RESEARCH / Focused target
The closed form of the higher-antibracket coefficients
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S3/HomologicalAlgebra/HigherAntibracketCoefficients.resultAbsent from this graph
Problem
The question
M. Manetti and G. Ricciardi, Universal Lie Formulas for Higher Antibrackets, SIGMA 12 (2016), 053, arXiv:1509.09032v3, Conjecture 2.4:
For every $n\ge 2$ the coefficients $c^n_i$ of Theorem~{\rm \ref{thm.standardform}} are given by the formula
$c_i^n=\frac{\displaystyle (-1)^{n} \prod_{j=2}^{i}\dfrac{n(n-1)-(j-1)(j-2)}{2}}{\displaystyle \sum_{h=2}^n h\left(\prod_{j=2}^{h}\dfrac{n(n-1)-(j-1)(j-2)}{2}\right)\left(\prod_{j=h}^{n-1}\dfrac{(1-j)(j+2)}{2}\right)}$,
where every empty product is intended to be equal to $1$. Moreover $(-1)^{n}c^n_i>0$ for every $n\ge i\ge 1$.
The coefficients of Theorem 2.3 are the unique rationals with $\Phi^{n+1}_f=(c_1^n\rho_1^n+c_2^n\rho_1^{n-2}\rho_2+\dots+c_n^n\rho_n)f$ for every linear operator $f$ of every graded commutative algebra. The paper proves Theorem 2.3 from Theorem 6.4, the same identity $\Phi^{n+1}=(c_1^n\rho_1^n+\dots+c_n^n\rho_n)\Phi^1$ for the operators $\Phi^{m,i}$ and $\rho_k$ on $\mathbb{Q}[x]$, through the $\mathfrak g$-module morphism of Lemma 6.3; uniqueness in Theorem 2.3 identifies the two sequences. The authors verified the formula for $n\le12$.
Motivation
Our foothold
Higher Koszul brackets (higher antibrackets) measure the failure of an
operator to be a derivation and govern the BV formalism. Theorem 2.3
reduces every bracket $\Phi^{n+1}_f$ to iterated commutators with the
multiplication operators, with only the monomials
$\rho_1^{n-i}\rho_i$; the conjecture makes these universal Lie formulas
explicit. The frozen declaration
D5/S3/HomologicalAlgebra/HigherAntibracketCoefficients.result proves it.
Gap
Missing bridges
Issue #12104 classifies the conjecture as Tier 1 and records the literature checks before any Lean:
- arXiv v3, the journal version, still states the conjecture;
- OEIS A272688 (revision 8, 2025-11-05) records the diagonal values $x_n^n=(-1)^n n!,c_n^n$ and calls the formula conjectural;
- the citing works arXiv:1512.05480, 2012.14812, 2206.02048, 2407.16348, 2409.09809, 2601.19698 and 1703.00290 do not address the coefficients;
- an adversarial search of equivalent formulations (Koszul brackets, higher derived brackets, loop homotopy algebras, Bessel and Legendre coefficient tables) and of the authors' later work found no settlement.
These are orchestrator-reported readings, not-found-in-searched-scope;
they do not establish exhaustive worldwide novelty.
Route
Proposed approach
Record the coefficient of $x^d/d!$ in $\Psi(x^s)$ as the lattice point $(d,s)$.
- From the definitions, $\rho_k(d,s)=\alpha_k(d),(d+k,s)-\binom{s+k}{k+1}(d,s+k)$ with $\alpha_k(d)=\binom{d+k}{k}-\binom{d+k}{k+1}$, and $\rho_i\Phi^1=(i,1)-(0,i+1)$.
- The two moves of $\rho_1$ commute, so $\rho_1^t$ of a point is a binomial sum over paths, weighted by products $A$ of $\alpha_1(u)=(u+1)(2-u)/2$ and $B$ of $-\binom{u+1}{2}$.
- The identity becomes $n+1$ linear equations, one for each point $(d,n+1-d)$. Since $\alpha_1(2)=0$, the equations with $d\ge3$ involve only $c_3,\dots,c_d$ and are triangular; the equations $d=0,1$ then fix $c_1,c_2$. This proves uniqueness.
- For $n\ge3$ put
$c_i=(-1)^n\dfrac{(n+i-2)!}{(n-i)!,2^{i-1}}\cdot\dfrac{2^{n-1}}{(n-2)!,(n+1)!}$.
Every equation reduces to
$\sum_{i}(-1)^{M-i}\binom{M}{i}\binom{i+a}{b}=\binom{a}{b-M}$ for
$M\le b$ and $0$ for $M>b$, the iterated forward difference of
$x\mapsto\binom{x}{b}$ given by Mathlib's
fwdDiff_iter_choose. - The printed product equals $(n+i-2)!/((n-i)!,2^{i-1})$, and by the same sum the printed denominator equals $(n-2)!(n+1)!/2^{n-1}$ for $n\ge3$.
- For $n=2$ the equations give $c_1=c_2=1/2$, which is the printed value.
- For $n=1$ the two equations give $c_1^1=-1$.
- The sign clause for $n\ge i\ge1$ follows from the factorial form and from $c_1^1=-1$.
Falsifier
What would falsify this route
The kernel-checked result states three parts: for every $n\ge2$ a
sequence satisfies the identity of Theorem 6.4 if and only if it agrees with
the printed formula on $1\le i\le n$; for $n=1$ it satisfies the identity if
and only if $c_1=-1$; and for every $n\ge1$ every solution has
$(-1)^n c_i>0$ for $1\le i\le n$. A different normalization of $\rho_k$, of $\Phi^{m,i}$, or of
the sign of $\Phi^m$ changes the question.
Evidence
Evidence to collect
An independent SymPy recomputation from the raw definitions (operators on $x^0,\dots,x^{n+1}$, exact rationals) finds a unique solution for each $n=2,\dots,13$ and no mismatch with the printed formula; it reproduces the paper's Table 1 (for example $4/5,24/5,24,72$ at $n=4$) and OEIS A272688, and a perturbed target $\Phi^6+\Phi^{6,1}$ has no solution (issue #12104).
The canonical source is
D5/S3/HomologicalAlgebra/HigherAntibracketCoefficients.lean. Its public
declarations are phi, koszul, rho, formula, claim and result.
The frozen module state has statement identity sha256:6998dc0e291c064a2c343c9a4821335d47a6c2dc056684232f87b75b7c720aad. The
result declaration has statement identity sha256:bee22258d60b9bfd38ae17b3a86112a02384da8fdd4fa30347e9519cd18b3b73. The Freeze
event is sha256:1f3e9e157898a2ab02b0c9f4da7ff57cf30523b12f743fc7376af7647af01c3c. It has no project-level frozen prerequisites
(pinned Mathlib only). The proof uses only the standard axioms propext,
Classical.choice and Quot.sound; no sorry, native_decide, or new
axiom.
Triage
Scope assessment
Tier 1 published conjecture; resolution Proved by
D5/S3/HomologicalAlgebra/HigherAntibracketCoefficients.result.
proof_shape: content; admission_basis: open-problem-resolution (issue
#12104). Utility kind none (a general theorem for all $n$).
What the settlement shows
- Proved by
result: for every $n\ge2$ the coefficients are the unique solution of the identity and equal the printed formula; for $n=1$ the unique solution is $c_1^1=-1$; and $(-1)^n c_i^n>0$ for every $n\ge i\ge1$. - Proved inside the proof of
result: for $n\ge3$, $c_i^n=(-1)^n,2^{n-i}\dfrac{(n+i-2)!}{(n-2)!,(n+1)!,(n-i)!}$, and the printed denominator equals $(n-2)!(n+1)!/2^{n-1}$; it equals $2$ for $n=2$. - Follows from the closed form (not stated in Lean): consecutive coefficients satisfy $c_{i+1}^n/c_i^n=(n-i)(n+i-1)/2$ for $n\ge3$ and $1\le i<n$.
- Mechanism: $\alpha_1(2)=0$, so $\rho_1$ cannot move a coordinate from $d=2$ to $d=3$. The paths from $(0,i+1)$ stop at $d\le2$, the equations with $d\ge3$ see only the paths from $(i,1)$ with $i\ge3$, and the coefficient system is triangular. All remaining sums are iterated forward differences $\sum_i(-1)^{M-i}\binom Mi\binom{i+a}b$, equal to $\binom a{b-M}$ for $M\le b$ and to $0$ for $M>b$ (Route, step 4); the printed denominator uses the case $M=b$, where the value is $1$.
- Follows from the proved closed form: the diagonal of OEIS A272688 is $x_n^n=(-1)^n n!,c_n^n=\dfrac{n!,(2n-2)!}{(n-2)!,(n+1)!}$ for $n\ge3$ (and $1$ at $n=2$).
- Unchanged: the paper's Theorems 2.1, 2.3 and 6.4 do not depend on the conjecture; with it, every higher antibracket $\Phi^{n+1}_f=\sum_i c_i^n\rho_1^{n-i}\rho_i f$ has explicit rational coefficients.
- Open here: closed forms for normal forms with other orderings of the $\rho_k$ (for example $\rho_i\rho_1^{n-i}$), and for the noncommutative brackets of the paper's Remark 5.6.
ASSUMED-UNVERIFIED
Unverified assumptions
The literature checks do not establish worldwide priority or absence of an independent answer; the full texts of Manetti's Lie Methods in Deformation Theory (2022) and of Dijoux (2020) were not accessible. The Lean kernel verifies the encoded statement and its axiom closure; its correspondence to the paper, including the reading of the coefficients of Theorem 2.3 as the unique solution of Theorem 6.4 and the statement inside $\mathrm{End}_{\mathbb Q}(\mathbb Q[x])$ rather than the subalgebra $\mathfrak a$, is checked by reading the source and the definitions.