Jacobian conjecture counterexample in three dimensions explained

Overview

The Jacobian conjecture asserts that a polynomial map with nowhere‑vanishing Jacobian determinant is globally invertible with a polynomial inverse. In July 2026 Terence Tao published a counterexample showing the conjecture fails in three dimensions (and therefore in any higher dimension). The post gives an explicit degree‑seven polynomial map F, verifies that its Jacobian is the constant −2, and exhibits three distinct points that F sends to the same value, proving non‑injectivity.

The Counterexample Polynomial

The map F : C^3 → C^3 is defined by

F(z₁,z₂,z₃) = ig(((1+z₁z₂)³ z₃ + z₂²(1+z₁z₂)(4+3z₁z₂)), z₂ + 3z₁(1+z₁z₂)² z₃ + 3z₁z₂²(4+3z₁z₂), 2z₁ − 3z₁²z₂ − z₁³z₃ig).

This is a polynomial of total degree seven in the three variables.

Verification of Jacobian and Non‑injectivity

A direct computation shows that the Jacobian determinant of F is the constant −2:

det DF = −2.

Because the Jacobian is a non‑zero constant, the map is locally invertible everywhere (inverse function theorem). However, F is not globally injective; the three points

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

all map to the same image (−1/4,0,0):

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

Thus the Jacobian conjecture fails in dimension three.

Geometric Reformulation via Symmetric Powers

To demystify the construction, Tao rephrases the problem using symmetric powers of C². Let Sym^k(V) denote the k‑th symmetric power of a vector space V. Consider the multiplication map

F : Sym¹(C²) × Sym²(C²) → Sym³(C²), F(L,Q) = L·Q,

where L is a linear polynomial and Q a quadratic polynomial in two variables. In coordinates, writing L(z,w)=az+bw and Q(z,w)=cz²+dz+ew, the map becomes

F((a,b),(c,d,e)) = (ac, ad+bc, ae+bd, be).

This map is polynomial and equivariant under the group C* × C* × SL₂(C) acting by scaling L and Q and by simultaneous change of variables in C².

Construction Using Resultant Normalization

The resultant Res(L,Q) = a²e − abd + cb² measures whether L and Q share a root. It is SL₂‑invariant and transforms under the scaling (L,Q) ↦ (λL, λ⁻¹Q) by a factor λ. Imposing the normalization Res(L,Q)=1 removes the scaling symmetry and yields a four‑dimensional variety

X = { (L,Q) : Res(L,Q)=1 } ⊂ Sym¹(C²)×Sym²(C²).

On X the map F remains polynomial and still enjoys the SL₂‑equivariance.

Local Injectivity Argument

If Res(L,Q)=1 ≠ 0 then L and Q have no common root. Using the SL₂ action we may send the unique root of L to infinity, which forces a=0 after a suitable change of variables. In this gauge the equations become b=1+O(a), c=1−(3/2)a y+O(a²), etc., showing that (L,Q) can be recovered uniquely from the cubic polynomial F(L,Q) up to arbitrarily small perturbations. Hence F is locally injective on X.

Global Non‑injectivity via Three‑to‑One Map

Even after fixing Res(L,Q)=1, the multiplication map F is not globally injective. A generic cubic polynomial C ∈ Sym³(C²) factors as a product of three linear polynomials C = L₁L₂L₃. The three pairs

(L₁, L₂L₃), (L₂, L₁L₃), (L₃, L₁L₂)

all map to the same C under F, and they are not related by the scaling symmetry (L,Q)↦(λL,λ⁻¹Q). Thus F is generically three‑to‑one on X, establishing property (b) of the reformulated counterexample: a locally injective polynomial map that fails to be globally injective.

Obtaining the Affine Variety Isomorphic to C³

The variety X defined by Res(L,Q)=1 is not isomorphic to affine C⁴. By intersecting X with a suitable affine hyperplane V ⊂ Sym³(C²) (chosen so that V avoids the origin and corresponds to the differential operator D = ∂_z²∂_w), one obtains a three‑dimensional variety

X_V = { (L,Q) : Res(L,Q)=1, F(L,Q)∈V }.

Choosing V as the hyperplane { C(z,w)=f z³ + g z² w + h z w² + i w³ : g=1 } leads to the explicit equations

X_V = { (a,b,c,d,e)∈C⁵ : a²e−abd+cb²=1, ad+bc=1 }.

When a≠0 the two equations can be solved for d and e as rational functions of a,b,c, giving a birational equivalence X_V \ {a=0} ≅ { (a,b,c)∈C³ : a≠0 }. The fibre over a=0 reduces to the equations cb²=1 and bc=1, which have the unique solution b=c=1, leaving d and e free. This shows that X_V is actually isomorphic to affine C³ via a polynomial change of variables.

Polynomial Change of Variables and Inverse Map

Introducing coordinates (a,y,z) by the ansatz

b = 1 + a y, c = 1 − (3/2) a y + a² z, d = (1−bc)/a = ½ y − a z + (3/2) a y² − a² y z, e = (1+abd−cb²)/a² = −2z + 4y² − 4a y z + 3a y³ − 2a² y² z,

provides a polynomial parametrization of X_V. The map F in these coordinates becomes

G(a,y,z) = ( a − (3/2)a²y + a³z, ½ y − 3a z + 6a y² − 6a² y z + (9/2)a² y³ − 3a³ y² z, −2z + 4y² − 6a y z + 7a y³ − 6a² y² z + 3a² y⁴ − 2a³ y³ z ).

Its Jacobian determinant is the constant −1, confirming the constant Jacobian property. The inverse map is also polynomial:

a = a, y = 2bd − ae, z = 2d² + ce + 6b d² + 3b c e − (9/2) e.

Thus the original counterexample F is recovered from G by a linear change of variables, completing the digestion of the example.

Remarks on Dimensions and Open Problems

The construction works in three dimensions; by composing with any polynomial embedding C³ → Cⁿ (n>3) one obtains counterexamples in all higher dimensions. The Jacobian conjecture remains open for maps C² → C² and is trivial in one dimension.

Community Reaction (Hacker News)

Readers highlighted the surprising nature of the cancellation that makes the Jacobian constant. One commentator vanderZwan noted:

"Sounds like the most interesting part would be learning what approaches the LLM did use to see if that’s reusable elsewhere."

Others remarked on the accessibility of the exposition:

"The introduction to this piece was easy to follow, but as soon as he got into recapitulating it with algebra he lost me (because I’m bad at math)." – @tptacek

Some reflected on the broader implications for problem‑solving:

"Finding a different way of thinking about a problem often leads to a breakthrough… I suspect many old problems will fall because of it." – @jmward01

A few joked about the difficulty of following the mathematics:

"After reading a quarter of the article I started wondering, is this what non coders feel when vibe coding software?" – @aayushdutt

These comments illustrate both the technical challenge and the enthusiasm generated by the counterexample among the audience.