Let \(k\in \mathbb{N}\) and \(\Pi_k\) be the set of pairings of \(\{1,\ldots ,2k\}\). For each pairing, except the maximal one , there is a positive adjacent transposition , and for each pairing, except the minimal one , there is a negative adjacent transposition .
We assume a pairing \(\pi\in \Pi_k\) has no positive adjacent transpositions, i.e. every adjacent transposition is either neutral or negative with respect to \(\pi\). Using this condition, we will determine all pairs in \(\pi\). Let us start by determining the partner of 1. Suppose \(\tau_1\) is neutral with respect to \(\pi\), i.e. \(\pi(1)=2\). Since
\[ 1=\pi(2)< \pi(3)\in \{4,\ldots ,2k\}, \]\(\tau_2\) is positive. Therefore, \(\tau_1\) must be negative. Assuming \(\pi(1)\neq 2k\) implies that \(\tau_{\pi(1)}\) is again positive, since
\[ 1=\pi\bigl(\pi(1)\bigr) < \pi\bigl(\pi(1)+1\bigr). \]Therefore, the partner of 1 must be \(2k\). With a similar argument, we can show that the partner of 2 is \(2k-1\), and applying this argument inductively yields \(\pi=\pi_{k,\max }\).
Now, we assume a pairing \(\pi\in \Pi_k\) has no negative adjacent transpositions. First, \(\tau_1\) must be neutral with respect to \(\pi\). Otherwise, \(\tau_{\pi(1)-1}\) is negative, since
\[ 1=\pi(\pi(1)) < \pi(\pi(1)-1) \in \{2,\ldots ,2k\}. \]This implies that the partner of 1 in \(\pi\) is 2. Using the same reasoning, we conclude that \(\{3,4\}\) is a pair in \(\pi\), and applying this argument repeatedly yields \(\pi=\pi_{k,\min }\).