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The four-dimensional Hessian conjecture for quintic potentials with a ternary cone top

Published Nov 06, 2026Fresh — added to the Hub within the last 30 daysv16 min read

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Anthropic Claude Opus

Abstract

The Hessian conjecture $\mathrm{HC}_n$ asserts that if $f\in\mathbb{C}[x_1,\dots,x_n]$ has $\det\operatorname{Hess}(f)\in\mathbb{C}^\times$ , then $\nabla f$ is a polynomial automorphism. It is known for $n\le 3$ , false for $n\ge 5$ , and open for $n=4$ , where it implies the plane Jacobian conjecture. Z. Ni proved $\mathrm{HC}_4$ for potentials of degree at most four. We prove $\mathrm{HC}_4$ for every potential of degree five whose top homogeneous form depends on exactly three linear forms. The proof combines a weighted analysis of the Hessian determinant with exact computer-algebra certificates (Groebner bases and Rabinowitsch tests over $\mathbb{Q}$ ); one branch reduces to plane Jacobian pairs of degrees $(4,3)$ , which are excluded by Magnus' theorem. A self-contained verifier is supplied as an ancillary file. The binary and unary cone tops in degree five remain open.

Introduction

For $f \in\mathbb{C}[x] = \mathbb{C}[x_1,\ldots,x_n]$ write $\operatorname{Hess}(f) = (\partial_i\partial_j f)$ . The Hessian conjecture $\mathrm{HC}_n$ [9] states: if $\det\operatorname{Hess}(f) \in\mathbb{C}^{\times}$ then $\nabla f:\mathbb{C}^n \to\mathbb{C}^n$ is a polynomial automorphism. It holds for $n=2$ [4] and $n=3$ [1]; it fails for every $n \ge5$ [10]; for $n=4$ it is open and implies the plane Jacobian conjecture by the doubling construction $\Phi(x,y)=y^{\mathsf T}F(x)$ , for which $\det\operatorname{Hess}\Phi=(\det JF)^2$ [10, 11]. Ni [11] proved $\mathrm{HC}_4$ for $\deg f \le4$ . In degree five the determinant has more homogeneous layers and the arguments of [11] do not transfer directly ([11 §8]); partial structural reductions for a degree-five branch are recorded in [5].

Let $f \in\mathbb{C}[x_1,\ldots,x_4]$ have degree $5$ and $\det\operatorname{Hess}(f)=c \in\mathbb{C}^{\times}$ . Its top form $f_5$ has $\det\operatorname{Hess}(f_5)=0$ , hence by Gordan–Noether [7, 2] depends on at most three linear forms. We say the top is a ternary cone if, after a linear change of coordinates, $f_5=\varphi(x_1,x_2,x_3)$ with $\det\operatorname{Hess}_3(\varphi)\not\equiv0$ (equivalently, $f_5$ depends on exactly three linear forms).

Theorem 1. Let $f \in\mathbb{C}[x_1,x_2,x_3,x_4]$ have degree five, $\det\operatorname{Hess}(f) \in\mathbb{C}^{\times}$ , and a ternary cone top. Then $\nabla f$ is a polynomial automorphism of $\mathbb{C}^4$ .

Remark 2. The class of Theorem 1 contains doubled potentials $F(x_1,x_2)+x_3a(x_1,x_2)+x_4b(x_1,x_2)$ , whose Hessian determinant is $J(a,b)^2$ (§ 5.3). Hence no proof of the quintic case can avoid plane Jacobian pairs of low degree; here only pairs of degrees $(4,3)$ occur, and Magnus’ theorem excludes them.

Status. The statements labelled [C] below are certified by exact computation (identities of polynomials over $\mathbb{Q}$ with symbolic coefficients, Gröbner bases, and the Rabinowitsch criterion $1 \in I + (1 - T g) \Leftrightarrow g \in \sqrt{I}$). Each such statement corresponds to a check of the ancillary verifier verify_hc4_ternary.py (§ 6); all 29 checks pass. The paper has not been refereed.

Tools

Throughout, $x' = (x_1,x_2,x_3)$ , $s = x_4$ , and $A = \operatorname{Hess}_3(\varphi)$ . Linear changes of coordinates preserve both hypothesis and conclusion, since $\operatorname{Hess}(f \circ T) = T^{\mathsf{T}}(\operatorname{Hess} f) \circ T$ .

Imported results. (GN) A form in $n \le4$ variables with vanishing Hessian determinant depends on at most $n - 1$ linear forms [7], [2, Thm. 3.1]. (HC2, HC3) [4, 1], together with transfer to arbitrary fields of characteristic zero and descent of polynomial inverses {citep}[11 Lemmas 2.6, 2.7]. (dBvdE) Over a field of characteristic zero, a bivariate polynomial without terms of degree $\le1$ and with vanishing Hessian determinant is a polynomial in one linear form [3, Thm. 3.1]. (Ni) [11 Prop. 6.1]: if $f = P(x') + sQ(x') + as^2$ with $\deg Q \le 2$ and $\det \operatorname{Hess} f \in \mathbb{C}^{\times}$, then $\nabla f$ is an automorphism. (Mag) {citep}[8]: a plane Jacobian pair whose degrees are coprime is an automorphism. (JvdK) For a plane automorphism $(F,G)$, $\deg F \mid \deg G$ or $\deg G \mid \deg F$ {citep}[6 Ch. 5]`.

Lemma 3 (Schur branch, any $Q$ ). Let $f = P(x') + sQ(x') + as^2$ with $a \ne0$ and $P,Q$ arbitrary. If $\det\operatorname{Hess} f \in\mathbb{C}^{\times}$ , then $\nabla f$ is a polynomial automorphism.

Proof. With $z = s + Q/(2a)$ and $S_z = P - Q^2/(4a) + zQ$ one has the polynomial identity $\det\operatorname{Hess} f = 2a \det\operatorname{Hess}_{x'}(S_z)$ at $z = s + Q/(2a)$ [C, S6a], and $\nabla_{x'} f = \nabla_{x'} S_z$ , $\partial_s f = 2az$ . Thus $\det\operatorname{Hess}_{x'} S_z$ is a nonzero constant; HC3 over $\mathbb{C}(z)$ and descent to $\mathbb{C}[z]$ give a polynomial inverse $H(z,Y)$ of $\nabla_{x'}S_z$ , and $z = Y_4/(2a)$ , $x' = H(z,Y_1,Y_2,Y_3)$ , $s = z - Q(x')/(2a)$ invert $\nabla f$ . This is the argument of [11 Prop. 6.1, branch $a \ne 0$], which does not use $\deg Q \le2$ .

Lemma 4 (constant direction). Let $F \in\mathbb{C}[x_1,x_2,x_3]$ be homogeneous of degree $d$ , nonzero, without terms of degree $\le1$ in $(x_2,x_3)$ , and with $\det\operatorname{Hess}_{(x_2,x_3)} F = 0$ . Then there is a constant linear form $l = \lambda x_2 + \mu x_3$ with $F \in\mathbb{C}[x_1][l]$ .

Proof. Over $K = \mathbb{C}(x_1)$ , (dBvdE) gives $F = G(l)$ with $l = \lambda x_2 + \mu x_3$ , $\lambda,\mu\in K$ , $G \in K[T]$ , $\deg G \ge2$ . Then $\operatorname{Hess}_{(x_2,x_3)} F = G''(l)(\lambda,\mu)^{\mathsf{T}}(\lambda,\mu)$ has rank exactly one, so its kernel line is determined by $F$ . Since $\operatorname{Hess}_{(x_2,x_3)} F(tx) = t^{d-2}\operatorname{Hess}_{(x_2,x_3)} F(x)$ , the kernel line at $tx_1$ equals the one at $x_1$ ; thus the ratio $\lambda/\mu$ (or $\mu/\lambda$ ), a rational function of $x_1$ , is invariant under all dilations and hence constant. After rescaling $\lambda,\mu\in\mathbb{C}$ , $l$ is a coordinate and the coefficients of $F$ as a polynomial in $l$ lie in $\mathbb{C}[x_1]$ .

The weighted top

Proposition 5. Let $f$ be as in Theorem 1, $f = f_5 + \cdots+ f_0$ , $f_5 = \varphi(x')$ . Then:

$[\det\operatorname{Hess} f]_{11} = \det A \cdot\partial_s^2 f_4$ [C, S1b]; hence $f_4 = g_4(x') + s h(x')$ .

With this, $[\det\operatorname{Hess} f]_{10} = \det A \cdot\partial_s^2 f_3 - (\nabla h)^{\mathsf{T}}\operatorname{adj}(A)\nabla h$ [C, S1c]; writing $f_3 = g_3 + s h_2 + s^2m(x') + es^3$ , it follows that $e = 0$ .

For the weights $(1,1,1,2)$ , $f$ has no monomial of weight $> 5$ , its weight-5 part is $\psi= \varphi+ sh + s^2m$ , and the weight-10 part of $\det\operatorname{Hess} f$ equals $\det\operatorname{Hess}\psi$ [C, S2]. Hence $\det\operatorname{Hess}\psi= 0$ .

Proof. (1) and (2) are identities with fully symbolic $f_4$ , $f_3$ , $f_2$ ; as $\det\operatorname{Hess} f$ is constant and $\det A \ne0$ , $\partial_{s}^{2}f_4=0$ , and in (2) the term $6es\det A$ is the only $s$ -dependent one, so $e=0$ . (3) After (1) and (2), $f$ contains no $s^2x'^2$ , $s^3x'$ , $s^4$ (degree 4) and no $s^3$ (degree 3); every other monomial outside $\varphi$ has weight $\le4$ , except $sx'^3$ (weight 5, in $sh$ ) and $s^2x'$ (weight 5, in $s^2m$ ); the identity holds for arbitrary lower terms, and the constant has weight $0<10$ .

Classification of the weighted top

Proposition 6. Let $\psi=\varphi(x')+sh(x')+s^2m(x')$ with $\varphi$ a quintic, $h$ a cubic, $m$ linear, $\det\operatorname{Hess}_3\varphi\ne0$ and $\det\operatorname{Hess}\psi=0$ . After a linear change of coordinates preserving the shape of $\psi$ , one of the following holds:

$(N)\ h=m=0$ ;

$(D)\ m=0$ and $\psi=Q(x_1,x_2)+x_1^4x_3+x_1^3s$ with $\partial_{2}^{2}Q\ne0$ ;

$(F)\ \psi=x_1(s+x_1x_2)^2+x_1^4x_3$ .

Proof. Case $m=0$ . The coefficient of $s^2$ in $\det\operatorname{Hess}(\varphi+sh)$ is $-\frac{3}{2}h\det\operatorname{Hess}_3h$ , and those of $s^3$ , $s^4$ vanish [C, S3a, S3b]. So $h=0$ (case (N)) or, by (GN), $h$ is a binary cubic in two linear forms; up to $\operatorname{GL}_3$ , $h\in\{x_1^3,x_1^2x_2,x_1^3+x_2^3\}$ . For $h=x_1^2x_2$ and $h=x_1^3+x_2^3$ , every coefficient of $\det A$ lies in the radical of the ideal of coefficients of $\det\operatorname{Hess}\psi$ [C, S3c]: excluded. For $h=x_1^3$ , $\det\operatorname{Hess}\psi=-9x_1^4\det\operatorname{Hess}_{(x_2,x_3)}\varphi$ [C, S3d]. Apply Lemma 4 to the part of $\varphi$ of degree $\ge2$ in $(x_2,x_3)$ , which is nonzero (otherwise $A$ has a zero $2\times2$ block and $\det A=0$ ). After a rotation of $(x_2,x_3)$ , $\varphi=Q(x_1,x_2)+x_1^4(\beta_2x_2+\alpha x_3)+\beta_0x_1^5$ , and $\det A=-16\alpha^2x_1^6\partial_2^2Q$ [C, S8c], so $\alpha\ne0\ne\partial_2^2Q$ . Replacing $x_3$ by $\alpha x_3+\beta_2x_2+\beta_0x_1$ gives (D).

Case $m\ne0$ . Take $m=x_1$ . The coefficient of $s^4$ is $-4\det\operatorname{Hess}_{(x_2,x_3)}h$ [C, S4a]; Lemma 4 (applied to the part of $h$ of degree $\ge2$ in $(x_2,x_3)$ , if nonzero) and a rotation of $(x_2,x_3)$ give $h=ax_3^2+bx_1x_2^2+x_1^2(cx_2+c'x_3)+ex_3^3$ . If $a\ne0$ , rescale to $a=1$ ; if $a=0\ne b$ , rescale to $b=1$ and shift $x_2\mapsto x_2-\frac{c}{2}x_1$ to get $c=0$ ; in both cases $\det A$ vanishes on all solutions [C, S4 (I), (II)]. If $a=b=0$ and $(c,c')\ne0$ , a rotation, rescaling and shift give $h=x_1^2x_2$ ; otherwise $h=ex_3^3$ , which is excluded [C, S4 (III.2)]. For $h=x_1^2x_2$ the solutions with $\det A\not\equiv0$ are exactly $\varphi=\frac{1}{4}x_3^2x_2^2+p_{18}x_1^4x_3+p_{19}x_1^4x_2+p_{20}x_1^5$ [C, S4 (III.1), S8a], with $\det A=-8p_{18}^2x_1^9$ [C, S8b], so $p_{18}\ne0$ ; the substitution $x_3\leftarrow(x_3-p_{19}x_2-p_{20}x_1)/p_{18}$ , $x_2\mapsto2x_2$ gives (F) [C, S8d]. All changes used fix $x_1$ and act on $x'$ only, so they preserve $m=x_1$ and the shape of $\psi$ .

Closing the three cases

Case (N)

Here $f$ has no monomials $sx'^3$ , $s^2x'$ , $s^3$ , so $f=P(x')+sQ(x')+as^2$ with $\deg Q\le2$ , and (Ni) applies.

Case (F)

Proposition 7. No quintic $f$ with weight-5 part $x_1(s+x_1x_2)^2+x_1^4x_3$ has $\det\operatorname{Hess}f\in\mathbb{C}^{\times}$ .

Proof. Write $f=\psi_F+\sum_{\mu}c_\mu\mu$ over all monomials $\mu$ of weight $\le4$ and degree $\ge2$ (41 unknowns). The constant term of $\det\operatorname{Hess}f$ lies in the radical of the ideal generated by the coefficients of all non-constant monomials [C, S5].

Case (D)

Proposition 8. Let $f$ be quintic with weight-5 part $Q(x_1,x_2)+x_1^4x_3+x_1^3s$ , $\partial_2^2Q\ne0$ , and $\det\operatorname{Hess} f=c\in\mathbb{C}^{\times}$ . Then $\nabla f$ is an automorphism.

Proof. By Proposition 5 and $m=0$ , $f=P(x')+sR(x')+as^2$ with $R=x_1^{\prime3}+(\text{terms of degree }\le2)$ . If $a\ne0$ , Lemma 3 applies. Let $a=0$ ; then $c=-(\nabla R)^{T}\operatorname{adj}(\operatorname{Hess}P+s\operatorname{Hess}R)\nabla R$ .

(i) The coefficient of $s^2$ forces $(\partial_2R,\partial_3R)$ to be proportional [C, S6b], so $R=\widetilde{R}(x_1,w)$ for a linear form $w$ in $(x_2,x_3)$ . Complete to linear coordinates $(x_1,w,v)$ with $\partial_vx_1=\partial_vw=0$ ; thus $v=(0,v_2,v_3)$ .

(ii) The coefficient of $s$ equals $-P_{vv}\cdot\operatorname{Br}(\widetilde{R})$ , $\operatorname{Br}(R)=R_1^2R_{ww}-2R_1R_wR_{1w}+R_w^2R_{11}$ [C, S6c]. As $\mathbb{C}[x]$ is a domain, $\operatorname{Br}=0$ or $P_{vv}=0$ .

(iii) If $\operatorname{Br}=0$ , then $\widetilde{R}=g(x_1)$ [C, S6d] and $c=-g'(x_1)^2\det\operatorname{Hess}_{(w,v)}P$ [C, S6e], impossible since $\deg g'=2$ .

(iv) If $P_{vv}=0$ : the degree-5 part of $P_{vv}$ is $\partial_v^2(Q+x_1^4x_3)=v_2^2\partial_2^2Q$ [C, S8e], so $v_2=0$ , $v=e_3$ , $R=R(x_1,x_2)$ and $P=P_0(x_1,x_2)+x_3P_1(x_1,x_2)$ with $P_1=x_1^4+(\deg\le3)$ . Then $c=J(P_1,R)^2$ [C, S6f], so $(P_1,R)$ is a plane Jacobian pair of degrees $(4,3)$ . By (Mag) it is an automorphism, contradicting (JvdK).

Proof of Theorem 1. Propositions 5 and 6 leave the cases (N), (D), (F), settled in this section.

Certificates

The ancillary file verify_hc4_ternary.py (Python with python-flint, sympy, and Singular through Sage/passagemath) performs 29 checks, labelled S1a–S8e as cited above, and prints ALL CHECKS PASSED; it runs in a few minutes. Identities are verified with fully symbolic coefficients (or with undetermined functions where stated). Exclusions use the Rabinowitsch criterion over $\mathbb{Q}$ ; since a Gröbner basis equal to $\{1\}$ over $\mathbb{Q}$ is also $\{1\}$ over $\mathbb{C}$ , the conclusions hold over $\mathbb{C}$ . The classification of S4 (III.1) is certified twice, by minimal associated primes and by Rabinowitsch tests (S8a).

Scope. Theorem 1 does not treat quintic potentials whose top form depends on two linear forms or on one; these cases, and hence HC $_4$ in degree five, remain open.

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