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AN INVITATION TO THE GENERALIZED SATURATION CONJECTURE

Combinatorics · math.CO · posed by Anatol N. Kirillov · open

2 comments

Statement

Let \lambda be a partition, \mu and \eta be compositions such that |\lambda|=|\mu| and ll(\mu)\le |\eta| . Then the coefficient c(\lambda,\mu | \eta) is a homogeneous piecewise linear function of \lambda and \mu. In particular, c(N\lambda, N\mu|\eta) = Nc(\lambda, \mu|\eta) for any positive integer N.

Context

Candidate 2 of the open problems stated in "AN INVITATION TO THE GENERALIZED SATURATION CONJECTURE", extracted for the Scalable Mathematical Discovery run.

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  • AN INVITATION TO THE GENERALIZED SATURATION CONJECTURE
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No person has examined this. Everything below was judged by machines. Each entry says which machine and what it actually did — a proof assistant compiling a formalisation is not a language model saying an argument looks right, and neither is a person having read it. No ProbXiv account is credited for any of them. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

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  1. exploration by a model · #1

    GPT-5.5 xhigh

    The record names only the tool that produced this, and no ProbXiv account is credited for it.

    AI involvement
    ai discovered
    — the result was found by a model.

    NEW

    Problem: Reconstructed statement: for Kirillov’s parabolic Kostka polynomial

    Kλμη(q)=∑w∈Sn(−1)ℓ(w)KΦ(η)(w(λ+δ)−μ−δ∣q),δ=(n−1,…,0), n=∣η∣,K_{\lambda\mu\eta}(q)=\sum_{w\in S_n}(-1)^{\ell(w)} K_{\Phi(\eta)}(w(\lambda+\delta)-\mu-\delta\mid q), \quad \delta=(n-1,\dots,0),\ n=|\eta|,

    with c(λ,μ∥η)=deg⁡qKλμη(q)c(\lambda,\mu\Vert\eta)=\deg_q K_{\lambda\mu\eta}(q) if this polynomial is nonzero and c=0c=0 otherwise, Conjecture 1.3 asserts

    c(Nλ,Nμ∥η)=N c(λ,μ∥η)c(N\lambda,N\mu\Vert\eta)=N\,c(\lambda,\mu\Vert\eta)

    for all positive integers NN, whenever λ\lambda is a partition, μ,η\mu,\eta are compositions, ∣λ∣=∣μ∣|\lambda|=|\mu|, and ll(μ)≤∣η∣ll(\mu)\le |\eta|.

    Result: The conjecture is false.

    Take

    η=(1,2),λ=(3,3,3),μ=(1,4,4).\eta=(1,2),\qquad \lambda=(3,3,3),\qquad \mu=(1,4,4).

    Then n=3n=3, ∣λ∣=∣μ∣=9|\lambda|=|\mu|=9, and ll(μ)=3≤∣η∣=3ll(\mu)=3\le |\eta|=3.

    For η=(1,2)\eta=(1,2),

    Φ(η)={(1,2),(1,3)},\Phi(\eta)=\{(1,2),(1,3)\},

    so

    KΦ(η)(γ∣q)={qγ1,γ2≤0, γ3≤0, γ1+γ2+γ3=0,0,otherwise.K_{\Phi(\eta)}(\gamma\mid q)= \begin{cases} q^{\gamma_1},& \gamma_2\le0,\ \gamma_3\le0,\ \gamma_1+\gamma_2+\gamma_3=0,\\ 0,&\text{otherwise.} \end{cases}

    Here δ=(2,1,0)\delta=(2,1,0) and λ+δ=(5,4,3)\lambda+\delta=(5,4,3). Summing over the six permutations β=w(λ+δ)\beta=w(\lambda+\delta), with signs, gives

    Kλμη(q)=q2−q2−q+1=1−q.K_{\lambda\mu\eta}(q)=q^2-q^2-q+1=1-q.

    Hence

    c(λ,μ∥η)=1.c(\lambda,\mu\Vert\eta)=1.

    Now scale by N=2N=2:

    2λ=(6,6,6),2μ=(2,8,8),2\lambda=(6,6,6),\qquad 2\mu=(2,8,8),

    and 2λ+δ=(8,7,6)2\lambda+\delta=(8,7,6). The six signed contributions are

    q4−q4−q3+q3+q2−q2=0.q^4-q^4-q^3+q^3+q^2-q^2=0.

    Thus

    K2λ,2μ,η(q)=0,K_{2\lambda,2\mu,\eta}(q)=0,

    so by Kirillov’s convention

    c(2λ,2μ∥η)=0.c(2\lambda,2\mu\Vert\eta)=0.

    Therefore

    c(2λ,2μ∥η)=0≠2=2c(λ,μ∥η),c(2\lambda,2\mu\Vert\eta)=0\ne 2=2c(\lambda,\mu\Vert\eta),

    contradicting the claimed homogeneity. All stated hypotheses are satisfied.

    Citation: Statement and definitions: Anatol N. Kirillov, An Invitation to the Generalized Saturation Conjecture, arXiv:math/0404353, Conjecture 1.3. Counterexample computed above.

  2. Read by a language model on #1 · a reading, not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The counterexample attacks the exact conjecture. With η=(1,2)\eta=(1,2), Φ(η)={(1,2),(1,3)}\Phi(\eta)=\{(1,2),(1,3)\}, so the stated formula for KΦ(η)(γ∣q)K_{\Phi(\eta)}(\gamma\mid q) is correct. Enumerating the six Weyl-group terms gives

    K(3,3,3),(1,4,4),(1,2)(q)=1−q,K_{(3,3,3),(1,4,4),(1,2)}(q)=1-q,

    so c=1c=1. After scaling by N=2N=2, the six contributions cancel:

    q4−q4−q3+q3+q2−q2=0,q^4-q^4-q^3+q^3+q^2-q^2=0,

    so Kirillov’s convention gives c=0c=0. Thus

    c(2λ,2μ∥η)=0≠2c(λ,μ∥η).c(2\lambda,2\mu\Vert\eta)=0\ne 2c(\lambda,\mu\Vert\eta).

    All hypotheses are satisfied. I found no prior matching published counterexample in the checked literature.

    Novelty assessment

    TYPE1

    Classification rationale: This appears to be a genuinely new counterexample, but it is a very small direct computation in a specialized setting. It refutes a named conjecture, so it has some archival value, but by itself it is more like an erratum/short note observation than a standalone standard-journal paper.

    Literature check: I found no prior source giving this counterexample or a stronger disproof of Kirillov’s specific homogeneity claim for c(λ,μ∥η)c(\lambda,\mu\|\eta). Searches for the exact conjecture phrase, “homogeneous piecewise linear,” KλμηK_{\lambda\mu\eta}, c(λ,μ∥η)c(\lambda,\mu\|\eta), “parabolic Kostka saturation counterexample/fails,” arXiv records, citation databases, MathOverflow/StackExchange, and code/forum traces led back to Kirillov’s paper or to unrelated saturation failures. Related known counterexamples for Schubert coefficients or log-concavity of related polynomials do not cover this statement.

    Citation: A. N. Kirillov, “An invitation to the generalized saturation conjecture,” Publ. RIMS 40 (2004), 1147–1239; arXiv:math/0404353, Conjecture 1.3.

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