As a follow-up of Allen's question [Coxeter exchanges in non-reduced words](https://mathoverflow.net/questions/223129), I wonder whether it is known that the Demazure product is well-defined in Artin groups. This is:


Let $(W,S)$ be a Coxeter system with corresponding Artin system $A(W)$ which is the group of words in $S$ up to the braid relations in $(W,S)$.


The **Demazure product** (or **greedy product**) $Dem(T)$ of a word $T$ in $S$ is given by the element in $W$ obtained by starting at the indentity $e \in W$ and going through $T$ from left to right, going upwards in weak order while skipping the letters that would take you downward.
(Example: $stuutut$ in $\{s=(12),t=(23),u=(34)\}$ in $S\_4$ is mapped to $Dem(stuutut) = stu\\_t\\_\\_ \in S\_4$ where the skips are underlined.)


**Question:** Has anyone seen the statement that the Demazure product is well-defined in Artin groups? I.e., is $Dem(T) = Dem(T')$ if $T$ and $T'$ differ in a braid move?


*The following was a false belief, see my comment:*
(I am rather sure I can prove that by showing that it only depends on the *inversion sequence* of a word, which is invariant under braid moves. But my impression would be that this is likely to be known.)


One immediate application would be that this would be another map $A(W) \rightarrow W$ (beside the "evaluation" map). For a finite Coxeter system, both would coincide as the identity map in the interval $[e,w\_\circ] \subseteq A(W)$, while the Demazure map would have the advantage that it preserves the weak order as $T \leq\_S T' \Rightarrow Dem(T) \leq\_S Dem(T')$. This is obviously not the case for the evaluation map. E.g., $ss >\_S s$ in $Weak(A(W))$ while for the evaluation $ss = e <\_S s$ in $Weak(W)$.



Demazure product in Coxeter and Artin groups
A complete combinatorial proof using Allen's comment:


Let $(W,S)$ be a Coxeter system, and let $Dem(T) \in W$ be the **Demazure product** or **greedy product** of a word $T$ in $S$.


**Claim:** $Dem(T)$ is the unique Bruhat maximal element in the set $\big\{ \prod Q : Q \subseteq T\big\}$ (where $Q\subseteq T$ means that $Q$ is a subsequence of $T$, and where $\prod Q$ means the product of the entries of $Q$ in the order in which they appear in $Q$).


The idea for the proof is to start with a subword $Q$ of $T$ and compare it with the subword $D$ of $T$ picked by the greedy product (you find the formal proof below). You scan through $Q$ from left to right and if you see a letter that is picked in $D$ but not in $Q$, you insert it into $Q$. If this goes up in Bruhat order, we are fine in doing so, and if you go down in Bruhat order, you find by the exchange condition a letter to its right that you can remove in exchange for the inserted letter.
By this procedure, you only go up in Bruhat order and we are done.


**Corollary:** The Demazure product is well-defined in Artin groups. This is, let $T$ be a word of $S$ and let $T'$ be obtained from $T$ by a braid move. Then $Dem(T) = Dem(T')$.


**Proof of corollary:** $T'$ is obtained from $T$ by replacing a consecutive substring $x = stst\ldots$ of length $m(s,t)$ by $y = tsts\ldots$. For any subword $Q$ of $T$, one can now choose the same subword in $T′$ as long as $Q$ does not contain all of $x$. But if this is the case, one can choose the subword $Q'$ of $T'$ where $Q'$ is obtained from $Q$ by using $y$ instead of $x$. $\square$


**Proof of Claim:**
This is a consequence of the following *lifting property* in Bruhat order as described in Proposition 2.2.7 of Björner-Brenti's [Combinatorics of Coxeter groups](http://www.math.washington.edu/~billey/classes/coxeter/EntireBook.pdf)


**Lemma 1** (lifting property)**:** Let $u < w$ in Bruhat order, and let $s$ be a right descent of $w$ but not of $u$. (Here, a *right descent* of an element $v \in W$ means a $t \in S$ satisfying $vt < v$.) Then $us \leq w$.


(Proof in Björner-Brenti, at least for the analogous statement about left descents; apply it to $u^{-1}$ and $w^{-1}$.)


**Lemma 2:** Let $u \leq w$ in Bruhat order, and let $s$ be a right ascent of $w$. (Here, a *right ascent* of an element $v \in W$ means a $t \in S$ satisfying $vt > v$.) Then $us \leq ws$.


**Proof:** Since $u \leq w$, we have that a reduced expression $a$ for $u$ which is a subword of a reduced expression $b$ for $w$. But since now $bs$ is a reduced expression for $ws$, it contains the expression $as$ (which might or might not be reduced) and we are done. $\square$


**Final induction to prove the Claim:** Let $T = t\_1\cdots t\_m$.
The case $m \in \{0,1\}$ is trivial, so assume $m>1$, let $T' = t\_1\cdots t\_{m-1}$ and we know that $Dem(T')$ is the unique Bruhat maximal element in $\{ \prod Q : Q \subseteq T'\}$.


Let $Q$ be a subword of $T$. 
If $Q$ is a subword of $T'$ we are done since by assumption $\prod Q \leq Dem(T') \leq Dem(T)$, so we only treat the case that $Q$ uses the last letter $t\_m$.


We have $Q \setminus t\_m$ is a subword of $T'$ so $\prod \left(Q\setminus t\_m\right) \leq Dem(T')$ by induction.


If $Dem(T) > Dem(T')$, we are in the situation of Lemma 2 and conclude $$\prod Q \leq Dem(T') \cdot t\_m = Dem(T).$$


If $Dem(T) = Dem(T')$, we are in the situation of Lemma 1 and conclude $$\prod Q \leq Dem(T') = Dem(T). \quad \square$$


(As usual with MO proofs, please let me know if something is unclear or plainly wrong.)


