Equations of parametric surfaces with base points via syzygies

Abstract

Let S be a parametrized surface in P 3 given as the image of φ : P 1 × P 1 → P 3 . This paper will show that the use of syzygies in the form of a combination of moving planes and moving quadrics provides a valid method for finding the implicit equation of S when certain base points are present. This work extends the algorithm provided by Cox [5] for when φ has no base points, and it is analogous to some of the results of Busé, Cox, and D'Andrea [2] for the case when φ : P 2 → P 3 has base points.

Matrix representations for toric parametrizations

“…T = P 2 , and [BCD03] does the same in the presence of base points. In [AHW05], square matrix representations of bihomogeneous parametrizations, i.e. T = P 1 × P 1 , are constructed with linear and quadratic syzygies, whereas [KD06] gives such a construction for parametrizations over toric varieties of dimension 2.…”

Implicitization of bihomogeneous parametrizations of algebraic surfaces via linear syzygies

“…This can be summarized by the following commutative diagram, which is the algebraic translation of the diagram (1).…”

“…In [1] a determinantal representation of the implicit equation of a bi-homogeneous parametrization is constructed with linear and quadratic relations, whereas [12] gives such a construction in the toric case.…”

Real equivalence of complex matrix pencils and complex projections of real Segre varieties

“…The linear transformation C ( +1)(m+1) → C n+1 (also denoted by P) induces a projection map P : CP m+ +m → CP n , which is well-defined for all elements in the domain except those lines in the kernel of P. It will be convenient to follow some of the recent literature on implicitization (e.g., [1,3,9]), and borrow some terminology from the classical (and not unrelated; see [15,§III.1]) theory of linear systems of quadrics. Definition 2.1.…”

“…Since the rank is invariant under real equivalence, we only need to inspect the list of canonical forms to find pairs (E, F) so that E + iF is a rank 1 complex matrix. Clearly, the proposition's cases 8, 9, and 12 span complex pencils which fall into the theorem's case (1). The proposition's cases 7 and 9 correspond to the theorem's cases (2) and (3), respectively.…”