Regular multi-types and the Bloom conjecture

TL;DR

This paper proves the Bloom conjecture for complex dimension three and verifies it for higher dimensions when s=n-2, advancing understanding of pseudoconvexity in complex analysis.

Contribution

It establishes the equality of vector field type and regular contact type, solving a long-standing open problem in complex geometry for specific cases.

Findings

Proves equality of vector field and contact types in complex dimension three.

Verifies Bloom conjecture for s=n-2 in higher dimensions.

Provides the first positive result in pseudoconvexity-sensitive cases since 1981.

Abstract

We prove equality of the vector field (iterated commutator) type and the regular contact type, which together with the Bloom theorem on equality of the Levi-form type and the regular contact type provides a complete solution of a long standing open problem of Bloom in the case of complex dimension three. For general dimensions, we verify the Bloom conjecture when $s=n-2$, which provides the first positive result in the pseudoconvexity sensitive case for a real hypersurface in ${\mathbb{C}}^n$ after his important work in 1981.

Equations

$$\begin{split}a^{(s)}(M,p)=\sup\limits_{X}\big{\{}r|\ &\exists\text{ an $s$-dimensional complex submanifold}\ X\\ &\text{whose order of contact with $M$ at $p$ is $r$}\big{\}}.\end{split}$$

$$\begin{array}{r}{t}^{(s)}(M,p)=\underset{B}{\sup }\{t(B,p)|B\text{is an s-dimensional subbundle of }{T}^{1,0}M\}.\end{array}$$

$${\text{tr}}_{V_B}{\mathcal{L}}_{M,p}=\sum _{j=1}^s⟨[L_j,\overline{L_j}],\partial \rho ⟩(p).$$

$$F_{m-3}\cdots F_{1}\big{(}\text{tr}_{V_{B}}\mathcal{L}_{M,p}\big{)}(p)=0$$

$$G_{m-2}\cdots G_{1}\big{(}\text{tr}_{V_{B}}\mathcal{L}_{M,p}\big{)}(p)\neq 0.$$

$$\begin{array}{r}{c}^{(s)}(M,p)=\underset{B}{\sup }\{c(B,p):B\text{ is an s-dimensional subbundle of}{T}^{1,0}M\}.\end{array}$$

$$t^{(n-1)}(M,p)=c^{(n-1)}(M,p)=a^{(n-1)}(M,p)forM\subset {\mathbb{C}}^n.$$

$$\begin{array}{r}{t}^{(s)}(M,p)=c^{(s)}(M,p)=a^{(s)}(M,p).\end{array}$$

$$\begin{array}{r}{t}^{(n-2)}(M,p)=a^{(n-2)}(M,p)=c^{(n-2)}(M,p).\end{array}$$

$$\begin{array}{r}{t}^{(1)}(M,p)=a^{(1)}(M,p)=c^{(1)}(M,p).\end{array}$$

$$\begin{array}{r}\rho =-2\text{Re}(w)+\chi (z,\overline{z},\text{Im}w),\chi (z,\overline{z},\text{Im}w)=O(|z^2|+|z\text{Im}w|).\end{array}$$

$$t^{(n-2)}(M,0),c^{(n-2)}(M,0)\ge a^{(n-2)}(M,0)=\infty$$

$$\begin{array}{r}\chi (z,0,0)=O(a^{(n-2)}(M,0)+1).\end{array}$$

$$\begin{split}L_{i}=\frac{\partial}{\partial z_{i}}-\frac{\partial\rho}{\partial z_{i}}\big{(}\frac{\partial\rho}{\partial w}\big{)}^{-1}\frac{\partial}{\partial w}\ \ \text{for}\ i=1,\cdots,n-1.\end{split}$$

$$S_j=\sum _{h=1}^{n-1}{a}_{jh}{L}_{h}with{a}_{jh}(0)={\delta }_{jh}for\; 1\le j,h\le n-2.$$

$$\begin{array}{r}{F}_{\ast }(L_i)=\sum _{j=1}^{n-1}\frac{\partial z_j^{\prime }}{\partial z_i}{L}_j^{\prime }\text{for}i=1,\cdots ,n-1.\end{array}$$

$$\begin{array}{rl}{l}_0^{\ast }=& \underset{1\le j\le n-2}{\min }\{k_j:k_j=\text{vanishing order of}{a}_{j(n-1)}(z_1,\cdots ,z_{n-2},0,\overline{z_1}.\cdots ,\overline{z_{n-2}},0)at\; 0\}.\end{array}$$

$$a_{j(n-1)}^{(l_0^{\ast })}(0,\cdots ,0,z_j,\cdots ,z_{n-2},0,\cdots ,0)=0\text{for all }1\le j\le n-2.$$

$$z_j^{\prime }=z_j\text{for}1\le j\le n-2,z_{n-1}^{\prime }=z_{n-1}-{\int }_0^{z_1}{a}_{1(n-1)}^{(l_0^{\ast })}(\xi ,z_2,\cdots ,z_{n-2},0,\cdots ,0)d\xi ,w=w^{\prime }.$$

$$\begin{array}{rl}& \frac{\partial }{\partial z_1}=\frac{\partial }{\partial z_1^{\prime }}-a_{1(n-1)}^{(l_0^{\ast })}(z_1,\cdots ,z_{n-2},0,\cdots ,0)\frac{\partial }{\partial z_{n-1}^{\prime }},\\ & \frac{\partial }{\partial z_j}=\frac{\partial }{\partial z_j^{\prime }}+O(l_0^{\ast })\frac{\partial }{\partial z_{n-1}^{\prime }}\text{for}2\le j\le n-2,\\ & \frac{\partial }{\partial z_{n-1}}=\frac{\partial }{\partial z_{n-1}^{\prime }}.\end{array}$$

$$\begin{array}{rl}{S}_1=& \sum _{h=1}^{n-1}{a}_{1h}{L}_h=a_{11}(L_1^{\prime }-a_{1(n-1)}^{(l_0^{\ast })}(z_1,\cdots ,z_{n-2},0,\cdots ,0)L_{n-1}^{\prime })\\ & +\sum _{h=2}^{n-2}{a}_{1h}(L_h^{\prime }+O(l_0^{\ast })L_{n-1}^{\prime })+a_{1(n-1)}{L}_{n-1}^{\prime }.\end{array}$$

$$-a_{11}{a}_{1(n-1)}^{(l_0^{\ast })}(z_1,\cdots ,z_{n-2},0,\cdots ,0)+\sum _{h=2}^{n-2}{a}_{1h}\cdot O(l_0^{\ast })+a_{1(n-1)}.$$

$$z_j^{\prime }=z_j,1\le j\le n-2,z_{n-1}^{\prime }=z_{n-1}-{\int }_0^{z_j}{a}_{j(n-1)}^{(l_0^{\ast })}(0,\cdots ,0,\xi ,z_{j+1},\cdots ,z_{n-2},0,\cdots ,0)d\xi .$$

$$a_{j(n-1)}^{(l_0^{\ast })}(0,\cdots ,0,z_j,\cdots ,z_{n-2},0,\cdots ,0)=0.$$

$$a_{h(n-1)}^{(l_0^{\ast })}(0,\cdots ,0,z_h,\cdots ,z_{n-2},0,\cdots ,0)=0\; for\; 1\le h\le j-1.$$

$$\begin{gathered} l_0:=\underset{1\le j\le n-2}{\min }\{k_j:k_j={\text{ord}}_{z=0}{a}_{j(n-1)}(z_1,\cdots ,z_{n-2},0,\overline{z_1}, \\ \cdots ,\overline{z_{n-2}},0)\}\ge a^{(n-2)}(M,0) \end{gathered}$$

$$a_{1(n-1)}^{(l_0)}(z_1,\cdots ,z_{n-2},0,\cdots ,0)=0.$$

$$a_{j(n-1)}^{(l_0)}(z_1,\cdots ,z_{n-2},0,\cdots ,0)\equiv 0$$

$$\begin{array}{r}{z}_1^{\prime }=z_1,z_j^{\prime }=z_j-{\alpha }_j{z}_1,\text{for}2\le j\le n-2,z_{n-1}^{\prime }=z_{n-1},w^{\prime }=w.\end{array}$$

$$\begin{array}{rl}& {\rho }^{[m]}(z_1,0,\cdots ,0,z_{n-1},0,\overline{z_1},0,\cdots ,0,\overline{z_{n-1}},0)\not\equiv 0,\\ & a_{1(n-1)}^{(l_0)}(z_1,0,\cdots ,0,\overline{z_1},0,\cdots ,0)\text{contains no non-trivial holomorphic terms}.\end{array}$$

Full text

1 Introduction

Let $D$ be a smoothly bounded pseudoconvex domain in ${\mathbb{C}}^{n}$ for $n\geq 2$. Many analytic and geometric properties of $D$ are determined by its boundary holomorphic invariants. To generalize his subelliptic estimate for the $\overline{\partial}$-Neumann problem from bounded strongly pseudoconvex domains [FK] to bounded weakly pseudoconvex domains in ${\mathbb{C}}^{2}$, Kohn in a fundamental paper [Kohn1] investigated three different boundary invariants for $D\subset{\mathbb{C}}^{2}$. These invariants describe, respectively, the maximum order of contact with smooth holomorphic curves at a boundary point, degeneracy of the Levi-form along the CR directions and the length of the iterated Lie brackets of boundary CR vector fields as well as their conjugates needed to recover the boundary contact direction. Kohn proved that all these invariants are in fact the same, called the type value of a point on $\partial D\subset{\mathbb{C}}^{2}$. When this type value is finite at each point, Kohn’s work in [Kohn1] together with that of Greiner [Gr] (see also Rothschild-Stein [RS]) gives the precise information of how much the subelliptic gain one obtains for the $\overline{\partial}$-Neumann problem for a smoothly bounded weakly pseudoconvex domain in ${\mathbb{C}}^{2}$. The finite type condition initiated by the work of Kohn has played fundamental roles in late studies of many problems. For instance, Bedford-Fornaess [BF] (see also the later work of Fornaess-Sibony [FS]) exploited peak functions over weakly pseudoconvex domains of finite type in ${\mathbb{C}}^{2}$ and discovered close connections of the type value of the boundary and the Hölder-continuity of the peak functions up to the boundary.

Generalization of Kohn’s notion of the boundary finite type condition to higher dimensions has been a subject under extensive investigations in the past 40 years in Several Complex Variables. Kohn later introduced a finite type condition in higher dimensions through the subelliptic multiplier ideals [Kohn2]. The understanding of this type has later revived to be a very active field of studies through the work of many people. (See Basyrov-Nicoara-Zaitsev[BNZ], Diederich-Fornaess [DF], Siu [Siu], Kim-Zaistev [KZ], Zaistev [Zai] and the reference therein.) Bloom [Bl1] and Bloom-Graham [BG1] established Kohn’s original notion of types in ${\mathbb{C}}^{2}$ to any dimensions which are called the regular multi-types. D’Angelo [DA1] introduced his important and famous notion of (D’Angelo) finite type conditions by considering the order of contact with not just smooth complex manifolds but possibly singular complex analytic varieties, which turns out to be equivalent to the existence of the subelliptic estimate by the work of Kohn [Kohn2], Diederich-Fornaess [DF] and Catlin [Cat2]. Catlin in [Cat1] studied his famous multitype condition as well as its connection with the boundary stratification in terms of the degeneracy of Levi forms. McNeal [Mc] and later Boas-Straube [BS] studied the the line type condition for convex domains and proved its equivalence with the D’Angelo type, which was further applied by Fu-Isav-Krantz [FIK] to prove the equivalence of the D’Angelo type with the regular contact type for Reinhardt domains.

All these type conditions mentioned above were introduced through different aspects of studies. Revealing the connections among them always brings our deeper understanding of the subject. For instance, proving that the Kohn multiplier ideal type is equivalent to the finite D’Angelo type would provide a new and much more direct solution of the $\overline{\partial}$-Neumann problem [Cat2].

In this paper, we will be concerned with the three multi-regular types. We will be especially interested in the question when all these types are equivalent, known as the Bloom problem. We will show that the vector field type (which is also called the Hörmander type in some other contents) coincides with the regular contact type in the case of dimension three. This result, together with the work of Bloom in 1981 [Bl2] on the equality of the Levi-form type with the contact type in ${\mathbb{C}}^{3}$, provides a complete solution of Bloom’s conjecture in the case of dimension three. In general dimensions, we will show that the first three pseudoconvexity-sensitive $(n-2)$-types all are the same. Our paper makes a progress along the lines of the long-standing Bloom conjecture after his striking work in 1981.

2 Statement of the main theorem

Let $M\subset\mathbb{C}^{n}$ be a smooth real hypersurface with $p\in M$. Then dim${}_{\mathbb{C}}T^{1,0}_{p}M=n-1$ for $p\in M$. For any $1\leq s\leq n-1$, we have the following three sets of important local holomorphic invariants ([Bl2]), used to describe the finite holomorphic non-degeneracy of $M$ at $p$.

(i): The $s$-contact type $a^{(s)}(M,p)$:

[TABLE]

Let $\rho$ be a defining function of $M$ near $p$, namely, $\rho\in C^{\infty}(U)$ with $U$ an open neighbourhood of $p\in\mathbb{C}^{n}$ and $U\cap M=\{\rho=0\}\cap U$, $d\rho|_{U\cap M}\neq 0$. Remark that the order of contact of $X$ with $M$ at $p$ is defined as the order of vanishing of $\rho|_{X}$ at $p$.

(ii) The $s$-vector field type $t^{(s)}(M,p)$:

Let $B$ be an $s$-dimensional subbundle of $T^{1,0}M$. We let $\mathcal{M}_{1}(B)$ be the $C^{\infty}(M)$-module spanned by the smooth tangential $(1,0)$ vector fields $L$ with $L|_{q}\in B|_{q}$ for each $q\in M$, together with the conjugate of these vector fields.

For $\mu\geq 1$, we let $\mathcal{M}_{\mu}(B)$ denote the $C^{\infty}(M)$-module spanned by commutators of length less than or equal to $\mu$ of vector fields from $\mathcal{M}_{1}(B)$. A commutator of length $\mu$ of vector fields in $\mathcal{M}_{1}(B)$ is a vector field of the following form: $[Y_{\mu},[Y_{\mu-1},\cdots,[Y_{2},Y_{1}]\cdots]$. Here $Y_{j}\in\mathcal{M}_{1}(B)$. Define $t^{(s)}(B,p)=m$ if $\langle F,\partial\rho\rangle(p)=0$ for any $F\in\mathcal{M}_{m-1}(B)$ but $\langle G,\partial\rho\rangle(p)\neq 0$ for a certain $G\in\mathcal{M}_{m}(B)$. Then

[TABLE]

$t^{(s)}(B,p)$ is the smallest length of the commutators by vector fields in $\mathcal{M}_{1}(B)$ to recover the complex contact direction in $\mathbb{C}T_{p}M$. $t^{(s)}(M,p)$ is the largest possible value among all $t^{(s)}(B,p)^{\prime}s$. Namely, $t^{(s)}(M,p)$ describes the degeneracy of the most degenerate $s$-subbundles of $T^{1,0}M$. Notice that it is intrinsically defined, independent of the ambient embedded space.

(iii) The $s$-Levi form tpype $c^{(s)}(M,p)$:

Let $B$ be as in (ii). Let $\mathcal{L}_{M,p}$ be a Levi form associated with a defining function $\rho$ near $p$ of $M$. For $V_{B}=\{L_{1},\cdots,L_{s}\}$, a basis of smooth sections of $B$ near $p$, we define the trace of $\mathcal{L}_{M,p}$ along $V_{B}$ by

[TABLE]

We define $c(B,p)=m$ if for any $m-3$ vector fields $F_{1},\cdots,F_{m-3}$ of $\mathcal{M}_{1}(B)$ and any basis $V_{B}$, it holds that

[TABLE]

and for a certain choice of $m-2$ vector fields $G_{1},\cdots,G_{m-2}$ of $\mathcal{M}_{1}(B)$ and a certain basis $V_{B}$, we have

[TABLE]

Then

[TABLE]

In his fundamental paper [Kohn1], when $n=2$, Kohn showed that $t^{(1)}(M,p)=c^{(1)}(M,p)=a^{(1)}(M,p)$. Bloom-Graham [BG2] and Bloom [Bl1] proved that

[TABLE]

And for any $1\leq s\leq n-2$, Bloom in [Bl2] observed that $a^{(s)}(M,p)\leq c^{(s)}(M,p)$ and $a^{(s)}(M,p)\leq t^{(s)}(M,p)$. For these results to hold there is no need to assume the pseudoconvexity of $M$. However, the following example of Bloom shows that for $n\geq 3$, when $M$ is not pesudoconvex, it may happen that $a^{(s)}(M,p)<c^{(s)}(M,p)$ and $a^{(s)}(M,p)<t^{(s)}(M,p)$ for $1\leq s\leq n-2$.

Example 2.1** (Bloom, [Bl2]).**

Let $\rho=2\text{Re}(w)+(z_{2}+\overline{z_{2}}+|z_{1}|^{2})^{2}$ and let $M=\{(z_{1},z_{2},w)\in\mathbb{C}^{3}|\ \rho=0\}$. Let $p=0$. Then $a^{(1)}(M,p)=4$ but $c^{(1)}(M,p)=t^{(1)}(M,p)=\infty$ .

With the pseudoconvexity assumption of $M$, Bloom in [Bl2] showed that when $M\subset{\mathbb{C}}^{3},$ $a^{(1)}(M,p)=c^{(1)}(M,p)$. Motivated by this result, Bloom in 1981 [Bl2] formulated the following conjecture:

Conjecture 2.2**.**

Let $M\subset\mathbb{C}^{n}$ be a pseudoconvex real hypersurface with $n\geq 3$. Then for any $1\leq s\leq n-2$ and $p\in M$,

[TABLE]

In this paper, we make a progress along the lines of the Bloom conjecture by presenting the proof of the following theorem:

Theorem 2.3**.**

Let $M\subset\mathbb{C}^{n}$ be a smooth pseudoconvex real hypersurface with $n\geq 3$. Then for $s=n-2$ and any $p\in M$, it holds that

[TABLE]

In particular, we obtain a proof of the remaining case of the Bloom conjecture in the case of complex dimension three by showing that when $M\subset{\mathbb{C}}^{3}$, we have $t^{(1)}(M,p)=a^{(1)}(M,p).$ ($s=n-2=1$ for $n=3$). This, together with the work of Bloom in 1981 on the equality $c^{(1)}(M,p)=a^{(1)}(M,p)$ for $M\subset{\mathbb{C}}^{3}$, finally provides a complete solution of the Bloom conjecture in the three dimensional case.

Theorem 2.4**.**

The Bloom conjecture holds in the case of complex dimension three. Namely, for a smooth pseudoconvex real hypersurface $M\subset\mathbb{C}^{3}$ and $p\in M$, it holds that

[TABLE]

Our proof of Theorem 2.3 is a combination of analytic and geometric arguments along the lines of CR geometry. (See the book by Baouendi-Ebenfelt-Rothschild [BER]). Our paper is mainly on commutators of vector fields. Commutators of vector fields are not just important in complex analysis but also play a fundamental role in many problems bordering complex analysis and sub-elliptic analysis. For instance, in the paper of Adwan-Berhanu [AB], the commutator type condition of vector fields is crucially applied to get various real analytic hypo-ellipticity results. See also the book of Berhanu-Cordaro-Hounie [BCH] and a paper of Derridj [Derr] for many references and historical discussions on this matter. In $\S 3$, we give a general set-up and provide a normalization of the related vector fields. In $\S 4$, we give a proof of Theorem 2.3 assuming Theorem 6.1. $\S 5$ and $\S 6$ are dedicated to the long proof of Theorem 6.1 on a sort of uniqueness of a complex linear PDE associated with a CR singular submanifold contained in a psuedoconvex hypersurface. Already from the work of Chern-Moser [CM], it is clear that a good weight system is always important to single out the boundary holomorphic invariants for real hypersurfaces in a complex Euclidean space. (See also [Bl2, BG2, Cat1, GS1, GS2, MW, HY, Kol1, Kol2] and the references therein concerning different weight systems used in different settings). In this work, we will adapt the weight system introduced by Bloom in [Bl2] to truncate the real hypersurface so that the singular Frobenius-Nagano theorem can be applied. Then we will derive contradictions if the theorem fails to be true by proving the non-existence of certain CR manifolds (through Proposition 4.5) and CR singular manifolds (through Theorem 6.1) generated by the truncated CR vector fields in the truncated hypersurface. To attack the Bloom conjecture, it is crucial to find a good use of the pseudoconvexity. In the case considered by Bloom [Bl2] for equality of $a^{1}(M,p)=c^{1}(M,p)$ of dimension three, the pseudoconvexity is used to fundamentally apply a neat result of Diederich-Fornaess [DF] (see also Freedman [Fre]), which says the Lie-bracket operation is closed for sections in the null space of Levi- form. This result can not be applied in the case we are considering. To handle the major difficulties in our consideration, the pseudoconvexity is used for the validity of the Hopf lemma (Proposition 4.5) and for obtaining the triviality of solutions of a complex linear equation with real part plurisubharmonic (Theorem 6.1). It is also interesting to notice the important role played by the Euler vector field in the course of our proof.

To finishing off this section, we would like to mention a result by D’Angelo [DA2] which shows that for a smooth pseudoconvex real hypersurface $M\subset\mathbb{C}^{n}$ $(n\geq 3)$ and for $p\in M$, if one of the two invariants $t^{(1)}(B_{1},p)\ \hbox{and}\ c^{(1)}(B_{1},p)$ is $4$ then they both are $4$. Here $B_{1}$ is a one dimensional smooth subbundle of $T^{(1,0)}M$. He also obtained some estimate of $c^{(1)}(B_{1},p)$ in terms of $t^{(1)}(B_{1},p)$. Here we mention that due to the pseudoconvexity, when $M$ at $p$ is not strongly pseudoconvex, then $t^{(1)}(B_{1},p)$ and $c^{(1)}(B_{1},p)$ are at least $4$. We also mention the paper by McNeal-Mernik ([MM]) and the paper by D’Angelo [DA4] on equality of the regular contact type with the D’Angelo type when either one is 4 under the pseudo convexity or even some weaker conditions.

3 Normalization of CR vector fields

Denote by $(z_{1},\cdots,z_{n-1},w)$ the coordinates in $\mathbb{C}^{n}$. Let $M\subset U$ be a smooth real hypersurface in ${\mathbb{C}}^{n}$ with $p\in M$ and let $\rho$ be a defining function of $M$ near $p$. After a holomorphic change of coordinates, we may assume that $p=0$ and $\rho$ takes the following form:

[TABLE]

We will assume that $a^{(n-2)}(M,0)<\infty$ in all that follows, for otherwise

[TABLE]

and thus all these invariants coincide. After a holomorphic change of coordinates of the form $(z^{\prime},w^{\prime})=(z,w+O(2))$, we assume that

[TABLE]

Shrinking $U$ if necessary, we assume $\frac{\partial\rho}{\partial w}(z,w)\neq 0$ for $(z,w)\in U$. For a defining function $\rho$ defined over $U$ as in (3.1), write

[TABLE]

Then $\{L_{i}\}_{i=1}^{n-1}$ forms a basis for the space of CR vector fields along $M$. Let $B$ be an $(n-2)$ dimensional subbundle of $T^{1,0}M$. Assume that the sections of $B$ are generated by a certain linearly independent smooth CR vector fields $S_{1},\cdots,S_{n-2}$ along $M$ near [math]. After a linear holomorphic change of coordinates, we assume that $S_{j}(0)=L_{j}(0)=\frac{\partial}{\partial z_{j}}|_{0}$ for $1\leq j\leq n-2$. Write

[TABLE]

We recall the following fact from [Bl2, Lemma 5.2], which gives the transformation law for $\{L_{j}^{\prime},\rho^{\prime}\}$ and $\{L_{j},\rho\}$ under a holomorphic change of coordinates $(z^{\prime},w^{\prime})=F(z,w)$ with $\rho=\rho^{\prime}\circ F$, $F(0)=0$.

Lemma 3.1**.**

Let $(z^{\prime},w^{\prime})=F(z,w)=(z_{1}^{\prime},\cdots,z_{n-1}^{\prime},w^{\prime})$ be a new holomorphic coordinate system where $z^{\prime}_{j}=z^{\prime}_{j}(z_{1},\cdots,z_{n-1})$ for $j=1,\cdots,n-1$, $w^{\prime}=w$ with $z^{\prime}(0)=0$. Then we have

[TABLE]

With $S_{j}$ and the frame $\{L_{j}\}$ being given as above, we define

[TABLE]

In this section, for a smooth function $A$, we write $A^{(\tau)}(z,\overline{z})$ for the sum of monomials of (ordinary) degree $\tau$ in its Taylor expansion at [math]; also when we mention a holomorphic change of coordinates, we refer to a special type of holomorphic maps of the form $(z^{\prime},w^{\prime})=F(z,w)$ as in Lemma 3.1.

Lemma 3.2**.**

Suppose $l_{0}^{*}\not=\infty$. After a holomorphic change of coordinates, we have

[TABLE]

Proof.

Let

[TABLE]

Then in the new coordinates $(z^{\prime},w^{\prime})$, we have

[TABLE]

In the new coordinates, by Lemma 3.1, we have

[TABLE]

Hence in the new coordinates, the coefficient $a_{1(n-1)}$ is changed to

[TABLE]

Recall that $a_{1j}=\delta_{1j}+o(1)$ for $1\leq j\leq n-2$. Hence in these new coordinates, which are still denoted by $(z,w)$, we have $a^{(l^{*}_{0})}_{1(n-1)}(z_{1},\cdots,z_{n-2},0,\cdots,0)=0$.

Suppose that we have achieved $a^{(l^{*}_{0})}_{h(n-1)}(0,\cdots,0,z_{h},\cdots,z_{n-2},0,\cdots,0)=0$ for $1\leq h\leq j-1$. We next show that we can make $a^{(l^{*}_{0})}_{j(n-1)}(0,\cdots,0,z_{j},\cdots,z_{n-2},0,\cdots,0)=0$ after a holomorphic change of coordinates. Set $w=w^{\prime}$ and

[TABLE]

By a similar argument as in the proof for $a^{(l^{*}_{0})}_{1(n-1)}(z_{1},\cdots,z_{n-2},0,\cdots,0)=0$, we have

[TABLE]

Notice that this transformation of coordinates preserves the property:

[TABLE]

By induction, this completes the proof of Lemma 3.2. ∎

Next, after the normalization as in (3.2), we either have

[TABLE]

or $l_{0}\leq a^{(n-2)}(M,0)-1.$ In the case of (3.9), we re-define $l_{0}$ to be $a^{(n-2)}(M,0)$.

Proposition 3.3**.**

Assume that $l_{0}\leq a^{(n-2)}(M,0)-1$. After a holomorphic change of coordinates we can normalize the coefficients of $\{S_{j}\}$ to further satisfy one of the following two normalization properties:

1. (I)

$a_{j(n-1)}^{(l_{0})}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)$* is holomorphic in $z_{1},\cdots,z_{n-2}$ ** for each** j, and there exits $j_{0}\in[2,n-2]$ such that $a^{(l_{0})}_{j(n-1)}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)=0$  for  $1\leq j\leq j_{0}-1,$ $a^{(l_{0})}_{j_{0}(n-1)}(0,\cdots,0,z_{j_{0}},\cdots,z_{n-2},0,\cdots,0)=0$, but *

$a^{(l_{0})}_{j_{0}(n-1)}(z_{1},\cdots,z_{n-2},0,\cdots,0)\not\equiv 0$. 2. (II)

$a^{(l_{0})}_{1(n-1)}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)$* is not a holomorphic polynomial *

and $a^{(l_{0})}_{1(n-1)}(z_{1},\cdots,z_{n-2},0,\cdots,0)=0$.

Proof.

(I): First, we assume that each $a^{(l_{0})}_{j(n-1)}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)$ is holomorphic, and each $a^{(l_{0})}_{j(n-1)}$ satisfies the properties as in Lemma 3.2. Then

[TABLE]

By the definition of $l_{0}$, we can find the smallest $j_{0}\in[2,n-2]$ such that

[TABLE]

for all $1\leq j\leq j_{0}-1$, but $a^{(l_{0})}_{j_{0}(n-1)}(z_{1},\cdots,z_{n-2},0,\cdots,0)\not\equiv 0$. By Lemma 3.2, this $j_{0}$ satisfies the property in part (I) of the proposition.

(II): Next, assume that $a^{(l_{0})}_{j(n-1)}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)$ is not holomorphic for a certain $j\in[1,n-2]$. Switching $j$ with the index $1$ and repeating the proof in Lemma 3.2, we can make $a^{(l_{0})}_{1(n-1)}(z_{1},\cdots,z_{n-2},0,\cdots,0)=0$ and achieve the other normalization properties as in Lemma 3.2. Notice that $l_{0}$ is not changed after this normalization procedure. This completes the proof of the proposition. ∎

Define the weight of $z_{j}$ and $\overline{z_{j}}$ for $1\leq j\leq n-2$ to be $1$. The weight of $z_{n-1}$ and $\overline{z_{n-1}}$ is defined to be $l_{0}+1$ and the weight of $w$ is defined to be $m$ that is the lowest weighted vanishing order of $\rho$ in the expansion of $\rho(z,0,\overline{z},0)$ at [math] with respect to the weights of $\{z_{1},\cdots,z_{n-2},z_{n-1}\}$ just defined. In what follows, for a smooth function $A$, we write $A^{[\sigma]}(z_{1},\cdots,z_{n-1},\overline{z_{1}},\cdots,\overline{z_{n-1}})$ for the weighted homogeneous part of weighted degree $\sigma$ with the weight system just defined in its Taylor expansion at [math]. Then we have the following

Proposition 3.4**.**

In the case of Proposition 3.3 (II), we can further apply a holomorphic transformation of coordinates and change the basis $\{S_{j}\}$ if needed to make the coefficients of $\{S_{j}\}$ in the expansion with respect to $\{L_{j}\}$ satisfy one of the following two normalizations:

1. (1)

$a^{(l_{0})}_{1(n-1)}(z_{1},0,\cdots,0,\overline{z_{1}},0,\cdots,0)\not\equiv 0$*, $a^{(l_{0})}_{1(n-1)}(z_{1},0,\cdots,0)=0$ *

($a^{(l_{0})}_{1(n-1)}(z_{1},\cdots,z_{n-2},0)=0$*, **in fact *), $\rho^{[m]}(z_{1},0,\cdots,0,z_{n-1},0,\overline{z_{1}},0,\cdots,0,\overline{z_{n-1}},0)$ is not identically zero (and contains no non-trivial holomorphic terms). 2. (2)

For a certain $j\in[1,n-2]$, $a^{(l_{0})}_{j(n-1)}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)$ is not holomorphic,

$\sum_{k=1}^{n-2}z_{k}a^{(l_{0})}_{k(n-1)}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)=0$,

$\rho^{[m]}(z_{1},\cdots,z_{n-1},0,\overline{z_{1}},\cdots,\overline{z_{n-1}},0)$* is not identically zero (and contains no non-trivial holomorphic terms).*

Proof.

Consider the following change of coordinates:

[TABLE]

We first give a sufficient condition under which, for a generic choice of $\alpha_{j}$ with $2\leq j\leq n-2$, we have

[TABLE]

Notice that

[TABLE]

The coefficient of ${z^{\prime}}_{1}^{t}{z^{\prime}}_{n-1}^{\mu}\overline{{z^{\prime}}_{1}}^{s}\overline{{z^{\prime}}_{n-1}}^{\nu}$ with $t+s+\mu+\nu=m$ in its Taylor expansion is

[TABLE]

Here $\alpha_{1}=1$, $\alpha=(\alpha_{1},\cdots,\alpha_{n-2})$, and $\rho^{[m]}_{(H\mu 0)(J\nu 0)}$ is the coefficient of

[TABLE]

in the Taylor expansion of $\rho$ at [math].

Notice that this term is [math] for a generic choice of $\alpha$ if and only if $\rho^{[m]}_{(H\mu 0)(J\nu 0)}=0$ for any pair $(H,J)$ with $\sum h_{\lambda}=t,\ \sum j_{\lambda}=s$. By our choice of the weight $m$, their exists a pair $(H\mu 0)(J\nu 0)$ with $|J|+\nu>0$ such that $\rho^{[m]}_{(H\mu 0)(J\nu 0)}\neq 0$. Thus for a generic choice of $\alpha_{j}^{\prime}s$, we have

[TABLE]

Since $\rho^{[m]}$ contains no holomorphic terms, so is $\rho^{\prime[m]}(z_{1}^{\prime},0,\cdots,0,z_{n-1}^{\prime},0,\overline{z_{1}^{\prime}},0,\cdots,0,\overline{z_{n-1}^{\prime}},0)$. Hence for a generic choice of $\alpha$, the statement in the first line of (LABEL:rhoan) holds.

Next notice that

[TABLE]

Thus by Lemma 3.1, we know

[TABLE]

Set $S_{1}^{\prime}=S_{1}+\sum_{\lambda=2}^{n-2}\alpha_{\lambda}S_{\lambda}$, $S_{j}^{\prime}=S_{j}$. Then

[TABLE]

Hence

[TABLE]

Then the coefficient of ${z_{1}}^{{}^{\prime}t}\overline{{z_{1}}^{{}^{\prime}s}}$ with $t+s=l_{0}$ in $a_{1(n-1)}^{\prime}(z_{1}^{\prime},0,\overline{z_{1}}^{\prime},0)$ is the following

[TABLE]

where $e_{\lambda}=(0,\cdots,0,1,0\cdots,0)$ with $1$ at the $\lambda$-th position. This term is [math] for a generic choice of $\alpha$ if and only if $\sum_{\lambda=1}^{n-2}(a^{(l_{0})}_{\lambda(n-1)})_{(H-e_{\lambda})J}=0$. We next proceed in two steps:

(1). First, we suppose that there exists a pair $(H,J)$ with $|J|\neq 0$ such that

[TABLE]

Then for a generic choice of $\alpha$, $a^{\prime(l_{0})}_{1(n-1)}(z_{1}^{\prime},0,\overline{z_{1}}^{\prime},0)$ contains no non-trivial holomorphic terms. Through the normalization procedure as in Lemma 3.2, we can make

[TABLE]

and thus, in particular, $a^{\prime(l_{0})}_{1(n-1)}(z_{1}^{\prime},0,\cdots,0)=0$. We point out that this transformation preserves the statement in the first line of (LABEL:rhoan). Then $a^{(l_{0})}_{1(n-1)}$ and $\rho^{[m]}$ satisfy the desired properties in (1) of Proposition 3.4. Next, we can repeat the same argument in Lemma 3.2 to normalize $a_{j(n-1)}^{(l_{0})}$ for $j\geq 2$ and thus obtain the normalization for $a_{j(n-1)}^{(l_{0})}$ with $j=2,\cdots,n-2$.

(2). We now suppose

[TABLE]

We will show that by a suitable change of coordinates of the form $z_{j}^{\prime}=z_{j}$, $z_{n-1}^{\prime}=z_{n-1}+g(z_{1},\cdots,z_{n-2})$, $w^{\prime}=w$, we can make

[TABLE]

Here $g(z_{1},\cdots,z_{n-2})$ is a homogeneous holomorphic polynomial of degree $l_{0}+1$.

In fact, under this transformation, we have

[TABLE]

Thus

[TABLE]

Hence

[TABLE]

Thus $\sum_{\lambda=1}^{n-2}(a^{\prime(l_{0})}_{\lambda(n-1)})_{(H-e_{\lambda})0}=0$ for any $H$ with $|H|=l_{0}+1$, which is equivalent to $\sum_{\lambda=1}^{n-2}z_{\lambda}a^{\prime(l_{0})}_{\lambda(n-1)}(z_{1},\cdots,z_{n-2},0)=0$, if and only if

[TABLE]

This is the well-known Euler equation and can be solved as follows:

Notice that if we write $g=\sum_{|J|=l_{0}+1}\Gamma_{J}z^{J}$, then

[TABLE]

Hence $g$ can be uniquely solved as $g=-\frac{1}{l_{0}+1}\sum_{\lambda=1}^{n-2}z_{\lambda}a^{(l_{0})}_{\lambda(n-1)}(z_{1},\cdots,z_{n-2},0)$. Thus we get the desired normalization property in (3.18). Notice that by (3.19), $(a^{(l_{0})}_{\lambda(n-1)})_{(H-e_{\lambda})J}$ with $|H|+|J|=l_{0}+1$, $|J|\neq 0$ is not changed under this transformation. Hence (3.17) still holds to be true.

Notice that (3.17) and (3.18) are equivalent to the normalization property in $(2)$ of Proposition 3.4. In fact,

[TABLE]

This completes the proof of Proposition 3.4. ∎

We summarize what we did in this section in the following:

Corollary 3.5**.**

Keep the same notations and definitions we have made so far. After a holomorphic change of coordinates and after choosing a suitable basis for $B$, we have one of the following three normalizations for $\{a_{j(n-1)}^{(l_{0})},\rho^{[m]}\}_{j=1}^{n-2}$ under the assumption that $l_{0}\leq a^{(n-2)}(M,0)-1$:

1. (I)

$a_{j(n-1)}^{(l_{0})}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)$* is holomorphic in $z_{1},\cdots,z_{n-2}$ ** for each** j, and there exits $j_{0}\in[2,n-2]$ such that $a^{(l_{0})}_{j(n-1)}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)=0$  for  $1\leq j\leq j_{0}-1,$ $a^{(l_{0})}_{j_{0}(n-1)}(0,\cdots,0,z_{j_{0}},\cdots,z_{n-2},0,\cdots,0)=0$, but *

$a^{(l_{0})}_{j_{0}(n-1)}(z_{1},\cdots,z_{n-2},0,\cdots,0)\not\equiv 0$. 2. (II)

$a^{(l_{0})}_{1(n-1)}(z_{1},0,\cdots,0,\overline{z_{1}},0,\cdots,0)\not\equiv 0$, $a^{(l_{0})}_{1(n-1)}(z_{1},0,\cdots,0)=0$,

$\rho^{[m]}(z_{1},0,\cdots,0,z_{n-1},0,\overline{z_{1}},0,\cdots,0,\overline{z_{n-1}},0)$* is not identically zero ( and contains no non-trivial holomorphic terms).* 3. (III)

For a certain $j\in[1,n-2]$, $a^{(l_{0})}_{j(n-1)}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)$ is not holomorphic, $\sum_{j=1}^{n-2}z_{j}a^{(l_{0})}_{j(n-1)}(z_{1},\cdots,z_{n-2},0,\overline{z_{1}},\cdots,\overline{z_{n-2}},0)=0$, and

[TABLE]

is not identically zero containing no non-trivial holomorphic terms.

4 Proof of Theorem 2.3

In this section, we present a proof of Theorem 2.3, assuming Lemma 5.2 and Theorem 6.1 whose proofs are long and will be given in $\S 5$ and $\S 6$.

Proof of the equality: $t^{(n-2)}(M,p)=a^{(n-2)}(M,p)$.

We keep the notations set up in $\S 2$ and $\S 3$. Assume that $M$ is defined as in (3.1) and (3.2). As we mentioned there, we assume that $a^{(n-2)}(M,p=0)<\infty$. Supposing that $t^{(n-2)}(M,0)>a^{(n-2)}(M,0)$, we will then seek a contradiction.

Let $B$ be an $(n-2)$-dimensional smooth vector subbundle of $T^{1,0}M$ such that $t^{(n-2)}(M,0)=t^{(n-2)}(B,0)$. By the assumption that $t^{(n-2)}(M,0)>a^{(n-2)}(M,0)$, for any $l\leq a^{(n-2)}(M,0)$ we have

[TABLE]

We also assume that the local sections of $B$ are generated by $S_{1},\cdots,S_{n-2}$.

Recall that the weight of $z_{j}$ for $1\leq j\leq n-2$ and their conjugates is $1$. Define the weight of $z_{n-1}$ and its conjugate to be $k=l_{0}+1$. Denote the weight of $w$ to be $m$, which is the lowest weighted vanishing order of $\rho(z,\overline{z},0)$ with respect to the weights just given. We also define

[TABLE]

By the definition of $a^{(n-2)}(M,0)$, when restricted to the $(n-2)$-manifold $\{(z,w):\ z_{n-1}=w=0\}$, the vanishing order of $\rho$ is bounded by $a^{(n-2)}(M,0)$. Thus $m\leq a^{(n-2)}(M,0)$. When $k\leq a^{(n-2)}(M,0)$, we assume that $S_{j}$ and $\rho$ are normalized as in Corollary 3.5.

Write

[TABLE]

Then $S_{j}^{0}$ is the sum of terms in $S_{j}$ of weighted degree $-1$.

Now, let $\mathcal{M}^{0}$ be the $C^{\infty}(M^{0})$-module spanned by $S_{j}^{0}$ and $\overline{S_{j}^{0}}$ for all $1\leq j\leq n-2$, where $M^{0}=\{(z,w):\ \rho^{[m]}=-2\text{Re}w+\chi^{[m]}(z,\overline{z},0)=0\}$ and $\mathcal{M}^{0}_{l}$ be the $C^{\infty}(M^{0})$ module formed by taking the Lie bracket of length $\leq l$ of sections from $\mathcal{M}^{0}$ for $l=2,\cdots$. $\mathcal{M}^{0}_{\infty}=\cup_{l\in\mathbb{N}}\mathcal{M}^{0}_{l}$. Now we need the following two lemmas.

Lemma 4.1**.**

It holds that $k<m$.

Proof.

Suppose that $k\geq m$. Then the weight of $z_{n-1}$ is no less than $m$. Hence $\chi^{[m]}(z,\overline{z},0)$ is independent of $z_{n-1}$. Write

[TABLE]

Since $S_{j}^{0}$ is tangent to $M^{0}$, whose defining function is independent of $z_{n-1}$, we see that $\widetilde{S_{j}^{0}}$ is tangent to $M^{0}$ and $\widetilde{a_{jn}}=\frac{\partial\chi^{[m]}}{\partial z_{j}}$. Hence $\widetilde{a_{jn}}$ is independent of $z_{n-1}$.

Regarding $M^{0}$ as a real hypersurface in ${\mathbb{C}}^{n-1}$. Let $\widetilde{\mathcal{M}}^{0}$ be the $C^{\infty}(M^{0})$ module spanned by $\widetilde{S_{j}^{0}}$ and $\overline{\widetilde{S_{j}^{0}}}$ for all $1\leq j\leq n-2$. Define $Q:{\mathcal{M}}^{0}\rightarrow\widetilde{\mathcal{M}}^{0}$ by sending $\sum_{j=1}^{n-1}d_{j}L_{j}^{0}\in{\mathcal{M}}^{0}$ to $\sum_{j=1}^{n-2}d_{j}L_{j}^{0}\in\widetilde{\mathcal{M}}^{0}$. Then by (4.1), for any $Z_{j}^{0}\in\widetilde{\mathcal{M}}^{0}$, there exists $Y_{j}^{0}\in{\mathcal{M}}^{0}$ with $Q(Y_{j}^{0})=Z_{j}^{0}$ such that

[TABLE]

(Indeed, we can simply take $Y_{j}^{0}$ to be $Z_{j}^{0}$, but regard it as a CR vector field of $M^{0}$ as a real hypersurface in ${\mathbb{C}}^{n}$.) Hence we have $t^{((n-1)-1)}(M^{0},0)>m$. However, by our construction, $a^{((n-1)-1)}(M^{0},0)=m$. This contradicts a result of Bloom-Graham for the $(n-1)$ types in [BG1], which says that $t^{((n-1)-1)}(M^{0},0)=a^{((n-1)-1)}(M^{0},0)$ for $M^{0}\subset{\mathbb{C}}^{n-1}$. ∎

Lemma 4.2**.**

For any $Y^{0}\in\mathcal{M}^{0}_{l}$, we have $\langle Y^{0},\partial\rho^{[m]}\rangle(0)=0$.

Proof.

Assume that $Y^{0}=[X_{l}^{0},\cdots,[X_{2}^{0},X_{1}^{0}]\cdots]$ with $X_{j}^{0}\in\mathcal{M}^{0}$. Write

[TABLE]

Here $Z_{j}^{0}$ is weighted homogeneous of degree $-1$ and wt$(B_{j})=\text{wt}(C_{j})=m-1$. A direct computation shows

[TABLE]

and by an induction,

[TABLE]

with $C_{l}^{0}$ a weighted homogeneous polynomial of weighted degree equal to $-l+m$. Hence $Y^{0}\equiv 0$ when $l>m$ and $Y^{0}|_{0}=0$ when $l<m$ mod $(\frac{\partial}{\partial z},\frac{\partial}{\partial\overline{z}},\frac{\partial}{\partial\overline{w}})$.

When $l=m$, Suppose that $Z_{j}\in\mathcal{M}_{1}$ such that $(Z_{j})^{0}=X_{j}^{0}$. Then $[Z_{j},Z_{k}]^{0}=[X_{j}^{0},X_{k}^{0}]$. Hence if $Z\in\mathcal{M}_{l}$ with $l=m$ such that $(Z)^{0}=Y^{0}$, then $Z=C_{l}^{0}Y^{0}+D_{l}\frac{\partial}{\partial w}$ mod $(\frac{\partial}{\partial z},\frac{\partial}{\partial\overline{z}},\frac{\partial}{\partial\overline{w}})$ with $\text{wt}(D_{l})>\text{wt}(C_{l}^{0})$. From (4.1), $Z|_{0}=0$ mod $(\frac{\partial}{\partial z},\frac{\partial}{\partial\overline{z}},\frac{\partial}{\partial\overline{w}})$. Thus we obtain $C_{m}^{0}\equiv 0$. Hence $\langle Y^{0},\partial\rho^{[m]}\rangle(0)=0$ for all $l\in\mathbb{N}$. ∎

Then we have $\frac{\partial}{\partial v}|_{0}\not\in\mathcal{M}^{0}_{\infty}$. By the Nagano theorem (see [BER], for instance), $\mathcal{M}^{0}_{\infty}$ gives a unique real analytic integral submanifold $N^{0}$ with $0\in N^{0}\subset M^{0}=\{-2\text{Re}w+\chi^{[m]}(z,\overline{z},0)=0\}$. Moreover, dim${}_{\mathbb{R}}N^{0}=\text{dim}_{\mathbb{R}}\text{Re}\mathcal{M}^{0}_{\infty}|_{N^{0}}$. Since $\frac{\partial}{\partial v}|_{0}\not\in T_{0}N^{0}$, $N^{0}$ is contained in the graph of $v=f_{1}(z,\overline{z},u)$ for a certain real analytic function $f_{1}$ near [math]. Since $u=\frac{1}{2}\chi^{[m]}(z,\overline{z},0)$, we conclude that $N^{0}$ is contained in the graph of

[TABLE]

We mention that from the pseudoconvexity of $M$, we immediately conclude the pseudoconvexity of $M^{0}$, which is equivalent to the plurisubharmonicity of $\hbox{Re}(f)=\chi^{[m]}(z,\overline{z},0)$.

Lemma 4.3**.**

The real dimension of $N^{0}$ is either $2n-3$ or $2n-2$.

Proof.

The proof is carried out in two steps according to the properties of $a_{j(n-1)}^{(k-1)}(z,\overline{z})$ in Proposition 3.3.

(1): Suppose we have the normalization in (I) of Proposition 3.3. We suppose that $(a_{j_{0}(n-1)}^{(k-1)})_{H+e_{\mu}}\neq 0$ with $H=(h_{1},\cdots,h_{n-2})$ and $1\leq\mu\leq j_{0}-1$. Then

[TABLE]

Write

[TABLE]

Then

[TABLE]

Since its conjugate is also in $\mathcal{M}_{\infty}^{0}$, we conclude that the dimension of $N^{0}$ is $2n-2$.

(2): Suppose we have the normalization in (II) of Proposition 3.3. Then there is a $(H,J)=(h_{1},\cdots,h_{n-2},j_{1},\cdots,j_{n-2})$ such that $(a_{1(n-1)}^{(k-1)})_{H(J+e_{\mu})}\neq 0$. Then

[TABLE]

Write $(X_{1},\cdots,X_{|H|})$ as in (LABEL:sh) and write

[TABLE]

Then

[TABLE]

Hence $Y_{HJ}\not\in\text{span}_{\mathbb{C}}\{S_{j}^{0},\overline{S_{j}^{0}},1\leq j\leq n-2\}$. Thus either Re$Y_{HJ}|_{0}\neq 0$ or Im$Y_{HJ}|_{0}\neq 0$.

Since $\frac{\partial}{\partial w}|_{0}$ and $\frac{\partial}{\partial\overline{w}}|_{0}$ are not tangent to $N^{0}$ at [math], the dimension of $N^{0}$ is either $2n-3$ or $2n-2$. ∎

Lemma 4.4**.**

When $N^{0}$ has real dimension $2n-2$, $f$ is a weighted homogeneous polynomial of weighted degree $m$.

Proof.

Let $X^{0}$ be a (weighted) homogeneous vector field from $\mathcal{M}^{0}_{\infty}$. Then from the equality that $X^{0}(-w+f)=\overline{X^{0}}({-w+f})\equiv 0$, it follows that $X^{0}(-w+f^{[m]})=X^{0}(\overline{-w+f^{[m]}})\equiv 0$. Hence the manifold defined by $w=f^{[m]}$ is also an integral manifold of the module $\mathcal{M}^{0}_{\infty}$ through [math]. By the uniqueness of the integrable manifold, we conclude that $f^{[m]}=f$. ∎

The rest of the argument is carried out according to the dimension of $N^{0}$. We remark that when the real dimension of $N^{0}$ is $2n-3$, it is a CR submanifold of hypersurface type, for it has a constant CR dimension $n-2$ everywhere. When its dimension is $2n-2$, it has CR dimension $n-1$ at the origin. Since it cannot be Levi-flat due to the fact that $\hbox{Re}(f)\not\equiv 0$, it is thus a codimension two CR singular submanifold.

Step I. In this step, we suppose $N^{0}$ is of real dimension $2n-2$. Since $\overline{S^{0}_{j}}$ is tangent to $N^{0}$, and since $N^{0}$ is defined by $w=f(z,\overline{z})$ for $z\approx 0$ in ${\mathbb{C}}^{n-1}$, we have

[TABLE]

By Lemma 4.1, we have $k<m\leq a^{(n-2)}(M,0)$. Our next discussions are divided into the following cases according to the normalization in Corollary 3.5.

Case (1): In this case, suppose that we have the normalization in (1) of Corollary 3.5. For $1\leq j\leq j_{0}-1$, $a_{j(n-1)}^{(k-1)}\equiv 0$. Thus (4.4) takes the form $\frac{\partial f}{\partial\overline{z_{j}}}=0$. Hence $f$ is holomorphic in $z_{1},\cdots,z_{j_{0}-1}$. By Lemma 5.2 to be proved in $\S 5$, since $\hbox{Re}(f)$ is plurisubharmonic and contains non-trivial holmorphic terms, $f$ is in fact independent of $z_{1},\cdots,z_{j_{0}-1}$. Setting $j=j_{0}$ in (4.4), we obtain

[TABLE]

Notice that the left hand side is independent of $z_{1},\cdots,z_{j_{0}-1}$. On the other hand, the right hand side is divisible by $\overline{a^{(k-1)}_{j_{0}(n-1)}}$, in which each term depends on $z_{1},\cdots,z_{j_{0}-1}$ or their conjugates. Thus $\frac{\partial f}{\partial\overline{z_{j_{0}}}}=\frac{\partial f}{\partial\overline{z_{n-1}}}=0$. Substituting this back to (4.4), we obtain $\frac{\partial f}{\partial\overline{z_{j}}}=0$ for each $1\leq j\leq n-2$. Thus $f$ is holomorphic in $z_{1},\cdots,z_{n-1}$.However, $\chi^{[m]}=\text{Re}(f)\neq 0$ does not contain any non-trivial holomorphic term. We thus reach a contraction.

Case (2): In this case, suppose we have the normalization in (2) of Corollary 3.5. Letting $j=1$ in (4.4) and restricting the equation to $z_{1}$ and $z_{n-1}$ spaces, we obtain:

[TABLE]

By our assumption, $a_{1(n-1)}^{(k-1)}(z_{1},0,\cdots,0,z_{n-2},\overline{z_{1}},0,\cdots,0,\overline{z_{n-2}})$ is not identically zero and contains no non-trivial holomorphic terms. By Theorem 6.1, we know $\chi^{[m]}=\text{Re}(f)=0$ when restricted to $z_{1}$ and $z_{n-1}$ spaces. This contradicts the last normalization in (2) of Corollary 3.5.

Case (3): In this case, suppose we have the normalization in (3) of Corollary 3.5. Then we have $\sum_{j=1}^{n-2}z_{j}a^{(k-1)}_{j(n-1)}(z_{1},\cdots,z_{n-2},\overline{z_{1}},\cdots,\overline{z_{n-2}})=0$. Since $f_{\overline{z_{j}}}+\overline{a^{(k-1)}_{j(n-1)}}f_{\overline{z_{n-1}}}=0$ and $a^{(k-1)}_{j(n-1)}$ is independent of $z_{n-1}$ and $w$, we get

[TABLE]

This is again the well-known Euler equation on $f$. Write $f(z,\overline{z})=\sum_{|\alpha|\geq 0}g_{\alpha}(z)\overline{z}^{\alpha}$, where $g(z)$ is holomorphic in $z$. Then

[TABLE]

Hence $g_{\alpha}(z)=0$ for $\sum_{j=1}^{n-2}|\alpha_{j}|>0$. Thus $f(z_{1},\cdots,z_{n-1},\overline{z_{1}},\cdots,\overline{z_{n-1}})$ is holomorphic in $z_{1},\cdots,z_{n-2}$. Hence $f_{\overline{z_{j}}}=0$ for each $1\leq j\leq n-2$. Substituting this back to $f_{\overline{z_{j}}}+\overline{a^{(k-1)}_{j(n-1)}}f_{\overline{z_{n-1}}}=0$, we know $\overline{a^{(k-1)}_{j(n-1)}}f_{\overline{z_{n-1}}}=0$. Recall that at least one $a^{(k-1)}_{j(n-1)}$ is not holomorphic and thus is nonzero. Thus $f_{\overline{z_{n-1}}}=0$. Hence $f(z_{1},\cdots,z_{n-1},\overline{z_{1}},\cdots,\overline{z_{n-1}})$ is holomorphic in $z_{1},\cdots,z_{n-1}$. Since Re$f$ contains no non-trivial holomorphic terms, we reach a contradiction.

Step II. In this step, we suppose $N$ is of real dimension $2n-3$.

Without loss of generality, we assume Re$Y_{HJ}|_{0}\neq 0$. Then

[TABLE]

Thus $N^{0}$ is a CR manifold of hypersurface type of finite type in the sense of Hömander-Bloom-Graham. With a rotation in $z_{n-1}$-variable, we can assume that Re$Y_{HJ}|_{0}=\frac{\partial}{\partial x_{n-1}}|_{0}$. Now, we define $\pi:N^{0}\rightarrow\mathbb{C}^{n-1}$ by sending $(z_{1},\cdots,z_{n-1},w)$ to $(z_{1},\cdots,z_{n-1})$. $\pi$ is a CR immersion near [math]. Write $\pi(N^{0})=\widetilde{N^{0}}\subset\mathbb{C}^{n-1}$. Then $\widetilde{N^{0}}$ is a real hypersurface in $\mathbb{C}^{n-1}$ and $\pi^{-1}:\widetilde{N^{0}}\rightarrow N^{0}$ is a local real analytic CR diffeomorphism with $\pi^{-1}(0)=0$. Write

[TABLE]

Since real analytic CR functions are restrictions of holomorphic functions, we can assume that $h(z_{1},\cdots,z_{n-1})$ is a holomorphic function. Notice that $h=O(|z|^{2})$ and define $(\xi_{1},\cdots,\xi_{n-1},\eta)=F(z_{1},\cdots,z_{n-1},w)=(z_{1},\cdots,z_{n-1},w-h(z_{1},\cdots,z_{n-1}))$. Then

[TABLE]

Also, $F(M^{0})$ is defined by $-2\text{Re}\eta+2\text{Re}h(\xi)+\chi^{[m]}(\xi,\overline{\xi},0)=0$ or $2\text{Re}\eta=2\text{Re}h(\xi)+\chi^{[m]}(\xi,\overline{\xi},0)=\widetilde{\rho}(\xi,\overline{\xi})$. Notice that $F(M^{0})$ is holomorphically equivalent to $M^{0}$. Hence $F(M^{0})$ is also pseudo-convex and of finite type in the sense of Hömander-Bloom-Graham. Notice that $\widetilde{N^{0}}=F(N^{0})\subset\widetilde{M_{0}}=F(M^{0})$. Hence, $\forall\xi\in\widetilde{N^{0}}$, $\widetilde{\rho}(\xi,\overline{\xi})=0$. Notice that $\widetilde{\rho}=O(|\xi|^{2})$ and is plurisubharmonic. By the following proposition, we reach a contradiction to the assumption that $2\text{Re}h(\xi)+\chi^{[m]}\not\equiv 0$.

Proposition 4.5**.**

Let $N$ be a real analytic hypersurface in $\mathbb{C}^{n-1}$ with $0\in N$ with $n\geq 3$. Let $\rho(z,\overline{z})$ be a real analytic plurisubharmonic function with $\rho=O(|z|^{2})$ as $z\rightarrow 0$ defined over a neighborhood of ${\mathbb{C}}^{n-1}$. Assume that $N$ is of finite type in the sense of Hömander–Bloom-Graham and $N\subset\{\rho=0\}$. Then $\rho\equiv 0$.

Proof.

Let $\phi:\Delta\rightarrow\mathbb{C}^{n-1}$ be a smooth small holomorphic disk attached to $N$ with $\phi(1)=0$. Namely, we assume that $\phi\in C^{\infty}(\overline{\Delta})\cap\text{Hol}(\Delta)$, $\phi(\partial\Delta)\subset N$, $\phi(1)=0$, $\phi(\overline{\Delta})$ is close to [math]. Since $\rho(\phi(\xi),\overline{\phi(\xi)})=0$ on $\partial\Delta$ and $\frac{\partial}{\partial\xi\partial\overline{\xi}}\rho(\phi(\xi),\overline{\phi(\xi)})\geq 0$ for $\xi\in\Delta$, $\rho(\phi(\xi),\overline{\phi(\xi)})$ is a subharmonic function in $\Delta$ smooth up to $\partial\Delta$. By the maximum principle, we have $\rho(\phi(\xi),\overline{\phi(\xi)})<0$ for $\xi\in\Delta$ unless $\rho(\phi(\xi),\overline{\phi(\xi)})\equiv 0$ for $\xi\in\Delta$. Now, we apply the Hopf Lemma to get

[TABLE]

and the equality holds if and only if $\rho(\phi(\xi),\overline{\phi(\xi)})\equiv 0$. On the other hand,

[TABLE]

as $\xi(\in(0,1))\rightarrow 1$. We conclude that $\rho(\phi(\xi),\overline{\phi(\xi)})\equiv 0$.

Next, by a result of Trépreau [Tr], since the union $\phi(\Delta)$ of all attached discs fill in at least one side of $N$ near [math], we see that $\rho\equiv 0$ in one side of $N$. Since we assumed that $\rho$ is real analytic, we conclude that $\rho\equiv 0$. This completes the proof of Proposition 4.5. ∎

We thus complete the proof of the equality that $t^{(n-2)}(M,p)=a^{(n-2)}(M,p)$. ∎

Proof of the equality: $c^{(n-2)}(M,p)=a^{(n-2)}(M,p)$.

We continue to use the notations and initial setups as in $\S 2$ and $\S 3$. By [Bl2], we have $c^{(n-2)}(M,p=0)\geq a^{(n-2)}(M,p=0)$. We will seek a contradiction supposing that $c^{(n-2)}(M,0)>a^{(n-2)}(M,0)$.

Let $B$ be an $(n-2)$-dimensional smooth subbundle of $T^{1,0}M$ such that $c^{(n-2)}(M,0)=c^{(n-2)}(B,0)$. Repeating the normalization procedures as in $\S 3$, we can find a basis $\{S_{j}\}$ of $B$ and a defining function $\rho$ that satisfy the normalization conditions as in Corollary 3.5. Since $c^{(n-2)}(M,0)>a^{(n-2)}(M,0)$, for any $l\leq a^{(n-2)}(M,0)$, we have

[TABLE]

As in the proof of $t^{(n-2)}(M,p)=a^{(n-2)}(M,p)$, we can similarly define the weights of $z_{1},\cdots,z_{n-1},w$, and define $S_{j}^{0}$, $\mathcal{M}^{0}$, $M^{0}$, $\mathcal{M}^{0}_{l}$, $\mathcal{M}^{0}_{\infty}$. By the same argument as that in Lemma 4.1, we have $k<m$. Similar to Lemma 4.2, we have the following:

Lemma 4.6**.**

For any $l$ and $Y^{0}_{1},\cdots,Y^{0}_{l-2}\in\mathcal{M}^{0}_{1}$, we have

[TABLE]

Proof.

First notice that $Y^{0}:=Y^{0}_{1}\cdots Y^{0}_{l-2}\sum_{j=1}^{n-2}\partial\overline{\partial}\rho^{[m]}(S^{0}_{j},\overline{S^{0}_{j}})$ is a weighted homogeneous polynomial of weighted degree $-l+m$. Hence $Y^{0}=0$ when $l>m$ and $Y^{0}|_{0}=0$ when $l<m$.

Next we suppose $l=m$. For any $1\leq j\leq l-2$, suppose $Z_{j}\in\mathcal{M}_{1}$ such that $(Z_{j})^{0}=Y_{j}^{0}$. By (4.6), we have

[TABLE]

Notice that

[TABLE]

We thus have $Y^{0}(0)=0$ for $l=m$. This completes the proof of Lemma 4.6. ∎

Now we similarly apply the Nagano theorem to conclude that $\mathcal{M}^{0}_{\infty}$ gives a unique real analytic integral submanifold $N^{0}$ with $N^{0}\subset M^{0}=\{-2\text{Re}w+\chi^{[m]}(z,\overline{z},0)=0\}$. Since the tangent space at each point of $N^{0}$ is generated by Re$\mathcal{M}^{0}_{\infty}$, by Lemma 4.6, we have

[TABLE]

for $\rho^{[m]}(S^{0}_{j},\overline{S^{0}_{j}})$ is real-analytic and it vanishes to infinite order at [math] along $N^{0}$. Since $\rho^{[m]}$ is plurisubharmonic, we have $\partial\overline{\partial}\rho^{[m]}(S^{0}_{j},\overline{S^{0}_{j}})\geq 0$ on $M^{0}$. Notice that $N^{0}\subset M^{0}$, we have $\partial\overline{\partial}\rho^{[m]}(S^{0}_{j},\overline{S^{0}_{j}})\equiv 0$ on $N^{0}$. Hence $\hbox{Re}(S_{j}^{0}),\hbox{Im}(S_{j}^{0})\in T^{N}(N^{0})$. By [DF, Proposition 2], for any vector field $Y^{0}$ in $\mathcal{M}^{0}_{j}$, $\hbox{Re}(Y^{0}),\ \hbox{Im}(Y^{0})\in T^{N}(N^{0})$ for each $j$. Hence for any $Y^{0}\in\mathcal{M}^{0}_{j}$, we have $\langle Y^{0},\partial\rho^{[m]}\rangle(0)=0$, for both the real part and the imaginary part of $Y^{0}|_{0}$ are in $\hbox{Re}(T_{0}^{(1,0)}N^{0})$. This then reduces the rest of the proof to that in the proof of the equality of $t^{(n-2)}(M,0)=a^{(n-2)}(M,0)$. The proof of the equality $c^{(n-2)}(M,p)=a^{(n-2)}(M,p)$ is now complete. ∎

5 Applications of positivity: Proofs of four lemmas

In this section, we prove four lemmas concerning a homogeneous polynomial whose real part is plurisubharmonic. These lemmas will be used for the proof of Theorem 6.1 in $\S 6$. (Lemma 5.2 was also used in $\S 3$). We begin with the following:

Lemma 5.1**.**

Let $h(\xi,\overline{\xi})$ be a homogeneous polynomial of $(\xi,\overline{\xi})\in\mathbb{C}\times\mathbb{C}$. Suppose that

[TABLE]

Then $h$ must be a monomial. Namely, $h=c\xi^{j}\overline{\xi}^{k}$ for a certain complex number $c$.

Proof.

This lemma may be known to experts. We give a simple proof here for convenience of a reader. Suppose that $h$ is not a monomial and takes the following form:

[TABLE]

Here and in what follows, we write $O(\xi^{k})$ for a homogeneous polynomial with degree in $\xi$ at least $k$. Then

[TABLE]

Thus

[TABLE]

On the other hand, $j+h=t+s$, $j<t$. Thus $j\neq t$ and $h\neq s$. Hence $ts+jh-th-js=(j-t)(h-s)\neq 0$. Thus $hh_{\xi\overline{\xi}}-h_{\xi}h_{\overline{\xi}}$ is not identically [math], which contradicts our hypothesis in (5.1). ∎

Lemma 5.2**.**

Let $f(z,\overline{z})$ be a weighted homogeneous polynomial (with any assigned positive weight on $z$) in $(z,\overline{z})\in\mathbb{C}^{n}\times\mathbb{C}^{n}$, which is holomorphic in its variable $z_{j}$ for each $j\in[1,k]$ with $k\leq n$. Assume that Re$f(z,\overline{z})$ is a plurisubharmonic function without non-trivial holomorphic terms. Then $f(z,\overline{z})$ is independent of $z_{1},\cdots,z_{k}$ and $\overline{z_{1}},\cdots\overline{z_{k}}$.

Proof.

We need only to prove the lemma with $k=1$ and the other case follows from an induction argument. Since Re$f(z,\overline{z})$ is plurisubharmonic, for each $j$ with $2\leq j\leq n$, we have

[TABLE]

Since $f(z,\overline{z})$ is holomorphic in $z_{1}$, we have

[TABLE]

Substituting these relations back to (5.2), we obtain $-|f_{z_{1}\overline{z_{j}}}|^{2}\geq 0$. Thus $f_{z_{1}\overline{z_{j}}}\equiv 0$. Since $f(z,\overline{z})$ is holomorphic in $z_{1}$, we see that

[TABLE]

is a holomorphic function. By our assumption,

[TABLE]

contains no non-trivial holomorphic terms. Hence $g(z,\overline{z})\equiv 0$, which implies that $f(z,\overline{z})$ is independent of $z_{1}$ and $\overline{z_{1}}$. ∎

Lemma 5.3**.**

Let $h(z,\overline{z})=\sum_{I\overline{J}}a_{I\overline{J}}z^{I}\overline{z^{J}}$ be a real nonzero plurisubharmonic polynomial in $(z,\overline{z})\in\mathbb{C}^{n}\times\mathbb{C}^{n},$ where $I=(i_{1},\cdots,i_{n}),J=(j_{1},\cdots,j_{n})$ with $i_{l}+j_{l}$ being a fixed positive integer (independent of $I,J$) denoted by $k_{l}$ for each $l\in[1,n]$. Assume that $h_{z_{1}\overline{z_{1}}}\not\equiv 0$. Then each $k_{l}$ is even and the coefficient of $\Pi_{l=1}^{n}|z_{l}|^{k_{l}}$ is positive.

Proof.

By the plurisubharmonicity of $h(z,\overline{z})$, we know $h_{z_{1}\overline{z_{1}}}\geq 0$. Since $h_{z_{1}\overline{z_{1}}}\not\equiv 0$, each $k_{j}$ is even. Write $z_{i}=r_{i}e^{i\theta_{i}}$, then for any $R_{i}\in(0,\infty)$, we have

[TABLE]

If the coefficient of $\Pi_{j=1}^{n}|z_{j}|^{k_{j}}$ is [math], then the above integral is [math]. Combining with $h_{z_{1}\overline{z_{1}}}\geq 0$, we obtain $h_{z_{1}\overline{z_{1}}}\equiv 0$. This contradicts our assumption that $h_{z_{1}\overline{z_{1}}}\not\equiv 0.$ This proves Lemma 5.3. ∎

Lemma 5.4**.**

Let $B(z_{1},\overline{z_{1}})$, $f(z_{2},\overline{z_{2}})$ and $g(z_{2},\overline{z_{2}})$ be three homogeneous polynomials of degree $k\geq 1$, $m\geq 1$ and $m\geq 1$, respectively, in the ordinary sense with $B(z_{1},\overline{z_{1}})\not\equiv 0$, $f(z_{2},\overline{z_{2}})\not\equiv 0$. Suppose that $B(z_{1},0)=B(0,\overline{z_{1}})=0$. Suppose that $F=Bf+z_{1}^{k}g$ with Re$F$ being a non zero plurisubharmonic polynomial without any non-trivial holomorphic term. Then $k$ and $m$ are even and Re$F=\alpha|z_{1}|^{k}|z_{2}|^{m}$ for some $\alpha>0$.

Proof.

By the assumption that Re$F$ is non-zero and plurisubharmonic, $\left(Re(F)\right)_{z_{1}\overline{z_{1}}}\geq 0$. Since $B(z_{1},0)=B(0,\overline{z_{1}})=0$ and Re$F$ contains no non-trivial holomorphic terms, one further concludes that $\left(Re(F)\right)_{z_{1}\overline{z_{1}}}$ is not identically [math]. By Lemma 5.3, $m$ and $k$ are even. Set $k=2k_{3}$ and $m=2m_{3}$. Write

[TABLE]

First we claim that $B_{k_{3}k_{3}}\neq 0$ and $f_{m_{3}m_{3}}\neq 0$. Otherwise the coefficient of the $|z_{1}|^{2k_{3}-2}|z_{2}|^{2m_{3}}$ in $\left(Re(F)\right)_{z_{1}\overline{z_{1}}}$ is zero, and thus by Lemma 5.3, we reach a contradiction. After writing $F=cB\cdot\frac{1}{c}f+z_{1}^{k}g$, we can assume that $B_{k_{3}k_{3}}=1$.

By the plurisubharmonicity of Re$(F)$, we have

[TABLE]

Notice that

[TABLE]

Thus

[TABLE]

The coefficients of $|z_{1}|^{2k-2}$ in $BB_{z_{1}\overline{z_{1}}}$ and $\overline{B}B_{z_{1}\overline{z_{1}}}$ are, respectively,

[TABLE]

The coefficients of $|z_{2}|^{2m-2}$ in $ff_{z_{2}\overline{z_{2}}}$ and $f\overline{f}_{z_{2}\overline{z_{2}}}$ are, respectively,

[TABLE]

Notice that $B_{z_{1}\overline{z_{1}}}f\cdot\text{Re}(z_{1}^{k}g_{z_{2}\overline{z_{2}}})$ is not divisible by $|z_{1}|^{2k-2}$ (unless is is identically zero). Hence the coefficient of $|z_{1}|^{2k-2}|z_{2}|^{2m-2}$ in $4(\text{Re}f)_{z_{1}\overline{z_{1}}}(\text{Re}f)_{z_{2}\overline{z_{2}}}$ is

[TABLE]

We similarly compute the coefficient of $|z_{1}|^{2k-2}|z_{2}|^{2m-2}$ in $4(\text{Re}F)_{z_{1}\overline{z_{2}}}(\text{Re}F)_{z_{2}\overline{z_{1}}}$ as follows:

[TABLE]

Thus

[TABLE]

The coefficients of $|z_{1}|^{2k-2}$ in $B_{z_{1}}B_{\overline{z_{1}}}$, $B_{z_{1}}\overline{B}_{\overline{z_{1}}}$, $B_{\overline{z_{1}}}\overline{B}_{{z_{1}}}$ and $\overline{B}_{z_{1}}\overline{B}_{\overline{z_{1}}}$ are, respectively

[TABLE]

The coefficients of $|z_{2}|^{2m-2}$ in $f_{\overline{z_{2}}}f_{z_{2}}$, $f_{\overline{z_{2}}}\overline{f}_{z_{2}}$, $f_{{z_{2}}}\overline{f}_{\overline{z_{2}}}$ and $\overline{f}_{\overline{z_{2}}}\overline{f}_{z_{2}}$ are, respectively,

[TABLE]

Notice that $kz_{1}^{k-1}g_{\overline{z_{2}}}(B_{\overline{z_{1}}}f_{{z_{2}}}+\overline{B}_{\overline{z_{1}}}\overline{f}_{{z_{2}}})$ is not divisible by $|z_{1}|^{2k-2}$ (when not identically zero). Hence the coefficient of $|z_{1}|^{2k-2}|z_{2}|^{2m-2}$ in $4(\text{Re}F)_{z_{1}\overline{z_{2}}}(\text{Re}F)_{z_{2}\overline{z_{1}}}$ is

[TABLE]

Hence the coefficient of $|z_{1}|^{2k-2}|z_{2}|^{2m-2}$ in $4(\text{Re}F)_{z_{1}\overline{z_{1}}}(\text{Re}F)_{z_{2}\overline{z_{2}}}-4(\text{Re}F)_{z_{1}\overline{z_{2}}}(\text{Re}F)_{z_{2}\overline{z_{1}}}$ is

[TABLE]

Here we have set

[TABLE]

Notice, by the Hölder inequality, that

[TABLE]

Thus we obtain the coefficient of $|z_{1}|^{2k-2}|z_{2}|^{2m-2}$ in

[TABLE]

is non positive. Furthermore, this coefficient is [math] if and only if for $h\leq j$, $t\leq s$ and for any $j^{*}+l^{*}=m-1$ with $l^{*}\not=0$:

[TABLE]

Repeating the argument in the proof of Lemma 5.3, we conclude that (5.11) holds and moreover

[TABLE]

Since $ReF$ and $Bf$ contain no non-trivial holomorphic terms, we see $g\equiv 0$. Setting $j=h=k_{3}$ in (5.11) and using the normalization that $B_{k_{3}k_{3}}=1$, we obtain $f_{ts}=-\overline{f_{st}}$ for $t\neq s$.

Now, if $f$ is of the form $f=f_{m_{3}m_{3}}|z_{2}|^{m}$ and Re$F=|z_{2}|^{m}p(z_{1},\overline{z_{1}})$, then (5.12) is equivalent to

[TABLE]

By Lemma 5.1, $p$ is a monomial. On the other hand, since $p$ is real valued, $p=\alpha|z_{1}|^{k}$ for some $\alpha>0$. Namely, Re$F=\alpha|z_{1}|^{k}|z_{2}|^{m}$. This proves the lemma.

For the rest of the proof, we suppose that $f$ is not of the form $f=f_{m_{3}m_{3}}|z_{2}|^{m}$. Since $f_{m_{3}m_{3}}\neq 0$, $f$ is not a monomial.

Now suppose

[TABLE]

Since $B(z_{1},0)=B(0,z_{1})=0$, we have $h,j\geq 1$. From (5.12), we get

[TABLE]

This gives

[TABLE]

which further forces $q$ to be a monomial.

(1) If $B_{hj}=0$ or $B_{jh}=0$, then $q=\frac{1}{2}\overline{B_{jh}}\overline{f}$ or $q=\frac{1}{2}B_{hj}{f}$, respectively. In either case, $q$ is not a monomial and thus we reach a contradiction.

(2) Assume that $B_{hj}\neq 0$, $B_{jh}\neq 0$ and $h<j$. In this case, $B_{hj}f_{m_{3}m_{3}}=-\overline{B_{jh}f_{m_{3}m_{3}}}$. Hence there is no term of the form $\gamma|z_{2}|^{m}$ in $q$. Setting $h=j$ in (5.11), we see $f_{ts}=-\overline{f_{st}}$ for $t\neq s$. Thus $B_{hj}f_{m_{3}m_{3}}|z_{2}|^{m}+\overline{B_{jh}f_{m_{3}m_{3}}|z_{2}|^{m}}=0$ and $f-f_{m_{3}m_{3}}|z_{2}|^{m}=-\overline{(f-f_{m_{3}m_{3}}|z_{2}|^{m})}$. Hence Re$F$ can be computed as follows:

[TABLE]

Thus we conclude that $q=\frac{1}{2}(B_{hj}-\overline{B_{jh}})(f-f_{m_{3}m_{3}}|z_{2}|^{m})$, which can not be a monomial and thus gives a contradiction.

Hence we must have $h\geq j$. But from the reality of Re$F$, we see that $B=|z_{1}|^{k}$ and Re$F$ takes the form $\frac{1}{2}|z_{1}|^{k}(f(z_{2},\overline{z_{2}})+\overline{f(z_{2},\overline{z_{2}})})$. Since $f_{ts}=-\overline{f_{st}}$ for $s\neq t$, we conclude that Re$F$ takes the form $\alpha|z_{1}|^{k}|z_{2}|^{m}$ with $\alpha>0$. This finally completes the proof of the lemma. ∎

6 Proof of Theorem 6.1

In this section, we provide a detailed proof of Theorem 6.1, which played a key role in the proof of our main theorem. We write $z=(z_{1},z_{2})$ for the coordinates in ${\mathbb{C}}^{2}$ in this section.

Theorem 6.1**.**

Define the weight of $z_{1}$ and $\overline{z_{1}}$ to be $1$, the weight of $z_{2}$ and $\overline{z_{2}}$ to be $k\in{\mathbb{N}}$ with $k>1$. Let $A=A(z_{1},\overline{z_{1}})$ be a homogenous polynomial of degree $k-1$ in $(z_{1},\overline{z_{1}})$ without holomorphic terms. Suppose that $f$ is a weighted homogeneous polynomial in $(z,\overline{z})$ of weighted degree $m>k$. Further assume that $\hbox{Re}(f)$ is plurisubharmonic, contains no non-trivial holomorphic terms and assume that $f$ satisfies the following equation:

[TABLE]

Then Re$(f)\equiv 0$.

Without the plurisubharmonicity on $\hbox{Re}(f)$, the above theorem can not be true as the following simple example demonstrates:

Example 6.2**.**

Let $L=\frac{\partial}{\partial z_{1}}-|z_{1}|^{2}\frac{\partial}{\partial z_{2}}$ and let $f=z_{1}\overline{z_{2}}+\frac{1}{2}|z_{1}|^{4}$. Then $\overline{L}(f)\equiv 0$. Notice that $\hbox{Re}(f)$ is not plurisubharmonic neither is [math]. Notice that $A=\overline{A}=-|z_{1}|^{2},\hbox{Re}(f)$ has no holomorphic terms. We also mention that in Theorem 6.1, we can not conclude $f\equiv 0$ as demenstrated by the following example: Let $L=\frac{\partial}{\partial z_{1}}+kz_{1}^{k}\overline{z^{k-1}_{1}}\frac{\partial}{\partial z_{2}}$ and $f=i(z_{2}+\overline{z_{2}}-|z_{1}|^{2k})^{2}$. Then $\overline{L}f=0$ and $\hbox{Re}(f)\equiv 0$. However $f\not\equiv 0$.

Proof of Theorem 6.1.

The proof of Theorem 6.1 is long. We will proceed according to the four different scenarios, two of which are reduced to CR equations along finite type hypersurfaces where Proposition 4.5 can be applied.

For $0\leq j\leq[\frac{m}{k}]:=m_{0}$, denoted by $f^{[j]}$ the sum of terms (monomial terms) in $f$ which has ordinary degree $j$ in $z_{2}$ and $\overline{z_{2}}$. Then

[TABLE]

In the course of the proof, for $j=1,2$, we write $O(|z_{j}|^{k})$ for a homogeneous polynomial with (the ordinary or un-weighted) degree in $z_{j}$ and $\overline{z_{j}}$ at least $k$. We also denote by $L(|z_{j}|^{k})$ a homogeneous polynomial with the un-weighted degree in $z_{j}$ and $\overline{z_{j}}$ at most $k$. For a homogeneous polynomial $P=\sum_{h+j=l}C_{hj}z_{1}^{h}\overline{z_{1}}^{j}$, we denote the integral of $P$ along $\overline{z_{1}}$ as

[TABLE]

We remark that after a transformation of the form: $(z_{1},z_{2})\rightarrow(z_{1},{\delta}^{-1}z_{2})$, $A$ and $f$, in the new coordinates still denoted by $(z_{1},z_{2})$, takes the form

[TABLE]

We will need this transformation to normalize certain coefficients in our proof.

Case I: In this case, we suppose $km_{0}<m$ or $km_{0}=m$, $f^{[m_{0}]}=0$.

Suppose $h$ is the largest integer such that $f^{[h]}\neq 0$. From (6.1), $f^{[h]}$ is holomorphic in $z_{1}$. We suppose that

[TABLE]

We then have $j\geq 1$. Since Re$f$ contains no holomorphic terms, $f_{h0}=0$. In particular, we see that we must have $h\geq 1$. In what follows, we set a term with a negative power to be zero.

First, we claim $f_{ts}=0$ for any $t\geq 1$. Since Re$f$ is plurisubharmonic, we obtain

[TABLE]

For $|z_{2}|\gg|z_{1}|$, we get

[TABLE]

Since $j\geq 1$, this is possible only when the left hand side is identically [math]. This implies that $f_{ts}=0$ for all $t\geq 1$. Thus

[TABLE]

Then

[TABLE]

Since Re$(f)$ is plurisubharmonic, we have

[TABLE]

Notice that

[TABLE]

Hence

[TABLE]

Again, by choosing $|z_{2}|\gg|z_{1}|$, we get $-j^{2}h^{2}|f_{0h}|^{2}|z_{1}|^{2j-2}|z_{2}|^{2h-2}\geq 0$. Hence $f_{0h}=0$, which means that $f^{[h]}\equiv 0$. This contradicts our assumption that $f^{[h]}\neq 0$. Hence Case I can only occur when Re$(f)=0$.

Case II:  We now assume that $km_{0}=m$, Re$(f^{[m_{0}]})\neq 0$.

Suppose

[TABLE]

Since Re$(f^{[m_{0}]})$ contains no non-trivial holomorphic terms, we have $f_{0m_{0}}=-\overline{f_{m_{0}0}}$. By the plurisubharmonicity of Re($f$), we get $($Re$(f^{[m_{0}]}))_{z_{2}\overline{z_{2}}}\geq 0$ and can not be identically zero. By Lemma 5.3, $m_{0}$ is even and Re$f_{m_{1}m_{1}}>0$. Here $m_{0}=2m_{1}$.

After a rotation transformation of the form $(z_{1},z_{2})\rightarrow(z_{1},\delta^{-1}z_{2})$ for some constant $\delta\not=0$, by (6.3), we can make

[TABLE]

We remark that this transformation does not change our original hypotheses in this case. Now (6.1) can be solved as

[TABLE]

In particular, we get

[TABLE]

By the plurisubharmonicty of Re$f$, we have $\text{Re}(f)_{z_{1}\overline{z_{1}}}\geq 0.$ Notice that $F(\overline{A})$ is divisible by $|z_{1}|^{2}$. Hence

[TABLE]

Hence

[TABLE]

Notice that the (ordinary) degree of $(\text{Re}f^{[m_{0}-1]})_{z_{1}\overline{z_{1}}}$ in $z_{2}$ and $\overline{z_{2}}$ is $m_{0}-1$ which is an odd number, we have

[TABLE]

Next, write $\overline{A}=\sum_{j+h=k-1,h\geq 1}\overline{A_{jh}}{z_{1}^{h}}\overline{z_{1}}^{j}.$ Then

[TABLE]

Hence

[TABLE]

Hence for $h+j=k$, $t+s=m_{0}-1$, we have

[TABLE]

Setting $t=m_{1}-1$, $s=m_{1}$ in the above equation and making use of (6.6), we get

[TABLE]

If $c=0$, then $A_{(h-1)j}=0$ for all $h+j=k,\ h\geq 1,\ j\geq 1$. This implies that $A\equiv 0$, which is impossible. Thus $c\neq 0$. From (6.10), we get

[TABLE]

Next, by (6.1) (6.7), we compute the following:

[TABLE]

We will compute the coefficient of $|z_{1}|^{2k}|z_{2}|^{m_{0}-2}$ in $f^{[m_{0}-2]}$. First, the coefficient of $|z_{2}|^{m_{0}-2}$ in $f^{[m_{0}]}_{\overline{z_{2}}^{2}}$ is $(m_{1}+1)m_{1}f_{(m_{1}-1)(m_{1}+1)}.$ Notice that

[TABLE]

Hence

[TABLE]

When $h+s=j+t+2,\ j+h=k-1,\ t+s=k-1,$ we have $j=k-1-h,\ t=h-1,\ s=k-h.$ Hence the coefficient in $F(\overline{A}F(\overline{A}))$ with the factor $|z_{1}|^{2k}$ is

[TABLE]

By (6.11), $H\leq 0$, also $H=0$ if and if $A_{(k-1-h)h}=0$ for all $h\geq 1$. This is equivalent to $A=0$, which is impossible. Thus $H<0$.

Notice that $\overline{A}$ is divisible by $z_{1}$, thus $F(\overline{A}z_{1}^{k})$ does not contain $|z_{1}|^{2k}$ term.

Thus the coefficient of $|z_{1}|^{2k}|z_{2}|^{m_{0}-2}$ in $f^{[m_{0}-2]}$ is $(m_{1}+1)m_{1}f_{(m_{1}-1)(m_{1}+1)}H.$ Recall that Re$f_{m_{1}m_{1}}>0$ and $c>0$. Together with (6.6), we get Re$f_{(m_{1}-1)(m_{1}+1)}>0$. Hence the real part of the coefficient of $|z_{1}|^{2k}|z_{2}|^{m_{0}-2}$ in $f^{[m_{0}-2]}$ must be negative. This contradicts the following

[TABLE]

which is true due to the fact that \Big{(}\text{Re}(f^{[m_{0}]})\Big{)}_{z_{1}\overline{z_{1}}}\ \hbox{and}\ \Big{(}\text{Re}(f^{[m_{0}-1]})\Big{)}_{z_{1}\overline{z_{1}}}=0.

Case III: $m=km_{0}$, $f^{[m_{0}]}\neq 0$, Re$(f^{[m_{0}]})=0$ and Re$(f^{[m_{0}-1]})\neq 0$.

Here, we reduce $f$ to the solution of a CR vector field of a real hypersurface of finite type in ${\mathbb{C}}^{2}$ and then apply Proposition 4.5 to reach a contradiction. Write $B:=-F(\overline{A})=\sum_{j+h=k}B_{jh}z_{1}^{j}\overline{z_{1}}^{h}$. By Lemma 5.3, both $k$ and $m_{0}-1$ are even. Define $k=2k_{2}$, $m_{0}=2m_{2}+1$. Then $B_{k_{2}k_{2}}\neq 0$ by Lemma 5.3, which implies that $A_{(k_{2}-1)k_{2}}\neq 0$. After a dilation transform of the form as in (6.3), we assume that $A_{(k_{2}-1)k_{2}}=-k_{2}$. Then $B_{k_{2}k_{2}}=1$. A direct computation shows

[TABLE]

From our assumption, Re$f^{[m_{0}-1]}$ is plurisubharmonic. By Lemma 5.4,

[TABLE]

Notice that $B_{k_{2}k_{2}}=1$ and $(m_{2}+1)\hbox{Re}(f_{m_{2}(m_{2}+1)})=\lambda\neq 0$. Since $\hbox{Re}(f^{[m_{0}]})=0$, we have $f_{(m_{2}+1)m_{2}}+\overline{f_{m_{2}(m_{2}+1)}}=0$. Notice that $f^{[m_{0}]}_{\overline{z_{2}}}-(m_{2}+1)f_{m_{2}(m_{2}+1)}|z_{2}|^{2m_{2}}$ has no term divisible by $|z_{2}|^{2m_{2}}$. Hence we conclude from (6.14)

[TABLE]

Collecting terms divisible by $z_{2}^{m_{2}+1}\overline{z_{2}}^{m_{2}-1}$ in (6.14), we get

[TABLE]

Hence $B$ is different from $\overline{B}$ by a constant. Since we normalized $B_{k_{2}k_{2}}=1$, we see that $B$ is real-valued. But $f^{[m_{0}]}_{\overline{z_{2}}}$ contains a term of the form $\mu|z_{2}|^{m_{0}-1}$ with Re$\mu\neq 0$. Thus $-F(\overline{A})=|z_{1}|^{k}$, namely, $A=-k_{2}z_{1}^{k_{2}-1}\overline{z_{1}}^{k_{2}}$.

Now, $L=\frac{\partial}{\partial z_{1}}+A(z_{1},\overline{z_{1}})\frac{\partial}{\partial z_{2}}$ forms a basis for the sections of CR vector fields along the real algebraic finite type hypersurface $M_{0}$ in ${\mathbb{C}}^{2}$ defined by $-z_{2}-\overline{z_{2}}=|z_{1}|^{2k_{2}}$ and $\overline{L}(f)\equiv 0.$ Thus $f$ is a CR polynomial on $M_{0}$ and $g=f(z_{1},\overline{z_{1}},z_{2},-z_{2}-|z_{1}|^{2k_{2}})$ is a weighted homogeneous holomorphic polynomial of degree $m>k$. Since $f-g\equiv 0$ over $M_{0}$, $M_{0}$ is contained in the zero set of the plurisubharmonic $\rho=\hbox{Re}(f-g)$ with $0\in M_{0}$. Notice that $\rho=O(|z|^{2})$, we conclude by Proposition 4.5 that $\rho\equiv 0$ or $\hbox{Re}(f)$ is pluriharmonic. This is a contradiction. Hence Case III cannot occur.

Case IV: $m=km_{0}$, $f^{[m_{0}]}\neq 0$ but Re$(f^{[m_{0}]})=$Re$(f^{[m_{0}-1]})=0$.

Write

[TABLE]

Then by our assumption that Re$(f^{[m_{0}]})=$Re$(f^{[m_{0}-1]})=0$ and $F(\overline{A})$ is divisible by $|z_{1}|^{2}$, we obtain as in Case (II) the following

[TABLE]

Hence for each pair $(h,j)$, if $B_{hj}\neq 0$, then $B_{jh}\neq 0$; for otherwise we get $f_{ts}=0$ for any $t+s=m_{0}$ and reach a contradiction. Since $B$ is nonzero, we can suppose there is a pair $(h_{0},j_{0})$ such that $B_{h_{0}j_{0}}\neq 0$ and thus $B_{j_{0}h_{0}}\neq 0$. Since $f^{[m_{0}]}\neq 0$, there is a certain $f_{t_{0}(s_{0}+1)}\neq 0$ and thus $f_{s_{0}(t_{0}+1)}\neq 0$. By (6.15), we have

[TABLE]

[TABLE]

Since $f_{t_{0}(s_{0}+1)}\neq 0$ and $f_{s_{0}(t_{0}+1)}\neq 0$, we have $|B_{h_{0}j_{0}}|=|B_{j_{0}h_{0}}|$. After a rotational transformation as in (6.3) with a suitable choice of $\delta$, we can assume that $B_{h_{0}j_{0}}=\overline{B_{j_{0}h_{0}}}$. Then by (6.15), we have

[TABLE]

By (6.16), $f^{[m_{0}]}_{{\overline{z_{2}}}}$ is pure imaginary. Also, it is not identically zero for the absolute value of each coefficient is a non-zero multiple of the others and at least one of them is non-zero. Now, by the first equation in $(\ref{tre})$, we easily conclude that $F(\overline{A})$ is a real-valued homogeneous polynomial divisible by $|z_{1}|^{2}$. Hence, $L=\frac{\partial}{\partial z_{1}}+A(z_{1},\overline{z_{1}})\frac{\partial}{\partial z_{2}}$ forms a basis for the sections of CR vector fields along the real algebraic finite type hypersurface $M_{0}$ in ${\mathbb{C}}^{2}$ defined by $z_{2}+\overline{z_{2}}=F(\overline{A})$ and $\overline{L}(f)\equiv 0.$ Now, following the same argument as in Case (III), we achieve a contradiction unless Re$(f)\equiv 0$.

Combining our arguments in Cases I-IV, we conclude the proof of Theorem 6.1. ∎