On Real Analytic Levi-Flat Hypersurfaces Associated with Milnor Fibrations

Abstract

Let G and X be germs of holomorphic vector fields at $0\in {\mathbb {C}}^n$. Consider the real analytic map $\psi _{G,X}:{\mathbb {C}}^n\rightarrow {\mathbb {C}}$ defined by $\psi _{G,X}(z)=\langle G(z),X(z)\rangle$, where $\langle \cdot ,\cdot \rangle$ represents the usual Hermitian product. In this paper, we investigate the following question: under which conditions on the germs of holomorphic vector fields G and X is the real analytic hypersurface $M=\{F(z)=2{\textrm{Re}}(\psi _{G,X}(z))=0\}$ Levi-flat? This problem was posed by Maria A. Soares Ruas.

1 Introduction and Statement of Results

Let M be a real analytic hypersurface at the origin $0\in {\mathbb {C}}^n$, $n\ge 2$, defined by the equation $F(z_1,\ldots ,z_n)=0$, where F is a real analytic function vanishing at 0. We say that M is nondegenerate if the Levi form

$$\begin{aligned} LF(z,\bar{z})=\sum _{1\le \alpha ,\beta \le n}g_{\alpha \bar{\beta }}z^{\alpha }\bar{z}^{\beta },\,\,\,\,\,g_{\alpha \bar{\beta }}=\left( \frac{\partial ^2 F}{\partial z^{\alpha }\partial \bar{z}^{\beta }}\right) _{0} \end{aligned}$$

(1)

is nondegenerate at 0. Otherwise, we say that M is Levi-flat. The purpose of this paper is to study the degeneracy of the Levi form of real analytic hypersurfaces obtained from real singularities with a Milnor fibration. More precisely, Milnor proved in Milnor (1968, Theorem 11.2) that if $f:({\mathbb {R}}^n,0)\rightarrow ({\mathbb {R}}^p,0)$, $n>p$, is a real analytic map whose derivative Df has rank p on a punctured neighborhood of $0\in {\mathbb {R}}^n$, then, for every sufficiently small sphere $\mathbb {S}_{\epsilon } \subset {\mathbb {R}}^n$, the mapping

$$\begin{aligned} \psi :=\frac{f}{\Vert f\Vert }: \mathbb {S}_{\epsilon }-N_K \rightarrow \mathbb {S}^{p-1} \end{aligned}$$

(2)

is a locally trivial fibration, where $K=f^{-1}(0) \cap \mathbb {S}_{\epsilon }$ is the singularity link, and $N_K$ is a tubular neighborhood of K in $\mathbb {S}_{\epsilon }$. The map $\psi$ can always be extended to $\mathbb {S}_{\epsilon }-K$ as the projection of a fibration, but this extension is not necessarily as $\dfrac{f}{\Vert f\Vert }$. Follows Ruas et al. (2002, Definition 1.1), we will say that $f:({\mathbb {R}}^n,0)\rightarrow ({\mathbb {R}}^p,0)$, $n>p$, satisfies the Milnor condition at 0 if Df has rank p on a punctured neighborhood of 0. When f satisfies the Milnor condition at 0, and furthermore, the map $\dfrac{f}{\Vert f\Vert }: \mathbb {S}_{\epsilon }-K \rightarrow \mathbb {S}^{p-1}$ is a fibration for every sufficiently small sphere $\mathbb {S}_{\epsilon } \subset {\mathbb {R}}^n$, we say that f satisfies the strong Milnor condition at 0, see for instance (Ruas et al. 2002, Definition 2.5). Maps of this type induce an open book decomposition on the sphere $\mathbb {S}_{\epsilon }$. Milnor pointed out in his book that is difficult to find examples satisfying the strong Milnor condition, (see Milnor 1968, p. 100). In Seade (1997) and Seade (1996), Seade presented a method for constructing families of nontrivial maps $f:{\mathbb {R}}^{2n} \rightarrow {\mathbb {R}}^2$ that satisfy the strong Milnor condition at 0. This construction is given as follows: let $\chi ({\mathbb {C}}^n,0)$ denote the space of all germs of holomorphic vector fields at $0\in {\mathbb {C}}^n$, and let G, X be elements in $\chi ({\mathbb {C}}^n,0)$. Consider the real analytic map

$$\begin{aligned}\psi _{G,X}:{\mathbb {C}}^{n}\cong {\mathbb {R}}^{2n}\rightarrow {\mathbb {C}}\cong {\mathbb {R}}^2\end{aligned}$$

defined by $\psi _{G,X}(z)=\langle G(z),X(z)\rangle$, where

$$\begin{aligned} \langle G(z),X(z)\rangle =\sum _{i=1}^{n}G_i(z)\cdot \overline{X}_i(z), \end{aligned}$$

(3)

is the usual Hermitian product. Note that the argument of $i\langle G(z), X(z)\rangle$ is the angle by which we rotate the field G so that it becomes orthogonal to the field X. Thus, the real analytic variety $\psi _{G,X}^{-1}(0)$, called the polar variety of G and X, is the set of points where G and X are orthogonal. Consequently, on the polar variety, the holomorphic foliations defined by the fields G and X are transversal, and their intersection gives rise to a foliation by real curves in $\psi _{G,X}^{-1}(0)$. In the particular case where X is the gradient field of a real analytic function $f:{\mathbb {R}}^{2n} \rightarrow {\mathbb {R}}^2$, the polar variety is the set of points where the foliations defined by the field G and the level curves of f are tangent. Furthermore, $\psi _{G,X}^{-1}(0)$ is a complete intersection defined by the equations

$$\begin{aligned} {\textrm{Re}}\ \langle G(z), X(z)\rangle = {\textrm{Im}}\ \langle G(z), X(z)\rangle =0. \end{aligned}$$

In Seade (1997), Seade proved that if $X=(z_1,\ldots ,z_n)$ is the radial field and $G=(\lambda _1z_1^{a_1}, \ldots ,\lambda _nz_n^{a_n})$, then $\psi _{G,X}$ satisfies the Milnor condition for any $\lambda _k \in {\mathbb {C}}^*$ and integers $a_k>1$. On the other hand, given $\sigma \in S_n$, a permutation of the set $\underline{n}:=\{1,\ldots ,n\}$, families of vector fields of the form $G=(\lambda _1z_{\sigma _1}^{a_1}, \ldots , \lambda _nz_{\sigma _n}^{a_n})$ and $X=(\beta _1z_1^{b_1}, \ldots , \beta _nz_n^{b_n})$ that satisfy the Milnor condition or the strong Milnor condition at the origin were classified by Ruas–Seade–Verjovsky (Ruas et al. 2002, Theorem 2.7).

In this paper, we consider $M=\{F(z)=0\}$ defined by

$$\begin{aligned} F(z):=2{\textrm{Re}}(\psi _{G,X}(z))=\psi _{G,X}(z)+\overline{\psi _{G,X}(z)}, \end{aligned}$$

(4)

where $G,X\in \chi ({\mathbb {C}}^n,0)$. A simple example in $({\mathbb {C}}^2,0)$ is when we consider $G(z_1,z_2)=(z_1,z_2)$ and $X=(z_2,-z_1)$. Then

$$\begin{aligned} M=\{F(z_1,z_2)=2{\textrm{Re}}(\psi _{G,X}(z))=z_1\bar{z}_2-z_2\bar{z}_1=0\} \end{aligned}$$

(5)

is a real analytic Levi-flat hypersurface at $0\in {\mathbb {C}}^2$ whose Levi foliation admits as leaves the complex curves $z_1=c\cdot z_2$, where $c\in {\mathbb {C}}$ (see Burns and Gong 2003, p. 51). Motivated by this, M.A. Soares Ruas have posed the following problem:

Problem 1

Under which conditions on the germs of holomorphic vector fields G and X is the real analytic hypersurface $M=\{F(z)=2{\textrm{Re}}(\psi _{G,X}(z))=0\}$ Levi-flat?

In order to answer Problem 1, we consider the vector fields G and X explored in Ruas et al. (2002, pp. 203–211). More specifically, our first result is as follows:

Theorem A

Let G and X be elements in $\chi ({\mathbb {C}}^ n,0)$, $n\ge 2$, of the form $G(z)=(z_1^{a_1},\ldots , z_n^{a_n})$ and $X(z)=(z_1^{b_1},\ldots , z_n^{b_n})$, where $a_k > b_k \ge 1$ are positive integers, for all $k=1,\ldots ,n$. Then $M=\{F(z)=2{\textrm{Re}}(\psi _{G,X}(z))=0\}$, i.e.,

$$\begin{aligned}M=\left\{ {\textrm{Re}}\left( \sum _{k=1}^{n} z_k^{a_k}\bar{z}_k^{b_k}\right) =0\right\} .\end{aligned}$$

is nondegenerate at $0\in {\mathbb {C}}^n$.

Our second and third theorems are motivated by the example given in (5):

Theorem B

Let G and X be elements in $\chi ({\mathbb {C}}^ n,0)$, $n\ge 2$, of the form $G(z)=(z_1^a,z_2^b)$ and $X(z)=(z_2^b,z_1^a)$ with a, b positive integers. Then $M=\{F(z)=2{\textrm{Re}}(\psi _{G,X}(z))=0\}$, i.e.,

$$\begin{aligned}M=\left\{ {\textrm{Re}}\left( z_1^{a}\bar{z}_2^{b}+z_2^{b}\bar{z}_1^{a}\right) =0\right\} \end{aligned}$$

is a Levi-flat hypersurface at $0\in {\mathbb {C}}^2$.

Theorem C

Let G and X be elements in $\chi ({\mathbb {C}}^ n,0)$, $n\ge 2$, of the form $G=(z_1^{a_1},z_2^{a_2})$ and $X=(z_2^{b_2},z_1^{b_1})$, where $a_1\ge b_1$ and $a_2 \ge b_2$ are positive integers satisfying $a_1b_2=a_2b_1$. Then $M=\{F(z)=2{\textrm{Re}}(\psi _{G,X}(z))=0\}$, i.e.,

$$\begin{aligned}M=\left\{ {\textrm{Re}}\left( z_1^{a_1}\bar{z}_2^{b_2}+z_2^{a_2}\bar{z}_1^{b_1}\right) =0\right\} \end{aligned}$$

is Levi-flat if, and only if, $a_1=b_1$ and $a_2=b_2$.

Finally, we consider a family of vector fields studied in Ruas et al. (2002, Theorem 2.1).

Theorem D

Let G and X be elements in $\chi ({\mathbb {C}}^ n,0)$, $n\ge 2$, of the form $G=(z_1^{a_1},\ldots , z_n^{a_n})$ and $X=(z_{\sigma _1}^{b_{\sigma _1}},\ldots , z_{\sigma _n}^{b_{\sigma _n}})$, where $a_k\ge b_k$ are positive integers. Let us assume that for some $\ell \in \underline{n}$, the integers $a_{\ell }, b_{\ell }, a_{\sigma _{\ell }}, b_{\sigma _{\ell }}$ satisfy the following conditions: $a_{\ell }>b_{\ell }$ and $a_{\ell }b_{\sigma _{\ell }}=b_{\ell }a_{\sigma _{\ell }}$. Then, $M=\{F(z)=2{\textrm{Re}}(\psi _{G,X}(z))=0\}$, i.e.,

$$\begin{aligned}M=\left\{ {\textrm{Re}}\left( \sum _{k=1}^{n} z_k^{a_k}\bar{z}_{\sigma _k}^{b_{\sigma _k}}\right) =0\right\} .\end{aligned}$$

is nondegenerate at $0\in {\mathbb {C}}^n$.

Following our results it seems that the property of M being Levi-flat is related to the property that the function $\psi _{G,X}$ does not satisfy the Milnor condition, see Ruas et al. (2002, Theorem 2.1).

The paper is organized as follows: In Sect. 2, we introduce the concept of real analytic Levi-flat hypersurfaces at $(\mathbb {C}^n, 0)$, shedding light on essential properties that will play a pivotal role throughout the paper. Section 3 is dedicated to proving Theorem A. In Sect. 4, we establish the validity of Theorems B and C, while Sect. 5 focuses on the proof of Theorem D. Finally, in Sect. 6, we provide examples for further illustration.

2 Levi-Flat Hypersurfaces

In this section, we will discuss real analytic Levi-flat hypersurfaces at $({\mathbb {C}}^n,0)$. These are real analytic hypersurfaces whose regular part is foliated by immersed complex submanifolds of codimension one. Levi-flat hypersurfaces naturally arise in the theory of foliations as invariant subsets. In general, germs of codimension one holomorphic foliations that leave invariant hypersurfaces of this type admit a meromorphic first integral (see Cerveau and Lins-Neto 2011, Theorem 1). On the other hand, there are examples of holomorphic webs that leave invariant Levi-flat hypersurfaces (see Da Silva and Fernández-Pérez 2023; Fernández-Pérez 2013; Shafikov and Sukhov 2015). Levi-flat hypersurfaces are a central focus of the development in this paper.

Let $M=\{F(z)=0\}$ be a germ of real analytic hypersurface at $0\in {\mathbb {C}}^n$, where $F:({\mathbb {C}}^n,0)\rightarrow (\mathbb {R},0)$ is a real analytic function at $0\in {\mathbb {C}}^n$. The singular set of M is denoted by ${\textsf {Sing}}(M)$ and defined by

$$\begin{aligned}{\textsf {Sing}}(M):=\{F(z)=0\}\cap \{dF(z)=0\}.\end{aligned}$$

We define the regular part of M as $M^{*}:=\{F(z)=0\}{\setminus }\{dF(z)=0\}$. In $M^{*}$, the Levi distribution is given by $L_p:=Ker(\partial F(p)) \subset T_p M^{*}$, where $p \in M^{*}$. Note that $L_p$ is the unique complex hyperplane contained in $T_p M^{*}$.

Definition 2.1

We say that M is Levi-flat if the Levi distribution on $M^{*}$ is integrable. In this case, the Levi distribution induces a foliation on $M^{*}$ called the Levi foliation, denoted by $\mathcal {L}$.

The Levi distribution can also be given by the 1-form $\eta =i(\partial {F}-\bar{\partial } F)$ called the Levi 1-form. Thus, the integrability of the Levi distribution is equivalent to the integrability of the form $\eta$ in the sense of Frobenius, that is, $\eta$ is integrable if and only if $\left. \eta \wedge d\eta \right| _{M^*} \equiv 0$.

The simplest example of a Levi-flat hypersurface is given below.

Example 2.1

In ${\mathbb {C}}^n$ with coordinates $(z_1,\ldots ,z_n)$, consider $M=\{Im(z_n)=0\}.$ Then M is a smooth Levi-flat real analytic hypersurface, meaning ${\textsf {Sing}}(M)=\emptyset$. The Levi distribution on M is given by $L_p = \textrm{Ker} (dz_n(p))$, $p \in M^*.$ The leaves of the Levi foliation on M are given by $\{z_n=c\}$ where $c\in {\mathbb {R}}$.

Let’s consider a slightly more elaborate example given by Brunella (2007, Example 1.2).

Example 2.2

With coordinates (z, w) in ${\mathbb {C}}^2$ such that $z=x+iy$ and $w=s+it$, the real analytic hypersurface M given by

$$\begin{aligned}M=\{(z,w) \in {\mathbb {C}}^2: t^2=4(y^2+s)y^2\}\end{aligned}$$

is Levi-flat, with singular set ${\textsf {Sing}}(M)=\{t=y=0\}$. The leaves of the Levi foliation on $M^*$ are given by $L_c=\{w=(z+c)^2: Im(z)\ne 0\}$ with $c \in {\mathbb {R}}$.

The next result provides the local form of a smooth Levi-flat hypersurface. Essentially, it tells us that, at regular points, every Levi-flat hypersurface is locally similar to the Example 2.1.

Theorem 2.1

[Cartan’s theorem (Cartan 1933)] Let $M \subset {\mathbb {C}}^n$ be a real analytic Levi-flat hypersurface. In a neighborhood of each point $p \in M^*$, there exists a holomorphic coordinate system $z=(z_1, \ldots , z_n)$ such that $M=\{Im(z_n)=0\}.$

A criterion for the integrability of the Levi form is given in the next proposition.

Proposition 2.2

Let $M=\{F(z)=0\}$ be a germ of a real analytic hypersurface at $0\in {\mathbb {C}}^n$. Then, M is Levi-flat if and only if $\partial F(p) \wedge \overline{\partial }F(p) \wedge \partial \overline{\partial } F(p) = 0$ for all $p \in M$.

Proof

Let $\eta =i(\partial F-\bar{\partial }F)$ denote the Levi 1-form of M. Assuming M is Levi-flat, this implies $\eta \wedge d\eta |_{M^{*}}=0$, which is equivalent to $(\partial {F}-\bar{\partial }F)\wedge \partial \bar{\partial } F|_{M^{*}}=0$. Consequently, we have $\bar{\partial }F\wedge \partial \bar{\partial } F|_{M^{*}}=\partial F\wedge \partial \bar{\partial } F|_{M^{*}}$. In particular, $\partial F(p) \wedge \overline{\partial }F(p) \wedge \partial \overline{\partial } F(p) = 0$ for all $p \in M$.

Conversely, the condition $\partial F(p) \wedge \overline{\partial }F(p) \wedge \partial \overline{\partial } F(p) = 0$ for all $p \in M$ is equivalent to $(\partial {F}-\bar{\partial }F)\wedge \partial \bar{\partial } F\wedge dF |_{M^{*}}=0$. Hence

$$\begin{aligned} (\partial {F}-\bar{\partial }F)\wedge \partial \bar{\partial } F\wedge dF=F\theta , \end{aligned}$$

(6)

where $\theta$ is a 4-form in some open subset of ${\mathbb {C}}^n$. Since $\theta \wedge dF=0$, we can express $\theta$ as $\theta =\beta \wedge dF$, where $\beta$ is a 3-form in some open subset of ${\mathbb {C}}^n$. Substituting this into Eq. (6), we obtain

$$\begin{aligned} \left[ (\partial {F}-\bar{\partial }F)\wedge \partial \bar{\partial } F-F\beta \right] \wedge dF=0.\end{aligned}$$

Thus, there exists a 2-form $\kappa$ such that

$$\begin{aligned}(\partial {F}-\bar{\partial }F)\wedge \partial \bar{\partial } F=F\beta +\kappa \wedge dF,\end{aligned}$$

this expression implies that $(\partial {F}-\bar{\partial }F)\wedge \partial \bar{\partial } F|_{M^{*}}=0$, leading to the integrability of $\eta$. $\square$

Now, let’s verify that the regular part of M is mapped to the regular part of $M'$ for M and $M'$ being biholomorphic (not necessarily Levi-flat).

Lemma 2.3

Let $u \in M'$. Then $z=z(u) \in {\textsf {Sing}}(M)$ if and only if $u \in {\textsf {Sing}}(M')$.

Proof

Denote $M'= \{G(u)=0\}$ where $G:({\mathbb {C}}^n, 0) \rightarrow ({\mathbb {R}}), 0$ is a real analytic function at $0 \in {\mathbb {C}}^n$.

We have $dG(u)=\partial G(u)+\overline{\partial }G(u)$, where

$$\begin{aligned} \partial G = \sum _{j=1}^n \frac{\partial G}{\partial u_j}du_j,\,\,\,\,\,\, \overline{\partial } G = \sum _{j=1}^n \frac{\partial G}{\partial \overline{u}_j}d\overline{u}_j. \end{aligned}$$

A point $u \in M'$ belongs to the singular set Sing($M'$) if and only if all partial derivatives $\dfrac{\partial G}{\partial u_j}(u), \dfrac{\partial G}{\partial \bar{u}_j}(u)$ are identically zero. By the chain rule, we obtain

$$\begin{aligned}\frac{\partial G}{\partial u_j}(u)=\sum _{\alpha =1}^n \frac{\partial G}{\partial z_{\alpha }}(z(u))\frac{\partial z_{\alpha }}{\partial u_j}(u),\end{aligned}$$

thus, it follows that

$$\begin{aligned} \left( \begin{array}{ccc} \frac{\partial G}{\partial u_1}(u)&\cdots&\frac{\partial G}{\partial u_n}(u) \end{array}\right) = \left( \begin{array}{ccc} \frac{\partial G}{\partial z_1}(z(u))&\cdots&\frac{\partial G}{\partial z_n}(z(u))\end{array}\right) \left( \begin{array}{ccc} \frac{\partial z_1}{\partial u_1}(u) &{} \cdots &{} \frac{\partial z_1}{\partial u_n}(u)\\ \vdots &{} \ddots &{} \vdots \\ \frac{\partial z_n}{\partial u_1}(u) &{} \cdots &{} \frac{\partial z_n}{\partial u_n}(u)\\ \end{array}\right) . \end{aligned}$$

Since $z=z(u)$ is a biholomorphism, the change of coordinates matrix is invertible, and therefore,

$$\begin{aligned}\left( \begin{array}{ccc} \frac{\partial G}{\partial u_1}(u)&\cdots&\frac{\partial G}{\partial u_n}(u) \end{array}\right) =0 \iff \left( \begin{array}{ccc} \frac{\partial G}{\partial z_1}(z(u))&\cdots&\frac{\partial G}{\partial z_n}(z(u))\end{array}\right) =0.\end{aligned}$$

$\square$

In the next proposition, we will see that the Levi-flat property is invariant under a change of coordinates. More specifically, we establish the following result.

Proposition 2.4

Let $G: V \subset {\mathbb {C}}^n \rightarrow {\mathbb {R}}$ be a real-analytic function in coordinates $z=(z_1,\ldots , z_n)$ and let $z=z(u)$ be a change of coordinates, that is, a biholomorphism from the open set $V \subset {\mathbb {C}}^n$ to an open set $U \subset {\mathbb {C}}^n$. Then, the hypersurface $M'=\{u \in U: G(u)=G(z(u))=0\}$ is Levi-flat if and only if $M=\{z \in V: G(z)=0\}$ is Levi-flat.

Proof

The partial derivatives of G in the coordinates $z=(z_1,\ldots ,z_n)$ are given by

$$\begin{aligned} \partial G = \sum _{\alpha =1}^n \frac{\partial G}{\partial z_{\alpha }}dz_{\alpha },\,\,\,\,\,\,\, \overline{\partial } G = \sum _{\beta =1}^n \frac{\partial G}{\partial \overline{z}_{\beta }}d\overline{z}_{\beta },\,\,\,\,\, \partial \overline{\partial } G =\!\!\sum _{\delta ,\beta =1}^n\frac{\partial ^2 G}{\partial z_{\delta }\partial \overline{z}_{\beta }} d z_{\delta } \wedge d\overline{z}_{\beta }. \end{aligned}$$

Therefore, in coordinates $z=(z_1, \ldots , z_n)$, we have

$$\begin{aligned} \partial G \wedge \overline{\partial } G\wedge \partial \overline{\partial } G = \sum _{\alpha , \beta , \gamma , \delta }\left( \frac{\partial G}{\partial z_{\alpha }}\frac{\partial G}{\partial \overline{z}_{\beta }}\frac{\partial ^2 G}{\partial z_{\delta } \partial \overline{z}_{\gamma }}\right) d z_{\alpha } \wedge d\overline{z}_{\beta } \wedge d z_{\delta } \wedge d\overline{z}_{\gamma }. \end{aligned}$$

(7)

Now, by making the change of coordinates $z=z(u)$, we obtain

$$\begin{aligned} dz_{\alpha }=\sum _{j=1}^n \frac{\partial z_{\alpha }}{\partial u_j}du_j,\,\,\,\,\,\,\,\, d\overline{z}_{\beta }=\sum _{k=1}^n \frac{\partial \overline{z}_{\beta }}{\partial \overline{u}_k}d\overline{u}_k, \end{aligned}$$

from which it follows that

$$\begin{aligned}d z_{\alpha } \wedge d\overline{z}_{\beta } \wedge d z_{\delta } \wedge d\overline{z}_{\gamma }=\sum _{j,k,\ell ,m} \frac{\partial z_{\alpha }}{\partial u_j}\frac{\partial \overline{z}_{\beta }}{\partial \overline{u}_k}\frac{\partial z_{\delta }}{\partial u_{\ell }}\frac{\partial \overline{z}_{\gamma }}{\partial \overline{u}_m}du_j \wedge d\overline{u}_k\wedge du_{\ell } \wedge d\overline{u}_m.\end{aligned}$$

Hence, in coordinates $u=(u_1,\ldots ,u_n)$,

$$\begin{aligned} \partial G \wedge \overline{\partial } G \wedge \partial \overline{\partial } G = \! \!\!\sum _{\alpha , \beta , \gamma , \delta }\!\left[ \!\!\left( \frac{\partial G}{\partial z_{\alpha }}\frac{\partial G}{\partial \overline{z}_{\beta }}\frac{\partial ^2 G}{\partial z_{\delta } \partial \overline{z}_{\gamma }}\right) \!\!\! \left( \sum _{j,k,\ell , m}\!\! \!\left( \frac{\partial z_{\alpha }}{\partial u_j}\frac{\partial \overline{z}_{\beta }}{\partial \overline{u}_k}\frac{\partial z_{\delta }}{\partial u_{\ell }}\frac{\partial \overline{z}_{\gamma }}{\partial \overline{u}_m}\right) \!d u_j \!\wedge \! d\overline{u}_k\! \wedge \! d u_{\ell }\! \wedge \! d\overline{u}_m\! \right) \!\!\right] . \end{aligned}$$

From the expression above, combined with Eq. (7), it follows that $M'$ is Levi-flat if and only if M is Levi-flat. This concludes the proof of the Proposition 2.4. $\square$

2.1 Complexification

Let $F:({\mathbb {C}}^n,0)\rightarrow (\mathbb {R},0)$ be a real analytic function at $0\in {\mathbb {C}}^n$. The complexification of F is defined by

$$\begin{aligned}F_{{\mathbb {C}}}(z,w)=\sum _{\mu ,\nu }F_{\mu ,\nu }z^{\mu }w^{\nu },\end{aligned}$$

where $F(z)=\sum _{\mu ,\nu }F_{\mu ,\nu }z^{\mu }\overline{z}^{\nu }$ is a power series of F convergent in a neighborhood of the origin. We observe that $F_{{\mathbb {C}}}$ is holomorphic at $({\mathbb {C}}^n\times {\mathbb {C}}^n,0)$. The complexification of M is defined as $M_{{\mathbb {C}}}=\{F_{{\mathbb {C}}}=0\},$ and the complexification of the Levi 1-form is given by

$$\begin{aligned}\eta _{{\mathbb {C}}}=i\sum _{k=1}^n\left( \frac{\partial F_{{\mathbb {C}}}}{\partial z_k}dz_k - \frac{\partial F_{{\mathbb {C}}}}{\partial w_k}dw_k\right) .\end{aligned}$$

Given that $\eta$ is integrable on $M^{*}$, it follows that $\eta _{{\mathbb {C}}}$ is also integrable on $M_{{\mathbb {C}}}^{*}$. It’s worth noting that we can express $dF_{{\mathbb {C}}}=\alpha + \beta$ and $\eta _{{\mathbb {C}}}=i(\alpha - \beta )$, where

$$\begin{aligned} \alpha =\sum _{k=1}^n\frac{\partial F_{{\mathbb {C}}}}{\partial z_k}dz_k \, \text {and} \, \beta =\sum _{k=1}^n\frac{\partial F_{{\mathbb {C}}}}{\partial w_k}dw_k. \end{aligned}$$

(8)

Furthermore, we observe that $\alpha$ and $\beta$ define the same foliation as $\eta _{{\mathbb {C}}}$ on $M^*_{{\mathbb {C}}}$. Thus, the integrability of $\eta _{{\mathbb {C}}}$ is equivalent to

$$\begin{aligned} \alpha (z,w)\wedge d\alpha (z,w)\wedge \beta (z,w)=0\,\,\,\,\,\text { for all}\,\,\,\, (z,w) \in M^{*}_{{\mathbb {C}}}. \end{aligned}$$

(9)

This condition will be used later to verify that the real part of a certain family of polar varieties is not Levi-flat.

3 Proof of Theorem A

Before proving Theorem A, we will make some considerations for the case where the vector fields G and X are given by $G=(z_1^{a_1},\ldots , z_n^{a_n})$ and $X=(z_1^{b_1},\ldots , z_n^{b_n})$, with $a_k, b_k$ positive integers such that $a_k\ge b_k$, for $k \in \underline{n}$. In this case, the Hermitian product of G and X is given by

$$\begin{aligned}\langle G(z), X(z) \rangle = \sum _{k=1}^{n} z_{k}^{a_k} \overline{z}_{k}^{b_k}\end{aligned}$$

and

$$\begin{aligned} F(z)=2{\textrm{Re}}\langle G(z), X(z) \rangle = \sum _{k=1}^{n}z_{k}^{a_k} \overline{z}_{k}^{b_k} + z_{k}^{b_k} \overline{z}_{k}^{a_k}. \end{aligned}$$

(10)

Note that, if $a_k=b_k$ for every $k \in \underline{n}$, then $M=\{F=0\}$ is a point. Indeed, if $a_k=b_k$ for every $k \in \underline{n}$, we have

$$\begin{aligned}F(z)=\sum _{k=1}^n2|z_k|^{a_k}\end{aligned}$$

and consequently, $M=\{0\}$. Therefore, in the statement of Theorem A, we do not consider $a_k=b_k$ for every $k \in \underline{n}$.

Regarding the singular set of M, we have the following proposition:

Proposition 3.1

Let $G=(z_1^{a_1}, \ldots , z_n^{a_n})$ and $X=(z_1^{b_1}, \ldots , z_n^{b_n})$, with $a_k\ge b_k\ge 1$, and $M=\{F=0\}$, where $F(z)=2{\textrm{Re}}\langle G(z), X(z) \rangle$. Then ${\textsf {Sing}}(M)=\{0\}$.

Proof

Taking the partial derivatives of F, we have

$$\begin{aligned}\partial F= \sum _{k=1}^{n} \left( a_{k}z_{k}^{(a_{k} -1)} \overline{z}_{k}^{b_k} + b_{k}z_{k}^{(b_{k} -1)} \overline{z}_{k}^{a_k} \right) dz_{k}\end{aligned}$$

and

$$\begin{aligned} \overline{\partial } F=\sum _{k=1}^{n} \left( b_{k}z_{k}^{a_{k}} \overline{z}_{k}^{(b_{k} -1)} + a_{k}z_{k}^{b_{k}} \overline{z}_{k}^{(a_{k}-1)} \right) dz_{k}.\end{aligned}$$

Then $dF(z)= \partial F(z) + \overline{\partial } F(z)=0$ if and only if

$$\begin{aligned} \left\{ \begin{array}{cc} a_{1}z_{1}^{(a_{1} -1)} \overline{z}_{1}^{b_1} + b_{1}z_{1}^{(b_1-1)} \overline{z}_{1}^{a_1} &{} =0 \\ b_{1}z_{1}^{a_{1}} \overline{z}_{1}^{(b_{1} -1)} + a_{1}z_{1}^{b_{1}} \overline{z}_{1}^{(a_{1}-1)} &{} =0 \\ \vdots \\ a_{n}z_{n}^{(a_{n} -1)} \overline{z}_{n}^{b_n} + b_{n}z_{n}^{(b_{n} -1)} \overline{z}_{n}^{a_n}&{} =0 \\ b_{n}z_{n}^{a_{n}} \overline{z}_{n}^{(b_{n} -1)} + a_{n}z_{n}^{b_{n}} \overline{z}_{n}^{(a_{n}-1)} &{} =0 \end{array} \right. \end{aligned}$$

Clearly, $0 \in \textrm{Sing}(M)$. Now let’s denote $I:=\{k \in \underline{n}; a_k=b_k\}$ and $J:=\{k \in \underline{n}; a_k>b_k\}$. We have $\underline{n}=I \cup J$ and $I \cap J=\emptyset$. Consider the following equations from the system above:

$$\begin{aligned} a_{k}z_{k}^{(a_{k} -1)} \overline{z}_{k}^{b_k} + b_{k}z_{k}^{(b_k-1)} \overline{z}_{k}^{a_k}&=0 \end{aligned}$$

(11)

$$\begin{aligned} b_{k}z_{k}^{a_{k}} \overline{z}_{k}^{(b_{k} -1)} + a_{k}z_{k}^{b_{k}} \overline{z}_{k}^{(a_{k}-1)}&=0 \end{aligned}$$

(12)

If $k \in I$, the equations above are rewritten as:

$$\begin{aligned} 2a_{k}z_{k}^{(a_{k} -1)} \overline{z}_{k}^{a_k}&=0 \\ 2a_{k}z_{k}^{a_{k}} \overline{z}_{k}^{(a_{k}-1)}&=0 \end{aligned}$$

and it follows that $z=(z_1,\ldots ,z_n) \in {\textsf {Sing}}(M)$ implies $z_k=0$, for all $k \in I$. Now, let’s assume that $z=(z_1,\ldots , z_n) \in {\textsf {Sing}}(M)$ and $z_k \ne 0$, for some $k \in J$. From (11), we obtain

$$\begin{aligned}a_{k}z_{k}^{(a_{k} -1)} \overline{z}_{k}^{b_k} = -b_{k}z_{k}^{(b_k-1)} \overline{z}_{k}^{a_k},\end{aligned}$$

that yields,

$$\begin{aligned}\frac{a_k}{b_k}=-\left( \frac{\overline{z}_k}{z_k}\right) ^{c_k}.\end{aligned}$$

From (12) we get

$$\begin{aligned}b_{k}z_{k}^{a_{k}} \overline{z}_{k}^{(b_{k} -1)} = -a_{k}z_{k}^{b_{k}} \overline{z}_{k}^{(a_{k}-1)},\end{aligned}$$

that is,

$$\begin{aligned}\frac{b_k}{a_k}=-\left( \frac{\overline{z}_k}{z_k}\right) ^{c_k}.\end{aligned}$$

Thus, from (11) and (12), we deduce $\frac{a_k}{b_k}= \frac{b_k}{a_k}$, but this is a contradiction, because $a_k > b_k$. Hence, ${\textsf {Sing}}(M) = \{0\}$ and $M^* = M - \{0\}.$ $\square$

Now, we prove a technical lemma that will be used in the proof of Theorem A.

Lemma 3.2

Let $M_{\mathbb {C}}$ be the complexification of $M=\left\{ F(z)=\displaystyle 2{\textrm{Re}}\left( \sum _{k=1}^{n}z_{k}^{a_k} \overline{z}_{k}^{b_k}\right) =0\right\}$, $n\ge 3$. Consider the functions

$$\begin{aligned} g_j(z,w)&=a_jw_j^{(a_j-1)}z_j^{b_j} + b_jw_j^{(b_j-1)}z_j^{a_j}\\ f_k(z,w)&=a_kz_k^{(a_k-1)}w_k^{b_k} + b_kz_k^{(b_k-1)}w_k^{a_k}\\ h_{\ell }(z,w)&=a_{\ell }b_{\ell }\left( z_{\ell }^{(a_{\ell }-1)}w_{\ell }^{(b_{\ell }-1)}+ z_{\ell }^{(b_{\ell }-1)}w_{\ell }^{(a_{\ell }-1)}\right) , \end{aligned}$$

where $a_m> b_m\ge 1$ are integers for each $m=1,\ldots ,n$, and $n\ge 3$. For each triple $(j, k, \ell )$ of indices $j,k,\ell =1,\ldots ,n$; there exists $(z_0,w_0) \in M_{{\mathbb {C}}}$ such that

$$\begin{aligned}f_k(z_0,w_0)g_j(z_0,w_0)h_{\ell }(z_0,w_0)\ne 0.\end{aligned}$$

Proof

In this case, the complexification of F is given by

$$\begin{aligned} F_{{\mathbb {C}}}(z,w)=\sum _{k=1}^n z_k^{a_k}w_k^{b_k}+w_k^{a_k}z_k^{b_k}. \end{aligned}$$

(13)

For each triple $(j, k, \ell )$, let’s choose $(z_0, w_0) = (z_1,\ldots , z_n, w_1,\ldots , w_n)$ satisfying:

• 1. $z_j=z_k=z_{\ell }=w_j=w_k=1$,

• 2. $w_{\ell }$ such that $w_{\ell }^{b_{\ell }}+w_{\ell }^{a_{\ell }}=-4$,

• 3. $z_m=w_m=0$ for the remaining indices.

We observe that $(z_0,w_0) \in M_{{\mathbb {C}}}$. In fact, we have

$$\begin{aligned} F_{{\mathbb {C}}}(z_0,w_0)&=z_j^{a_j}w_j^{b_j}+w_j^{a_j}z_j^{b_j}+z_k^{a_k}w_k^{b_k}+w_k^{a_k}z_k^{b_k}+ z_{\ell }^{a_{\ell }}w_{\ell }^{b_{\ell }}+w_{\ell }^{a_{\ell }}z_{\ell }^{b_{\ell }}\\&=4+w_{\ell }^{b_{\ell }}+w_{\ell }^{a_{\ell }} \end{aligned}$$

then $F_{{\mathbb {C}}}(z_0,w_0)=0$ on $M_{{\mathbb {C}}}^*$. From item (3) above, we obtain

$$\begin{aligned}w_{\ell }^{b_{\ell }-1}+w_{\ell }^{a_{\ell }-1}=-\frac{4}{w_{\ell }}.\end{aligned}$$

Furthermore, we have

$$\begin{aligned}g_j(z_0,w_0)=a_j+b_j,\end{aligned}$$

$$\begin{aligned}f_k(z_0,w_0)=a_k+b_k,\end{aligned}$$

$$\begin{aligned}h_{\ell }(z_0,w_0){} & {} =a_{\ell }b_{\ell }\left( w_{\ell }^{(b_{\ell }-1)}+w_{\ell }^{(a_{\ell }-1)}\right) \\{} & {} =-\frac{4a_{\ell }b_{\ell }}{w_{\ell }}.\end{aligned}$$

Hence

$$\begin{aligned}f_k(z_0,w_0)g_j(z_0,w_0)h_{\ell }(z_0,w_0){} & {} =-\frac{4a_{\ell }b_{\ell }}{w_{\ell }}(a_j+b_j)(a_k+b_k)\\{} & {} \ne 0.\end{aligned}$$

$\square$

Now, let’s restate Theorem A for completeness.

Theorem A

Let G and X be elements in $\chi ({\mathbb {C}}^ n,0)$, $n\ge 2$, of the form $G(z)=(z_1^{a_1},\ldots , z_n^{a_n})$ and $X(z)=(z_1^{b_1},\ldots , z_n^{b_n})$, where $a_k > b_k \ge 1$ are positive integers, for all $k=1,\ldots ,n$. Then $M=\{F(z)=2{\textrm{Re}}(\psi _{G,X}(z))=0\}$, i.e.,

$$\begin{aligned}M=\left\{ {\textrm{Re}}\left( \sum _{k=1}^{n} z_k^{a_k}\bar{z}_k^{b_k}\right) =0\right\} .\end{aligned}$$

is nondegenerate at $0\in {\mathbb {C}}^n$.

Proof

We divide the proof into two cases $n=2$ and $n\ge 3$. For $n=2$, we examine the Eq. (9) associated to the complexification of $M=\{F=0\}$, where

$$\begin{aligned} F(z)=z_1^{a_1}\overline{z}_1^{b_1}+z_2^{a_2}\overline{z}_2^{b_2}+\overline{z}_1^{a_1}z_1^{b_1}+\overline{z}_2^{a_2}z_2^{b_2}. \end{aligned}$$

(14)

Note that, the complexification of F is given by

$$\begin{aligned} F_{{\mathbb {C}}}(z,w)=z_1^{a_1}w_1^{b_1}+z_2^{a_2}w_2^{b_2}+w_1^{a_1}z_1^{b_1}+w_2^{a_2}z_2^{b_2} \end{aligned}$$

(15)

then $dF_{{\mathbb {C}}}=\alpha + \beta$, where

$$\begin{aligned} \alpha&=(a_1z_1^{(a_1-1)}w_1^{b_1} + b_1z_1^{(b_1-1)}w_1^{a_1})dz_1 + (a_2z_2^{(a_2-1)}w_2^{b_2}+b_2z_2^{(b_2-1)}w_2^{a_2})dz_2, \\ \beta&=(a_1w_1^{(a_1-1)}z_1^{b_1} + b_1w_1^{(b_1-1)}z_1^{a_1})dw_1 + (a_2w_2^{(a_2-1)}z_2^{b_2}+b_2w_2^{(b_2-1)}z_2^{a_2})dw_2. \end{aligned}$$

Now we have

$$\begin{aligned}d\alpha= & {} a_1b_1\left( z_1^{(a_1-1)}w_1^{(b_1-1)}+ z_1^{(b_1-1)}w_1^{(a_1-1)})dw_1\wedge dz_1\right. \\{} & {} \left. + a_2b_2(z_2^{(a_2-1)}w_2^{(b_2-1)}+z_2^{(b_2-1)}w_2^{(a_2-1)}\right) dw_2\wedge dz_2.\end{aligned}$$

We use the following notations

$$\begin{aligned} \alpha _1&= \left( a_1z_1^{(a_1-1)}w_1^{b_1} + b_1z_1^{(b_1-1)}w_1^{a_1}\right) ,\\ \alpha _2&= \left( a_2z_2^{(a_2-1)}w_2^{b_2}+b_2z_2^{(b_2-1)}w_2^{a_2}\right) ,\\ \beta _1&= \left( a_1w_1^{(a_1-1)}z_1^{b_1} + b_1w_1^{(b_1-1)}z_1^{a_1}\right) ,\\ \beta _2&= \left( a_2w_2^{(a_2-1)}z_2^{b_2}+b_2w_2^{(b_2-1)}z_2^{a_2}\right) ,\\ g_1&= a_1b_1\left( z_1^{(a_1-1)}w_1^{(b_1-1)}+ z_1^{(b_1-1)}w_1^{(a_1-1)}\right) ,\\ g_2&= a_2b_2\left( z_2^{(a_2-1)}w_2^{(b_2-1)}+z_2^{(b_2-1)}w_2^{(a_2-1)}\right) . \end{aligned}$$

So, we have $\alpha \wedge d\alpha \wedge \beta = (\alpha _1\beta _1 g_2 +\alpha _2 \beta _2 g_1) dz_1\wedge dz_2 \wedge dw_1 \wedge dw_2.$ Therefore,

$$\begin{aligned} \alpha _1\beta _1 g_2&= a_2b_2\left( a_1^2+b_1^2\right) \left( z_2^{(a_2-1)}w_2^{(b_2-1)}+z_2^{(b_2-1)}w_2^{(a_2-1)}\right) z_1^{(a_1+b_1-1)}w_1^{(a_1+b_1-1)}\\&\quad +a_1b_1a_2b_2\left( z_1^{(2a_1-1)}w_1^{(2b_1-1)}+z_1^{(2b_1-1)}w_1^{(2a_1-1)}\right) \left( z_2^{(a_2-1)}w_2^{(b_2-1)}+z_2^{(b_2-1)}w_2^{(a_2-1)}\right) ,\\ \alpha _2\beta _2 g_1&= a_1b_1\left( a_2^2+b_2^2\right) \left( z_1^{(a_1-1)}w_1^{(b_1-1)}+z_1^{(b_1-1)}w_1^{(a_1-1)}\right) z_2^{(a_2+b_2-1)}w_2^{(a_2+b_2-1)}\\&\quad +a_1b_1a_2b_2\left( z_2^{(2a_2-1)}w_2^{(2b_2-1)}+z_2^{(2b_2-1)}w_2^{(2a_2-1)}\right) \left( z_1^{(a_1-1)}w_1^{(b_1-1)}+z_1^{(b_1-1)}w_1^{(a_1-1)}\right) . \end{aligned}$$

Now, using the expression (15), we obtain the following relationship on $M_{{\mathbb {C}}}^*$:

$$\begin{aligned} \left( z_2^{(a_2-1)}w_2^{(b_2-1)}+z_2^{(b_2-1)}w_2^{(a_2-1)}\right)&= -\frac{z_1w_1}{z_2w_2}\left( z_1^{(a_1-1)}w_1^{(b_1-1)}+z_1^{(b_1-1)}w_1^{(a_1-1)}\right) . \end{aligned}$$

With this, we can rewrite

$$\begin{aligned} \alpha _1\beta _1 g_2&= \left( \frac{z_1^{(a_1-1)}w_1^{(b_1-1)}+z_1^{(b_1-1)}w_1^{(a_1-1)}}{z_2w_2}\right) \left[ -a_2b_2\left( a_1^2+b_1^2\right) z_1^{(a_1+b_1)}w_1^{(a_1+b_1)}\right. \\&\quad -\,\left. a_1b_1a_2b_2\left( z_1^{2a_1}w_1^{2b_1}+z_1^{2b_1}w_1^{2a_1}\right) \right] ,\\ \alpha _2\beta _2 g_1&= \left( \frac{z_1^{(a_1-1)}w_1^{(b_1-1)}+z_1^{(b_1-1)}w_1^{(a_1-1)}}{z_2w_2}\right) \left[ a_1b_1\left( a_2^2+b_2^2\right) z_2^{(a_2+b_2)}w_2^{(a_2+b_2)}\right. \\&\quad +\,\left. a_1b_1a_2b_2\left( z_2^{2a_2}w_2^{2b_2}+z_2^{2b_2}w_2^{2a_2}\right) \right] . \end{aligned}$$

In this way, we get

$$\begin{aligned} \alpha _2\beta _2 g_1 + \alpha _1\beta _1 g_2&= \left( \frac{z_1^{(a_1-1)}w_1^{(b_1-1)}+z_1^{(b_1-1)}w_1^{(a_1-1)}}{z_2w_2}\right) \left[ a_1b_1\left( a_2^2+b_2^2\right) z_2^{(a_2+b_2)}w_2^{(a_2+b_2)}\right. \\&\quad + a_1b_1a_2b_2\left( z_2^{2a_2}w_2^{2b_2}+z_2^{2b_2}w_2^{2a_2}-z_1^{2a_1}w_1^{2b_1}+z_1^{2b_1}w_1^{2a_1}\right) \\&\quad -\,\left. a_2b_2\left( a_1^2+b_1^2\right) z_1^{(a_1+b_1)}w_1^{(a_1+b_1)}\right] . \end{aligned}$$

Again using (15), we obtain

$$\begin{aligned}\left( z_2^{2a_2}w_2^{2b_2}+z_2^{2b_2}w_2^{2a_2}-z_1^{2a_1}w_1^{2b_1}-z_1^{2b_1}w_1^{2a_1}\right) =2\left( z_1^{(a_1+b_1)}w_1^{(a_1+b_1)}-z_2^{(a_2+b_2)}w_2^{(a_2+b_2)}\right) .\end{aligned}$$

Hence

$$\begin{aligned} \alpha _2\beta _2 g_1 + \alpha _1\beta _1 g_2&=\left( \frac{z_1^{(a_1-1)}w_1^{(b_1-1)}+z_1^{(b_1-1)}w_1^{(a_1-1)}}{z_2w_2}\right) \left[ a_1b_1\left( a_2^2+b_2^2\right) z_2^{(a_2+b_2)}w_2^{(a_2+b_2)}\right. \\&\quad + 2a_1b_1a_2b_2\left( z_1^{(a_1+b_1)}w_1^{(a_1+b_1)}-z_2^{(a_2+b_2)}w_2^{(a_2+b_2)}\right) \\&\quad -\,\left. a_2b_2\left( a_1^2+b_1^2\right) z_1^{(a_1+b_1)}w_1^{(a_1+b_1)}\right] , \end{aligned}$$

that is,

$$\begin{aligned} \alpha _2\beta _2 g_1 + \alpha _1\beta _1 g_2&= \left( \frac{z_1^{(a_1-1)}w_1^{(b_1-1)}+z_1^{(b_1-1)}w_1^{(a_1-1)}}{z_2w_2}\right) \left[ a_1b_1(a_2-b_2)^2\left( z_2^{(a_2+b_2)}w_2^{(a_2+b_2)}\right) \right. \\&\quad -\,\left. a_2b_2(a_1-b_1)^2\left( z_1^{(a_1+b_1)}w_1^{(a_1+b_1)}\right) \right] . \end{aligned}$$

Therefore, by (9), we conclude that $\eta |_{M^{*}}$ is integrable if and only if $a_1b_1(a_2-b_2)^2=0$ and $a_2b_2(a_1-b_1)^2=0$, that is, $a_2=b_2$ and $a_1=b_1$. This completes the proof for $n=2$.

Now, we consider $n\ge 3$. With the notation introduced in Lemma 3.2, we can rewrite

$$\begin{aligned} \alpha = \sum _{k=1}^n f_kdz_k,\,\,\,\,\,\,\,\,\,\,\, \beta = \sum _{k=1}^n g_kdw_k. \end{aligned}$$

Then $d\alpha =\sum _{\ell =1}^n h_{\ell }dw_{\ell }\wedge dz_{\ell }.$ Indeed, we have

$$\begin{aligned}d\alpha =\sum _{k,\ell =1}^n\frac{\partial f_k}{\partial z_{\ell }}dz_{\ell }\wedge dz_k+\sum _{k,\ell =1}^n\frac{\partial f_k}{\partial w_{\ell }}dw_{\ell }\wedge dz_k.\end{aligned}$$

Furthermore, we have the following relationships:

$$\begin{aligned} \frac{\partial f_k}{\partial z_{\ell }}&=0,\, \text{ for }\, k\ne \ell ,\\ \frac{\partial f_k}{\partial w_{\ell }}&=0,\, \text{ for }\, k\ne \ell ,\\ \frac{\partial f_{\ell }}{\partial w_{\ell }}&=a_{\ell }b_{\ell }\left( z_{\ell }^{(a_{\ell }-1)}w_{\ell }^{(b_{\ell }-1)}+ z_{\ell }^{(b_{\ell }-1)}w_{\ell }^{(a_{\ell }-1)}\right) . \end{aligned}$$

Hence, it follows that

$$\begin{aligned}d\alpha =\sum _{\ell =1}^n a_{\ell }b_{\ell }\left( z_{\ell }^{(a_{\ell }-1)}w_{\ell }^{(b_{\ell }-1)}+ z_{\ell }^{(b_{\ell }-1)}w_{\ell }^{(a_{\ell }-1)}\right) dw_{\ell }\wedge dz_{\ell }=\sum _{\ell =1}^n h_{\ell }\,dw_{\ell }\wedge dz_{\ell }.\end{aligned}$$

Thus, we have

$$\begin{aligned} \alpha \wedge d\alpha \wedge \beta = \sum _{\ell \ne j,k}f_k g_j h_{\ell }\ dz_k\wedge dw_{\ell } \wedge dz_{\ell } \wedge dw_j. \end{aligned}$$

(16)

We observe that for $j \ne k$, the coefficient of the term $dz_k\wedge dw_{\ell } \wedge dz_{\ell } \wedge dw_j$ in 4-form $\alpha \wedge d\alpha \wedge \beta$ is exactly $f_k g_j h_{\ell }$. In other words, if $j \ne k$,

$$\begin{aligned}f_k g_j h_{\ell }\ dz_k\wedge dw_{\ell } \wedge dz_{\ell } \wedge dw_j\end{aligned}$$

is the only term in Eq. (16) that is of the form $\rho \ dz_k\wedge dw_{\ell } \wedge dz_{\ell } \wedge dw_j.$ Thus, given a triple $(j, k, \ell )$ of pairwise distinct indices, by Lemma (3.2) there exists $(z_0,w_0) \in M_{{\mathbb {C}}}$ such that

$$\begin{aligned}\alpha (z_0,w_0)\wedge d\alpha (z_0,w_0) \wedge \beta (z_0,w_0) \ne 0.\end{aligned}$$

Therefore, M is not Levi-flat, and we conclude the proof of Theorem A. $\square$

4 The Case of Vector Fields $G=(z_1^{a_1}, z_2^{a_2})$ and $X=(z_{\sigma _1}^{b_{\sigma _1}}, z_{\sigma _2}^{b_{\sigma _2}})$

In this section, we will explore some results regarding the singular set of the hypersurface. $M=\{F(z)=0\}$, where $F(z)=2{\textrm{Re}}\langle G(z),X(z) \rangle$, $G=(z_1^{a_1}, z_2^{a_2})$, and $X=(z_{\sigma _1}^{b_{\sigma _1}}, z_{\sigma _2}^{b_{\sigma _2}})$, where $\sigma \in \mathcal {S}_n$ is a permutation of the set $\{1, \dots , n\}$. Let’s start by considering the particular case where $G=(z_1^a,z_2^b)$ and $X=(z_2^b,z_1^a)$ with a, b being positive integers. In this case, the Hermitian product of G with X is given by

$$\begin{aligned}\langle G,X \rangle = z_1^a \overline{z}_2^b + \overline{z}_1^a z_2^b.\end{aligned}$$

We observe that $\langle G,X \rangle \in {\mathbb {R}}$. Therefore, let’s consider

$$\begin{aligned} M=\{F(z_1,z_2)=z_1^a \overline{z}_2^b + \overline{z}_1^a z_2^b=0\}. \end{aligned}$$

(17)

We observe that for these fields, the map $\psi _{G,X}$ does not satisfy the Milnor condition, as stated in theorem (Ruas et al. 2002, Theorem 2.7). In this case, the hypersurface M will be Levi-flat, as we will show in Theorem B. First, we will verify the following result regarding the singular set of M.

Proposition 4.1

Let M be the hypersurface described in Eq. (17). Then we have the following options for $\textrm{Sing}(M)$:

• 1. $\textrm{Sing}(M)=\{0\}$, if $b=a=1$,

• 2. $\textrm{Sing}(M) =\{z_1=0\}$, if $b=1$ and $a>1$,

• 3. $\textrm{Sing}(M)=\{z_2=0\}$, if $a=1$ and $b>1$,

• 4. $\textrm{Sing}(M)=\{z_1=0\} \cup \{z_2=0\}$, if $a,b>1$.

Proof

The partial derivatives of F are given by

$$\begin{aligned} \partial F&= az_1^{(a-1)}\overline{z}_2^b dz_1+b\overline{z}_1^az_2^{(b-1)}dz_2,\\ \overline{ \partial }F&= a\overline{z}_1^{(a-1)}z_2^b d\overline{z}_1+ bz_1^a\overline{z}_2^{(b-1)}d\overline{z}_2. \end{aligned}$$

Thus, $(z_1,z_2) \in \textrm{Sing}(M)$ if and only if the following equations are satisfied:

$$\begin{aligned} az_1^{(a-1)}\overline{z}_2^b&= 0\\ b\overline{z}_1^az_2^{(b-1)}&= 0\\ a\overline{z}_1^{(a-1)}z_2^b&= 0\\ bz_1^a\overline{z}_2^{(b-1)}&=0 \end{aligned}$$

If $a, b > 1$, it follows that $\textrm{Sing}(M)=\{z_1=0\} \cup \{z_2=0\}$. If $a=1$ or $b=1$, we will have $\textrm{Sing}(M)=\{z_2=0\}$ or $\textrm{Sing}(M)=\{z_1=0\}$, respectively. Finally, if $a=b=1$, we obtain $\textrm{Sing}(M)=\{0\}$. $\square$

Let’s restate Theorem B for completeness.

Theorem B

Let G and X be elements in $\chi ({\mathbb {C}}^ n,0)$, $n\ge 2$, of the form $G(z)=(z_1^a,z_2^b)$ and $X(z)=(z_2^b,z_1^a)$ with a, b positive integers. Then $M=\{F(z)=2{\textrm{Re}}(\psi _{G,X}(z))=0\}$, i.e.,

$$\begin{aligned}M=\left\{ z_1^{a}\bar{z}_2^{b}+z_2^{b}\bar{z}_1^{a}=0\right\} \end{aligned}$$

is a Levi-flat hypersurface at $0\in {\mathbb {C}}^2$.

Proof

In this case, F is given by $F(z)=z_1^a \overline{z}_2^b + \overline{z}_1^a z_2^b$ and its complexification will be given by

$$\begin{aligned} F_{{\mathbb {C}}}(z,w)=z_1^a w_2^b + w_1^a z_2^b. \end{aligned}$$

(18)

Therefore, we have on $M_{{\mathbb {C}}}^*$

$$\begin{aligned} \alpha&= az_1^{(a-1)}w_2^b dz_1+bw_1^az_2^{(b-1)}dz_2,\\ \beta&= aw_1^{(a-1)}z_2^b dw_1+ bz_1^aw_2^{(b-1)}dw_2,\\ d\alpha&= abz_1^{(a-1)}w_2^{(b-1)} dw_2 \wedge dz_1+abw_1^{(a-1)}z_2^{(b-1)}dw_1\wedge dz_2. \end{aligned}$$

Hence

$$\begin{aligned} \alpha \wedge d\alpha \wedge \beta&= \left( a^2b^2 z_1^{(2a-1)}w_2^{(2b-1)}w_1^{(a-1)}z_2^{(b-1)}\right. \\&\quad +\, \left. a^2b^2 w_1^{(2a-1)}z_2^{(2b-1)}z_1^{(a-1)}w_2^{(b-1)}\right) dz_1 \wedge dw_1 \wedge dz_2 \wedge dw_2. \end{aligned}$$

From (18), we get

$$\begin{aligned}z_1^{(a-1)}w_2^{(b-1)}= - \left( \frac{w_1z_2}{z_1w_2}\right) w_1^{(a-1)}z_2^{(b-1)},\end{aligned}$$

which implies that $\alpha \wedge d\alpha \wedge \beta$ is equal to

$$\begin{aligned}{} & {} \left[ a^2b^2z_1^{(2a-1)}w_2^{(2b-1)}w_1^{(a-1)}z_2^{(b-1)}\right. \\{} & {} \quad - \,\left. a^2b^2\left( \frac{w_1z_2}{z_1w_2}\right) w_1^{(2a-1)}z_2^{(2b-1)}w_1^{(a-1)}z_2^{(b-1)})dz_1\right] dz_1 \wedge dw_1 \wedge dz_2 \wedge dw_2,\end{aligned}$$

that is,

$$\begin{aligned}\alpha \wedge d\alpha \wedge \beta =\left[ a^2b^2 \frac{w_1^{(a-1)}z_2^{(b-1)}}{z_1w_2} \left( z_1^{2a}w_2^{2b}-w_1^{2a}z_2^{2b}\right) \right] dz_1 \wedge dw_1 \wedge dz_2 \wedge dw_2.\end{aligned}$$

Again, from (18), we have $z_1^aw_2^b=-w_1^az_2^b$, which yields $z_1^{2a}w_2^{2b}=w_1^{2a}z_2^{2b}$ in $M_{{\mathbb {C}}}$. Thus, M is Levi-flat. $\square$

Now, we consider

$$\begin{aligned} M=\{F(z)={\textrm{Re}}(z_1^{a_1}\bar{z}_2^{b_2}+z_2^{a_2}\bar{z}^{b_1}_{1})=0\} \end{aligned}$$

(19)

In Proposition 4.1, we saw that $\textrm{Sing}(M)=\{z_1=0\} \cup \{z_2=0\}$ in the case where $a_1=b_1>1$ and $a_2=b_2>1$. Now, let’s verify that this also occurs in the case where $a_1>b_1>1$ and $a_2>b_2>1$ (even without the assumption $a_1b_2=a_2b_1$).

Proposition 4.2

Let M be the hypersurface described in Eq. (19). Then, we have the following options for $\textrm{Sing}(M)$:

• 1. $\textrm{Sing}(M)=\{0\}$, if $b_1=b_2=1$;

• 2. $\textrm{Sing}(M)=\{z_1=0\}$, if $b_2=1$ and $b_1>1$;

• 3. $\textrm{Sing}(M)=\{z_2=0\}$, if $b_1=1$ and $b_2>1$;

• 4. $\textrm{Sing}(M)=\{z_1=0\} \cup \{z_2=0\}$ if $b_1,b_2>1$.

Proof

We observe that F is given by

$$\begin{aligned}F(z)=z_1^{a_1}\overline{z}_2^{b_2}+\overline{z}_1^{a_1}z_2^{b_2}+z_2^{a_2}\overline{z}_1^{b_1}+\overline{z}_2^{a_2}z_1^{b_1 },\end{aligned}$$

and its partial derivatives are given by

$$\begin{aligned} \partial F&= \frac{1}{2}\left[ \left( a_1z_1^{(a_1-1)}\overline{z}_2^{b_2}+b_1z_1^{(b_1-1)}\overline{z}_2^{a_2}\right) dz_1+ \left( a_2z_2^{(a_2-1)}\overline{z}_1^{b_1}+b_2z_2^{(b_2-1)}\overline{z}_1^{a_1}\right) dz_2\right] ,\\ \overline{ \partial }F&= \frac{1}{2}\left[ \left( a_1\overline{z}_1^{(a_1-1)}z_2^{b_2}+b_1\overline{z}_1^{(b_1-1)}z_2^{a_2}\right) d\overline{z}_1+ \left( a_2\overline{z}_2^{(a_2-1)}z_1^{b_1}+b_2\overline{z}_2^{(b_2-1)}z_1^{a_1}\right) d\overline{z}_2\right] . \end{aligned}$$

Therefore, $(z_1,z_2) \in M$ belongs to the singular set if it satisfies the equations

$$\begin{aligned} a_1z_1^{(a_1-1)}\overline{z}_2^{b_2}+b_1z_1^{(b_1-1)}\overline{z}_2^{a_2}&= 0 \end{aligned}$$

(20)

$$\begin{aligned} a_1\overline{z}_1^{(a_1-1)}z_2^{b_2}+b_1\overline{z}_1^{(b_1-1)}z_2^{a_2}&= 0\end{aligned}$$

(21)

$$\begin{aligned} a_2z_2^{(a_2-1)}\overline{z}_1^{b_1}+b_2z_2^{(b_2-1)}\overline{z}_1^{a_1}&= 0\end{aligned}$$

(22)

$$\begin{aligned} a_2\overline{z}_2^{(a_2-1)}z_1^{b_1}+b_2\overline{z}_2^{(b_2-1)}z_1^{a_1}&= 0 \end{aligned}$$

(23)

Let’s assume that $b_1, b_2 > 1$. Then we see that $\{z_1=0\} \cup \{z_2=0\} \subset \textrm{Sing}(M)$. Now, suppose by contradiction that $(z_1,z_2) \in \textrm{Sing}(M)$ with $z_1 \ne 0$ and $z_2 \ne 0$. From Eq. (20), we obtain

$$\begin{aligned}\frac{a_1}{b_1}=-\frac{z_1^{(b_1-1)}\overline{z}_2^{a_2}}{z_1^{(a_1-1)}\overline{z}_2^{b_2}}=-\frac{\overline{z}_2^{(a_2-b_2)}}{z_1^{(a_1-b_1)}},\end{aligned}$$

and from Eq. (23), we obtain

$$\begin{aligned}\frac{a_2}{b_2}=-\frac{\overline{z}_2^{(b_2-1)}z_1^{a_1}}{\overline{z}_2^{(a_2-1)}z_1^{b_1}}=-\frac{z_1^{(a_1-b_1)}}{\overline{z}_2^{(a_2-b_2)}},\end{aligned}$$

which implies

$$\begin{aligned}\frac{a_1}{b_1}=\frac{b_2}{a_2},\end{aligned}$$

In other words, $a_1a_2=b_1b_2$. This is absurd, since $a_1>b_1$ and $a_2>b_2$ by assumption. We conclude that ${\textsf {Sing}}(M)=\{z_1=0\} \cup \{z_2=0\}$, if $b_1,b_2 >1$. Now let’s assume $b_1=1$ and $b_2>1$. Then, the equations for the singular set are given by

$$\begin{aligned} a_1z_1^{(a_1-1)}\overline{z}_2^{b_2}+\overline{z}_2^{a_2}&= 0 \end{aligned}$$

(24)

$$\begin{aligned} a_1\overline{z}_1^{(a_1-1)}z_2^{b_2}+z_2^{a_2}&= 0 \end{aligned}$$

(25)

$$\begin{aligned} a_2z_2^{(a_2-1)}\overline{z}_1+b_2z_2^{(b_2-1)}\overline{z}_1^{a_1}&= 0 \end{aligned}$$

(26)

$$\begin{aligned} a_2\overline{z}_2^{(a_2-1)}z_1+b_2\overline{z}_2^{(b_2-1)}z_1^{a_1}&= 0 \end{aligned}$$

(27)

Clearly, $\{z_2=0\} \subset \textrm{Sing}(M)$. Now suppose $z_2\ne 0$, then we necessarily have $z_1\ne 0$ by Eq. (24). Thus, it follows from Eqs. (24) and (27):

$$\begin{aligned} a_1&=-\frac{\overline{z}_2^{(a_2-b_2)}}{z_1^{(a_1-1)}},\\ \frac{a_2}{b_2}&=-\frac{z_1^{(a_1-1)}}{\overline{z}_2^{(a_2-b_2)}}, \end{aligned}$$

which implies $a_1=\frac{b_2}{a_2}$, that is, $a_1a_2=b_2$. However, this contradicts $a_2>b_2$ and $a_1>1$. Therefore, ${\textsf {Sing}}(M)=\{z_2=0\}$ if $b_1=1$ and $b_2>1$. In a similar way we obtain ${\textsf {Sing}}(M)=\{z_1=0\}$ if $b_2=1$ and $b_1>1$, and we also obtain ${\textsf {Sing}}(M)=\{ 0\}$ if $b_1=b_2=1$. $\square$

To prove Theorem C, we will use the following lemma

Lemma 4.3

Let $M_{{\mathbb {C}}}$ be the complexification of

$$\begin{aligned}M=\{F(z)=z_1^{a_1}\overline{z}_2^{b_2}+\overline{z}_1^{a_1}z_2^{b_2}+z_2^{a_2}\overline{z}_1^{b_1}+\overline{z}_2^{a_2}z_1^{b_1 }=0\},\end{aligned}$$

we have $\alpha \wedge d\alpha \wedge \beta = h\ dz_1 \wedge dw_1 \wedge dz_2 \wedge dw_2$, where

$$\begin{aligned}h&= c(a_1a_2+b_1b_1-2c)\left( z_1^{a_1+b_1}w_2^{a_2+b_2}\right. \\&\quad -\,\left. z_2^{(a_2+b_2)}w_1^{(a_1+b_1)}\right) \left( \frac{z_2^{(a_2-1)}w_1^{(b_1-1)}+z_2^{(b_2-1)}w_1^{(a_1-1)}}{z_1w_2}\right) \end{aligned}$$

and $c=a_1b_2=a_2b_1$.

Proof

The complexification of F is given by

$$\begin{aligned}F_{{\mathbb {C}}}=z_1^{a_1}w_2^{b_2}+w_1^{a_1}z_2^{b_2}+z_2^{a_2}w_1^{b_1}+w_2^{a_2}z_1^{b_1}.\end{aligned}$$

Therefore, we have

$$\begin{aligned} \alpha&=\left( a_1z_1^{(a_1-1)}w_2^{b_2}+b_1z_1^{(b_1-1)}w_2^{a_2}\right) dz_1+ \left( a_2z_2^{(a_2-1)}w_1^{b_1}+b_2z_2^{(b_2-1)}w_1^{a_1}\right) dz_2,\\ \beta&= \left( a_1w_1^{(a_1-1)}z_2^{b_2}+b_1w_1^{(b_1-1)}z_2^{a_2}\right) dw_1+ \left( a_2w_2^{(a_2-1)}z_1^{b_1}+b_2w_2^{(b_2-1)}z_1^{a_1}\right) dw_2,\\ d\alpha&= \left( a_1b_2z_1^{(a_1-1)}w_2^{(b_2-1)}+a_2b_1z_1^{(b_1-1)}w_2^{(a_2-1)}\right) dw_2 \wedge dz_1\\&\quad +\left( a_2b_1z_2^{(a_2-1)}w_1^{(b_1-1)}+a_1b_2z_2^{(b_2-1)}w_1^{(a_1-1)}\right) dw_1 \wedge dz_2. \end{aligned}$$

We will use the notations

$$\begin{aligned} \alpha _1&= a_1z_1^{(a_1-1)}w_2^{b_2}+b_1z_1^{(b_1-1)}w_2^{a_2};\\ \alpha _2&= a_2w_1^{b_1}z_2^{(a_2-1)}+b_2w_1^{a_1}z_2^{(b_2-1)};\\ \beta _1&= a_1w_1^{(a_1-1)}z_2^{b_2}+b_1w_1^{(b_1-1)}z_2^{a_2};\\ \beta _2&= a_2z_1^{b_1}w_2^{(a_2-1)}+b_2z_1^{a_1}w_2^{(b_2-1)};\\ g_1&= a_1b_2z_1^{(a_1-1)}w_2^{(b_2-1)}+a_2b_1z_1^{(b_1-1)}w_2^{(a_2-1)};\\ g_2&= a_2b_1w_1^{(b_1-1)}z_2^{(a_2-1)}+a_1b_2w_1^{(a_1-1)}z_2^{(b_2-1)}. \end{aligned}$$

so that we can write:

$$\begin{aligned} \alpha&= \alpha _1dz_1+\alpha _2dz_2;\\ \beta&= \beta _1dw_1+\beta _2dw_2;\\ d\alpha&= g_1 dw_2 \wedge dz_1+g_2 dw_1 \wedge dz_2. \end{aligned}$$

With these notations, we get

$$\begin{aligned}\alpha \wedge d\alpha \wedge \beta = (\alpha _1 \beta _2g_2+\alpha _2\beta _1g_1)dz_1 \wedge dw_1 \wedge dz_2 \wedge dw_2.\end{aligned}$$

In $M_{{\mathbb {C}}}$, we have

$$\begin{aligned} z_1^{a_1}w_2^{b_2}+z_1^{b_1}w_2^{a_2}=-(w_1^{a_1}z_2^{b_2}+w_1^{b_1}z_2^{a_2}), \end{aligned}$$

(28)

which implies on $M_{{\mathbb {C}}}^*$

$$\begin{aligned} z_1^{(a_1-1)}w_2^{(b_2-1)}+z_1^{(b_1-1)}w_2^{(a_2-1)} =-\left( \frac{w_1z_2}{z_1w_2}\right) (w_1^{(a_1-1)}z_2^{(b_2-1)}+w_1^{(b_1-1)}z_2^{(a_2-1)}).\end{aligned}$$

Taking $c=a_1b_2=b_1a_2$, we have

$$\begin{aligned}g_1=c( z_1^{(a_1-1)}w_2^{(b_2-1)}+z_1^{(b_1-1)}w_2^{(a_2-1)})=-c\left( \frac{w_1z_2}{z_1w_2}\right) (w_1^{(a_1-1)}z_2^{(b_2-1)}+w_1^{(b_1-1)}z_2^{(a_2-1)}),\end{aligned}$$

and it follows that

$$\begin{aligned} g_1=-\left( \frac{w_1z_2}{z_1w_2}\right) g_2. \end{aligned}$$

Therefore, on $M_{{\mathbb {C}}}^*$, we obtain

$$\begin{aligned}\alpha _1 \beta _2g_2+\alpha _2\beta _1g_1=\alpha _1 \beta _2g_2-\alpha _2\beta _1\left( \frac{w_1z_2}{z_1w_2}\right) g_2 =\frac{g_2}{z_1w_2}(z_1w_2\alpha _1 \beta _2-w_1z_2\alpha _2\beta _1),\end{aligned}$$

thus

$$\begin{aligned} \alpha \wedge d\alpha \wedge \beta =\left[ \frac{g_2}{z_1w_2}(z_1w_2\alpha _1 \beta _2-w_1z_2\alpha _2\beta _1)\right] dz_1 \wedge dw_1 \wedge dz_2 \wedge dw_2. \end{aligned}$$

(29)

Now, we observe

$$\begin{aligned} \alpha _1 \beta _2&=\left( a_1z_1^{(a_1-1)}w_2^{b_2}+b_1z_1^{(b_1-1)}w_2^{a_2}\right) \left( a_2z_1^{b_1}w_2^{(a_2-1)}+b_2z_1^{a_1}w_2^{(b_2-1)}\right) \\&=\left( a_1a_2+b_1b_2\right) z_1^{(a_1+b_1-1)}w_2^{(a_2+b_2-1)}+c\left( z_1^{(2a_1-1)}w_2^{(2b_2-1)}+z_1^{(2b_1-1)}w_2^{(2a_2-1)}\right) , \end{aligned}$$

which implies

$$\begin{aligned} z_1w_2\alpha _1 \beta _2= \left( a_1a_2+b_1b_2\right) z_1^{(a_1+b_1)}w_2^{(a_2+b_2)}+c\left( z_1^{2a_1}w_2^{2b_2}+z_1^{2b_1}w_2^{2a_2}\right) . \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} w_1z_2\alpha _2\beta _1=(a_1a_2+b_1b_2)w_1^{(a_1+b_1)}z_2^{(a_2+b_2)}+c(w_1^{2b_1}z_2^{2a_2}+w_1^{2a_1}z_2^{2b_2}), \end{aligned}$$

so that $z_1w_2\alpha _1 \beta _2-w_1z_2\alpha _2\beta _1$ is equal to

$$\begin{aligned}&(a_1a_2+b_1b_2)\left( z_1^{(a_1+b_1)}w_2^{(a_2+b_2)}-w_1^{(a_1+b_1)}z_2^{(a_2+b_2)}\right) \\&\quad +c\left( z_1^{2a_1}w_2^{2b_2}+z_1^{2b_1}w_2^{2a_2}-w_1^{2b_1}z_2^{2a_2}-w_1^{2a_1}z_2^{2b_2}\right) .\end{aligned}$$

Again, from Eq. (28), on $M_{{\mathbb {C}}}^*$ we have:

$$\begin{aligned}(z_1^{a_1}w_2^{b_2}+z_1^{b_1}w_2^{a_2})^2=(w_1^{a_1}z_2^{b_2}+w_1^{b_1}z_2^{a_2})^2,\end{aligned}$$

that is,

$$\begin{aligned}2(w_1^{(a_1\!+\!b_1)}z_2^{(a_2\!+\!b_2)}-z_1^{(a_1\!+\!b_1)}w_2^{(a_2\!+\!b_2)})=z_1^{2a_1}w_2^{2b_2}\!+\!z_1^{2b_1}w_2^{2a_2}\!-\!w_1^{2b_1}z_2^{2a_2}\!-\!w_1^{2a_1}z_2^{2b_2},\end{aligned}$$

which implies that $z_1w_2\alpha _1 \beta _2-w_1z_2\alpha _2\beta _1$ is equal to

$$\begin{aligned}&(a_1a_2+b_1b_2)\left( z_1^{(a_1+b_1)}w_2^{(a_2+b_2)}-w_1^{(a_1+b_1)}z_2^{(a_2+b_2)}\right) \\&\quad +2c\left( w_1^{(a_1+b_1)}z_2^{(a_2+b_2)}-z_1^{(a_1+b_1)}w_2^{(a_2+b_2)}\right) . \end{aligned}$$

So, $z_1w_2\alpha _1 \beta _2-w_1z_2\alpha _2\beta _1$ is equal to

$$\begin{aligned}(a_1a_2+b_1b_2-2c)\left( z_1^{(a_1+b_1)}w_2^{(a_2+b_2)}-w_1^{(a_1+b_1)}z_2^{(a_2+b_2)}\right) .\end{aligned}$$

Finally, we conclude

$$\begin{aligned} \frac{g_2}{z_1w_2}\left( z_1w_2\alpha _1 \beta _2-w_1z_2\alpha _2\beta _1\right){} & {} = \frac{g_2}{z_1w_2}\left( a_1a_2+b_1b_2-2c\right) \\{} & {} \quad \left( z_1^{(a_1+b_1)}w_2^{(a_2+b_2)}-w_1^{(a_1+b_1)}z_2^{(a_2+b_2)}\right) \end{aligned}$$

and the lemma follows by substituting the above expression and the expression for $g_2$ into Eq. (29). $\square$

Now, we prove Theorem C.

Theorem C

Let G and X be elements in $\chi ({\mathbb {C}}^ n,0)$, $n\ge 2$, of the form $G=(z_1^{a_1},z_2^{a_2})$ and $X=(z_2^{b_2},z_1^{b_1})$, where $a_1\ge b_1$ and $a_2 \ge b_2$ are positive integers satisfying $a_1b_2=a_2b_1$. Then $M=\{F(z)=2{\textrm{Re}}(\psi _{G,X}(z))=0\}$, i.e.,

$$\begin{aligned}M=\left\{ {\textrm{Re}}\left( z_1^{a_1}\bar{z}_2^{b_2}+z_2^{a_2}\bar{z}_1^{b_1}\right) =0\right\} \end{aligned}$$

is Levi-flat if, and only if, $a_1=b_1$ and $a_2=b_2$.

Proof

From Lemma 4.3, we have $\alpha (z,w)\wedge \beta (z,w)\wedge d\alpha (z,w)=0$ for all $(z,w) \in M_{{\mathbb {C}}}$ if, and only if, $a_1a_2+b_1b_2-2c=0$, i.e., $2c=a_1a_2+b_1b_2$. Using the fact $c=a_1b_2=b_1a_2$, it follows that

$$\begin{aligned}2c\!=\!a_1a_2\!+\!b_1b_2\iff a_1b_2\!+\!b_1a_2\!=\!a_1a_2\!+\!b_1b_2\iff b_1(a_2\!-\!b_2)\!=\!a_1(a_2-b_2).\end{aligned}$$

Therefore, $M_{{\mathbb {C}}}$ is Levi-flat if, and only if, $a_1=b_1$ and $a_2=b_2$. This concludes the proof of Theorem C. $\square$

Remark 4.1

From the above result, we conclude that for $n=2$, the hypersurface M will not Levi-flat precisely when the map $\psi _{G,X}$ satisfies the Milnor condition at the origin.

5 Proof of Theorem D

Now, we study the case with permutations in higher dimensions. Consider $\sigma \in S_n$ a permutation on the set $\underline{n}:=\{1,\ldots ,n\}$, and let’s use the notation $\sigma _k:=\sigma (k)$. We employ a transversality argument, along with the dimension 2 case (see Theorem C), to obtain the following result:

Theorem D

Let G and X be elements in $\chi ({\mathbb {C}}^ n,0)$, $n\ge 2$, of the form $G=(z_1^{a_1},\ldots , z_n^{a_n})$ and $X=(z_{\sigma _1}^{b_{\sigma _1}},\ldots , z_{\sigma _n}^{b_{\sigma _n}})$, where $a_k\ge b_k$ are positive integers. Let us assume that for some $\ell \in \underline{n}$, the integers $a_{\ell }, b_{\ell }, a_{\sigma _{\ell }}, b_{\sigma _{\ell }}$ satisfy the following conditions: $a_{\ell }>b_{\ell }$ and $a_{\ell }b_{\sigma _{\ell }}=b_{\ell }a_{\sigma _{\ell }}$. Then, $M=\{F(z)=2{\textrm{Re}}(\psi _{G,X}(z))=0\}$, i.e.,

$$\begin{aligned}M=\left\{ {\textrm{Re}}\left( \sum _{k=1}^{n} z_k^{a_k}\bar{z}_{\sigma _k}^{b_{\sigma _k}}\right) =0\right\} .\end{aligned}$$

is nondegenerate at $0\in {\mathbb {C}}^n$.

Proof

Let’s assume by contradiction that M is Levi-flat. Therefore, the regular part $M^*$ is foliated by complex submanifolds of (complex) dimension $n-1$. Thus, we can choose $i: {\mathbb {C}}^2 \hookrightarrow {\mathbb {C}}^n$ to be a transversal embedding to M (see Cerveau and Lins-Neto 2011, Corollary 3.3), so that the regular part of $i^{-1}(M)$ is also foliated by complex submanifolds of dimension 1 (Riemann surfaces). Without loss of generality, we can assume that $\sigma (1) = 2$ and that $a_1, b_1, a_2, b_2$ are the integers satisfying the conditions in the statement. Making a change of coordinates and using the fact that the Levi-flat property is invariant under biholomorphisms (Proposition 2.4), we can assume that the embedding $i:{\mathbb {C}}^2\rightarrow {\mathbb {C}}^n$ is given by $i: (z_1, z_2) \mapsto (z_1, z_2, 0,\ldots ,0).$ We observe

$$\begin{aligned}i^{-1}(M)=\{(z_1,z_2) \in {\mathbb {C}}^2: z_1^{a_1}\overline{z}_2^{b_2}+\overline{z}_1^{a_1}z_2^{b_2}+z_2^{a_2}\overline{z}_1^{b_1}+\overline{z}_2^{a_2}z_1^{b_1}=0\}.\end{aligned}$$

But by Theorem C, $i^{-1}(M)$ does not have the regular part foliated by complex submanifolds. Therefore, M is not Levi-flat. $\square$

6 Examples

In this section, we will explore examples where the fields G and X do not satisfy the Milnor condition. In these examples, our hypersurfaces are all Levi-flat.

Example 6.1

Given the Pham–Brieskorn polynomial $f:{\mathbb {C}}^2\rightarrow {\mathbb {C}}$ defined by $f(z)=z_1^p+z_2^q$, where $p,q>2$. Consider the holomorphic vector field

$$\begin{aligned}G(z)=\left( \frac{\partial f(z)}{\partial z_2},-\frac{\partial f(z)}{\partial z_1}\right) =\left( qz_2^{(q-1)}, -pz_1^{(p-1)}\right) \end{aligned}$$

whose solutions represent the fibers of f. Also, take the vector field $X=(a_1,a_2)$. Thus, the Hermitian product of G and X is given by

$$\begin{aligned}\Psi _{G,X}(z)=\langle G(z),X(z) \rangle =\overline{a}_1qz_2^{(q-1)}-\overline{a_2}pz_1^{(p-1)}.\end{aligned}$$

Clearly, $\Psi _{G,X}$ does not satisfy the Milnor fibration condition at the origin. Now we take

$$\begin{aligned}M=\{F(z)=2{\textrm{Re}}{\Psi _{G,X}(z)}=0\}.\end{aligned}$$

Since $\Psi _{G,X}$ is a holomorphic function, then M is Levi-flat.

Example 6.2

Let $G=(z_1,iz_2,(-1-i)z_3)$ and $X=(z_1,z_2,z_3)$. According to Seade (1997, Theorem 1), $\psi _{G,X}$ does not satisfy the Milnor fibration condition at the origin. The Hermitian product of G and X is given by

$$\begin{aligned}\langle G,X \rangle =z_1\overline{z}_1+iz_2\overline{z}_2+(-1-i)z_3\overline{z}_3.\end{aligned}$$

Consider $M=\{F(z)=2{\textrm{Re}}\ \langle G(z),X(z) \rangle =z_1\overline{z}_1-z_3\overline{z}_3=0\}$. Let’s us verify that M is Levi-flat. First, we note that $F(z)=z_1\overline{z}_1-z_3\overline{z}_3$, and the partial derivatives of F are given by

$$\begin{aligned} \partial F&= \overline{z}_1dz_1-\overline{z}_3dz_3,\\ \overline{\partial }F&= z_1d\overline{z}_1-z_3d\overline{z}_3. \end{aligned}$$

Then, ${\textsf {Sing}}(M)=\{(0,z_2,0) \in {\mathbb {C}}^3:z_2\in {\mathbb {C}}\}$. Clearly, $\dim _{{\mathbb {R}}}{\textsf {Sing}}(M)=2$. Moreover,

$$\begin{aligned}M^*=\{(z_1,z_2,z_3) \in {\mathbb {C}}^3: z_1\overline{z}_1- z_3\overline{z}_3=0, z_1\ne 0 \;\text{ and }\; z_3\ne 0\}.\end{aligned}$$

We have $\partial \overline{\partial }F= dz_1\wedge d\overline{z}_1-dz_3\wedge d\overline{z}_3,$ and

$$\begin{aligned} \partial F \wedge \overline{\partial } F&= z_1\overline{z}_1dz_1\wedge d\overline{z}_1 - \overline{z}_1z_3 dz_1 \wedge d\overline{z}_3 -\overline{z}_3z_1 dz_3 \wedge d\overline{z}_1 + z_3\overline{z}_3 dz_3\wedge d\overline{z}_3, \end{aligned}$$

that yields $\partial F \wedge \overline{\partial } F \wedge \partial \overline{\partial } F=(z_3\overline{z}_3-z_1\overline{z}_1)dz_1\wedge d\overline{z}_1\wedge dz_3\wedge d\overline{z}_3.$ Thus, $\partial F(p) \wedge \overline{\partial }F (p)\wedge \partial \overline{\partial }F(p)=0$ for all $p \in M$ and therefore, M is Levi-flat.

Example 6.3

Let $G(z)=(z_1,z_2)$ and $X(z)=(-iz_2,iz_1)$. Then

$$\begin{aligned}M=\{F(z)=2{\textrm{Re}}\ \langle G(z),X(z)\rangle =i(z_2\overline{z}_1-z_1\overline{z}_2)=0\}\end{aligned}$$

is clearly Levi-flat whose singular set is ${\textsf {Sing}}(M)=\{0\}$.

In general, when $G(z)=(z_1,z_2)$ is the radial vector field, $X=(\lambda _1z_1, \lambda _2z_2)$, and if ${\textrm{Re}}(\lambda _k)\ne 0$ for $k=1,2$, then the singular set of $M=\{2{\textrm{Re}}\ \langle G(z),X(z)\rangle =0\}$ is just the origin $0 \in {\mathbb {C}}^2$, and M is Levi-flat. However, in this case, the map $\psi _{G,X}$ also does not satisfy the Milnor fibration condition at the origin, see Seade (1996, Example 3.4).