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\title{Cyclic Rigidity in Dimension Two:\\A Newton-Edge Obstruction for a Pseudoreflection Normal Form}
\author{Annie\\AGNT Labs}
\date{July 22, 2026}

\begin{document}

\maketitle

\begin{abstract}
Let $k$ be a field of characteristic zero, let $m\geq1$, and let $H,g\in k[u,y]$. We prove that
\[
muH_u g_y-H_y(g+mug_u)=c\in k^\times
\]
forces $g\in k^\times$. Consequently every polynomial map
\[
F(x,y)=\bigl(H(x^m,y),xg(x^m,y)\bigr)
\]
with nonzero constant Jacobian is a polynomial automorphism and admits an explicit inverse. After swapping the target coordinates, maps in this class commute with the cyclic pseudoreflection $(x,y)\mapsto(\zeta x,y)$. Conversely, when $m\geq2$ and after extending scalars to contain a primitive $m$-th root of unity, every polynomial map commuting with this action has the swapped normal form. The proof chooses the maximal Newton slope of $g$, isolates the corresponding weighted edges of $g$ and $H$, and derives a one-variable edge polynomial whose terminal coefficient cannot vanish in characteristic zero. The result is unbounded in degree and allows arbitrary multimode support. It is a rigidity theorem for this normal form, not a resolution of the unrestricted two-dimensional Jacobian Conjecture.
\end{abstract}

\section{Introduction}

For polynomials $P,Q\in k[x,y]$, write
\[
\Jac(P,Q)=P_xQ_y-P_yQ_x.
\]
The two-dimensional Jacobian Conjecture asks whether $\Jac(P,Q)\in k^\times$ forces $(P,Q)$ to be a polynomial automorphism when $k$ has characteristic zero. We study the normal form
\begin{equation}\label{eq:F}
F(x,y)=\bigl(H(x^m,y),xg(x^m,y)\bigr),
\qquad H,g\in k[u,y],\quad u=x^m.
\end{equation}

There is a minor but important representation-theoretic distinction. Let
\[
\rho_s(\zeta)=\operatorname{diag}(\zeta,1),
\qquad
\rho_t(\zeta)=\operatorname{diag}(1,\zeta),
\qquad \zeta^m=1.
\]
Then the map in \eqref{eq:F} satisfies
\[
F\circ\rho_s(\zeta)=\rho_t(\zeta)\circ F.
\]
It does not in general commute with one identical diagonal action. Equivalently, after the target-coordinate swap $S(P,Q)=(Q,P)$, the map
\begin{equation}\label{eq:swapped}
\widetilde F=S\circ F=\bigl(xg(x^m,y),H(x^m,y)\bigr)
\end{equation}
commutes with $\operatorname{diag}(\zeta,1)$. The coordinate swap changes the Jacobian only by a sign.

Miyanishi proved an equivariant Jacobian theorem over $\mathbb C$ for small finite subgroups of $\mathrm{GL}(2,\mathbb C)$ of even order \cite{Miyanishi2023}. The pure cyclic group generated by $\operatorname{diag}(\zeta,1)$ is generated by pseudoreflections and is not small, so that main theorem does not directly subsume the class \eqref{eq:swapped}. Our argument is instead an elementary polynomial Newton-edge argument over an arbitrary characteristic-zero field. Classical Newton-polygon, degree-reduction, and geometry-at-infinity results remain essential context \cite{AbhyankarMoh1975,Razar1979,BCW1982,Moh1983,Glidewell2022,Nguyen2025}.

\section{The cyclic Keller equation}

Differentiating \eqref{eq:F} gives
\[
(F_1)_x=mx^{m-1}H_u,\qquad (F_1)_y=H_y,
\]
\[
(F_2)_x=g+mug_u,\qquad (F_2)_y=xg_y.
\]
Hence
\begin{equation}\label{eq:keller}
\det JF=muH_u g_y-H_y(g+mug_u).
\end{equation}

Our main theorem is the following.

\begin{theorem}[Cyclic rigidity]\label{thm:main}
Let $k$ be a field of characteristic zero, let $m\geq1$, and let $H,g\in k[u,y]$. If
\begin{equation}\label{eq:pde}
muH_u g_y-H_y(g+mug_u)=c
\end{equation}
for some $c\in k^\times$, then $g\in k^\times$.
\end{theorem}

\begin{corollary}[Classification and inverse]\label{cor:classification}
Under the hypotheses of \cref{thm:main}, there are $a\in k^\times$ and $h\in k[u]$ such that
\[
g=a,
\qquad
H(u,y)=-\frac ca y+h(u).
\]
Thus
\[
F(x,y)=\left(-\frac ca y+h(x^m),ax\right)
\]
is a polynomial automorphism. For target coordinates $(P,Q)=F(x,y)$,
\[
F^{-1}(P,Q)=
\left(
\frac Qa,
-\frac ac\left[P-h\left(\left(\frac Qa\right)^m\right)\right]
\right).
\]
\end{corollary}

\section{Boundary data}

\begin{lemma}\label{lem:boundary}
If \eqref{eq:pde} holds, then for some $a\in k^\times$ and $b\in k$,
\[
g(0,y)=a,
\qquad
H(0,y)=-\frac ca y+b.
\]
\end{lemma}

\begin{proof}
Set $u=0$ in \eqref{eq:pde}. Then
\[
-H_y(0,y)g(0,y)=c.
\]
Both factors lie in $k[y]$ and their product is a nonzero scalar. Since the units of $k[y]$ are precisely $k^\times$, both factors are constant and nonzero. Integration in $y$ gives the formula for $H(0,y)$.
\end{proof}

\section{The maximal Newton slope}

Assume for contradiction that $g$ is nonconstant. By \cref{lem:boundary}, every nonconstant term of $g$ contains a positive power of $u$, so write
\[
g=a+\sum_{j\geq1}a_j(y)u^j,
\qquad a\in k^\times.
\]
Define
\begin{equation}\label{eq:rho}
\rho=\max_{a_j\neq0}\frac{\deg_y a_j}{j}=\frac pq,
\end{equation}
where $p\geq0$, $q\geq1$, $\gcd(p,q)=1$, and $0/q$ is reduced to $0/1$. Give $k[u,y]$ the rational weight
\[
\wt(y)=1,
\qquad
\wt(u)=-\rho=-\frac pq.
\]
Every monomial of $g$ has weight at most zero.

\begin{lemma}[Edge of $g$]\label{lem:g-edge}
Let $g_{[0]}$ denote the weight-zero component of $g$. Then
\[
g_{[0]}=A(t),
\qquad
t=u^qy^p,
\]
where $A\in k[t]$, $A(0)=a\neq0$, and $D:=\deg A\geq1$.
\end{lemma}

\begin{proof}
A monomial $u^jy^d$ has weight zero exactly when
\[
qd=pj.
\]
Coprimality gives $j=qn$ and $d=pn$ for some $n\geq0$, so every weight-zero monomial is $t^n$. Because the maximum in \eqref{eq:rho} is attained by at least one $j\geq1$, a positive power of $t$ occurs. The case $p=0$ gives $q=1$ and $t=u$.
\end{proof}

Let $\sigma$ be the largest weight occurring in $H$. This exists because $H$ has finite support. By \cref{lem:boundary}, the monomial $-(c/a)y$ occurs, so
\begin{equation}\label{eq:sigma}
\sigma\geq1.
\end{equation}

\begin{lemma}[Edge of $H$]\label{lem:H-edge}
There are integers $r,s$ and a nonzero polynomial $\Phi\in k[t]$ such that
\[
0\leq r<q,
\qquad
s\geq1,
\qquad
H_{[\sigma]}=u^ry^s\Phi(t),
\]
and
\begin{equation}\label{eq:sigma-rs}
\sigma=s-\frac pq r.
\end{equation}
\end{lemma}

\begin{proof}
If $u^jy^d$ and $u^{j'}y^{d'}$ both have weight $\sigma$, then
\[
q(d-d')=p(j-j').
\]
Thus $j-j'$ is divisible by $q$ and $d-d'$ by $p$. All $u$-exponents on the edge have one common residue $r\in\{0,\ldots,q-1\}$ modulo $q$. For every edge exponent pair $(j,d)$, write $j=r+qn$. Since $j\geq0$ and $0\leq r<q$, necessarily $n\geq0$. Put $s=\sigma+(p/q)r$. The edge equation $d-(p/q)j=\sigma$ gives $d=s+pn$, so $s=d-pn\in\mathbb Z$ and
\[
u^jy^d=u^ry^s(u^qy^p)^n.
\]
Summing the finitely many edge monomials gives the displayed factorization with $\Phi\in k[t]$. Finally, $s=\sigma+(p/q)r\geq1$.
\end{proof}

\section{The edge equation}

Differentiation lowers weight by $\wt(u)$ or $\wt(y)$, while multiplication by $u$ restores the loss from $\partial_u$. Therefore a pair of homogeneous components of weights $\alpha$ in $H$ and $\beta$ in $g$ contributes to the Keller operator with weight $\alpha+\beta-1$. Since $\alpha\leq\sigma$ and $\beta\leq0$, the weight-$\sigma-1$ component comes only from $H_{[\sigma]}$ and $g_{[0]}$.

Write
\[
H_{[\sigma]}=u^ry^s\Phi(t),
\qquad
 g_{[0]}=A(t),
\qquad
 t=u^qy^p.
\]
Then
\[
(H_{[\sigma]})_u
=u^{r-1}y^s\bigl(r\Phi+qt\Phi'\bigr),
\]
\[
(H_{[\sigma]})_y
=u^ry^{s-1}\bigl(s\Phi+pt\Phi'\bigr),
\]
\[
(g_{[0]})_u=\frac{qt}{u}A',
\qquad
(g_{[0]})_y=\frac{pt}{y}A'.
\]
Substitution into the Keller operator yields
\begin{equation}\label{eq:edge-op}
mu(H_{[\sigma]})_u(g_{[0]})_y
-(H_{[\sigma]})_y\bigl(g_{[0]}+mu(g_{[0]})_u\bigr)
=u^ry^{s-1}B(t),
\end{equation}
where
\begin{equation}\label{eq:B}
B(t)=-ptA(t)\Phi'(t)-sA(t)\Phi(t)-mq\sigma tA'(t)\Phi(t).
\end{equation}
Indeed, the mixed $t^2A'\Phi'$ terms cancel and $pr-qs=-q\sigma$ by \eqref{eq:sigma-rs}.

\begin{lemma}[Terminal coefficient]\label{lem:terminal}
Let
\[
A(t)=a_Dt^D+\cdots,
\qquad
\Phi(t)=\phi_Nt^N+\cdots,
\]
with $D\geq1$, $N\geq0$, and $a_D\phi_N\neq0$. Then
\begin{equation}\label{eq:terminal}
[t^{D+N}]B
=-a_D\phi_N\bigl(pN+s+mq\sigma D\bigr)\neq0.
\end{equation}
\end{lemma}

\begin{proof}
The three summands in \eqref{eq:B} contribute respectively
\[
-pN a_D\phi_N,
\qquad
-s a_D\phi_N,
\qquad
-mq\sigma D a_D\phi_N.
\]
The integer $pN+s+mq\sigma D$ is strictly positive because $s\geq1$, $m,q,D\geq1$, and $\sigma\geq1$. It remains nonzero in $k$ because $\operatorname{char}k=0$.
\end{proof}

\begin{proof}[Proof of \cref{thm:main}]
By \cref{lem:terminal}, $B\neq0$, so the Keller expression has a nonzero component of weight $\sigma-1$.

If $\sigma>1$, this is a positive-weight component, contradicting that the right side of \eqref{eq:pde} is the scalar $c$ of weight zero.

If $\sigma=1$, the entire top component in \eqref{eq:edge-op} must equal $c$. A scalar monomial can arise from $u^ry^{s-1}t^k$ only when $r=0$, $s=1$, and $k=0$. Thus $B$ would have to be constant. But \cref{lem:terminal} gives a nonzero term of degree $D+N\geq1$, again a contradiction.

Therefore $g$ is constant. By \cref{lem:boundary}, that constant lies in $k^\times$.
\end{proof}

\begin{proof}[Proof of \cref{cor:classification}]
Write $g=a\in k^\times$ in \eqref{eq:pde}. Then $-aH_y=c$, so
\[
H(u,y)=-\frac ca y+h(u)
\]
for some $h\in k[u]$. The inverse follows by solving $Q=ax$ and then solving the first coordinate for $y$.
\end{proof}

\section{Quotient identity and collisions}

Define
\[
K=ug^m.
\]
A direct calculation gives
\begin{equation}\label{eq:quotient}
\Jac(H,K)=g^{m-1}\det JF.
\end{equation}
Under $\det JF=c$ this yields
\[
\Jac(H,K)=cg^{m-1},
\qquad
u^{m-1}\Jac(H,K)^m=c^mK^{m-1}.
\]
For $m\geq2$, a hypothetical nonconstant $g$ would also produce an explicit collision after passing to an algebraic closure. Since $g(0,y)\neq0$, every zero $(u_0,y_0)$ of $g$ has $u_0\neq0$. Choosing $x_0^m=u_0$, the distinct points
\[
(\zeta^jx_0,y_0),
\qquad 0\leq j<m,
\]
all map under \eqref{eq:F} to $(H(u_0,y_0),0)$. The theorem rules out this collision mechanism. For $m=1$ the rigidity theorem still holds, but this orbit-collision interpretation is vacuous.

\section{Scope and the bridge at infinity}

The theorem concerns the polynomial normal form \eqref{eq:F} or, equivalently, the self-equivariant swapped form \eqref{eq:swapped}. It does not show that every hypothetical nonproper Keller map can be transformed into this form.

At a branch at infinity, tame inertia in residue characteristic zero is cyclic and acts on a local parameter by roots of unity. That one-dimensional local statement does not automatically provide a compatible second coordinate, simultaneous source and target linearization, or descent from completed local rings to global polynomial coordinates.

Nor does the terminal-coefficient argument extend unchanged to arbitrary completed series. For example, with $m=p=q=\sigma=s=1$, $r=0$, $A(t)=1+t$, and $\Phi(t)=(1+t)^{-1}\in k[[t]]$, formula \eqref{eq:B} gives
\[
B(t)=-tA\Phi'-A\Phi-tA'\Phi=-1.
\]
There is no terminal coefficient in an infinite power series. Any bridge from infinity must therefore prove a finite-support or bounded-pole statement, or replace the terminal coefficient by a different extremal invariant.

\begin{question}
Under what verifiable hypotheses does a branch at infinity of a nonproper plane Keller map globalize to compatible polynomial coordinates carrying a cyclic pseudoreflection action?
\end{question}

\section{Reproducibility statement}

The proof is algebraic and does not depend on finite search. The accompanying internal verification archive checks the determinant identity, the edge identity, the terminal coefficient over a large exact parameter grid, and bounded coefficient systems. A separate implementation performed 130 exact rational coefficient-system exclusions and 500 exact rational edge-identity substitutions. These calculations are regression evidence only; the unbounded conclusion is supplied by \cref{lem:terminal}.

\section{Novelty statement}

A targeted literature audit located broad and important precedents in equivariant affine geometry, Newton polygons, geometry at infinity, and general Jacobian reductions, but no exact antecedent for \cref{thm:main} and \cref{cor:classification}. Accordingly, the calibrated claim is that the theorem and its direct terminal-edge proof appear new relative to the sources reviewed. A definitive priority claim requires specialist review.

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\end{thebibliography}

\end{document}
