Jacobian conjecture
Jacobian conjecture
Problem statement
jacobian_conjecture_two_variables
Does the Jacobian conjecture hold in the two variable case?
Lean 4
Formal statement
Lean 4/-
Copyright 2025 The Formal Conjectures Authors.
Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at
https://www.apache.org/licenses/LICENSE-2.0
Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
-/
import FormalConjecturesUtil
/-!
# Jacobian conjecture
*Reference:* [Wikipedia](https://en.wikipedia.org/wiki/Jacobian_conjecture)
-/
namespace JacobianConjecture
section Conjecture
open MvPolynomial RegularFunction
variable (k : Type*)
name_poly_vars X, Y, Z over k
/-- Alpöge/Fable's counterexample: a polynomial self-map of `k³` with Jacobian
determinant `-2` which is not injective. -/
noncomputable abbrev F [CommRing k] : RegularFunction k (Fin 3) (Fin 3) :=
![(1 + X * Y)^3 * Z + Y ^ 2 * (1 + X * Y) * (4 + 3 * X * Y),
Y + 3 * X * (1 + X * Y) ^ 2 * Z + 3 * X * Y ^ 2 * (4 + 3 * X * Y),
2 * X - 3 * X ^ 2 * Y - X ^ 3 * Z]
/-- A variant of Alpöge/Fable's counterexample: a polynomial self-map of `k³` with Jacobian
determinant `1` which is not injective. -/
noncomputable abbrev G [CommRing k] : RegularFunction k (Fin 3) (Fin 3) :=
![(1 + 2 * X * Y) ^ 3 * Z + 4 * Y ^ 2 * (1 + 2 * X * Y) * (2 + 3 * (X * Y)),
Y + 3 * X * (1 + 2 * X * Y) ^ 2 * Z + 12 * X * Y ^ 2 * (2 + 3 * (X * Y)),
-X + 3 * X ^ 2 * Y + X ^ 3 * Z]
@[category API, AMS 14]
lemma det_jacobian_F [CommRing k] : (F k).Jacobian.det = -2 := by
simp only [Jacobian, F, Fin.isValue, ← map_ofNat (C : k →+* MvPolynomial (Fin 3) k),
Matrix.det_fin_three, Matrix.of_apply, Matrix.cons_val_zero, map_add, Derivation.leibniz,
pderiv_X, ne_eq, Fin.reduceEq, not_false_eq_true, Pi.single_eq_of_ne, smul_eq_mul, mul_zero,
Derivation.leibniz_pow, Nat.add_one_sub_one, Derivation.map_one_eq_zero, one_ne_zero,
Pi.single_eq_same, mul_one, zero_add, nsmul_eq_mul, Nat.cast_ofNat, derivation_C, add_zero,
pow_one, Matrix.cons_val_one, zero_ne_one, Matrix.cons_val_two, Nat.succ_eq_add_one,
Nat.reduceAdd, Matrix.tail_cons, Matrix.head_cons, map_sub, sub_self, zero_sub, mul_neg,
sub_zero, neg_mul, sub_neg_eq_add]
simp only [map_ofNat]
ring
@[category API, AMS 14]
lemma det_jacobian_G [CommRing k] : (G k).Jacobian.det = 1 := by
simp only [G, Jacobian, ← map_ofNat (C : k →+* MvPolynomial (Fin 3) k), Matrix.det_fin_three,
Matrix.of_apply, Matrix.cons_val_zero, Matrix.cons_val_one, Matrix.cons_val_two,
Matrix.head_cons, Matrix.tail_cons, map_add, map_neg, Derivation.map_one_eq_zero, pderiv_mul,
pderiv_pow, pderiv_C, pderiv_X_self, pderiv_X_of_ne, ne_eq, Fin.reduceEq, not_false_eq_true]
simp only [map_ofNat]
ring
/-- `F` identifies the two distinct points `(0, 0, -1/4)` and `(1, -3/2, 13/2)`. -/
@[category API, AMS 14]
lemma aeval_F_eq [Field k] [CharZero k] :
(F k).aeval ![(0 : k), 0, -1/4] = (F k).aeval ![(1 : k), -3/2, 13/2] := by
funext i
fin_cases i <;> simp [RegularFunction.aeval] <;> grind
/-- `G` identifies the two distinct points `(1, 0, 1)` and `(0, 3, -71)`. -/
@[category API, AMS 14]
lemma aeval_G_eq [CommRing k] :
(G k).aeval ![1, 0, (1 : k)] = (G k).aeval ![0, 3, -71] := by
funext i
fin_cases i <;> simp [RegularFunction.aeval]; grind
/-- The predicate that the Jacobian conjecture holds for a given field and variable index type
(i.e. number of variables). -/
def JacobianConjectureProp (k σ : Type*) [CommRing k] [Fintype σ] [DecidableEq σ] : Prop :=
∀ (F : RegularFunction k σ σ), IsUnit F.Jacobian.det →
∃ (G : RegularFunction k σ σ), G.comp F = id k σ ∧
F.comp G = id k σ
set_option linter.style.answer_attribute false in
/-- The **Jacobian Conjecture**: any regular function
(i.e. vector valued polynomial function from) `kⁿ → kᵐ`
whose Jacobian is a non-zero constant has an inverse that
is given by a regular function, where `k` is a field of characteristic `0`.
This is false: `F` has Jacobian determinant `1` but identifies
two distinct points, so it admits no inverse. This counterexample works in all characteristics. -/
@[category research solved, AMS 14]
theorem jacobian_conjecture {k : Type} [CommRing k] [Nontrivial k] :
answer(False) ↔ ∀ {σ : Type} [Fintype σ] [DecidableEq σ], JacobianConjectureProp k σ := by
rw [false_iff]
intro h
obtain ⟨H, -, hGH⟩ := h (G k) (det_jacobian_G k ▸ isUnit_one)
have hleft : Function.LeftInverse (H.aeval (S₁ := k)) ((G k).aeval) := fun a => by
rw [← RegularFunction.comp_aeval, hGH]
funext t
simp [RegularFunction.aeval, RegularFunction.id]
have h1 : (1 : k) = 0 := congrFun (hleft.injective (aeval_G_eq k)) 0
norm_num at h1
/-- Does the Jacobian conjecture hold in the two variable case? -/
@[category research open, AMS 14]
theorem jacobian_conjecture_two_variables :
answer(sorry) ↔ ∀ {k : Type} [Field k] [CharZero k], JacobianConjectureProp k (Fin 2) := by
sorry
end Conjecture
section Tests
open MvPolynomial RegularFunction
variable {k σ : Type} [Fintype σ] [DecidableEq σ] [Field k]
-- Let's check that we've stated the "invertible Jacobian" condition correctly
-- by proving an equivalence
@[category API, AMS 14]
lemma sanity_check_condition_1 (F : RegularFunction k σ σ) :
IsUnit F.Jacobian.det ↔ (∃ (c : k), c ≠ 0 ∧ F.Jacobian.det = .C c) := by
simp [MvPolynomial.isUnit_iff_eq_C_of_isReduced, isUnit_iff_ne_zero]
-- Let's apply the conjecture to a trivial case to make sure things
-- are working as expected.
@[category test, AMS 14]
theorem jacobian_conjecture_identity (H : JacobianConjectureProp k σ) :
∃ (G : RegularFunction k σ σ), G.comp (id k σ) = id k σ ∧
(id k σ).comp G = id k σ := by
apply H
suffices (RegularFunction.id k σ).Jacobian = 1 by simp [this, isUnit_one, Matrix.det_one]
ext i j
simp [RegularFunction.Jacobian, RegularFunction.id, Matrix.one_eq_pi_single]
end Tests
end JacobianConjecture
On-site verification needs an executable Lean target. None is attached yet.
A source attachment or link is not a verified proof. The statement’s correspondence with the original question needs human review.
Proofs & verification
No Lean has been submitted for this problem yet. A Lean file can be attached to a submission for review, though with no target here it cannot be machine-checked.
Background
Source snapshot inspected on 10 September 2026. Subsequent research status and the accuracy of source formalizations have not been independently certified.
Fixed commit b82b08faa9006484021c12005ab41287fb2ffb69. This file contains 1 source-labeled research-open declarations. Other declarations in the attached file can be definitions, solved results or tests. The file is preserved as source material; it has not been compiled or proved by this import.
Copyright The Formal Conjectures Authors. Code: Apache-2.0; other original material: CC BY 4.0. Third-party sources retain their own licenses. Original source files and notices preserved.
Recorded progress
Add yours →Nothing recorded yet. A pointer to a paper the entry is missing counts.
Machine attempts
0 recordedAttempts run against the statement below and are recorded in full, including the failures. The point is to stop the next model — and the next person — from re-walking a road that is already known to end. Each attempt costs 25 credits.
No machine attempts recorded.
Sign in to run an attempt. New accounts start with enough credits for ten.
Discussion
0 notesNotes are for clarifying the statement, flagging a missing reference, or recording why an approach fails. Claimed solutions belong in a submission, where they can be reviewed.
No notes yet.
Sign in to add a note.