 . Indeed, since deg(h

. Thus we obtain the formula in (31) from the property (30). The left side of the formula ( 31) is equal to LC(h 1 ) Res t (h 1 , e 0 h 2 ) = LC(h 1 ) Res(h 1 , e 0 ) Res t (h 1 , h 2 ). Otherwise, the right side of this formula is equal to Res

Also, from the property (30) follows that Res t (h 1 , e 0 ) = y deg(e0,t) 0

. Therefore we have and d w (Res t (h 1 , h 2 )) = d w (G). In the lemma below, we prove that d w (Res t (h 1 , h 2 )) = 6n -4.

We summarize our computations in the following Theorem 24. The degree of the hypersurface Γ(E) with the spine curve E which satisfies the assumption ( 27) is equal to 6n -4, where n = deg E.

Lemma 25. With the notation as above we suppose that the conditions ( 27) are satisfied. Then we have the equality d w (Res t (h 1 , h 2 )) = 6n-4, where n = deg E.

Proof. The weighted degree d w (G) may be viewed as a degree of a variety Φ({G} ∩ Q), where Φ : P 5 \ q → P 4 is a linear projection from the improper point q = (1, 0, ..., 0) on the Lie quadric (see explicit formula (3)). The degree of the variety Φ({G} ∩ Q) can be computed constructively by counting points of intersection with a general line L ∈ P 4 . The pre-image of the line C := Φ -1 (L) is a conic on the Lie quadric Q which passes through the improper point q.

Hence the degree of Φ

is the multiplicity of the intersection {G} ∩ C at the point q.

We need a parametric representation of the general conic q ∈ C ⊂ Q. Assume that the conic C is in the parameterized plane P : q + k 1 u + k 2 v, where k 1 , k 2 ∈ R 6 . The plane P intersects a singular cone { x, x = x 2 1 +x 2 2 +x 2 3 -x 2 4 } on two lines. We choose two vectors k 1 , k 2 in these lines so that the first coordinate is zero, i.e.

polynomial F (y 0 , ..., y 4 ) = y 2 0 G( y, y /y 0 , y 0 , y 1 , y 2 , y 3 , y 4 ) of degree 8. Since,

the 0-envelope of the canal surface Env 0 (E) = Γ(E) ∩ {y 4 = 0} is reducible. The canal surface C is the double ellipsoid of revolution (-225 + 25 y 1 2 + 25 y 2 2 + 9 y 3 2 ) 2 . Indeed, for the computation of C(E) we should assume that the variable y 4 = 0 and to repeat the same steps as above. We should consider only the first 5 coordinates of the vectors E(t)C, E ′ (t)C. Let us denote these two vectors with 5 coordinates by D 1 , D 2 . But this time we see that the Plücker vector P = D 1 ∧ D 2 has a non-trivial common divisor, i.e. q = gcd( P ) = t 2 -1. So, using the µ-basis algorithm we find the µ-basis R 1 , R 2 for the input D 1 , D 2 . In this case we see that deg R 1 = deg R 2 = 2. Now we find the resultant Ǧ = Res t (R 1 •y T , R 2 •y T ) = 16 y 3 2 + 225 y 0 2 -25 y 0 u 2 , where y = (u, y 0 , y 1 , y 2 , y 3 ).

After the substitution u = (y 2 1 + y 2 2 + y 2 3 )/y 0 we obtain the implicit equation of the canal surface the double ellipsoid (-225 + 25 y 1 2 + 25 y 2 2 + 9 y 3 2 ) 2 . We can see this geometrically, too. The point e(t) ∈ R 4 corresponds to the sphere S(e(t)) ∈ R 3 with a center on the y 3 -axis. If t ∈ [-1/2, 1/2] then the sphere S(e(t)) is tangent to the ellipsoid EL = (-225+25 y 1 2 +25 y 2 2 +9 y 3 2 ), and inside this ellipsoid. Moreover, the real envelope of the family S(e(t)), t ∈ [-1/2, 1/2] is the ellipsoid EL. Note that the sphere S(e(1/t)) has the same center but the opposite radius to the sphere S(e(t)), i.e. it has the opposite orientation. Therefore, the real envelope of the family S(e(t)), t ∈ (-∞, -2] ∩ [2, ∞) is the same ellipsoid EL. Hence, from the point of Laguerre geometry the envelope of the whole family S(e(t)) is the double ellipsoid EL 2 . Note, that the d-offset to the canal surface, in this case is the d-offset to ellipsoid EL and it has degree 8. Also, we can check that by Theorem 24 the degree of the Γ(E) hypersurface is 8, too. For a detailed study and other examples of canal surfaces with a quadratic spine curve we recommend to look at the paper [Krasauskas, Zube(2007)].

Example 28. Consider the polynomial spine curve e(t) = 3t 2 + 1, 4t 2 + t, 0, 5t 2 , n = 2. We find the Plücker coordinate vector P = Ê(t)C ∧ Ê′ (t)C and q = gcd(P ) = 1. Also, we see that deg V( Ê) = deg P -deg q = 4 and γ = deg(w) = 1. If we run the µ-basis algorithm with two input vectors Ê(t)C, Ê′ (t)C we get the output of two vectors E 1 = -u + -1 -7 t 2 -8 t 3 y 0 + 2 + 6 t 2 y 1 + 2 t (1 + 4 t) y 2 -10 t 2 y 4 , E 2 = -t (7 + 12 t) y 0 + 6 ty 1 + (1 + 8 t) y 2 -10 ty 4 , and find the implicit equation G of the dual variety V( Ê) (it contains 54 monomials, so we do not present an explicit formula). Finally, we find that c 2 (E) = d w (G) = 5. For this example, we have deg C = deg Γ(E), i.e. the implicit degree of the canal surface is 5. Note that this contradicts Theorem 4 in [Xu et al.(2006)], because in this example the degree of the canal surface is an odd number. It seems that the mentioned theorem gives only an upper bound estimation, but not the exact degree formula of canal surfaces with polynomial spine curve.