The multiplicity of the largest distance Laplacian eigenvalue in a connected graph.
The distance Laplacian of a graph is the (symmetric) matrix with row and column sums $0$, which has the negative distances between two vertices as its off-diagonal entries. This statistic is the largest multiplicity of an eigenvalue.
For example, the cycle on four vertices has distance Laplacian
$$ \left(\begin{array}{rrrr} 4 & -1 & -2 & -1 \\ -1 & 4 & -1 & -2 \\ -2 & -1 & 4 & -1 \\ -1 & -2 & -1 & 4 \end{array}\right). $$
Its eigenvalues are $0,4,4,6$, so the statistic is $1$.
The path on four vertices has eigenvalues $0, 4.7\dots, 6, 9.2\dots$ and therefore also statistic $1$.
The graphs with statistic $n-1$, $n-2$ and $n-3$ have been characterised, see [1].

The graph of inversions of a permutation.
For a permutation of $\{1,\dots,n\}$, this is the graph with vertices $\{1,\dots,n\}$, where $(i,j)$ is an edge if and only if it is an inversion of the permutation.

Let $\sigma = (i_{11}\cdots i_{1k_1})\cdots(i_{\ell 1}\cdots i_{\ell k_\ell})$ be a permutation given by cycle notation such that every cycle starts with its maximal entry, and all cycles are ordered increasingly by these maximal entries.
Maps $\sigma$ to the permutation $[i_{11},\ldots,i_{1k_1},\ldots,i_{\ell 1},\ldots,i_{\ell k_\ell}]$ in one-line notation.
In other words, this map sends the maximal entries of the cycles to the left-to-right maxima, and the sequences between two left-to-right maxima are given by the cycles.

The lower 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 $u$.