2
$\begingroup$

For $n\ge 3$. Let $s_1\cdots s_n$ be a reduced expression of $x$. Suppose $s_1\cdots s_{n-1}\le w$ and $s_2\cdots s_{n}\le w$.

Does this imply $x\le w$?

$\endgroup$
  • 2
    $\begingroup$ Have you tried looking at some small Coxeter groups to see if this is true? $\endgroup$ – user44191 Aug 13 at 3:12
4
$\begingroup$

Let $W = S_5$ (as a side note, we could let $W = S_3 \times S_2$). Let $w = (132)(45), x = (123)(45) \in W$; we write $x = (12)(45)(23)$ as a reduced expression. Then the subexpressions written in the question are $(12)(45)$ and $(45)(23) = (23)(45)$. The reduced expressions for $w$ are $(23)(12)(45), (23)(45)(12), (45)(23)(12)$; for each of those, the appearance of both $(12)(45) = (45)(12), (45)(23) = (23)(45)$ as outputs for subexpressions is clear, so $(12)(45), (23)(45) \leq w$, satisfying the hypothesis of the theorem, but $x \nleq w$, so the answer to the question is no.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.