Jacobian conjecture counterexample in three dimensions explained via symmetric powers

The Jacobian conjecture is false in dimension three

The Jacobian conjecture asserts that a polynomial map F: C^n → C^n whose Jacobian determinant is a non‑zero constant must have a polynomial inverse. Terry Tao’s 2026 blog post shows that this statement fails for n = 3 by exhibiting an explicit polynomial map with constant Jacobian –2 that is not injective, thus providing a concrete counterexample.

Explicit counterexample and its verification

The map is defined by F(z1, z2, z3) = ((1+z1 z2)^3 z3 + z2^2 (1+z1z2) (4+3z1z2),\nz2 + 3 z1 (1+z1z2)^2 z3 + 3 z1 z2^2 (4+3z1z2),\n2 z1 - 3 z1^2 z2 - z1^3 z3). A direct calculation gives det(DF) = –2, a non‑zero constant. Moreover, F(0,0,–1/4) = F(1,–3/2,13/2) = F(–1,3/2,13/2) = (–1/4,0,0), showing that three distinct points map to the same image, so F is not globally invertible.

Geometric digestion: reformulation via symmetric powers

To demystify the miraculous cancellation, Tao rephrases the problem. Consider the multiplication map F: Sym^1(C^2) × Sym^2(C^2) → Sym^3(C^2), F(L,Q) = LQ, where L is a linear polynomial and Q a quadratic polynomial in two variables. This map is polynomial and equivariant under the action of C^× × C^× × SL_2(C).

By imposing the normalization Res(L,Q) = 1, where the resultant Res(L,Q) = a^2 e – abd + c b^2 for L(z,w)=az+bw and Q(z,w)=cz^2+dz+ew, the domain becomes a four‑dimensional variety X = {(L,Q) : Res(L,Q)=1} ⊂ Sym^1(C^2)×Sym^2(C^2). The restricted map F: X → Sym^3(C^2) remains locally injective but is generically three‑to‑one because a generic cubic factors as L1 L2 L3, giving three distinct pairs (Li, Lj Lk) that map to the same cubic and are not related by the scaling symmetry.

Taking a three‑dimensional affine slice V ⊂ Sym^3(C^2) defined by the differential operator D = ∂_z^2 ∂_w (i.e., the condition g = 1 in the cubic C(z,w)=f z^3+ g z^2 w+ h z w^2+ i w^3) yields the variety {(a,b,c,d,e) ∈ C^5 : a^2 e – abd + c b^2 = 1, ad + bc = 1}. When a ≠ 0 the equations can be solved polynomially for d and e, giving a birational equivalence with {(a,b,c) ∈ C^3 : a ≠ 0}. The fibre over a = 0 reduces to the unique point (b,c) = (1,1) with free d,e, which is affine. A further polynomial change of variables (a,y,z) ↦ (a, 1+ay, 1 – (3/2)ay + a^2 z, (1–bc)/a, (1+abd–cb^2)/a^2) provides a global polynomial parametrisation of the whole variety by three coordinates, establishing that X is isomorphic to C^3. Composing this isomorphism with the multiplication map yields the original counterexample F.

Significance and open problems

The counterexample settles the Jacobian conjecture in dimension three (and hence in all higher dimensions by taking products with identity maps). The conjecture remains open for n = 2 and is trivial for n = 1. The geometric interpretation highlights that the failure stems from the non‑injectivity of the multiplication map after imposing a resultant normalization, a phenomenon visible already at the level of symmetric powers.

Community reaction on Hacker News

Readers expressed astonishment at the algebraic miracle and curiosity about the AI‑assisted discovery. One comment noted: "While this is an extremely quick verification, the construction presented in this fashion appears like a massive miracle… So finding such a polynomial looks highly unlikely to be located by brute force." Another asked: "Sounds like the most interesting part would be learning what approaches the LLM did use to see if that’s reusable elsewhere." A user remarked on the conversational tone with the AI: "Everything Tao said was constantly followed by praise: “That’s exactly the right way to think about it.”, “Yes, you are exactly right.” … Seems like sycophancy is still an issue." Questions were raised about the underlying proof that a non‑zero Jacobian polynomial must be constant, and about whether the counterexample might have been present in the model’s training data.

Conclusion

The Jacobian conjecture’s failure in dimension three is now established via an explicit polynomial map whose constant Jacobian and non‑injectivity follow from a transparent geometric construction using symmetric powers, resultants, and a carefully chosen affine slice. This work not only resolves a long‑standing open problem in one direction but also illustrates how AI‑assisted exploration can guide the discovery of intricate algebraic examples.