Detecting Affine Equivalences Between Implicit Planar Algebraic Curves

Abstract

We present a complete algorithm for computing the affine equivalences between two implicit planar algebraic curves. We provide evidence of the efficiency of the algorithm, implemented in Maple, and compare its performance with existing algorithms. As a part of the process for developing the algorithm, we characterize planar algebraic curves, possibly singular, possibly reducible, invariant under infinitely many affine equivalences.

Appendix I: Characterizing the Algebraic Curves with Infinitely Many Self-Affine Equivalences

In this appendix we will prove Theorem 4 in Sect. 4.1, and we will provide some examples of this type of curves. Thus, let ${\mathcal {C}}_{f}$ be a planar algebraic curve defined by $f$, not containing any line. We want to look for the self-affine equivalences of ${\mathcal {C}}_{f}$ fixing the origin, assuming that we have already performed the step described in Sect. 3. By writing $f(x,y)$ as sum of homogeneous forms $f_{i}(x,y)$ like in Eq. (17), if ${\mathcal {T}}(x,y)=(a_{11}x+a_{12}y,a_{21}x+a_{22}y)$ is a self-equivalence of the curve ${\mathcal {C}}_{f}$ defined by $f$ then $\lambda f=f\circ \mathcal {T}$, with $\lambda \in {\mathbb{R}}$, and each form $f_{i}$ also satisfies that $\lambda f_{i}=f_{i}\circ \mathcal {T}$ (see also Sect. 4.2). By applying if necessary an affine change of coordinates to $f$, the form $f_{i}(x,y)$ can be written as

$$ f_{i}(x,y)=b_{i}(x-\xi _{1} y)^{r_{1}}\cdots (x-\xi _{m} y)^{r_{m}}, $$

(23)

where $r_{1}+\cdots +r_{m}=i$, and $\xi _{i}\in {\mathbb{C}}$. We can also write

$$\begin{aligned} &f_{i}\circ \mathcal {T}=b_{i}(a_{11}-\xi _{1} a_{21})^{r_{1}}\cdots (a_{11}- \xi _{m} a_{21})^{r_{m}}\cdot \left [x- \dfrac{-a_{12}+\xi _{1} a_{22}}{a_{11}-\xi _{1} a_{21}} y\right ]^{r_{1}} \cdots\\ &\quad \left [x- \dfrac{-a_{12}+\xi _{m} a_{22}}{a_{11}-\xi _{m} a_{21}}y\right ]^{r_{m}} \end{aligned}$$

Since $\lambda f_{i}=f_{i}\circ \mathcal {T}$, we get that

$$ b_{i}(a_{11}-\xi _{1} a_{21})^{r_{1}}\cdots (a_{11}-\xi _{m} a_{21})^{r_{m}}= \lambda , $$

(24)

and that each matrix $A=\left (a_{ij}\right )$ is in bijective correspondence with the Möbius transformation

$$ \varphi (\xi )=\dfrac{-a_{12}+a_{22}\xi}{a_{11}-a_{21}\xi} $$

(25)

that maps the set $\{\xi _{i}\}_{i=1,\ldots ,m}$ onto itself; take into account here that if $B=\kappa A$, with $\kappa \in {\mathbb{R}}$, then the Möbius transformation in Eq. (25) associated with both $A$, $B$ is the same. However, if $A$ is a self-equivalence of ${\mathcal {C}}_{f}$, then $\kappa A$ defines a homothety, and the only algebraic curves which are invariant under homotheties are the unions of intersecting lines. Since we are assuming that ${\mathcal {C}}_{f}$ does not contain any line, we can safely establish that there is a $1:1$ correspondence between the matrices $A$ associated with self-equivalences of ${\mathcal {C}}_{f}$, and the Möbius transformations in Eq. (25).

Now a first conclusion is the following.

Lemma 6

If ${\mathcal {C}}_{f}$ has finitely many self-equivalences, $m\leq 2$.

Proof

If ${\mathcal {C}}_{f}$ has finitely many self-equivalences, this also holds for the homogeneous form $f_{i}(x,y)$ in Eq. (23). Thus, any self-equivalence of $f_{i}$ corresponds to a Möbius transformation as in Eq. (25) mapping the set $\{\xi _{i}\}_{i=1,\ldots ,m}$ onto itself, and this correspondence is bijective. Since a Möbius transformation is completely fixed when we know the image $\varphi (\xi )$ for three different values of $\xi$, if $m\geq 3$ then there are finitely many Möbius transformations mapping the set $\{\xi _{i}\}_{i=1,\ldots ,m}$ onto itself. Since the matrices $A$ associated with self-equivalences of ${\mathcal {C}}_{f}$ are in $1:1$ correspondence with this set of Möbius transformations, the result follows. □

So we deduce that each homogeneous form $f_{i}$ has at most two factors, and therefore at most two $\xi _{i}$. In fact, there must be exactly two different $\xi _{1}$, $\xi _{2}$, since otherwise the curve ${\mathcal {C}}_{f}$ is a union of lines, which is a case we are excluding. Furthermore, for each $\xi _{i}$, $i=1,2$, we have a condition $\varphi (\xi _{i})=\xi _{j}$ as in Eq. (25). Thus, we have two conditions, in total, for the entries of the matrix $A$, that can be written as

$$ \textstyle\begin{array}{r@{\quad}c@{\quad}l} -a_{12}+a_{22}\xi _{1}-\eta _{1}a_{11}+\eta _{1}\xi _{1} a_{21}&=&0, \\ -a_{12}+a_{22}\xi _{2}-\eta _{2}a_{11}+\eta _{2}\xi _{2} a_{21}&=&0, \end{array} $$

(26)

with $\xi _{1}\neq \xi _{2}$ and $\eta _{1}\neq \eta _{2}$; furthermore, either $\eta _{1}=\xi _{1}$, $\eta _{2}=\xi _{2}$ or $\eta _{1}=\xi _{2}$, $\eta _{2}=\xi _{1}$ depending on whether or not $\varphi (\xi _{i})=\xi _{j}$ holds for $i=j$ or for $i\neq j$.

Now we are finally ready to prove Theorem 4.

Proof of Theorem 4

The implication $(\Rightarrow )$ is a consequence of the arguments in this section. So let us see $(\Leftarrow )$. We know that the homogeneous forms of $f$ must be of the type

$$ c_{i}(x-\xi _{1} y)^{r_{1}}(x-\xi _{2} y)^{r_{2}}, $$

where $r_{1},r_{2}\geq 0$, and at least one homogeneous form satisfies that $r_{1}+r_{2}>0$. Now assume that we have two different homogeneous forms of this type, $f_{i}(x,y)=c_{i}(x-\xi _{1} y)^{r_{1}}(x-\xi _{2} y)^{r_{2}}$, $f_{\ell}(x,y)=c_{\ell}(x-\xi _{1} y)^{s_{1}}(x-\xi _{2} y)^{s_{2}}$, $0< r_{1}+r_{2}< s_{1}+s_{2}$. Suppose now that $r_{1}r_{2}\neq 0$; we will address the case $r_{1}r_{2}=0$ later. Then each of the forms $f_{i}$, $f_{\ell}$ must satisfy the condition in Eq. (24) for the same $\eta$. Applying easy manipulations on these conditions, we derive the conditions

$$ (a_{11}-\xi _{2} a_{21})^{r_{2}s_{1}-s_{2}r_{1}}=\gamma ^{s_{1}-r_{1}}, \text{ }(a_{11}-\xi _{1} a_{21})^{r_{1}s_{2}-r_{2}s_{1}}=\gamma ^{s_{2}-r_{2}}. $$

If $r_{2}s_{1}-s_{2}r_{1}\neq 0$, in turn we derive two conditions of the type

$$ a_{11}-\xi _{2} a_{21}=\gamma _{1},\text{ }a_{11}-\xi _{1} a_{21}= \gamma _{2} $$

(27)

with $\gamma _{1},\gamma _{2}\neq 0$. Thus, collecting Eq. (26) and Eq. (27), if $r_{2}s_{1}-s_{2}r_{1}\neq 0$ we get a linear system for the $a_{ij}$ where the coefficient matrix is

$$ \begin{pmatrix} -1 & \xi _{1} & -\eta _{1} & \eta _{1}\xi _{1} \\ -1 & \xi _{2} & -\eta _{2} & \eta _{2}\xi _{2} \\ 0 & 0 & 1 & -\xi _{2} \\ 0 & 0 & 1 & -\xi _{1} \end{pmatrix} $$

Since $\xi _{1}\neq \xi _{2}$, one can check that the rank of the above matrix is 4, so there are at most finitely many solutions for the $a_{ij}$. Thus, we conclude that $r_{2}s_{1}-s_{2}r_{1}=0$. Now if $r_{1},r_{2}\neq 0$ then $s_{1}/r_{1}=s_{2}/r_{2}=k$, and therefore $s_{1}=kr_{1}$, $s_{2}=kr_{2}$, so the form $f_{\ell}$ is a power of $f_{i}$, multiplied by a constant.

Finally, if $r_{1}r_{2}=0$, since not both $r_{1}$, $r_{2}$ are zero we can assume $r_{2}\neq 0$, $r_{1}=0$. Then we can also assume $s_{1}\neq 0$, because if all the homogeneous forms of $f$ depend only on $\xi _{2}$, the curve ${\mathcal {C}}_{f}$ is a union of lines. In this case since $r_{1}=0$, from Eq. (24) we directly derive a condition $a_{11}-\xi _{2} a_{21}=\gamma _{1}$. Taking this into account and Eq. (24) for the homogeneous form $f_{\ell}$, we get another condition $a_{11}-\xi _{1} a_{21}=\gamma _{2}$. Then, arguing as in the case $r_{1}r_{2}\neq 0$ we get a contradiction with the fact that ${\mathcal {C}}_{f}$ has infinitely many self-equivalences.

Thus, we conclude that all the homogeneous forms of $f$ are powers of $(x-\xi _{1} y)^{r_{1}}(x-\xi _{2} y)^{r_{2}}$, for $r_{1},r_{2}>0$, and therefore the theorem is proven. □

Remark 2

The curves $(x-\xi _{1} y)^{r_{1}}(x-\xi _{2} y)^{r_{2}}=c$, with $c\in {\mathbb{C}}$, are always either rational, or reducible curves with rational components. Indeed, with $\xi _{1}=\xi _{2}$, we have a collection of lines. If $\xi _{1}\neq \xi _{2}$, the change $X:=x-\xi _{1} y$, $Y:=y$ yields

$$ X^{r_{1}}(X+aY)^{r_{2}}=c, $$

(28)

with $a=\xi _{1}+\xi _{2}$. If $a=0$ or $c=0$ we get again a collection of lines. If $a,c\neq 0$ and $\text{gcd}(r_{1},r_{2})=1$, Eq. (28) corresponds to a rational planar curve parametrized by

$$ \left (t^{r_{2}},\dfrac{1}{a}\left (\dfrac{c^{1/r_{2}}}{t^{r_{1}}}-t^{r_{2}} \right )\right ), $$

(29)

so $(x-\xi _{1} y)^{r_{1}}(x-\xi _{2} y)^{r_{2}}=c$ is rational as well. If $\text{gcd}(r_{1},r_{2})\neq 1$, Eq. (28) is reducible and its components are rational curves, with parametrizations similar to Eq. (29).

Example 1

Circles with the same center

Let

$$ f(x,y)=(x^{2}+y^{2}-a_{1})\cdots (x^{2}+y^{2}-a_{n}). $$

The polynomial $f$ can be written as a polynomial in $x^{2}+y^{2}=(x+iy)(x-iy)$, where $i^{2}=-1$, of degree $n$. The curve $f(x,y)=0$, which is the union of $n$ circles centered at the origin, is invariant under the family of rotations fixing the origin, defined by the matrix

$$ A= \begin{pmatrix} \text{cos}(\theta ) & -\text{sin}(\theta ) \\ \text{sin}(\theta ) & \text{cos}(\theta )\end{pmatrix} . $$

Example 2

Hyperbolas with the same center

Let

$$ f(x,y)=(x^{2}-y^{2}-b_{1})\cdots (x^{2}-y^{2}-b_{n}). $$

The polynomial $f$ can be written as a polynomial in $x^{2}-y^{2}=(x+y)(x-y)$. The curve $f(x,y)=0$, which is the union of $n$ hyperbolas centered at the origin, is invariant under the family of hyperbolic rotations fixing the origin, defined by the matrix

$$ A= \begin{pmatrix} \text{cosh}(\alpha ) & \text{sinh}(\alpha ) \\ \text{sinh}(\alpha ) & \text{cosh}(\alpha )\end{pmatrix} . $$

Example 3

Consider the irreducible quartic defined by

$$ f(x,y)=x^{3}(x-y)-1. $$

This curve admits the rational parametrization $\left (t,t-\dfrac{1}{t^{3}}\right )$, so it is a rational quartic. One can check that this curve is invariant by the family of affine transformations, fixing the origin, defined by

$$ A= \begin{pmatrix} \beta & 0 \\ \dfrac{\beta ^{4}-1}{\beta ^{3}} & \dfrac{1}{\beta ^{3}}\end{pmatrix} . $$