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Let $W$ be the symmetric group on $n+1$ letters. Let $\ell$ be the length function on $W$.

As the title says, can we characterize all $v\in W$ such that there exists a $w\in W$ such that for all simple transpositions $s=(i,i+1)$ ($1\leq i\leq n$) which satisfy $\ell(sv)=\ell(v)+1$ we have $w=sv$.

This amounts to say that $v$ has at most one successor (or a unique successor if we exclude the longest element of $W$) in the covering relation of the weak Bruhat order.

Usually, the weak Bruhat order is defined by multiplying with simple transpositions from the right. For my question, it is more convenient to multiply from the left. The difference is not very relevant.

Thank you for your help in advance.

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    $\begingroup$ These are the permutations $\sigma$ such that at most one $i \in \left\{1,2,\ldots,n\right\}$ satisfies $\sigma^{-1}\left(i\right) < \sigma^{-1}\left(i+1\right)$. In other words, these are the permutations that lie in the shuffle of $\left(n+1\right) n \cdots \left(u+1\right)$ with $u \left(u-1\right) \cdots 1$ for some $u \in \left\{0,1,\ldots,n+1\right\}$. $\endgroup$
    – darij grinberg
    Commented Sep 18, 2017 at 16:11
  • $\begingroup$ Thank you @darij grinberg ! The question turns out to pretty simple. Still, I was not able to work out the shuffle description on my own. If $u\neq 0$, then $u$ is the unique $i$ such that $\sigma^{-1}(i)<\sigma^{-1}(i+1)$. $\endgroup$
    – Christoph Mark
    Commented Sep 19, 2017 at 11:44

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