Exact obstructions to descending the cubic construction

The threefold counterexample is affine-linear in its extra variable and is a forgetful map for a marked binary cubic. Every direct way we found to remove that variable can be excluded uniformly, not merely by a bounded search.

Affine-linear coordinate theorem

Theorem. Let \(P,Q\in\mathbb C[x,y]\) satisfy \(J(P,Q)=\kappa\ne0\). If one coordinate is affine-linear in one source variable, then \((P,Q)\) is a polynomial automorphism.

Write

\[ P=a(x)y+b(x),\qquad Q=\sum_{i=0}^m q_i(x)y^i. \]

For \(a\ne0\), the coefficient of \(y^m\) in the Jacobian is

\[ m a'q_m-aq_m'=0, \]

so \(q_m=\lambda a^m\). Subtracting \(\lambda P^m\) from \(Q\) and repeating gives

\[ Q=h(P)+e(x). \]

Therefore

\[ \kappa=J(P,Q-h(P))=-a(x)e'(x). \]

Both factors are units of \(\mathbb C[x]\), hence constants, and the resulting map is triangular after a target shear. If \(a=0\), then \(\kappa=b'(x)Q_y\), and the same unit argument again gives a triangular map.

This pinpoints the dimensional break: “linear in the extra variable” is a rich threefold ansatz, but a trivial plane ansatz.

Marked affine planes of binary forms

The obstruction survives every degree. Let

\[ Q_{a,b}=Q_0+aQ_1+bQ_2\]

be a genuine affine two-plane of binary forms, dehomogenize by \(q_i(w)=Q_i(w,1)\), and use the same projective-root chart as the threefold example,

\[ w=\frac{1+xy}{x}=y+\frac1x. \]

Suppose polynomial outputs \(A,B\) satisfy

\[ q_0(w)+Aq_1(w)+Bq_2(w)=0. \]

Put

\[ s=q_0'+Aq_1'+Bq_2',\qquad D=q_1q_2'-q_1'q_2. \]

Since \((x,w)\) has the same area form as \((x,y)\), differentiation at fixed \(w\) and at fixed \(x\) gives

\[ q_1A_x+q_2B_x=0, \quad q_1'A_x+q_2'B_x=s_x, \quad q_1A_w+q_2B_w=-s. \]

Taking the indicated determinants yields the exact identity

\[ \boxed{D(w)J(A,B)=s\,s_x.} \]

If \(J(A,B)=\kappa\ne0\), integration in \(x\) gives

\[ s^2=2\kappa D(w)x+H(w). \]

This is impossible in \(\mathbb C(w)(x)\): the right side has valuation one at its linear factor in \(x\), whereas every rational square has even valuations. Finally \(D=0\) would imply \(q_2/q_1\in\mathbb C\), contradicting independence of the two direction forms.

Consequence. No affine two-plane of binary cubics, quartics, sextics, or forms of any degree can reproduce the marked-root construction in this chart. The third free coefficient in dimension three is essential.

Polynomial graphs and linear target projections

Another tempting descent is to choose a polynomial graph \(z=g(x,y)\), perhaps one passing through all three colliding source points, and retain two linear combinations of the three outputs. This also fails for every polynomial \(g\).

The highest homogeneous parts of the restricted outputs are

\[ (F_{1,\rm top},F_{2,\rm top},F_{3,\rm top}) =(y^3R,3y^2R,-R), \]

where

\[\begin{split} R= \begin{cases} x^3g_d,&\deg g=d\ge1,\\ x^2(cx+3y),&g=c. \end{cases} \end{split}\]

In either case \(R_x\ne0\), and the three leading Jacobians are

\[\begin{split} \begin{aligned} J(y^3R,3y^2R)&=-3y^4RR_x,\\ J(y^3R,-R)&=3y^2RR_x,\\ J(3y^2R,-R)&=6yRR_x. \end{aligned} \end{split}\]

Row-reducing any rank-two linear target projection gives one of these three pivot cases. The gaps between the coordinate degrees make its displayed term the unique highest Jacobian term, so it cannot be a nonzero constant.

This includes the simple interpolating graph

\[ z=\frac{27x^2-1}{4}, \]

which contains all three known colliding points. The exact quadratic coefficient search in research/n2/search_graph_sections.py independently returns the unit ideal in every target-projection chart.

Arbitrary target polynomials on the collision graph

The same quadratic graph admits a stronger, purely local obstruction. Put

\[ (A,B,C)=F\left(x,y,\frac{27x^2-1}{4}\right) \]

and order the tangent minors as \(\bigl(J(A,B),J(A,C),J(B,C)\bigr)\). At the three colliding points they form the exact matrix

\[\begin{split} M= \begin{pmatrix} 0&0&-2\\ -81/8&-3/2&106\\ 81/8&3/2&106 \end{pmatrix}. \end{split}\]

Its rows satisfy

\[ M_+ + M_- +106M_0=0. \]

Now let \(R,S\in\mathbb C[a,b,c]\) be arbitrary target polynomials, not necessarily linear. All three source points have the same target value, so the coefficients of the target two-form \(dR\wedge dS\) are the same vector \(\omega\in\mathbb C^3\) at all three. The chain rule says that the three restricted Jacobians are \(M_i\omega\). If they were one nonzero constant \(\kappa\), the row relation would give

\[ (1+1+106)\kappa=108\kappa=0, \]

a contradiction. Thus this natural collision graph cannot be repaired by any nonlinear polynomial change of target functions. The matrix and row relation are part of the exact SymPy certificate.

Nonlinear equivariant target slices

There is one more tempting escape from the linear-slice theorem. The target weights \((-2,-1,1)\) permit triangular homogeneous coordinates such as

\[ R_\phi=A+B^2\phi(BC), \qquad S_\psi=B+AC\psi(AC^2). \]

Could \(R_\phi(F)\) or \(S_\psi(F)\) become a source coordinate even though no linear output does? They cannot.

Recall the source invariants \(u=xy\), \(v=x^2z\) and the semi-invariant numerators \(a,b,c\) from Plane near-models. If \(r=R_\phi(F)\) were a coordinate, then \(V(r)\cong\mathbb A^2\) would carry the restricted \(\mathbb C^*\)-action. Its tangent weights at the origin are \((1,-1)\). Every algebraic torus action on the affine plane is linearizable [Gutwirth, 1962], so its invariant ring would be a polynomial ring in one variable. Exactness of invariants for the reductive group \(\mathbb C^*\), together with the fact that every weight-two source polynomial is \(x^2\mathbb C[u,v]\), instead gives

\[ \mathbb C[V(r)]^{\mathbb C^*} =\mathbb C[u,v]/(k_\phi), \qquad k_\phi=a+b^2\phi(bc). \]

This curve is not \(\mathbb A^1\). If \(\phi\) has degree \(m\) and nonzero leading coefficient \(\ell\), then

\[ (k_\phi)_{\rm top} =9\ell(-3)^m u^{2m+4}(v+3u)^{2m+2}. \]

It has two distinct points at infinity, whereas \(\mathbb A^1\) has one. For \(\phi=0\), already \(a=(1+u)(3u^3+u^2v+4u^2+2uv+v)\) is reducible. Hence no \(R_\phi(F)\) is a coordinate.

The weight-minus-one family is even more rigid. If \(s=S_\psi(F)\) were a coordinate, the same argument with tangent weights \((1,-2)\) would give

\[ \mathbb C[V(s)]^{\mathbb C^*} =\mathbb C[u,v]/\bigl(b+ac\psi(ac^2)\bigr)\cong\mathbb C[t]. \]

But at \(u=-1\) one has \(a=0\) and \(b=2\). Therefore \(1+u\) is a visibly nonconstant unit on this quotient curve, impossible in \(\mathbb C[t]\).

The closest special case is \(R=A+\lambda B^2\). For \(k=a+\lambda b^2\),

\[ k(-1,v)=4\lambda. \]

Thus \(1+u\) is a nonconstant unit when \(\lambda\ne0\), while \(\lambda=0\) returns the reducible \(F_1\) slice. Even the unusually simple value \(\lambda=-1/12\) does not escape: using \(a=(b-u)(1+u)/3\) gives

\[ 12k=4(1+u)-\bigl(b-2(1+u)\bigr)^2, \]

but the apparent affine-line parameter misses \(1+u=0\) and is a punctured line. This rules out the most natural nonlinear slice that fills the missing divisor of the \(F_1=-1/4\) fiber. It does not classify every nonlinear target coordinate, so genuinely nonequivariant slices remain an open route.

No linear output can be the lowering coordinate

Let \(L=\alpha F_1+\beta F_2+\gamma F_3=h(x,y)z+g(x,y)\). Its \(z\)-coefficient is

\[ h=\alpha(1+xy)^3+3\beta x(1+xy)^2-\gamma x^3. \]

On a component \(1+xy=rx\) of \(h=0\), where

\[ \alpha r^3+3\beta r^2-\gamma=0, \]

the constant term reduces exactly to

\[ g=r(\alpha r+4\beta)+\frac{\alpha r+2\beta}{x}. \]

Each such component is a copy of \(\mathbb C^*\). Additivity of the compactly supported Euler characteristic shows that a generic fiber of \(L\) has

\[ \chi_c=1+N, \]

where \(N\) counts the components on which the displayed Laurent polynomial is nonconstant, with the elementary \(x=0\) correction when \(\alpha=0\). Unless \(L\) is proportional to \(F_1\), one has \(N\ge1\), so the generic fiber is not \(\mathbb A^2\). The remaining cases are also impossible:

• \(F_1\) is reducible as a polynomial, so it is not a coordinate;

• if \(\alpha=0,\beta\ne0\), the generic Euler characteristic is at least two;

• \(F_3=x(2-3xy-x^2z)\) has generic fiber with a nonconstant unit and Euler characteristic zero.

Thus no nonzero linear combination of the three outputs is a source coordinate. Linear target slicing cannot turn the threefold collision into a plane Keller collision.