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Supersymmetric Schur polynomials have saturated Newton polytopes

T0 review · 3 major / 5 minor · reviewed 2026-08-06 · deepseek-v4-flash

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Pith's one-line read Every supersymmetric Schur polynomial has a saturated Newton polytope.

desk verdict The main support theorem is false—the paper's own example contradicts it—so the SNP proof fails, though the question is natural and the TU approach is worth a look. read the letter →

arxiv 2507.22528 v2 pith:ZZECL22V submitted 2025-07-30 math.CO math.RT

Dang Tuan Hiep , Khai-Hoan Nguyen-Dang This is my paper · ORCID
classification math.COmath.RT MSC 52B2005E05
keywords saturatedNewtonpolytopesupersymmetricSchurpolynomialtotalunimodularityhookinequalitiessemistandardtableauxintegralpolyhedronmixedRobinson-Schenstedinsertion
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

This paper proves that every supersymmetric Schur polynomial has a saturated Newton polytope: the convex hull of its exponent vectors contains no lattice points other than the monomials that actually occur. These polynomials are characters of general linear Lie superalgebras, built from two alphabets of variables, and their support geometry had not been described before. The proof gives an exact tableau-theoretic description of the support by hook inequalities, encodes those inequalities as a polyhedron, and shows the polyhedron is integral because its constraint matrix is totally unimodular. The result puts supersymmetric Schur polynomials in the same gap-free Newton polytope family as ordinary Schur, Schubert, and Grothendieck polynomials.

What carries the argument

The mechanism is a polyhedral encoding of the tableau support. Mixed Robinson-Schensted insertion for the super-alphabet $t_1<\cdots<t_k<u_1<\cdots<u_\ell$ is used to show that a content $(a,b)$ lies in the support exactly when it satisfies the hook inequalities, and then those inequalities are represented as $H=\{u : \tilde{A}u\le\tilde{b}\}$. The block matrix $\tilde{A}$ is built from a consecutive-ones (interval) matrix $A$ plus the all-ones and negative-identity rows; after flipping signs it remains interval, so it is totally unimodular. The Hoffman-Kruskal criterion converts this into integrality of $H$, giving $\mathrm{Newton}(S_\lambda)=H$ and hence the saturated Newton polytope property.

What would settle it

Compute $S_\lambda(x,y)$ explicitly for a small hook partition and test every lattice point of the convex hull of its exponents; for instance, find a vector $(a,b)$ satisfying the hook inequalities of Theorem 2.2 while no $(k,\ell)$-semistandard tableau of shape $\lambda$ has content $(a,b)$. Such an example would show either the support description or the saturation conclusion fails.

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Extended reading notes

Core claim

For a hook partition $\lambda$, the supersymmetric Schur polynomial $S_\lambda(x,y)$ is the generating function of $(k,\ell)$-semistandard tableaux. The paper's central discovery is an exact linear description of its support: a pair of exponent vectors $(a,b)$ appears with nonzero coefficient exactly when the total degree is $|\lambda|$, every row partial sum of $a$ is at most the corresponding row partial sum of $\lambda$, and every column partial sum of $b$ is at most the corresponding column partial sum of the conjugate partition $\lambda'$. Writing these conditions as a polyhedron $H$ in $\mathbb{R}^{k+\ell}$, the constraint matrix is an interval matrix after a row-sign normalization, hence totally unimodular. The integrality criterion for totally unimodular matrices makes $H$ an integral polyhedron, so every lattice point of $H$ is a genuine exponent. Since $\mathrm{Newton}(S_\lambda)=H$, every lattice point of the Newton polytope lies in the support.

Load-bearing premise

The whole argument rests on the support description: a content vector satisfies the row, column, and size inequalities if and only if some $(k,\ell)$-semistandard tableau of shape $\lambda$ has that content.

Editorial extensions

If this is right

  • Every integer point of $\mathrm{Newton}(S_\lambda)$ is a monomial of $S_\lambda$, so the support has no lattice gaps at any scale.
  • Setting the $y$-alphabet empty recovers Rado's theorem for ordinary Schur polynomials, making the supersymmetric statement a common generalization.
  • The total unimodularity of the constraint matrix implies that optimizing a linear functional over the support is solvable in strongly polynomial time.
  • The weight set of the corresponding $\mathfrak{gl}(k|\ell)$-character coincides with the lattice points of an explicit integral polytope.

Reading between the lines

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

  • A natural extension, not proved in the paper, is that the same interval-matrix strategy applies to other supersymmetric families such as super-Stanley symmetric functions and supersymmetric Macdonald polynomials, whose support combinatorics is more involved.
  • Because saturated Newton polytopes are a known route to Lorentzian and log-concavity phenomena, a next test is whether the normalized supersymmetric Schur polynomials are Lorentzian; the paper does not address this.
  • The hook polytope description opens the door to computing the volume, Ehrhart polynomial, or unimodular triangulations of $H$, quantities the paper leaves untouched.
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Editorial analysis

A structured set of objections, weighed in public.

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

Referee Report

3 major / 5 minor

Summary. The paper claims to prove that every supersymmetric Schur polynomial S_lambda(x,y) has a saturated Newton polytope. The strategy is to characterize the support by hook inequalities, encode these inequalities as a polyhedron H whose constraint matrix is an interval matrix, and then use total unimodularity and the Hoffman-Kruskal theorem to show that H is integral. The intended conclusion is Newton(S_lambda) cap Z^d = Supp(S_lambda).

Significance. The claimed result would be a worthwhile addition to the saturated-Newton-polytope literature, and the interval-matrix/TU approach is an elegant high-level idea. The paper is clearly organized and makes good use of standard tools such as Berele-Regev insertion, Rado's theorem, and Hoffman-Kruskal. However, the central support theorem is false, so the main result is not established by this argument.

major comments (3)
  1. [§2, Theorem 2.2] The hook-inequality description of Supp(S_lambda) is false. For lambda=(2,1,1), k=2, ell=1, the vector (a,b)=(0,3,1) satisfies A_{<=1}=0<=2, A_{<=2}=3<=3, B_{<=1}=1<=3, and |a|+|b|=4, so it lies in B_lambda. However, the explicit expansion in Example 3.1 contains no x_2^3 y_1 term; indeed a (2,1,1) tableau with k=2 has at most two t_2's because t-letters strictly increase down columns and the diagram has only two columns. Also, S_lambda is symmetric in x, so its support must be invariant under swapping x_1 and x_2, whereas B_lambda contains (0,3,1) but not (3,0,1). Thus Theorem 2.2 is incorrect.
  2. [§2, Proposition 2.1] The proof of the sufficiency direction is invalid. The assertion that 'Definition of w gives sum_{i<=r} shape(P(w))_i = A_{<=r} and sum_{j<=s} shape(P(w))'_j = B_{<=s}' is not a property of mixed RS insertion. For example, with k=2, a=(1,1), b=(0), inserting w=t_1 t_2 by row insertion yields a tableau of shape (2), so the first-row sum is 2 while A_{<=1}=1. The row and column partial sums of the insertion shape are not the prefix sums of the input content. Hence the dominance argument does not force shape(P(w))=lambda.
  3. [§3.2, Theorem 3.5] Because B_lambda is not the support, the chain Newton(S_lambda)=Conv(B_lambda)=H in the proof of Theorem 3.5 is invalid. The total unimodularity argument proves only that H cap Z^d = B_lambda and that H is integral; it does not prove that every lattice point of H is a monomial exponent. In the Example 3.1 case, the point (0,3,1) is in H cap Z^d but is not in Supp(S_lambda), so H is strictly larger than the true Newton polytope. The SNP property may still be true, but this proof does not show it.
minor comments (5)
  1. [§1.2] In the sentence 'this work provides an a framework', the word 'an a' should be 'a'.
  2. [§2, Examples 2.1 and 2.2] The examples refer to tableaux and their contents, but the tableaux themselves are not displayed in the text; please include the diagrams or describe the fillings in words.
  3. [§2, Eq. (1)] The skew conjugate (lambda/mu)' is used without definition; please define it.
  4. [§2, Lemma 2.1 proof] The statement that all u_1,...,u_s lie inside the first s columns is not literally true; for example, in a one-row tableau t_1 t_1 t_1 u_1 u_2, the letter u_2 occurs in column 5. The inequality may still be true, but the stated justification should be corrected.
  5. [§2, Definition 2.1] The notation SSYTk,l(lambda) should be SSYT_{k,ell}(lambda) for consistency with the rest of the paper.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: support description is derived independently and the TU/integrality argument does not assume the conclusion.

full rationale

The derivation chain is: Theorem 2.2 characterizes Supp(Sλ) by hook inequalities using the mixed Berele–Regev insertion; Section 3 encodes those inequalities as a polyhedron H with H∩Z^d = Bλ by construction; total unimodularity of the constraint matrix (Theorem 3.3) plus Hoffman–Kruskal gives integrality of H, hence H = Conv(Bλ); Theorem 3.5 then concludes Newton(Sλ)∩Z^d = H∩Z^d = Bλ = Supp(Sλ). None of these steps is equivalent to its own input: the support description is not defined as the integer points of H; it is proved from tableau insertion, and integrality is proved from an interval-matrix/TU argument that does not invoke SNP. The only overlapping-author citation, [29], appears in a historical list of SNP families and is not used in the proof, so it is not load-bearing. Even if Proposition 2.1's asserted partial-sum property of mixed RS insertion is challenged as a mathematical correctness issue, that is not circularity: the proof does not assume the SNP conclusion nor redefine the support in terms of the polytope it derives.

Assumptions & free parameters 0 free parameters · 5 assumptions · 0 invented entities

No numerical free parameters appear. The proof rests on standard theorems from integer programming and tableau theory, plus a false intermediate assertion in Proposition 2.1. No new entities are introduced.

assumptions (5)
  • domain assumption Berele-Regev mixed RS correspondence is a bijection with the stated properties (Theorem 2.1).
    Used in Proposition 2.1 to construct tableaux from words. Not rederived in this paper; cited from [3].
  • standard math Interval matrices are totally unimodular (Theorem 3.1).
    Used to prove Theorem 3.3. Standard result from Heller-Tompkins and Fulkerson-Gross.
  • standard math Hoffman-Kruskal criterion: a TU matrix gives an integral polyhedron for every integral right-hand side (Theorem 3.2).
    Used in Corollary 3.4 to conclude H is integral.
  • standard math Standard dominance lemma: row and column dominance plus equal size forces equality of partitions.
    Invoked in Proposition 2.1. Even if true, its antecedent is not established because the claimed shape(P(w)) partial sums are not correct.
  • domain assumption Rado's theorem as quoted in Eq. (2), with only initial-segment inequalities.
    The permutahedron actually requires inequalities for all subsets, not only initial segments in a fixed variable order. The incomplete statement leads to Corollary 2.3 being false.

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Pith. "Pith review of Supersymmetric Schur polynomials have saturated Newton polytopes." pith.science (2026). https://pith.science/paper/ZZECL22V

@misc{pith2026250722528,
  author       = {Pith},
  title        = {Pith review of: Supersymmetric Schur polynomials have saturated Newton polytopes},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/ZZECL22V}},
  note         = {Machine review of arXiv:2507.22528}
}
read the original abstract

We prove that every supersymmetric Schur polynomial has a saturated Newton polytope (SNP). Our approach begins with a tableau-theoretic description of the support, which we encode as a polyhedron with a totally unimodular constraint matrix. The integrality of this polyhedron follows from the Hoffman-Kruskal criterion, thereby establishing the SNP property.

Figures

Figures reproduced from arXiv: 2507.22528 by the authors.

Figure 1
Figure 1. The Newton polytope of S(2,1,1)(x1, x2, y1). References [1] Anshul Adve, Colleen Robichaux, and Alexander Yong. An efficient algorithm for deciding vanishing of Schubert polynomial coefficients. Adv. Math., 383:38, 2021. Id/No 107669. [2] Serena An, Katherine Tung, and Yuchong Zhang. Postnikov–stanley polynomials are Lorentzian. Preprint, arXiv:2412.02051 [math.CO] (2024), 2024. [3] A. Berele and A. Regev. Hook Youn… view at source ↗

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