Square-root cancellation for the signs of Latin squares

TL;DR

This paper proves that the difference between the counts of even and odd Latin squares of even order is at most on the order of the square root of the total number of Latin squares, supporting the Alon-Tarsi conjecture asymptotically.

Contribution

It establishes an upper bound on the difference between even and odd Latin squares, providing the first asymptotic evidence for the Alon-Tarsi conjecture.

Findings

The difference between even and odd Latin squares is at most $L(n)^{1/2 + o(1)}$.

The proof uses differential operators applied to exponential integrals over SU(n).

Two different proofs are provided, inspired by Kumar-Landsberg's work.

Abstract

Let $L(n)$ be the number of Latin squares of order $n$, and let $L^{\textrm{even}}(n)$ and $L^{\textrm{odd}}(n)$ be the number of even and odd such squares, so that $L(n) = L^{\textrm{even}}(n) + L^{\textrm{odd}}(n)$. The Alon-Tarsi conjecture states that $L^{\textrm{even}}(n)\neq L^{\textrm{odd}}(n)$ when $n$ is even (when $n$ is odd the two are equal for very simple reasons). In this short note we prove that $|L^{\textrm{even}}(n) - L^{\textrm{odd}}(n)|\leq L(n)^{\frac{1}{2} + o(1)},$ thus establishing the conjecture that the number of even and odd Latin squares, while conjecturally not equal in even dimensions, are equal to leading order asymptotically. Two proofs are given: both proceed by applying a differential operator to an exponential integral over $\mathrm{SU}(n)$. The method is inspired by a recent result of Kumar-Landsberg.