Exact Poisson obstructions on the exceptional pseudo-plane

Consider

\[ T=\operatorname{Spec}R, \qquad R=\mathbb C[a,b,c]/(b^2-a-a^2c) \]

with the Poisson bracket induced by the scaffold \(a=x^2\), \(b=x(1+x^2y)\), \(c=2y+x^2y^2\):

\[ \{a,b\}=2a^2, \qquad \{a,c\}=4b, \qquad \{b,c\}=2+4ac. \]

A pair \(U,V\in R\) satisfying \(\{U,V\}=1\) would pull back to a noninjective plane Keller map. No such pair is produced here. This note proves several exact no-go theorems which sharply constrain where one could occur. Every finite polynomial calculation is independently checked by certify_pseudoplane_poisson.py.

1. Global symplectic form and two Darboux charts

Put \(F=b^2-a-a^2c\). The bracket is the hypersurface Jacobian bracket

\[ \{f,g\}=-2\det\frac{\partial(F,f,g)}{\partial(a,b,c)}, \]

and its nowhere-vanishing two-form is

\[ \omega=-\frac12\operatorname{Res}_T \frac{da\wedge db\wedge dc}{F}. \]

There are two particularly simple rational Darboux charts. On \(D(a)\) set

\[ p=-\frac1{2a},\qquad q=b. \]

Then \(\{p,q\}=1\) and

\[ c=2p+4p^2q^2. \]

On \(D(1+ac)\) set

\[ P=\frac b2, \qquad Q=\frac{c}{1+ac}. \]

Again \(\{P,Q\}=1\), and

\[ a=\frac{4P^2}{1+4P^2Q},\qquad b=2P,\qquad c=Q(1+4P^2Q). \]

The two opens cover \(T\). Their overlap has \(q\ne0\) and transition

\[ \boxed{P=\frac q2,\qquad Q=-2p-q^{-2}.} \]

This transition is symplectic. The term \(q^{-2}\) is the global obstruction seen by all of the elementary Darboux attempts below: local coordinates are easy, but a local mate regular on one component of \(b=0\) develops a pole on the other.

2. The two components of \(b=0\)

Let

\[ D_0=\{a=b=0\}, \qquad D_\infty=\{b=0,\ 1+ac=0\}. \]

Then

\[ \operatorname{div}(a)=2D_0, \quad \operatorname{div}(b)=D_0+D_\infty, \quad \operatorname{div}(1+ac)=2D_\infty. \]

Near \(D_0\), \((b,c)\) are regular coordinates and

\[ a=b^2-cb^4+2c^2b^6-5c^3b^8+\cdots, \qquad 1+2ac=\sqrt{1+4b^2c}. \]

Consequently

\[ \{f,g\}=2(1+2ac)(f_b g_c-f_c g_b). \]

Writing \(U=U_0(c)+bU_1(c)+O(b^2)\) and similarly for \(V\), a necessary first-jet condition is

\[ \boxed{2\bigl(U_1V_0'-U_0'V_1\bigr)=1.} \]

Thus the divisor itself causes no local obstruction. The obstruction is the incompatibility of regularity at both \(D_0\) and \(D_\infty\).

3. None of the generators has a polynomial mate

On \(D(a)\) the ring is \(\mathbb C[a^{\pm1},b]\) and

\[ \{a,-\}=2a^2\partial_b, \qquad \{b,-\}=-2a^2\partial_a. \]

If \(\{a,V\}=1\), integration in the function field gives

\[ V=\frac{b}{2a^2}+h(a). \]

Along \(D_0\), the first term has valuation \(1-4=-3\), while every term in \(h(a)\) has even valuation. It cannot be regular.

If \(\{b,V\}=1\), then

\[ V=\frac1{2a}+h(b). \]

Regularity at \(D_\infty\), where \(a\) is a unit and \(b=0\), forbids a pole of \(h\) at \(b=0\). In the second Darboux chart the same expression is

\[ V=\frac Q2+\frac1{8P^2}+h(2P), \]

which cannot then be regular at \(D_0\), where \(P=0\). This proves the all-degree no-go for \(b\) without a bounded ansatz.

For \(c\), work over \(K=\mathbb C(c)\) and put

\[ s=1+2ac,\qquad z=s+2\sqrt c\,b. \]

The relation gives \(s^2-4cb^2=1\), so \(K(\sqrt c)(T)=K(\sqrt c)(z)\), while

\[ \{c,z\}=-4\sqrt c\,z. \]

The equation \(\{c,V\}=1\) would therefore require a rational antiderivative of \(dz/z\). This is impossible: \(dz/z\) has nonzero residues, whereas an exact rational differential has zero residues. Hence \(c\) has no rational mate, and a fortiori no polynomial mate.

4. No homogeneous Darboux pair

The hyperbolic \(\mathbb C^*\)-action has weights

\[ \operatorname{wt}(a,b,c)=(2,1,-2), \]

and the Poisson bracket has degree \(+1\). Put \(t=ac\). Since \(R=\mathbb C[a,c]\oplus b\mathbb C[a,c]\), its homogeneous pieces are explicit:

\[\begin{split} R_{2k}= \begin{cases} a^k\mathbb C[t],&k\ge0,\\ c^{-k}\mathbb C[t],&k<0, \end{cases} \qquad R_{2k+1}=bR_{2k}. \end{split}\]

If homogeneous \(U,V\) have constant bracket, their weights sum to \(-1\); one is even and the other odd. Up to swapping the pair, for \(k\ge0\) write

\[ U=a^kP(t),\qquad V=bc^{k+1}Q(t). \]

Direct differentiation gives

\[\begin{split} \begin{aligned} \{U,V\}=2t^k\bigl(&k(2(k+1)+(2k+3)t)PQ\\ &+(2k+1)t(1+t)P'Q+2kt(1+t)PQ'\bigr). \end{aligned} \end{split}\]

For \(k\ge1\) this is divisible by \(t\). For \(k=0\) it is \(2t(1+t)P'Q\). Neither can be a nonzero constant.

For the negative weights put \(k=-h\), \(h\ge1\), and write

\[ U=c^hP(t),\qquad V=ba^{h-1}Q(t). \]

Then

\[\begin{split} \begin{aligned} \{U,V\}=-2t^{h-1}\bigl(&h((2h-1)+2ht)PQ\\ &+(2h-1)t(1+t)P'Q+2ht(1+t)PQ'\bigr). \end{aligned} \end{split}\]

The factor \(t^{h-1}\) excludes \(h\ge2\). At \(h=1\) this becomes

\[ \{U,V\}=-\frac2Q\frac d{dt}\bigl(t(1+t)PQ^2\bigr). \]

If it were \(1\), then

\[ \frac d{dt}\bigl(t(1+t)PQ^2\bigr)=-\frac12Q. \]

For nonzero polynomials \(P,Q\), the two sides have degrees \(\deg P+2\deg Q+1\) and \(\deg Q\), respectively, a contradiction. Therefore

\[\begin{split} \boxed{\begin{gathered} \text{There is no }\mathbb C^*\text{-homogeneous pair }U,V\text{ with}\\ \{U,V\}=1. \end{gathered}} \end{split}\]

5. No pair of ambient total degree at most two

This stronger bounded theorem allows both coordinates to be nonhomogeneous. Modulo \(b^2=a+a^2c\), the images of ambient polynomials of total degree at most two have nonconstant basis

\[ a,b,c,a^2,ab,ac,bc,c^2,a^2c. \]

The constant coefficient of \(\{U,V\}=1\) says that the determinant of the linear \((b,c)\)-coefficient matrix is \(1/2\). An \(\mathrm{SL}_2\) change of target coordinates therefore reduces the general pair to

\[\begin{split} \begin{aligned} U={}&\frac b2+Aa+Da^2+Eab+Fac+Gbc+Hc^2+Ia^2c,\\ V={}&c+A'a+D'a^2+E'ab+F'ac+G'bc+H'c^2+I'a^2c. \end{aligned} \end{split}\]

For coefficient-pairs \(X=(X,X')\), put \(\Delta(X,Y)=XY'-X'Y\). Comparing coefficients in \(\{U,V\}-1\) gives, among others,

\[ 8\Delta(E,H)+\Delta(F,G)=0, \qquad 6\Delta(E,H)+\Delta(F,G)+1=0. \]

Hence \(\Delta(E,H)=1/2\), so \(E,H\) are independent. The equations

\[ \Delta(E,I)=\Delta(H,I)=0 \]

force \(I=0\). Next

\[ 10\Delta(E,G)-4\Delta(F,I)=0, \qquad \Delta(G,H)=0 \]

force \(G=0\). The remaining equations successively give

\[ H'=0,\quad A=0,\quad E=F'=0,\quad -E'H=\frac12,\quad D=0, \]

then

\[ F=2A'H,\qquad A'F=0,\qquad A'=F=0,\qquad E'=0, \]

contradicting \(-E'H=1/2\). Thus

\[ \boxed{\text{No two ambient quadratics have Poisson bracket }1\text{ in }R.} \]

The exact coefficient ideal has 23 generators in 14 unknowns after the normalization above; SymPy computes its reduced Groebner basis as \([1]\). This is notably stronger than the earlier linear no-go. It is also sharp in a different direction: the identity

\[ 1=\frac12\{b,c\}+\frac12\{ab,c^2\}-4\{ac,bc\} \]

expresses \(1\) as a sum of three brackets of quadratics, but this section proves that it cannot be compressed into one quadratic bracket.

6. Locally nilpotent and self-similar directions

The Hamiltonian derivation

\[ \delta=\frac12\{a,-\} =a^2\partial_b+2b\partial_c \]

is locally nilpotent, with flow

\[ (a,b,c)\longmapsto (a,b+\lambda a^2,c+2\lambda b+\lambda^2a^2). \]

Its image lies in the proper ideal \((a^2,b)\), so it has no slice. More generally, the classification for generalized Danielewski surfaces applies with \(f(a)=a^2\) and \(\varphi(a,b)=b^2-a\) [Bianchi and Veloso, 2017]; it says

\[ \operatorname{ML}(R)=\mathbb C[a], \qquad \operatorname{LND}(R)=\{f(a)\delta:f\in\mathbb C[a]\}. \]

External-input flag. The displayed classification of all locally nilpotent derivations is Corollary 8 of the cited paper, not a result proved by the finite certificate script. The formula for \(\delta\), its local nilpotence, and its lack of a slice are proved directly here and checked exactly. None of the generator, homogeneous, or quadratic no-go theorems depends on the external classification.

Finally, the known nonproper etale self-map is, in \((a,b,c)\) order,

\[ \eta(a,b,c)=\bigl(b^2,b(1+2ac),4c\bigr). \]

It is conformally Poisson with factor four:

\[ \{\eta^*f,\eta^*g\}=4\eta^*\{f,g\}. \]

For \(t=ac\) and \(s=1+2t\) it induces the Chebyshev dynamics

\[ t\longmapsto4t(1+t), \qquad s\longmapsto2s^2-1=T_2(s). \]

The homogeneous theorem rules out every equivariant Darboux pair. A viable construction must therefore be mixed-weight, nontriangular, and globally compatible with the \(q^{-2}\) Darboux transition. The non-equivariant deformations of \(\eta\) are the most structured remaining source of such mixed-weight ansatzes.