关于秩二泊松猜想的一个明确反例
An Explicit Counterexample to the Rank-Two Poisson Conjecture
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中文总结 AI 辅助
本文构造了秩二泊松猜想的明确反例，给出满足特定泊松括号关系的多项式，证明对应泊松自同态非自同构，反驳了泊松猜想，并构造了第四外尔代数的非自同构自同态。
中文摘要 AI 辅助
设𝒫₂=ℂ[x,q,p,z]具有由{p,x}={z,q}=1及不同生成元间其余括号均为零所确定的典范泊松括号。本文中，“秩二”指典范泊松代数标准索引下的两个典范对，故存在四个多项式生成元，且泊松张量的几何秩为四。我们给出满足{D,R}=1、{S,T}=1、{R,S}={R,T}={D,S}={D,T}=0且R=x(2−3xq)的明确多项式R,T,D,S∈ℚ[x,q,p,z]。由此，映射(x,q,p,z)↦(R,T,D,S)定义了𝒫₂的一个泊松自同态，但该自同态并非自同构。这反驳了两个典范对的泊松猜想，进而也反驳了所有数量≥2个典范对的泊松猜想。对应的𝔸⁴多项式映射保持典范辛形式，雅可比行列式为1，且其明确纤维恰好由三个点构成。证明采用多项式源坐标系，其中辛恒等式简化为三个显示的系数恒等式。一个独立附录利用相同的四个输出多项式及其哈密顿对偶，构造了第四外尔代数的一个明确非自同构自同态。
英文摘要
Let \[ {\mathcal P}_2={\mathbb C}[x,q,p,z] \] carry the canonical Poisson bracket determined by $ \{p,x\}=\{z,q\}=1 $ and by the vanishing of the other brackets between distinct generators. Here and throughout, ``rank two'' means two canonical pairs in the standard indexing of the canonical Poisson algebras; thus there are four polynomial generators and the Poisson tensor has geometric rank four. We give explicit polynomials \[ R,T,D,S\in{\mathbb Q}[x,q,p,z] \] satisfying \[ \{D,R\}=1,\qquad \{S,T\}=1, \qquad \{R,S\}=\{R,T\}=\{D,S\}=\{D,T\}=0, \] while \[ R=x(2-3xq). \] Consequently, the assignment $ (x,q,p,z)\mapsto(R,T,D,S) $ defines a Poisson endomorphism of ${\mathcal P}_2$ that is not an automorphism. This disproves the Poisson Conjecture for two canonical pairs, and hence for every number of canonical pairs at least two. The associated polynomial map of ${\mathbb A}^4$ preserves the canonical symplectic form, has Jacobian determinant one, and has an explicit fiber consisting of exactly three points. The proof uses a polynomial source coordinate system in which the symplectic identity reduces to three displayed coefficient identities. A separate appendix uses the same four output polynomials and their Hamiltonian duals to construct an explicit nonautomorphic endomorphism of the fourth Weyl algebra.
