A hands-on approach to tensor product surfaces
Alexandra Seceleanu
joint with H. Schenck, J. Validashti
April 1st 2012
AMS Sectional Meeting, Lawrence KS
Surface modeling
Spline surfaces closely following a control grid embedded in R3 are
standard tools of today’s CAD systems.
Figure: A surface following a control point grid
Tensor product surface
Definition
A tensor product surface of bidegree (d1 , d2 ) is a piecewise
polynomial parametric surface surface that is built by connecting
several polynomial patches in a smooth manner:
f : [0, 1] × [0, 1] −→ A3
(u, s) 7→
t
X
Pij Bi (u)Bj0 (s)
i,j=1
The control points Pij ∈ A3 define the control grid of the spline
surface. The polynomials Bi (u), Bj0 (s) are called blending
functions.
Tensor product surface - homogeneous version
I
Instead of thinking about tensor product surfaces as affine
maps:
I
we homogenize to get a regular map defined by four
polynomials with no common zeros on P1 × P1 :
φ : P1 × P1 −→ P3
((s, t), (u, v )) 7→ (p0 (s, t, u, v ) : p1 (s, t, u, v ) : p2 (s, t, u, v ) : p3 (s, t, u, v ))
Main problems in geometric modeling
1. Implicitize: We assume the parametrization φ known but the
implicit equation F of the surface not known!
2. Find singular locus: In general one does not want singular
points in the interior of the patch, but singular points on the
surface decrease the degree of the implicit equation.
Main idea:
I
study these problem using the information contained in the
resolution of the ideal I = (p0 : p1 : p2 : p3 ).
Example
φ : P1 × P1 −→ P3 , ((s, t), (u, v )) 7→ (s 2 u, s 2 v , t 2 u, t 2 v + stv )
I = (s 2 u, s 2 v , t 2 u, t 2 v + stv )
I
the bigraded resolution
R(−2, −2)
⊕
0 ← I ← R(−2, −1)4 ← R(−3, −2)2 ← R(−4, −2)2 ← 0
⊕
R(−4, −1)2
I
the implicit equation:
X = V(x0 x12 x2 − x12 x22 + 2x0 x1 x2 x3 − x02 x32 ).
I
the reduced codimension one singular locus of X is:
V(x0 , x2 ) ∪ V(x1 , x3 ) ∪ V(x0 , x1 ).
The surface in the example
The Segre-Veronese variety
From now on I = four-generated bigraded ideal of bidegree (2,1).
Main idea:
I
study the resolution of I by inspecting how the plane
Span(p0 , p1 , p2 , p3 ) ⊂ Span(s 2 u, s 2 v , stu, stv , t 2 u, t 2 v ) = P5
meets the image Σ2,1 of the Segre map
P(H 0 (OP1 (2))) × P(H 0 (OP1 (1)))−→P(H 0 (OP1 ×P1 (2, 1)))
The Segre-Veronese variety
Main idea:
I
study the resolution of I by inspecting how the plane
P(U) = Span(p0 , p1 , p2 , p3 ) meets the image Σ2,1 of the
Segre map
σ2,1
P2 × P1 −→ P5 .
Recall at each point of Σ2,1 there are 2 types of fibers:
I
lines P1
I
planes P2
Linear syzygies and lines on Σ2,1
P(U) = Span(p0 , p1 , p2 , p3 ).
Proposition
The ideal I = (p0 , p1 , p2 , p3 )
1. has a unique linear syzygy of bidegree (0, 1) iff
F ⊆ P(U) ∩ Σ2,1 , where F is a P1 fiber of Σ2,1 .
2. has a pair of linear syzygies of bidegree (0, 1) iff
P(U) ∩ Σ2,1 = Σ1,1 .
3. has a unique linear syzygy of bidegree (1, 0) iff F ⊆ P(U) ∩ Q,
where F is a P1 fiber of Q.
Degree (0,2) syzygies and quadrics on Σ2,1
Proposition
There is a minimal first syzygy on I of bidegree (0, 2) iff there
exists P(W ) ' P2 ⊆ P(U) such that P(W ) ∩ Σ2,1 is a smooth
conic.
Degree (0,2) syzygies and quadrics on Σ2,1
Proposition
There is a minimal first syzygy on I of bidegree (0, 2) iff there
exists P(W ) ' P2 ⊆ P(U) such that P(W ) ∩ Σ2,1 is a smooth
conic.
Proposition
Only minimal first syzygies of bidegrees (1, 0) and (0, 2) are
compatible.
Main result
Theorem
There are exactly 6 resolutions for ideals generated by four
bidegree (2,1) forms without basepoints :
The six Betti types
Type
6
5
4
Bigraded Minimal Free Resolution of I
R(−2, −2)2
⊕
0 ← I ← R(−2, −1)4 ←
← R(−4, −2) ← 0
R(−4, −1)2
R(−2, −2)
⊕
0 ← I ← R(−2, −1)4 ← R(−3, −2)2 ← R(−4, −2)2 ← 0
⊕
R(−4, −1)2
R(−2, −3)
⊕
R(−3, −1)
R(−3, −3)
⊕
⊕
0 ← I ← R(−2, −1)4 ← R(−3, −2)2 ← R(−4, −3) ← R(−5, −3) ← 0
⊕
⊕
R(−4, −2)
R(−5, −2)2
⊕
R(−5, −1)
The six Betti types - continued
Type
3
2
1
Bigraded Minimal Free Resolution of I
R(−2, −4)
⊕
R(−3, −1)
⊕
R(−3, −4)2
2
⊕
R(−4, −4)
R(−3, −2)
⊕
⊕
0 ← I ← R(−2, −1)4 ←
← R(−4, −3)2 ←
←0
R(−5, −3)
R(−3, −3)
⊕
⊕
R(−5, −2)2
R(−4, −2)
⊕
R(−5, −1)
R(−2, −3)
⊕
R(−3, −3)2
⊕
0 ← I ← R(−2, −1)4 ← R(−3, −2)4 ←
← R(−4, −3) ← 0
⊕
R(−4, −2)3
R(−4, −1)2
R(−2, −4)
⊕
R(−3, −4)2
⊕
0 ← I ← R(−2, −1)4 ← R(−3, −2)4 ←
← R(−4, −4) ← 0
⊕
R(−4, −2)3
R(−4, −1)2
Table: The six Betti types for bidegree (2,1) four-generated ideals
Relations between syzygies and singularity types
Type
1
2
3
4
5a
5b
6
Lin. Syz. Emb. Pri. Sing. Loc.
Example
none
m
T
(s 2 u+stv , t 2 u, s 2 v +stu, t 2 v +stv)
none
m, P1
C ∪ L1
(s 2 u, t 2 u, s 2 v + stu, t 2 v + stv )
1 type (1, 0)
m
L1
(s 2 u + stv , t 2 u, s 2 v , t 2 v + stu)
1 type (1, 0) m, P1
L1
(stv , t 2 v , s 2 v − t 2 u, s 2 u)
1 type (0, 1) P1 , P2 L1 ∪ L2 ∪ L3
(s 2 u, s 2 v , t 2 u, t 2 v + stv )
1 type (0, 1)
P1
L1 ∪ L2
(s 2 u, s 2 v , t 2 u, t 2 v + stu)
2 type (0, 1) none
∅
(s 2 u, s 2 v , t 2 u, t 2 v )
Table: The primary decomposition and singularities for the six Betti types
I
T = twisted cubic curve, C = smooth plane conic Li = lines
I
m = hs, t, u, v i, Pi = hli , s, ti, li = linear form of bidegree (0, 1)