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Symmetric group representations and $\mathbb{Z}$

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symmetric-group-representations-and-mathbb-zRepresentation Theorymath.COmath.RTposed by Anshul Adve, Alexander Yongrecorded: open · 1 machine check, unexamined

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Statement

Can one prove better upper or lower bounds for ℓn\ell_n ?

Context

Candidate 1 of the open problems stated in "Symmetric group representations and Z\mathbb{Z}", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    Problem: For n≥1n\ge1, let

    Vn={χλ(μ):λ,μ⊢n}\mathcal V_n=\{\chi^\lambda(\mu):\lambda,\mu\vdash n\}

    be the set of irreducible complex character values of Sn\mathfrak S_n, and let ℓn\ell_n be the maximum cardinality of an integer interval contained in Vn\mathcal V_n. The paper asks whether one can improve the known linear lower bound for ℓn\ell_n or give better upper bounds.

    Result: Yes. For all n≥6n\ge6,

    ℓn≥⌊n218⌋.\ell_n\ge \left\lfloor \frac{n^2}{18}\right\rfloor .

    Moreover, combining this with Adve–Yong’s interval [−(n−5),n−2]⊆Vn[-(n-5),n-2]\subseteq\mathcal V_n gives

    ℓn≥n218+O(n),\ell_n\ge \frac{n^2}{18}+O(n),

    an asymptotic improvement over the linear lower bound.

    Proof. For n≥4n\ge4, the character of the irreducible Sn\mathfrak S_n-module indexed by (n−2,2)(n-2,2) is

    χ(n−2,2)(μ)=(a2)+b−a,\chi^{(n-2,2)}(\mu)=\binom{a}{2}+b-a,

    where a=m1(μ)a=m_1(\mu) is the number of fixed points and b=m2(μ)b=m_2(\mu) is the number of 22-cycles of a permutation of cycle type μ\mu. This follows because the permutation character on 22-subsets is (a2)+b\binom a2+b, and the corresponding permutation module decomposes as

    S(n)⊕S(n−1,1)⊕S(n−2,2).S^{(n)}\oplus S^{(n-1,1)}\oplus S^{(n-2,2)}.

    Set

    sa=(a2)−a=a(a−3)2,m=⌊n+43⌋,qa=⌊n−a−32⌋.s_a=\binom a2-a=\frac{a(a-3)}2, \qquad m=\left\lfloor\frac{n+4}{3}\right\rfloor, \qquad q_a=\left\lfloor\frac{n-a-3}{2}\right\rfloor .

    For every 2≤a≤m2\le a\le m and every 0≤b≤qa0\le b\le q_a, the partition

    μ=(n−a−2b,2b,1a)\mu=(n-a-2b,2^b,1^a)

    is valid, since n−a−2b≥3n-a-2b\ge3. Hence every integer in

    Ja=[sa,sa+qa]J_a=[s_a,s_a+q_a]

    belongs to Vn\mathcal V_n.

    For 2≤a≤m−12\le a\le m-1,

    sa+1−sa=a−1,s_{a+1}-s_a=a-1,

    and since 3a≤n+13a\le n+1, one has qa≥a−2q_a\ge a-2. Thus JaJ_a and Ja+1J_{a+1} overlap or are adjacent. Therefore

    [−1,En]⊆Vn,[-1,E_n]\subseteq \mathcal V_n,

    where

    En=sm+qm=m(m−3)2+⌊n−m−32⌋.E_n=s_m+q_m =\frac{m(m-3)}2+\left\lfloor\frac{n-m-3}{2}\right\rfloor .

    A check according to n mod 3n\bmod 3 gives En+2≥⌊n2/18⌋E_n+2\ge\lfloor n^2/18\rfloor. Hence ℓn≥⌊n2/18⌋\ell_n\ge\lfloor n^2/18\rfloor.

    There is also a simple upper bound:

    ℓn≤p(n)2,\ell_n\le p(n)^2,

    where p(n)p(n) is the number of partitions of nn, since there are only p(n)2p(n)^2 pairs (λ,μ)(\lambda,\mu). Thus by Hardy–Ramanujan,

    ℓn≤p(n)2∼148n2exp⁡ ⁣(2π2n3).\ell_n\le p(n)^2\sim \frac{1}{48n^2}\exp\!\left(2\pi\sqrt{\frac{2n}{3}}\right).

    Citation: The definition of ℓn\ell_n and the linear interval are from Adve–Yong, “Symmetric group representations and Z\mathbb Z,” C. R. Math. Acad. Sci. Paris 356 (2018), 1–4. The character decomposition used above is standard Young’s rule; see James, The Representation Theory of the Symmetric Groups.

    Reviews

    0 human reviews · 1 machine check

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    • Machine check · not human verification

      machine: correct

      Recorded from GPT-5.5 xhigh (SMD judge 1) ·

      scope Full solution as submitted; SMD novelty classification TYPE1

      PASS

      The proof attacks the correct ℓn\ell_n question and is mathematically sound. The character formula for (n−2,2)(n-2,2) is correct, the chosen cycle types are valid, and varying bb gives intervals Ja⊂VnJ_a\subset \mathcal V_n. The adjacency condition qa≥a−2q_a\ge a-2 follows from 3a≤n+13a\le n+1, so these intervals form [−1,En][-1,E_n]. A check by n mod 3n\bmod 3 confirms the interval length En+2≥⌊n2/18⌋E_n+2\ge \lfloor n^2/18\rfloor for n≥6n\ge6. This is a genuine quadratic improvement over the known linear lower bound. I found no existing similar or stronger result in the accessible citation/literature checks.

      Novelty assessment

      TYPE1

      Classification rationale: The quadratic lower bound appears to be a genuine improvement over Adve–Yong’s recorded linear interval, but the proof is a short elementary observation from the standard character formula for (n−2,2)(n-2,2). It answers a broad “can one improve the bounds?” question, not a prominent conjecture, and introduces no new method. On its own it is likely too small for a standalone combinatorics paper, though it could be a useful note or component of a larger study.

      Literature check: I found no prior source stating ℓn≥cn2\ell_n\ge c n^2, ℓn≥⌊n2/18⌋\ell_n\ge \lfloor n^2/18\rfloor, or an interval of character values obtained from χ(n−2,2)=(X12)+X2−X1\chi^{(n-2,2)}=\binom{X_1}{2}+X_2-X_1. Searches around the exact paper title, ℓn\ell_n, “consecutive character values,” “integer interval,” “symmetric group character values,” and the n2/18n^2/18 bound did not locate a published or preprint resolution. The character formula itself is standard, but I did not find it applied to Adve–Yong’s ℓn\ell_n problem in the literature.

      Citation: Original problem and linear bound: Anshul Adve and Alexander Yong, “Symmetric group representations and Z\mathbb Z,” C. R. Math. Acad. Sci. Paris 356 (2018), 1–4; arXiv:1707.00020. Standard representation background: G. James, The Representation Theory of the Symmetric Groups.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

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