This question is related to my other question [Sign of the permutation which brings a subsequence back to its original form](https://mathoverflow.net/questions/424340/sign-of-the-permutation-which-brings-a-subsequence-back-to-its-original-form). Suppose I have a complete ordered set $\{a\_{1},\dotsc,a\_{2n}\}$ and take $\pi$ to be a permutation of $\{1,\dotsc,2n\}$ such that $a\_{\pi(1)}\le a\_{\pi(2)}\le \dotsb \le a\_{\pi(2n)}$. If we throw away the element $a\_{1}$ and some other element, say, $a\_{j}$, we end up with a complete ordered set which has now $2n-2$ elements. Suppose $\pi'$ is the permutation which brings it to an increasing order again, that is $a\_{\pi'(2)}\le a\_{\pi'(3)}\le\dotsb \le a\_{\pi'(2n-2)}$. What is the relation between the sign of $\pi$ and the sign of $\pi'$?



Given $\pi$ permutation on $\{1,\dotsc,n\}$, what is the sign of a permutation of $\{2,\dotsc,\hat\jmath,\dotsc,n\}$?
$\DeclareMathOperator\sgn{sgn}$This is almost the same as your previous question, just with the order of the operations switched—whether you think of $\pi$ as ordering or disordering is just a matter of taking the inverse.


You previously needed $\pi^{-1}(1) - 1$ transpositions to bring the element $a\_1$ to its new position $a\_{\pi(\pi^{-1}(1))}$. Instead, simply throw away $a\_1$. Separately, you would have needed $\lvert\pi^{-1}(j) - j\rvert$ transpositions to bring the element $a\_j$ to its new position $a\_{\pi(\pi^{-1}(j))}$. This is unaffected by throwing away $a\_1$, unless $\pi^{-1}(j)$ equals $1$, in which case it should be reduced by $1$. Thus, throwing away both $a\_1$ and $a\_j$ gives you a new permutation $\pi'$ whose sign is $\sgn(\pi') = \sgn(\pi)\cdot(-1)^{(\pi^{-1}(1) - 1) + \lvert\pi^{-1}(j) - j\rvert}$ unless $\pi^{-1}(j) = 1$, in which case it is $\sgn(\pi') = \sgn(\pi)\cdot(-1)^{\pi^{-1}(1) + \lvert\pi^{-1}(j) - j\rvert}$.


