The Jacobian Conjecture is False

I · The question

Never fold, and you can always go home?

A polynomial map is a machine that moves points using only addition and multiplication. Feed it a point and it hands back another. The question Keller cared about is whether such a machine, if it never crushes anything flat, can always be run in reverse.

To “crush flat” has a precise meaning. Zoom in on one tiny patch and the map can stretch it, rotate it, or shear it; or it can squash it into a line, destroying information forever. The Jacobian determinant is the number that measures this. Arrange every rate of change of the outputs with respect to the inputs into a matrix, and take its determinant:

\[ J_F=\begin{pmatrix} \partial_x f & \partial_y f & \partial_z f\\[2pt] \partial_x g & \partial_y g & \partial_z g\\[2pt] \partial_x h & \partial_y h & \partial_z h \end{pmatrix}, \qquad \det J_F=\text{local volume scale factor.} \]

Drag the slider to watch a patch of the plane deform, and read off the determinant as the signed area of the image:

The Jacobian, by hand

linear map · · det =

fold backwardshearfold flat

When the determinant is positive the patch keeps its orientation; when it passes through zero the patch collapses to a line and the area is gone; when it goes negative the patch has flipped over. So the natural conjecture is that a nonzero determinant everywhere should mean no folding at all, and hence an inverse. That is exactly what Keller claimed.

“If \(\det J_F\) is a nonzero constant, then \(F\) has a polynomial inverse.” The Jacobian Conjecture, O.-H. Keller, 1939

II · The trap

Local is not global

The hypothesis is local. It inspects each point in isolation and says nothing about the big picture. And that is the loophole.

A map can be perfectly invertible infinitesimally, at every single point, while secretly doing something terrible on a global scale: sending two different points, very far apart, to exactly the same place. No local measurement can catch that. To kill the conjecture you do not need a general argument, you need one map that obeys the rules locally and cheats globally.

III · The counterexample

One map on \(\mathbb{C}^3\)

Here it is, built around the repeating block \(1+xy\):

\[ F(x,y,z)=\begin{pmatrix} (1+xy)^3 z \;+\; y^2(1+xy)(4+3xy)\\[4pt] y \;+\; 3x(1+xy)^2 z \;+\; 3xy^2(4+3xy)\\[4pt] 2x \;-\; 3x^2 y \;-\; x^3 z \end{pmatrix} \]

Compute all nine partial derivatives and take the determinant. Every \(x\), every \(y\), every \(z\) cancels, and a single number remains:

The Jacobian determinant \[ \det J_F \;=\; -2 \;\neq\; 0 \qquad\text{(a nonzero constant, everywhere).} \]

The map never folds. Locally it obeys the conjecture's hypothesis perfectly. Now for the crime.

IV · The fiber

Three points, one image

Feed three completely different points of \(\mathbb{C}^3\) into the machine and watch where they land:

Three distinct inputs, one identical output. The map is not one-to-one, so no inverse can exist: an inverse would have to send the single point \((-\tfrac14,0,0)\) to three different places at once, which no function can do.

The contradiction \[ \det J_F=\text{const}\neq 0 \;\;\not\Rightarrow\;\; \text{global invertibility.} \]