math.AG — Algebraic Geometry
Contributed by Roy van Rijn
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\title{The Generalized Vanishing Conjecture:\\
The Two-Variable Theorem and the First Failing Dimension}
\author{Roy van Rijn\\{\small Independent researcher}}
\date{4 August 2026}
\hypersetup{
pdftitle={The Generalized Vanishing Conjecture: The Two-Variable Theorem and the First Failing Dimension},
pdfauthor={Roy van Rijn},
pdfsubject={The exact dimensional boundary of the Generalized Vanishing Conjecture},
pdfkeywords={Generalized Vanishing Conjecture, constant-coefficient differential operators,
Newton polygons, Hall's theorem, Gaussian moments, counterexample}
}
\begin{document}
\maketitle
\begin{abstract}
We determine the exact dimensional range of the Generalized Vanishing
Conjecture for constant-coefficient differential operators over a field of
characteristic zero. In two variables we prove that
\[
\Lambda^m(P^m)=0\quad(m\geq1)
\qquad\Longrightarrow\qquad
\Lambda^m(QP^m)=0\quad(m\gg0)
\]
for every fixed polynomial Q. The proof follows two moving faces. Hall's
marriage theorem puts the lowest operator face strictly to the right of the
highest polynomial face near ordinary degree. A prime-dilation argument
shows that, while the two faces have different weights, their horizontal
Newton intervals cannot overlap. Since the supports are finite, the two
piecewise-linear weight envelopes must eventually meet. Their first meeting
is an unequal-weight common threshold, where a fixed multiplier cannot repair
the resulting linear derivative deficit. The only deep external theorem is
the constant-term theorem of Duistermaat and van der Kallen. We then give a
homogeneous
three-variable counterexample. With
ρ=t2+xy, A=ρ+x2,
C=yρ2−2xt2ρ−x3t2, and
Δ=4∂x∂y+∂t2, the pair
\[
\Lambda=\Delta^6,\qquad P=AC^2
\]
satisfies Λm(Pm)=0 while
Λm(x2Pm)=0 for every m≥1. The construction is a
homogeneous endpoint-contact lift of Long's three-dimensional Gaussian
counterexample. It is the smallest member of an explicit cusp-profile
family with arbitrary winding, profile, and radial suspension; in
particular every Δk, k≥6, has such a failure. Consequently
unrestricted, and already homogeneous, GVC holds in dimension n exactly
when n≤2.
\end{abstract}
\section{Introduction}
For a polynomial λ(X,Y), write
\[
\Lambda=\lambda(\partial_x,\partial_y).
\]
The Generalized Vanishing Conjecture asks whether, for all polynomials
P,Q over a characteristic-zero field,
\begin{equation}\label{eq:gvc}
\Lambda^m(P^m)=0\quad\text{for every }m\geq1
\quad\Longrightarrow\quad
\Lambda^m(QP^m)=0\quad\text{for all sufficiently large }m.
\end{equation}
It grew out of the Vanishing Conjecture and its relation to the Jacobian
Conjecture; see Zhao~\cite{Zhao} and van den Essen, Willems, and
Zhao~\cite{EWZ}.
There were several particularly important precursors. De Bondt proved GVC
for every product of linear differential forms, and therefore for every
homogeneous operator in two variables~\cite{deBondt}. Feng and Sun proved
the nonhomogeneous binary case
\[
\Lambda=(\partial_x-\Phi(\partial_y))\partial_y
\]
for an arbitrary one-variable polynomial Φ~\cite{FengSun}. Wilson's
independent July 2026 preprint proves the two-variable Gaussian Moments
Conjecture by a related Newton-face and prime-isolation method
~\cite{WilsonGMC}. Wilson treats the Gaussian functional rather than an
arbitrary constant-coefficient operator. In the present argument, Hall
localization and separation on shifted output rays are the additional steps
that pass from that philosophy to unrestricted binary GVC.
Also in 2026, Long found a five-term quartic in three real Gaussian variables for which
all pure Gaussian moments vanish while one fixed mixed moment equals m!
at every power~\cite{LongGMC}. Long's example disproves the Gaussian
Moments Conjecture in every dimension at least three, but it is not itself a
constant-coefficient GVC counterexample.
This paper closes the gap on both sides. First we prove \eqref{eq:gvc} for
arbitrary, possibly nonhomogeneous, operators in two variables. We then
lift the endpoint-coefficient mechanism behind Long's example to a
homogeneous polynomial and a homogeneous differential operator in three
variables. The two results meet at the exact boundary.
\begin{theorem}[Binary Generalized Vanishing Conjecture]\label{thm:main}
Let k be a field of characteristic zero. Every constant-coefficient
differential operator Λ on k[x,y] satisfies
\eqref{eq:gvc}.
\end{theorem}
The global proof is organized as follows. Give both
the operator variables and the polynomial variables weight (s,1). Let
L(s) be the least operator weight and U(s) the greatest polynomial
weight. Hall's theorem orders the corresponding faces just to the right of
s=1. Prime-dilated endpoint separation prevents the faces from passing
one another while U(s)−L(s)>0. Finite support then forces this difference
to reach zero. At the first zero all operator terms lie above one threshold
and all polynomial terms lie below it, which gives eventual mixed
vanishing.
For orientation, the load-bearing dependencies in the binary argument are
\[
\begin{array}{c}
\text{Duistermaat--van der Kallen}+\text{split-symbol polarization}
\longrightarrow \text{Proposition~\ref{prop:cutoff}},\\[2pt]
\text{Duistermaat--van der Kallen}+\text{Hall localization}
\longrightarrow \text{Lemma~\ref{lem:hall}},\\[2pt]
\text{rational-point exposure}+\text{prime isolation}
\longrightarrow \boxed{\text{Proposition~\ref{prop:shifted}}}
\longrightarrow \text{Corollary~\ref{cor:horizontal}},\\[2pt]
\text{Lemma~\ref{lem:hall}}+\text{Corollary~\ref{cor:horizontal}}
\longrightarrow \text{first common threshold}
\longrightarrow \text{Proposition~\ref{prop:threshold}}
\longrightarrow \text{Theorem~\ref{thm:main}}.
\end{array}
\]
Thus Proposition~\ref{prop:shifted} is the single arithmetic bridge from
the local Hall gap to the global envelope argument.
\section{Two standard reductions}
We use the Newton polytope of a Laurent polynomial F,
\[
\Newt(F)=\operatorname{conv}(\Supp(F)).
\]
The only substantial external input is the following theorem.
\begin{theorem}[Duistermaat--van der Kallen~\cite{DK}]\label{thm:dk}
Let F be a Laurent polynomial over a characteristic-zero field. If
$\CT(F^m)=0$ for every m≥1, then
$0\notin\Newt(F)$.
\end{theorem}
We will also use the following immediate consequence.
\begin{lemma}[Exposing a rational point]\label{lem:expose}
Let F be a nonzero Laurent polynomial and let
$\rho\in\Newt(F)\cap\Q^n$. Then there is an integer a≥1 such that
\[
[z^{a\rho}]F^a\ne0.
\]
For finitely many polynomials and rational points, the same a may be
chosen for all of them.
\end{lemma}
\begin{proof}
Choose N with Nρ∈Zn and apply
Theorem~\ref{thm:dk} contrapositively to
z−NρFN. To use one exponent for several polynomials, place them
in disjoint sets of Laurent variables, multiply them, and apply the same
argument to the product.
\end{proof}
We next dispose of the range in which the degree of P is no larger than
the first positive order of Λ. This familiar split-symbol argument
is included so that the main theorem has no hidden degree case.
\begin{proposition}[Binary degree cutoff]\label{prop:cut
A mix of human-written and AI-generated textHuman understanding: some partsGeneralized Vanishing Conjecture