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Powers of the Vandermonde determinant are eventually non-SNP

T0 review · reviewed 2026-07-30 · grok-4.5

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Pith's one-line read Every fixed power of the Vandermonde determinant eventually misses a lattice monomial inside its Newton polytope.

desk verdict They close MTY Conjecture 2.25 with an explicit even-k≥4 hole and a short Dyson–Jack vanishing argument that actually checks out. read the letter →

arxiv 2607.23828 v1 pith:4GSQC7FA submitted 2026-07-26 math.CO

Thien Le , Melanie Weber This is my paper · ORCID
classification math.CO MSC 05E0552B2013P15
keywords VandermondedeterminantNewtonpolytopesaturatedDysonconstanttermJackpolynomialsmajorizationpermutahedron
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

A polynomial has a saturated Newton polytope when every lattice point inside the polytope appears with a nonzero coefficient. This paper proves that for every fixed positive integer power k, the k-th power of the Vandermonde determinant fails that property once the number of variables is large enough. Odd powers fail already in three variables by skew-symmetry, the square was already known to fail from five variables onward, and the new work supplies an explicit missing exponent for every even power k at least 4. The missing coefficient is forced by a Dyson-type constant-term identity obtained from Jack polynomial orthogonality. The result gives a uniform asymptotic picture of support gaps for all fixed powers and shows that geometric membership alone cannot recover the monomial support in high dimension.

What carries the argument

A Dyson constant-term lemma: the ratio of constant terms of D_{s,m} f / e_s over D_{s,m} equals a rising-factorial multiple of the specialization ε_{−1/m}(f). Applied to an explicit degree-s symmetric polynomial R built from elementary generating functions, the specialization vanishes, forcing the target monomial coefficient to zero.

What would settle it

For the smallest even case k = 4 (so n = 4, H = 9, L = 3), expand a_δ4^4 by computer algebra and check whether the coefficient of x1^9 x2^9 x3^3 x4^3 is exactly zero while the exponent still majorizes 4·(3,2,1,0).

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Extended reading notes

Core claim

For every integer k ≥ 1 there exists N_k such that the k-th power of the Vandermonde determinant in n variables is non-SNP for all n ≥ N_k. Concretely, when k = 2m ≥ 4 the lattice point with two large equal coordinates H = (k−1)^2 and the remaining coordinates equal to L = (m−1)(k−1) lies in the Newton polytope yet has coefficient zero.

Load-bearing premise

The argument stands or falls on the claim that Jack specialization at the negative parameter −1/m kills every Jack summand except the elementary one, so the constant-term ratio really equals that specialization of R.

Editorial extensions

If this is right

  • For every fixed k the support of the Vandermonde power is eventually a proper subset of the lattice points of its permutahedron multiple.
  • Geometric membership tests alone cannot decide coefficient nonvanishing for these powers in high dimension.
  • The same explicit two-level exponents and Jack–Dyson reduction give concrete missing monomials once n reaches k (even k ≥ 4), 5 (k = 2), or 3 (odd k).
  • Propagation from the seed dimension immediately yields non-SNP for all larger numbers of variables.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • Similar Jack-specialization cancellations may produce missing lattice points for other classical discriminants or resultants whose Newton polytopes are permutahedra or their Minkowski sums.
  • The explicit even-power holes suggest a systematic search for the minimal N_k by testing only the two-block majorization boundary rather than the full support.
  • Because the vanishing reduces to a one-variable negative-binomial identity after specialization, the same pattern may extend to q-Vandermonde or Macdonald analogues.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Circularity Check

0 steps flagged · score 0.0 of 10

No significant circularity: vanishing is an explicit elementary evaluation after standard external Jack/Macdonald input, not a restatement of the claim.

full rationale

The load-bearing chain is: (i) Newton(a_δn^k)=kP(δn) and Rado majorization (external); (ii) an explicit even-power lattice point α whose membership is checked by direct inequalities; (iii) coefficient extraction reducing [x^α] to a Dyson constant term of a degree-s form R; (iv) Lemma 2.1, proved in-paper from Macdonald’s finite-variable Jack orthogonality and specialization formula (Mac95), converting that CT ratio into ε_{-1/m}(R); (v) the elementary identity ε_{-1/m}(R)=−(−1)^s N(N−1)(1−1)^{N−2}=0 for N≥4. Nothing in (i)–(v) defines the target non-SNP statement in terms of itself, fits a parameter to the claim, or imports a uniqueness/ansatz result from the present authors. Odd k and k=2 are handled by alternation and an external MTY citation plus propagation; those are independent external inputs. AI provenance and the Lean repo are verification/process notes, not load-bearing self-citations of the conjecture. The derivation is therefore self-contained against its stated external benchmarks.

Assumptions & free parameters 0 free parameters · 6 assumptions · 2 invented entities

Pure mathematics proof. Load-bearing inputs are standard theorems from Macdonald’s Jack theory, Rado’s majorization characterization of permutahedra, Newton-polytope Minkowski sums, and the MTY propagation lemma. No empirical fits. The only paper-specific objects are the explicit exponent α and the auxiliary symmetric functions E_r, R used to name intermediate coefficients.

assumptions (6)
  • standard math Rado’s theorem: α ∈ kP(n−1,…,0) iff α ⪯ k(n−1,…,0) (majorization).
    Used in §2.1 and Part 2 of Theorem 4.1 to place α in the Newton polytope.
  • standard math Newton(fg)=Newton(f)+Newton(g); Newton(a_δn)=P(δn).
    Gives Newton(a_δn^k)=kP(δn); cited via Sturmfels/Postnikov.
  • standard math Finite-variable Jack polynomials P_λ^{(τ)} are orthogonal for the CT scalar product ⟨·,·⟩'_{s,τ} and P_{(1^s)}^{(τ)}=e_s.
    Mac95 VI.10; core of Lemma 2.1 proof in §5.
  • standard math Macdonald specialization formula ε_{−τ}(P_λ^{(τ)}) = ∏_u (−τ+τa'(u)−ℓ'(u))/(τa(u)+ℓ(u)+1).
    Mac95 VI.10 (10.20); forces vanishing for λ≠(1^s) and evaluates the (1^s) case.
  • domain assumption MTY19 Lemma 2.24: non-SNP propagates from n to n+1 by appending kn on the new variable.
    Invoked to upgrade the seed n=N_k to all larger dimensions; authors sketch the extremal x_{n+1} argument in the appendix proof.
  • standard math CT(D_{s,m})>0 via nonnegative torus integrand strictly positive on a positive-measure set.
    Ensures the denominator in Lemma 2.1 is nonzero (§5, display (41)).
invented entities (2)
  • Even-power witness α=(H,H,L^{s}) with H=q^2, L=(m−1)q, N=2m, s=N−2, q=N−1 independent evidence
    purpose: Explicit lattice point claimed to lie in Newton(a_δN^N) with vanishing coefficient for every even k=N≥4.
    Paper-specific combinatorial construction; membership and vanishing are proved, not postulated.
  • Auxiliary symmetric functions E_r and R=−∑ (−1)^r C(N,r+1) E_r E_{s−r} independent evidence
    purpose: Package the [x1^H x2^H] extraction so the remaining coefficient is a single Dyson CT of R/e_s.
    Definitional bookkeeping inside the proof; not a physical or ontological posit.

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Pith. "Pith review of Powers of the Vandermonde determinant are eventually non-SNP." pith.science (2026). https://pith.science/paper/4GSQC7FA

@misc{pith2026260723828,
  author       = {Pith},
  title        = {Pith review of: Powers of the Vandermonde determinant are eventually non-SNP},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/4GSQC7FA}},
  note         = {Machine review of arXiv:2607.23828}
}
abstract

We prove a conjecture of Monical, Tokcan, and Yong that every fixed positive power of the Vandermonde determinant is non-SNP in all sufficiently many variables, where a polynomial is non-SNP if there is a lattice point in its Newton polytope that does not appear with nonzero coefficient. This means our result proves that for every even power $k\geq4$, there is always such a missing lattice monomial in large enough dimensions. The odd case follows from alternation, and the quadratic case was previously known. For every even power $k\geq4$, we exhibit an explicit lattice point in the Newton polytope of $a_{\delta_k}^k$ whose coefficient vanishes. The vanishing is obtained from a Dyson constant-term identity, proved using the finite-variable Jack scalar product and Macdonald's specialization formula. The key even-power construction and proof strategy arose from prompting with OpenAI Codex (GPT Sol 5.6 Extra High), a large language model; the complete transcript appears in the appendix. The authors subsequently checked and organized the argument. The accompanying Lean formalization is available at https://github.com/steven-le-thien/vandermonde-snp.

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Works this paper leans on

37 extracted references · 1 linked inside Pith

  1. [1]

    Assume that a complete affirmative proof exists

  2. [2]

    Use as many as 64 agents dynamically

  3. [3]

    Begin with independent, genuinely different approaches

  4. [4]

    Keep a registry of approach families and their exact gaps

  5. [5]

    Prevent early convergence on an attractive idea

  6. [6]

    Require concrete lemmas, constructions, equations, or counterexamplesnot progress reports

  7. [7]

    Mark theorem-strength gaps as blocked

  8. [8]

    Assign independent adversarial agents to attack every proposed proof

Show all 37 references
  1. [9]

    Check known failure modes and circular reductions explicitly

  2. [10]

    Continue launching new rounds after failures

  3. [11]

    Spend at least eight hours before considering giving up

  4. [12]

    29 THE PROBLEM For n 1, define _n = (n1,n2,...,1,0) and the Vandermonde polynomial a_{_n}(x_1,...,x_n) = _{1i<jn}(x_ix_j)

    Return only a complete proof that survives audit. 29 THE PROBLEM For n 1, define _n = (n1,n2,...,1,0) and the Vandermonde polynomial a_{_n}(x_1,...,x_n) = _{1i<jn}(x_ix_j). A polynomial f has a saturated Newton polytope, abbreviated SNP, when supp(f) = Newton(f) Z^n. Prove: Fo...

    arXiv

  5. [13]

    Newton polytope: Newton(a_{_n}^k) = kP_{_n}, where P_{_n} is the permutahedron generated by the permutations of _n. 30

  6. [14]

    Majorization test: If is rearranged into weakly decreasing order, then kP_{_n} Z^n exactly when is integral, _{i=1}^n _i = k n(n1)/2, and, for every 1 r < n, _{i=1}^r _i k(rn r(r+1)/2)

  7. [15]

    Propagation: Check every detail of Lemma 2.24, including why the coefficient with new exponent kn in x_{n+1} can only arise by choosing the x_{n+1}^k term from every new factor

  8. [16]

    Reproduce these cases by exact arithmetic; do not trust undocumented computations

    Computed cases reported in the source: N_2 = 5, N_4 = 4, N_6 = 4, N_8 = 3, and N_{2j1}=3 was checked there for 1j4. Reproduce these cases by exact arithmetic; do not trust undocumented computations. MANDATORY ODD-POWER REDUCTION Immediately verify the following observation. Fo...

  9. [17]

    It proves only finitely many values of k

  10. [18]

    It proves only that a coefficient vanishes modulo p

  11. [19]

    It relies on floating-point cancellation

  12. [20]

    It finds a zero coefficient without proving the exponent lies in kP_{_n}

  13. [21]

    It proves polytope membership by assuming full monomial support

  14. [22]

    It proves a Schur coefficient is zero and silently treats this as a monomial coefficient being zero. 33

  15. [23]

    It constructs an involution without proving it is defined everywhere, fixed-point-free, sign-reversing, and weight-preserving

  16. [24]

    It invokes a constant-term identity outside its hypotheses

  17. [25]

    It divides by a quantity that may vanish for some k

  18. [26]

    It leaves an uncovered congruence class of even k

  19. [27]

    It invokes Lemma 2.24 without checking the extremal x_{n+1}-coefficient argument

  20. [28]

    It confuses finding some threshold N_k with proving that N_k is minimal

  21. [29]

    It uses a recurrence without proving all boundary values and that the recurrence reaches every required parameter

  22. [30]

    It assumes that a candidate exponent is in the Newton polytope merely because its coordinates have the correct total sum

  23. [31]

    It hides the original conjecture inside phrases such as generic cancellation, standard representation theory, or a routine coefficient calculation. ADVERSARIAL REVIEW Every proposed final proof must be attacked by independent agents assigned to: verify every quantifier; recomp...

  24. [32]

    A finite integer n(k)

  25. [33]

    An explicit integer vector (k), possibly defined casewise

  26. [34]

    A proof that (k) kP_{_{n(k)}}

  27. [35]

    A characteristic-zero proof that [x^{(k)}]a_{_{n(k)}}^k=0

  28. [36]

    A rigorous use of Lemma 2.24 showing non-SNP for every nn(k)

  29. [37]

    The final response must be a self-contained mathematical proof

    A check that all positive integers k are covered. The final response must be a self-contained mathematical proof. Include definitions and every essential lemma. Computations may guide discovery but may not replace uniform arguments. Do not return a research diary, registry, pa...

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