Demazure product in Coxeter and Artin groups

As a follow-up of Allen's question Coxeter exchanges in non-reduced words, 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 $Weak(W)$.

• An alternative definition (in the Coxeter group case) is $Dem(T) =$ the unique Bruhat-maximal element of $\{\prod Q\ :\ Q \subseteq T\}$. I don't know a combinatorial proof in the Coxeter case that there is a unique maximum, but I doubt it's hard, and perhaps could be extended to the Artin case. (In the Weyl case, the geometric proof is that the Bott-Samelson manifold is irreducible, hence its image in $G/B$ must be a $B$-orbit closure, whose $T$-fixed points have a unique Bruhat maximum.) – Allen Knutson Nov 12 '15 at 11:36
• Great, starting from your comment, it is actually trivial to get that the Demazure product is stable under braid moves: replace a consecutive substring $stst...$ of length $m(s,t)$ by $tsts...$ to get $T'$. For any subword $Q$ I pick in $T$, I can pick the same subword in $T'$ as long as I have not used all of $stst...$. But if that is the case, I can as well have picked $tsts...$ in $T'$, and I am done. (I don't see an immediate combinatorial argument for your claim either at the moment, though.) – Christian Stump Nov 12 '15 at 16:22
• Trivial counterexample to the belief that the greedy product only depends on the inversion set: $inv(sstt) = inv(ttss)$ in $A_2$, but $Dem(sstt) = st \neq ts = Dem(ttss)$. – Christian Stump Nov 17 '15 at 13:31

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

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.)

• Note: Your Lemma 1 also follows from Corollary 2.5 (a) in Lusztig's arXiv:math/0208154v2, applied to $y = u^{-1}$ and $z = w^{-1}$. Notice that the requirement that $s$ not be a right descent of $u$ is not needed. – darij grinberg Apr 15 '16 at 23:42
• Actually, I think that when you say you're using Lemma 1, you actually use the more general version where $s$ is allowed to be a right descent of $u$ and where the condition $u < w$ is replaced by $u \leq w$ (because you apply it to $u = \prod \left(Q \setminus t_m\right)$ and $w = Dem(T')$). This version easily follows from your Lemma 1 (indeed, it is easy to prove $us \leq w$ both in the case $u=w$ and in the case $us<u$), but unless you replace Lemma 1 by this version I think you cannot claim that "we are in the situation of Lemma 1". – darij grinberg Apr 15 '16 at 23:52
• (Note that the more general version of Lemma 1 directly follows from Lusztig's Corollary 2.5 (a), whereas Björner-Brenti's Proposition 2.2.7 only gives the weaker version; things like this make Lusztig my go-to place for lemmas about Coxeter groups.) Other than this, thanks for the nice proof! It sort-of proves the claim of mathoverflow.net/questions/81539 as well, I believe. – darij grinberg Apr 15 '16 at 23:53
• I agree, thanks for pointing that out -- I was implicitly excluding the other cases $s$ being a right descent of $w$ and $u = w$, but should make that explicit. – Christian Stump Apr 17 '16 at 9:35