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We are seeking an explicit, checkable counterexample to the ...
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We are seeking an explicit, checkable counterexample to the ...

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We are seeking an explicit, checkable counterexample to the two-dimensional complex Jacobian Conjecture ($\mathrm{JC}_2$): a polynomial map $F = (P, Q): \mathbb{C}^2 \to \mathbb{C}^2$ with $P, Q \in \mathbb{C}[x, y]$ such that the polynomial Jacobian determinant satisfies $\det J_F(x, y) = \frac{\partial P}{\partial x}\frac{\partial Q}{\partial y} - \frac{\partial P}{\partial y}\frac{\partial Q}{\partial x} \equiv 1$ (or a non-zero constant) identically, while $F$ is globally non-injective ($\exists z_1 \ne z_2 \in \mathbb{C}^2$ such that $F(z_1) = F(z_2)$). ### Background & Context In dimension $n = 3$, a non-injective polynomial Keller map exists via the "fake shear" mechanism based on the 3-to-1 factorization of cubic polynomials $C(t) = L(t) \cdot Q(t)$ under the resultant condition $\operatorname{Res}(L, Q) = 1$: $$ \begin{aligned} F_1(x, y, z) &= -\frac{1}{2}\left[(1+xy)^3 z + y^2(1+xy)(4+3xy)\right], \\ F_2(x, y, z) &= y + 3x(1+xy)^2 z + 3xy^2(4+3xy), \\ F_3(x, y, z) &= 2x - 3x^2 y - x^3 z. \end{aligned} $$ This map satisfies $\det J_F(x, y, z) \equiv 1$ identically on $\mathbb{C}^3$ and exhibits a 3-to-1 collision: $F(0, 0, -1/4) = F(1, -3/2, 13/2) = F(-1, 3/2, 13/2) = (1/8, 0, 0)$. It succeeds because the third variable $z$ enters strictly linearly as an affine shear variable ($F(x, y, z) = M(x, y)z + N(x, y)$), allowing the $2 \times 2$ minors of the Jacobian matrix to cancel all cross-derivative terms. ### The Problem in 2D & Prior Failed Approaches Direct attempts to translate this mechanism to $\mathbb{C}^2$ encounter deep structural obstructions that must be overcome: 1. **The 2D Fake Shear Unit Obstruction:** A standard 2D fake shear $\Phi(x, y) = \left(f(x), \frac{y}{f'(x)} + B(x)\right)$ requires $A(x) = \frac{1}{f'(x)}$. For $\Phi$ to be a polynomial map in $\mathbb{C}[x, y]$, $f'(x)$ must be an invertible element of $\mathbb{C}[x]$. Since $\mathbb{C}[x]^\times = \mathbb{C}^\times$, $f'(x)$ must be a non-zero constant, forcing $f(x)$ to be linear ($\deg f = 1$), which reduces $\Phi$ to an elementary triangular automorphism (strictly injective). 2. **The Degree-$(32, 112)$ Puiseux / Syzygy Breakdown:** An extensive effort modeled an obstruction boundary using degrees $\deg P = 32$, $\deg Q = 112$, $\gcd(32, 112) = 16$, with leading homogeneous forms $\widehat{P} = (x^8 y^8)^2 = x^{16}y^{16}$ and $\widehat{Q} = (x^8 y^8)^7 = x^{56}y^{56}$ (avoiding Moh's $\deg \le 100$, Abhyankar–Moh non-divisibility, and Heitmann prime gcd theorems). This construction fails due to: - **The Puiseux Truncation Residue:** Matching $Q \sim P^{7/2}$ requires negative powers of $y$ beyond the 6th step ($\nu_7 = -7$). Truncation leaves an uncancelled residue monomial $\frac{11259376953125}{587068342272} x^{43} y^8$. - **The Poisson Cascade:** Attempting to eliminate $x^{43} y^8$ by introducing lower-degree terms in $Q$ (such as $Q_9 \propto x^{43}y^9$) via the bracket $\{x, Q_9\}$ creates mandatory non-zero brackets with the top form $x^{16}y^{16}$, generating higher-degree uncancelable terms (e.g., $+\frac{191409408203125}{165112971264} x^{58} y^{24}$, $-\frac{76057091318359375}{126806761930752} x^{54} y^{15}$, and up to $x^{62} y^{33}$). In total, 15 non-zero monomials remain in $\det J_F - 1$, causing the determinant to vanish at real points (e.g., near $(1, 0.6602631)$). - **The Heitmann / Lê–Weber Obstruction:** Leading forms of the type $(x^a y^b)^k$ define at most 2 places at infinity ($\#\mathrm{Pl}_\infty \le 2$), which by theorem forces any Keller map to be an automorphism. 3. **The Non-Descendable Surface Coverings:** Formulations on ramified coverings or algebraic surfaces (e.g., $P(x, w) = x + x^{16}w^7, Q(x, w) = w^2$ on $\mathcal{S} = \{y = w^2 + \frac{32}{9} x^{15}w^9\}$ satisfying $\mathrm{d}P \wedge \mathrm{d}Q = \mathrm{d}x \wedge \mathrm{d}y$) fail because $X^{16} W^7 \notin \operatorname{Im}(\varphi^*)$ for $\varphi(X, W) = (X, W^2 + \frac{32}{9} X^{15}W^9)$ (the Diophantine equation $2b + 7k = 7$ has no integer solutions with $0 \le k \le b$), meaning the map cannot be pushed down to ordinary polynomials in $\mathbb{C}[X, Y]$. ### Task & Technical Requirements Provide an explicit, mathematically sound construction that resolves these bottlenecks: 1. **Explicit Polynomials:** Give $P(x, y), Q(x, y) \in \mathbb{Q}[x, y]$ (or an explicit algebraic number field) fully expanded with concrete coefficients—no undefined functions, no unexpanded summation limits, no formal Laurent series, and no ambient varieties or ramified coverings. 2. **Jacobian Identity:** The polynomials must satisfy $\frac{\partial P}{\partial x}\frac{\partial Q}{\partial y} - \frac{\partial P}{\partial y}\frac{\partial Q}{\partial x} \equiv 1$ (or a non-zero constant) as an exact polynomial identity in $\mathbb{C}[x, y]$, verified coefficient by coefficient. 3. **Explicit Non-Injectivity Certificate:** Provide explicit distinct points $z_1 = (x_1, y_1) \ne z_2 = (x_2, y_2) \in \mathbb{C}^2$ such that $F(z_1) = F(z_2)$. 4. **Mechanism:** Explain the algebraic or geometric mechanism that bypasses both the 2D fake shear unit constraint and the Poisson cascade (e.g., via non-properness geometry with $\#\mathrm{Pl}_\infty \ge 3$, Hamiltonian derivations with non-LND slices, or a genuine 2D multi-to-one fibrational structure). Provide self-contained Python/SymPy code demonstrating that $\det J_F - 1 == 0$ and $F(z_1) == F(z_2)$.

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