A digestion of the Jacobian conjecture counterexample

21 July, 2026 in math.AG | Tags: Jacobian conjecture, polynomials | by Terence Tao

The notorious Jacobian conjecture can be formulated concretely over the complex numbers as follows.

Conjecture 1 (Jacobian Conjecture) Let ${F:{\bf C}^n \rightarrow {\bf C}^n}$ be a polynomial map in ${n}$ complex variables, whose Jacobian ${\mathrm{det} DF}$ is a non-zero constant. Then ${F}$ is invertible (with polynomial inverse).

The condition that the Jacobian ${\mathrm{det} DF}$ is non-zero is equivalent to ${F}$ being locally invertible. (The implication of local invertibility from non-vanishing Jacobian follows from the inverse function theorem; the converse implication can be derived from the Weierstrass preparation theorem, but is omitted here.) Also, from the fundamental theorem of algebra, once the Jacobian polynomial ${\mathrm{det} DF}$ is non-zero, it must be constant. So the hypothesis “Jacobian ${\mathrm{det} DF}$ is a non-zero constant” can be replaced with “ ${F}$ is locally invertible”. So the Jacobian conjecture can be viewed as an assertion that local invertibility implies global invertibility. The complex numbers can be easily replaced with other fields of characteristic zero by the Lefschetz principle, but I prefer to work in the concrete setting of the complex numbers.

It was recently shown (using the Fable AI) that the conjecture is false in three dimensions (and thus in higher dimensions as well):

Theorem 2 (Counterexample to conjecture) There exists a polynomial ${F : {\bf C}^3 \rightarrow {\bf C}^3}$ which has non-zero constant Jacobian, but is not invertible.

The conjecture remains open in two dimensions, and is easy to establish in one dimension.

The example can be stated completely explicitly: one can take

$\displaystyle F(z_1,z_2,z_3) = \Big((1+z_1 z_2)^3 z_3 + z_2^2 (1+z_1z_2) (4+3z_1z_2), \ \ \ \ \ (1)$

$\displaystyle z_2 + 3 z_1 (1+z_1z_2)^2 z_3 + 3 z_1 z_2^2 (4+3z_1z_2),$

$\displaystyle 2 z_1 - 3 z_1^2 z_2 - z_1^3 z_3\Big)$

and one can verify by a brief calculation that

$\displaystyle \mathrm{det} DF = -2$

and

$\displaystyle F(0,0,-1/4) = F(1,-3/2, 13/2) = F(-1,3/2,13/2)$

$\displaystyle = (-1/4,0,0).$

While this is an extremely quick verification, the construction presented in this fashion appears like a massive miracle. The polynomial ${F}$ has degree seven, so a priori the Jacobian ${\mathrm{det} DF}$ ought to be a polynomial in three variables of degree as large as ${3 \times 6 = 18}$ , so the fact that all non-constant coefficients of this polynomial vanish looks like a massive cancellation involving ${\binom{18+3}{3}-1 = 1299}$ equations, which is much larger than the ${\binom{7+3}{3} = 120}$ degrees of freedom for a generic degree seven polynomial of three variables. So finding such a polynomial looks highly unlikely to be located by brute force.

The example has since been retroactively explained in more geometric terms. As a “digestion” exercise to myself, I sought to write this explanation with relatively little use of algebraic geometry, in a manner that minimizes the amount of “miracles” required, although there are still a few places were some remarkable phenomena occur.

It is convenient to use the local injectivity formulation, and to generalize the domain ${{\bf C}^3}$ to an equivalent affine variety. Namely, we will show

Theorem 3 (Counterexample, reformulated) There exists an affine variety ${X \subset {\bf C}^5}$ that is isomorphic to ${{\bf C}^3}$ by polynomial changes of variable, and a polynomial map ${F : X \rightarrow {\bf C}^3}$ which is locally injective, but not globally injective.

Clearly one can get from Theorem 3 to Theorem 2 by composing with the isomorphism ${X \cong {\bf C}^3}$ and using the previously mentioned fact that local injectivity implies non-zero constant Jacobian. Our objective is now to find data ${X}$ , ${F : X \rightarrow {\bf C}^3}$ that obeys three separate properties:

(a) ${F}$ is locally injective on ${X}$ .
(b) ${F}$ is not globally injective on ${X}$ .
(c) ${X}$ is isomorphic to ${{\bf C}^3}$ by polynomial changes of variable.

It turns out that ${F}$ and ${X}$ can be built out of the operation of multiplication of low degree polynomials. Namely, consider the following three simple affine spaces:

The space ${\mathrm{Sym}^1({\bf C}^2)}$ of linear homogeneous polynomials ${L(z,w) = az + bw}$ of two complex variables ${z,w}$ .
The space ${\mathrm{Sym}^2({\bf C}^2)}$ of quadratic homogeneous polynomials ${Q(z,w) = cz^2 + dzw + ew^2}$ of two complex variables ${z,w}$ .
The space ${\mathrm{Sym}^3({\bf C}^2)}$ of cubic homogeneous polynomials ${C(z,w) = fz^3 + gz^2w + hzw^2 + iw^3}$ of two complex variables ${z,w}$ .

${\mathrm{Sym}^k(V)}$

${k^{th}}$

symmetric power

${V}$

${{\bf C}^2, {\bf C}^3, {\bf C}^4}$

${F : \mathrm{Sym}^1({\bf C}^2) \times \mathrm{Sym}^2({\bf C}^2) \rightarrow \mathrm{Sym}^3({\bf C}^2)}$

${(L,Q)}$

${L}$

${Q}$

$\displaystyle F(L,Q) := LQ.$

(Right now, the domain and range of this map ${F}$ is larger dimensional than the target of three; we will cut the dimensions down to three as the argument progresses.)

The map ${F}$ , essentially a map from ${{\bf C}^5}$ to ${{\bf C}^4}$ , is clearly polynomial; it is given explicitly in coordinates as

$\displaystyle F( (a,b), (c,d,e) ) = (ac, ad + bc, ae + bd, be). \ \ \ \ \ (2)$

The map ${F}$ also enjoys two basic (and commuting) symmetries:

If one applies a scaling ${(L,Q) \mapsto (\lambda_1 L, \lambda_2 Q)}$ for some non-zero complex numbers ${\lambda_1, \lambda_2}$ , then the product ${LQ}$ is scaled by ${C \mapsto \lambda_1 \lambda_2 C}$ : ${F( \lambda_1 L, \lambda_2 Q) = \lambda_1 \lambda_2 F(L,Q)}$ .
If one applies a change of variables ${(L, Q) \mapsto (L \circ T, Q \circ T)}$ for some invertible linear transformation ${T \in \mathrm{SL}_2({\bf C})}$ , then the product ${LQ}$ is transformed by ${C \mapsto C \circ T}$ : ${F(L \circ T, Q \circ T) = F(L,Q) \circ T}$ .

${{\bf C}^\times \times {\bf C}^\times \times \mathrm{SL}_2({\bf C})}$

The five-dimensional domain ${\mathrm{Sym}^1({\bf C}^2) \times \mathrm{Sym}^2({\bf C}^2)}$ is of course larger than the four-dimensional range ${\mathrm{Sym}^3({\bf C}^2)}$ , so the map ${F}$ clearly cannot be injective. This can already be seen from the scaling symmetry, as the specific scalings

$\displaystyle (L, Q) \mapsto (\lambda L, \lambda^{-1} Q) \ \ \ \ \ (3)$

for ${\lambda \in {\bf C}^\times}$ modify the linear and quadratic polynomials ${L,Q}$ but not their product ${C = LQ}$ . But even if one quotients out by this symmetry (3) to cut the dimension of the domain down to four, the map ${F}$ is still not injective for the following basic reason. A generically chosen cubic polynomial ${C}$ will split into the product ${C = L_1 L_2 L_3}$ of three independent linear polynomials. Then there are three pairs

$\displaystyle (L_1, L_2 L_3), (L_2, L_1 L_3), (L_3, L_1 L_2) \ \ \ \ \ (4)$

which all map to the same cubic polynomial

$\displaystyle F(L_1, L_2 L_3) = F(L_2, L_1 L_3) = F(L_3, L_1 L_2) = C$

under the multiplication map ${F}$ , but are not related to each other by scaling symmetry (3). Thus, we see that even after quotienting out by the scaling symmetry (3), the multiplication map ${F}$ is generically non-injective in a three-to-one fashion. Thus we already have achieved something resembling goal (b)!

It will be convenient to “spend” the scaling symmetry ${(L, Q) \mapsto (\lambda L, \lambda^{-1} Q)}$ to obtain a useful normalization. If ${L(z,w) = az+bw}$ is a linear polynomial and ${Q(z,w)}$ is a quadratic polynomial, the resultant ${\mathrm{Res}(L,Q)}$ can be defined by the determinant

$\displaystyle \mathrm{Res}(L,Q) = \begin{vmatrix} a & b & 0 \\ 0 & a & b \\ c & d & e \end{vmatrix} = a^2 e - abd + c b^2. \ \ \ \ \ (5)$

If we have a factoring

$\displaystyle L(z,w) = a (z - \alpha w), \quad Q(z,w) = c (z - \beta_1 w)(z - \beta_2 w)$

then the resultant can also be described as

$\displaystyle \mathrm{Res}(L,Q) = a^2 c (\alpha - \beta_1) (\alpha - \beta_2).$

Thus the resultant measures whether the linear polynomial ${L}$ and the quadratic polynomial ${Q}$ share a common root. A fundamental fact about resultants is that they are ${SL_2}$ \-invariant: for any ${T \in SL_2({\bf C})}$ , we have

$\displaystyle \mathrm{Res}(L \circ T, Q \circ T) = \mathrm{Res}(L,Q).$

One way to see this is to check it first for translations ${(z,w) \mapsto (z, w + h z)}$ (which translate the roots ${\alpha,\beta_1,\beta_2}$ by ${h}$ while leaving ${a,c}$ unchanged) and for inversions ${(z,w) \mapsto (w,z)}$ (which map ${\alpha,\beta_1,\beta_2}$ to ${1/\alpha, 1/\beta_1, 1/\beta_2}$ while mapping ${a,c}$ to ${-a\alpha}$ and ${c\beta_1 \beta_2}$ respectively), and then noting that these transformations generate all of ${SL_2({\bf C})}$ . They also interact very nicely with scaling:

$\displaystyle \mathrm{Res}(\lambda_1 L, \lambda_2 Q) = \lambda_1^2 \lambda_2 \mathrm{Res}(L,Q).$

In particular, the scaling symmetry (3) multiplies ${\mathrm{Res}(L,Q)}$ by ${\lambda}$ :

$\displaystyle \mathrm{Res}(\lambda L, \lambda^{-1} Q) = \lambda \mathrm{Res}(L,Q). \ \ \ \ \ (6)$

Thus, we can (generically) normalize away this scaling symmetry by imposing the condition

$\displaystyle \mathrm{Res}(L,Q) = 1. \ \ \ \ \ (7)$

We now have a restricted multiplication map (which by abuse of notation we will continue to call ${F}$ ) from the four-dimensional variety

$\displaystyle \{ (L,Q) \in \mathrm{Sym}^1({\bf C}^2) \times \mathrm{Sym}^2({\bf C}^2) : \mathrm{Res}(L,Q) = 1\} \ \ \ \ \ (8)$

to the four-dimensional space ${\mathrm{Sym}^3({\bf C}^2)}$ . This map ${F}$ is still not globally injective, as we can take the three pairs in (4) from before and apply the scaling (3) separately to each of the three pairs to obtain the normalization (7). So we have kept property (b). Furthermore, this map retains the ${SL_2}$ \-equivariance (and also one remaining scaling symmetry, though we will not make much further use of that symmetry).

But we now also have property (a)! Suppose we want to show the local injectivity of ${F}$ in the neighborhood of a pair ${(L,Q)}$ with ${\mathrm{Res}(L,Q) = 1}$ . As the resultant is non-vanishing, the root ${\alpha}$ of ${L}$ (which exists in the Riemann sphere, or projective line if you prefer) is distinct from the two roots ${\beta_1, \beta_2}$ of ${Q}$ (though the latter two roots could be equal to each other). Applying the ${SL_2}$ action (which performs Möbius transforms on the roots), one can assume without loss of generality that ${\alpha}$ is the point at infinity (or equivalently ${a=0}$ ), thus ${L(z,w) = b w}$ for some complex number ${b}$ and ${Q(z,w) = c (z - \beta_1 w)(z - \beta_2 w)}$ for some complex numbers ${c, \beta_1, \beta_2}$ , with the resultant condition (7) simplifies to ${cb^2 = 1}$ (so in particular ${c,b}$ are also non-zero). It is then clear that if one perturbs ${L}$ and ${Q}$ by a small amount (say, modifying each coefficient by ${O(\varepsilon)}$ ), then the root ${\alpha=\infty}$ of ${L}$ will perturb to something large ( ${\gg 1/\varepsilon}$ ), while the roots ${\beta_1,\beta_2}$ of ${Q}$ stay bounded. Thus, just from knowledge of the product ${F(L,Q)}$ , one can reconstruct which of the three roots of this cubic polynomial will be the perturbed root of ${L}$ , and which two will be the perturbed roots of ${Q}$ ; from this and (6), (7) we can also reconstruct the leading coefficient ${c}$ of ${Q}$ , and this completely determines both ${L}$ and ${Q}$ . This establishes the local injectivity property (a). (In fact it is étale, but we will not need the machinery of étale maps here.)

Unfortunately, (the four-dimensional analogue of) condition (c) fails: the quadric hypersurface (8) is not isomorphic to the affine space ${{\bf C}^4}$ . But we can try to get around this by passing to a three-dimensional slice. Let ${V}$ be some three-dimensional affine plane of ${\mathrm{Sym}^3({\bf C}^2)}$ (which we will take to avoid the origin for technical reasons), then we can restrict ${F}$ as a map from the set

$\displaystyle \{ (L,Q) \in \mathrm{Sym}^1({\bf C}^2) \times \mathrm{Sym}^2({\bf C}^2) : \mathrm{Res}(L,Q) = 1; \ \ \ \ \ (9)$

$\displaystyle F(L,Q) \in V\}$

to ${V}$ . The latter is clearly identifiable (by linear changes of coordinate) to ${{\bf C}^3}$ . As ${F}$ was already locally invertible, it remains locally invertible under restriction; and because generic cubic polynomials ${C}$ had three preimages under ${F}$ in (8), this continues to be the case after restricting to (9) (unless ${V}$ was somehow so degenerate that it had no generic elements, but this turns out to be impossible). So we have retained properties (a) and (b). The miracle is that, with a good choice of ${V}$ , we can also obtain (c) and obtain the desired counterexample to the Jacobian conjecture: despite appearances, the variety (9) is in fact equivalent to the affine space ${{\bf C}^3}$ by polynomial changes of variable!

Let’s see how. The affine hyperplanes in ${\mathrm{Sym}^3({\bf C}^2)}$ avoiding the origin are parameterized by the dual space of ${\mathrm{Sym}^3({\bf C}^2)}$ avoiding the origin, which one can think of as the non-zero third order homogeneous differential operators ${D = j \partial_z^3 + k \partial_z^2 \partial_w + l \partial_z \partial_w^2 + m \partial_w^3}$ in two variables. Indeed, every such operator ${D}$ generates an affine hyperplane ${\{ C \in \mathrm{Sym}^3({\bf C}^2) : D(C) = 1\}}$ that avoids the origin, and conversely by duality every affine hyperplane avoiding the origin arises in this form uniquely. Just as the cubic polynomials in ${\mathrm{Sym}^3({\bf C}^2)}$ can be factored into three linear polynomials, the differential operators in the dual space ${\mathrm{Sym}^3({\bf C}^2)^*}$ can also be factored into three linear differential operators, e.g.,

$\displaystyle D = j (\partial_z - \gamma_1 \partial_w) (\partial_z - \gamma_2 \partial_w) (\partial_z - \gamma_3 \partial_w)$

in the case that ${j}$ is non-zero. The ${SL_2}$ action moves the roots ${\gamma_1,\gamma_2,\gamma_3}$ around the Riemann sphere by Möbius transformations. As these transformations are ${3}$ \-transitive, the actual selection of such roots is not too important (and the scaling symmetry similarly makes the choice of leading coefficient ${j}$ unimportant); the only thing to keep track of is whether the roots repeat. Up to the symmetries, there are in fact just three different equivalence classes of differential operator ${D}$ (and thus of affine hyperplane ${V}$ ) to consider:

Operators where the three roots ${\gamma_1,\gamma_2,\gamma_3}$ are all distinct, thus ${D = D_1 D_2 D_3}$ for independent first-order operators ${D_1,D_2,D_3}$ .
Operators where two roots coincide and one is distinct, thus ${D = D_1^2 D_2}$ for independent first-order operators ${D_1,D_2}$ .
Operators where all three roots coincide, thus ${D = D_1^3}$ for some first-order operator ${D_1}$ .

It turns out that the affine miracle for (9) occurs precisely in the second case, when ${D}$ has two identical roots. I do not have a completely satisfactory geometric explanation for this miracle, but one can verify it by the following coordinate computation.

By applying the ${SL_2}$ action, we can normalize so that ${D = \partial_z^2 \partial_w}$ , thus ${V}$ is now the affine hyperplane of cubic polynomials ${C(z,w) = f z^3 + g z^2 w + h z w^2 + i w^3}$ with ${g=1}$ . Using (2) and (5), the variety (9) can now be described explicitly in coordinates as

$\displaystyle \{ (a,b,c,d,e) \in {\bf C}^5 : a^2 e - abd + cb^2 = 1; ad + bc = 1 \}. \ \ \ \ \ (10)$

At first glance this seems to be a generic-looking variety cut out by a cubic equation and a quadratic equation – hardly a candidate to be affine! But observe that if ${a}$ is non-zero, then the second equation ${ad+bc = 1}$ can be solved for ${d}$ ,

$\displaystyle d = \frac{1 - bc}{a} \ \ \ \ \ (11)$

and the first equation ${a^2 e - abd + cb^2 = 1}$ can be solved for ${e}$ ,

$\displaystyle e = \frac{1 + abd - cb^2}{a^2}. \ \ \ \ \ (12)$

Putting these two equations together, we see that as long as one removes the case ${a=0}$ , the quintuple ${(a,b,c,d,e)}$ is uniquely determined by ${(a,b,c)}$ by a change of variables which is Laurent in ${a}$ and polynomial in ${b,c}$ . Thus we have a nice birational equivalence

$\displaystyle \{ (a,b,c,d,e) \in {\bf C}^5 : a^2 e - abd + cb^2 = 1; ad + bc = 1; a \neq 0 \}$

$\displaystyle \cong \{ (a,b,c) \in {\bf C}^3 : a \neq 0 \}.$

Thus we have already almost established property (c): the variety (9) becomes birationally equivalent to ${{\bf C}^3}$ after cutting out the ${a=0}$ subvariety. In particular, for each fixed non-zero value ${a_0}$ of ${a}$ , the corresponding fiber

$\displaystyle \{ (a,b,c,d,e) \in {\bf C}^5 : a^2 e - abd + cb^2 = 1; ad + bc = 1; a = a_0 \}$

of (10) is equivalent to ${{\bf C}^2}$ by polynomial changes of variable, since we can reconstruct ${d,e}$ from the coordinates ${b,c}$ by the polynomial formulae

$\displaystyle d = \frac{1-bc}{a_0}; \quad e = \frac{1 + a_0 d b - c b^2}{a_0^2}.$

So we just need to glue back in the ${a=0}$ fiber. Indeed, from (10) we see that the fiber at ${0}$ is just

$\displaystyle \{ (0,b,c,d,e) \in {\bf C}^5 : cb^2 = 1; bc = 1 \}.$

Now we observe a key miracle: the cubic equation ${cb^2 = 1}$ and quadratic equation ${bc = 1}$ have a unique affine solution ${b=c=1}$ (as opposed to the six possible solutions that Bezout’s theorem might suggest – the other five solutions live on the line at infinity). So the fiber here is also affine:

$\displaystyle \{ (0,1,1,d,e) \in {\bf C}^5 : d, e \in {\bf C} \}.$

This is extremely encouraging for the purposes of establishing property (c), as it strongly suggests that the variety (10) has the structure of an ${{\bf C}^2}$ \-bundle over ${{\bf C}^1}$ , which is already extremely close to being isomorphic to the affine space ${{\bf C}^3}$ . The main remaining task is to make sure that nothing singular happens in the limit ${a \rightarrow 0}$ , and that a global polynomial coordinate chart for (10) that covers both the ${a \neq 0}$ and ${a = 0}$ fibers can be constructed.

The standard way to proceed here is to manipulate various tangent spaces using the modern machinery of algebraic geometry and commutative algebra, but given my own background, I prefer to adopt the language of analysis, and in particular big-O notation (in place of the ideals used in algebraic geometry), in order to investigate the limit ${a \rightarrow 0}$ by hand. On the variety (10), let us use ${O(X)}$ to denote any multiple of ${X}$ by a polynomial expression in ${a,b,c,d,e}$ . Thus, for instance, the equation ${ad + bc = 1}$ implies that

$\displaystyle bc = 1 + O(a) \ \ \ \ \ (13)$

while the equation ${a^2 e - abd + cb^2 = 1}$ implies that

$\displaystyle cb^2 = 1 + O(a) \ \ \ \ \ (14)$

as well as the more refined estimate

$\displaystyle cb^2 = 1 + abd + O(a^2). \ \ \ \ \ (15)$

In the ${a=0}$ case we could conclude that ${b=c=1}$ . Now we perturb this observation. Multiplying (13) by ${b}$ we have ${b^2 c = b + O(a)}$ , which on substitution into (14) gives ${b = 1 + O(a)}$ ; substituting this back into either (13) or (14) also gives ${c = 1 + O(a)}$ .

We can get some more precise asymptotics by also taking advantage of (15). Substituting ${ad+bc=1}$ into (15), we obtain after some algebra

$\displaystyle 2 cb^2 = 1 + b + O(a^2).$

So if we write ${b = 1+O(a)}$ more explicitly as ${b = 1 + a y}$ , then we have

$\displaystyle 2 c (1 + 2ay + O(a^2)) = 2 + ay + O(a^2)$

and thus

$\displaystyle c = 1 - \frac{3}{2} ay + O(a^2). \ \ \ \ \ (16)$

Substituting this back into (11) gives an asymptotic for ${d}$ :

$\displaystyle d = \frac{1 - bc}{a}$

$\displaystyle = \frac{1 - (1 + ay) (1 - \frac{3}{2} ay + O(a^2))}{a}$

$\displaystyle =\frac{1}{2} y + O(a).$

Finally, one can insert these estimates into (12), although one only gets a trivial bound in this case:

$\displaystyle e = \frac{1 + abd - cb^2}{a^2}$

$\displaystyle = \frac{1 + a (1+O(a)) (\frac{1}{2} y + O(a)) - (1 - \frac{3}{2} ay + O(a^2)) (1 + ay)^2}{a^2}$

$\displaystyle = O(1).$

Expanding the ${O(a^2)}$ error term in (16) as ${a^2 z}$ , and doing a little more algebra, we thus have a polynomial change of variables

$\displaystyle a = a$

$\displaystyle b = 1 + ay$

$\displaystyle c = 1 - \frac{3}{2} ay + a^2 z$

$\displaystyle d = \frac{1-bc}{a} = \frac{1}{2} y - az + \frac{3}{2} a y^2 - a^2 yz$

$\displaystyle e = \frac{1 + abd - cb^2}{a^2} = -2z + 4y^2 - 4ayz + 3ay^3 - 2a^2 y^2 z$

which completely parameterizes the variety (10) by polynomial combinations of three coordinates ${a,y,z}$ . This already gives (a) and thus completes the proof of Theorem 3.

The previous computations, when expanded out, also gives polynomial inverse maps:

$\displaystyle a = a$

$\displaystyle y = 2bd - ae$

$\displaystyle z = 2d^2 + ce + 6bd^2 + 3bce - \frac{9}{2} e$

The map from ${(a,y,z)}$ to the ${(f,h,i)}$ coefficients of ${F(L,Q)}$ (dropping the ${g}$ coefficient which is constrained to equal ${1}$ ), we obtain a polynomial map

$\displaystyle (a,y,z) \mapsto (G_1(a,y,z), G_2(a,y,z), G_3(a,y,z))$

with

$\displaystyle \begin{array}{rl} G_1(a,y,z) &= a - \frac{3}{2} a^2 y + a^3 z \\ G_2(a,y,z) &= \frac{1}{2} y - 3az + 6ay^2 - 6a^2 yz + \frac{9}{2} a^2 y^3 - 3a^3 y^2 z \\ G_3(a,y,z) &= -2z + 4y^2 - 6ayz + 7ay^3 - 6a^2 y^2 z + 3a^2 y^4 \\ & \quad - 2 a^3 y^3 z \end{array}$

which theory predicts to have a constant Jacobian, and indeed one can calculate that the Jacobian is ${-1}$ . This is essentially the original example up to trivial changes of variable; indeed, one can check that the map

$\displaystyle (a, y, -2z) \mapsto (G_3(a,y,z), 2G_2(a,y,z), 2G_1(a,y,z))$

is exactly the map ${F}$ given in (1).

AI disclosure: I used an AI chatbot to discuss various aspects of this problem and to confirm several of the calculations made here.

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26 comments

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21 July, 2026 at 2:39 pm

Anonymous

Beautiful explanation. Two very small corrections:

1. At one point, you write “in coordinates it is given explicitly in coordinates as”; you probably want this to be either “in coordinates it is” or “it is given explicitly in coordinates as”. 2. You write “the map F is still not surjective…” but I think you meant injective.

\[Corrected, thanks – T.\]

21 July, 2026 at 5:24 pm

Jim Balter

Likewise, “Recently, it was recently shown” is redundant.

\[Corrected, thanks – T.\]

21 July, 2026 at 4:16 pm

swiftlytacod41af927bc

Is there some representation-theoretic or moduli-theoretic intuition — perhaps something about the interaction of the SL₂-orbit stratification of the dual space with the discriminant locus — that would allow one to foresee ahead of time that the double-root stratum (and only that stratum) gives rise to an affine slice, instead of needing to directly check that ? Would one expect this to be a low-dimensional curiosity, or is there some general principle for when a GIT-type quotient construction like this results in affine varieties?”

21 July, 2026 at 4:42 pm

Terence Tao

In my chatbot discussion I tried to pursue this line of thinking. As far as I could tell, there does not seem to be any plausible known set of computable invariants (including those that somehow take advantage of equivariant structure) that would force affine-ness, though perhaps some weaker “pseudo-affine” structure might be detectable by such a soft approach. But perhaps there are experts in algebraic geometry who will be able to give a more authoritative answer.

21 July, 2026 at 4:22 pm

Anonymous

Also, from the fundamental theorem of algebra, once the Jacobian polynomial det(DF) is non-zero, it must be constant.

I might be missing something obvious—I see how the above follows via the Nullstellensatz, but not via the fundamental theorem of algebra per se (save insofar as the latter follows from the former).

21 July, 2026 at 4:39 pm

Terence Tao

The fundamental theorem of algebra tells us that any one-dimensional polynomial is either constant, or has at least one zero (over the complex numbers). By induction, the same is true in higher dimensions; so if a polynomial $P : {\mathbf C}^n \to {\mathbf C}$ has no zeroes, it must be constant.

21 July, 2026 at 4:30 pm

Ariel

This feels like a too beautiful counterexample, almost too beautiful to be discovered by a human. Polynomial multiplication is the most natural algebraic object that is locally injective but globally non-injective.

21 July, 2026 at 5:06 pm

Sam Hopkins

There is also some discussion of this here: <https://sbseminar.wordpress.com/2026/07/20/the-new-counterexample-to-the-jacobian-conjecture/>

22 July, 2026 at 2:46 am

Will Sawin

Let me state in the language of this post something from that thread: Given the equations $a^2e-abd+cb^2=1$ and $ad+bc=1$ , the key step seems to be to subtract $b$ times the second equation from the first to cancel the $cb^2$ term, obtaining $a( ae-2bd)=1-b$ , which motivates using coordinates $x=a$ and $y=2bd-ae$ since $b$ can be expressed as a polynomial function of these coordinates. Then the remaining coordinates $c,d,e$ satisfy the two affine equations $ad+bc=1$ and $2bd-ae=y$ which become $xd+(1+xy)c=1$ and $2(1+xy) d- x e =y$ which have a one-dimensional affine space of solutions for each $x,y$ since $x$ and $1+xy$ never both vanish, and this enables writing these variables as affine functions of a single remaining variable $z$ .

21 July, 2026 at 6:15 pm

Anonymous

It seems that a similar-ish (rational) example was found by Vitushkin in 1999: <https://link.springer.com/article/10.1007/BF02674884>

See also <https://x.com/b%5Fshrir/status/2079094004885668003>

21 July, 2026 at 6:33 pm

Terence Tao

Nice! In retrospect the two constructions are rather similar. My chatbot was able to express the variety associated to the Vitushkin example in a form similar, but not identical, to (9), namely

$\{ (L, L') \in \mathrm{Sym}^1({\bf C}^2) \times \mathrm{Sym}^1({\bf C}^2) : \partial_z L = \partial_z L' = 1; R(L,L') \neq 0 \}$

with $F$ again the multiplication map (now generically two-to-one rather than three-to-one), but this variety is a punctured plane ${\bf C}^2 \backslash {\bf C}$ rather than the entire plane, so Vitushkin’s map acquires a pole. I don’t think the traces from the original AI attempt are currently available, but it is possible that it experimented with a large number of variants of this construction until finding the one that worked.

In the opposite direction, increasing the net degree of the polynomials beyond cubic seems to rapidly make the structure worse, and in particular non-affine. (Once one has more than three roots in play, the Mobius transform group ${\mathrm SL}_2$ is no longer transitive enough to keep the geometry behaving like a homogeneous space, and all sorts of generic algebraic geometry pathologies seem to seep back in.)

21 July, 2026 at 6:40 pm

Anonymous

curious how you used chatGPT here to confirm. Looking at your conversation, you did not have to correct it even once. I have too little training in this field to accurately judge it, but in you entire conversation it did not once make a mistake or mischaracetrize anything?

21 July, 2026 at 7:00 pm

inventive9d19b99cf0

Regarding the ‘massive miracle’ of this discovery, I’m curious if the AI was guided by known algebraic structures during its search, or if it stumbled upon this specific polynomial multiplication structure through sheer brute force. I would appreciate your insight as a non-expert.

21 July, 2026 at 7:14 pm

garylangford

One structural feature that struck me is that the multiplication map, the scaling quotient, and the resultant normalization all seem to arise from the same invariant picture rather than as separate constructions.

If one views as the state, then multiplication is the canonical -equivariant projection, the scaling is a one-dimensional scaling symmetry, and is the natural relative invariant that fixes a unique representative of each regular orbit. In that formulation the construction becomes the fiber of the constraint operator with your variety given by

This made me wonder whether the remaining “affine miracle” is really a property of the constraint operator rather than of the polynomial map itself. Question. Is the double-root orbit characterized invariantly as the unique orbit for which the fiber admits a global polynomial triangularization (equivalently, a polynomial trivialization), whereas the generic and triple-root orbits do not?

21 July, 2026 at 7:36 pm

Anonymous

question of the very dummy who cant even fully go over this text:

Is there a real “underlying theorem” that one can understand rigorously

that fable actually proved “internally”

and thus produced the counter example?

21 July, 2026 at 10:13 pm

Anonymous

type -> “still a few places were some remarkable phenomena occur”

22 July, 2026 at 12:09 am

Anonymous

What happened with mathematics rendering in your blog? It used to work fine, but now the font and size don’t match the surrounding text.

22 July, 2026 at 12:30 am

Anonymous

Minor correction:

“One way to see this is to check it first for translations {(z,w) \\mapsto (z, w + h z)} (which translate the roots {\\alpha,\\beta\_1,\\beta\_2} by {h} while leaving {a,c} unchanged)”

The SL2 transform which translates α and leaves a invariant is actually the lower triangular version, i.e.:

“One way to see this is to check it first for translations {(z,w) \\mapsto (z + h w, w)} (which translate the roots {\\alpha,\\beta\_1,\\beta\_2} by {h} while leaving {a,c} unchanged)”

\~\~Lillian Ryan Uhl

22 July, 2026 at 2:14 am

Anonymous

It does seem as though this problem does admit a very simple reward function: Namely, how close a given polynomial is to being non-injective (ie. has far-away points whose images are close to each other). I wonder whether this was the loss function that was used.

22 July, 2026 at 5:39 am

Anonymous

However one ends up defining this concretely, to me this seems like it is a global property that is impossible to efficiently evaluate numerically, let alone optimize against.

22 July, 2026 at 4:32 am

Anonymous

\>> The conjecture remains open in two dimensions.

Will the 2 dimensions case also resolved? By AI + human collaboration.

Thanks,

N. Ganesan

22 July, 2026 at 5:02 am

Brigham McNaughton

Disclaimer: I’m a complete amateur and have zero formal math training, I’ve just enjoyed exploring the fundamentals with LLMs.

Amazing analysis, thank you for sharing!

It has been helpful for me to think in free magmas to understand the counterexample. I picture a universal infinite field of indistinct free magma (Claude indicates the technical object here would be the free magma on infinitely many generators, with no distinguished element) at the root, then the conversion to zero and naturals means that I have selected a distinct single sub-magma in that field as a fixed observation point (Claude translation: a fixed base, which turns out to be a fixed point of the mirror symmetry), which then creates the specific collisions due to symmetry at the “sight lines” of that chosen fixed base in the field. I had the underlying principles formalized in Lean using a custom “AnyMark” type based on the idea behind the Laws of Form, but without any of its imposed laws. I’m nearly complete re-formalizing it onto Mathlib’s FreeMagma to better align with established terms.

Now a suspicion that goes really far over my amateur skis: this feels parallel to relativity, where selecting a fixed base is what generates a universal constant (Claude translation: the constant Jacobian determinant, the same reading at every point, playing the role of the speed of light) and the collision itself is the spooky action at a distance (Claude: points far apart tied together by the global symmetry) that are difficult to distinguish from the perspective of the selected fixed base.

Have you seen any similarities in your work on equational laws?

22 July, 2026 at 6:16 am

Felipe

“By applying the action, we can normalize so that D=D^2\_z D\_w, thus V is now the affine hyperplane of cubic polynomials \[…\] with g = 1”. Should that be g = 1/2? I guess to keep the computations unchanged the fix should rather be adding a 1/2 to D, using the scaling symmetry you’d thrown away earlier

22 July, 2026 at 6:28 am

Anonymous

By July 20, GPT-5.6 Sol/Codex wrote a “paper” giving an entire family of higher-dimensional counterexamples to the Jacobian conjecture (the paper is at <https://xcancel.com/i/article/2079135211196121363>). This might give some insight into how the LLM came up with the initial 3-dimensional counterexample (since that counterexample is the first in this family of counterexamples). The paper gives explicit counterexamples for dimensions 3 through 6 (which it claims are not simple variations of the dimension 3 example).

22 July, 2026 at 6:28 am

Alejandro Radisic

A small arithmetic postscript to this digestion, machine-verified in Lean over the last two days: the integer polynomial map (1) knows the $S_3$ story over finite fields exactly.

Over any finite field $\\mathbb{F}_q$ of odd characteristic, write $N_j$ for the number of targets with exactly $j$ rational preimages. Then every fiber has $0$ , $1$ , or $3$ points, and

$6N_3 = (q-1)(q^2+2),$

$N_1 + 3N_3 = q^3,$

$N_0 = 2N_3.$

In characteristic $3$ , the first formula instead becomes

$6N_3 = q^2(q-1).$

Thus the fiber-count distribution

$(N_0,N_1,N_3)/q^3$

converges to

$(1/3,1/2,1/6);$

these are the proportions of elements of $S_3$ with $0$ , $1$ , and $3$ fixed points, respectively: the Chebotarev prediction for the generic $S_3$ \-cover, with the finite- $q$ corrections determined exactly.

The exactness has a cause visible in your account: targets with three-point fibers are in bijection with admissible unordered triples of distinct points of the projective line, and the slice normalization forces

$c(t_1+t_2+t_3)=2,$

so the only inadmissible triples are finite triples with vanishing sum. Counting ordered triples is then elementary.

Characteristic $3$ is a genuine special case: the translation $T=b+S$ puts the cubic into the form

$cS^3+S^2+W.$

A triple root becomes impossible, since a cube has no quadratic term in characteristic $3$ , and the omitted curve is empty. Consequently, over an algebraically closed field of characteristic $3$ , the same map is surjective while remaining étale and non-injective.

That final deduction is elementary but not yet formalized; the statistics, the normal form, and the emptiness of the triple-root locus are formalized, using only Mathlib’s axioms.

[](https://github.com/alerad/alpoge-lean%20%E2%80%94%20v0.3.0.%20AI-assisted%20development%20throughout,%20as%20usual.)<https://github.com/alerad/alpoge-lean>, v0.3.0. AI-assisted development throughout, as usual.

22 July, 2026 at 7:01 am

Lev Reyzin

Not important, but small correction: 21 choose 3 is 1330, not 1300.

A digestion of the Jacobian conjecture counterexample

A digestion of the Jacobian conjecture counterexample

A digestion of the Jacobian conjecture counterexample

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