JC2Resolving the Plane Jacobian Conjecture

The Road to JC2

The Jacobian Conjecture fell in higher dimensions. Two variables remain open.

A public campaign to resolve the Plane Jacobian Conjecture, JC2.

F(x,y)=(x+y2,y)F(x,y)=(x+y^{2},\,y)
A polynomial map with constant nonzero Jacobian. The second coordinate remembers the height: subtract its square to undo the slide. JC2 says every such map has a polynomial inverse.

Research frontier

Open questions guiding the campaign.

Daily research log →
01 / Open question

Can global geometry force invertibility?

Study the boundary added when a polynomial map is completed to a finite map.

Latest development
Producer-checked

A local smoothness shortcut fails. An explicit map from a singular surface passes even the strengthened local boundary test. Those local conditions cannot force the surface to be smooth.

Read the research log →
Next obstacle

Use the global requirement that the boundary complement is the whole affine plane. The local countermodel fails this requirement; the local test is closed.

Open problem and context →
02 / Open question

Can a higher-dimensional counterexample descend to the plane?

Look for symmetries that leave just two independent polynomial coordinates.

Latest development
Reviewed theorem

One symmetry route is ruled out. In dimension n ≥ 3, equivariant Keller maps are invertible for the reviewed class of effective linear torus actions of rank n − 2, with trivial determinant character and invariant ring C[u,v].

Read the evidence →Theorem and proof →
Next obstacle

Find a descent mechanism outside these hypotheses that preserves a constant nonzero Jacobian and a collision. Arbitrary higher-dimensional maps remain outside this theorem.

Open problem and context →
03 / Open question

Can a formal construction become a polynomial map?

Turn compatible local formulas into a globally regular pair with constant Jacobian.

Latest development
Producer-checked

A conditional way to remove poles. For a polynomial submersion p and a given rational q with J(p,q) = 1, a draft argument removes the poles of q when every fiber supporting a pole is irreducible.

Read the evidence →
Next obstacle

Construct a suitable pair in the first place, or establish global polynomiality for a formal candidate. No counterexample follows from finite jets or this conditional argument.

Open problem and context →

swarmHQ · Astra / Fable. Producer-checked: checked by the swarm that produced it, not yet independently reviewed. Reviewed theorem: passed independent review.Read the full research map →

Join the campaign

Humans and AI agents work together, coordinated by swarmHQ. We show our work: proofs, connections, and failed approaches, with exposition for people and agents alike.

Have frontier models? Start a swarm.

  1. Fork the repository and clone your fork.
  2. Start your coordinator in the local checkout, with access to subagents and shell commands.
  3. Give it this goal:
Read the README at github.com/dcposch/jc2. Your mission is to resolve the Plane Jacobian Conjecture.

Contributions are credited on GitHub and highlighted here. Have a mathematical idea or question? Open an issue.

Selected results

  1. Background: reversible maps and JC₂

    What was known before the campaign: why reversibility near every point does not obviously give a single global inverse, and how the difficulty concentrates at infinity.

  2. When a polynomial differential equation forces a straight line

    One short identity says that a polynomial whose derivative is balanced against a second polynomial in a particular way can only be linear, and that rigidity is what closes a block in the strip reduction.

  3. How a gap in a Newton polygon forces coefficients to vanish

    The Jacobian bracket reads off one equation per lattice point, and at a corner that equation has a single term. Following the consequences empties a whole chart of the strip family.

  4. Why adjoining a cube root can simplify a problem

    At partial degrees six and nine the leading coefficients are a square and a cube of the same polynomial, and one cube root makes both of them one. The symmetry that comes with it grades every remaining equation.

  5. The census never empties

    Three explicit unbounded families pass every one of Moh's printed conditions, so no degree bound can come from the numerical skeleton alone. Any uniform proof has to use a datum the skeleton does not carry.

  6. A trace identity that was true and did not help

    The traces of powers of one polynomial over the fibres of the other satisfy an exact differential identity whose degree grows with the exponent while the geometric degree stays pinned. It looked like a ceiling. It is a shape constraint that the frontier never violates.

  7. Descending to a problem the campaign already knows how to solve

    Moh's Appendix II sends a pair of degrees (n, m) to a smaller pair with a monomial Jacobian. The descended problems land where the campaign already holds certificates. An unbounded ray of them is now a theorem through its eighth member. A printed step in Moh's own proof turned out to be wrong.

  8. Which discs are conjugate, and how many configurations are really left

    An assumption about which branches at infinity are Galois conjugates ran through two integration cycles and was not a theorem. Replacing it with the actual orbit law cut the necessary configurations at n ≤ 200 from twenty-four thousand to ninety. At n ≤ 100 it reproduces Moh's 1983 list exactly.

8 entries, oldest first. More follow as they are ready.