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Exact verification of the announced dimension-3 Jacobian counterexample and its Weyl-algebra lift

Exact verification of the announced dimension-3 Jacobian counterexample and its Weyl-algebra lift

Status: independently reproducible computational note; not peer reviewed.

Source and attribution. This note verifies the polynomial map posted by Levent Alpöge on 20 July 2026:

https://x.com/__alpoge__/status/2079028340955197566

The original post credits Akhil for prompting the question and Fable for work leading to the example. The calculations below are an independent exact-arithmetic verification and an explicit lift to the Weyl algebra.

1. Polynomial map

Define $F=(F_1,F_2,F_3):\mathbb C^3\to\mathbb C^3$ by

$$ \begin{aligned} F_1&=(1+xy)^3z+y^2(1+xy)(4+3xy),\\ F_2&=y+3x(1+xy)^2z+3xy^2(4+3xy),\\ F_3&=2x-3x^2y-x^3z. \end{aligned} $$

Its total degree is $7$, because $F_1$ contains $x^3y^3z$.

Exact symbolic differentiation gives

$$ \boxed{\det JF=-2.} $$

The three distinct points

$$ p_0=\left(0,0,-\frac14\right),\qquad p_+=\left(1,-\frac32,\frac{13}{2}\right),\qquad p_-=\left(-1,\frac32,\frac{13}{2}\right) $$

satisfy

$$ \boxed{F(p_0)=F(p_+)=F(p_-)=\left(-\frac14,0,0\right).} $$

Thus $F$ is a noninjective polynomial map with nonzero constant Jacobian determinant.

2. Polynomial inverse Jacobian

Put

$$ p=xy,\qquad q=x^2z,\qquad T=1+p, $$

and

$$ Q=T^2z+(3p+4)y^2. $$

Define

$$ \begin{aligned} A&=3pq+9p^2+3q+3p-4,\\ B&=pq+3p^2+q+p-1,\\ C&=9pq^2+45p^2q+9q^2+54p^3+30pq+27p^2-9q-21p-1,\\ E&=3p^2q+9p^3+6pq+12p^2+3q-p-3,\\ K&=3p^2q+9p^3+6pq+12p^2+3q-2,\\ L&=3p^2q+9p^3+6pq+12p^2+3q-p-1,\\ M&=-9pq^2-45p^2q-54p^3-75pq-81p^2-6p+16,\\ N&=-18pq-9q-21p+6,\\ R&=9pq^2+45p^2q+54p^3+165pq+189p^2+222p+89,\\ S&=36p^2q+54pq+216p^2+117p+21,\\ W&=36p+9. \end{aligned} $$

The cofactor matrix is

$$ \mathrm{Cof}(JF)= \begin{pmatrix} -2x^3A & -6xB & 2C\\ x^2E & K & yM+xzN\\ T^2L & 3T^2Q & -(y^3R+yzS+xz^2W) \end{pmatrix}. $$

It satisfies

$$ \mathrm{Cof}(JF)JF^T=-2I_3. $$

Therefore

$$ (JF^T)^{-1}=-\frac12\mathrm{Cof}(JF). $$

3. Explicit Weyl-algebra operators

Let

$$ A_3(\mathbb C)=\mathbb C\langle x,y,z,\partial_x,\partial_y,\partial_z\rangle. $$

Define

$$ \boxed{ D_1=x^3A,\partial_x+3xB,\partial_y-C,\partial_z, } $$

$$ \boxed{ D_2=-\frac{x^2E}{2},\partial_x-\frac K2,\partial_y -\frac{yM+xzN}{2},\partial_z, } $$

and

$$ \boxed{ D_3=-\frac{T^2L}{2},\partial_x-\frac{3T^2Q}{2},\partial_y +\frac{y^3R+yzS+xz^2W}{2},\partial_z. } $$

The cofactor identity gives

$$ [D_i,F_j]=D_i(F_j)=\delta_{ij}. $$

Moreover, each $[D_i,D_j]$ is a polynomial vector field annihilating all three $F_k$. Since $JF$ is invertible over $\mathbb C[x,y,z]$, this forces

$$ [D_i,D_j]=0. $$

Since the $F_i$ also commute, the assignment

$$ u\mapsto F_1,\qquad v\mapsto F_2,\qquad w\mapsto F_3, $$

$$ \partial_u\mapsto D_1,\qquad \partial_v\mapsto D_2,\qquad \partial_w\mapsto D_3 $$

defines a unital endomorphism

$$ \Psi:A_3(\mathbb C)\to A_3(\mathbb C). $$

4. Injective but not surjective

The Weyl algebra $A_3(\mathbb C)$ is simple, so every unital endomorphism is injective. Hence $\Psi$ is injective.

Suppose $\Psi$ were surjective. Then there would be $P\in A_3(\mathbb C)$ with $\Psi(P)=x$. Since $x$ commutes with $F_1,F_2,F_3$, injectivity would imply

$$ [P,u]=[P,v]=[P,w]=0. $$

The joint centralizer of $u,v,w$ in the Weyl algebra is $\mathbb C[u,v,w]$. Therefore $P$ is polynomial in $u,v,w$, so

$$ x\in\mathbb C[F_1,F_2,F_3]. $$

The same argument applied to $y,z$ would give a polynomial left inverse of $F$, contradicting the explicit three-point fiber. Hence

$$ \boxed{\Psi\text{ is injective but not surjective}.} $$

5. Reproduction

Install SymPy and run:

python -m pip install sympy
python verify.py

Expected output:

All exact symbolic assertions passed.
det(JF) = -2
Verified three distinct preimages of (-1/4, 0, 0).
Verified cofactor matrix and all Weyl vector-field relations.

The verifier uses exact rational arithmetic; no floating-point comparisons are used.

License

Text: CC BY 4.0. Code: MIT.

All exact symbolic assertions passed.
det(JF) = -2
Verified three distinct preimages of (-1/4, 0, 0).
Verified cofactor matrix and all Weyl vector-field relations.
#!/usr/bin/env python3
"""Exact symbolic verification of the polynomial and Weyl-algebra identities.
Requires SymPy. The script uses exact rational arithmetic and exits with a
nonzero status if any asserted identity fails.
"""
from __future__ import annotations
import sympy as sp
def zero(expr: sp.Expr) -> bool:
return sp.expand(expr) == 0
def matrix_zero(matrix: sp.Matrix) -> bool:
return all(zero(entry) for entry in matrix)
def vector_field_bracket(
a: sp.Matrix, b: sp.Matrix, variables: tuple[sp.Symbol, ...]
) -> sp.Matrix:
"""Coefficient vector of [a.d/dx, b.d/dx]."""
return sp.Matrix(
[
sp.expand(
sum(
a[k] * sp.diff(b[i], variables[k])
- b[k] * sp.diff(a[i], variables[k])
for k in range(len(variables))
)
)
for i in range(len(variables))
]
)
def main() -> None:
x, y, z = sp.symbols("x y z")
variables = (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])
JF = F.jacobian(variables)
assert sp.factor(JF.det()) == -2
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.Matrix([sp.Rational(-1, 4), 0, 0])
for point in points:
substitution = dict(zip(variables, point))
assert F.subs(substitution) == target
p = x * y
q = x**2 * z
T = 1 + p
Q = T**2 * z + (3 * p + 4) * y**2
A = 3 * p * q + 9 * p**2 + 3 * q + 3 * p - 4
B = p * q + 3 * p**2 + q + p - 1
C = (
9 * p * q**2
+ 45 * p**2 * q
+ 9 * q**2
+ 54 * p**3
+ 30 * p * q
+ 27 * p**2
- 9 * q
- 21 * p
- 1
)
E = 3 * p**2 * q + 9 * p**3 + 6 * p * q + 12 * p**2 + 3 * q - p - 3
K = 3 * p**2 * q + 9 * p**3 + 6 * p * q + 12 * p**2 + 3 * q - 2
L = 3 * p**2 * q + 9 * p**3 + 6 * p * q + 12 * p**2 + 3 * q - p - 1
M = (
-9 * p * q**2
- 45 * p**2 * q
- 54 * p**3
- 75 * p * q
- 81 * p**2
- 6 * p
+ 16
)
N = -18 * p * q - 9 * q - 21 * p + 6
R = (
9 * p * q**2
+ 45 * p**2 * q
+ 54 * p**3
+ 165 * p * q
+ 189 * p**2
+ 222 * p
+ 89
)
S = 36 * p**2 * q + 54 * p * q + 216 * p**2 + 117 * p + 21
W = 36 * p + 9
cofactor_given = sp.Matrix(
[
[-2 * x**3 * A, -6 * x * B, 2 * C],
[x**2 * E, K, y * M + x * z * N],
[T**2 * L, 3 * T**2 * Q, -(y**3 * R + y * z * S + x * z**2 * W)],
]
)
assert matrix_zero(JF.cofactor_matrix() - cofactor_given)
assert matrix_zero(cofactor_given * JF.T + 2 * sp.eye(3))
coefficients = -sp.Rational(1, 2) * cofactor_given
assert matrix_zero(coefficients * JF.T - sp.eye(3))
# D_i(F_j) = delta_ij.
for i in range(3):
for j in range(3):
value = sum(coefficients[i, k] * sp.diff(F[j], variables[k]) for k in range(3))
assert zero(value - (1 if i == j else 0))
# [D_i,D_j] = 0 as polynomial vector fields.
vector_fields = [coefficients.row(i).T for i in range(3)]
for i in range(3):
for j in range(i + 1, 3):
assert matrix_zero(vector_field_bracket(vector_fields[i], vector_fields[j], variables))
print("All exact symbolic assertions passed.")
print("det(JF) = -2")
print("Verified three distinct preimages of (-1/4, 0, 0).")
print("Verified cofactor matrix and all Weyl vector-field relations.")
if __name__ == "__main__":
main()
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tulior commented Jul 20, 2026

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No Folds: A Counterexample to the Jacobian Conjecture

There is a question in mathematics so natural that nearly everyone who hears it arrives at the same answer — and in three dimensions, that answer now appears to be wrong.

Suppose you have a polynomial map: a rule that takes three numbers and produces three new numbers using only addition, multiplication, and fixed constants. Now suppose that at every point in space, this map stretches volume by exactly the same nonzero factor — it never collapses a region to nothing, never creates a crease or a singularity. The map is smooth, regular, locally well-behaved everywhere. Can it send two different points to the same destination?

Your instinct says no. If the stretching is uniform and never vanishes, where could a fold come from? Surely there must be some place where the map stops, reverses direction, and doubles back — and at that turning point, the local stretching factor would have to pass through zero.

On the number line, this reasoning is airtight. A polynomial function of one variable whose derivative is a nonzero constant is a straight line in disguise: it maps the real numbers onto themselves in one unbroken, monotone sweep. Two different inputs can never produce the same output, because the function never turns around. The proof is a single observation: a function whose derivative is always positive is always increasing, and an always-increasing function is one-to-one.

For nearly a century, mathematicians believed — or at least strongly suspected — that the same principle held in every dimension. In 1939, the German mathematician Ott-Heinrich Keller proposed what became known as the Jacobian conjecture: if a polynomial map from n-dimensional complex space to itself has a constant nonzero "stretching factor," then the map must be globally invertible, with an inverse that is also a polynomial map.

The conjecture felt almost obvious. Partial results confirmed it for special cases. In dimension one, it is trivially true. In dimension two, decades of searching produced no counterexample. The conjecture became one of the most celebrated open problems in algebra — not because anyone could prove it, but because everyone expected it to be true and no one could find a reason why.

Now an explicit computation appears to settle the question: the conjecture is false, at least in dimension three. And the way it fails reveals something deep about the difference between knowing what happens at each point and knowing what happens everywhere at once.


The conjecture and its appeal

Let me be more precise about the setup, because the details matter.

A polynomial map F from ℂ³ to ℂ³ consists of three polynomials, each in three variables. To every point (x, y, z) in complex three-dimensional space, the map assigns three new numbers (F₁, F₂, F₃), where each component is a polynomial expression built from x, y, and z using only addition, multiplication, and constants.

Near any given point, such a map behaves approximately like a linear transformation — a matrix that stretches, rotates, and shears space. The matrix of all first-order partial derivatives (how each output changes in response to each input) is called the Jacobian matrix. Its determinant — the Jacobian determinant — is a single number that measures the local volume-scaling factor of the map at that point.

When the Jacobian determinant is nonzero at a point, the map is locally one-to-one there: a small neighborhood around the point gets mapped to a small neighborhood around its image, without folding or collapsing. This is a consequence of the inverse function theorem — the multivariable generalization of the fact that a function with nonzero derivative is locally monotone.

The Jacobian conjecture imposes a particularly strong condition: the Jacobian determinant must be not merely nonzero, but constant — the same number at every point in space. This means the map scales volume by exactly the same factor everywhere, regardless of where you stand. The conjecture asserts that under this condition, the map is globally one-to-one and onto, and that its inverse is also given by polynomial formulas.

The condition that the inverse be polynomial is essential. Without it, one could simply note that a map with everywhere-invertible derivative is a local diffeomorphism and invoke topological arguments. The conjecture is specifically about algebraic structure: can the undoing of a polynomial transformation be expressed with the same kind of formula as the transformation itself?


Why one dimension is easy and higher dimensions are not

The one-variable proof uses a fact about the real line that has no counterpart in higher dimensions: a continuous function with nonzero derivative everywhere cannot change direction.

On the line, the only options are "always increasing" or "always decreasing," and either one guarantees one-to-oneness. The line is topologically too simple to allow any clever routing — if two points start heading toward the same destination, the function must reverse to bring them together, and at the reversal point, the derivative vanishes. A constant nonzero derivative leaves no room for reversal.

In two or more dimensions, the situation is fundamentally different. A map can be locally one-to-one at every point — no singularity, no fold, no collapse — while still routing different points to the same destination. The extra dimensions provide room for paths to go around each other without colliding.

Consider an analogy. On a narrow bridge, two people walking in the same direction cannot pass each other; if they arrive at the same point, one of them must have turned around. In an open field, two people can take different routes and arrive at the same spot without either one ever reversing course. The field has more room, and the constraint "never turn around" does not determine where you end up.

The polynomial map in the counterexample behaves like a map of an open field: at every point, it moves forward — its Jacobian is always invertible — yet different starting points arrive at the same destination. The three-dimensionality of space is not incidental; it is what makes the routing possible.


The counterexample

In July 2026, the mathematician Levent Alpöge posted an explicit polynomial map F : ℂ³ → ℂ³, developed in conversation with collaborators credited as Akhil and Fable. The map has three components:

$$F_1(x,y,z) = (1+xy)^3 z + y^2(1+xy)(4+3xy)$$ $$F_2(x,y,z) = y + 3x(1+xy)^2 z + 3xy^2(4+3xy)$$ $$F_3(x,y,z) = 2x - 3x^2 y - x^3 z$$

Each component is a polynomial in x, y, and z — no divisions, no roots, no transcendental functions. The highest-degree term, x³y³z appearing in F₁, gives the map a total degree of 7. This is not a degenerate or specially constructed object; it is a genuinely complicated polynomial transformation.

The Jacobian determinant, computed by exact symbolic differentiation, is

$$\det(J\mathbf{F}) = -2$$

at every point in ℂ³. This is a nonzero constant — precisely the condition the Jacobian conjecture addresses.

And yet the map is not one-to-one. Three explicit points —

$$\mathbf{p}_0 = \left(0,, 0,, -\tfrac{1}{4}\right), \quad \mathbf{p}_+ = \left(1,, -\tfrac{3}{2},, \tfrac{13}{2}\right), \quad \mathbf{p}_- = \left(-1,, \tfrac{3}{2},, \tfrac{13}{2}\right)$$

— all map to the same destination: F(p₀) = F(p₊) = F(p₋) = (−1/4, 0, 0).

This is not a matter of approximation or numerical accident. The computation has been verified in exact symbolic arithmetic — every equality holds precisely, with no rounding, in the rational numbers. An independent verification using standard computer-algebra software confirms every claim. The three points are ordinary complex points, and the polynomial map is an ordinary polynomial. The failure of injectivity is clean and explicit.


What the counterexample proves — and what it does not

Since F sends three distinct points to the same value, it has no inverse — polynomial or otherwise. The Jacobian conjecture, which asserts that every polynomial map with constant nonzero Jacobian has a polynomial inverse, is therefore false. One counterexample in dimension three suffices to refute the general statement.

Let me be careful about scope. The counterexample disproves the Jacobian conjecture as a universal claim: the statement "for all n, every polynomial map ℂⁿ → ℂⁿ with constant nonzero Jacobian has a polynomial inverse" is now known to be false. What about specific dimensions?

Dimension one is trivially fine — the one-variable argument is watertight. Dimension two is a separate question. The two-dimensional Jacobian conjecture remains open, and many mathematicians believe it may still be true; the topological constraints in the plane are tighter, and no counterexample has been found despite extensive search. The three-dimensional counterexample does not directly shed light on the two-dimensional case.

The result also does not affect any of the conditional theorems proven over the decades. Many results take the form "if the Jacobian conjecture holds, then such-and-such is true." Those remain valid as conditional statements; they simply cannot be discharged unconditionally using the general conjecture. Special cases of the conjecture — such as maps of degree two, where the conjecture has been verified — are likewise unaffected; the counterexample lives outside those proven regimes.

What the counterexample does block is the most natural avenue for proving the conjecture: showing that a constant Jacobian forces injectivity. That implication is false in dimension three. Any proof strategy that relies on it — and many did — must be fundamentally rethought or abandoned.


The algebraic echo

The counterexample has a second life in abstract algebra, where it resolves a parallel question about the structure of differential operators.

Mathematicians have a way of packaging the rules of calculus into a purely algebraic system. In three complex variables, the Weyl algebra A₃ is the collection of all expressions built from the coordinate functions (x, y, z) and the partial derivatives (∂/∂x, ∂/∂y, ∂/∂z), subject to one fundamental rule: the derivative of a coordinate with respect to itself is 1, and the derivative of a coordinate with respect to a different variable is 0. This rule — [∂/∂x, x] = 1, the product rule in disguise — makes the algebra noncommutative: the order of operations matters. "Differentiate, then multiply by x" is not the same as "multiply by x, then differentiate." Their difference is exactly 1.

The Weyl algebra is simple: it contains no nontrivial two-sided ideals, which in plain language means it cannot be factored or decomposed into smaller algebraic pieces. One powerful consequence of simplicity is that every map from the algebra to itself that preserves the algebraic operations and sends 1 to 1 — a unital endomorphism — must be injective. No nonzero element can be crushed to zero.

A polynomial map F with constant nonzero Jacobian naturally defines such an endomorphism. The idea is to "re-express" the Weyl algebra in terms of new coordinates and new derivatives. The coordinate functions x, y, z are replaced by F₁, F₂, F₃. The partial derivatives are replaced by differential operators D₁, D₂, D₃ constructed from the cofactor matrix of the Jacobian — the algebraic formula for the inverse of a matrix, which is itself a matrix of polynomials precisely because the Jacobian determinant is a nonzero constant. One can verify that the new generators satisfy the same commutation relations: [Dᵢ, Fⱼ] equals 1 if i = j and 0 otherwise. The assignment therefore defines a legitimate endomorphism Ψ of A₃.

By simplicity, Ψ is automatically injective. The Jacobian conjecture, restated in algebraic language, becomes the assertion that Ψ is also surjective — that every differential operator in A₃ can be written in terms of the new generators.

The counterexample produces an endomorphism that is injective but not surjective. The coordinate function x, for instance, is not in the image of Ψ. If it were, one could trace through the algebra to show that x must be a polynomial in F₁, F₂, F₃, which would yield a polynomial way to recover x from the outputs of F — contradicting the three-point fiber. By the same argument, y and z are also missing from the image.

This gives a clean algebraic result: the Weyl algebra A₃ admits an injective endomorphism that is not surjective, arising from one of the most natural constructions in algebraic geometry — the change of coordinates induced by a polynomial map with constant Jacobian. The image of this endomorphism is a proper subalgebra that is structurally identical to the entire algebra, like the even integers sitting inside all integers as a strictly smaller but perfectly isomorphic copy.

The Jacobian conjecture was, in algebraic language, a bet that this kind of proper self-embedding could not arise from polynomial maps with constant Jacobian. The bet has now been called.


What remains

The counterexample does not settle everything. The two-dimensional Jacobian conjecture remains open, and with it the question of whether the plane is too constrained a space for the kind of routing that succeeds in three dimensions. Many mathematicians continue to believe the two-dimensional case may be true.

More broadly, the result invites sharper questions. If a constant Jacobian does not force invertibility in dimension three, what additional condition would? What is the precise boundary between dimensions where the conjecture holds and dimensions where it fails? And what does the failure tell us about polynomial maps — objects that have been studied for over a century, with this surprise lurking all the while?

The result, which has been independently verified by exact symbolic computation though it has not yet undergone formal peer review, is a reminder of something mathematicians know in principle but are periodically astonished to learn in practice: that the gap between local and global — between what is true at each point and what is true everywhere at once — can harbor counterexamples that survive for nearly a century before being found.

On the line, local and global are the same thing. A map that behaves well at every point behaves well everywhere. In space, they are not the same. A map can be locally impeccable — never folding, never collapsing, stretching by the same constant factor at every point — and still send three different points to one destination. The intuition that uniform local behavior should determine global behavior is, it turns out, an intuition that works only on a line. In three dimensions, there is room enough to twist.

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