A characterization and implementation of corank one map germs from $2$-space to $3$-space in the computer algebra system SINGULAR

1.   Introduction

It is very interesting to discuss the recognition problem for the classification of singularities due to some equivalence relations. Classification for map germs under some equivalence relation, means finding a list of map germs and showing that all map germs satisfying certain conditions are equivalent to a map germ in the list. Recognition means finding some criteria that describe when a given map germ is equivalent to a map germ belonging to some class or list. Classification and recognition of singularities of map germs are well understood terms and have been subjects of a large number of investigations in the literature ^{[1,3,4,5,6,7,8,9,10,11,12,13,14]}.

Throughout this paper we deal with $\mathcal{A}$-finite map germs from plane to space of corank at most one at origin. Any map germ ${f}$ from $({\mathbb{C}}^n, 0) \rightarrow ({\mathbb{C}}^p, 0)$ has a corank one if its Jacobian matrix at $0$ has rank equal to min$(n, p) -1$. Let $m_2$ be a maximal ideal in the ring of holomorphic function germ in two variables, denoted by $\mathcal{O}_2$, and $p, q \in m_2^2$ be function germs. A map germ having corank at most $1$ is $\mathcal{A}$-equivalent to $(x, p(x, y), q(x, y))$. Mond ^[1,2] studied the geometry and $\mathcal{A}$-classification of corank one map germs from surfaces to 3-spaces. He classified the map germs of corank at most $1$ from $({\mathbb{C}}^2, 0) \rightarrow ({\mathbb{C}}^3, 0)$ having codimension less than or equal to $6.$ In ^[2], three analytic invariants $C(f), T(f)$ and $N(f)$ are associated to a map germ from $({\mathbb{C}}^2, 0)$ to $({\mathbb{C}}^3, 0)$, and together they constitute a complete set of invariants for the $\mathcal{A}$-classification of map germs in Table 1 of ^[1]. In this paper, we present extensive results to characterize the $\mathcal{A}$-finite map germs of corank at most one given in Table 1 of ^[2].

2.   Preliminaries and some invariants

We use the notation $\mathbb{C}$ to denote the set of complex numbers. Let $A(2, 3) = < x, y > \mathbb{C}[[x, y]]^3$ be the set of map germs from plane to space at origin. Let $\mathcal{A} = Aut_{ {\mathbb C}}({\mathbb C}^2, 0)\times Aut_{ {\mathbb C}}({\mathbb C}^3, 0)$. Then, we have a canonical action of $\mathcal{A}$ on $A(2, 3)$ defined by

satisfying $((\varphi, \psi), f)\mapsto \psi\circ f\circ \varphi^{-1}.$

Definition 2.1. Let $f_1, f_2 \in A(2, 3)$. Then, $f_1$ is said to be $\mathcal{A}$-equivalent to $f_2$, if they lie in the same orbit under the action of $\mathcal{A}$, i.e., there exist two diffeomorphisms, $\psi : (\mathbb{C}^3, 0) \rightarrow (\mathbb{C}^3, 0)$ and $\varphi : (\mathbb{C}^2, 0) \rightarrow (\mathbb{C}^2, 0)$, satisfying $\psi \circ f_1 = f_2 \circ \varphi$. We use the notation $f_1 \sim_{\mathcal{A} } f_2$, if $f_1$ is $\mathcal{A}$-equivalent to $f_2$.

Definition 2.2. Let $f \in A(2, 3)$. The orbit map $\theta_f : \mathcal{A} \rightarrow A(2, 3)$ is defined as $\theta_f(\varphi, \psi) = \psi \circ f \circ \varphi^{-1}$. Exceptionally, we have $\theta_f(id) = f$. Let ${\mathcal{A}}_f : = Im(\theta_f)$, and then the orbit of $f$ under the action of $\mathcal{A}$ is the image of $\theta_f$. The extended tangent space to the orbit at $f, \, \mathcal{T}_{ {\mathcal{A}}_{{f}}^e, {f}}$, is defined as

Definition 2.3. Let $f \in A(2, 3)$ with tangent space $\mathcal{T}_{ {\mathcal{A}}_{{f}}^e, {f}}$. The extended $\mathcal{A}$-codimension at $f$ is defined as:

$c_e(f)$ exist with the restriction that $f$ should be $\mathcal{A}$-finite and has been implemented in the computer algebra system SINGULAR (see library $\mathtt{ "classifyMapGerms.lib"})$.

Lemma 2.4. Let $f$ be a map germ from the plane to space at origin. If the corank of $f$ is at most $1$ at $0$, then $f \sim _{\mathcal{A}}(x, p(x, y), q(x, y))$ with $p(x, 0) = 0, q(x, 0) = 0.$

Definition 2.5. Let $f$ be a map germ. Then, the multiplicity $m(f)$ of $f$ is defined by

If $f \sim _{\mathcal{A}} (x, p(x, y), q(x, y))$, then

Definition 2.6. Let $f$ be a map germ from $({\mathbb{C}}^2, 0) \rightarrow ({\mathbb{C}}^3, 0)$ of corank $1$ at $0$. We define $C(f)$ to be the number of crosscaps which appear on the image of a stable perturbation. If $f \sim _{\mathcal{A}} (x, p(x, y), q(x, y))$, then (cf. ^[16])

Definition 2.7. Let $f$ be a map germ from $({\mathbb{C}}^2, 0) \rightarrow ({\mathbb{C}}^3, 0)$ of corank $1$ at $0$. We define $T(f)$ to be the number of triple points which appear on the image of a stable perturbation. If $f \sim _{\mathcal{A}} (x, p(x, y), q(x, y))$, then (cf. ^[16])

where $I_3(f)$ is the ideal in $\mathbb{C} \times \mathbb{C}^3 ( = \{(x, y, u, v): x, y, u, v \in \mathbb{C} \}$ generated by the following $4$ functions:

3.   Characterization of map germs from $({\mathbb{C}}^2, 0)$ to $({\mathbb{C}}^3, 0)$ of corank at most $1$

In this section we characterize simple map germs of corank $1$ in terms of numerical invariants. Table 1 depicts all map germs from plane to space of corank at most $1$, codimension $\leq 6$ at origin, plus some others (e.g, $S_k, B_k, C_k$ and $H_k$, where $k > 6$). The following definition can be found in ^[1].

Definition 3.1. A map germ $f$ from $({\mathbb{C}}^2, 0) \rightarrow ({\mathbb{C}}^3, 0)$ is said to be $\mathcal{A}$-simple if there is a finite number of equivalence classes such that if $f$ is embedded in any family $F: ({\mathbb C}^2\times p, (0, p_0)) \rightarrow ({\mathbb{C}}^3, 0)$, then, for every $(0, p)$ in a sufficiently small neighbourhood of $(0, p_0)$, the germ of $F(x, y, p)$ lies in one of these equivalence classes.

Proposition 3.2. The map germs $f$ from plane to space of corank $\leq 1$ at origin of codimension $\leq 6$, plus some others (e.g, $S_k, B_k, C_k$ and $H_k$, where $k > 6$), are given in Table 1.

Proof. For the proof, see ^[1,2].

Proposition 3.3. Let $f$ be a map germ from plane to space of corank $1$ at origin. Then, $j^2f$ is $\mathcal{A}$-equivalent to one of the following type:

Proof. For the proof, see Proposition 4:2 ^[1].

The extended $\mathcal{A}$-codimension, number of crosscaps and number of triple points play an important role in the characterization of map germs from plane to space. These invariants are given in Table 2.

In the following propositions, we characterize the $2$-jets of the map germ $f$ from plane to space with corank $\leq 1$ at origin in terms of number of crosscaps, $C(f)$, and multiplicity, $m(f)$.

Proposition 3.4. Let $f$ be a map germ from plane to space with corank $0$ at origin. If $C(f) = 0$, then $j^2f$ is of type $(x, y, 0).$

Proof. Let $f = (x, p(x, y), q(x, y))$. If

then $\frac{\partial p}{\partial y}$ or $\frac{\partial q}{\partial y}$ is a unit. We may assume that $\frac{\partial p}{\partial y}$ is a unit. Using the inverse function theorem to the map ${\mathbb{C}}^2 \rightarrow {\mathbb{C}}^2$ defined by $(x, p)$, we may assume that $p(x, y) = y$. Then, obviously, $(x, y, q(x, y)) {\sim}_{\mathcal{A}} (x, y, 0).$

vLemma 3.5. Let $f$ be a map germ from plane to space with corank $1$ at origin.

(1) If $C(f) = 1$, then $m(f) = 2$.

(2) If $m(f) > 2$, then $C(f) > 2$.

Proof. We may assume that $f = (x, p(x, y), q(x, y))$ with

and

If $b_{01}\neq 0$ or $c_{01}\neq 0$, then $\frac{\partial p}{\partial y}$ resp. $\frac{\partial q}{\partial y}$ is a unit. This implies

Since $C(f) = 1$, therefore we have $< \frac{\partial p}{\partial y}, \frac{\partial q}{\partial y} > = < x, y >$. This implies that $b_{02}\neq 0$ or $c_{02}\neq 0$. Assume that $b_{02}\neq 0$, and then $p(0, y) = b_{02}y^2+$ terms of higher order. This implies that

The second statement can be proved similarly.

Proposition 3.6. Let $f$ be a map germ from plane to space with corank $1$ at origin. Then,

(1) if $C(f) = 1$, then type of $j^2f$ is $(x, y^2, xy)$;

(2) if $C(f) \geq 2$ and $m(f) = 2$, then type of $j^2f$ is $(x, y^2, 0);$

(3) if $C(f) \geq 2$ and $m(f) > 2$, then type of $j^2f$ is $(x, xy, 0)$ or $(x, 0, 0).$^*

^*If $C(f) > 1$, then the image of a stable perturbation is not a cone on its boundary. Thus, the stable perturbation is not equivalent to a representative of the original germ, and hence the original germ was not stable.

Proof. Let $f$ be a map germ from plane to space with corank $1$ at origin. Then, $f$ can be written as

By using the left coordinate change $Y_1 = Y -b_{i0}X, Z_1 = Z- c_{i0}X$, we get

$(1)$ If $m(f) = 2$, then we may assume that $b_{01} = c_{01} = 0$ and $b_{02} \neq 0$. By using the left coordinate change $Z_2 = Z_1- \frac{c_{02}}{b_{02}}Y_1$, we get

where, $\alpha = c_{11}-\frac{b_{11}c_{02}}{b_{02}}.$ Now, if $C(f) = 1$, then $\begin{vmatrix} b_{11} & b_{02} \\ c_{11} & c_{02} \\ \end{vmatrix}; \neq 0$ otherwise, $C(f) \neq 1.$ This gives $c_{11}\neq \frac{b_{11}c_{02}}{b_{02}}$ and therefore $\alpha \neq 0$. Now, by using the left coordinate change $Y_3 = Y_2- \frac{b_{11}}{\alpha}Z_2$, we get

This implies $j^2f$ is of type $(x, y^2, xy).$

$(2)$ If $m(f) = 2$, then from $(1)$ we get

Now, if $C(f) \geq 2$, then $\begin{vmatrix} b_{11} & b_{02} \\ c_{11} & c_{02} \\ \end{vmatrix}; = 0$ otherwise, $C(f) < 2.$ This gives $c_{11} = \frac{b_{11}c_{02}}{b_{02}}$, so we have

This can be written as

By using the transformation $y \rightarrow y - \frac{b_{11}}{2}x$, we get

and this gives that $j^2f$ is of type $(x, y^2, 0).$

$(3)$ If $m(f) > 2$, then we may assume that $b_{01} = c_{01} = b_{02} = c_{02} = 0$, and thus

If $b_{11}\neq 0$, then, the left coordinate change $Z_1 = Z- \frac{c_{11}}{b_{11}} Y$ gives,

This implies that, $j^2f$ is of type $(x, xy, 0)$. If $c_{11} = b_{11} = 0$, then $j^2f$ is of type $(x, 0, 0).$

Proposition 3.7. Let $f$ be a map germ from plane to space of corank $0$ at origin. If $j^2f {\sim}_{\mathcal{A}} (x, y, 0)$, then $f$ is of type $S$.

Proof. This can be proved in a similar way as we proved Proposition $3.4$.

Proposition 3.8. Let $f$ be a map germ from plane to space with corank $1$ at origin and $j^2f {\sim}_{\mathcal{A}} (x, y^2, xy).$ Then, $f$ is of type $S_{\circ}.$

Proof. If $j^2f {\sim}_{\mathcal{A}} (x, y^2, xy)$, then

Since $f$ is $2$-determined (see Theorem $4:3$ ^[1]),

Proposition 3.9. Let $f$ be a map germ from plane to space with corank $1$ at origin, $j^2f {\sim}_{\mathcal{A}} (x, y^2, 0)$, and $C(f) = 2$. Then,

(1) if $c_e(f) = 1$, then type of $f$ is $S_1$;^†

^†Proposition $4.1:16 (i)$ of ^[1] implies that $c_e(f) = 1$ implies that $f$ is of type $S_1$. Here, the assumption $C(f) = 2$ is not needed.

$(2)$ if $c_e(f) = k, \, k \geq 2$, then type of $f$ is $B_k$

Proof. If $j^2f {\sim}_{\mathcal{A}} (x, y^2, 0)$, then by Theorem $4.1:1$ ^[1],

We can write it as

If $C(f) = 2$, then $a_{2, 0}\neq 0$, otherwise, $C(f)\neq 2$. Take $a_{2, 0} = 1$, and then

$(1)$ If $c_e(f) = 1$, then $a_{0, 1}\neq 0$, otherwise, $c_e(f) \neq 1$. Since $f$ is $3$-determined, $f(x, y) \sim_{\mathcal{A}} (x, y^2, y^3+ x^2y).$

$(2)$ If $c_e(f) \geq 2$, then $a_{0, 1} = 0$. This gives,

Now, if $c_e(f) = k < \infty$ and $j^3f \sim_{\mathcal{A}} (x, y^2, x^2y)$, then it follows from Mond's classification that,

Proposition 3.10. Let $f$ be a map germ from plane to space with corank $1$ at origin, $j^2f {\sim}_{\mathcal{A}} (x, y^2, 0)$, and $C(f) = 3$. Then,

(1) if $c_e(f) = 2$, then type of $f$ is $S_2$;

(2) if $c_e(f) = 3$, then type of $f$ is $C_3$;

(3) if $c_e(f) = 4$, then type of $f$ is $F_4$.

Proof. If $j^2f {\sim}_{\mathcal{A}} (x, y^2, 0)$, then by Theorem $4.1:1$ ^[1], we can write it as

We can write it as

If $C(f) = 3$, then $a_{2, 0} = 0$ and $a_{3, 0}\neq 0$, otherwise, $C(f)\neq 3$. So,

$(1)$ If $c_e(f) = 2$, then $a_{0, 1}\neq 0$. So, by using left coordinate change $\overline{X} = X, \, \, \, \overline{Y} = Y, \, \, \, \overline{Z} = Z-\frac{a_{1, 1}}{a_{0, 1}}XZ,$ we get

Since $f$ is $4$-determined,

$(2)$ If $c_e(f) = 3$, then $a_{0, 1} = 0$ and $a_{1, 1}\neq 0$. Also, $a_{3, 0} \neq 0$, otherwise, $f$ will not be $\mathcal{A}$-finite. So, we have

Since $f$ is $4$-determined, $f(x, y) \sim_{\mathcal{A}} (x, y^2, xy^3+x^3y).$

$(3)$ If $c_e(f) = 4$, then $a_{0, 1} = a_{1, 1} = 0$ and $a_{0, 2}\neq 0$. Also, $a_{3, 0} \neq 0$, otherwise, $f$ will not be $\mathcal{A}$-finite. So, we have

Since $f$ is $5$-determined, $f(x, y) \sim_{\mathcal{A}} (x, y^2, x^3y+y^5).$

Proposition 3.11. Let $f$ be a map germ from plane to space with corank $1$ at origin, $j^2f {\sim}_{\mathcal{A}} (x, y^2, 0)$, and $C(f) = k, 3 < k < \infty$. Then,

(1) if $c_e(f) = C(f)-1$, then $f$ is of type $S_{k-1}$;

(2) if $c_e(f) = C(f)$, then $f$ is of type $C_k.$

Proof. Since $j^2f \sim_{\mathcal{A}} (x, y^2, 0)$,

We can write it as

If $C(f) = k$, then $a_{2, 0} = \dots = a_{k-1, 0} = 0$ and $a_{k, 0}\neq 0.$ Now, if $c_e(f) = C(f)-1$, then $a_{0, 1} \neq 0$, otherwise, $f$ will not be $\mathcal{A}$-finite. This gives $j^3f \sim_{\mathcal{A}} (x, y^2, y^3).$ Then, by Mond's classification $f$ is of type $f(x, y) \sim_{\mathcal{A}} (x, y^2, y^3+x^{k+1}y).$

If $c_e(f) = C(f)$, then $a_{0, 1} = 0$ and $a_{1, 1}\neq 0$, otherwise, $f$ will not be $\mathcal{A}$-finite. This gives $j^3f \sim_{\mathcal{A}} (x, y^2, 0).$ Thus, it follows from Mond's classification that $f$ is of type $f(x, y) \sim_{\mathcal{A}} (x, y^2, xy^3+x^ky).$

Proposition 3.12. Let $f$ be a map germ from plane to space with corank $1$ at origin, $j^2f {\sim}_{\mathcal{A}} (x, xy, 0)$, and $C(f) = 2$. If $c_e(f) = k, \, k \geq 2$, then $f$ is of type $H_k.$

Proof. Since $j^2f \sim_{\mathcal{A}} (x, xy, 0)$,

By using suitable left coordinate changes, we get

If $C(f) = 2$, then $b_{0, 3}\neq 0$. Take $b_{0, 3} = 1$, we have

It is easy to see that

Now, by using transformation $y\rightarrow y-\frac{b_{1, 2}}{3}x$, we get

This gives $j^3f {\sim}_{\mathcal{A}} (x, xy, y^3)$. If $c_e(f) \geq 2$, then by using Theorem $4.2.1:2(a)$ ^[1], we have

Proposition 3.13. Let $f$ be a map germ from plane to space with corank $1$ at origin, $j^2f {\sim}_{\mathcal{A}} (x, xy, 0)$, and $C(f) = 3$. Then,

(1) if $c_e(f) = 4$ and $T(f) = 1$, then $f$ is of type $P_3$ or $P_4(\frac{1}{2})$ or $T_4$;

(2) if $c_e(f) = 4$ and $T(f) = 2$, then $f$ is of type $P_4(1)$ or $Q_4$;

(3) if $c_e(f) = 5$ and $T(f) = 3$, then $f$ is of type $Q_5$;

(4) if $c_e(f) = 6$ and $T(f) = 4$, then $f$ is of type $Q_6$ or $R_4$;

(5) if $c_e(f) = k$ and $T(f) = k-2$, then $f$ is of type $Q_k, \, \, k \geq 7$.

Proof. Since $j^2f \sim_{\mathcal{A}} (x, xy, 0)$,

If $c_e(f) = 4$, then $b_{0, 3} = 0$. Then, the coordinate changes $\overline{X} = X, \overline{Y} = Y-a_{2, 1}XY, \overline{Z} = Z- b_{2, 1}XY$ gives

If $b_{1, 2} = 0$, then the transformation $x \rightarrow, \, \, y \rightarrow y+a_{1, 2}y^2$, gives

Thus, $j^3f \sim_{\mathcal{A}} (x, xy+a_{0, 3}y^3, 0).$ Now, if $C(f) = 3$, then it follows from the Proposition $4.2.4:1$ ^[1] $f$ must have a $4$-jet equivalent to $(x, xy+y^3, y^4).$ Since $f$ is $4$-determined (see Theorem $4.2.4:2$ ^[1]),

If $b_{1, 2}\neq 0$, then the coordinate change $\overline{X} = X, \overline{Y} = Y-(\frac{a_{1, 2}}{b_{1, 2}})Z, \overline{Z} = Z$ gives

This implies

Now, by the Lemma $4.2.2:1$ ^[1], this type of $f$ must have a $4$-jet equivalent to $(x, xy+y^3, xy^2+cy^4)$.

$(1)$ If $c_e(f) = 4$ and $T(f) = 1$, then $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+cy^4), \, \, c\neq 0, \frac{1}{2}, 1, \frac{3}{2}$ or $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+\frac{1}{2}y^4+y^5)$. These two types can be differentiated by computing their normal forms of order $4$ and finding $c$^‡. If $c\neq 0, \frac{1}{2}, 1, \frac{3}{2}$, then $f$ is of type $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+cy^4)$, and if $c = \frac{1}{2}$, then $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+\frac{1}{2}y^4+y^5).$

^‡This can be done by using Algorithm $1$.

$(2)$ If $c_e(f) = 4$ and $T(f) = 2$, then $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+y^4+y^6)$ or $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+y^7).$ These two types can be differentiated by computing their normal forms of order $4$ and finding $c$. If $c = 1$, then $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+y^4+y^6)$ and, if $c = 0$, then $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+y^7).$

$(3)$ It is straightforward to prove.

$(4)$ We have $j^4f \sim_{\mathcal{A}} (x, xy+y^3, xy^2)$. If $c_e(f) = 6$, then $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+y^{13})$ or $f(x, y) \sim_{\mathcal{A}} (x, xy+y^6+by^7, xy^2+y^4+cy^6).$ These two types can be differentiated by computing their normal forms of order $3$. If $j^3f \sim_{\mathcal{A}} (x, xy+y^3, xy^2)$, then $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+y^{13})$, and if $j^3f \sim_{\mathcal{A}} (x, xy, xy^2)$, then $f(x, y) \sim_{\mathcal{A}} (x, xy+y^6+by^7, xy^2+y^4+cy^6).$

$(5)$ Now if $c_e(f) > 6$, and $T(f) > 4$, then from Theorem $4.2.2:7$ ^[1] $f(x, y) \sim_{\mathcal{A}} (x, xy+y^3, xy^2+y^{3k-5}).$

The next algorithm is useful to differentiate the types of $P_3$ and $P_4(\frac{1}{2})$.

Proposition 3.14. Let $f$ be a map germ from the plane to space with corank $1$ at origin, $j^2f \sim_{\mathcal{A}} (x, xy, 0)$, and $C(f) = 4$. If $c_e(f) = 4$ and $T(f) = 1$, then $f$ is of type $P_4(\frac{3}{2}).$

Proof. The proof is similar to the Proposition 3.13(2).

Proposition 3.15. Let $f$ be a map germ from plane to space with corank $1$ at origin, $j^2f \sim_{\mathcal{A}} (x, 0, 0)$, and $C(f) = 4$. If $c_e(f) = 4$ and $T(f) = 1$, then $f$ is of type $X_4.$

Proof. Since $j^2f \sim_{\mathcal{A}} (x, 0, 0)$,

If $C(f) = 4$ and $T(f) = 1$, then $b_{0, 3} = 0$; thus,

By using the coordinate changes $\overline{X} = X, \overline{Y} = Y-Z, \overline{Z} = Z$, we get

Thus,

Now, if $T(f) = 1$, then by using Proposition $4.3:2$ ^[1], we get $j^4f \sim_{\mathcal{A}} (x, y^3, xy^2+xy^2+y^4).$ As $f$ is $4$-determined, $f(x, y) \sim_{\mathcal{A}} (x, y^3, xy^2+xy^2+y^4).$

4.   Singular examples

We have implemented the characterization in the computer algebra system Singular ^[15]. The code can be downloaded from https://www.mathcity.org/files/ahsan/Proc-classifycoRank1Maps.txt. We give some examples. The examples are constructed from the normals form by applying a generic $\mathcal{A}$-equivalence.

In the first example we have as an input the map $f(x, y) = (f_1, f_2, f_3)$, where

In SINGULAR this can be written as

To compute the required type of map germs, we use the procedure:

In the second example we have as an input the map $f(x, y) = (f_1, f_2, f_3)$, where

In SINGULAR this can be written as

To compute the required type of map germs, we use the procedure:

5.   Conclusions

A classifier for map germs from plane to space in terms of codimension, crosscaps and triple points has been given. Moveover, this classifier is implemented in the computer algebra system SINGULAR.

Availability of data and materials

The code used in this paper can be downloaded from: https://www.mathcity.org/files/ahsan/ProcclassifycoRank1Maps.txt.

Acknowledgements

This work was supported by the National Natural Science Foundation of China (Nos. 62172116, 61972109) and the Guangzhou Academician and Expert Workstation (No. 20200115-9).

The research of the second and third authors is supported by Higher Education Commission of Pakistan by the project No. 7495 /Punjab/NRPU/R&D/HEC/2017.

Conflict of interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.