---
title: "Square-root cancellation for the signs of Latin squares"
authors:
  - "Levent Alpoge"
year: 2014
venue: "Combinatorica (from CV)"
arxiv: "1412.7574"
doi: null
source: "https://arxiv.org/abs/1412.7574"
topics:
  - "number-theory"
seed_rank: 1212
seed_batch: "alpoge-math-2026-08-24"
reviewed: "2026-08-24"
pool: "maths-foundations"
relevance_score: 8
lineage: arithmetic-geometry
cites:
  []
see:
  []
---

# Square-root cancellation for the signs of Latin squares

## One-sentence takeaway

|L^even(n)-L^odd(n)| <= L(n)^{1/2+o(1)}: even and odd Latin squares agree to leading order (Alon-Tarsi still open for the inequality), via a differential operator on an exponential integral over SU(n).

## Why it's lovely

Undergraduate combinatorics with representation-theoretic flavour. Kumar-Landsberg inspired.

## Problem

Alon-Tarsi: L^even != L^odd for even n. Weaker conjecture: they agree asymptotically.

## Design

Two proofs; both apply a differential operator to an integral over SU(n).

## Evidence

arXiv:1412.7574, 23 Dec 2014. CV lists Combinatorica; DOI not confirmed — cite arXiv plus venue name.

## Limitations

Does not prove Alon-Tarsi. Short note.

## Implications for Broadside

Early Alpoge; keep for corpus completeness.

## Bottom line

Mint 1412.7574.

## Links

- arXiv: [1412.7574](https://arxiv.org/abs/1412.7574)