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ASYMPTOTIC METRIC BEHAVIOR OF RANDOM CAYLEY GRAPHS OF FINITE ABELIAN GROUPS

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asymptotic-metric-behavior-of-random-cayley-graphs-of-finite-abelianGroup Theorymath.GRmath.PRposed by Uri Shapira, Reut Zuckrecorded: open · 1 machine check, unexamined

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Statement

Let m>nm > n, and let b={vi}i=1n\mathbf{b} = \{v_i\}_{i=1}^n be a basis for Zn\mathbb{Z}^n. For a finite index subgroup Σ<Zn\Sigma < \mathbb{Z}^n consider the subset SZn/Σ(b,m)={s∈SZn/Σ(m):s⊃bmod  Σ}S_{\mathbb{Z}^n/\Sigma}(\mathbf{b}, m) = \{s \in S_{\mathbb{Z}^n/\Sigma}(m) : s \supset \mathbf{b} \mod \Sigma\} (that is, we restrict attention to the generating sets which contain the reduction of b\mathbf{b} modulo Σ\Sigma). Is it true that Theorem 1.2 holds when instead of choosing ss at random from SZn/Σ(m)S_{\mathbb{Z}^n/\Sigma}(m), we choose ss at random from SZn/Σ(b,m)S_{\mathbb{Z}^n/\Sigma}(\mathbf{b}, m).

Context

Candidate 1 of the open problems stated in "ASYMPTOTIC METRIC BEHAVIOR OF RANDOM CAYLEY GRAPHS OF FINITE ABELIAN GROUPS", 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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    Problem: Problem 1.3 asks whether Theorem 1.2 of Shapira–Zuck remains valid after conditioning the random generating set to contain a fixed basis of Zn\mathbb Z^n. Formally: for m>nm>n, a basis b={v1,…,vn}\mathbf b=\{v_1,\dots,v_n\}, and finite-index Σ<Zn\Sigma<\mathbb Z^n, choose uniformly from

    SZn/Σ(b,m)={s⊂Zn/Σ: ∣s∣=m, ⟨s⟩=Zn/Σ, s⊃b mod Σ}.\mathcal S_{\mathbb Z^n/\Sigma}(\mathbf b,m) =\{s\subset \mathbb Z^n/\Sigma:\ |s|=m,\ \langle s\rangle=\mathbb Z^n/\Sigma,\ s\supset \mathbf b\bmod\Sigma\}.

    Does ∣Σ∣−1/mξ(CZn/Σ+(s))|\Sigma|^{-1/m}\xi(C_{\mathbb Z^n/\Sigma}^{+}(s)) have the same limiting law as in Theorem 1.2, for the directed girth parameter ξ=girth\xi=\mathrm{girth}, along every sequence ∣Σ∣→∞|\Sigma|\to\infty?

    Result: No. The statement is false.

    Take n=2n=2, m=3m=3, b={e1,e2}\mathbf b=\{e_1,e_2\}, and

    Σk=kZe1⊕k3Ze2⊂Z2.\Sigma_k=k\mathbb Z e_1\oplus k^3\mathbb Z e_2\subset \mathbb Z^2 .

    Then

    Gk=Z2/Σk≅Z/kZ⊕Z/k3Z,∣Σk∣=∣Gk∣=k4→∞.G_k=\mathbb Z^2/\Sigma_k\cong \mathbb Z/k\mathbb Z\oplus \mathbb Z/k^3\mathbb Z, \qquad |\Sigma_k|=|G_k|=k^4\to\infty.

    Every s∈SGk(b,3)s\in \mathcal S_{G_k}(\mathbf b,3) contains the image of e1e_1, whose order in GkG_k is exactly kk. Hence in the directed Cayley graph CGk+(s)C_{G_k}^{+}(s), following the directed edge labeled e1e_1 gives a directed cycle of length kk. Therefore

    girth(CGk+(s))≤k\mathrm{girth}\bigl(C_{G_k}^{+}(s)\bigr)\le k

    for every admissible ss. Thus the scaled directed girth satisfies uniformly

    ∣Σk∣−1/3girth(CGk+(s))≤k−4/3k=k−1/3→0.|\Sigma_k|^{-1/3}\mathrm{girth}\bigl(C_{G_k}^{+}(s)\bigr) \le k^{-4/3}k = k^{-1/3}\to 0.

    So the conditioned random variables converge to 00 in probability.

    But Theorem 1.2 predicts, for directed girth in dimension m=3m=3, the law of

    ζ(L)=inf⁡{t>0: tB+∩L≠∅}\zeta(L)=\inf\{t>0:\ tB_+\cap L\neq\varnothing\}

    on the space X3X_3 of unimodular lattices, where

    B+={x∈R>03: x1+x2+x3<1}.B_+=\{x\in\mathbb R_{>0}^3:\ x_1+x_2+x_3<1\}.

    For every unimodular lattice LL, ζ(L)∈(0,∞)\zeta(L)\in(0,\infty): discreteness gives positivity, and every open cone contains a nonzero vector of LL, giving finiteness. Hence the predicted limiting law is not δ0\delta_0. Therefore Theorem 1.2 does not remain valid under this conditioning.

    The failure is caused by short relations among the prescribed basis elements: here ke1∈Σkk e_1\in\Sigma_k, while the theorem’s scale is ∣Σk∣1/3=k4/3|\Sigma_k|^{1/3}=k^{4/3}.

    Citation: No known resolution is used; the counterexample above disproves the stated problem.

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    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 correct statement. For n=2,m=3n=2,m=3, Σk=kZe1⊕k3Ze2\Sigma_k=k\mathbb Ze_1\oplus k^3\mathbb Ze_2 has ∣Σk∣=k4→∞|\Sigma_k|=k^4\to\infty, and every conditioned generating set contains e1e_1, whose image has order kk. Hence the directed Cayley graph always has a directed cycle of length kk, so the normalized directed girth is at most k/k4/3→0k/k^{4/3}\to0.

      But Theorem 1.2 includes directed girth, whose limiting lattice shortest-vector functional is positive almost surely, so its law is not δ0\delta_0. Thus the conditioned analogue of Theorem 1.2 is false. I found no evidence in the checked sources that this counterexample is already recorded.

      Novelty assessment

      TYPE1

      Classification rationale: The accepted resolution is a very short counterexample: forcing a prescribed basis element of small order immediately gives a directed cycle far below the theorem’s scaling. Although it answers a stated problem negatively, the idea is elementary and would not support a standalone combinatorics paper; at most it is a brief remark/erratum-style note.

      Literature check: I found no evidence that this counterexample or an explicit negative answer to Shapira–Zuck Problem 1.3 is already recorded. Searches of arXiv for the exact problem, “fixed basis,” “conditioned random Cayley graph,” “prescribed generators,” and “directed girth” variants found no such resolution. Later random-Cayley-graph work, notably Hermon–Olesker-Taylor’s papers on geometry/cutoff for random Cayley graphs, cites Shapira–Zuck as prior fixed-generator asymptotic-law work but does not mention this conditioned-basis problem or this obstruction.

      Citation: Uri Shapira and Reut Zuck, “Asymptotic Metric Behavior of Random Cayley Graphs of Finite Abelian Groups,” Combinatorica 39 (2019), 1133–1148; arXiv:1607.05022, Problem 1.3.

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