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Let YY and ZZ be tt-cross-intersecting sets in GnG_n whose sizes meet the bound in Theorem 1.2. If nn is sufficiently large compared to tt, then Y=ZY = Z and YY or YTY^T is a tt-coset.

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  • INTERSECTION THEOREMS FOR FINITE GENERAL LINEAR GROUPS
  • FAR
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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: For a fixed prime power qq, let Gn=GLn(Fq)G_n=\mathrm{GL}_n(\mathbb F_q). Two matrices x,y∈Gnx,y\in G_n are tt-intersecting if

    dim⁡ker⁡(x−y)≥t\dim\ker(x-y)\ge t

    equivalently they agree on tt linearly independent vectors. Sets Y,Z⊆GnY,Z\subseteq G_n are tt-cross-intersecting if every (y,z)∈Y×Z(y,z)\in Y\times Z is tt-intersecting. A tt-coset is a set

    {g∈Gn:gui=vi (1≤i≤t)},\{g\in G_n: gu_i=v_i\ (1\le i\le t)\},

    where u1,…,utu_1,\dots,u_t and v1,…,vtv_1,\dots,v_t are linearly independent. Its size is

    M=∏i=tn−1(qn−qi).M=\prod_{i=t}^{n-1}(q^n-q^i).

    The reconstructed conjecture is: if nn is sufficiently large compared with tt, and Y,Z⊆GnY,Z\subseteq G_n are tt-cross-intersecting with equality in Ernst–Schmidt Theorem 1.2,

    ∣Y∣∣Z∣=M,\sqrt{|Y||Z|}=M,

    then Y=ZY=Z, and YY or YTY^T is a tt-coset. Here YT={yT:y∈Y}Y^T=\{y^T:y\in Y\}. The phrase “sizes meet the bound” is interpreted as equality in Theorem 1.2.

    Result: The conjecture is true.

    Let

    N=∏i=0t−1(qn−qi),∣Gn∣=MN.N=\prod_{i=0}^{t-1}(q^n-q^i),\qquad |G_n|=MN.

    In the proof of Ernst–Schmidt Theorem 5.3, for all sufficiently large nn, there is a real symmetric weighted adjacency operator AA, supported only on non-tt-intersecting pairs, such that with respect to the normalized inner product on R(Gn)\mathbb R(G_n),

    • A1=1A\mathbf 1=\mathbf 1;
    • on the relevant nontrivial eigenspace the eigenvalue is
    η=−1N−1;\eta=-\frac1{N-1};
    • every other nontrivial eigenvalue has absolute value strictly less than ρ=1/(N−1)\rho=1/(N-1).

    Let f=1Yf=1_Y, g=1Zg=1_Z, and write

    f=a1+f0,g=b1+g0,f=a\mathbf 1+f_0,\qquad g=b\mathbf 1+g_0,

    with f0,g0⊥1f_0,g_0\perp \mathbf 1. Since Y,ZY,Z are cross-intersecting and AA is supported on non-tt-intersecting pairs,

    0=⟨f,Ag⟩=ab+⟨f0,Ag0⟩.0=\langle f,Ag\rangle=ab+\langle f_0,Ag_0\rangle .

    Equality in the product bound gives

    ab=∣Y∣∣Z∣∣Gn∣2=M2M2N2=1N2.ab=\frac{|Y||Z|}{|G_n|^2}=\frac{M^2}{M^2N^2}=\frac1{N^2}.

    By the spectral bound,

    ab≤ρ∥f0∥∥g0∥=1N−1a(1−a)b(1−b).ab\le \rho\|f_0\|\|g_0\| =\frac1{N-1}\sqrt{a(1-a)b(1-b)}.

    Substituting ab=N−2ab=N^{-2} gives

    (1−a)(1−b)≥(1−1N)2.(1-a)(1-b)\ge \left(1-\frac1N\right)^2.

    But ab=N−2ab=N^{-2}, so AM–GM gives a+b≥2/Na+b\ge 2/N, hence

    (1−a)(1−b)=1−a−b+1N2≤(1−1N)2.(1-a)(1-b)=1-a-b+\frac1{N^2}\le \left(1-\frac1N\right)^2.

    Thus equality holds throughout, so a=b=1/Na=b=1/N, i.e.

    ∣Y∣=∣Z∣=M.|Y|=|Z|=M.

    Moreover equality in the spectral inequality forces f0,g0f_0,g_0 to lie in the η=−1/(N−1)\eta=-1/(N-1) eigenspace, and equality in Cauchy–Schwarz gives f0=g0f_0=g_0. Hence f=gf=g, so Y=ZY=Z.

    Now Y=ZY=Z and cross-intersection imply that YY is tt-intersecting. Since ∣Y∣=M|Y|=M, the Ellis–Kindler–Lifshitz equality theorem for forbidden intersections in GLn(Fq)\mathrm{GL}_n(\mathbb F_q) applies: for sufficiently large nn, an extremal tt-intersecting family in GnG_n is either a tt-coset or the transpose of one. Therefore YY or YTY^T is a tt-coset.

    This proves Conjecture 1.4.

    Citation: The spectral operator and eigenvalue facts are from Ernst–Schmidt, Intersection theorems for finite general linear groups, Math. Proc. Cambridge Philos. Soc. 175 (2023), 129–160, especially Theorem 5.3 and its proof. The extremal tt-intersecting classification is from Ellis–Kindler–Lifshitz, Forbidden intersection problems for families of linear maps, Discrete Analysis 2023:19, arXiv:2208.04674.

  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 argument attacks the correct Conjecture 1.4. The equality refinement of the Ernst–Schmidt weighted Hoffman proof is valid: product equality forces ∣Y∣=∣Z∣=M|Y|=|Z|=M, and the strict spectral gap plus Cauchy–Schwarz equality forces the centered characteristic vectors to coincide, hence Y=ZY=Z. Then YY is an extremal tt-intersecting family, and the Ellis–Kindler–Lifshitz equality theorem gives that YY, or YTY^T, is a tt-coset. The cited spectral facts match the Ernst–Schmidt proof. I found no separate published statement explicitly resolving the cross-intersecting equality classification, though the proof is a corollary of known results.

    Novelty assessment

    TYPE1

    Classification rationale: The statement appears not to be explicitly published, but it is a very short equality-case refinement of the already published Ernst–Schmidt cross-Hoffman argument plus the Ellis–Kindler–Lifshitz extremal classification. It would be a useful remark/addendum, not a standalone publishable contribution.

    Literature check: I found Conjecture 1.4 stated in Ernst–Schmidt, Math. Proc. Camb. Philos. Soc. 175 (2023), 129–160, and the paper notes only that Ellis–Kindler–Lifshitz proved Conjecture 1.3. The paper’s Theorem 1.2/5.3 gives the cross-intersection bound and an eigenspace equality condition, but not Y=ZY=Z or the coset classification. Cambridge’s cited-by list and targeted searches for “Conjecture 1.4”, “t-cross-intersecting”, “t-coset”, and the relevant GL(n,q)(n,q) phrases did not reveal a later explicit resolution. The supplied arXiv URL is unrelated; the relevant open-access source is the Cambridge article.

    Citation: Alena Ernst and Kai-Uwe Schmidt, “Intersection theorems for finite general linear groups,” Math. Proc. Cambridge Philos. Soc. 175 (2023), 129–160.
    David Ellis, Guy Kindler, and Noam Lifshitz, “Forbidden intersection problems for families of linear maps,” Discrete Analysis 2023:19, arXiv:2208.04674.

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