{"id":"0ad1299e-b1ab-4337-a631-03e940b14f1a","arxiv_id":"2607.23828","paper_version":1,"verdict":"ACCEPT","confidence":"HIGH","novelty_score":7.0,"correctness_risk":"low","formal_verification":"partial","parameter_count":0,"one_line_summary":"For every k≥1, a_δn^k is non-SNP for all sufficiently large n, via an explicit even-power hole from a Dyson–Jack constant-term identity.","lead":"Every fixed power of the Vandermonde determinant eventually has a lattice point in its Newton polytope with coefficient zero. The proof settles a 2019 conjecture and gives an explicit missing monomial for every even power k≥4.","discovery_kind":"extension","skeptic_critique":{"model":"moonshotai/kimi-k3","headline":"No significant objection identified. The proof's load-bearing steps survive line-by-line re-derivation; the only soft spot is transcription risk in the standard Jack specialization behind Lemma 2.1, and the central claim admits an independent finite exact check at k=4.","rationale":"The reader's verdict (ACCEPT, HIGH confidence) and weakest-assumption identification (Lemma 2.1's reliance on Macdonald's specialization at τ=1/m) match my own read. I attempted to falsify the argument at its most sensitive points — the sign bookkeeping in R, the binomial identity (34), the majorization boundary r=2, and the specialization ε_{−1/m}(E_r) — and each re-derives cleanly. The m=1 boundary behavior (where (1−1)^0=1 would give a nonzero coefficient) is consistent with the known k=2 exception, which is good evidence the construction is not accidentally vacuous. Lemma 2.1 is standard finite-variable Jack theory and passes two independent spot checks (f=e_s in general; f=p_2 at s=2,m=1 by direct expansion). The residual risks are (a) undetected transcription error in Macdonald's (10.20) normalization and (b) the Lean formalization being partial and unpinned (no commit hash). Both are mitigated by the fact that the final claim for each fixed k is a finite exact computation; the k=4 case is small enough to serve as a complete independent certificate, and the Codex transcript reports exact modular verification at m=2,3,4. This is a correctness-risk observation, not a circularity or soundness defect, and it does not rise to the level of moving the verdict. ACCEPT stands.","tokens_in":22151,"tokens_out":5107,"duration_ms":121615,"concrete_test":"Independently of all Jack machinery, compute the k=4 instance exactly: using the edge-label model, sum (−1)^{Σt_ij}∏C(4,t_ij) over t_ij∈{0..4} on the 6 edges of K_4 subject to the exponent equations for α=(9,9,3,3), in exact integer arithmetic. Verify the sum is 0 and that α∈4P_{(3,2,1,0)} via majorization. This exercises the entire chain (membership + characteristic-zero vanishing) in the smallest even case; a nonzero result would refute Theorem 4.1 at its root. As a second step, clone the Lean repo and build to confirm the formalization covers Lemma 2.1 and Theorem 4.1 rather than only auxiliary lemmas.","verdict_should_be":"UNCHANGED","load_bearing_attack":"I re-derived the load-bearing chain and could not find a break. (i) Polytope membership: after centering by the mean mq, the majorization inequalities reduce to q(k−r) ≤ mr(k−r) for 2≤r<k, which holds since q=2m−1≤mr; the centered RHS mr(k−r) correctly equals k(rk−r(r+1)/2)−r·k(k−1)/2. (ii) The coefficient extraction (22*)–(23*): the sign product (−1)^p(−1)^{p−1}(−1)^{N−p−1}=(−1)^p (N even) yields R=−Σ(−1)^r C(N,r+1)E_rE_{s−r}, homogeneous of degree s as required for Lemma 2.1. (iii) The specialization ε_{−1/m}(E_r)=(−1)^r(r+1) follows from N/m=2, and identity (34) C(N,r+1)(r+1)(N−r−1)=N(N−1)C(N−2,r) is exact, so ε(R) collapses to N(N−1)(1−1)^{N−2}=0 precisely when m≥2 — correctly failing at m=1, consistent with k=2 being SNP up to n=4 and handled separately. (iv) Lemma 2.1 rests on Macdonald VI.10 orthogonality and (10.20); the reader flagged this as weakest, and I agree it is the only place a normalization error could hide, but it passes internal consistency checks: with f=e_s the formula returns 1 on both sides (using ε_{−1/m}(e_s)=(−1)^s(1/m)_s/s!), and a direct hand computation at s=2,m=1 with f=p_2 gives CT ratio −1, matching (−1)^2·2!/(1)_2·ε_{−1}(p_2)=−1. The LLM provenance raises baseline transcription risk, but the argument is short, parameter-free, and partially Lean-checked. No circularity: nothing about SNP is assumed in deriving the vanishing.","agreement_with_reader":"agree"},"referee_report":null,"author_rebuttal":null,"desk_editor":{"model":"grok-4.5","letter":"Punchline: the missing uniform even-power case is done. For every even k=2m≥4 they give an explicit lattice point α=(H,H,L^{k-2}) in Newton(a_δk^k) and prove its coefficient is zero by reducing to a Dyson constant term that specializes to (1−1)^{N−2}=0. Odd k and k=2 were already settled; the new content is that construction plus the write-up of Lemma 2.1.\n\nWhat works: the argument is short and parameter-free once you accept standard Macdonald material. Majorization membership is elementary after centering. The coefficient extraction to the symmetric R of degree s is written out in §6 with signs tracked. The specialization ε_{−1/m}(E_r)=(−1)^r(r+1) is clean because N/m=2, and the binomial identity that kills ε(R) is correct and correctly fails at m=1, which matches the known k=2 threshold. They ship a Lean repo and print the full Codex transcript; that is real reproducibility hygiene, not window dressing. Circularity is essentially nil—nothing about SNP is assumed in the vanishing.\n\nSoft spot, in proportion: the load-bearing external input is Macdonald’s finite-variable Jack orthogonality and specialization (VI.10). That is standard, and internal consistency checks (f=e_s, a small s=2 hand check) pass, but it is still the place a normalization slip could hide. LLM provenance raises baseline transcription risk; the authors say they checked and organized, and Lean helps, but a referee should still line-audit the Jack normalizations and the GitHub commit, not just the prose. Significance is real inside the SNP/Newton-polytope and Vandermonde-coefficient circle; it does not reorganize a larger field.\n\nWho it is for: people who care about support of structured polynomials, permutahedra, or constant-term identities. Worth a serious referee. I would bring it to reading group and would cite the theorem if I work near this area. Send it out.","headline":"They close MTY Conjecture 2.25 with an explicit even-k≥4 hole and a short Dyson–Jack vanishing argument that actually checks out.","tokens_in":24116,"tokens_out":534,"would_cite":true,"duration_ms":16882,"reading_group":"yes","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["05E05","52B20","13P15"],"pacs":[],"model":"grok-4.5","headline":"Every fixed power of the Vandermonde determinant eventually misses a lattice monomial inside its Newton polytope.","keywords":["Vandermonde determinant","Newton polytope","saturated Newton polytope","Dyson constant term","Jack polynomials","majorization","permutahedron"],"falsifier":"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).","tokens_in":23698,"feed_emoji":"∅","tokens_out":914,"duration_ms":20328,"temperature":0.7,"pith_summary":"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.","feed_headline":"Vandermonde powers eventually miss lattice monomials","feed_subtitle":"Every fixed power fails Newton-polytope saturation once the number of variables is large enough","key_machinery":"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.","core_discovery":"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.","pith_inferences":["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."],"forward_implications":["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."],"fun_headline_variants":["Even Vandermonde powers miss an explicit lattice monomial","Vandermonde^k is non-SNP past N_k via a zero-coeff lattice point","Dyson identity kills a Newton-polytope lattice monomial for k≥4","Fixed Vandermonde powers eventually fail Newton saturation","Explicit H,L point in Newton polytope of a_δ^k has coeff 0"],"cache_read_input_tokens":16512,"weakest_assumption_plain":"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.","fun_headline_variants_meta":{"raw":{"variants":["Even Vandermonde powers miss an explicit lattice monomial","Vandermonde^k is non-SNP past N_k via a zero-coeff lattice point","Dyson identity kills a Newton-polytope lattice monomial for k≥4","Fixed Vandermonde powers eventually fail Newton saturation","Explicit H,L point in Newton polytope of a_δ^k has coeff 0"]},"model":"grok-4.5","effort":"low","cost_usd":0.001767,"raw_usage":{"total_tokens":887,"prompt_tokens":780,"num_sources_used":0,"completion_tokens":87,"cost_in_usd_ticks":17668000,"prompt_tokens_details":{"text_tokens":780,"audio_tokens":0,"image_tokens":0,"cached_tokens":256},"completion_tokens_details":{"audio_tokens":0,"reasoning_tokens":20,"accepted_prediction_tokens":0,"rejected_prediction_tokens":0}},"tokens_in":780,"tokens_out":87,"duration_ms":3050,"temperature":1.0,"reasoning_tokens":20,"cache_read_input_tokens":256,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-07-30T11:07:45.219995+00:00","model_set":{"reader":"grok-4.5"},"falsifier":"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).","supporting_citations":[],"review_version":1}