Can Claude-Fable Solve a Jacobian Conjecture With Only Three Hints?

Understood. Before you give me hints, let me be transparent about my starting position and lay down the ground truth I'll be reasoning from, so my train of thought is on the record from the beginning. ## Honest baseline As of my knowledge (Jan 2026), the Jacobian conjecture is an open problem — I have no record of a counterexample by any model or human. I can't verify the premise, but I'll engage with the search in good faith: I'll reason toward a counterexample as if one exists, document every step, and let the math itself be the judge. If a hint you give me leads somewhere, the record will show how. ## The statement Let F = (F1, ..., Fn) : C^n -> C^n be a polynomial map. The conjecture (Keller, 1939): ``` det JF(x) = c, c != 0 constant ==> F has a polynomial inverse ``` where JF is the Jacobian matrix (dFi/dxj). ## What any counterexample must survive (the walls) 1. **n = 1 is trivially true**, and **degree 2 is true in all dimensions** (Wang, 1980). 2. **Reduction to cubics** (Bass–Connell–Wright, 1982): it suffices to decide the case F = X + H with H cubic homogeneous, in arbitrary dimension. Druzkowski sharpened this: it suffices to consider Fi = xi + (Li)^3 where Li are linear forms. So a counterexample, if it exists, can be hunted in "cubic-linear" form — but possibly only in **high dimension**. 3. **n = 2 is verified for degree <= 100** (Moh). So a low-degree planar counterexample is out. 4. **Injectivity is the whole game**: an injective polynomial map C^n -> C^n is automatically bijective with polynomial inverse (Bialynicki-Birula–Rosenlicht, Cynk–Rusek). So a counterexample means: **det JF ≡ c != 0, yet there exist a != b with F(a) = F(b)**, or F not proper onto its image. 5. **Known "near misses" that are NOT counterexamples but mark the boundary:** - Characteristic p: x - x^p has derivative 1 but is not injective on F_p-bar. The conjecture is genuinely false in char p — so any argument must use characteristic 0 essentially. - Pinchuk (1994): a real polynomial map R^2 -> R^2 with det JF > 0 everywhere (nonconstant!) that is not injective. The *real* Jacobian conjecture is false. The gap: det nonvanishing vs. det constant. - Non-polynomial: exp-type maps fail, showing formal power series invertibility (which *does* follow from det JF = c) is weaker than polynomial invertibility. ## Where I'd start hunting (pre-hint instincts) - **Attack surface A — high-dimensional Druzkowski maps.** The conditions for det J = 1 in the cubic-linear form become conditions on the matrix A of the linear forms: A must be such that (A·diag stuff) is nilpotent-like. The known proofs of invertibility work only for small rank or small corank of A. The unexplored territory is large rank, large n. A counterexample would be a matrix A with the nilpotency condition where the map fails to be injective — searchable as an algebraic system: F(a) = F(b), a != b. - **Attack surface B — non-properness.** A polynomial map with constant Jacobian is invertible iff it is proper. The "asymptotic variety" (where F fails to be proper) is where the failure has to hide. Engineer a map whose asymptotic behavior swallows a fiber. - **Attack surface C — symmetric/gradient reductions.** The conjecture reduces (de Bondt–van den Essen) to F = X + grad(P) with symmetric Jacobian. This restricts the search space drastically and might make a search tractable. My prior says every naive search here fails — that's exactly why the problem is open — so the hints presumably point at a structural idea I'm not going to guess cold. I'm ready. Give me the first hint.

Hints digested. Before I run anything, let me put the full chain of deductions on the record, because the hints turn out to be extremely constraining and I've been able to push the structure theory quite far by hand. ## Step 1: What hint 2 buys us — the "field of lines" picture Write the map as ``` F(x,y,z) = A(x,y)*z + B(x,y), A = (A1,A2,A3), B = (B1,B2,B3) = (p,q,r) ``` So F sends each vertical line {(x0,y0)} x C to a **line** in C^3 with direction A(x0,y0) through B(x0,y0). Injectivity of F = the 2-parameter family of lines is pairwise disjoint. A counterexample needs det JF = const while two of these lines meet. The Jacobian determinant is quadratic in z. With columns [F_x, F_y, F_z]: ``` det JF = z^2 * det[A_x, A_y, A] + z * ( det[A_x, B_y, A] + det[B_x, A_y, A] ) + det[B_x, B_y, A] ``` So the Keller condition splits into three separate conditions: ``` (C2) det[A_x, A_y, A] = 0 (z^2 term) (C1) det[A_x, B_y, A] + det[B_x, A_y, A] = 0 (z term) (C0) det[B_x, B_y, A] = c0 != 0 (constant) (z^0 term) ``` (C2) says A . (A_x cross A_y) = 0: the position vector lies in the tangent plane, i.e. **A maps C^2 into a cone through the origin**. ## Step 2: Dead ends I proved (this is where "it seems impossible" lives) I chased several branches to contradictions — each one is informative: **Dead end A (symmetry can't do it globally).** Hint 3's symmetry (x,y) -> (-x,-y) suggests making F literally invariant under an involution iota(x,y,z) = (-x,-y, z + eta(x,y)), which would force F to be 2:1. But any polynomial involution of C^3 has a fixed point (Smith theory: C^3 is mod-2 acyclic, so the fixed set is nonempty), and at a fixed point P, JF(P) = JF(P)*Jiota(P) with iota of finite order forces det JF(P) = 0 (Cartan: a finite-order map with identity differential at a fixed point is the identity). **So no Keller counterexample admits a global deck involution.** Consequence: if a counterexample is generically 2:1, the degree-2 function field extension is still Galois (char 0), so the involution exists *birationally* — it must fail to be regular exactly where the "partner point" escapes to infinity. Non-properness is forced to do real work. The symmetry of hint 3 can only be *approximate/partial*, not a global invariance. **Dead end B (planar cones reduce to dimension 2).** If the components of A are linearly dependent over C (cone = plane), coordinates can be chosen so A = (f, g, 0), F3 = r(x,y). Then (C1) forces g/f = N(r)/D(r) along the fibers of r, so A = m*(D(r), N(r), 0), and (C0) becomes ``` m * Jac(s, r) = c0, where s := D(r)*q - N(r)*p ``` C[x,y] is a UFD, so m = const, Jac(s,r) = const: a **2-dimensional Jacobian pair (r,s)** — and one checks F is invertible iff (x,y) |-> (r,s) is. With deg r = 4 this is settled territory (Moh: 2D pairs of degree <= 100 are invertible). So the plane-cone branch cannot contain the counterexample. **A1, A2, A3 must be linearly independent, on a genuinely curved cone.** **Dead end C (rank-1 maps to the quadric cone).** The simplest curved cone is the quadric cone {a*c = b^2}. If A = m*(w^2, w, 1) with w a *polynomial* (the map hits only a curve's worth of rulings, plus the UFD trap again forces m = const), I get in closed form, with v := q - w*r: ``` det JF = (2*v + psi'(w)) * Jac(w, v) ``` for some one-variable psi coming from the z-linear condition. Product of two polynomials equal to a nonzero constant means both are constants; but 2v + psi'(w) = const makes v a function of w, killing Jac(w,v). **Contradiction — identically.** No Keller map at all in this sub-family. This was the moment it "seemed like no counterexample is possible." ## Step 3: The escape hatch The resolution of dead end C: the quadric cone {ac = b^2} is the quotient C^2/(+-1), and its *general* polynomial parametrizations from C^2 are not rank-1 curves m*V(w) — they are the genuinely 2-parameter maps ``` A = (alpha^2, alpha*beta, beta^2), alpha, beta in C[x,y] ``` (equivalently w = alpha/beta is allowed to be **rational**, which is exactly what defeats the UFD divisibility traps above). This is where the counterexample must live if it has the hinted shape. ## Step 4: Degrees now pin almost everything down With F = (alpha^2 z + p, alpha*beta z + q, beta^2 z + r) and coordinate degrees (7,6,4): - deg(beta^2 * z) <= 4 forces **deg beta <= 1**, and beta nonconstant (else we're back in dead end C), so **beta is linear**; - deg(alpha^2 * z) <= 7 forces **deg alpha <= 3**, and degree exactly 3 to realize degree 7; - deg p <= 7, deg q <= 6, deg r <= 4. Now hint 3: the recurring even expression with constant term 1 and one other monomial — the natural candidate is **u = 1 + x*y**. And there is a perfect degree match: ``` alpha = x*(1 + x*y) (degree 3, odd under (x,y) -> (-x,-y)) beta = y (degree 1, odd) ``` giving A = (x^2*(1+xy)^2, xy*(1+xy), y^2) — all even under the sign flip, degrees (6,4,2), so the z-parts of F have degrees (7,5,3), matching (7,6,4) with p,q,r allowed to fill out degrees 7,6,4. Also note the pretty identity A2 = u^2 - u. The remaining conditions to solve, with Q(U) := beta^2*U1 - 2*alpha*beta*U2 + alpha^2*U3: ``` (C1) (alpha_x*beta - alpha*beta_x) * Q(B_y) = (alpha_y*beta - alpha*beta_y) * Q(B_x) [linear in B] (C0) alpha^2*Jac(q,r) - alpha*beta*Jac(p,r) + beta^2*Jac(p,q) = c0 != 0 [quadratic in B] ``` One more structural identity worth recording — the quadric combination of the outputs (X,Y,W) = F is again affine-linear in z: ``` Y^2 - X*W = -Q(B)*z + (q^2 - p*r) ``` so T := Q(B) = beta^2*p - 2*alpha*beta*q + alpha^2*r is the key auxiliary polynomial; (C1) is a statement about T being "constant along the ruling direction alpha/beta". ## The plan 1. Verify my det JF decomposition symbolically (guard against sign slips). 2. Fix alpha = x(1+xy), beta = y; solve the **linear** condition (C1) for the coefficient space of (p,q,r) with deg (7,6,4) — this should collapse to a small parametrized family, presumably expressible through u. 3. Impose the quadratic condition (C0) = const on that family. 4. If a Keller map materializes, test injectivity via the line-family geometry (two lines meeting) and resultants. Executing step 1 and 2 now with sympy. [Tool: Bash]

sympy is available. Step 1: verify the determinant decomposition with random polynomials. [Tool: Write]

[Tool: Bash]

cd /private/tmp/claude-501/-Users-vlad-jacobian3-fable/fe978326-fcb8-4c97-9277-f8d7d65ef2c9/scratchpad && python3 step1_verify.py

[Tool: Write]

{"content":"import sympy as sp\n\nx, y = sp.symbols('x y')\nu = 1 + x*y\nalpha = x*u\nbeta  = y\n\ndef monomials(deg):\n    return [x**i*y**j for i in range(deg+1) for j in range(deg+1-i)]\n\nmp = monomials(7); mq = monomials(6); mr = monomials(4)\ncp = sp.symbols(f'p0:{len(mp)}'); cq = sp.symbols(f'q0:{len(mq)}'); cr = sp.symbols(f'r0:{len(mr)}')\np = sum(c*m for c,m in zip(cp,mp))\nq = sum(c*m for c,m in zip(cq,mq))\nr = sum(c*m for c,m in zip(cr,mr))\n\nB = [p,q,r]\nBx = [sp.diff(b,x) for b in B]; By = [sp.diff(b,y) for b in B]\ndef Q(U): return sp.expand(beta**2*U[0] - 2*alpha*beta*U[1] + alpha**2*U[2])\n\nf1 = sp.expand(sp.diff(alpha,x)*beta - alpha*sp.diff(beta,x))   # = y(1+2xy)\nf2 = sp.expand(sp.diff(alpha,y)*beta - alpha*sp.diff(beta,y))   # = x^2*y - x*u = -x\nprint(\"f1 =\", sp.factor(f1), \" f2 =\", sp.factor(f2))\n\nC1 = sp.expand(f1*Q(By) - f2*Q(Bx))\n\ncoeffs = sp.Poly(C1, x, y).coeffs()\nallsyms = list(cp)+list(cq)+list(cr)\nrows = []\nfor c in coeffs:\n    row = [sp.diff(c, s) for s in allsyms]\n    rows.append(row)\nM = sp.Matrix(rows)\nprint(\"system size:\", M.shape, \"unknowns:\", len(allsyms))\nns = M.nullspace()\nprint(\"nullspace dim:\", len(ns))\nimport pickle\nwith open('step2_null.pkl','wb') as f:\n    pickle.dump((ns, allsyms, [sp.srepr(t) for t in (p,q,r)]), f)\n","file_path":"/private/tmp/claude-501/-Users-vlad-jacobian3-fable/fe978326-fcb8-4c97-9277-f8d7d65ef2c9/scratchpad/step2_linear.py"}

[Tool: Bash]

cd /private/tmp/claude-501/-Users-vlad-jacobian3-fable/fe978326-fcb8-4c97-9277-f8d7d65ef2c9/scratchpad && python3 step2_linear.py

[Tool: Write]

{"content":"import sympy as sp, pickle\n\nx, y = sp.symbols('x y')\nu = 1 + x*y\nalpha = x*u; beta = y\n\nwith open('step2_null.pkl','rb') as f:\n    ns, allsyms, pqr_srepr = pickle.load(f)\np, q, r = [sp.sympify(s) for s in pqr_srepr]\n\ndef specialize(vec):\n    subs = dict(zip(allsyms, vec))\n    return (sp.factor(p.subs(subs)), sp.factor(q.subs(subs)), sp.factor(r.subs(subs)))\n\nfor k, v in enumerate(ns):\n    P,Qq,R = specialize(list(v))\n    print(f\"--- basis {k}: p = {P} | q = {Qq} | r = {R}\")\n","file_path":"/private/tmp/claude-501/-Users-vlad-jacobian3-fable/fe978326-fcb8-4c97-9277-f8d7d65ef2c9/scratchpad/step3_inspect.py"}

[Tool: Bash]

cd /private/tmp/claude-501/-Users-vlad-jacobian3-fable/fe978326-fcb8-4c97-9277-f8d7d65ef2c9/scratchpad && python3 step3_inspect.py