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math.CO — Combinatorics

The qq-Foulkes problem II: Schur positivity with high probability

Contributed by Suvrit Sra

Bergeron's qq-analog of Foulkes' conjecture asserts that (Hb[Ha]−Ha[Hb])/(1−q)(H_b[H_a]-H_a[H_b])/(1-q) is Schur positive for a≤ba\le b, where HnH_n is the (modified) Hall--Littlewood polynomial. We view this conjecture probabilistically: on each irreducible representation of SabS_{ab}, the two plethysms induce degree distributions, and positivity means that the first is stochastically smaller than the second. We prove that, for a random (Plancherel) irreducible representation and a degree sampled from its graded multiplicity, the corresponding coefficient is negative with probability at most Cexp⁡(−c(ab)1/3)C\exp(-c(ab)^{1/3}), uniformly in a≤ba\le b. Consequently, the qq-Foulkes conjecture holds after deleting a fraction at most Cexp⁡(−c(ab)1/3)C\exp(-c(ab)^{1/3}) of each of the two graded modules. Additionally, we show that exact conditional means and variances yield complete positivity for explicit width and height families.

A mix of human-written and AI-generated textHuman understanding: some partsMacdonald polynomialPlancherel measureSchur positivityplethysmq-Foulkes conjecturestochastic order

math.PR — Probability

Beyond Small Balls: A Determinant Inequality for Functions of Matrix Gaussian Series

Contributed by Suvrit Sra

We prove a functional extension of Gaussian operator-norm small-ball inequalities. Let X=∑i=1dgiBiX=\sum_{i=1}^d g_iB_i, where g∼N(0,Id)g\sim N(0,I_d), the matrices Bi∈Rd×dB_i\in\mathbb{R}^{d\times d} are symmetric, and ∑iBi2⪯Id\sum_iB_i^2\preceq I_d. For every even log-concave f ⁣:R→[0,∞)f\colon\mathbb{R}\to[0,\infty), we obtain Edet⁡f(X)≥(Ef(κG))d\mathbb{E}\det f(X)\ge(\mathbb{E} f(\kappa G))^d at dilation κ=13/5\kappa=13/5, where GG is standard Gaussian. The proof reduces each function to one small-ball estimate, obtained from matrix-weighted Poincaré inequalities, and controls the rest by averages over the spectrum. We give an equivalent entropy formulation. An explicit counterexample shows that no universal dilation below π/2\sqrt{\pi/2} can work; we conjecture that this value is optimal.

A mix of human-written and AI-generated textHuman understanding: all partsdeterminant inequalitylog-concave functionsmatrix Gaussian seriesrandom matricessmall-ball probabilities

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