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A Koteljanskii inequality for permanents

A mix of human-written and AI-generated textHuman understanding: all partsmath.RA — Rings and Algebrasmath.CO — Combinatoricsmath.PR — Probability

Contributed by Suvrit Sra ↗

Version 1 / Nov 02, 2026/ arXiv: 2609.39979v1 / CC BY 4.0

Abstract

We prove a permanental analogue of Koteljanskii's inequality. If AA is an inverse MM-matrix that becomes symmetric after a positive diagonal similarity, then per(AS∪T) per(AS∩T)≥per(AS) per(AT)\mathrm{per}(A_{S\cup T})\,\mathrm{per}(A_{S\cap T})\ge\mathrm{per}(A_S)\,\mathrm{per}(A_T) for all S,T⊆[n]S,T\subseteq[n], where ASA_S is the principal submatrix indexed by SS. The proof expresses permanents as moments of a complex Gaussian vector and uses Ginibre's correlation inequality. Finally, an explicit counterexample shows that symmetry cannot be dropped.

Provenance statement

The conjecture was formulated by the author in April 2015, with partial proofs. After a sequence of back and forth discussions with GPT-5.5 (Pro) the first proof was attained. Subsequently, an audit with GPT-6 (Astra) revealed a few small errors, that were fixed. AI assisted with the drafting of the paper (GPT-6, and Claude Opus 5.5 and Fable 5.1), while the author strove to rewrite the entire paper by hand to simplify, polish, and compactify.

Tools used

OpenAI
GPT-5.5 Pro
OpenAI
GPT-6 Astra
Anthropic
Claude Opus 5.5 and Fable 5.1

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Version history

  1. v1Submitted by Suvrit SraInitial depositCurrentOct 31, 2026