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Distance graphs and rigidity

Algebra · math.CO · math.GR · posed by Hiroshi Maehara · open

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Statement

Is Cτ(A+)C_{\tau}(A_{+}) always isomorphic to Cπ(A+)C_{\pi}(A_{+}) ? In other words, is there a bijection f:Cτ→Cπf:C_{\tau}\to C_{\pi} that satisfies the condition d(x,y)∈A+⇔d(x,y)\in A_{+}\Leftrightarrow d(f(x),f(y))∈A+?d(f(x),f(y))\in A_{+}?

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No person has examined this. Everything below was judged by machines. say whether it holds →

  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: Let A+\mathbb A_+ be the set of positive real algebraic numbers. For r>0r>0, let CrC_r be the Euclidean circle of radius rr, and let Cr(A+)C_r(\mathbb A_+) be the graph on CrC_r in which x∼yx\sim y iff d(x,y)∈A+d(x,y)\in\mathbb A_+. Reconstructed Problem 7.1 asks: for every positive transcendental τ\tau, is

    Cτ(A+)≅Cπ(A+)?C_\tau(\mathbb A_+)\cong C_\pi(\mathbb A_+)?

    Result: Yes.

    Parametrize CrC_r by angles modulo 2π2\pi. For a∈A+a\in\mathbb A_+ with a<2ra<2r, put

    αa(r)=2arcsin⁡a2r∈(0,π).\alpha_a(r)=2\arcsin\frac{a}{2r}\in(0,\pi).

    Then two points of CrC_r are adjacent exactly when their angular difference is ±αa(r)\pm \alpha_a(r) for some algebraic a∈(0,2r)a\in(0,2r).

    Key lemma: if rr is transcendental, the set

    {αa(r) mod 2π:a∈A+, a<2r}\{\alpha_a(r)\bmod 2\pi: a\in\mathbb A_+,\ a<2r\}

    is Z\mathbb Z-linearly independent in R/2πZ\mathbb R/2\pi\mathbb Z.

    Proof of lemma. Suppose

    ∑j=1nmjαaj(r)∈2πZ\sum_{j=1}^n m_j\alpha_{a_j}(r)\in 2\pi\mathbb Z

    with distinct aj∈A+a_j\in\mathbb A_+ and integers mjm_j. For variable t>max⁡(aj/2)t>\max(a_j/2), the function

    ∏jexp⁡(imjαaj(t))−1\prod_j \exp(i m_j\alpha_{a_j}(t))-1

    is algebraic over Q‾(t)\overline{\mathbb Q}(t). Since it vanishes at the transcendental value t=rt=r, it vanishes identically. Hence

    ∑jmjαaj(t)\sum_j m_j\alpha_{a_j}(t)

    is constantly in 2πZ2\pi\mathbb Z; as t→∞t\to\infty, it tends to 00, so it is identically 00. Expanding

    αa(t)=2arcsin⁡(a/2t)=∑k≥0cka2k+1t−(2k+1),ck≠0,\alpha_a(t)=2\arcsin(a/2t)=\sum_{k\ge0} c_k a^{2k+1}t^{-(2k+1)},\qquad c_k\ne0,

    gives

    ∑jmjaj2k+1=0for all k≥0.\sum_j m_j a_j^{2k+1}=0\quad\text{for all }k\ge0.

    For k=0,…,n−1k=0,\dots,n-1, this is a Vandermonde system in the distinct numbers aj2a_j^2, so mjaj=0m_j a_j=0, hence every mj=0m_j=0. The lemma follows.

    Therefore each connected component of Cr(A+)C_r(\mathbb A_+), for transcendental rr, is the Cayley graph of the free abelian group on countably many generators, with respect to its standard basis. This component graph is independent of rr.

    Also, every vertex has countable degree, so every connected component is countable. Since CrC_r has cardinality 2ℵ02^{\aleph_0}, the graph has 2ℵ02^{\aleph_0} connected components. Rotations show all components are mutually isomorphic.

    Thus for any two positive transcendental radii r,sr,s, including r=τr=\tau and s=πs=\pi, both graphs are disjoint unions of continuum many copies of the same countable component graph. Hence

    Cτ(A+)≅Cπ(A+).C_\tau(\mathbb A_+)\cong C_\pi(\mathbb A_+).

    Citation: No known literature resolution used; proof given above.

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

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

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The proof attacks the correct statement for positive transcendental radii. The key independence lemma is valid: a nonzero algebraic function over Q‾(t)\overline{\mathbb Q}(t) cannot vanish at a transcendental tt, and the expansion at infinity plus Vandermonde argument forces all integer coefficients to vanish. Hence each component is the same standard Cayley graph of Z(N)\mathbb Z^{(\mathbb N)}, and both graphs have continuum many such components. I found only related algebraic-distance/chromatic-number literature, not a prior comparable isomorphism result.

    Novelty assessment

    TYPE1

    Classification rationale: The result appears genuinely new, but it is a short, elementary resolution of a very niche problem. The proof is a clean application of specialization at a transcendental parameter, a power-series/Vandermonde independence argument, and routine Cayley-graph/component counting. It resolves an explicit problem of Maehara, but the problem has very low visibility and the result seems unlikely to sustain a substantial standalone combinatorics paper without further generalizations.

    Literature check: I found no prior resolution of Maehara’s Problem 7.1. The closest prior work is Homma–Maehara’s 1990 paper, which proves the chromatic-number dichotomy for algebraic-distance graphs on circles, and Maehara’s 2004 survey/problem paper, where this isomorphism question is posed. Citation searches for Maehara’s 2004 paper show only a few later citations, mainly concerning rigidity/unit-bar frameworks, not the circle isomorphism problem. Searches for the exact terminology “algebraic-distance graphs on circles,” “C_tau,” “C_pi,” “Problem 7.1,” and related phrases did not reveal an open-access note, forum post, or paper containing this classification.

    Citation: H. Maehara, “Distance graphs and rigidity,” Contemporary Mathematics 342 (2004), 149–168, DOI: 10.1090/conm/342/06139.
    M. Homma and H. Maehara, “Algebraic Distance Graphs and Rigidity,” Trans. Amer. Math. Soc. 319 (1990), 561–572.

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