(in contrary to what I thought in my commentfirst,) here is a proof that the exchange condition"exchange condition" holds: in the following sense.
It is based on the root configuration in http://arxiv.org/abs/1111.3349 [1].
Let $(W,S)$ be a Coxeter system, $Q \in S^*$ a word in $S$, $w \in W$, $P$ a subword of $Q$ (given. We everywhere consider subwords as being indexed by the positions of its letters)in $Q$, andso if $Q = ss$, then the two subwords $s{-}$ und ${-}s$ are considered to be different.
Let $(Q,w,P)$ be the set of subwords $X$ of $Q \setminus P$ such that the complementcomplement $R = Q \setminus X$ has greedy product $Dem(R) = w$ and the skips in the greedy product are in exactly the positions in $P$, we call these positions the greedy skips.
Let (So this is the situation Allen introduced in the question, except that he did not mention the part about the complement.)
Let $D(Q,w,P)$ be the simplicial complex with facets given by $(Q,w,P)$ and observe that it. This complex is clearly a pure complexsince every facet contains exactly $len(Q)-\ell_S(w)-len(P)$ many letters.
I believeObservation 1: For every facet $X$ of $D(Q,w,P)$, the disjoined union $X \cup (P\setminus P')$ is also a facet of $D(Q,w,P')$ for $P' \subseteq P$.
Construction 2: The root configuration in Definition 3.1 in [1] is given for a facet $X$ of $D(Q,w,\{\})$ as follows. Associate to each letter $q_i$ in $Q$ a root $R(X,q_i)$ by applying the prefix up to position $q_{i-1}$ of the word $Q \setminus X$ of $w$ to the simple root $\alpha_{q_i}$.
In my example $Q = tsstst$, the element $w = sts = tst$ and the facet $t{-}s{--}t$ with complement ${-}s{-}ts{-}$ which is a reduced word for $w$, the root configuration is
$$
R(X,\cdot) =\beta,\alpha,s(\alpha),s(\beta),st(\alpha),sts(\beta)
$$
where $\alpha = \alpha_s, \beta = \alpha_t$ which is equal to
$$
23,12,-12,13,23,-12
$$
where I can showwrite $12$ for $\alpha$, $23$ for $\beta$, and $13$ for $\alpha+\beta$.
Observation 3: The collections of roots of the root configuration of the complement of $X$ is exactly the inversion set of $w$ (in particular, that's all positive roots).
Observation 4: The negative roots in the root configuration could be used as greedy skips. This also means that picking a a facet $X$ of $D(Q,w,\{\})$, and a subset $P$ of the letters for which the root configuration is negative results in a facet $X \setminus P$ of $D(Q,w,P)$.
Observation 5: If a root in the root configuration is negative, than all appearances of the same root "to the right" are also negative. Analogously, if a root there is positive then all appearances "to the left" are positive.
Statement 6: The simplicial complex $D(Q,w,P)$ is vertex-decomposable by showing.
Proof: I show that the deletion and the link of the first letter $q_1$ in $Q$ are again complexes of the same form. The argument is almost the same as in Allen and Ezra's mentioned paper.
The crucial step isAs before, the followinglink of (which$q_1$ is indeed depending on some notions we recently introduced, but which are still fairly elementary):given by $D(Q\setminus q_1,w,P)$.
Let $q$ beAlso the first letter in $Q$ and let $Q' = Q \setminus q$, and we assumesituation that $q$$q_1$ is not a left descent of $w$ (this is, there the same since this link is a word for $w$ that starts with $q$).
Then we have to showthen the exchange conditionsame as the deletion.
This
The last step is the "exchange condition": If $\ell_S(q_1w) < \ell_S(w)$, we haveaim to show that if there is for any facet $X$ of $D(Q,w,P)$ such that this first letter $q$ is in $X$,there is there exists a facet $Y$ containing $X \setminus \{q_1\}$.
For this, consider $X' = X \cup P$ as a facet of $D(Q,w,P)$ such$D(Q,w,\{\})$ as in Observation 1.
We now know from Observation 4 that $q$$P$ is a subset of the negative roots in the symmetric differenceroot configuration of $X$ and $Y$$X'$.
In other words, there must be a facetWe know from $Y$$\ell_S(q_1w) < \ell_S(w)$ that contains all letters$\alpha_{q_1}$ is in the inversion set of $X$ except this initial$w$ which is by Observation 3 equal to the root configuration of the complement of $q$$X$.
The point is that this
This provides to obtain Z by "flipping" $Y$ exists$q_1$ in the complex $D(Q,w,\{\})$ because of$X'$ to the exchange conditionunique position not in $(W,S)$$X'$ for which the root configuration is given by the same simple root $\alpha_{q_1}$.
But the "flip" fromCall this position $X$$q_i$ and we have $Y' = (X' \setminus q_1 ) \cup q_i$.
It remains to show that $Y$ in$Y'$ contains $D(Q,w,\{\})$ changes$P$ and that all roots in the root configuration as describedof $Y'$ at positions in $P$ are negative, since Observation 1 then tells us that $Y = Y' \setminus P$ is the desired facet of $D(Q,w,P)$.
Lemma 3.3(3) in my paper http://arxiv.org/abs/1111.3349 with Vincent Pilaud.
Therefore[1] tells us how the root configuration gets changed by applyingchanges when doing the simple generatorflip $q \in W$$X'$ to all entries$Y'$.
This is indeed easy to see: the roots in the root configuration betweenafter $q_i$ are not changed, while we apply $s_{q_1}$ to all the first positionroots before that,
$R(Y',j) = R(X',j)$ for $j>i$, and the position of the unique letter in $Y \setminus X$$R(Y',j) = s_{q_1}\big(R(X',j)\big)$ for $j \leq i$.
But this simple generator cannot change
Since we only care about the sign of any root insigns, observe that $s_{q_1}$ does only affect the root configuration exceptsign of the simple root $\alpha_q$$\alpha_{q_1}$, while all other signs are unchanged.
But the simple rootsince $\alpha_q$ can only appear with a negative sign$R(X',i)$ is positive (and ind.eed in equal to $\alpha_{q_1}$), Observations 4 and 5 finally tell us that $P$ is a subset of the letters for which the root configuration after the position of the letter in $Y\setminus X$
Finally$Y'$ is negative. As desired, we conclude that $Y$$Y=Y' \setminus P$ is also a facet of $D(Q,w,P)$ using item 1 in my lengthy comment-as-an-answer.
(I have to run now and will add some details tomorrow. I hope this is already somewhat clear -- sorry for being sloppy!)
