Equality of the Singularity Critical Locus Dimension and the Newton Polyhedron Combinatorial Dimension

Abstract

The purpose of this paper is to determine the local dimension of the critical locus of a generic singularity. We use combinatorial methods to calculate this dimension in terms of a convex object associated with the singularity, called the Newton polyhedron. In this article, we prove that the local dimension of the critical locus of a generic singularity $f:({\mathbb{C}}^n,0)⟶(\mathbb{C},0),$ $n\le 4,$ is equal to the combinatorial dimension of the Newton polyhedron of the gradient mapping $\nabla f.$ Therefore, there is some symmetry between combinatorial properties of the Newton polyhedron of a generic singularity and geometric properties of its critical locus.

1. Introduction

In our research, we study discrete invariants of complex analytic singularities. More precisely, we try to read these invariants from a certain combinatorial convex object in real space, associated with a singularity, called the Newton polyhedron. A singularity is represented by a holomorphic function defined in some neighborhood of a critical point. In the sixties and seventies of the twentieth century, Vladmir I. Arnold posed the following questions related to the Newton polyhedron of a singularity (see [1]):

1968-2 

What topological characteristics of a real (complex) polynomial are computable from the Newton polyhedron (and the signs of the coefficients)?

1975-1 

Every interesting discrete invariant of a generic singularity with a Newton polyhedron $\Gamma$ is an interesting function of the polyhedron. Study: the signature, the number of moduli, the singularity index, the integral monodromy, the variation, the Bernstein polynomial, and ${\mu }_i$ (for a generic section).

1975-21 

Express the main numerical invariants of a typical singularity with a given Newton polyhedron (e.g., the signature, the genus of the 1-dimensional Milnor fiber) in terms of the polyhedron.

So far, many invariants have been read from the Newton polyhedron of a generic singularity (i.e., singularity with generic coefficients). The most important of them is the Milnor number [2]. Also, the bifurcation set of a polynomial function [3], zeta-function of monodromy [4], Lojasiewicz exponent [5,6,7,8,9], Lê numbers [10], and local Euler obstruction of isolated determinantal singularities [11,12] are determined by the Newton polyhedron. So, we see that many geometric and topological properties of a generic singularity are reflected in corresponding combinatorial properties of its Newton polyhedron. Thus, we can say that there is some kind of symmetry between a generic singularity and its Newton polyhedron.

In this paper, we study the dimension of the critical locus of a generic singularity and we want to show that this is determined by the Newton polyhedron. It is known that if singularities have the same Newton polyhedron and if their Lê numbers exist, then their critical loci have the same dimensions ([10], Corollary 5.1). It seems that the assumption about the existence of the Lê numbers is unnecessary. This is how the hypothesis was born that critical loci of generic singularities with the same Newton polyhedron have the same dimension. Firstly, we studied the case of an isolated singularity, i.e., its critical locus has a zero dimension; in [13], we gave combinatorial conditions in terms of the Newton polyhedron to determine when a generic singularity is an isolated singularity. In [14], we generalized this result to the case of non-isolated singularity and we gave a formula for the dimension of the critical locus of a generic singularity, $f:({\mathbb{C}}^n,0)⟶(\mathbb{C},0),$ $n\le 3$, in terms of the Newton polyhedron of $f.$ The main result of our article is to extend this result to the case of $n=4.$ More precisely, we prove that the dimension of the critical locus of a generic singularity, $f:({\mathbb{C}}^4,0)⟶(\mathbb{C},0),$ is equal to the combinatorial dimension of the Newton polyhedron of the gradient mapping $\nabla f$ (see Theorem 3, Corollary 1). Our result confirms Arnold’s Conjecture in this case. It is possible to compute a dimension of an analytic set by Gröbner basis ([15]). However, the complexity of the Gröbner basis computations may by exponential, and our combinatorial methods could be more effective in many cases. To achieve singularities with generic coefficients, we use Kushnirenko nondegeneracy (see Preliminaries). Also, C.T.C. Wall gave some similar nondegeneracy conditions, but his conditions are too strong and imply that the nondegenerate singularity (in his sense) has to be an isolated singularity (see [16]).The concept of the nondegeneracy of singularity is studied in detail in books by Oka [17] and Mondal [18], also in the article [19]. Another aspect of our paper is finite determinacy. Recall that an analytic function f is finitely determined if its topology is determined by its Taylor polynomial of some degree. It is known that f is finitely determined if and only if it has an isolated singularity [20,21]. We show that the dimension of the critical locus of Kushnirenko nondegenerate singularity is finitely determined (see Corollaries 2 and 3). Hence, calculating a dimension of the critical locus is easier in this case (see Example 4).

2. Preliminaries

Add $\mathbb{N}=\{0,1,2,\dots \}.$ Let $f:({\mathbb{C}}^n,0)⟶(\mathbb{C},0)$ be a singularity, i.e., a germ of a holomorphic function having critical point at $0.$ Denote the germ of the critical locus of f by $\Sigma \left(f\right)$. If $dim\Sigma \left(f\right)=0$, then we say that f is an isolated singularity. The germ f is represented by a convergent power series:

$$f\left(z\right)=\sum _{\alpha }{c}_{\alpha }{z}^{\alpha },\alpha =({\alpha }_1,\dots ,{\alpha }_n)\in {\mathbb{N}}^n,c_{\alpha }\in \mathbb{C},z^{\alpha }=z_1^{{\alpha }_1}\cdots z_n^{{\alpha }_n}.$$

(1)

We will now give some definitions following Kushnirenko’s famous paper [2]

• $suppf=\{\alpha \in {\mathbb{N}}^n:c_{\alpha }\ne 0\}$—support of f;
• ${\Gamma }_{+}\left(f\right)$—convex hull of $\alpha +{\mathbb{R}}_{+}^n,\alpha \in suppf$—Newton polyhedron of f (see Figure 1);
• $\Gamma \left(f\right)$—family of compact faces of ${\Gamma }_{+}\left(f\right)$—Newton boundary of f;
• $f_{\Delta }\left(z\right):={\sum }_{\alpha \in \Delta }{c}_{\alpha }{z}^{\alpha },\Delta \in \Gamma \left(f\right)$;
• f—Kushnirenko nondegenerate on $\Delta$ if the following system of equations

  $${\displaystyle \frac{\partial f_{\Delta }}{\partial z_1}}\left(z\right)=\cdots ={\displaystyle \frac{\partial f_{\Delta }}{\partial z_n}}\left(z\right)=0$$

  (2)

  has no solution in ${(\mathbb{C}\setminus \left\{0\right\})}^n$;
• f—Kushnirenko nondegenerate, if f nondegenerate on each face $\Delta \in \Gamma \left(f\right)$.

If we start with a given subset, $A\subset {\mathbb{N}}^n,$ $(0,\dots ,0)\notin A,$ we can also define an abstract Newton polyhedron ${\Gamma }_{+}\left(A\right)$ as a convex hull of sets $\alpha +{\mathbb{R}}_{+}^n,\alpha \in A.$ Then, we say that ${\Gamma }_{+}\left(A\right)$ is generated by $A.$

Now, we pass to the case of the mapping. Let $f=(f_1,\dots ,f_m):({\mathbb{C}}^n,0)⟶({\mathbb{C}}^m,0)$ be a germ of a holomorphic mapping and $\mathcal{A}=(A_1,\dots ,A_m)$ be a tuple of subsets of ${\mathbb{N}}^n.$ Now, we will introduce the following definitions.

Definition 1.

We define a tuple

$$suppf=(supp{f}_1,\dots ,supp{f}_m)$$

(3)

and call it the support of f.

Definition 2.

We define a tuple

$${\Gamma }_{+}\left(f\right)=\left({\Gamma }_{+}\left(f_1\right),\dots ,{\Gamma }_{+}\left(f_m\right)\right)$$

(4)

and call it the Newton polyhedron of f.

We define a tuple

$${\Gamma }_{+}\left(\mathcal{A}\right)=\left({\Gamma }_{+}\left(A_1\right),\dots ,{\Gamma }_{+}\left(A_m\right)\right)$$

(5)

and call it an abstract Newton polyhedron generated by $\mathcal{A}$.

Let $I\subset \{1,2,\dots ,n\}$; we add

$$O{X}_I=\{x\in {\mathbb{R}}^n:x_i=0\mathrm{for}i\notin I\},$$

(6)

so $O{X}_I$ is the coordinate subspace spanned by axes $O{X}_i,i\in I.$

Definition 3.

We say $\mathcal{A}$ that satisfies the $\left(k\right)$-Kushnirenko condition (simply (k)-condition) if, for each $I\subset \{1,\dots ,n\}$, there are at least $\left|I\right|-k$ nonempty sets among the following sets:

$$A_1\cap O{X}_I,\dots ,A_m\cap O{X}_I.$$

(7)

Definition 4.

We say that f satisfies the $\left(k\right)$-Kushnirenko condition (simply (k)-condition) if $suppf$ satisfies the $\left(k\right)$-Kushnirenko condition.

Remark 1.

For $k=0$, we will succinctly write “the Kushnirenko condition” instead of “$\left(0\right)$-Kushnirenko condition”. It seems that Kushnirenko was the first to give such a condition [22]. If f is a function (not a mapping), then Definition 4 is different from ([14], Definition 2.2). In this case, the old definition of the Kushnirenko condition ([14], Definition 2.2) corresponds to the condition $\nabla f$, satisfying the $\left(k\right)$-Kushnirenko condition in the sense of our new Definition 4. Hertling and Kurbel collected conditions equivalent to the Kushnirenko condition in the case of a quasihomogeneous polynomial ([23], Lemma 2.1), but this lemma is also true without the assumption of quasihomogeneity.

Definition 5.

We define a combinatorial dimension of $\mathcal{A}:$

$$dim\mathcal{A}=min\{k\in \mathbb{N}:\mathcal{A}satisfies\left(k\right)-condition\}$$

(8)

Example 1.

Let $\mathcal{A}=\left(\right\{(1,0,0)\},\{(0,1,0)\},\{(0,0,1)\left\}\right).$ It is easy to see that for each $I\subset \{1,2,3\}$, there is exactly $\left|I\right|$ nonempty subsets among $A_i\cap O{X}_I,$ $i=1,2,3.$ Hence, “a density” of $\mathcal{A}$ on each coordinate subspace I is maximal and equal to $\left|I\right|.$ Therefore, the combinatorial dimension is minimal, $dim\mathcal{A}=0.$

Example 2.

Let $\mathcal{A}=\left(\right\{(1,0,0),(0,1,1)\},\{(0,1,0),(1,0,1)\},\{(1,1,0)\left\}\right).$ Observe that all $A_i\cap O{X}_3,$ $i=1,2,3,$ are empty sets and “a density” on axis $O{X}_3$ is minimal. Hence, $dim\mathcal{A}\ge 1.$ It easy to check that for each $I\subset \{1,2,3\}$, there is at least $\left|I\right|-1$ nonempty subsets among $A_i\cap O{X}_I,$ $i=1,2,3.$ Summing up $dim\mathcal{A}=1.$

Remark 2.

Since $supp{f}_i\cap O{X}_I\ne \varnothing$ if and only if ${\Gamma }_{+}\left(f_i\right)\cap O{X}_I\ne \varnothing$ for each $I\subset \{1,\dots ,n\}$ we get that

$$dim(suppf)=dim{\Gamma }_{+}\left(f\right)$$

(9)

Remark 3.

It is also easy to observe that the following conditions are equivalent:

• $dim\mathcal{A}=k.$
• $\mathcal{A}$ satisfies k-condition and it does not satisfy $(k-1)$- condition.

3. Main Results

In the paper [14], we prove the following theorem:

Theorem 1.

Let $f:({\mathbb{C}}^n,0)⟶(\mathbb{C},0),$ $n\le 3$ be a singularity. If f is Kushnirenko nondegenerate, then the following conditions are equivalent:

(i) 

$dim\Sigma \left(f\right)=d$;

(ii) 

$supp\nabla f$ satisfies d-condition and it does not satisfy $(d-1)$- condition,

for each $0\le d\le n.$

By Remarks 1–3, we can reformulate the above theorem as follows.

Theorem 2.

Let $f:({\mathbb{C}}^n,0)⟶(\mathbb{C},0),$ $n\le 3$ be a singularity. If f is Kushnirenko nondegenerate, then

$$dim\Sigma \left(f\right)=dim{\Gamma }_{+}(\nabla f)$$

(10)

Roughly speaking, $dim{\Gamma }_{+}(\nabla f)$ is a measure of the density of supports $f_{z_i}^{\prime }$ on the coordinate subsystems. If this density increases in all coordinate subsystems, the dimension of the critical locus decreases. If this density is maximal, then the singularity has an isolated critical point at $0.$

Therefore, we may put forward the following conjecture.

Hypothesis 1.

Let $f:({\mathbb{C}}^n,0)⟶(\mathbb{C},0),$ be a singularity. If f is Kushnirenko nondegenerate, then

$$dim\Sigma \left(f\right)=dim{\Gamma }_{+}(\nabla f)$$

(11)

Now, we give the main result of the paper, which confirms our conjecture for $n=4$.

Theorem 3.

Let $f:({\mathbb{C}}^4,0)⟶(\mathbb{C},0)$ be a singularity. If f is Kushnirenko nondegenerate, then

$$dim\Sigma \left(f\right)=dim{\Gamma }_{+}(\nabla f)$$

(12)

Example 3.

Let

$$f(x,y,z,w)=xyz+xyw+wz.$$

We can easily confirm that f is Kushnirenko nondegenerate. Add $I=\left\{1\right\}.$ We see that all supports in $supp\nabla f$ are disjoint with $O{X}_I.$ Therefore, f does not satisfy the 0 - Kushnirenko condition. On the other hand, it is easy to see that f satisfies the 1-Kushnirenko condition. Hence,

$$dim\Sigma \left(f\right)=dim{\Gamma }_{+}(\nabla f)=1.$$

As a direct corollaries of the main result, we get the following:

Corollary 1.

Let $f:({\mathbb{C}}^n,0)⟶(\mathbb{C},0),$ $n\le 4$ be a singularity. If f is Kushnirenko nondegenerate, then

$$dim\Sigma \left(f\right)=dim{\Gamma }_{+}(\nabla f)$$

(13)

Corollary 2.

Let $f,g:({\mathbb{C}}^n,0)⟶(\mathbb{C},0),$ $n\le 4$ be Kushnirenko nondegenerate singularities. If ${\Gamma }_{+}(\nabla f)={\Gamma }_{+}(\nabla g),$ then

$$dim\Sigma \left(f\right)=dim\Sigma \left(g\right)$$

(14)

Since the Newton polyhedron is determined by a finite number of vertices, then as a direct consequence of Corollary 2, we have the following.

Corollary 3.

The dimension of the critical locus is finitely determined in the class of Kushnirenko nondegenerate singularities of n-variables, $n\le 4.$

Remark 4.

Finite determinacy means that this dimension is uniquely determined by a finite numbers of terms in the Taylor series of singularity.

Example 4.

Consider the polynomial function

$$f(x,y,z)=x^2{y}^2+y^4+z^4+x^2{y}^3+y^2{x}^3+y^4{z}^2+y^4{z}^4+x^2{y}^2{z}^4$$

The Newton diagram $\Gamma \left(f\right)$ of f is nothing but the triangle in ${\mathbb{R}}_{+}^3$ (with coordinates $(x,y,z)$), defined by the vertices $A=(2,2,0)$, $B=(0,4,0)$ and $C=(0,0,4)$. We can easily confirm that f is Kushnirenko nondegenerate singularity. Now, add

$$g(x,y,z)=x^2{y}^2+y^4+z^4$$

It is easy to confirm that g is Kushnirenko nondegenerate, as $\nabla f$ and $\nabla g$ have the same Newton polyhedron. We can easily calculate $dim\Sigma \left(g\right)=1.$ Hence, by Corollary 2, we also get $dim\Sigma \left(f\right)=1.$

4. The Proof of the Main Results

We will imitate the proof of [14], Theorem 3.2. However, the proof in the case of $dim\Sigma \left(f\right)=2$ requires more effort, which is shown in the following lemma.

Lemma 1.

Let $f:({\mathbb{C}}^4,0)⟶(\mathbb{C},0)$ be a nondegenerate singularity. If $dim\Sigma \left(f\right)=2,$ then $supp\nabla f$ does not satisfy the $\left(1\right)$- Kushnirenko condition.

Proof. 

Since f is Kushnirenko nondegenerate, by ([14], Proposition 4.3), we get

$$\Sigma \left(f\right)\subset \{z_1{z}_2{z}_3{z}_4=0\}$$

(15)

Therefore, without loss of a generality, we can assume that $dim\Sigma \left(f\right)\cap V\left(z_1\right)=2$. Now, let us expand f with respect to $z_1$

$$f(z_1,z_2,z_3,z_4)=g_0(z_2,z_3,z_4)+z_1{g}_1(z_2,z_3,z_4)+z_1^2{g}_2(z_2,z_3,z_4)+\dots$$

(16)

Hence, $dim\Sigma \left(g_0\right)\cap V\left(g_1\right)=2.$ Consider these cases:

• $g_0\equiv 0.$ Then, $z_1|f.$ Add $I=\{2,3,4\}.$ Then, $supp{f}_{z_i}^{\prime }\cap O{X}_I,i=2,3,4$ are empty sets. Hence, $supp\nabla f$ does not satisfy the (1) - Kushnirenko condition.
• $g_0\not\equiv 0.$ Since f is nondegenerate, then $g_0$ is also nondegenerate and

  $$\Sigma \left(g_0\right)\subset \{z_2{z}_3{z}_4=0\}$$

  (17)
  Therefore, without loss of a generality, we can assume that

  $$dim(\Sigma \left(g_0\right)\cap V\left(g_1\right)\cap V\left(z_2\right))=2.$$

  (18)
  Now, let us expand $g_0$ and $g_1$ with respect to $z_2$

  $$g_0(z_2,z_3,z_4)=g_{00}(z_3,z_4)+z_2{g}_{01}(z_3,z_4)+z_2^2{g}_{02}(z_3,z_4)+\dots$$

  (19)

  $$g_1(z_2,z_3,z_4)=g_{10}(z_3,z_4)+z_2{g}_{11}(z_3,z_4)+z_2^2{g}_{12}(z_3,z_4)+\dots$$

  (20)
  Hence, $dim(\Sigma g_{00}\cap V\left(g_{01}\right)\cap V\left(g_{10}\right)=2.$ Therefore, functions $g_{00},g_{01},g_{10}$ are identically equal to 0, and f has a form:

  $$f(z_1,z_2,z_3,z_4)=z_2^{2}h(z_2,z_3,z_4)+z_1{z}_{2}k(z_2,z_3,z_4)+z_1^{2}l(z_2,z_3,z_4),$$

  (21)
  $h\not\equiv 0,$ $k\not\equiv 0$ or $l\not\equiv 0.$ Add $I=\{3,4\}.$ Then, $supp{f}_{z_i}^{\prime }\cap O{X}_I,i=1,2,3,4,$ are empty sets. Hence, $supp\nabla f$ does not satisfy the (1)—Kushnirenko condition.

This concludes the proof. □

Proof of Theorem 3. 

Let $0\le d\le 4.$ By Remark 3, it is enough to prove the equivalence of conditions (i) and (ii). Since the conditions (ii) are disjoint for different $d,$ it is enough to prove only the implication from (i) to (ii). By [14], Proposition 4.1, $supp\nabla f$ satisfies the $\left(d\right)$—Kushnirenko condition. It is enough to show that $supp\nabla f$ does not satisfy the $(d-1)$—Kushnirenko condition. Let us consider these cases:

• $d=4$. Since $f\not\equiv 0$ and $ordf\ge 2,$ this case is impossible.
• $d=3.$ It is a consequence of [14], Proposition 4.5.
• $d=2.$ It is a consequence of Lemma 1.
• $d=1.$ It is a consequence of the main result of [13].

This concludes the proof. □