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 $x\le w$?
MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.
Sign up to join this communityFor $n\ge 3$. 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 $x\le w$?
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.