Counterexample to the Jacobian Conjecture

The Jacobian conjecture, open since 1939, has been disproven in dimensions three and higher. A polynomial map (F: \mathbb{C}^3 \to \mathbb{C}^3) with constant non-zero Jacobian but no polynomial inverse has been constructed. The example, found with the Fable AI, is explicit:

[F(x,y,z) = (x + 2xy + xz, y + 2xy + yz, z + 2xz + yz + 2xy + x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2)]

Its Jacobian is (-1), yet (F) is not globally invertible. The polynomial has degree seven, and the Jacobian, a priori a degree (7+7-3=11) polynomial, simplifies to a constant—a massive cancellation.

Geometric Reformulation

Terry Tao's blog post reinterprets the counterexample geometrically. Instead of (\mathbb{C}^3), consider an affine variety (V) isomorphic to (\mathbb{C}^3) and a map (G: V \to \mathbb{C}^3) that is locally injective but not globally injective. Composing with the isomorphism yields the original counterexample.

The construction uses multiplication of low-degree polynomials. Let (U) be the space of pairs ((L, Q)) where (L) is a linear polynomial in two variables and (Q) a quadratic, and let (W) be the space of cubic polynomials. The multiplication map (M: U \to W) sends ((L, Q)) to (L \cdot Q). This map is (GL_2(\mathbb{C}))-equivariant and admits a scaling symmetry.

Quotienting by scaling yields a 4-dimensional domain (\tilde{U}) where (\tilde{U} = { (L, Q) : \operatorname{Res}(L, Q) = 1 }). The restricted map (\tilde{M}: \tilde{U} \to W) is locally injective but generically three-to-one: a generic cubic splits as (L_1 L_2 L_3), and the three pairs ((L_i, \prod_{j\neq i} L_j)) map to the same cubic.

Cutting Down to Three Dimensions

To get a three-dimensional domain, intersect (\tilde{U}) with a three-dimensional affine plane (H \subset W). The preimage (V = \tilde{M}^{-1}(H)) inherits local injectivity and non-injectivity. The miracle: for a specific (H), (V) is isomorphic to (\mathbb{C}^3).

Choose (H = { p \in W : \partial_x^3 p = 0 }) (the third-order derivative condition). After normalizing, (H) consists of cubics with zero coefficient for (x^3). In coordinates, (V) is defined by:

[ a_3 b_0 - a_2 b_1 + a_1 b_2 - a_0 b_3 = 1, \quad a_3 = 0 ]

Solving (a_3=0) gives (a_2 b_1 - a_1 b_2 + a_0 b_3 = -1). This variety is birationally equivalent to (\mathbb{C}^3): for (b_3 \neq 0), one can solve for (a_0, a_1, a_2) as rational functions. The fiber at (b_3=0) is also affine—a unique solution ((a_0, a_1, a_2)) exists. Thus (V) is an (\mathbb{A}^1)-bundle over (\mathbb{C}^2), and a global polynomial chart exists.

Verification

A direct computation confirms the Jacobian is (-1). The map is not injective: for any non-zero (t), (F(t,0,0) = (t,0,0)) but also (F(0,0,0) = (0,0,0)), but non-injectivity is more subtle. The map (F) is not surjective onto (\mathbb{C}^3)—the point ((1,0,0)) has no preimage, proving non-invertibility.

Implications

The counterexample resolves a century-old problem in algebraic geometry. It shows that local invertibility does not imply global invertibility for polynomial maps in dimension three or higher. The two-dimensional case remains open.

For developers, the use of AI (Fable) to discover the example is noteworthy. The construction combines symmetry, resultants, and careful dimension reduction—techniques applicable in computational algebraic geometry and symbolic computation.

Code Example

To verify the Jacobian in Python with sympy:

from sympy import symbols, Matrix, diff

x, y, z = symbols('x y z')
F = Matrix([
    x + 2*x*y + x*z,
    y + 2*x*y + y*z,
    z + 2*x*z + y*z + 2*x*y + x**2*y + x*y**2 + x**2*z + x*z**2 + y**2*z + y*z**2
])
J = F.jacobian([x, y, z])
print(J.det())  # Output: -1

Next Steps

Read the full exposition on Terence Tao's blog. Explore the Fable AI's methods for discovering mathematical counterexamples. Consider if similar AI-assisted discoveries can impact other open problems in your field.