The number of maximal entries in the last diagonal of the monotone triangle.
Consider the alternating sign matrix
$$ \left(\begin{array}{rrrrr} 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & -1 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{array}\right). $$
The corresponding monotone triangle is
$$ \begin{array}{ccccccccc} 5 & & 4 & & 3 & & 2 & & 1 \\ & 5 & & 4 & & 3 & & 1 & \\ & & 5 & & 3 & & 1 & & \\ & & & 5 & & 3 & & & \\ & & & & 4 & & & & \end{array} $$
The first entry $1$ in the last diagonal is maximal, because rows are strictly decreasing and its left neighbour is $2$. Also, the entry $3$ in the last diagonal is maximal, because diagonals from north-west to south-east are weakly decreasing, and its north-west neighbour is also $3$. All other entries in the last diagonal are non-maximal, thus the statistic on this matrix is $2$.
Conjecturally, this statistic is equidistributed with St000066The column of the unique '1' in the first row of the alternating sign matrix..

Sends the permutation $\pi \in \mathfrak{S}_n$ to the permutation $\pi^{-1}c$ where $c = (1,\ldots,n)$ is the long cycle.

Maps a permutation to its permutation matrix as an alternating sign matrix.

The upper bound in the Grassmann interval corresponding to the decorated permutation.
Let $I$ be the anti-exceedance set of a decorated permutation $w$. Let $v$ be the $k$-Grassmannian permutation determined by $v[k] = w^{-1}(I)$ and let $u$ be the permutation satisfying $u = wv$. Then $[u, v]$ is the Grassmann interval corresponding to $w$.
This map returns $v$.