Jacobian Conjecture

A map is a rule that sends each point to a new point. For example, the rule $(x, y) \to (x + y^{2}, y)$ sends the point $(1, 2)$ to $(5, 2)$ . A polynomial is a formula that only adds and multiplies numbers and variables, so this is a polynomial map.

A map is reversible if you can always undo it. The map above is reversible: the rule $(u, v) \to (u - v^{2}, v)$ sends $(5, 2)$ back to $(1, 2)$ . The map $(x, y) \to (x^{2}, y)$ is not reversible. It sends both $(1, 5)$ and $(- 1, 5)$ to $(1, 5)$ , so nothing can tell you which one you started from.

A map stretches or squeezes small regions. The Jacobian determinant is a number, calculated from the rule at each point, that says how much the area of a tiny region gets multiplied. If it is 2 at a point, tiny shapes there double in area. If it is 0, tiny shapes there are flattened. The map $(x^{2}, y)$ has determinant $2 x$ , so it flattens along the line $x = 0$ . That is where it folds over and the two sides collide.

Map	Determinant	Can you undo it
$(x + y^{2}, y)$	$1$	Yes
$(x^{2}, y)$	$2 x$	No: $x$ and $- x$ collide
$(e^{x} \cos y, e^{x} \sin y)$	$e^{2 x}$	No: it repeats

If the determinant is not zero at a point, then the map has no collisions close to that point. But points far apart can still collide. The map $(e^{x} \cos y, e^{x} \sin y)$ never flattens anything, yet it gives the same output each time $y$ grows by $2 π$ . That map is not a polynomial map. It uses $e$ , a fixed number about 2.718, and cosine and sine, which repeat as a point moves around a circle.

The Jacobian conjecture said that polynomial maps cannot do this. If a polynomial map has the same nonzero determinant at every point, then it is reversible, and its undo rule is also a polynomial map. With one variable this is easy to see, because the map must be $c x + d$ , for fixed numbers $c$ and $d$ . With more variables the question stayed open.

The conjecture is false with 3 or more variables. Levent Alpöge posted this map and credited the AI model Claude Fable 5. The three lines give the three output numbers:

\begin{matrix} F_{1} & = {(1 + x y)}^{3} z + y^{2} (1 + x y) (4 + 3 x y) \\ F_{2} & = y + 3 x {(1 + x y)}^{2} z + 3 x y^{2} (4 + 3 x y) \\ F_{3} & = 2 x - 3 x^{2} y - x^{3} z \end{matrix}

Its determinant is $- 2$ at every point, so it never flattens anything. Yet three different points have the same output:

\begin{matrix} F (0, 0, - \frac{1}{4}) & = (- \frac{1}{4}, 0, 0) \\ F (1, - \frac{3}{2}, \frac{13}{2}) & = (- \frac{1}{4}, 0, 0) \\ F (- 1, \frac{3}{2}, \frac{13}{2}) & = (- \frac{1}{4}, 0, 0) \end{matrix}

You can check the first line by hand. With $x = 0$ and $y = 0$ , the term $1 + x y$ is 1, so $F_{1} = z = - 1 / 4$ . The other two formulas give 0. The second and third lines take more arithmetic, but they are also exact.

Three inputs with one output cannot be undone, so the conjecture fails. To get an example with 4 variables, add a fourth coordinate that the map leaves unchanged. The same step works for any larger number. The case of exactly 2 variables remains open.