Square-root cancellation for the signs of Latin squares

Abstract

Let L(n) be the number of Latin squares of order n, and let L even(n) and L odd(n) be the number of even and odd such squares, so that L(n)=L even(n)+L odd(n). The Alon-Tarsi conjecture states that L even(n) ≠ L 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

$$\left| {{L^{even}}\left( n \right) - {L^{odd}}\left( n \right)} \right| \leqslant L{\left( n \right)^{\frac{1}{2} + o\left( 1 \right)}}$$

, 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.

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Correspondence to Levent Alpoge.

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Alpoge, L. Square-root cancellation for the signs of Latin squares. Combinatorica 37, 137–142 (2017). https://doi.org/10.1007/s00493-015-3373-7

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Mathematics Subject Classification (2000)

  • 05B15
  • 05A16