Minimal dominant permutation in weak order

Question

Consider $S_\infty$ as a Coxeter group with Coxeter generators the adjacent transpositions $s_i$, $i\geq 1$. We view elements of $S_\infty$ as functions $u:\mathbb{N}\to\mathbb{N}$. Recall the Lehmer code $\mathrm{code}(u)$ is the sequence defined by $$\mathrm{code}_i(u)=| \{j>i\mid u(i)>u(j)\} |$$ A permutation $u$ is said to be dominant if $\mathrm{code}_i(u)\geq\mathrm{code}_{i+1}(u)$ for all $i$.

For a permutation $u$, associate a dominant permutation $\mu(u)$ recursively, by declaring that if $u$ is dominant, set $\mu(u)=u$, and if $u$ is not dominant, let $i$ be the maximal index such that $\mathrm{code}_i(u)<\mathrm{code}_{i+1}(u)$, and define $\mu(u)=\mu(us_i)$.

I want to prove that for any $u$, if $\mu$ is a dominant permutation such that $u\leq_R \mu$ (right weak order), then $\mu(u)\leq_R\mu$. Any ideas?

Answer 1

I found a proof. An alternative characterization of dominant permutations is that they are $132$-avoiding. Also, an alternative characterization of right weak order is that $u\leq_R v$ if and only if all left inversions that occur in $u$ also occur in $v$.

Suppose $u\leq_R\mu$ for $\mu$ dominant. If $u$ is dominant, then clearly $\mu(u) \leq_R\mu$. Otherwise we use induction on $\ell(u, \mu(u)) $. Let $i$ be the maximal index such that $\mathrm{code}_i(u) <\mathrm{code}_{i+1}(u)$. Then there exists an index $j>i+1$ such that there is a $132$ pattern at $i, i+1,j$. This pattern must not occur in $\mu$, but the existing inversions must remain. Thus $u(i+1)$ must precede $u(i) $ in $\mu$. Since this is the only left inversion that is added as we pass from $u$ to $us_i$, it follows that $us_i\leq_R\mu$. Therefore by the induction hypothesis $$\mu(u) =\mu(us_i) \leq_R \mu$$ and the result follows by induction.