On a Generic Dimension of the Critical Locus

Let f:(Cn,0)→C,0,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f:({\mathbb {C}}^n,0)\rightarrow \left( {\mathbb {C}},0\right) ,$$\end{document}n≤3,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\le 3,$$\end{document} be a nondegenerate singularity. In this article we give a combinatorial characterization of the dimension of the critical locus of f in terms of its support. We also show that this dimension can be read off from the Newton diagram of f, which solves one of Arnold’s problems in this case.


Introduction
In 1968 and 1975 Vladimir I. Arnold posed the following problems (see [1]): 1968-2 What topological characteristics of a real (complex) polynomial are computable from the Newton diagram (and the signs of the coefficients)?

Preliminaries
Let f : (C n , 0) −→ (C, 0) be a nonzero holomorphic function in an open neighborhood of 0 ∈ C n . We say that f is a singularity if f (0) = 0, grad f (0) = 0, where grad f = (f z1 , . . . , f zn ). We say that f is an isolated singularity if f is a singularity, which has an isolated critical point in the origin i.e. additionally grad f (z) = 0 for z = 0 near 0. We note N = {0, 1, 2, . . .}. Let ν∈N n a ν z ν be the Taylor expansion of f at 0. We define the set supp f = {ν ∈ N n : a ν = 0} and call it the support of f. We define We say that S ⊂ R n is a face of Γ + (f ) if S = Δ(u, Γ + (f )) for some u ∈ R n + \{0}. The vector u is called a primitive vector of S. It is easy to see that S is a closed and convex set and S ⊂ Fr(Γ + (f )), where Fr(A) denotes the boundary of A. One can prove that a face S ⊂ Γ + (f ) is compact if and only if all coordinates of its primitive vector u are positive. We call the family of all compact faces of Γ + (f ) the Newton boundary of f and denote it by Γ(f ). For every compact face S ∈ Γ(f ) we define the quasihomogeneous polynomial f S = ν∈S a ν z ν . We say that f is nondegenerate on the face S ∈ Γ(f ) if the system of equations We say that f is nondegenerate in the sense of Kouchnirenko Observe that OX I is the hyperplane spanned by axes OX i , i ∈ I. If d = 0 instead of the (0)-Kouchnirenko condition we will write simply the Kouchnirenko condition.

Main Results
In this section we give the main results of this paper. The following result was proved in [2]. The aim of this article is to move the above theorem to the case of a non-isolated singularity. Precisely we show that the dimension of the critical locus of a nondegenerate singularity is determined by its support in the case n ≤ 3. To compute this dimension it is enough to check simple combinatorial conditions imposed on the support. Let n ≤ 3.
The second result shows that the dimension of the critical locus of a nondegenerate singularity depends only on its Newton diagram.
As a direct consequence of Theorems 3.2 and 3.3 we get the following.

Proof of the Main Results
We start with the following.
Vol. 75 (2020) On a Generic Dimension of the Critical Locus Page 5 of 9 59 for some h i ∈ O n . Substitute z i = 0 for i / ∈ I to the system of equations: ∂f ∂z j1 = · · · = ∂f ∂z jp = 0.
We get a system of p equations with |I| variables. Therefore by (1) and Corollary 8 in [3, p. 81] we get which contradicts the assumption that dim 0 Σf ≤ d.
Remark 4.2. The proof of the above proposition is analogous to the proof "in one side" of the main result in [6] in the case of an isolated singularity. See also Corollary 3.12 in [9].
It turns out that the critical locus of a nondegenerate singularity lies in the sum of the coordinate hyperplanes.   Using Proposition 4.5 we give a simple characterization of a nondegenerate singularity, when its critical locus has codimension one.  Proof. Since the conditions (ii) are disjoint for different d, it is enough to prove only the implication from (i) to (ii). The case n = 1 is trivial. Assume that (i) holds and n > 1. Then by Proposition 4.1 supp f satisfies the (d)-Kouchnirenko condition. Now, we show that supp f does not satisfy the (d − 1)-Kouchnirenko condition. Consider the cases: • d = n. Then f ≡ 0 and supp f does not satisfy the (n−1)-Kouchnirenko condition. • d = n − 1. It follows from Proposition 4.5 • d = 0. It is easy to check that supp f does not satisfy the (−1)-Kouchnirenko condition. It finishes the proof for n = 2. If n = 3 and d = 1 by the main result of [2] we get supp f does not satisfy the Kouchnirenko condition. It finishes the proof for n = 3.