雅可比猜想被证伪：Fable AI 发现三维多项式反例

臭名昭著的雅可比猜想：https://en.wikipedia.org/wiki/Jacobian_conjecture 可以在复数域上具体表述如下。

雅可比行列式非零的条件等价于局部可逆。（由逆函数定理可以得出非零雅可比行列式意味着局部可逆；相反的推论可以从魏尔斯特拉斯准备定理推导：https://en.wikipedia.org/wiki/Weierstrass_preparation_theorem，但这里省略。）此外，根据代数基本定理，一旦雅可比多项式非零，它必须是常数。因此假设“雅可比行列式是非零常数”可以替换为“局部可逆”。所以雅可比猜想可以被视为局部可逆性必然导致全局可逆性的断言。复数可以很容易地被特征为零的其他域替换，依据是勒夫谢茨原理：https://en.wikipedia.org/wiki/Algebraic_geometry_and_analytic_geometry#The_Lefschetz_principle，但我更喜欢在复数的具体环境下工作。

最近有人证明：https://www.newscientist.com/article/2580374-ais-solution-to-87-year-old-riddle-takes-mathematicians-by-surprise/（使用 Fable AI）该猜想在三维中是假的（因此在更高维中也是假）：

该猜想在二维中仍未解决，在一维中容易证明。

这个例子可以完全显式地说明：可以采用

该例子后来被用更几何的术语回溯性解释：https://x.com/davikrehalt/status/2079175065695035442。作为对自己的“消化”练习，我试图用相对少量代数几何写出该解释，以一种最小化“奇迹”出现次数的方式，尽管仍有一些地方出现了某些显著现象。

使用局部单射形式并将定义域推广到等价的仿射簇是方便的。也就是说，我们将展示

显然，可以通过与同构体组合，并利用前面提到的局部单射性意味着非零常数雅可比这个事实，从定理3：#jac2 推导出定理2：#jac1。我们现在的目标是找到满足三个独立性质的数据：

事实证明，和可以通过低阶多项式的乘法操作构建。也就是说，考虑以下三个简单的仿射空间：

映射，本质上是从 映射到 的映射，显然是多项式的；在坐标中明确表示为

五维域当然大于四维值域，所以该映射显然不可能是单射的。这从缩放对称性就可以看出，具体缩放为

“利用”缩放对称性来获得一个有用的归一化将是方便的。如果 是线性多项式而 是二次多项式，则判别式：https://en.wikipedia.org/wiki/Resultant 可以通过行列式定义

我们现在有一个受限的乘法映射（出于符号方便仍称为 ）从四维仿射簇

不幸的是，（四维的）条件(c) 失败：二次型超曲面(8)：#quad 并不与仿射空间 同构。但我们可以尝试通过取三维切片来绕过这一问题。设 为 的某个三维仿射平面（出于技术原因我们取其避免原点），则我们可以将 限制为从集合 的映射

看看如何做。在 避免原点的仿射超平面由 避免原点的对偶空间参数化，可以将其看作是两个变量的非零三阶同次微分算子。事实上，每个这样的算子生成一个避免原点的仿射超平面，反之，通过对偶性，每个避免原点的仿射超平面都以这种形式唯一出现。正如 中的三次多项式可以分解为三个线性多项式， 对偶空间中的微分算子也可以分解为三个线性微分算子，例如，

事实证明，对于 (9)：#quad-v 的仿射奇迹恰好发生在第二种情况，即当有两个相同的根时。我没有一个完全令人满意的几何解释，但可以通过以下坐标计算来验证它。

通过应用该作用，我们可以将 归一化，从而 现在是满足 的三次多项式的仿射超平面。使用 (2)：#mult 和 (5)：#reslq，代数簇 (9)：#quad-v 现在可以用坐标明确描述为

所以我们只需要把纤维重新粘合进去。实际上，从 (10)：#explicit 我们看到 在的纤维只是

在这里进行下一步的标准方法是使用现代代数几何和交换代数的工具操作各种切空间，但鉴于我自己的背景，我更喜欢采用分析语言，特别是大 O 符号（代替代数几何中使用的理想）来手工研究极限。在代数簇 (10)：#explicit 上，让我们用 表示 乘以 一个在 中的多项式表达式。因此，例如，该方程意味着

我们也可以利用 (15)：#bc-3 得到一些更精确的渐近式。将 代入 (15)：#bc-3，经过一些代数运算，我们得到

将 (16)：#c-eq 中的误差项展开为 , 再进行一些代数运算，我们得到一个多项式变元替换

之前的计算，展开后，也给出了多项式的逆映射：

AI 披露：我使用了 AI 聊天机器人：https://chatgpt.com/share/6a5fdc7a-d6f8-83e8-bbea-8deb42cfed56 来讨论这个问题的各个方面，并确认了这里进行的几个计算。

The notorious Jacobian conjecture：https://en.wikipedia.org/wiki/Jacobian_conjecture can be formulated concretely over the complex numbers as follows.

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：https://en.wikipedia.org/wiki/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：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.

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):

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

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 domain to an equivalent affine variety. Namely, we will show

Clearly one can get from Theorem 3：#jac2 to Theorem 2：#jac1 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:

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:

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

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

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：https://en.wikipedia.org/wiki/Resultant can be defined by the determinant

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

Unfortunately, (the four-dimensional analogue of) condition (c) fails: the quadric hypersurface (8)：#quad 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

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 an 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.,

It turns out that the affine miracle for (9)：#quad-v 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.

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

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

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)：#explicit, let us use to denote any multiple of by a polynomial expression in . Thus, for instance, the equation implies that

We can get some more precise asymptotics by also taking advantage of (15)：#bc-3. Substituting into (15)：#bc-3, we obtain after some algebra

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

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

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.