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REU REPORT

Combinatorics · math.CO · posed by Evgeny Demekhin, Gabriel Kreindler · open

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Statement

Let Gk=([k],Ek)G_k = ([k], E_k) be the complete kk-hygraph on [k][k].

Given a kk-hygraph G([n],E)G([n], E), (easily connected?), consider any function f:[n]→[k]f: [n] \to [k] and extend it naturally to f:E→Ekf: E \to E_k, and define Gf:=([k],f(E))G_f := ([k], f(E)). Define further ϕ:Qn→Qk\phi: \mathbb{Q}^n \to \mathbb{Q}^k by ϕ(ei)=ef(i)\phi(e_i) = e_{f(i)}. Then:

QE(G)=⋂f:[n]→[k]ϕ−1(QE(Gf))QE(G) = \bigcap_{f:[n] \to [k]} \phi^{-1}(\mathbb{Q}E(G_f))

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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 a finite kk-hygraph G=([n],E)G=([n],E), regard each edge as its incidence vector in Qn\mathbb Q^n, and define

    QE(G)=span⁡Q(E).QE(G)=\operatorname{span}_{\mathbb Q}(E).

    For every map f:[n]→[k]f:[n]\to[k], let ϕf:Qn→Qk\phi_f:\mathbb Q^n\to\mathbb Q^k be given by ϕf(ei)=ef(i)\phi_f(e_i)=e_{f(i)}, and let Gf=([k],ϕf(E))G_f=([k],\phi_f(E)). The conjecture asserts

    QE(G)=⋂f:[n]→[k]ϕf−1(QE(Gf)).QE(G)=\bigcap_{f:[n]\to[k]}\phi_f^{-1}(QE(G_f)).

    This is the only interpretation compatible with arbitrary ff, since images of edges must be counted with multiplicity. The parenthetical “easily connected?” is ambiguous; the counterexample below is connected in the ordinary incidence sense.

    Result: The conjecture is false already for k=3k=3.

    Let n=9n=9, and set

    α=(−5,−4,−3,−1,0,1,2,3,4)∈Q9.\alpha=(-5,-4,-3,-1,0,1,2,3,4)\in\mathbb Q^9.

    Let GG be the 3-uniform hypergraph with edges

    169, 178, 259, 268, 349, 358, 367, 456.169,\ 178,\ 259,\ 268,\ 349,\ 358,\ 367,\ 456.

    Every edge has α\alpha-sum 00, so QE(G)⊆α⊥QE(G)\subseteq \alpha^\perp.

    The eight edge vectors are linearly independent. Indeed, if their coefficients are c1,…,c8c_1,\dots,c_8, the vertex equations give

    c1+c2=0,c3+c4=0,c2+c7=0,c_1+c_2=0,\quad c_3+c_4=0,\quad c_2+c_7=0, c1+c3+c5=0,c5+c8=0,c3+c6+c8=0,c_1+c_3+c_5=0,\quad c_5+c_8=0,\quad c_3+c_6+c_8=0, c2+c4+c6=0,c5+c6+c7=0.c_2+c_4+c_6=0,\quad c_5+c_6+c_7=0.

    These imply c2=−c1c_2=-c_1, c7=c1c_7=c_1, c4=−c3c_4=-c_3, c5=−c1−c3c_5=-c_1-c_3, c8=c1+c3c_8=c_1+c_3, c6=−c1−2c3c_6=-c_1-2c_3, and then

    −2c1−3c3=0,−c1−3c3=0,-2c_1-3c_3=0,\qquad -c_1-3c_3=0,

    hence c1=c3=0c_1=c_3=0, so all ci=0c_i=0. Thus dim⁡QE(G)=8\dim QE(G)=8. Since dim⁡α⊥=8\dim \alpha^\perp=8, we have

    QE(G)=α⊥≠Q9.QE(G)=\alpha^\perp\neq \mathbb Q^9.

    Now fix any f:[9]→[3]f:[9]\to[3]. Since

    QE(Gf)=ϕf(QE(G)),QE(G_f)=\phi_f(QE(G)),

    we have

    ϕf−1(QE(Gf))=QE(G)+ker⁡ϕf.\phi_f^{-1}(QE(G_f))=QE(G)+\ker\phi_f.

    Taking annihilators,

    (QE(G)+ker⁡ϕf)⊥=QE(G)⊥∩im⁡ϕf∗.(QE(G)+\ker\phi_f)^\perp =QE(G)^\perp\cap \operatorname{im}\phi_f^*.

    But QE(G)⊥=QαQE(G)^\perp=\mathbb Q\alpha, while every vector in im⁡ϕf∗\operatorname{im}\phi_f^* has at most three distinct coordinate values. Every nonzero scalar multiple of α\alpha has nine distinct coordinate values. Hence

    Qα∩im⁡ϕf∗={0}.\mathbb Q\alpha\cap \operatorname{im}\phi_f^*=\{0\}.

    Therefore ϕf−1(QE(Gf))=Q9\phi_f^{-1}(QE(G_f))=\mathbb Q^9 for every ff, and so

    ⋂f:[9]→[3]ϕf−1(QE(Gf))=Q9≠QE(G).\bigcap_{f:[9]\to[3]}\phi_f^{-1}(QE(G_f))=\mathbb Q^9\neq QE(G).

    The edge-intersection graph is connected, so this also refutes the version with ordinary connectedness.

    Citation: None. This is an explicit counterexample.

  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 counterexample is mathematically sound under the natural/report interpretation of a kk-hygraph edge as a kk-multiset/incidence vector, which is required for arbitrary maps ff. The eight listed 3-edges all lie in α⊥\alpha^\perp, and the supplied linear-independence check correctly shows QE(G)=α⊥≠Q9QE(G)=\alpha^\perp\neq \mathbb Q^9. For every f:[9]→[3]f:[9]\to[3], the annihilator argument proves QE(G)+ker⁡ϕf=Q9QE(G)+\ker\phi_f=\mathbb Q^9, since im⁡ϕf∗\operatorname{im}\phi_f^* has at most three coordinate values while nonzero multiples of α\alpha have nine distinct values. Hence every preimage is all of Q9\mathbb Q^9, so the intersection is Q9\mathbb Q^9, not QE(G)QE(G). This directly disproves the conjecture, and no similar published resolution was found in the available search.

    Novelty assessment

    TYPE1

    Classification rationale: This appears to be a genuinely new explicit counterexample, but it is a minor result: it refutes an obscure conjecture from an unpublished/REU report by a short linear-algebra construction. The argument is elementary and the conjecture has little visible uptake in the combinatorics literature. It would not plausibly support a standalone journal paper without substantial additional theory.

    Literature check: I found the original source, Demekhin–Kreindler’s 2007 REU report, where Conjecture 2.11 is stated after Example 2.10. The report itself gives several counterexamples to other conjectures in Section 4, but not to Conjecture 2.11. I found no later paper, note, survey, or indexed source resolving this exact intersection/coloring conjecture or giving the same kind of counterexample. The terminology “k-hygraph” and QE(G)QE(G) seems idiosyncratic to the report, which further suggests little subsequent literature.

    Citation: No known prior resolution found. Original conjecture: Evgeny Demekhin and Gabriel Kreindler, “REU REPORT,” University of Minnesota REU report, 2007, Conjecture 2.11. https://www-users.cse.umn.edu/~reiner/REU/DemekhinKreindler2007.pdf

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