How often are two permutations comparable?

Abstract

Two permutations of $\left [ n \right ]$ are comparable in the Bruhat order if one is closer, in a natural way, to the identity permutation, $1 2 \cdots n$, than the other. We show that the number of comparable pairs is of order $\left (n!\right )^2/n^2$ at most, and $\left (n!\right )^2\left (0.708\right )^n$ at least. For the related weak order, the corresponding bounds are $\left (n!\right )^2\left (0.362\right )^n$ and $\left (n!\right )^2\prod _{i=1}^n \left (H\left (i\right )/i\right )$, where $H\left (i\right ):=\sum _{j=1}^i 1/j$. In light of numerical experiments, we conjecture that for each order the upper bound is qualitatively close to the actual number of comparable pairs.