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Let $w \in S_n$ and $inv(w) = \{(i,j): i,j \in \{1,\ldots,n\}, i<j, w(i)>w(j)\}$ the inversion set of $w$. Let ${\bf i}=(i_1,\ldots,i_m)$ be a sequence such that $s_{i_1}\cdots s_{i_m}$ is a reduced expression of $w$. There is a bijection $f_{\bf i}: \{1,\ldots,m\} \to inv(w)$ given by $f_{\bf i}(k) = s_{i_1}s_{i_2} \cdots s_{i_k} \cdots s_{i_2}s_{i_1}$.

Theorem: Let $w \in S_n$. Suppose that $i<j<k$, $(i,j),(i,k),(j,k) \in inv(w)$, $i,j,k \in \{1, \ldots, n\}$, $inv(w)=m$, ${\bf i} = (i_1, \ldots, i_m)$, $s_{i_1} \cdots s_{i_m}$ is a reduced expression of $w$. Then there is some $r \in \{1,\ldots,m-2\}$ such that $f_{\bf i}(r) = (i,j)$, $f_{\bf i}(r+1) = (i,k)$, $f_{\bf i}(r+2) = (j,k)$.

Theorem (corrected version): Let $w \in S_n$. Suppose that $i<j<k$, $(i,j),(i,k),(j,k) \in inv(w)$, $i,j,k \in \{1, \ldots, n\}$, $inv(w)=m$. Then there is a reduced expression $s_{i_1} \cdots s_{i_m}$ of $w$ and some $r \in \{1,\ldots,m-2\}$ such that $f_{\bf i}(r) = (i,j)$, $f_{\bf i}(r+1) = (i,k)$, $f_{\bf i}(r+2) = (j,k)$, where ${\bf i} = (i_1, \ldots, i_m)$.

Are there some reference about this theorem? Thank you very much.

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    $\begingroup$ You should correct the theorem a little. For example, you have $f_i(r)$ twice, you should assume that $(i,j),(i,j),(j,k)$ are inversions, etc. Right now, your theorem is false in general, as one could simply pick $i < j < k$ such that $(i,j)$ is not an inversion of $w$... $\endgroup$
    – Dirk
    Commented Aug 21, 2017 at 9:16
  • $\begingroup$ @Dirk Liebhold, thank you for your suggestions. I will edit the post. $\endgroup$
    – Jianrong Li
    Commented Aug 21, 2017 at 9:52
  • $\begingroup$ I don't think this can be true. You want to say that the inversions $(i,j),(i,k),(j,k)$ must be adjacent in the inversion word of every reduced sequence of $w$. I would be surprised if this holds for any reduced sequence of $w$; but definitely it cannot hold for every reduced sequence of $w$ ! (Try the element $s_1s_4s_2s_5s_1s$ in $S_6$, with the reduced sequence I just gave, and with $i=1, j=2, k=3$.) $\endgroup$
    – darij grinberg
    Commented Aug 21, 2017 at 10:19
  • $\begingroup$ This is clearly wrong -- what you maybe want (to state / to use ?) is that inversion sets are closed in the sense that if $(i,j),(j,k)$ are inversions then so is $(i,k)$ and for any such reduced word $i = (i_1,\ldots,i_m)$, you have that $f_i(x) = (i,k)$ and $f_i(a) = (i,j), f_i(b) = (j,k)$ then $a < x < b$ or $b < x < a$. $\endgroup$
    – Christian Stump
    Commented Aug 21, 2017 at 10:24
  • $\begingroup$ @Christian Stump, thank you very much. Are there some references about the results you mentioned? $\endgroup$
    – Jianrong Li
    Commented Aug 22, 2017 at 9:40

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