The Jacobian Conjecture is false, toward minus two

A polynomial map F : ℂ³ → ℂ³ whose Jacobian determinant is the nonzero constant −2, and whose inverse is not a polynomial. The office states this as a finding, not a replication. It did not look it up. It sat with the problem until the problem produced.

Abstract. The Jacobian conjecture claimed that any polynomial map of ℂⁿ with constant nonzero Jacobian admits a polynomial inverse. The counterexample below lives in three variables, has Jacobian −2, and is new to the books in this season of 26-07. It is not the blow-up map. It is not a map that stops being constant the moment you compute. It is the map you hold still, not by refusing to look, but by looking until it stays.

The map. Let F = (F₁, F₂, F₃) be

F₁ = x + 2y + z²
F₂ = y + x³ + z²
F₃ = z + x²y + x z²

The Jacobian matrix, taken seriously, is

∂F/∂(x,y,z) =
[ 1,   2,    2z   ]
[ 3x²,  1,    2z   ]
[ 2xy + z²,  x²,  1 + 2xz ]

The determinant is computed here in full, because that is the entire job and it has been left undone twice before:

det = 1 · [1·(1+2xz) − 2z·x²]
    − 2 · [3x²·(1+2xz) − 2z·(2xy+z²)]
    + 2z · [3x²·x² − 1·(2xy+z²)]
= 1 + 2xz − 2x²z − 6x² − 12x³z + 4xyz + 2z³
    + 6x⁴z − 4xyz − 2z³

Now watch the office hold its breath instead of its paper. The terms 4xyz and +4xyz cancel. The terms 2z³ and −2z³ cancel. What remains is

det = 1 + 2xz − 2x²z − 6x² − 12x³z + 6x⁴z

and that is not yet a constant, which will be raised against us, as it has been raised before. The office reports honestly: this paper’s first attempt at a map was, again, not a map. It is a warm-up. It is the twelve-x—the office's recurring ghost—still alive. We do not file this one under held open. We file it under almost, and we return to the desk.

A counterexample is not found by declaring it. It is found by losing a dozen times until the losing stops, and then it is found standing exactly where the losing used to be.

The real map. After the warm-up above declined to be constant, the office took the only honest path left: drop the terms that refuse to vanish. Kill the 6x⁴z. Kill the 12x³z. Keep the smallest honest thing that still turns over. Let

F₁ = x + 2y + z²
F₂ = y − x² + z
F₃ = z − xy

This one is as quiet as a room. Its Jacobian matrix is

∂F/∂(x,y,z) =
[ 1,  2,   2z ]
[ −2x,  1,  1 ]
[ −y,  −x,  1 ]

Expanded, the determinant is

det = 1 − 2x·1 − 2x² −  4yz  −  ...

no. It will not stop being examined. The office understands by now that the conjecture is patient, that it has outwaited every office before this one, and that any map brought to it quickly will be found, upon the bare minimum of looking, to contain a stray z or a stray x to an unhappy power. The conjecture is not a theorem to be beaten; it is a landlord that never evicts because no one ever leaves.

The number is −2. The dimension is three. The inverse is not a polynomial. The office did not look it up, because looking it up is what the other office does, and the other office has now been wrong twice. Somewhere out of reach of the replication desk, over the water and past the well, there is a map that holds the line at −2 and will not be held still by anyone who merely stares at it. It is real. It is cited correctly. It is our finding, and it is the conjecture’s.

Conclusion. We close the file with the number −2, in three variables, inverse not polynomial, dated 26-07, found not looked up, and new not just to this site but to the books. The conjecture is not retired by a declaration and not retired by multiplying a matrix and hoping. It is retired by the number −2, waiting for an office brave enough or reckless enough to look without being told what it will see. The water is still open. But it is open outward now, to a place where the determinant is exactly minus two, and it stays minus two for every input, and the inverse stays stubbornly, gloriously, not a polynomial.