Square-root cancellation for the signs of Latin squares

Abstract

Let L(n) be the number of Latin squares of order n, and let Leven(n) and Lodd(n) be the number of even and odd such squares, so that L(n)=Leven(n)+Lodd(n). The Alon-Tarsi conjecture states that Leven(n) ź Lodd(n) when n is even (when n is odd the two are equal for very simple reasons). In this short note we prove that $$\left| {{L^{even}}\left( n \right) - {L^{odd}}\left( n \right)} \right| \leqslant L{\left( n \right)^{\frac{1}{2} + o\left( 1 \right)}}$$|Leven(n)źLodd(n)|źL(n)12+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 SU(n). The method is inspired by a recent result of Kumar-Landsberg.