Jacobian Conjecture Counterexample

The Jacobian conjecture predicts that a polynomial map F:ℂn⟹ℂn with constant nonzero Jacobian determinant must have a polynomial inverse. Since an invertible map must be injective, a counterexample can be proved by finding a polynomial map with constant nonzero Jacobian determinant that sends two distinct inputs to the same output.

The example below does exactly that in dimension 3. The verification is deliberately written in a computational form so the algebra can be checked directly.

Define F:ℂ3⟹ℂ3 by

F(x,y,z)=((F_1)(x,y,z),(F_2)(x,y,z),(F_3)(x,y,z))

where

(F_1)(x,y,z)=(1+x*y)3*z+y2*(1+x*y)*(4+3*x*y)

(F_2)(x,y,z)=y+3*x*(1+x*y)2*z+3*x*y2*(4+3*x*y)

(F_3)(x,y,z)=2*x−3*x2*y−x3*z

The computation splits into two independent checks.

First, the Jacobian determinant is constant and nonzero:

det(J)*F=−2

Second, the map sends three distinct points to the same output:

F(0,0,−1/4)=F(1,−3/2,13/2)=F(−1,3/2,13/2)=(−1/4,0,0)

The first condition is the hypothesis of the Jacobian conjecture. The second condition contradicts injectivity. Together, these two facts show that this map is a counterexample if the displayed formula is correct.

The following setup block defines the map once. The later checks use these definitions.

import sympy as sp
x, y, z = sp.symbols("x y z")
F1 = (1 + x*y)**3*z + y**2*(1 + x*y)*(4 + 3*x*y)
F2 = y + 3*x*(1 + x*y)**2*z + 3*x*y**2*(4 + 3*x*y)
F3 = 2*x - 3*x**2*y - x**3*z
F = sp.Matrix([F1, F2, F3])
F

The next check computes the Jacobian matrix and its determinant symbolically. The expected output is the constant −2.

J = F.jacobian([x, y, z])
detJ = sp.factor(J.det())
detJ

The collision check can be written directly as three formula evaluations:

F(0,0,−1/4)=(−1/4,0,0)

F(1,−3/2,13/2)=(−1/4,0,0)

F(−1,3/2,13/2)=(−1/4,0,0)

The three inputs are distinct, so this proves that F is not injective.

For a runnable check of the same claim, use the following code block. It returns True exactly when all three points map to the same target.

points = [
    (0, 0, sp.Rational(-1, 4)),
    (1, sp.Rational(-3, 2), sp.Rational(13, 2)),
    (-1, sp.Rational(3, 2), sp.Rational(13, 2)),
]
target = (sp.Rational(-1, 4), 0, 0)
images = [tuple(F.subs({x: a, y: b, z: c})) for a, b, c in points]
len(set(points)) == 3 and all(image == target for image in images)