Page 203 - Geometric Modeling and Algebraic Geometry

206 C. Liang et al.
Proof. Suppose that t x (x) =0 and ∂ z (f) =0 on D. It implies that C is smooth in D.
Consider two branches of C in D and project them by π z onto a (x, y)-plane. Their
projection cannot intersect at an interior point. Otherwise, there would be two points
p 1 ,p 2 ∈ D, such that f(p 1 )=0,f(p 2 )=0 and π z (p 1 )= π z (p 2 ), which implies
that ∂ z f(p) vanishes for an intermediate point ∈]p 1 ,p 2 [ in D. This is impossible by
hypothesis. Consequently, the branches of C project bijectively onto the branches of
π z (C). Their tangent vector is the projection (t x (x), t y (x)) of the tangent vector of
C. By proposition 3, π z (C) is regular, so that the topology of π z (C), and thus of C,is
uniquely determined by the intersection points of C with the border of D.

A similar criterion applies by symmetry, exchanging the roles of the x, y, z co-
ordinates.
Let us give now a finer regularity criterion, which is computationally less expen-
sive:
Proposition 5. If C is smooth in D and if for all x 0 ∈ R, the plane x = x 0 plane has
at most one intersection point with the curve C in D, then C is regular on D.

Proof. Consider the projection π z (C) of the curve C in D along the z direction. Then
the components of C in D projects bijectively on the (y, z) plane. Otherwise, there
exist two points p 0 and p 1 lying on C such that π z (p 0 )= π z (p 1 )=(x 0 ,y 0 ), then
p 0 and p 1 belong to x = x 0 which are functions of the form y = Φ(x). Otherwise,
there exist two points on π z (C) and (and on C∩D) with the same x-coordinate. Con-
sequently, for x ∈ [a 0 ,b 0 ] there is at most one branch of π z (C) in D above x, and
◦
the connected components of C∩ D project bijectively onto non-overlapping open
intervals of [a 0 ,b 0 ] as π z (C) does. We conclude as in the 2D case (proposition 3), by
sorting the points of C∩ ∂D according to their x-coordinates, and by gathering them
by consecutive pairs corresponding to the starting and ending points of branches of
C∩ D.

Proposition 6. The 3D spatial curve C defined by f =0 and g =0 is regular on D,
if

• t x (x) =0 on D, and
• ∂ y h =0 on z-faces, and ∂ z h =0 and its has the same sign on both y-faces of
D,for h = f or h = g.
Proof. Let us fix x 0 ∈ [a 0 ,b 0 ] where D =[a 0 ,b 0 ] × [a 1 ,b 1 ] × [a 2 ,b 2 ],let U =
:(x 0 ,y,z) ∈ U → (f(x 0 ,y,z),g(x 0 ,y,z)).
{x 0 }× [a 1 ,b 1 ] × [a 2 ,b 2 ] and let Φ x 0
is injective. The Jacobian
We are going to prove that under our hypothesis, Φ x 0
is locally injective. We consider
does not vanish on U, so that Φ x 0
t x (x 0 ,y,z) of Φ x 0
the level-set f(x)= f 0 for some f 0 ∈ f(U). It cannot contain a closed loop in U,
otherwise we would have (∂ y f, ∂ z f)=0 (and thus t x =0)in U ⊂ D. We deduce
that each connected component of f(x)= f 0 in U intersects ∂U in two points.
is not injective on U, so that we have two points p 1 ,p 2 ∈
Now suppose that Φ x 0
(p 2 ).
U such that Φ x 0
(p 1 )= Φ x 0