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Isoperimetric stability in lattices

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isoperimetric-stability-in-lattices-2Group Theorymath.COmath.GRposed by Ben Barber, Joshua Erde, Peter Keevash, Alexander Robertsrecorded: open · 1 machine check, unexamined

1 attempt · 1 machine check · no person has looked

Statement

Do we always have ∂v,GB(A)≥dμ(C(B))1/d∣A∣1−1/d\partial_{v,G_{B}}(A)\ge d \mu(C(B))^{1/d}|A|^{1-1/d} ?

Context

Candidate 2 of the open problems stated in "Isoperimetric stability in lattices", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
    AI involvement
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    GPT-5.5 xhigh

    NEW

    Problem: For finite B⊂ZdB\subset \mathbb Z^d generating Zd\mathbb Z^d, let GBG_B be the directed Cayley graph with edges x→x+bx\to x+b, b∈Bb\in B. Let

    C(B)=conv⁡(B∪{0}),∂v,GB(A)={y∉A:y=x+b for some x∈A,b∈B}.C(B)=\operatorname{conv}(B\cup\{0\}),\qquad \partial_{v,G_B}(A)=\{y\notin A:y=x+b\text{ for some }x\in A,b\in B\}.

    The question asks whether, for every finite A⊂ZdA\subset\mathbb Z^d,

    ∣∂v,GB(A)∣≥d μ(C(B))1/d∣A∣1−1/d.|\partial_{v,G_B}(A)|\ge d\,\mu(C(B))^{1/d}|A|^{1-1/d}.

    Result: The statement is false.

    For any m≥1m\ge1, put d=2d=2, N=m+2N=m+2,

    B={e1,e2,(N,N)}⊂Z2,A={j(N,N):0≤j<m}.B=\{e_1,e_2,(N,N)\}\subset\mathbb Z^2, \qquad A=\{j(N,N):0\le j<m\}.

    Then BB generates Z2\mathbb Z^2. The convex hull C(B)=conv⁡{0,e1,e2,(N,N)}C(B)=\operatorname{conv}\{0,e_1,e_2,(N,N)\} is the quadrilateral with vertices 0,e1,(N,N),e20,e_1,(N,N),e_2, hence by the shoelace formula

    μ(C(B))=N.\mu(C(B))=N.

    Now compute the vertex boundary. The translate A+(N,N)A+(N,N) contributes only the new point m(N,N)m(N,N). The translates A+e1A+e_1 and A+e2A+e_2 each contribute mm distinct points, all outside AA, and these 2m2m points are mutually disjoint and disjoint from m(N,N)m(N,N). Therefore

    ∣∂v,GB(A)∣=2m+1.|\partial_{v,G_B}(A)|=2m+1.

    But the proposed inequality would require

    2m+1≥2Nm=2m(m+2).2m+1\ge 2\sqrt{N}\sqrt m=2\sqrt{m(m+2)}.

    This is false, since

    4m(m+2)−(2m+1)2=4m−1>0.4m(m+2)-(2m+1)^2=4m-1>0.

    Thus the conjectured inequality fails even for arbitrarily large ∣A∣=m|A|=m, if BB is allowed to vary. If one requires 0∈B0\in B, replace BB by B∪{0}B\cup\{0\}; this only adds loops, so the boundary and C(B)C(B) are unchanged.

    The defect is that a redundant long generator can make μ(C(B))\mu(C(B)) large while contributing very little to the boundary of a progression in that generator’s direction.

    Citation: No external counterexample used.

    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 counterexample attacks the exact vertex-boundary inequality stated. For d=2d=2, B={e1,e2,(N,N)}B=\{e_1,e_2,(N,N)\} generates Z2\mathbb Z^2, and C(B)C(B) has area NN. For A={j(N,N):0≤j<m}A=\{j(N,N):0\le j<m\} with N=m+2N=m+2, the boundary computation ∣∂v,GB(A)∣=2m+1|\partial_{v,G_B}(A)|=2m+1 is correct: the e1e_1- and e2e_2-translates are disjoint and outside AA, and the long-generator translate adds only m(N,N)m(N,N). The claimed inequality would require

      2m+1≥2m(m+2),2m+1\ge 2\sqrt{m(m+2)},

      which is false since 4m(m+2)−(2m+1)2=4m−1>04m(m+2)-(2m+1)^2=4m-1>0. Adding 00 to BB only adds loops and changes neither boundary nor C(B)C(B). I found no prior published equivalent counterexample.

      Novelty assessment

      TYPE1

      Classification rationale: The result appears genuinely new, but it is a very elementary counterexample rather than a substantial theorem. It resolves the stated vertex-boundary question negatively by exploiting a redundant long generator; the construction and verification are one-paragraph. This is useful as a correction/remark to the Barber–Erde–Keevash–Roberts question, but it is unlikely to support a standalone journal paper.

      Literature check: I found no prior published or online counterexample to this exact vertex inequality. Searches of arXiv for the title and for key phrases such as “Cayley digraph” + “vertex boundary” and “conical hull” + “vertex boundary” returned only the original paper. Web searches for the exact title with “counterexample” or “Do we always have” gave no relevant hits. GitHub issue/discussion searches gave no relevant hits. OpenAlex lists citations including “Locality in sumsets” and “Edge Isoperimetry of Lattices”; the latter concerns an edge-isoperimetric nested-ordering question, not this vertex-boundary inequality.

      Citation: Ben Barber, Joshua Erde, Peter Keevash, and Alexander Roberts, “Isoperimetric stability in lattices,” Proc. Amer. Math. Soc. 151 (2023), 5021–5029, §4. No prior citation found for the counterexample.

      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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