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\title[The four-variable Hessian and planar Jacobian conjectures]{Equivalence of the Four-Variable Hessian and Planar Jacobian Conjectures}
\author{Qiyue Tang}
\date{}
\subjclass[2020]{Primary 14R15. Secondary 14R10}
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\keywords{Hessian conjecture, Jacobian conjecture, weighted leading forms}
\begin{document}
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\begin{abstract}
We prove that the four-variable Hessian conjecture and the planar Jacobian conjecture are equivalent over the complex numbers.
\end{abstract}
\maketitle
\section{Introduction}
Kraus \cite{kraus} stated the planar Jacobian assertion in 1884. As Rodr\'iguez D\'iaz \cite[Section~2.1]{rodriguez} explains, the argument breaks down at infinity, where a coordinate function has a pole and its derivative at the point in question is undefined. In 1939 Keller \cite{keller} asked whether a polynomial transformation with integer coefficients and Jacobian determinant one has a polynomial inverse. He also established the birational case \cite[Satz~3]{keller}, in which a rational inverse is already given. For an integer \(n\ge1\), write \(x=(x_1,\ldots,x_n)^{\mathsf T}\) for the coordinate column on \(\C^n\). For a polynomial map \(G=(G_1,\ldots,G_n)\colon\C^n\to\C^n\), its Jacobian matrix is \(DG=(\partial G_i/\partial x_j)_{1\le i,j\le n}\). Call \(G\) a Keller map if \(\det DG\) is a nonzero constant. The Jacobian conjecture \(\mathrm{JC}_n\) asserts that every Keller map of \(\C^n\) is a polynomial automorphism, that is, it has a two-sided polynomial inverse.

Bass, Connell and Wright \cite{bcw} reduced the family of Jacobian conjectures to maps \(x+H(x)\) with \(H\) cubic homogeneous, allowing the dimension to increase. De Bondt and van den Essen \cite{symmetric} further reduced this family to the case in which \(DH\) is symmetric. The symmetry identity \(\partial H_i/\partial x_j=\partial H_j/\partial x_i\) gives
\[
 \frac{\partial}{\partial x_j}\left(\frac14\sum_{i=1}^n x_i H_i(x)\right)
 =\frac14\left(H_j(x)+\sum_{i=1}^n x_i\frac{\partial H_j}{\partial x_i}(x)\right)
 =H_j(x),
\]
by homogeneity of \(H_j\). The symmetric reduction therefore concerns gradient maps in a dimension that may exceed that of the original map.

For a polynomial \(f\) on \(\C^n\), let \(\grad f=(\partial f/\partial x_i)_{i=1}^n\) be its gradient column and let \(\Hess f=(\partial^2 f/\partial x_i\partial x_j)_{i,j=1}^n\). The Hessian conjecture \(\mathrm{HC}_n\) is the restriction of \(\mathrm{JC}_n\) to \(G=\grad f\). Meng \cite{meng} formulated the conjecture through the formal Legendre transform. De Bondt and van den Essen formulated it independently in terms of gradient maps \cite[Remark~4.5]{adamus}. Subtract the affine part of \(f\) so that \(\grad f(0)=0\). This changes the gradient by a constant translation. For \(\det\Hess f\in\C^\times=\C\setminus\{0\}\), the linear term of \(\grad f\) is \(\Hess f(0)x\) with invertible matrix. Write \(p=(p_1,\ldots,p_n)^{\mathsf T}\) for the target coordinate column. The formal inverse \(X(p)\) starts with \(\Hess f(0)^{-1}p\). At each total degree \(k\ge2\), the degree-\(k\) part of \(\grad f(X(p))=p\) is linear in the degree-\(k\) part of \(X\), with coefficient matrix \(\Hess f(0)\). All other terms are determined at lower degrees because the nonlinear part of \(\grad f\) has degree at least two in its inputs. This recursion gives a unique right inverse. A left inverse is obtained by the same degree comparison for \(X(\grad f(x))=x\): composition of each homogeneous unknown with \(\Hess f(0)x\) is an invertible linear operation. Associativity makes the two inverses equal.

Meng's formal Legendre transform is
\[
 f^*(p)=p^{\mathsf T}X(p)-f(X(p)).
\]
For \(1\le i\le n\), differentiation gives
\[
 \frac{\partial f^*}{\partial p_i}
 =X_i(p)+\sum_{j=1}^n
 \left(p_j-\frac{\partial f}{\partial x_j}(X(p))\right)
 \frac{\partial X_j}{\partial p_i}(p)
 =X_i(p).
\]
Thus \(f^*\) is polynomial exactly when \(X\) is polynomial: one direction follows by differentiation and the other by the formula for \(f^*\). A polynomial \(X\) satisfies both inverse identities as polynomial identities. Since \(D(\grad f)=\Hess f\), this recovers the gradient formulation. The doubling construction in Lemma~\ref{lem:lift} gives \(\mathrm{HC}_{2n}\Rightarrow\mathrm{JC}_n\), while the gradient formulation gives \(\mathrm{JC}_n\Rightarrow\mathrm{HC}_n\).

Dillen \cite{dillen} proved the Hessian conjecture in dimension two, and de Bondt \cite{hess3} proved it in dimension three. De Bondt's proof uses a linear change of variables and a positive weighting whose leading form has nonzero Hessian determinant. In July 2026, Alp\"oge announced a noninjective Keller map in dimension three. Its Jacobian determinant is \(-2\) and three distinct points have the same image \cite{alpoge,afp}. Adjoining identity coordinates preserves both the determinant and the collision, so the example refutes \(\mathrm{JC}_n\) for every \(n\ge3\). Meng and Yang \cite[Theorem~6.1]{mengyang} used doubling followed by a one-variable partial Legendre transform to obtain a degree-fourteen polynomial in five variables with Hessian determinant \(128\) and a noninjective gradient. Adjoining a term \(x_{n+1}^2/2\) adds an identity gradient coordinate and a Hessian block of determinant one, so this example refutes \(\mathrm{HC}_n\) for every \(n\ge5\). The one-variable assertions follow from the fact that a polynomial with nonzero constant derivative is affine. The two remaining dimension-specific assertions are therefore \(\mathrm{JC}_2\) and \(\mathrm{HC}_4\).

For \(\mathrm{HC}_4\), Ni \cite[Theorem~1.2]{ni} treats quartic potentials by splitting the highest homogeneous part according to its Hessian rank. Adamus and Hajto \cite[Theorem~4.17 and Remark~4.5]{adamus} develop a Legendre-potential criterion for invertibility and distinguish the known implication \(\mathrm{HC}_4\Rightarrow\mathrm{JC}_2\) from its converse. The converse follows from the following description of four-variable potentials.

\begin{theorem}\label{thm:main}
Let $f\in\C[x,y,z,w]$ have nonzero constant Hessian determinant. Either $\grad f$ is a polynomial automorphism or, in some constant linear coordinates, $f$ has the form
\[
 f=R(x,y)+zA(x,y)+wB(x,y),
 \qquad \det D(A,B)\in\C^\times.
\]
Here \(R,A,B\in\C[x,y]\) and \(D(A,B)\) is the Jacobian matrix with respect to \(x,y\). Consequently \(\mathrm{HC}_4\) and \(\mathrm{JC}_2\) are equivalent.
\end{theorem}

For a homogeneous polynomial of degree at least two in four variables, the Gordan--Noether theorem gives a nonzero constant vector in the Hessian kernel when the Hessian determinant vanishes \cite{gordannoether,bfp}. The polynomial \(y^2+x^2z+x^3w\), for example, is multiplied by \(t^6\) under the substitution \((x,y,z,w)\mapsto(tx,t^3y,t^4z,t^3w)\), so it is homogeneous for weights \((1,3,4,3)\). Its Hessian is singular because the \(z,w\) rows are proportional over \(\C(x)\). Its gradient components \(2xz+3x^2w,2y,x^2,x^3\) vanish at the origin and are linearly independent over \(\C\). A constant vector in the Hessian kernel would give a linear combination of these components that is constant, hence zero, so that vector must vanish. A translation direction of a Zariski closed set \(Z\subseteq\C^4\) is a nonzero vector \(d\) such that \(a+td\in Z\) for all \(a\in Z\) and \(t\in\C\). In this example, fixing \(x\ne0\) lets the first two gradient components vary freely, with the last two fixed at \((x^2,x^3)\). The gradient-image closure is the product of \(\C^2\) with the closure of the curve parametrized by \((x^2,x^3)\), and translations in the first two coordinates preserve it. De Bondt asked for such a direction for the gradient-image hypersurface of an arbitrary four-variable polynomial with Hessian rank three \cite[Section~6, Problem~1]{quasi}. Theorem~\ref{thm:r4} supplies the direction, and Corollary~\ref{cor:debondt-problem-one} answers the relation question in all singular ranks. De Bondt's low-rank classification \cite[Corollary~7.3]{debondt} then determines the possible leading forms as weights change.

\begin{lemma}\label{lem:lift}
Let \(b=(b_1,\ldots,b_n)^{\mathsf T}\) and \(v=(v_1,\ldots,v_n)^{\mathsf T}\) be two blocks of \(n\) variables. For a polynomial \(R(b)\) and a polynomial map \(G\colon\C^n\to\C^n\), set \(f(b,v)=R(b)+v^{\mathsf T}G(b)\). Then
\[
 \det\Hess f=(-1)^n(\det DG)^2.
\]
If this determinant is a nonzero constant, a polynomial change of source coordinates and a permutation of the target coordinates carry \(\grad f\) to the map \((b,v)\mapsto(G(b),v)\). In this case \(\grad f\) is a polynomial automorphism if and only if \(G\) is.
\end{lemma}
\begin{proof}
With \(\Hess_b f\) denoting the matrix of second derivatives in \(b\) with \(v\) fixed, the two blocks of \(\grad f\) and the Hessian are
\[
 \grad f(b,v)=\bigl(\grad R(b)+DG(b)^{\mathsf T}v,\,G(b)\bigr),
 \qquad
 \Hess f=\begin{pmatrix}
 \Hess_b f & DG^{\mathsf T}\\
 DG & 0
 \end{pmatrix}.
\]
Moving the last \(n\) rows above the first \(n\) uses \(n^2\) row transpositions. The resulting block triangular matrix has diagonal blocks \(DG,DG^{\mathsf T}\), giving the determinant formula.

If \(\det\Hess f\) is a nonzero constant, so is \(\det DG\), since its square is a unit in \(\C[b_1,\ldots,b_n]\). Use the source coordinate \(v'=\grad R(b)+DG(b)^{\mathsf T}v\) in place of \(v\), fixing \(b\). The inverse substitution is
\[
 v=(DG(b)^{\mathsf T})^{-1}\bigl(v'-\grad R(b)\bigr),
\]
which is polynomial by the adjugate formula for the inverse matrix. The gradient in these source coordinates sends \((b,v')\) to \((v',G(b))\). Exchanging its output blocks gives \((b,v')\mapsto(G(b),v')\). If \(G\) is invertible, the inverse of this product is \((b,v')\mapsto(G^{-1}(b),v')\). Conversely, the second block of any inverse of the product is \(v'\), since the product fixes that block. Restricting both composition identities to \(v'=0\) shows that the first block of the inverse on that slice is a two-sided inverse of \(G\).
\end{proof}

\input{hessian_r4}
\input{hessian_weights}
\input{hessian_entrances}

\section{Proof of the main theorem}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
If \(f\) is quadratic, its gradient is affine with invertible linear part \(\Hess f\). The determinant condition excludes degrees below two, and Proposition~\ref{prop:ordinary-top} gives the two alternatives for degree at least three. Under \(\mathrm{JC}_2\), the map \((A,B)\) in the second alternative has a polynomial inverse. Lemma~\ref{lem:lift}, applied with \(n=2\), then gives a polynomial inverse of the gradient. This proves \(\mathrm{JC}_2\Rightarrow\mathrm{HC}_4\).

Conversely, given a planar Keller map \(G=(A,B)\), apply Lemma~\ref{lem:lift} to \(f=zA+wB\). Under \(\mathrm{HC}_4\), the gradient is a polynomial automorphism, so the lemma gives a polynomial inverse of \(G\).
\end{proof}

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\end{thebibliography}
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\begingroup\small
\noindent\textsc{Zhili College, Tsinghua University, Beijing 100084, China}\par
\noindent\textit{Email address}\enspace
\href{mailto:tangqy24@mails.tsinghua.edu.cn}{\texttt{tangqy24@mails.tsinghua.edu.cn}}\par
\endgroup
\end{document}