The largest eigenvalue of a graph if it is integral.
If a graph is $d$-regular, then its largest eigenvalue equals $d$. One can show that the largest eigenvalue always lies between the average degree and the maximal degree.
This statistic is undefined if the largest eigenvalue of the graph is not integral.

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 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 $v$.

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.