A digestion of the Jacobian conjecture counterexample

# A digestion of the Jacobian conjecture counterexample Source: [https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/) The notorious[Jacobian conjecture](https://en.wikipedia.org/wiki/Jacobian_conjecture)can be formulated concretely over the complex numbers as follows\. > **Conjecture 1 \(Jacobian Conjecture\)**Letbe a polynomial map incomplex variables, whose Jacobianis a non\-zero constant\. Thenis invertible \(with polynomial inverse\)\. The condition that the Jacobianis non\-zero is equivalent tobeing locally invertible\. \(The implication of local invertibility from non\-vanishing Jacobian follows from the inverse function theorem; the converse implication can be derived from the[Weierstrass preparation theorem](https://en.wikipedia.org/wiki/Weierstrass_preparation_theorem), but is omitted here\.\) Also, from the fundamental theorem of algebra, once the Jacobian polynomialis non\-zero, it must be constant\. So the hypothesis “Jacobianis a non\-zero constant” can be replaced with “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](https://en.wikipedia.org/wiki/Algebraic_geometry_and_analytic_geometry#The_Lefschetz_principle), but I prefer to work in the concrete setting of the complex numbers\. Recently, it was[recently shown](https://www.newscientist.com/article/2580374-ais-solution-to-87-year-old-riddle-takes-mathematicians-by-surprise/)\(using the Fable AI\) that the conjecture is false in three dimensions \(and thus in higher dimensions as well\): > **Theorem 2 \(Counterexample to conjecture\)**There exists a polynomialwhich 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    and one can verify by a brief calculation that  and   While this is an extremely quick verification, the construction presented in this fashion appears like a massive miracle\. The polynomialhas degree seven, so*a priori*the Jacobianought to be a polynomial in three variables of degree as large as, so the fact that all non\-constant coefficients of this polynomial vanish looks like a massive cancellation involvingcoefficients, which is much larger than thedegrees 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](https://x.com/davikrehalt/status/2079175065695035442)\. 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 domainto an equivalent affine variety\. Namely, we will show > **Theorem 3 \(Counterexample, reformulated\)**There exists an affine varietythat is isomorphic toby polynomial changes of variable, and a polynomial mapwhich is locally injective, but not globally injective\. Clearly one can get from Theorem[3](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#jac2)to Theorem[2](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#jac1)by composing with the isomorphismand using the previously mentioned fact that local injectivity implies non\-zero constant Jacobian\. Our objective is now to find data,that obeys three separate properties: The advantage of splitting the problem in to these three components is that we can build towards each of them separately\. It turns out thatandcan be built out of the operation of multiplication of low degree polynomials\. Namely, consider the following three simple affine spaces: \(The notationhere refers to the[symmetric power](https://en.wikipedia.org/wiki/Symmetric_power)of a vector space\.\) Clearly these spaces are isomorphic torespectively\. Furthermore, we have a multiplication map, mapping a pairof a linear polynomialand a quadratic polynomialto a cubic polynomial  \(Right now, the domain and range of this mapis larger dimensional than the target of three; we will cut the dimensions down to three as the argument progresses\.\) The map, essentially a map fromto, is clearly polynomial; in coordinates it is given explicitly in coordinates as  The mapalso enjoys two basic \(and commuting\) symmetries: So this map enjoys a huge amount of equivariance, basically with respect to an action of the five\-dimensional group\. The five\-dimensional domainis of course larger than the four\-dimensional range, so the mapclearly cannot be injective\. This can already be seen from the scaling symmetry, as the specific scalings  formodify the linear and quadratic polynomialsbut not their product\. But even if one quotients out by this symmetry[\(3\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#specific)to cut the dimension of the domain down to four, the mapis still not surjective for the following basic reason\. A generically chosen cubic polynomialwill split into the productof three independent linear polynomials\. Then there are three pairs which all map to the same cubic polynomial  under the multiplication map, but are not related to each other by scaling symmetry[\(3\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#specific)\. Thus, we see that even after quotienting out by the scaling symmetry[\(3\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#specific), the multiplication mapis 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 symmetryto obtain a useful normalization\. Ifis a linear polynomial andis a quadratic polynomial, the[resultant](https://en.wikipedia.org/wiki/Resultant)can be defined by the determinant  If we have a factoring  then the resultant can also be described as  Thus the resultant measures whether the linear polynomialand the quadratic polynomialshare a common root\. A fundamental fact about resultants is that they are\-invariant: for any, we have  One way to see this is to check it first for translations\(which translate the rootsbywhile leavingunchanged\) and for inversions\(which maptowhile mappingtoandrespectively\), and then noting that these transformations generate all of\. They also interact very nicely with scaling:  In particular, the scaling symmetry[\(3\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#specific)multipliesby: Thus, we can \(generically\) normalize away this scaling symmetry by imposing the condition We now have a restricted multiplication map \(which by abuse of notation we will continue to call\) from the four\-dimensional variety  to the four\-dimensional space\. This mapis still not globally injective, as we can take the three pairs in[\(4\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#l123)from before and apply the scaling[\(3\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#specific)separately to each of the three pairs to obtain the normalization[\(7\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#res1)\. So we have kept property \(b\)\. Furthermore, this map retains the\-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 ofin the neighborhood of a pairwith\. As the resultant is non\-vanishing, the rootof\(which exists in the Riemann sphere, or projective line if you prefer\) is distinct from the two rootsof\(though the latter two roots could be equal to each other\)\. Applying theaction \(which performs Möbius transforms on the roots\), one can assume without loss of generality thatis the point at infinity \(or equivalently\), thusfor some complex numberandfor some complex numbers, with the resultant condition[\(7\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#res1)simplifies to\(so in particularare also non\-zero\)\. It is then clear that if one perturbsandby a small amount \(say, modifying each coefficient by\), then the rootofwill perturb to something large \(\), while the rootsofstay bounded\. Thus, just from knowledge of the product, one can reconstruct which of the three roots of this cubic polynomial will be the perturbed root of, and which two will be the perturbed roots of; from this and[\(6\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#scal),[\(7\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#res1)we can also reconstruct the leading coefficientof, and this completely determines bothand\. This establishes the local injectivity property \(a\)\. \(In fact it is[étale](https://en.wikipedia.org/wiki/Formally_%C3%A9tale_morphism), but we will not need the machinery of étale maps here\.\) Unfortunately, \(the four\-dimensional analogue of\) condition \(c\) fails: the quadric hypersurface[\(8\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#quad)is not isomorphic to the affine space\. But we can try to get around this by passing to a three\-dimensional slice\. Letbe some three\-dimensional affine plane of\(which we will take to avoid the origin for technical reasons\), then we can restrictas a map from the set   to\. The latter is clearly identifiable \(by linear changes of coordinate\) to\. Aswas already locally invertible, it remains locally invertible under restriction; and because generic cubic polynomialshad three preimages underin[\(8\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#quad), this continues to be the case after restricting to[\(9\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#quad-v)\(unlesswas somehow so degenerate that it had no generic elements, but this turns out to be impossible\)\. So we have retained properties \(a\) and \(b\)\. The miracle is that, with a good choice of, we can also obtain \(c\) and obtain the desired counterexample to the Jacobian conjecture: despite appearances, the variety[\(9\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#quad-v)is in fact equivalent to the affine spaceby polynomial changes of variable\! Let’s see how\. The affine hyperplanes inavoiding the origin are parameterized by the dual space ofavoiding the origin, which one can think of as the non\-zero third order homogeneous differential operatorsin two variables\. Indeed, every such operatorgenerates affine hyperplanethat avoids the origin, and conversely by duality every affine hyperplane avoiding the origin arises in this form uniquely\. Just as the cubic polynomials incan be factored into three linear polynomials, the differential operators in the dual spacecan also be factored into three linear differential operators, e\.g\.,  in the case thatis non\-zero\. Theaction moves the rootsaround the Riemann sphere by Möbius transformations\. As these transformations are\-transitive, the actual selection of such roots is not too important \(and the scaling symmetry similarly makes the choice of leading coefficientunimportant\); 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\(and thus of affine hyperplane\) to consider: It turns out that the affine miracle for[\(9\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#quad-v)occurs precisely in the second case, whenhas 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 theaction, we can normalize so that, thusis now the affine hyperplane of cubic polynomialswith\. Using[\(2\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#mult)and[\(5\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#reslq), the variety[\(9\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#quad-v)can now be described explicitly in coordinates as  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 ifis non\-zero, then the second equationcan be solved for, and the first equationcan be solved for, Putting these two equations together, we see that as long as one removes the case, the quintupleis uniquely determined byby a change of variables which is Laurent inand polynomial in\. Thus we have a nice birational equivalence   Thus we have already*almost*established property \(c\): the variety[\(9\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#quad-v)becomes birationally equivalent toafter cutting out thesubvariety\. In particular, for each fixed non\-zero valueof, the corresponding fiber  of[\(10\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#explicit)is equivalent toby polynomial changes of variable, since we can reconstructfrom the coordinatesby the polynomial formulae  So we just need to glue back in thefiber\. Indeed, from[\(10\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#explicit)we see that the fiber atis just  Now we observe a key miracle: the cubic equationand quadratic equationhave a unique affine solution\(as opposed to the six possible solutions that Bezout’s theorem might suggest\)\. So the fiber here is also affine:  This is extremely encouraging for the purposes of establishing property \(c\), as it strongly suggests that the variety[\(10\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#explicit)has the structure of an\-bundle over, which is already extremely close to being isomorphic to the affine space\. The main remaining task is to make sure that nothing singular happens in the limit, and that a global polynomial coordinate chart for[\(10\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#explicit)that covers both theandfibers 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 limitby hand\. On the variety[\(10\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#explicit), let us useto denote any multiple ofby a polynomial expression in\. Thus, for instance, the equationimplies that  while the equationimplies that as well as the more refined estimate In thecase we could conclude that\. Now we perturb this observation\. Multiplying[\(13\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#bc-1)bywe have, which on substitution into[\(14\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#bc-2)gives; substituting this back into either[\(13\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#bc-1)or[\(14\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#bc-2)also gives\. We can get some more precise asymptotics by also taking advantage of[\(15\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#bc-3)\. Substitutinginto[\(15\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#bc-3), we obtain after some algebra  So if we writemore explicitly as, then we have  and thus Substituting this back into[\(11\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#d-eq)gives an asymptotic for:    Finally, one can insert these estimates into[\(12\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#e-eq), although one only gets a trivial bound in this case:    Expanding theerror term in[\(16\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#c-eq)as, and doing a little more algebra, we thus have a polynomial change of variables      which completely parameterizes the variety[\(10\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#explicit)by polynomial combinations of three coordinates\. This already gives \(a\) and thus completes the proof of Theorem[3](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#jac2)\. The previous computations, when expanded out, also gives polynomial inverse maps:    The map fromto thecoefficients of\(dropping thecoefficient which is constrained to equal\), we obtain a polynomial map  with  which theory predicts to have a constant Jacobian, and indeed one can calculate that the Jacobian is\. This is essentially the original example up to trivial changes of variable; indeed, one can check that the map  is exactly the mapgiven in[\(1\)](https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/#explicit-ex)\. AI disclosure: I[used an AI chatbot](https://chatgpt.com/share/6a5fdc7a-d6f8-83e8-bbea-8deb42cfed56)to discuss various aspects of this problem and to confirm several of the calculations made here\.