1 Introduction↩︎

plus 2.0pt minus 2.0pt plus 2.0pt minus 2.0pt Implicitization of parametric equations of curves and surfaces is an important problem of long history in computer graphics and computer-aided geometric design. Several tools that are used to implicitize parametric curves and surfaces are resultant, Gröbner basis and Wu’s method. In 1995, Sederberg and Chen proposed a new implicitization method, the method of moving curves and moving surfaces [1]. On this basis, Cox et al. first introduced the concept of \(\mu\)-basis, which provides a new approach to computing the implicit equation of a planar rational parametric curve [2]. Due to its algebraic and geometric properties, \(\mu\)-basis has become a bridge connecting the parametric representation and the implicit representation of curves and surfaces. By using \(\mu\)-basis, new methods have been developed for studying geometric properties of curves, including point inversion, reparametrization and singularity computation [3]. Subsequent studies have extended the concept of \(\mu\)-bases to rational ruled surfaces, tensor-product rational surfaces, and various types of special surfaces [4]–[11].

Meanwhile, research into methods for computing \(\mu\)-bases has continued to advance. In 2003, Chen and Wang applied vector elimination to the module of moving lines of curves and proposed an efficient algorithm for computing \(\mu\)-bases of plane rational parametric curves [12]. Song and Goldman generalized this method to compute \(\mu\)-bases of rational parametric curves in arbitrary dimensions [13].Subsequently, Hong et al. proposed another efficient algorithm based on the echelon form computation of numerical matrices [14]. Huang and Chen extended the notion of \(\mu\)-bases to univariate polynomial matrices and presented an algorithm for computing \(\mu\)-bases via polynomial matrix factorization [15]. Wang observed that these algorithms do not focus on the explicit forms of \(\mu\)-bases, and in [16] gave explicit formulas for the \(\mu\)-bases of plane rational quadratic and cubic curves.

Motivated by this, with the aid of the vector polynomials \(\mathbf{Q}_i\) defined by Goldman et al. in [17], we establish the relationship between \(\mathbf{Q}_i \times \mathbf{F}(t)\) and the explicit formulas for \(\mu\)-bases. In Section 3, we present explicit formulas for the \(\mu\)-bases of plane rational quartic curves in the form of theorems. When the degree is two or three, the explicit formulas for the \(\mu\)-bases coincide with the results of Wang [16].

2 Basic knowledge↩︎

2.1 \(\mu\)-basis↩︎

In the real plane \(\mathbb{R}^2\), a rational parametric curve is defined by \[(x,y)=\left(\frac{f_1(t)}{f_0(t)},\frac{f_2(t)}{f_0(t)}\right),\] where \(f_0,f_1,f_2\) are polynomials with real coefficients and \(\mathrm{gcd}(f_0,f_1,f_2)=1\). For convenience, we often consider its homogeneous form in the projective plane \(\mathbb{PR}^2\): \[\label{eq:2461} \mathbf{F}(t)=(f_0(t),f_1(t),f_2(t)).\tag{1}\]

The degree of \(\mathbf{F}(t)\) is defined as \[\mathrm{deg}(\mathbf{F}(t))\triangleq \mathrm{max}\bigl(\mathrm{deg}(f_0),\mathrm{deg}(f_1),\mathrm{deg}(f_2)\bigr).\]

A moving line \(L(x,y,w;t)=0\) in \(\mathbb{PR}^2\) is a family of lines corresponding to the parameter \(t\): \[\label{eq:2462} L(x,y,w;t)=A(t)x+B(t)y+C(t)w=0,\tag{2}\] where \(A(t),B(t),C(t)\in\mathbb{R}[t]\) and \[\mathrm{deg}(L(x,y,w;t))= \mathrm{max}\bigl(\mathrm{deg}(A(t)),\mathrm{deg}(B(t)),\mathrm{deg}(C(t))\bigr).\] For convenience, the moving line in 2 is denoted by \(\mathbf{L}(t)=(A(t),B(t),C(t))\). If \[\label{eq:2463} \mathbf{L}(t)\cdot\mathbf{F}(t) =A(t)f_0(t)+B(t)f_1(t)+C(t)f_2(t)\equiv0,\tag{3}\] then \(\mathbf{L}(t)\) is said to follow \(\mathbf{F}(t)\). From an algebraic point of view, \(\mathbf{L}(t)\) is a syzygy of \(\mathbf{F}(t)\). We denote by \(\mathbf{L}\) the set of all moving lines following \(\mathbf{F}(t)\). Then \(\mathbf{L}\) is the syzygy module of \(\mathbf{F}(t)\). The \(\mu\)-basis of a plane rational curve is defined as follows [16].

2.2 Vector Polynomials \(\mathbf{Q}_k(t)\)↩︎

First, we write \(\eqref{eq:2461}\) in the form of a vector polynomial, namely \(\mathbf{F}(t)=\displaystyle\sum_{0}^{n}\boldsymbol{p}_i t^i\), where \(\boldsymbol{p}_i\in\mathbb{R}^3\) and \(\boldsymbol{p}_0,\boldsymbol{p}_n\ne \boldsymbol{0}.\)

For \(k=0,1,\cdots,n-1\), \(\mathbf{F}(t)\) can be decomposed as follows: \[\mathbf{F}(t)=t^{n-k}(\mathbf{p}_nt^k+\cdots+\mathbf{p}_{n-k})+(\mathbf{p}_{n-k-1}t^{n-k-1}+\cdots+\mathbf{p}_0),\] Let \(\mathbf{Q}_k(t)=\mathbf{p}_nt^k+\cdots+\mathbf{p}_{n-k}\). By the properties of the cross product, we have \[\mathbf{Q}_k(t)\times\mathbf{F}(t)=(\mathbf{p}_n\times\mathbf{p}_{n-k-1})t^{n-1}+\text{lower-degree terms},\] and \(\left(\mathbf{Q}_k\times\mathbf{F}(t)\right)\cdot\mathbf{F}(t)=0\), which shows that \(\mathbf{Q}_0\times\mathbf{F}(t)\)，\(\mathbf{Q}_1\times\mathbf{F}(t)\)， \(\cdots\)，\(\mathbf{Q}_{n-1}\times\mathbf{F}(t)\)are \(n\) moving lines of degree \(n-1\) following \(\mathbf{F}(t)\).

In order to find a \(\mu\)-basis of the curve, we consider \(\mathbf{Q}_0\times\mathbf{F}(t)\) and \(\mathbf{Q}_{n-1}\times\mathbf{F}(t)\).

Using the vector cross product identity \((\mathbf{A}\times\mathbf{B})\times(\mathbf{C}\times\mathbf{D})=\mathbf{C}((\mathbf{A}\times\mathbf{B})\cdot\mathbf{D})-\mathbf{D}((\mathbf{A}\times\mathbf{B})\cdot\mathbf{C})\), we have \[\label{eq:2464} \bigl(\mathbf{Q}_{n-1}\times\mathbf{F}(t)\bigr)\times\bigl(\mathbf{Q}_0\times\mathbf{F}(t)\bigr)=\left[\frac{(\mathbf{p}_n\times\mathbf{p}_0)\cdot\mathbf{F}(t)}{t}\right]\mathbf{F}(t).\tag{4}\]

) When \(\mathbf{p}_n\times\mathbf{p}_0=\mathbf{0}\), \(\mathbf{Q}_{n-1}\times\mathbf{F}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\) are linearly dependent over \(\mathbb{R}[t]\).
) When \(\mathbf{p}_n\times\mathbf{p}_0\ne\mathbf{0}\), \(\mathbf{Q}_{n-1}\times\mathbf{F}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\) are linearly independent over \(\mathbb{R}[t]\), where \(\displaystyle\frac{(\mathbf{p}_n\times\mathbf{p}_0)\cdot\mathbf{F}(t)}{t}\) is a scalar polynomial of degree \(n-2\), and it can be rewritten as \[\frac{(\mathbf{p}_n\times\mathbf{p}_0)\cdot\mathbf{F}(t)}{t}=-\frac{(\mathbf{F}(t)\times\mathbf{p}_0)\cdot\mathbf{p}_n}{t}=-\left(\mathbf{Q}_{n-1}\times\mathbf{F}(t)\right)\cdot \mathbf{p}_n.\] If \(\mathbf{Q}_{n-1}\times\mathbf{F}(t)\) can be factorized as \[\mathbf{Q}_{n-1}\times\mathbf{F}(t)=-\left[\frac{(\mathbf{p}_n\times\mathbf{p}_0)\cdot\mathbf{F}(t)}{t}\right]\cdot\mathbf{g}(t),\] where \(\mathbf{g}(t)\) is a vector polynomial of degree one satisfying \(\mathbf{g}(t)\cdot\mathbf{p}_n=1\), then \(\mathbf{g}(t)\times\left(\mathbf{Q}_0\times\mathbf{F}(t)\right)=\mathbf{F}(t)\) and \(\mathrm{deg}(\mathbf{g}(t))+\mathrm{deg}\left(\mathbf{Q}_0\times\mathbf{F}(t)\right)=\mathrm{deg}(\mathbf{F}(t))\), which shows that \(\mathbf{g}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\) form a \(\mu\)-basis of the curve \(\mathbf{F}(t)\).

3 Explicit Formulas for the \(\mu\)-Bases of Quartic Curves↩︎

For a quartic non-degenerate curve \(\mathbf{F}(t)=\mathbf{p}_0+\mathbf{p}_1t+\mathbf{p}_2t^2+\mathbf{p}_3t^3+\mathbf{p}_4t^4\), where \(\mathbf{p}_i\in\mathbb{R}^3,i=0,1,2,3,4\), \(\mathbf{p}_0\ne\mathbf{0}\) and \(\mathbf{p}_4\ne\mathbf{0}\). In what follows, \(|\mathbf{p}_i\mathbf{p}_j\mathbf{p}_k|\) denotes the determinant formed by the vectors \(\mathbf{p}_i^\mathrm{T}\), \(\mathbf{p}_j^\mathrm{T}\), and \({\mathbf{p}}_k^\mathrm{T}\).

As stated in Section 2.2, let \(\mathbf{Q}_0=\mathbf{p}_4,\mathbf{Q}_1=\mathbf{p}_3+\mathbf{p}_4t,\mathbf{Q}_2=\mathbf{p}_2+\mathbf{p}_3t+\mathbf{p}_4t^2,\mathbf{Q}_3=\mathbf{p}_1+\mathbf{p}_2t+\mathbf{p}_3t^2+\mathbf{p}_4t^3\), then we have: \[\begin{align} \mathbf{Q}_0\times\mathbf{F}(t)&=&(\mathbf{p}_4\times \mathbf{p}_0)+(\mathbf{p}_4\times\mathbf{p}_1)t+(\mathbf{p}_4\times\mathbf{p}_2)t^2+(\mathbf{p}_4\times\mathbf{p}_3)t^3,\\ \mathbf{Q}_1\times\mathbf{F}(t)&=&(\mathbf{p}_3\times\mathbf{p}_0)+\big[(\mathbf{p}_3\times\mathbf{p}_1)+(\mathbf{p}_4\times\mathbf{p}_0)\big]t+\big[(\mathbf{p}_3\times\mathbf{p}_2)+(\mathbf{p}_4\times\mathbf{p}_1)\big]t^2+(\mathbf{p}_4\times\mathbf{p}_2)t^3,\\ \mathbf{Q}_2\times\mathbf{F}(t)&=&(\mathbf{p}_2\times\mathbf{p}_0)+\big[(\mathbf{p}_2\times\mathbf{p}_1)+(\mathbf{p}_3\times\mathbf{p}_0)\big]t+\big[(\mathbf{p}_3\times\mathbf{p}_1)+(\mathbf{p}_4\times\mathbf{p}_0)\big]t^2+(\mathbf{p}_4\times\mathbf{p}_1)t^3,\\ \mathbf{Q}_3\times\mathbf{F}(t)&=&(\mathbf{p}_1\times\mathbf{p}_0)+(\mathbf{p}_2\times\mathbf{p}_0)t+(\mathbf{p}_3\times \mathbf{p}_0)t^2+(\mathbf{p}_4\times\mathbf{p}_0)t^3. \end{align}\] In the following, we classify the discussion according to whether \(\mathbf{p}_4\times\mathbf{p}_0\) is the zero vector.

3.1 The Case \(\mathbf{p}_4\times\mathbf{p}_0=\boldsymbol{0}\)↩︎

Since in this case \(\mathbf{Q}_3\times\mathbf{F}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\) are linearly dependent over \(\mathbb{R}[t]\), we consider the simplified \(\displaystyle\frac{\mathbf{Q}_0\times\mathbf{F}(t)}{t}\) and \(\mathbf{Q}_1\times\mathbf{F}(t)\), and have \[\displaystyle\frac{\mathbf{Q}_0\times\mathbf{F}(t)}{t}=(\mathbf{p}_4\times\mathbf{p}_1)+(\mathbf{p}_4\times\mathbf{p}_2)t+(\mathbf{p}_4\times\mathbf{p}_3)t^2,\] \[\displaystyle\frac{\mathbf{Q}_0\times\mathbf{F}(t)}{t}\times(\mathbf{Q}_1\times\mathbf{F}(t))=-(|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|+|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|t)\cdot\mathbf{F}(t).\] and when \(|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|^2-|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|\cdot|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|=0\), \(\displaystyle\frac{\mathbf{Q}_0\times\mathbf{F}(t)}{t}\) can be decomposed as follows \[\displaystyle\frac{\mathbf{Q}_0\times\mathbf{F}(t)}{t}=\left(|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|+|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|t\right)\left(\displaystyle\frac{\mathbf{p}_4\times\mathbf{p}_1}{|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|}+\displaystyle\frac{\mathbf{p}_4\times\mathbf{p}_3}{|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|}t\right).\] Therefore, let \(\mathbf{L}(t)=\displaystyle\frac{\mathbf{p}_4\times\mathbf{p}_1}{|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|}+\displaystyle\frac{\mathbf{p}_4\times\mathbf{p}_3}{|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|}t\), then we have \(\mathbf{L}(t)\times(\mathbf{Q}_1\times\mathbf{F}(t))=-\mathbf{F}(t)\), and the degrees of \(\mathbf{L}(t)\) and \(\mathbf{Q}_1\times\mathbf{F}(t)\) are 1 and 3, respectively.

When \(|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|^2-|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|\cdot|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|\ne0\), consider the vector identity in \(\mathbb{R}^3\) \[|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|\cdot\mathbf{p}_1-|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|\cdot\mathbf{p}_2+|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|\cdot\mathbf{p}_3-|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_3|\cdot\mathbf{p}_4=\mathbf{0},\] taking the cross product with \(\mathbf{p}_4\) on both sides, we obtain \[\label{eq:3461} |\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|\cdot(\mathbf{p}_4\times\mathbf{p}_1)-|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|\cdot(\mathbf{p}_4\times\mathbf{p}_2)+|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|\cdot(\mathbf{p}_4\times\mathbf{p}_3)=\mathbf{0}.\tag{5}\] Using the above formula, we can construct a polynomial of degree 2 \[\label{eq:3462} \mathbf{L}(t)=|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|\cdot(\mathbf{Q}_2\times\mathbf{F}(t))-|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|\cdot(\mathbf{Q}_1\times\mathbf{F}(t))+|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|\cdot(\mathbf{Q}_0\times\mathbf{F}(t))\tag{6}\] It is easy to see that \(\mathbf{L}(t)\cdot\mathbf{F}(t)\equiv0\), and \[\begin{align} \displaystyle\frac{\mathbf{Q}_0\times\mathbf{F}(t)}{t}\times\mathbf{L}(t)&=&\displaystyle\frac{\mathbf{L}(t)\cdot\mathbf{p}_4}{t}\mathbf{F}(t)\\ &=&(|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|^2-|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|\cdot|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|)\mathbf{F}(t), \end{align}\] At this point, \(\displaystyle\frac{\mathbf{Q}_0\times\mathbf{F}(t)}{t}\) and \(\mathbf{L}(t)\) form a \(\mu\)-basis of the curve \(\mathbf{F}(t)\). In summary, we have

This curve satisfies \(\mathbf{p}_4 \times\mathbf{p}_0 = \boldsymbol{0}\), \(|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|^2-|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|\cdot|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|=8\ne0\), and further computation gives \[\begin{align} \square&=&|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|\left(\mathbf{p}_2\times\mathbf{p}_0\right)-|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|\left(\mathbf{p}_3\times\mathbf{p}_0\right)=(0,-4,2),\\ \triangle&=&||\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|\left[\left(\mathbf{p}_2\times\mathbf{p}_1\right)+\left(\mathbf{p}_3\times\mathbf{p}_0\right)\right]-|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|\left(\mathbf{p}_3\times\mathbf{p}_1\right)+|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|\left(\mathbf{p}_4\times\mathbf{p}_1\right)=(4,-2,4),\\ \star&=&|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|\left(\mathbf{p}_3\times\mathbf{p}_1\right)-|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|\left[\left(\mathbf{p}_3\times\mathbf{p}_2\right)+\left(\mathbf{p}_4\times\mathbf{p}_1\right)\right]+|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|\left(\mathbf{p}_4\times\mathbf{p}_2\right)=(0,-2,-6). \end{align}\] Therefore, we can directly obtain \[\displaystyle\mathbf{u}(t) =\displaystyle\frac{\mathbf{Q}_0 \times \mathbf{F}(t)}{t} = (0,-2t^2-2t,2t^2+2),\] \[\mathbf{L}(t)=\square+\triangle t+\star t^2=(4t,-2t^2-2t-4,-6t^2+4t+2).\] Verification gives \(\mathbf{u}(t) \times \mathbf{L}(t) = 8F(t)\), so \(\mathbf{u}(t)\) and \(\mathbf{L}(t)\) form a \(\mu\)-basis of the curve.

Furthermore, the implicit equation of the curve \(\mathbf{F}(t)\) is \[\begin{align} F(x,y,w) &= \det\begin{pmatrix} -4y+2w & 0 & 2w & 0 \\ 4x-2y+4w & -4y+2w & -2y & 2w\\ -2y-6w & 4x-2y+4w & -2y+2w &-2y\\ 0 & -2y-6w & 0 & -2y+2w \end{pmatrix} \\ &= \begin{aligned}[t] &320w^4+128w^3x-448w^3y+64w^2x^2-256w^2xy+256w^2y^2 \\ &-64wx^2y-64wxy^2+64wy^3+64xy^3+64y^4=0. \end{aligned} \end{align}\]

3.2 The Case \(\mathbf{p}_4\times\mathbf{p}_0\ne\boldsymbol{0}\)↩︎

Similarly to formula \(\eqref{eq:3461}\), we can obtain: \[\label{eq:3463} |\mathbf{p}_b\mathbf{p}_c\mathbf{p}_4|\cdot(\mathbf{p}_4\times\mathbf{p}_a)-|\mathbf{p}_a\mathbf{p}_c\mathbf{p}_4|\cdot(\mathbf{p}_4\times\mathbf{p}_b)+|\mathbf{p}_a\mathbf{p}_b\mathbf{p}_4|\cdot(\mathbf{p}_4\times\mathbf{p}_c)=\mathbf{0}.\tag{7}\]

Using the above formula, we obtain four quadratic moving lines following \(\mathbf{F}(t)\), \[\begin{align} \mathbf{u}_0&=&|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|(\mathbf{Q}_0\times\mathbf{F}(t))-|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|(\mathbf{Q}_1\times\mathbf{F}(t))+|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|(\mathbf{Q}_2\times\mathbf{F}(t)),\\ \mathbf{u}_1&=&|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_4|(\mathbf{Q}_0\times\mathbf{F}(t))-|\mathbf{p}_0\mathbf{p}_3\mathbf{p}_4|(\mathbf{Q}_1\times\mathbf{F}(t))+|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|(\mathbf{Q}_3\times\mathbf{F}(t)),\\ \mathbf{u}_2&=&|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_4|(\mathbf{Q}_0\times\mathbf{F}(t))-|\mathbf{p}_0\mathbf{p}_3\mathbf{p}_4|(\mathbf{Q}_2\times\mathbf{F}(t))+|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|(\mathbf{Q}_3\times\mathbf{F}(t)),\\ \mathbf{u}_3&=&|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_4|(\mathbf{Q}_1\times\mathbf{F}(t))-|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_4|(\mathbf{Q}_2\times\mathbf{F}(t))+|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|(\mathbf{Q}_3\times\mathbf{F}(t)). \end{align}\] Taking the cross product of any \(\mathbf{u}_i,\mathbf{u}_j,(i,j=0,1,2,3)\), and simplifying by using the Grassmann–Plücker relations, we obtain: \[\label{eq:3464} \mathbf{u}_i\times\mathbf{u}_j=|\mathbf{p}_a\mathbf{p}_b\mathbf{p}_4|\cdot E\cdot\mathbf{F}(t),\tag{8}\] where \(0\leq a<b\leq3\) is a pair of indices distinct from \(i\) and \(j\), and \[\begin{align} E&=&|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_4|^2-(|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|+|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_4|)\cdot|\mathbf{p}_0\mathbf{p}_3\mathbf{p}_4|\\ &&+\,|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|\cdot|\mathbf{p}_1\mathbf{p}_3\mathbf{p}_4|-|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_4|\cdot|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_4|-|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_2|\cdot|\mathbf{p}_2\mathbf{p}_3\mathbf{p}_4|. \end{align}\]

The non-degeneracy of the curve guarantees that there exists some \(|\mathbf{p}_a\mathbf{p}_b\mathbf{p}_4|\ne0\), and hence, when \(E\ne0\), according to the definition of a \(\mu\)-basis, \(\mathbf{u}_i\) and \(\mathbf{u}_j\) form a \(\mu\)-basis of the curve \(\mathbf{F}(t)\).

In order to consider the case when \(E=0\), we first give the following lemma:

First, since \[\begin{align} \mathbf{u}_i(t)\cdot\mathbf{F}(t)&=&\mathbf{u}_i(t)\mathbf{F}(t)^T=\mathbf{u'}_i(t)S^{-T}\mathbf{F}(t)^T=\mathbf{u'}_i(t)(\mathbf{F}(t)S^{-1})^T\\ &=&\mathbf{u'}_i(t)\mathbf{F'}(t)^T=\mathbf{u'}_i(t)\cdot\mathbf{F'}(t)=0, \end{align}\] so we have that \(\mathbf{u}_1(t)\) and \(\mathbf{u}_2(t)\) follow \(\mathbf{F}(t)\).

Since a linear transformation does not change the degree of a polynomial, it remains only to prove that \(\mathbf{u}_1(t)\times\mathbf{u}_2(t)=\lambda\mathbf{F}(t)\).

Consider the cross product extended to the polynomial vector space \(\mathbb{R}[t]^3\). For any \(\mathbf{h}_1(t),\mathbf{h}_2(t)\in\mathbb{R}[t]^3\) and any invertible matrix \(B\in GL_3(\mathbb{R})\), we have \[\label{eq:3465} (\mathbf{h}_1(t)B)\times(\mathbf{h}_2(t)B) = \det(B)(\mathbf{h}_1(t)\times\mathbf{h}_2(t))B^{-T}.\tag{9}\]

Using \(\eqref{eq:3465}\) and \(\mathbf{u'}_1(t)\times\mathbf{u'}_2(t)=\lambda'\mathbf{F'}(t)\)，\(\lambda'\in\mathbb{R}^*\)，we obtain \[\begin{align} \mathbf{u}_1(t)\times\mathbf{u}_2(t)&=&(\mathbf{u'}_1(t)S^{-T})\times(\mathbf{u'}_2(t)S^{-T})=\det(S^{-T})(\mathbf{u'}_1(t)\times\mathbf{u'}_2(t))(S^{-T})^{-T}\\ &=&\frac{1}{\det(S)}\lambda'\mathbf{F'}(t)S=\frac{1}{\det(S)}\lambda'\mathbf{F}(t)S^{-1}S=\lambda\mathbf{F}(t),\;\text{where}\;\lambda=\frac{1}{\det(S)}\lambda'. \end{align}\] This completes the proof.

Thus, when \(E=0\), without loss of generality, assume that \(|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_4|\ne0\). Let the invertible matrix \(S=(\mathbf{p}_0^T, \mathbf{p}_1^T, \mathbf{p}_4^T)^T\) and \(\mathbf{F'}(t)=\mathbf{F}(t)S^{-1}\). Then among \(\mathbf{p'}_i=\mathbf{p}_iS^{-1}(i=0,1,2,3,4)\), \(\mathbf{p'}_0\), \(\mathbf{p'}_1\), and \(\mathbf{p'}_4\) form a standard basis of \(\mathbb{R}^3\).

Suppose \(\mathbf{p'}_2=(a,b,c)\)、\(\mathbf{p'}_3=(d,e,f)\)，then \[\mathbf{F'}(t)=(1+at^2+dt^3,t+bt^2+et^3,ct^2+ft^3+t^4).\] Substituting \(\mathbf{p'}_i\) into \(E\), we obtain. \[E=b^2-(bf-ce+1)e-df+a-c(ae-bd).\] Let the degree-one moving line following \(\mathbf{F'}(t)\) be \(\mathbf{L}(t)=\mathbf{A}+\mathbf{B}t=(x_0,y_0,z_0)+(x_1,y_1,z_1)t\), then we have \[\mathbf{L}(t)\cdot\mathbf{F'}(t)\equiv0 \iff \left\{ \begin{array}{l} \left. \begin{array}{l} x_0 = 0 \\ z_1 = 0 \\ y_0 = -x_1 \\ y_1 = b x_1 - c z_0 \end{array} \right\} \implies \mathbf{L}(t) = \left( x_1 t,\;-x_1 + (b x_1 - c z_0) t,\;z_0 \right) \\[15pt] \begin{bmatrix} a + b^2 - e & f - bc \\ d + eb & 1 - ce \end{bmatrix} \begin{bmatrix} x_1 \\ z_0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \end{array} \right.\] then \(\det \begin{bmatrix} a + b^2 - e & f - bc \\ d + eb & 1 - ce \end{bmatrix}=E\)，show: \[\small E=0 \Longleftrightarrow (x_1,z_0)\ne(0,0) \Longleftrightarrow \exists\;\mathbf{L}(t)\text{ of degree one following }\mathbf{F'}(t) \Longleftrightarrow Syz(\mathbf{F'})=R(-1)\oplus R(-3).\]

Using \(\mathbf{L}(t)\cdot\mathbf{F'}(t)=(\mathbf{A}+\mathbf{B}t)\cdot(\mathbf{p}_0+\mathbf{p}_1t+\mathbf{p}_2t^2+\mathbf{p}_3t^3+\mathbf{p}_4t^4)=0\)，it can be simplified \[\begin{align} (\mathbf{Q}_0\times\mathbf{F'}(t))\times\mathbf{L}(t)&=&(\mathbf{p}_{4}\cdot\mathbf{A})\cdot\mathbf{F'}(t)=z_0\cdot\mathbf{F'}(t),\\ (\mathbf{Q}_1\times\mathbf{F'}(t))\times\mathbf{L}(t)&=&(\mathbf{p}_{3}\cdot\mathbf{A})\cdot\mathbf{F'}(t)=(fz_0-ex_1)\cdot\mathbf{F'}(t),\\ (\mathbf{Q}_2\times\mathbf{F'}(t))\times\mathbf{L}(t)&=&(\mathbf{p}_{2}\cdot\mathbf{A})\cdot\mathbf{F'}(t)=(cz_0-bx_1)\cdot\mathbf{F'}(t),\\ (\mathbf{Q}_3\times\mathbf{F'}(t))\times\mathbf{L}(t)&=&(\mathbf{p}_{1}\cdot\mathbf{A})\cdot\mathbf{F'}(t)=-x_1\cdot\mathbf{F'}(t). \end{align}\] Since \((x_1,z_0)\ne(0,0)\), if \((\mathbf{Q}_0\times\mathbf{F'}(t))\times\mathbf{L}(t)=\mathbf{0}\), then \((\mathbf{Q}_3\times\mathbf{F'}(t))\times\mathbf{L}(t)\ne\mathbf{0}\), and vice versa. Thus, when \(E=0\), the curve \(\mathbf{F}^{'}(t)\) has a \(\mu\)-basis whose degrees are 1 and 3, respectively. According to Lemma 2, the original curve \(\mathbf{F}(t)\) also has such a \(\mu\)-basis. In summary, we have

When \(E=0\), all nonzero \(\mathbf{u}_i(t)\) are pairwise parallel in \(\mathbb{R}^3[t]\). Take \[\mathbf{L}(t)=(x_0+x_1t,y_0+y_1t,z_0+z_1t)=\gcd(\mathbf{u}_i)\] as a degree-one moving line. If \(z_0=0\), then the cubic moving line is \(\mathbf{Q}_3\times\mathbf{F}(t)\); if \(x_1=0\), then the cubic moving line is \(\mathbf{Q}_0\times\mathbf{F}(t)\).

Therefore, when a non-degenerate curve is given, we can obtain a \(\mu\)-basis of the curve according to the following procedure.

This curve satisfies \(\mathbf{p}_4 \times\mathbf{p}_0 \ne \boldsymbol{0}\) and \(E=0\). Computing \(\mathbf{u}_i\) gives: \[\begin{align} \mathbf{u}_0&=&(0,t^2,-t)=t(0,t,-1),\\ \mathbf{u}_1&=&(0,-t,1)=-(0,t,-1),\\ \mathbf{u}_2&=&(0,-t^2-t,t+1)=-(t+1)(0,t,-1),\\ \mathbf{u}_3&=&(0,-t^2-t,t+1)=-(t+1)(0,t,-1). \end{align}\] Let \(\mathbf{L}(t)=(0,t,-1)\). Computing \(\mathbf{Q}_i\times\mathbf{F}(t)\) gives: \[\begin{align} \mathbf{Q}_0\times\mathbf{F}(t)&=&(-t^2-t^3,1+t+t^2,0),\\ \mathbf{Q}_1\times\mathbf{F}(t)&=&(-t^2-t^3,1+2t+2t^2+t^3,-1-t-t^2),\\ \mathbf{Q}_2\times\mathbf{F}(t)&=&(0,t+2t^2+t^3,-1-2t-t^2)=(1+t)^2\mathbf{L}(t),\\ \mathbf{Q}_3\times\mathbf{F}(t)&=&(0,t^2+t^3,-t-t^2)=t(1+t)\mathbf{L}(t). \end{align}\] and \[\begin{align} (\mathbf{Q}_0\times\mathbf{F}(t))\times\mathbf{L}(t)&=&-\mathbf{F}(t),\\ (\mathbf{Q}_1\times\mathbf{F}(t))\times\mathbf{L}(t)&=&-\mathbf{F}(t). \end{align}\] Thus, according to the definition of a \(\mu\)-basis, both \(\mathbf{L}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\), and \(\mathbf{L}(t)\) and \(\mathbf{Q}_1\times\mathbf{F}(t)\), form a \(\mu\)-basis of the curve \(\mathbf{F}(t)\). Using \(\mathbf{Q}_0\times\mathbf{F}(t)\) and \(\mathbf{L}(t)\), the implicit equation of the curve \(\mathbf{F}(t)\) is \[\begin{align} F(x,y,w) &= \det\begin{pmatrix} y & -w & 0 & 0 \\ 0 & y & -w & 0 \\ 0 & 0 & y & -w \\ -x & y-x & y & y \end{pmatrix} \\ &= \begin{aligned}[t] &y^{4} + y^{3}w + y(y-x)w^2 -xw^3= 0. \end{aligned} \end{align}\]

4 Results for Quadratic and Cubic Curves↩︎

4.1 Quadratic Non-degenerate Curves↩︎

Let \(\mathbf{F}(t)=\mathbf{p}_0+\mathbf{p}_1t+\mathbf{p}_2t^2\) be a curve, where \(\mathbf{p}_i\in\mathbb{R}^3,i=0,1,2,3\), \(\mathbf{p}_0\ne0\) and \(\mathbf{p}_2\ne0\). Let \(M=(\mathbf{p}_0^T,\mathbf{p}_1^T,\mathbf{p}_2^T)\) be a \(3\times3\) matrix. Then \(|M|\ne0\).

Let \(\mathbf{Q}_0=\mathbf{p}_2\) and \(\mathbf{Q}_1=\mathbf{p}_1+\mathbf{p}_2t\). Taking the cross product of \(\mathbf{Q}_i\) and \(\mathbf{F}(t)\) gives \(\mathbf{Q}_i\times\mathbf{F}(t)\), \(i=0,1\). Considering the cross product of \(\mathbf{Q}_1\times\mathbf{F}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\), we have \[\bigl(\mathbf{Q}_1\times\mathbf{F}(t)\bigr)\times\bigl(\mathbf{Q}_0\times\mathbf{F}(t)\bigr)=|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_2|\mathbf{F}(t),\] and \(\mathrm{deg}\left(\mathbf{Q}_1\times\mathbf{F}(t)\right)+\mathrm{deg}\left(\mathbf{Q}_0\times\mathbf{F}(t)\right)=\mathrm{deg}(\mathbf{F}(t))\). Therefore, \(\mathbf{Q}_1\times\mathbf{F}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\) form a \(\mu\)-basis of the curve \(\mathbf{F}(t)\), which is consistent with the result of Wang [16].

4.2 Cubic Non-degenerate Curves↩︎

Let \(\mathbf{F}(t)=\mathbf{p}_0+\mathbf{p}_1t+\mathbf{p}_2t^2+\mathbf{p}_3t^3\) be a curve, where \(\mathbf{p}_i\in\mathbb{R}^3,i=0,1,2,3\), \(\mathbf{p}_0\ne0\) and \(\mathbf{p}_3\ne0\). Let \(M=(\mathbf{p}_0^T,\mathbf{p}_1^T,\mathbf{p}_2^T,\mathbf{p}_3^T)^T\) be a \(4\times3\) matrix, and \(Rank(M)=3\).

Let \(\mathbf{Q}_0=\mathbf{p}_3\), \(\mathbf{Q}_1=\mathbf{p}_2+\mathbf{p}_3t\), and \(\mathbf{Q}_2=\mathbf{p}_1+\mathbf{p}_2t+\mathbf{p}_3t^2\). Taking the cross product of \(\mathbf{Q}_i\) and \(\mathbf{F}(t)\) gives \(\mathbf{Q}_i\times\mathbf{F}(t)\), \(i=0,1,2\). Considering the cross product of \(\mathbf{Q}_2\times\mathbf{F}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\), we have \[\begin{align} \label{eq:4461} \bigl(\mathbf{Q}_2\times\mathbf{F}(t)\bigr)\times\bigl(\mathbf{Q}_0\times\mathbf{F}(t)\bigr)=\left[\frac{(\mathbf{p}_3\times\mathbf{p}_0)\cdot\mathbf{F}(t)}{t}\right]\mathbf{F}(t)=|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|+|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|t. \end{align}\tag{10}\]

1)When \(\mathbf{p}_3\times\mathbf{p}_0=\mathbf{0}\)，\(\mathbf{Q}_2\times\mathbf{F}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\) are linearly dependent over \(\mathbb{R}[t]\). In this case, considering the cross product of \(\mathbf{Q}_1\times\mathbf{F}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\), we have \[\bigl(\mathbf{Q}_1\times\mathbf{F}(t)\bigr)\times\bigl(\mathbf{Q}_0\times\mathbf{F}(t)\bigr)=|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_3|\mathbf{F}(t),\] and \(\mathrm{deg}\left(\mathbf{Q}_1\times\mathbf{F}(t)\right)+\mathrm{deg}\left(\mathbf{Q}_0\times\mathbf{F}(t)\right)=\mathrm{deg}(\mathbf{F}(t))\)， So \(\mathbf{Q}_1\times\mathbf{F}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\) form a \(\mu\)-basis of the curve \(\mathbf{F}(t)\).

2)When \(\mathbf{p}_3\times\mathbf{p}_0\ne\mathbf{0}\)，\(\eqref{eq:4461}\) holds, and \(\mathbf{Q}_2\times\mathbf{F}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\) are linearly independent over \(\mathbb{R}[t]\). If \(\mathbf{Q}_2\times\mathbf{F}(t)\) can be decomposed as \[\mathbf{Q}_2\times\mathbf{F}(t)=\left(|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|+|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|t\right)\cdot\left(\frac{\mathbf{p}_1\times\mathbf{p}_0}{|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|}+\frac{\mathbf{p}_3\times\mathbf{p}_0}{|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|}t\right),\] it is required that \(\displaystyle\frac{|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|^2}{|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|}=|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_2|\). In this case, let \(\mathbf{L}(t)=\displaystyle\frac{\mathbf{p}_1\times\mathbf{p}_0}{|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|}+\frac{\mathbf{p}_3\times\mathbf{p}_0}{|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|}t\), then we have \[\mathbf{L}(t)\times\bigl(\mathbf{Q}_0\times\mathbf{F}(t)\bigr)=\mathbf{F}(t)\,\,\,,\,\,\mathrm{deg}\left(\mathbf{L}(t)\right)+\mathrm{deg}\left(\mathbf{Q}_0\times\mathbf{F}(t)\right)=\mathrm{deg}(\mathbf{F}(t)),\] So \(\mathbf{L}(t)\) and \(\mathbf{Q}_0\times\mathbf{F}(t)\) form a \(\mu\)-basis of the curve \(\mathbf{F}(t)\).

When \(\displaystyle\frac{|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|^2}{|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|}\ne|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_2|\)，according to \[|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_3|(\mathbf{p}_3\times \mathbf{p}_0)-|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|(\mathbf{p}_3\times \mathbf{p}_1)+|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|(\mathbf{p}_3\times \mathbf{p}_2)=0,\] we construct \(\mathbf{L}(t)=|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_3|(\mathbf{Q}_2\times \mathbf{F}(t))-|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|(\mathbf{Q}_1\times \mathbf{F}(t))+|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|(\mathbf{Q}_0\times \mathbf{F}(t))\),Expanding it gives \(\mathbf{L}(t)=\mathbf{L}_0+\mathbf{L}_1 t\),where \[\begin{align} \mathbf{L}_0&=&|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_3|(\mathbf{p}_1\times \mathbf{p}_0)-|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|(\mathbf{p}_2\times \mathbf{p}_0)+|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|(\mathbf{p}_3\times \mathbf{p}_0),\\ \mathbf{L}_1&=&-|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|(\mathbf{p}_3\times \mathbf{p}_0)+|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|(\mathbf{p}_3\times \mathbf{p}_1)-|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_2|(\mathbf{p}_3\times \mathbf{p}_2). \end{align}\] It satisfies \[\begin{cases} \mathbf{L}(t)\cdot\mathbf{F}(t)=0,\\ \left(\mathbf{Q}_2\times\mathbf{F}(t)\right)\times\mathbf{L}(t)=\left(|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|^2-|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_2|\cdot|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|\right)\cdot \mathbf{F}(t). \end{cases}\] Therefore \[\begin{align} \mathbf{L}(t)&=&-|\mathbf{p}_1\mathbf{p}_2\mathbf{p}_3|(\mathbf{p}_1\times \mathbf{p}_0)+|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|(\mathbf{p}_2\times \mathbf{p}_0)-|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|(\mathbf{p}_3\times \mathbf{p}_0)\\ &&+\left[|\mathbf{p}_0\mathbf{p}_2\mathbf{p}_3|(\mathbf{p}_3\times \mathbf{p}_0)-|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_3|(\mathbf{p}_3\times \mathbf{p}_1)+|\mathbf{p}_0\mathbf{p}_1\mathbf{p}_2|(\mathbf{p}_3\times \mathbf{p}_2)\right]t,\\ \mathbf{Q}_2\times\mathbf{F}(t)&=&(\mathbf{p}_1\times\mathbf{p}_0)+(\mathbf{p}_2\times \mathbf{p}_0)t+(\mathbf{p}_3\times\mathbf{p}_0)t^2. \end{align}\] form a \(\mu\)-basis of the curve \(\mathbf{F}(t)\), which is consistent with the result of Wang [16].

5 Further discussion↩︎

In this paper, by constructing the vector polynomials \(\mathbf{Q}_i\times\mathbf{F}(t)\), we give explicit formulas for the \(\mu\)-bases of quartic non-degenerate real plane rational curves, and use these \(\mu\)-bases to derive closed-form expressions for their implicit equations. In the future, how to derive explicit formulas for the \(\mu\)-bases of plane rational curves of arbitrary degree will be a problem that we need to further study and solve.

References↩︎