Arnold's problem on monotonicity of the Newton number for surface singularities

TL;DR

This paper provides a complete characterization of when the Newton number remains monotonic for surface singularities, solving Arnold's problem from 1982 by relating Newton polyhedra and their associated invariants.

Contribution

It introduces a simple condition based on Newton polyhedra that characterizes the equality of Newton numbers for surface singularities, addressing Arnold's longstanding problem.

Findings

Characterization of Newton number equality in terms of Newton polyhedra

Complete solution to Arnold's problem for surface singularities

Conditions for monotonicity of the Newton number

Abstract

According to the Kouchnirenko theorem, for a generic (precisely non-degenerate in the Kouchnirenko sense) isolated singularity $f$ its Milnor number $\mu (f)$ is equal to the Newton number $\nu (\Gamma_{+}(f))$ of a combinatorial object associated to $f$, the Newton polyhedron $\Gamma_+ (f)$. We give a simple condition characterising, in terms of $\Gamma_+ (f)$ and $\Gamma_+ (g)$, the equality $\nu (\Gamma_{+}(f)) = \nu (\Gamma_{+}(g))$, for any surface singularities $f$ and $g$ satisfying $\Gamma_+ (f) \subset \Gamma_+ (g)$. This is a complete solution to an Arnold's problem (1982-16) in this case.

Equations

$$\nu ({\Gamma }_{+}(f)):=n!V_n-(n-1)!V_{n-1}+\dots +(-1{)}^{n-1}1!V_1+(-1{)}^n{V}_0,$$

$$\text{mathbbm}P-\text{mathbbm}Q:=\overline{\text{mathbbm}P\setminus \text{mathbbm}Q}.$$

$${\Gamma }_{+}=\mathrm{conv}\left(\bigcup _{\text{mathbbm}i\in A}(\text{mathbbm}i+\text{mathbbm}{R}_{⩾0}^n)\right).$$

$${\Gamma }_{-}:=\text{mathbbm}{R}_{⩾0}^n-{\Gamma }_{+}.$$

$$\nu ({\Gamma }_{+}):=n!V_n-(n-1)!V_{n-1}+\dots +(-1{)}^{n-1}1!V_1+(-1{)}^n{V}_0,$$

$${\mathrm{vol}}_n\text{mathbbm}P(B,Q)=\frac{{\mathrm{vol}}_{n-1}(B)\mathrm{dist}(Q,H)}{n}.$$

$${\tilde{\Gamma }}_{+}=\mathrm{conv}({\Gamma }_{+}\cup \{P_1,\dots ,P_k\}),$$

$$\nu ({\Gamma }_{+})-\nu ({\Gamma }_{+}+\{P_1,\dots ,P_k\})=\sum _{1⩽i⩽k}(\nu ({\Gamma }_{+}+\{P_1,\dots ,P_{i-1}\})-\nu ({\Gamma }_{+}+\{P_1,\dots ,P_i\})),$$

$$\nu ({\Gamma }_{+})-\nu ({\Gamma }_{+}^P)=\frac{3!{\mathrm{vol}}_2(W)\cdot 1}{3}-2!{\mathrm{vol}}_2(W)=0.$$

$$\nu ({\Gamma }_{+})-\nu ({\Gamma }_{+}^P)=\frac{3!{\mathrm{vol}}_2(W)\cdot 1}{3}-2!{\mathrm{vol}}_2(W)-\frac{2!{\mathrm{vol}}_1(L)\cdot 1}{2}+{\mathrm{vol}}_1(L)=0.$$

$$\nu ({\Gamma }_{+})-\nu ({\Gamma }_{+}^P)=\nu ({\Gamma }_{+})=\frac{3!{\mathrm{vol}}_2(W_{xy})\cdot 1}{3}-2!{\mathrm{vol}}_2(W_{xy})-2!{\mathrm{vol}}_2(W_{xz})-2!{\mathrm{vol}}_2(W_{yz})+$$

$$+{\mathrm{vol}}_1(L_x)+{\mathrm{vol}}_1(L_y)+{\mathrm{vol}}_1(L_z)-1=0.$$

$$\nu ({\Gamma }_{+})-\nu ({\Gamma }_{+}^P)=3!{\mathrm{vol}}_3({\Gamma }_{-}-{\Gamma }_{-}^P)>0.$$

$$\nu ({\Gamma }_{+})-\nu ({\Gamma }_{+}^P)=\frac{3!{\mathrm{vol}}_2(W)h}{3}-2!{\mathrm{vol}}_2(W)=2{\mathrm{vol}}_2(W)(h-1)>0.$$

$$\nu ({\Gamma }_{+})-\nu ({\Gamma }_{+}^P)$$

$$=3!{\mathrm{vol}}_3(V)+2!\sum _{1⩽i⩽k}{\mathrm{vol}}_2(W_i)(h_i-1)⩾3!{\mathrm{vol}}_3(V)>0.$$

$$\nu ({\Gamma }_{+})-\nu ({\Gamma }_{+}^P)$$

$$\displaystyle=\!\begin{multlined}3!\,\operatorname{vol}_{3}(V)+2\,\sum_{1\leqslant i\leqslant k}\operatorname{vol}_{2}(W_{i})\,(h_{i}-1)+{}\\ +2\,\sum_{1\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{2}(\widetilde{W}_{j})\,(\widetilde{h}_{j}-1)+\operatorname{vol}_{1}(L)\geqslant\operatorname{vol}_{1}(L)>0.\end{multlined}3!\,\operatorname{vol}_{3}(V)+2\,\sum_{1\leqslant i\leqslant k}\operatorname{vol}_{2}(W_{i})\,(h_{i}-1)+{}\\ +2\,\sum_{1\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{2}(\widetilde{W}_{j})\,(\widetilde{h}_{j}-1)+\operatorname{vol}_{1}(L)\geqslant\operatorname{vol}_{1}(L)>0.$$

$$\nu ({\Gamma }_{+})-\nu ({\Gamma }_{+}^P)=$$

$$\displaystyle=\!\begin{multlined}3!\,\operatorname{vol}_{3}(V)+3!\,\sum_{1\leqslant i\leqslant k}\operatorname{vol}_{3}(\mathbbm{P}_{i})+3!\,\sum_{2\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{3}(\widetilde{\mathbbm{P}}_{j})+{}\\ -2!\,\sum_{1\leqslant i\leqslant k}\operatorname{vol}_{2}(W_{i})-2!\,\sum_{1\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{2}(\widetilde{W}_{j})+1!\,\operatorname{vol}_{1}(L)=\end{multlined}3!\,\operatorname{vol}_{3}(V)+3!\,\sum_{1\leqslant i\leqslant k}\operatorname{vol}_{3}(\mathbbm{P}_{i})+3!\,\sum_{2\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{3}(\widetilde{\mathbbm{P}}_{j})+{}\\ -2!\,\sum_{1\leqslant i\leqslant k}\operatorname{vol}_{2}(W_{i})-2!\,\sum_{1\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{2}(\widetilde{W}_{j})+1!\,\operatorname{vol}_{1}(L)=$$

$$\displaystyle=\!\begin{multlined}3!\,\operatorname{vol}_{3}(V)+2\,\sum_{2\leqslant i\leqslant k}\operatorname{vol}_{2}(W_{i})\,(h_{i}-1)+2\,\sum_{2\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{2}(\widetilde{W}_{j})\,(\widetilde{h}_{j}-1)+{}\\ +2\,\operatorname{vol}_{2}(W_{1})\,(h_{1}-1)-2\,\operatorname{vol}_{2}(\widetilde{W}_{1})+\operatorname{vol}_{1}(L).\end{multlined}3!\,\operatorname{vol}_{3}(V)+2\,\sum_{2\leqslant i\leqslant k}\operatorname{vol}_{2}(W_{i})\,(h_{i}-1)+2\,\sum_{2\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{2}(\widetilde{W}_{j})\,(\widetilde{h}_{j}-1)+{}\\ +2\,\operatorname{vol}_{2}(W_{1})\,(h_{1}-1)-2\,\operatorname{vol}_{2}(\widetilde{W}_{1})+\operatorname{vol}_{1}(L).$$

$$\displaystyle=\!\begin{multlined}3!\,\operatorname{vol}_{3}(V)+2\,\sum_{2\leqslant i\leqslant k}\operatorname{vol}_{2}(W_{i})\,(h_{i}-1)+2\,\sum_{2\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{2}(\widetilde{W}_{j})\,(\widetilde{h}_{j}-1){}+2\,\frac{\operatorname{vol}_{1}(L)}{2}\,\widetilde{h}_{1}\,(h_{1}-1)+\\ -2\,\frac{\operatorname{vol}_{1}(L)}{2}\,h_{1}+\operatorname{vol}_{1}(L)=\\ \end{multlined}3!\,\operatorname{vol}_{3}(V)+2\,\sum_{2\leqslant i\leqslant k}\operatorname{vol}_{2}(W_{i})\,(h_{i}-1)+2\,\sum_{2\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{2}(\widetilde{W}_{j})\,(\widetilde{h}_{j}-1){}+2\,\frac{\operatorname{vol}_{1}(L)}{2}\,\widetilde{h}_{1}\,(h_{1}-1)+\\ -2\,\frac{\operatorname{vol}_{1}(L)}{2}\,h_{1}+\operatorname{vol}_{1}(L)=\\$$

$$\displaystyle=\!\begin{multlined}3!\,\operatorname{vol}_{3}(V)+2\,\sum_{2\leqslant i\leqslant k}\operatorname{vol}_{2}(W_{i})\,(h_{i}-1)+2\,\sum_{2\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{2}(\widetilde{W}_{j})\,(\widetilde{h}_{j}-1)+{}\operatorname{vol}_{1}(L)\,(\widetilde{h}_{1}-1)\,(h_{1}-1).\end{multlined}3!\,\operatorname{vol}_{3}(V)+2\,\sum_{2\leqslant i\leqslant k}\operatorname{vol}_{2}(W_{i})\,(h_{i}-1)+2\,\sum_{2\leqslant j\leqslant\widetilde{k}}\operatorname{vol}_{2}(\widetilde{W}_{j})\,(\widetilde{h}_{j}-1)+{}\operatorname{vol}_{1}(L)\,(\widetilde{h}_{1}-1)\,(h_{1}-1).$$

Full text

1. Introduction

Let $f:(\mathbbm{C}^{n},0)\rightarrow(\mathbbm{C},0)$ be an isolated singularity (that is $f$ possesses an isolated critical point at $0\in\mathbbm{C}^{n}$), in the sequel: a singularity, in short. The Milnor number $\mu(f)$ (see [9]) of a generic $f$ can be expressed, as proved by Kouchnirenko [7], using a combinatorial object associated to $f$, the Newton polyhedron $\Gamma_{+}(f)\subset\mathbbm{R}^{n}_{\geqslant 0}$. More precisely, under an appropriate non-degeneracy condition imposed on $f$, it holds $\mu(f)=\nu(\Gamma_{+}(f))$, where $\nu(\Gamma_{+}(f))$ is the Newton number of $\Gamma_{+}(f)$. For $\Gamma_{+}(f)$ convenient (which means that the Newton polyhedron contains a point on each coordinate axis) the latter number is equal to

[TABLE]

where $V_{n}$ is the $n$-dimensional volume of the (usually non-convex) polyhedron ‘under’ $\Gamma_{+}(f)$, $V_{n-1}$ is the sum of $(n-1)$-dimensional volumes of the polyhedra ‘under’ $\Gamma_{+}(f)$ on all hyperplanes $\{x_{i}=0\}$, $V_{n-2}$ is the sum of $(n-2)$-dimensional volumes of the polyhedra ‘under’ $\Gamma_{+}(f)$ on all hyperplanes $\{x_{i}=x_{j}=0\}$, and so on.

In his acclaimed list of problems, V. I. Arnold posed the following ([1, 1982-16]):

‘Consider a Newton polyhedron $\Delta$ in $\mathbbm{R}^{n}$ and the number $\mu(\Delta)=n!V-\Sigma(n-1)!V_{i}+\Sigma(n-2)!V_{ij}-\cdots$, where $V$ is the volume under $\Delta$, $V_{i}$ is the volume under $\Delta$ on the hyperplane $x_{i}=0$, $V_{ij}$ is the volume under $\Delta$ on the hyperplane $x_{i}=x_{j}=0$, and so on.

Then $\mu(\Delta)$ grows (non strictly monotonically) as $\Delta$ grows (whenever $\Delta$ remains coconvex and integer?). There is no elementary proof even for $n=2$.’

(here, Arnold’s terminology slightly differs from ours: $\Delta$ should be understood as $\mathbbm{R}^{n}_{\geqslant 0}\leavevmode\nobreak\ \setminus\leavevmode\nobreak\ \Gamma_{+}(f)$ for a singularity $f$, and then $\mu(\Delta)=\nu(\Gamma_{+}(f))$). In the comments to the problem, S. K. Lando wrote that the monotonicity of $\mu(\Delta)$ follows from the semi-continuity of the spectrum of a singularity, proved independently by A. N. Varchenko [13] and by J. Steenbrink [12], and that he himself had given an elementary proof for $n=2$ (unpublished). Such a proof (for $n=2$) was eventually published by A. Lenarcik [8]. In the case of an arbitrary $n$, other proofs were offered by M. Furuya [4], J. Gwoździewicz [6] and C. Bivià-Ausina [3].

In the present paper we essentially complete the solution of the problem for surface singularities, i.e. for $n=3$. More specifically, we prove not only the monotonicity property of the Newton number, but also we give a simple geometrical condition characterising the situations in which this monotonicity is strict. We may describe this condition in the following intuitive way (for the precise statement see Theorem 1): for any $f,g:(\mathbbm{C}^{3},0)\rightarrow(\mathbbm{C},0)$ such that $\Gamma_{+}(f)\subset\Gamma_{+}(g)$ one has $\nu(\Gamma_{+}(f))=\nu(\Gamma_{+}(g))$ if, and only if, $\Gamma_{+}(f)$ and $\Gamma_{+}(g)$ differ by (possibly several) pyramids with bases in the coordinate planes and heights equal to $1$. The proof we propose is purely geometrical and elementary. We believe that a similar result should be valid in the $n$-dimensional case, but, if one simply tries to mimic the proof offered here, the amount of new combinatorial complications seems to increase enormously. We also expect that our result (and its potential multidimensional generalization) will have interesting applications in many aspects of effective singularity theory, e.g.: computation of the Łojasiewicz exponent, jumps of the Milnor numbers in deformations of singularities, searching for tropisms of "partial" gradient ideals $\left(\frac{\partial f}{\partial z_{1}},\ldots,\widehat{\frac{\partial f}{\partial z_{i}}},\ldots,\frac{\partial f}{\partial z_{n}}\right)\mathcal{O}_{n}$ of an isolated singularity $f$, etc.

Similar problem characterizing $f$ for which $\mu(f)$ is minimal (and equal to $\nu(\Gamma_{+}(f))$) among all singularities with the same Newton polyhedron $\Gamma_{+}(f)$ is given in the recent paper by P. Mondal [10].

2. Polyhedra

According to the standard definitions (see e.g. M. Berger [2]), a convex $n$-polyhedron in $\mathbbm{R}^{n}$ is an intersection of a finite family of closed half-spaces of $\mathbbm{R}^{n}$, having non-empty interior. An $n$-polyhedron in $\mathbbm{R}^{n}$ is a union of finitely many convex $n$-polyhedra in $\mathbbm{R}^{n}$. Let $k\leqslant n$; a $k$-polyhedron in $\mathbbm{R}^{n}$ is a finite union of $k$-polyhedra in $k$-dimensional affine subspaces of $\mathbbm{R}^{n}$. A compact connected $k$-polyhedron in $\mathbbm{R}^{n}$ is called a $k$-polytope in $\mathbbm{R}^{n}$.

For convenience, we introduce the following notations. Let $\mathbbm{P},\mathbbm{Q}\subset\mathbbm{R}^{n}$ be two $k$-polyhedra. The polyhedral difference (p-difference) of $\mathbbm{P}$ and $\mathbbm{Q}$ is the closure of their set-theoretical difference, in symbols

[TABLE]

One can check that $\mathbbm{P}-\mathbbm{Q}$ is also a $k$-polyhedron in $\mathbbm{R}^{n}$, or an empty set. We say that $\mathbbm{P}$ and $\mathbbm{Q}$ are relatively disjoint, if their relative interiors are disjoint (in appropriate $k$-dimensional affine subspaces). In particular, two $n$-polyhedra in $\mathbbm{R}^{n}$ are relatively disjoint if their interiors are disjoint.

We define the Newton polyhedra in an abstract way without any relation to singularities. A subset $\Gamma_{+}\subset\mathbbm{R}_{\geqslant 0}^{n}$ is said to be a Newton polyhedron when there exists a subset $A\subset\mathbbm{N}_{0}^{n}$ such that

[TABLE]

For such an $A$ we will write $\Gamma_{+}=\Gamma_{+}(A)$. In the sequel we will assume that are no superfluous points in $A$, implying $A$ is precisely the set of all the vertices of $\Gamma_{+}$.

Remark 1**.**

In the context of singularity theory, we take $A=\operatorname{supp}f$, where $f=\sum_{\mathbbm{i}\in\mathbbm{N}^{n}_{0}}a_{\mathbbm{i}}z^{\mathbbm{i}}$ around [math] and $\operatorname{supp}f:=\{\mathbbm{i}\in\mathbbm{N}_{0}^{n}:a_{\mathbbm{i}}\neq 0\}$.

A Newton polyhedron $\Gamma_{+}$ is called convenient if $\Gamma_{+}$ intersects all coordinate axes of $\mathbbm{R}^{n}$. Since $\mathbbm{N}_{0}^{n}$ is a lattice in $\mathbbm{R}^{n}_{\geqslant 0}$, the boundary of a convenient polyhedron $\Gamma_{+}$ is a finite union of convex $(n-1)$-polytopes (compact facets) and a finite union of convex unbounded $(n-1)$-polyhedra (unbounded facets) lying in coordinate hyperplanes. By $\Gamma$ we will denote the set of these compact facets, and sometimes – depending on the context – also their set-theoretic union. The closure of the complement of $\Gamma_{+}$ in $\mathbbm{R}^{n}_{\geqslant 0}$ will be denoted by $\Gamma_{-}$, i.e.

[TABLE]

It is an $n$-polytope in $\mathbbm{R}^{n}$ provided $\Gamma_{-}\neq\varnothing$. Hence, $\Gamma_{-}$ has finite $n$-volume. Similarly, for any $\varnothing\neq I\subset\{1,\ldots,n\}$, $\Gamma_{-}$ restricted to the coordinate hyperplane $\mathbbm{R}^{I}_{\geqslant 0}:=\{(x_{1},\ldots,x_{n})\in\mathbbm{R}^{n}_{\geqslant 0}:\hskip 2.5ptx_{i}=0\text{ for }i\not\in I\}$, that is $\Gamma_{-}^{I}:=\Gamma_{-}\cap\mathbbm{R}^{I}_{\geqslant 0}$, has finite $(\#I)$-volume. Consequently, we may define the Newton number $\nu(\Gamma_{+})$ of convenient $\Gamma_{+}$ by the formula

[TABLE]

where $V_{i}$ denotes the sum of $i$-volumes of $\Gamma_{-}^{I}$, for all $\varnothing\neq I\subset\{1,\ldots,n\}$ satisfying $\#I=i$. Note that $V_{0}=1$ if $\Gamma_{-}\neq\varnothing$ and $V_{0}=0$ if $\Gamma_{-}=\varnothing$. Hence $\nu(\mathbb{R}^{n}_{\geqslant 0})=0$. Clearly, we may also extend the domain of this definition to any $n$-polytope $\mathbbm{P}$ in $\mathbbm{R}^{n}$; thus $\nu(\mathbbm{P})$ makes sense. Then for any Newton polyhedron $\Gamma_{+}$ we have $\nu(\Gamma_{+})=\nu(\Gamma_{-})$. We will use both notations interchangeably.

The following notions will be useful in our proof. Let $B$ be a compact $(n-1)$-polyhedron in an $(n-1)$-dimensional hyperplane $H\subset\mathbbm{R}^{n}$ and $Q\in\mathbbm{R}^{n}\setminus H$. A pyramid $\mathbbm{P}(B,Q)$ with apex $Q$ and base $B$ is by definition the cone, $\operatorname{cone}(B,Q)$, with vertex $Q$ and base $B$.

By ([2, 12.2.2, p. 13]), the $n$-volume of $\mathbbm{P}(B,Q)$ can be computed using the elementary formula

[TABLE]

Finally, if $P_{0},\ldots,P_{k}\in\mathbbm{R}^{n}$ are linearly independent points, then by $\Delta(P_{0},\ldots,P_{k})$ we denote the $k$-simplex with vertices $P_{0},\ldots,P_{k}$.

3. The main Theorem

Let $\Gamma_{+}$, $\widetilde{\Gamma}_{+}$ be two convenient Newton polyhedra such that $\Gamma_{+}\varsubsetneq\widetilde{\Gamma}_{+}$. Then

[TABLE]

for some points $P_{1},\ldots,P_{k}$ lying under $\Gamma_{+}$, i.e. $P_{i}\in\mathbbm{N}_{0}^{n}\setminus\Gamma_{+}$. In such situation $\widetilde{\Gamma}_{+}$ will also be denoted by $\Gamma_{+}+\{P_{1},\ldots,P_{k}\}$ or $\Gamma_{+}^{P_{1},\ldots,P_{k}}$. Clearly, $\Gamma_{+}+\{P_{1},\ldots,P_{k}\}=(\Gamma_{+}+\{P_{1},\ldots,P_{k-1}\})+P_{k}$ and hence $\nu(\Gamma_{+}+\{P_{1},\ldots,P_{k}\})=\nu((\Gamma_{+}+\{P_{1},\ldots,P_{k-1}\})+P_{k})$. Since moreover

[TABLE]

it suffices to consider the monotonicity of the Newton number for polyhedra defined by sets which differ in one point only, i.e. for Newton polyhedra $\Gamma_{+}$ and $\Gamma_{+}^{P}$, for some $P\in\mathbbm{N}_{0}^{n}\setminus\Gamma_{+}$.

Theorem 1**.**

Let $\Gamma_{+}$ be a convenient Newton polyhedron in $\mathbbm{R}^{3}_{\geqslant 0}$ and let a lattice point $P$ lie under $\Gamma_{+}$ i.e. $P\in\mathbbm{N}_{0}^{3}\setminus\Gamma_{+}$. Then

1. (1)

$\nu(\Gamma_{+}^{P})\leqslant\nu(\Gamma_{+})$, 2. (2)

$\nu(\Gamma_{+}^{P})=\nu(\Gamma_{+})$* if, and only if, there exists a coordinate plane $H$ such that $P\in H$ and $\Gamma_{+}^{P}-\Gamma_{+}$ is a pyramid with base $(\Gamma_{+}^{P}-\Gamma_{+})\cap H$ and of height equal to $1$.*

Remark 2**.**

We believe that the same theorem is true, mutatis mutandis, in the $n$-dimensional case. In the simpler case $n=2$, the theorem is well-known ([6], [8] or [5]).

Remark 3**.**

In the particular case when $\Gamma_{+}^{P}-\Gamma_{+}$ is a $3$-dimensional simplex item 2 follows from Lemma (2.2) in [11].

Example 1**.**

Let us illustrate the second item of the theorem with some figures. Let $P$ lying under $\Gamma_{+}$ be such that $\nu(\Gamma_{+}^{P})=\nu(\Gamma_{+})$. Up to permutation of the variables, we have the following, essentially different, possible locations for $P$:

1. (1)

$P$* lies in the plane $\{z=0\}$ and not on axes (Figure 2$(a)$),* 2. (2)

$P$* lies in the plane $\{z=0\}$ and on the axis ${Ox}$ (Figure 2$(b)$).*

Remark 4**.**

Item 2 of Theorem 1 can be equivalently stated as follows:

1. 2’.

$\nu(\Gamma_{+}^{P})<\nu(\Gamma_{+})$* if, and only if, one of the following two conditions is satisfied:*

1. (a)

$P$* lies in the interior of $\Gamma_{-}$ i.e. $P\in\operatorname{Int}(\Gamma_{-})$,* 2. (b)

for each coordinate plane $H$ such that $P\in H$ the p-difference $\Gamma_{+}^{P}-\Gamma_{+}$ is either a pyramid with base $(\Gamma_{+}^{P}-\Gamma_{+})\cap H$ and of height greater or equal to $2$, or an $n$-polytope with at least two vertices outside of $H$.

Example 2**.**

The (weaker) requirement that the p-difference $\Gamma_{+}^{P}-\Gamma_{+}$ should lie in ‘a wall’ of thickness $1$ around a coordinate plane is not sufficient for the equality $\nu(\Gamma_{+}^{P})=\nu(\Gamma_{+})$. In fact, if $\Gamma_{+}$ is the Newton polyhedron of the surface singularity $f(x,y,z):=x^{6}+2y^{6}+z(x^{2}+y^{2})+z^{4}$ and $P=(3,2,0)$, then:

1. (1)

$\nu(\Gamma_{+})=15$, $\nu(\Gamma_{+}^{P})=13$, 2. (2)

$\Gamma_{+}^{P}-\Gamma_{+}$* is a $3$-polytope with ‘base’ $(\Gamma_{+}^{P}-\Gamma_{+})\cap{Oxy}$ and of height $1$, but it is not a pyramid; it has two vertices above ${Oxy}$.*

4. Proof of the Theorem

Let $\Gamma_{+}$ be a convenient Newton polyhedron in $\mathbbm{R}^{3}_{\geqslant 0}$. Let $P\in\mathbbm{N}_{0}^{3}\setminus\Gamma_{+}$ denote a lattice point under $\Gamma_{+}$.

Item 1 of the theorem will be proved in the course of the proof of item 2, because we will in fact show that the negation of the combinatorial condition in item 2 implies the strict inequality $\nu(\Gamma_{+}^{P})<\nu(\Gamma_{+})$.

We first prove that the combinatorial condition in item 2 implies the equality $\nu(\Gamma_{+})=\nu(\Gamma_{+}^{P})$. Without loss of generality we may assume that, having fixed coordinates $(x,y,z)$ in $\mathbbm{R}^{3}$, we have: $H=\{z=0\}$, $P\in H$ and $\Gamma_{+}^{P}-\Gamma_{+}$ is a pyramid with base $(\Gamma_{+}^{P}-\Gamma_{+})\cap H$ and of height equal to $1$. We must show that $\nu(\Gamma_{+}^{P})=\nu(\Gamma_{+})$. Since $\Gamma_{+}^{P}-\Gamma_{+}=\Gamma_{-}-\Gamma^{P}_{-}$, but the latter is a p-difference of two polytopes, we prefer to reason in terms of $\Gamma_{-}$ and $\Gamma_{-}^{P}$ instead.

We have three possibilities:

1.   $P$ does not lie on any axis, that is $P\not\in{Ox}$ and $P\not\in{Oy}$. Then the polytopes $\Gamma_{-}$ and $\Gamma_{-}^{P}$ are identical on ${Ox}$, ${Oy}$, ${Oz}$, ${Oxz}$ and ${Oyz}$. Denoting by $W$ the p-difference polygon of $\Gamma_{-}$ and $\Gamma^{P}_{-}$ in ${Oxy}$, we have by (1)

[TABLE]

2.   $P$ lies on ${Ox}$ or ${Oy}$ and $P\neq 0$. Up to renaming of the variables, we may assume that $P\in{Ox}$. Hence and by the assumption that $\Gamma_{+}$ is convenient the apex of the pyramid must lie in the plane $Oxz$. Denoting $W:=(\Gamma_{-}-\Gamma^{P}_{-})\cap{Oxy}$ and $L:=(\Gamma_{-}-\Gamma^{P}_{-})\cap{Ox}$, we have

[TABLE]

3. $P=0$. Then $\Gamma_{-}^{P}=\varnothing$. By the assumption that $\Gamma_{+}$ is convenient the apex of the pyramid must be $(0,0,1)$. Hence, if we denote by $L_{x}$, $L_{y}$, $L_{z}$, $W_{xy}$, $W_{xz}$, $W_{yz}$ the intersections of $\Gamma_{-}$ with coordinate axes and plane, respectively, then

[TABLE]

Let us pass to the proof of the inverse implication in item 2 of the theorem. Assume to the contrary, that the combinatorial condition in item 2 does not hold. Consider possible cases:

1.   $P$ does not lie in any coordinate plane. Then the p-difference $\Gamma_{-}-\Gamma^{P}_{-}$ is an $3$-polytope, disjoint from all the coordinate planes. Hence,

[TABLE]

2.   $P$ lies in a coordinate plane, but not on any axis. Without loss of generality, we may assume that $P\in{Oxy}\backslash({Ox}\cup{Oy})$. Then the p-difference $3$-polytope $\Gamma_{-}-\Gamma^{P}_{-}$ is disjoint from the planes ${Oxz}$ and ${Oyz}$, but $W:=(\Gamma_{-}-\Gamma^{P}_{-})\cap{Oxy}\neq\varnothing$ (Figure 3$(a)$). According to Remark 4, we should examine the following possibilities:

a)   $\Gamma_{-}-\Gamma^{P}_{-}$ is a pyramid with base $W$ and of height $h\geqslant 2$. We have

[TABLE]

b)   There are at least two vertices of $\Gamma_{-}-\Gamma^{P}_{-}$ lying above ${Oxy}$. Denote them by $Q_{1},\ldots,Q_{r}$, where $r\geqslant 2$. $W$ itself is a polygon in ${Oxy}$ of the shape depicted in Figure 3$(a)$.

Naturally, $W$ is a union of relatively disjoint triangles $W_{1},\ldots,W_{k}$, $k\geqslant 1$, with one common vertex $P$. Let us enumerate all the other vertices of $W$ according to their increasing $y$-coordinate and denote them by $R_{1},\ldots,R_{k+1}$; then we may put $W_{1}=\Delta(R_{1},R_{2},P)$, $W_{2}=\Delta(R_{2},R_{3},P)$, $\ldots$, $W_{k}=\Delta(R_{k},R_{k+1},P)$ (see Figure 3$(b)$).

We claim that each $W_{i}$ is the base of a pyramid $\mathbbm{P}_{i}$ with apex in $\{Q_{1},\ldots,Q_{r}\}$ such that $\mathbbm{P}_{i}\subset(\Gamma_{-}-\Gamma^{P}_{-})$ and, moreover, all the $\mathbbm{P}_{i}$ are pairwise relatively disjoint. In fact, each side $R_{i}R_{i+1}$ is an edge of a 2-dimensional facet of $\Gamma$, say $F_{i}$ $(i=1,\ldots,k)$. Clearly, the facets $F_{i}$ are pairwise relatively disjoint and their vertices above ${Oxy}$ are among $Q_{1},\ldots,Q_{r}$. The pyramids $\mathbbm{P}(F_{i},P)$ $(i=1,\ldots,k)$ are also pairwise relatively disjoint and lie in $\Gamma_{-}-\Gamma_{-}^{P}$. Moreover, the pyramids are all convex, $F_{i}$ being convex. For every $F_{i}$ choose arbitrarily $Q_{j_{i}}$ such that $Q_{j_{i}}$ is a vertex of $F_{i}$ $(i=1,\ldots,k)$. Define new pyramids $\mathbbm{P}_{i}:=\mathbbm{P}(W_{i},Q_{j_{i}})$. Since $W_{i}$ is a facet of $\mathbbm{P}(F_{i},P)$, we have $\mathbbm{P}_{i}\subset\mathbbm{P}(F_{i},P)$ and so $\mathbbm{P}_{i}$ are pairwise relatively disjoint and lie in $\Gamma_{-}-\Gamma_{-}^{P}$ $(i=1,\ldots,k)$, as desired.

Let $h_{i}$ be the height of $\mathbbm{P}_{i}$. Clearly, $h_{i}\geqslant 1$ $(i=1,\ldots,k)$. Moreover, the union $\bigcup_{1\leqslant i\leqslant k}\mathbbm{P}_{i}$ is not equal to $\Gamma_{-}-\Gamma_{-}^{P}$ because any edge joining a pair of vertices from $Q_{1},\ldots,Q_{r}$ (such edges always exist) does not belong to any of the pyramids $\mathbbm{P}_{1},\ldots,\mathbbm{P}_{k}$, but such an edge is an edge of the $3$-polytope $\Gamma_{-}-\Gamma_{-}^{P}$. Consequently, $V:=(\Gamma_{-}-\Gamma_{-}^{P})-\bigcup_{1\leqslant i\leqslant k}\mathbbm{P}_{i}$ is a non-empty compact $3$-polyhedron. We have

[TABLE]

3.   $P$ lies on an axis and $P\neq 0$. For definiteness, let $P\in{Ox}$. Then the p-difference $3$-polytope $\Gamma_{-}-\Gamma^{P}_{-}$ is disjoint from the plane ${Oyz}$ and from the axes ${Oy}$, ${Oz}$. Similarly as in 2(b) we divide the polygons $W:=(\Gamma_{-}-\Gamma^{P}_{-})\cap{Oxy}$ and $\widetilde{W}:=(\Gamma_{-}-\Gamma^{P}_{-})\cap{Oxz}$ into triangles $W_{1},\ldots,W_{k}$, $k\geqslant 1$, and $\widetilde{W}_{1},\ldots,\widetilde{W}_{\widetilde{k}}$, $\widetilde{k}\geqslant 1$, respectively, all of them having $P$ as the (only) common vertex.

We also arrange them such that $W_{i}=\Delta(R_{i},R_{i+1},P)$ (resp. $\widetilde{W}_{j}=\Delta(\widetilde{R}_{j},\widetilde{R}_{j+1},P)$), where the points $R_{1},\ldots,R_{k+1}$ (resp. $\widetilde{R}_{1},\ldots,\widetilde{R}_{\widetilde{k}+1}$) are enumerated according to their increasing $y$- (resp. $z$-) coordinates (see Figure 4). Observe that, in particular, $R_{1}=\widetilde{R}_{1}$.

Claim**.**

$W_{1},\ldots,W_{k}$* and $\widetilde{W}_{1},\ldots,\widetilde{W}_{\widetilde{k}}$ are bases of some $3$-pyramids $\mathbbm{P}_{1},\ldots,\mathbbm{P}_{k}$ and $\widetilde{\mathbbm{P}}_{1},\ldots,\widetilde{\mathbbm{P}}_{\widetilde{k}}$, respectively, all of which are pairwise relatively disjoint, possibly except for the pair $\mathbbm{P}_{1}$ and $\widetilde{\mathbbm{P}}_{1}$, lie in $\Gamma_{-}-\Gamma_{-}^{P}$ and whose vertices are taken from the set of vertices of $\Gamma_{+}$. Moreover, if $\mathbbm{P}_{1}$ and $\widetilde{\mathbbm{P}}_{1}$ are not relatively disjoint, then the triangle $\Delta(R_{1},R_{2},\widetilde{R}_{2})$ is a facet of $\Gamma$ and $\mathbbm{P}_{1}=\widetilde{\mathbbm{P}}_{1}=\Delta(R_{1},R_{2},\widetilde{R}_{2},P)$.*

Proof of Claim.  The segments $R_{1}R_{2}$, $R_{2}R_{3}$, $\ldots$ are edges of uniquely determined facets, say $F_{1},\ldots,\allowbreak F_{k}$, of $\Gamma$. Clearly, $F_{1},\ldots,F_{k}$ are pairwise relatively disjoint. Similarly, $\widetilde{R}_{1}\widetilde{R}_{2}$, $\widetilde{R}_{2}\widetilde{R}_{3}$, $\ldots$ are edges of uniquely determined facets, say $\widetilde{F}_{1},\ldots,\widetilde{F}_{\widetilde{k}}$, of $\Gamma$. It may happen that some $F_{i}$ are equal to some $\widetilde{F}_{j}$ (see Figure 5$(a)$).

Removing the duplicated facets from the sequence $F_{1},\ldots,F_{k},\widetilde{F}_{1},\ldots,\widetilde{F}_{\widetilde{k}}$, we obtain a new sequence $\mathbbm{F}_{1},\ldots,\mathbbm{F}_{l}$ of pairwise relatively disjoint facets of $\Gamma$ (Figure 5$(b)$) having the following properties:

• •

each side $R_{i}R_{i+1}$ $(i=1,\ldots,k)$ is an edge of a unique $\mathbbm{F}_{s}$ and, similarly, each side $\widetilde{R}_{j}\widetilde{R}_{j+1}$ $(j=1,\ldots,\widetilde{k})$ is an edge of a unique $\mathbbm{F}_{\widetilde{s}}$,  $s,\widetilde{s}\in\{1,\ldots,l\}$,

• •

each $\mathbbm{F}_{s}$ $(s=1,\ldots,l)$ has either one or at most two of its edges among $R_{1}R_{2}$, $\ldots,R_{k}R_{k+1}$, $\widetilde{R}_{1}\widetilde{R}_{2}$, $\ldots$, $\widetilde{R}_{\widetilde{k}}\widetilde{R}_{\widetilde{k}+1}$.

Upon renaming the $\mathbbm{F}_{s}$, we may assume that $\mathbbm{F}_{1}$ has $R_{1}R_{2}$ as one of its edges. For each $s\in\{1,\ldots,l\}$ we build the $3$-pyramid $\mathbbm{P}(\mathbbm{F}_{s},P)$. These pyramids are convex, pairwise relatively disjoint, lie in $\Gamma_{-}-\Gamma_{-}^{P}$ and each of them has either one or at most two of its edges among $R_{1}R_{2}$, $\ldots,R_{k}R_{k+1}$, $\widetilde{R}_{1}\widetilde{R}_{2}$, $\ldots$, $\widetilde{R}_{\widetilde{k}}\widetilde{R}_{\widetilde{k}+1}$. Hence, each $\mathbbm{P}(\mathbbm{F}_{s},P)$ $(s=1,\ldots,l)$ has either one or at most two of its facets among $W_{1}$, $\ldots,W_{k}$, $\widetilde{W}_{1}$, $\ldots$, $\widetilde{W}_{\widetilde{k}}$. In order to prove the claim, we only need to build the $\mathbbm{P}_{1},\ldots,\mathbbm{P}_{k}$, $\widetilde{\mathbbm{P}}_{1},\ldots,\widetilde{\mathbbm{P}}_{\widetilde{k}}$ as appropriately chosen pyramids hiding inside the larger $\mathbbm{P}(\mathbbm{F}_{s},P)$ $(s=1,\ldots,l)$.

Fix $s_{0}\in\{1,\ldots,l\}$. If $\mathbbm{P}(\mathbbm{F}_{s_{0}},P)$ has only one facet among $W_{1}$, $\ldots,W_{k}$, $\widetilde{W}_{1}$, $\ldots$, $\widetilde{W}_{\widetilde{k}}$, then we simply consider a pyramid whose base is equal to this facet and whose apex is chosen as any other vertex of $\mathbbm{F}_{s_{0}}$; such pyramid is then contained in $\mathbbm{P}(\mathbbm{F}_{s_{0}},P)$.

If $\mathbbm{P}(\mathbbm{F}_{s_{0}},P)$ has two facets among $W_{1}$, $\ldots,W_{k}$, $\widetilde{W}_{1}$, $\ldots$, $\widetilde{W}_{\widetilde{k}}$, then, by our construction, one of them is equal to some $W_{p}$, where $p\in\{1,\ldots,k\}$, and the other one is equal to some $\widetilde{W}_{q}$, where $q\in\{1,\ldots,\widetilde{k}\}$. Excluding $s_{0}=1$, we notice that there exist two relatively disjoint pyramids contained in $\mathbbm{P}(\mathbbm{F}_{s_{0}},P)$: one has its base equal to $W_{p}$, the other one has base $\widetilde{W}_{q}$, and both of them have their apices appropriately chosen from $\mathbbm{F}_{s_{0}}$. This follows from the fact that if $s_{0}\geqslant 2$, then the facet $\mathbbm{F}_{s_{0}}$ has at least four vertices (see Figure 5$(b)$), $R_{p}$, $R_{p+1}$, $\widetilde{R}_{q},\widetilde{R}_{q+1}$, which allows us to construct relatively disjoint pyramids $\mathbbm{P}_{p}:=\Delta(P,R_{p},R_{p+1},\widetilde{R}_{q})$ of base $W_{p}$ and $\widetilde{\mathbbm{P}}_{q}:=\Delta(P,\widetilde{R}_{q},\widetilde{R}_{q+1},R_{p+1})$ of base $\widetilde{W}_{q}$, both of them contained inside the pyramid $\mathbbm{P}(\mathbbm{F}_{s_{0}},P)$ (see Figure 6).

Now, let us treat the case $s_{0}=1$ and $\mathbbm{F}_{1}$ has two of its facets among $W_{1}$, $\ldots,W_{k}$, $\widetilde{W}_{1}$, $\ldots$, $\widetilde{W}_{\widetilde{k}}$. According to our earlier arrangements, one of these facets is equal to $W_{1}$ and the other one has to be equal to $\widetilde{W}_{1}$. If $\mathbbm{F}_{1}$ is not equal to the triangle $\Delta(R_{1},R_{2},\widetilde{R}_{2})$, i.e. $\mathbbm{F}_{1}$ has a fourth vertex, say $Q$, we may take $\mathbbm{P}_{1}:=\mathbbm{P}(W_{1},Q)$, $\widetilde{\mathbbm{P}}_{1}:=\mathbbm{P}(\widetilde{W}_{1},Q)$. These pyramids are relatively disjoint and lie in $\mathbbm{P}(\mathbbm{F}_{1},P)$. Such choice of pyramids is impossible if $\mathbbm{F}_{1}=\Delta(R_{1},R_{2},\widetilde{R}_{2})$, hence in this case we simply put $\mathbbm{P}_{1}=\widetilde{\mathbbm{P}}_{1}:=\Delta(R_{1},R_{2},\widetilde{R}_{2},P)=\mathbbm{P}(W_{1},\widetilde{R}_{2})=\mathbbm{P}(\widetilde{W}_{1},R_{2})$.

Since each $W_{1},\ldots,W_{k}$, $\widetilde{W}_{1},\ldots,\widetilde{W}_{\widetilde{k}}$ is a facet of some $\mathbbm{P}(\mathbbm{F}_{s},P)$, where $s\in\{1,\ldots,l\}$, we have constructed pyramids $\mathbbm{P}_{1},\ldots,\mathbbm{P}_{k},\widetilde{\mathbbm{P}}_{1},\ldots,\widetilde{\mathbbm{P}}_{\widetilde{k}}$ with all the required properties.  $\Diamond$

Denote by $h_{i}$ and $\widetilde{h}_{j}$ the heights of $\mathbbm{P}_{i}$ and $\widetilde{\mathbbm{P}}_{j}$, respectively. They are positive integers. Take the $p$-difference $V:=(\Gamma_{-}-\Gamma_{-}^{P})-\bigcup_{1\leqslant i\leqslant k}\mathbbm{P}_{i}-\bigcup_{1\leqslant j\leqslant\widetilde{k}}\widetilde{\mathbbm{P}}_{j}$. Then $V$ is either empty or it is a compact $3$-polyhedron. Let $L:=R_{1}P=(\Gamma_{-}-\Gamma_{-}^{P})\cap{Ox}$, that is $L$ be the p-difference $(\Gamma_{-}-\Gamma_{-}^{P})$ on the axis ${Ox}$.

According to the above claim, we have two possibilities to consider:

a)   $\mathbbm{P}_{1}\neq\widetilde{\mathbbm{P}}_{1}$. Using formula (1), we compute

[TABLE]

This gives the required inequality. Notice that in this case we actually do not use the assumption that $\Gamma_{-}-\Gamma_{-}^{P}$ is not a pyramid of height one.

b)   $\mathbbm{P}_{1}=\widetilde{\mathbbm{P}}_{1}=\Delta(R_{1},R_{2},\widetilde{R}_{2},P)$ and $\Delta(R_{1},R_{2},\widetilde{R}_{2})$ is a face of $\Gamma$. We have

[TABLE]

Since $W_{1}=\mathbbm{P}(L,R_{2})$, $\widetilde{W}_{1}=\mathbbm{P}(L,\widetilde{R}_{2})$ are two perpendicular facets of $\Delta(R_{1},R_{2},\widetilde{R}_{2},P)$, we can continue the above equality

[TABLE]

This shows that, whatever the situation, it holds $\nu(\Gamma_{+})-\nu(\Gamma^{P}_{+})\geqslant 0$ in this case. Moreover, $\nu(\Gamma_{+})-\nu(\Gamma^{P}_{+})>0$ if, and only if, one of the following three conditions is satisfied:

1. (i)

$V\neq\varnothing$, 2. (ii)

there exists $h_{i}$, $i\in\{2,\ldots,k\}$ or $\widetilde{h}_{j}$, $j\in\{2,\ldots,\widetilde{k}\}$ greater or equal to $2$, 3. (iii)

both $h_{1}$ and $\widetilde{h}_{1}$ are greater or equal to $2$.

Hence, the proof of the theorem will be finished once we show that the assumption ‘$\Gamma_{-}-\Gamma_{-}^{P}$ is not a pyramid of height one’ implies at least one of these conditions.

To this end assume that (iii) is not satisfied i.e. $h_{1}=1$ or $\widetilde{h}_{1}=1$, say $h_{1}=1$ (see Figure 7).

Since $\mathbbm{P}_{1}=\mathbbm{P}(W_{1},\widetilde{R}_{2})$ is a pyramid with base $W_{1}$ contained in a coordinate plane and of height $1$, our assumption implies that $\Gamma_{-}-\Gamma^{P}_{-}\neq\mathbbm{P}_{1}$. Hence, there exists a facet $F\in\Gamma$ of $\Gamma_{-}-\Gamma^{P}_{-}$ sharing the edge $R_{2}\widetilde{R}_{2}$ in common with the facet $\mathbbm{F}_{1}=\Delta(R_{1},R_{2},\widetilde{R}_{2})$.

If $F$ has some vertex $Q$ outside of ${Oxy}\cup{Oxz}$ (see Figure 8), then (i) holds because $F$ is not entirely covered by the pyramids $\mathbbm{P}_{2},\ldots,\mathbbm{P}_{k},\widetilde{\mathbbm{P}}_{2},\ldots,\widetilde{\mathbbm{P}}_{\widetilde{k}}$ (see explanations below) which implies $F-\bigcup_{2\leqslant i\leqslant k}\mathbbm{P}_{i}-\bigcup_{2\leqslant j\leqslant\widetilde{k}}\widetilde{\mathbbm{P}}_{j}$ lies in the boundary of $(\Gamma_{-}-\Gamma_{-}^{P})-\bigcup_{1\leqslant i\leqslant k}\mathbbm{P}_{i}-\bigcup_{1\leqslant j\leqslant\widetilde{k}}\widetilde{\mathbbm{P}}_{j}$ i.e. in the boundary of $V$. Hence $V\neq\varnothing$ i.e. (i) holds. To see $F$ is not entirely covered by the pyramids $\mathbbm{P}_{2},\ldots,\mathbbm{P}_{k},\widetilde{\mathbbm{P}}_{2},\ldots,\widetilde{\mathbbm{P}}_{\widetilde{k}}$ we consider the three possible cases:

1. $F$ has no edges in ${Oxy}\cup{Oxz}$. Then $F$ is not equal to any $F_{2},\ldots,F_{k},\widetilde{F}_{2},\ldots,\widetilde{F}_{\widetilde{k}}$ considered it the proof of Claim. Hence by construction of $\mathbbm{P}_{2},\ldots,\mathbbm{P}_{k},\widetilde{\mathbbm{P}}_{2},\ldots,\widetilde{\mathbbm{P}}_{\widetilde{k}}$ the facet $F$ is not covered by them. Precisely $F-\bigcup_{2\leqslant i\leqslant k}\mathbbm{P}_{i}-\bigcup_{2\leqslant j\leqslant\widetilde{k}}\widetilde{\mathbbm{P}}_{j}=F$.

2. $F$ has only one edge in ${Oxy}\cup{Oxz}$. Then $F$ is equal either to $F_{2}$ or to $\widetilde{F}_{2}$. Since in this case $F$ has at least $4$ vertices the pyramid either $\mathbbm{P}_{2}$ or $\widetilde{\mathbbm{P}}_{2}$, inscribed in $\mathbbm{P}(F,P)$ according to our construction of $\mathbbm{P}_{1},\ldots,\mathbbm{P}_{k},\widetilde{\mathbbm{P}}_{1},\ldots,\widetilde{\mathbbm{P}}_{\widetilde{k}}$, does not cover entirely $F$ (notice $\mathbbm{P}_{2}$ and $\widetilde{\mathbbm{P}}_{2}$ has only triangles as facets). Consequently $F-\mathbbm{P}_{2}$ or $F-\widetilde{\mathbbm{P}}_{2}$ is not empty.

3. $F$ has two edges in ${Oxy}\cup{Oxz}$, necessarily one in ${Oxy}$ and one in ${Oxz}$. Then $F=\mathbbm{F}_{i}$ for some $i\in\{2,\ldots,l\}$. Since in this case $F$ has at least $5$ vertices and two pyramids $\mathbbm{P}_{i^{\prime}}$ and $\widetilde{\mathbbm{P}}_{j}$ (for some $i^{\prime}\in\{2,\ldots,k\}$ and $j\in\{2,\ldots,\widetilde{k}\}$), inscribed in $\mathbbm{P}(\mathbbm{F}_{i},P)$, have triangles as facets then $F-\mathbbm{P}_{i^{\prime}}-\widetilde{\mathbbm{P}}_{j}$ is not empty.

If all the vertices of $F$ belong to ${Oxy}\cup{Oxz}$, then there are only three possible scenarios:

• $(\star)$

$F$ is a convex quadrilateral; then apart from $R_{2}$ and $\widetilde{R}_{2}$, it has only two further vertices, necessarily $R_{3}$ and $\widetilde{R}_{3}$ (see Figure 9$(a)$).

• $(\star\star)$

$F$ is a triangle with the third vertex lying in ${Oxy}$; this vertex is necessarily $R_{3}$ (see Figure 9(b)).

• $({\star\star}\star)$

$F$ is a triangle with the third vertex lying in ${Oxz}$; this vertex is necessarily $\widetilde{R}_{3}$ (see Figure 9(c)).

In case ($\star$) our construction of $\mathbbm{P}_{i}$, $\widetilde{\mathbbm{P}}_{j}$ shows that either $\mathbbm{P}_{2}$ or $\widetilde{\mathbbm{P}}_{2}$ has its height greater than or equal to $2$. Hence, (ii) is satisfied.

In case ($\star\star$) we are in a position to repeat the above construction of $F$, this time for the edge $R_{3}\widetilde{R}_{2}$ instead of $R_{2}\widetilde{R}_{2}$, which leads to some new facet $F^{\prime}$. To this facet $F^{\prime}$ also applies the same case analysis as the one previously performed for $F$.

In case (${\star\star}\star$) we have two options: the first one $\widetilde{h}_{1}\geqslant 2$; then our construction of $\widetilde{\mathbbm{P}}_{2}$ shows that we must have $\widetilde{h}_{2}=\widetilde{h}_{1}\geqslant 2$ so we get condition (ii). If, however, $\widetilde{h}_{1}=1$, we are in the position to repeat the above construction of $F$, this time for the edge $R_{2}\widetilde{R}_{3}$ instead of $R_{2}\widetilde{R}_{2}$, which leads to some new facet $F^{\prime}$. To this facet $F^{\prime}$ also applies the same case analysis as the one previously performed for $F$.

We see that, after finitely many steps, we either find a facet $F\in\Gamma$ of $\Gamma_{-}-\Gamma^{P}_{-}$ with one of its vertices outside of ${Oxy}\cup{Oxz}$ (then $V\neq\varnothing$ and then (i) holds), or we discover that (ii) holds.

4. $P=0$. Then $\Gamma_{-}^{P}=\varnothing$ and consequently $\nu(\Gamma_{+})-\nu(\Gamma_{+}^{P})=\nu(\Gamma_{-})-\nu(\Gamma_{-}^{P})=\nu(\Gamma_{-})$. By assumption $\Gamma_{-}$ is not a pyramid of base in a coordinate plane and height $1$. Hence $\Gamma_{+}$ intersects axes at points with coordinates greater or equal to $2$. Take auxiliary point $\widetilde{P}=(1,0,0)$ on axis $Ox$ liying under $\Gamma_{+}$. Then $\Gamma_{-}-\Gamma_{-}^{\widetilde{P}}$ is also not a pyramid of height $1$. Hence by 3 $\nu(\Gamma_{+})-\nu(\Gamma_{+}^{\widetilde{P}})>0$. Since $\Gamma_{-}^{\widetilde{P}}$ is a pyramid of height $1$ then $\nu(\Gamma_{-}^{\widetilde{P}})=0$. But $\Gamma_{-}=(\Gamma_{-}-\Gamma_{-}^{\widetilde{P}})\cup\Gamma_{-}^{\widetilde{P}}$ and of course $\nu((\Gamma_{-}-\Gamma_{-}^{\widetilde{P}})\cup\Gamma_{-}^{\widetilde{P}})=\nu(\Gamma_{-}-\Gamma_{-}^{\widetilde{P}})+\nu(\Gamma_{-}^{\widetilde{P}})$. Hence $\nu(\Gamma_{-})=\nu(\Gamma_{-}-\Gamma_{-}^{\widetilde{P}})=\nu(\Gamma_{-})-\nu(\Gamma_{-}^{\widetilde{P}})=\nu(\Gamma_{+})-\nu(\Gamma_{+}^{\widehat{P}})>0$. This ends the proof.

Corollary**.**

Let $\Gamma_{+}$ be a convenient Newton polyhedron in $\mathbb{R}^{3}_{\geqslant 0}$. Then $\nu(\Gamma_{+})\geqslant 0$. Moreover, $\nu(\Gamma_{+})=0$ if and only if $\Gamma_{+}=\mathbb{R}^{3}_{\geqslant 0}$ or $\Gamma_{+}$ intersects one of axes at the point with coordinate equal to $1$.

Remark 5**.**

The last corollary was proved by M. Furuya [4], Corollary $2.4$, in $n$-dimensional case.