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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.

Jacobian conjecture

${F:{\bf C}^n \rightarrow {\bf C}^n}$

{n}

${\mathrm{det} DF}$

{F}

The condition that the Jacobian is non-zero is equivalent to 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 is non-zero, it must be constant. So the hypothesis “Jacobian is 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, but I prefer to work in the concrete setting of the complex numbers.

${\mathrm{det} DF}$

{F}

Weierstrass preparation theorem

${\mathrm{det} DF}$

{F}

Lefschetz principle

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

recently shown

${F : {\bf C}^3 \rightarrow {\bf C}^3}$

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

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

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

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

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

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

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

{F}

${\mathrm{det} DF}$

${3 \times 6 = 18}$

${\binom{18+3}{3}-1 = 1329}$

${\binom{7+3}{3} = 120}$

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.

explained in more geometric terms

It is convenient to use the local injectivity formulation, and to generalize the domain to an equivalent affine variety. Namely, we will show

${{\bf C}^3}$

${X \subset {\bf C}^5}$

${{\bf C}^3}$

${F : X \rightarrow {\bf C}^3}$

Clearly one can get from Theorem 3 to Theorem 2 by composing with the isomorphism and 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:

${X \cong {\bf C}^3}$

{X}

${F : X \rightarrow {\bf C}^3}$

(a) is locally injective on .

{F}

{X}

(b) is not globally injective on .

{F}

{X}

${{\bf C}^3}$

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

{F}

{X}

The space of linear homogeneous polynomials of two complex variables .

${\mathrm{Sym}^1({\bf C}^2)}$

{L(z,w) = az + bw}

{z,w}

The space of quadratic homogeneous polynomials of two complex variables .

${\mathrm{Sym}^2({\bf C}^2)}$

${Q(z,w) = cz^2 + dzw + ew^2}$

{z,w}

The space of cubic homogeneous polynomials of two complex variables .

${\mathrm{Sym}^3({\bf C}^2)}$

${C(z,w) = fz^3 + gz^2w + hzw^2 + iw^3}$

{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}

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

{F}

The map , essentially a map from to , is clearly polynomial; in coordinates it is given explicitly in coordinates as

{F}

${{\bf C}^5}$

${{\bf C}^4}$

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

{F}

If one applies a scaling for some non-zero complex numbers , then the product is scaled by : .

${(L,Q) \mapsto (\lambda_1 L, \lambda_2 Q)}$

${\lambda_1, \lambda_2}$

{LQ}

${C \mapsto \lambda_1 \lambda_2 C}$

${F( \lambda_1 L, \lambda_2 Q) = \lambda_1 \lambda_2 F(L,Q)}$

If one applies a change of variables for some invertible linear transformation , then the product is transformed by : .

${(L, Q) \mapsto (L \circ T, Q \circ T)}$

${T \in \mathrm{SL}_2({\bf C})}$

{LQ}

${C \mapsto C \circ T}$

${F(L \circ T, Q \circ T) = F(L,Q) \circ T}$

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

The five-dimensional domain is of course larger than the four-dimensional range , so the map clearly cannot be injective. This can already be seen from the scaling symmetry, as the specific scalings

${\mathrm{Sym}^1({\bf C}^2) \times \mathrm{Sym}^2({\bf C}^2)}$

${\mathrm{Sym}^3({\bf C}^2)}$

{F}

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

${\lambda \in {\bf C}^\times}$

{L,Q}

{C = LQ}

(3)

{F}

{C}

${C = L_1 L_2 L_3}$

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

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

{F}

(3)

{F}

It will be convenient to “spend” the scaling symmetry to obtain a useful normalization. If is a linear polynomial and is a quadratic polynomial, the resultant can be defined by the determinant

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

{L(z,w) = az+bw}

{Q(z,w)}

resultant

${\mathrm{Res}(L,Q)}$

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

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

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

{L}

{Q}

${SL_2}$

${T \in SL_2({\bf C})}$

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

${(z,w) \mapsto (z, w + h z)}$

${\alpha,\beta_1,\beta_2}$

{h}

{a,c}

${(z,w) \mapsto (w,z)}$

${\alpha,\beta_1,\beta_2}$

${1/\alpha, 1/\beta_1, 1/\beta_2}$

{a,c}

${-a\alpha}$

${c\beta_1 \beta_2}$

${SL_2({\bf C})}$

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

(3)

${\mathrm{Res}(L,Q)}$

${\lambda}$

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

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

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

{F}

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

${\mathrm{Sym}^3({\bf C}^2)}$

{F}

(4)

(3)

(7)

${SL_2}$

But we now also have property (a)! Suppose we want to show the local injectivity of in the neighborhood of a pair with . As the resultant is non-vanishing, the root of (which exists in the Riemann sphere, or projective line if you prefer) is distinct from the two roots of (though the latter two roots could be equal to each other). Applying the action (which performs Möbius transforms on the roots), one can assume without loss of generality that is the point at infinity (or equivalently ), thus for some complex number and for some complex numbers , with the resultant condition (7) simplifies to (so in particular are also non-zero). It is then clear that if one perturbs and by a small amount (say, modifying each coefficient by ), then the root of will perturb to something large (), while the roots of stay 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), (7) we can also reconstruct the leading coefficient of , and this completely determines both and . This establishes the local injectivity property (a). (In fact it is étale, but we will not need the machinery of étale maps here.)

{F}

{(L,Q)}

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

${\alpha}$

{L}

${\beta_1, \beta_2}$

{Q}

${SL_2}$

${\alpha}$

{a=0}

{L(z,w) = b w}

{b}

${Q(z,w) = c (z - \beta_1 w)(z - \beta_2 w)}$

${c, \beta_1, \beta_2}$

(7)

${cb^2 = 1}$

{c,b}

{L}

{Q}

${O(\varepsilon)}$

${\alpha=\infty}$

{L}

${\gg 1/\varepsilon}$

${\beta_1,\beta_2}$

{Q}

{F(L,Q)}

{L}

{Q}

(6)

(7)

{c}

{Q}

{L}

{Q}

étale

Unfortunately, (the four-dimensional analogue of) condition (c) fails: the quadric hypersurface (8) is not isomorphic to the affine space . But we can try to get around this by passing to a three-dimensional slice. Let be some three-dimensional affine plane of (which we will take to avoid the origin for technical reasons), then we can restrict as a map from the set

(8)

${{\bf C}^4}$

{V}

${\mathrm{Sym}^3({\bf C}^2)}$

{F}

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

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

{V}

${{\bf C}^3}$

{F}

{C}

{F}

(8)

(9)

{V}

(9)

${{\bf C}^3}$

Let’s see how. The affine hyperplanes in avoiding the origin are parameterized by the dual space of avoiding the origin, which one can think of as the non-zero third order homogeneous differential operators in two variables. Indeed, every such operator generates affine hyperplane 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 can be factored into three linear polynomials, the differential operators in the dual space can also be factored into three linear differential operators, e.g.,

${\mathrm{Sym}^3({\bf C}^2)}$

${D = j \partial_z^3 + k \partial_z^2 \partial_w + l \partial_z \partial_w^2 + m \partial_w^3}$

{D}

${\{ C \in \mathrm{Sym}^3({\bf C}^2) : D(C) = 1\}}$

${\mathrm{Sym}^3({\bf C}^2)}$

${\mathrm{Sym}^3({\bf C}^2)^*}$

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

{j}

${SL_2}$

${\gamma_1,\gamma_2,\gamma_3}$

{3}

{j}

{D}

{V}

Operators where the three roots are all distinct, thus for independent first-order operators .

${\gamma_1,\gamma_2,\gamma_3}$

${D = D_1 D_2 D_3}$

${D_1,D_2,D_3}$

Operators where two roots coincide and one is distinct, thus for independent first-order operators .

${D = D_1^2 D_2}$

${D_1,D_2}$

Operators where all three roots coincide, thus for some first-order operator .

${D = D_1^3}$

${D_1}$

It turns out that the affine miracle for (9) occurs precisely in the second case, when 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.

(9)

{D}

By applying the action, we can normalize so that , thus is now the affine hyperplane of cubic polynomials with . Using (2) and (5), the variety (9) can now be described explicitly in coordinates as

${SL_2}$

${D = \frac{1}{2} \partial_z^2 \partial_w}$

{V}

${C(z,w) = f z^3 + g z^2 w + h z w^2 + i w^3}$

{g=1}

(2)

(5)

(9)

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

{a}

{ad+bc = 1}

{d}

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

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

{e}

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

{a=0}

{(a,b,c,d,e)}

{(a,b,c)}

{a}

{b,c}

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

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

(9)

${{\bf C}^3}$

{a=0}

${a_0}$

{a}

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

(10)

${{\bf C}^2}$

{d,e}

{b,c}

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

So we just need to glue back in the fiber. Indeed, from (10) we see that the fiber at is just

{a=0}

(10)

{0}

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

${cb^2 = 1}$

{bc = 1}

{b=c=1}

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

(10)

${{\bf C}^2}$

${{\bf C}^1}$

${{\bf C}^3}$

${a \rightarrow 0}$

(10)

${a \neq 0}$

{a = 0}

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 by hand. On the variety (10), let us use to denote any multiple of by a polynomial expression in . Thus, for instance, the equation implies that

${a \rightarrow 0}$

(10)

{O(X)}

{X}

{a,b,c,d,e}

{ad + bc = 1}

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

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

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

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

{a=0}

{b=c=1}

(13)

{b}

${b^2 c = b + O(a)}$

(14)

{b = 1 + O(a)}

(13)

(14)

{c = 1 + O(a)}

We can get some more precise asymptotics by also taking advantage of (15). Substituting into (15), we obtain after some algebra

(15)

{ad+bc=1}

(15)

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

{b = 1+O(a)}

{b = 1 + a y}

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

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

(11)

{d}

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

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

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

(12)

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

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

$\displaystyle = O(1).$

Expanding the error term in (16) as , and doing a little more algebra, we thus have a polynomial change of variables

${O(a^2)}$

(16)

${a^2 z}$

$\displaystyle a = a$

$\displaystyle b = 1 + ay$

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

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

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

(10)

{a,y,z}

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

$\displaystyle a = a$

$\displaystyle y = 2bd - ae$

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

{(a,y,z)}

{(f,h,i)}

{F(L,Q)}

{g}

{1}

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

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