equation; the other equalities hold by the definition of $E_{i}$ and Part (3).
Finally, Part (6) follows from Remark 3.21. ∎

We now present the data $w_{i},n_{i},N_{ij}$ defined earlier in this
subsection, starting with the description of the equivalence classes of $\sim$
in terms of the set $\\{w_{1}:n_{1},w_{2}:n_{2},\dots,w_{d}:n_{d}\\}$.

######  Proposition 4.16.

Let $(W,S)$ be a nontrivially $\mathbf{a}(2)$-finite Coxeter system of type
$X$, and let $\mathcal{S}(W)$ be the set of left $\mathbf{a}(2)$-stubs in $W$,
denoted in the same way as in 3.7. Let $\beta_{n}=\binom{n}{2}$ for all $n\geq
2$. Then the equivalence classes of $\sim$ are described by Table 7.

$X$ |  representatives and cardinalities of $\approx$-classes
---|---
$A_{n},n\geq 3$ |  $\\{13:\beta_{n}-1\\}$
$B_{3}$ |  $\\{13:3\\}$
$B_{n},n>3$ |  $\\{13:n,\quad 24:\beta_{n-1}-1\\}$
$\tilde{C}_{n-1},n\geq 5$ |  $\\{13:n-1,\;24:\beta_{n-2}-1,\;(n-2)n:n-1,\;1n:1\\}$
$E_{1,1}=D_{4}$ |  $\\{(-1)v:3,1v:3,\;(-1)1:3\\}$
$E_{q,r},r>q=1$ |  $\\{(-1)v:2+r=n-1,\;1v:\beta_{n}\\}$
$E_{q,r},r\geq q\geq 2$ |  $\\{(-1)1:\beta_{q+r+3}-1=\beta_{n+1}-1\\}$
$F_{4}$ |  $\\{13:9\\}$
$F_{n},n>4$ |  $\\{13:3n-3,\;35:\beta_{n-2}-1\\}$
$H_{3}$ |  $\\{13:5\\}$
$H_{n},n>3$ |  $\\{13:2n-1,\;24:\beta_{n-1}-1\\}$
Table 7. The slide equivalence classes of $\mathcal{S}(W)$

###### Proof.

Let $G$ be the Coxeter diagram of $(W,S)$. Recall that the number
$\lvert\mathcal{S}(W)\rvert$ is given in 3.7.(7), which we will use without
comment from now on.

When $(W,S)$ is of type $A_{n}(n\geq 3)$ or $E_{q,r}(r\geq q\geq 1)$, all
edges of $G$ are simple, so the relations $\approx$ and $\sim$ coincide. It
follows from 3.30 and 3.7.(7) that Table 7 gives the correct information in
these types. In particular, in type $E_{1,r}$ where $r>1$, counting the set
$C_{1}$ in 3.30.(2).(b) yields $n_{1}=\lvert C_{1}\rvert=r+2=n-1$, therefore
$n_{2}=\lvert C_{2}\rvert=\lvert\mathcal{S}(W)\rvert-\lvert
C_{1}\rvert=(\beta_{n+1}-1)-(n-1)=\beta_{n}$.

For the Coxeter systems of types $B_{n}(n\geq 3),\tilde{C}_{n-1}(n\geq
5),F_{n}(n\geq 4)$ and $H_{n}(n\geq 3)$, we sketch the key ideas for obtaining
the set $\\{w_{1}:n_{1},\dots,w_{d}:n_{2}\\}$. Let $Z$ be the set of all short
stubs of $\mathcal{S}(W)$. By Remark 3.25.(1), we may choose all the class
representatives $w_{1},\dots,w_{d}$ from $Z$, and to understand the
$\sim$-classes of $\mathcal{S}(W)$ it suffices to understand how the short
stubs in $Z$ fall into different $\sim$-classes. To this end, note that
removing the heavy edge or heavy edges in $G$ results in $r=2$ or $r=3$ linear
subgraphs of $G$ whose edges are simple and whose vertex sets partition $S$:
these are the sets $S_{1}=\\{1\\},S_{2}=\\{2,\dots,n-1\\},S_{3}=\\{n\\}$ in
type $\tilde{C}_{n-1}$, the sets $S_{1}=\\{1,2\\},S_{2}=\\{3,\dots,n\\}$ in
type $F_{n}(n\geq 4)$, and the sets $S_{1}=\\{1\\},S_{2}=\\{2,\dots,n\\}$
otherwise. Associate a tuple
$\mathrm{dist}(z):=(\operatorname{\mathrm{Supp}}(z)\cap S_{i})_{1\leq i\leq
r}$ to each stub $z\in Z$ to record how its support is distributed among these
subsets, and note that by the definition of simple slides, we have $z\sim
z^{\prime}$ if and only if $\mathrm{dist}(z)=\mathrm{dist}(z^{\prime})$ for
all $z,z^{\prime}\in Z$. Observe that in every Coxeter type we have included
in the list $w_{1},\dots,w_{d}$ one short stub corresponding to every possible
distribution. It follows that $w_{1},\dots,w_{d}$ is a complete, irredundant
list of class representatives of $\sim$ in $\mathcal{S}(W)$.

To obtain the sizes $n_{1},\dots,n_{d}$ of the equivalence classes, start with
the class represented by the short stub whose support lies entirely in the set
$S_{2}$ specified earlier. This stub is always $w_{2}$ in our list of
representatives. If $\lvert S_{2}\rvert=k$ then the size of its $\sim$-class
in $\mathcal{S}(W)$ is $n_{2}=\beta_{k}-1$ because $S_{2}$ generates a
parabolic subgroup of $W$ isomorphic to $A_{k}$ and
$\mathcal{S}(A_{k})=\beta_{k}-1$. Three $\sim$-classes remain in type
$\tilde{C}_{n-1}$ and one class remains otherwise. In the latter case we can
obtain cardinality $n_{1}$ of the remaining class as
$n_{1}=\lvert\mathcal{S}(W)\rvert-n_{2}$. In the former case, we note that the
element $w_{4}=1n$ admits no slides and is the only stub with first layer $1n$
by Corollary 3.4, so $\\{1n\\}$ is a $\sim$-class and $n_{4}=1$. By the
symmetry in $G$, it further follows that
$n_{1}=n_{3}=(\lvert\mathcal{S}(W)\rvert-n_{2}-n_{4})/2=(n-1)$. Alternatively,
in both cases we may list the elements of the remaining equivalence classes
and count them without difficulty with the help of Corollary 3.4 and 3.7, but
we omit the details. ∎

Our next result describes selected 0-cells and will lead to the cardinalities
$N_{ij}$ for all $1\leq i,j\leq d$. We postpone its proof to Section 5, where
we will refer to the cells given in the theorem _representative 0-cells_.

######  Theorem 4.17.

Let $(W,S)$ be a nontrivially $\mathbf{a}(2)$-finite Coxeter system of type
$X$. Then the following set equalities hold in $W$.

  1. (1)

If $X=A_{n}$ where $n\geq 3$, then $I(13,13)=\\{13\\}$.

  2. (2)

If $X=B_{3}$, then $I(13,13)=\\{13\\}$.

  3. (3)

If $X=B_{n}$ where $n>3$, then $I(24,24)=\\{24,2124\\}$.

  4. (4)

If $X=\tilde{C}_{n-1}$ where $n\geq 5$, then $I(24,24)=\\{24,2124,2\cdot
z,212\cdot z\\}$ where $z=45\cdots(n-1)n(n-1)\cdots 54$.

  5. (5)

If $X=E_{q,r}$ where $r\geq q\geq 1$, then for all $x\in\\{(-1)v,1v,(-1)1\\}$
we have $I(x,x)=\\{x\\}$.

  6. (6)

If $X=F_{4}$, then $I(24,24)=\\{24\\}$.

  7. (7)

If $X=F_{n}$ where $n>4$, then $I(24,24)=\\{24,243524\\}$.

  8. (8)

If $X=H_{3}$, then $I(13,13)=\\{13\\}$.

  9. (9)

If $X=H_{n}$ where $n>3$, then $I(24,24)=\\{24,2124\\}$.

######  Proposition 4.18.

Maintain the notation of 4.16. Then the numbers $N_{ij}$ where $1\leq i\leq
j\leq d$ are given by Table 8.

$X$ |  Values of $N_{ij}\,(1\leq i<j\leq d)$
---|---
$A_{n},n\geq 3$ |  $N_{11}=1$
$B_{3}$ |  $N_{11}=1$
$B_{n},n>3$ |  $N_{11}=N_{22}=2,N_{12}=1$
$\tilde{C}_{n},n\geq 5$ |  $N_{11}=N_{22}=N_{33}=N_{44}=4,\newline N_{12}=N_{23}=N_{14}=N_{34}=2,N_{13}=N_{24}=1$
$E_{q,r},r>q=1$ |  $N_{ii}=1$ for all $1\leq i\leq d$, $N_{ij}=0$ whenever $i\neq j$
$E_{q,r},r>q\geq 2$ |  $N_{11}=1$
$F_{4}$ |  $N_{11}=1$
$F_{n},n>4$ |  $N_{11}=N_{22}=2,N_{12}=1$
$H_{3}$ |  $N_{11}=1$
$H_{n},n>3$ |  $N_{ij}=2$ for all $1\leq i\leq j\leq 2$.
Table 8. Sizes of 0-cells of $\mathbf{a}$-value 2

###### Proof.

We sketch the proof. In 4.17 we have described $I_{ij}=I(w_{i},w_{j})$ for
some particular $i,j$. Taking the cardinality of this set shows that the
number $N_{ij}$ given in Table 8 is correct. For every other pair $(k,l)$ with
$1\leq k\leq l\leq d$, we can obtain $I_{kl}$ from $I_{ij}$ as explained in
Section 4.2 and check that that $N_{kl}=\lvert I_{kl}\rvert$ agrees with the
value given by Table 8. ∎

######  Remark 4.19.

In the above proof we mentioned computing $N_{kl}$ by computing the 0-cell
$I_{kl}$ and taking its cardinality, but the computation of $N_{kl}$ can often
be simplified. Since we are not interested in the set $I_{kl}=I(w_{k},w_{l})$
per se, we do not have to compute $I_{kl}$ from the set $I_{ij}$. Instead, it
is often convenient to compute a 0-cell $I^{\prime}_{kl}:=I(a,b)$ from a
0-cell $I^{\prime}_{ij}:=I(c,d)$ where $a,b,c,d$ are short stubs such that
$a\sim w_{k},b\sim w_{l},c\sim w_{i},d\sim w_{j}$ and such that we can relate
$c$ to $a$, and $d$ to $b$, with a small number of slides. Here the set
$I^{\prime}_{ij}$ has $N_{ij}$ elements by 4.12 and $N_{ij}$ is at most 4 by
Table 8, so in practice it is not difficult to obtain $I_{ij}$ by finding
$N_{ij}$ elements that are obviously in it. The point of using the stubs
$a,b,c,d$ is that if we select them carefully then we may obtain
$I^{\prime}_{kl}$ from $I^{\prime}_{ij}$ using a smaller number of setwise
star operations than we need for obtaining $I_{kl}$ from $I_{ij}$. For
example, in the group $\tilde{C}_{n-1}$ we have $w_{1}=13,w_{2}=24,w_{4}=1n$
by 4.16 and $N_{22}=\lvert I_{22}\rvert=4$ by 4.17.(4), and to compute
$N_{14}$ it is convenient to use the stubs $a:=14\sim w_{1},b:=1n\sim
w_{4},c:=24\sim w_{2},d:=2(n-1)\sim w_{2}$ as follows: the set
$I^{\prime}_{22}=I(c,d)=I(24,2(n-1))$ should contain $N_{22}=4$ elements, so
once we note that the set

$I^{\prime}:=\\{2\cdot z^{\prime},2\cdot z^{\prime}\cdot n(n-1),212\cdot
z^{\prime},212\cdot z^{\prime}\cdot n(n-1)\,|\,z^{\prime}=45\cdots(n-1)\\}$

is contained in $I^{\prime}_{22}$ we can conclude that
$I^{\prime}_{22}=I^{\prime}$. Since $a$ can be obtained from $c$ with one
slide along the edge $\\{1,2\\}$, applying the corresponding setwise left star
operations to $I^{\prime}_{22}=I(c,d)$ yields

$I(a,d)=\\{12\cdot z^{\prime},12\cdot z^{\prime}\cdot
n(n-1)\,|\,z^{\prime}=45\cdots(n-1)\\}.$

Similarly, since $b$ can be obtained from $d$ with two slides, along the edges
$\\{1,2\\}$ and $\\{n-1,n\\}$, applying the two corresponding setwise right
star operations in succession to $I(a,d)$ yields

$I^{\prime}_{14}=I(a,b)=\\{1\cdot z^{\prime\prime},121\cdot
z^{\prime\prime}\,|\,z^{\prime\prime}=45\cdots n\\},$

therefore $N_{14}=2$. Note that we only need three setwise star operations,
even if $n$ is large. $n$. In contrast, if we computed $I_{14}$ from $I_{22}$
by working with $w_{1}=13,w_{2}=24$ and $w_{4}=(n-1)n$ directly, then we need
many more setwise star operations to obtain $I_{14}$ when $n$ is large.

###  4.4. Sizes of $\mathbf{a}(2)$-cells

We are ready to compute the sizes of all left, right, and two-sided
Kazhdan–Lusztig cells of $\mathbf{a}$-value 2 in all $\mathbf{a}(2)$-finite
Coxeter groups:

######  Theorem 4.20.

Let $(W,S)$ be a nontrivially $\mathbf{a}(2)$-finite Coxeter system of type
$X$ and maintain the notation of § 4.3. In particular, let
$\beta_{n}=\binom{n}{2}$ for all $n\geq 2$. Then the sizes of 1-cells and
2-cells in $W_{2}$ are given by Table 9. When $W_{2}$ is a single two-sided
cell of size $N$, we simply write $N$ instead of $\lvert W_{2}\rvert=N$ in the
corresponding box in the table.

$X$ |  sizes of 1-cells |  sizes of 2-cells
---|---|---
$A_{n},n\geq 3$ |  $N_{1}=\beta_{n}-1$ |  $(\beta_{n}-1)^{2}$
$B_{3}$ |  $N_{1}=3$ |  $9$
$B_{n},n>3$ |  $N_{1}=\beta_{n+1},N_{2}=n^{2}-2n$ |  ${n^{4}}/{2}-2n^{3}+{7n^{2}}/{2}$
$\tilde{C}_{n-1},n\geq 5$ |  $N_{1}=N_{3}=n^{2}+1,\newline N_{2}=2n^{2}-6n+5,\newline N_{4}=n^{2}/2+{3}n/2+2$ |  $n^{4}-6n^{3}+20n^{2}-21n+10$
$E_{11}=D_{4}$ |  $N_{1}=N_{2}=N_{3}=3$ |  $\lvert C_{1}\rvert=\lvert C_{2}\rvert=\lvert C_{3}\rvert=9$
$E_{q,r},r>q=1$ |  $N_{1}=n-1,\newline N_{2}=\beta_{n}$ |  $\lvert C_{1}\rvert=(n-1)^{2},\newline \lvert C_{2}\rvert=\beta_{n}^{2}$
$E_{q,r},r\geq q\geq 2$ |  $N_{1}=\beta_{n+1}-1$ |  $N_{1}=(\beta_{n+1}-1)^{2}$
$F_{4}$ |  $N_{1}=9$ |  $81$
$F_{n},n>4$ |  $N_{1}=n^{2}/2+{7}n/2-4,\newline N_{2}=(n-1)^{2}$ |  $n^{4}/2-2n^{3}+{33}n^{2}/2-29n+14$
$H_{3}$ |  $N_{1}=5$ |  $25$
$H_{n},n>3$ |  $N_{1}=N_{2}=2(\beta_{n+1}-1)$ |  $2(\beta_{n+1}-1)^{2}$
Table 9. Sizes of 1-cells and 2-cells of $\mathbf{a}$-value 2

###### Proof.

We know the numbers $n_{i}$ and $N_{ij}$ for all $1\leq i,j\leq d$ by
Propositions 4.16, 4.18 and 4.15.(6), so the sizes of all 1-cells and 2-cells
can be calculated by directly applying the formulas from 4.15.(3)–(5). The
calculations are straightforward and similar for all $\mathbf{a}(2)$-finite
Coxeter types, so instead of carrrying out the computations in detail for all
types, we only illustrate below how to count the cells in
$\tilde{C}_{n-1}(n\geq 5)$, where the number $d$ of $\sim$-classes in
$\mathcal{S}(W)$ is the largest and the computations are the most complex.
Recall from 4.16 that $\mathcal{S}(W)$ contains $d=4$ slide equivalence
classes and their sizes are $n_{1}=n-1,n_{2}=\beta_{n-2}-1,n_{3}=n-1$ and
$n_{4}=1$. Consider the following $d\times d$ table, where the rows and
columns are labeled by $n_{i}$ and the entry in the $i$-th row, $j$-th column
is the number $N_{ij}$ from Table 8.

| $n-1$ | $\beta_{n-2}-1$ | $n-1$ | $1$
---|---|---|---|---
$n-1$ | 4 | 2 | 1 | 2
$\beta_{n-2}-1$ | 2 | 4 | 2 | 1
$n-1$ | 1 | 2 | 4 | 2
$1$ | 2 | 1 | 2 | 4
Table 10.

By 4.15.(2), for each $1\leq i\leq d$ we can compute $N_{i}$ as a weighted sum
of the $i$-th row in the above table, with the column labels as weights. For
example, we have

$N_{1}=\sum_{j=1}^{d}n_{j}N_{ij}=4\cdot(n-1)+2\cdot(\beta_{n-2}-1)+1\cdot(n-1)+2\cdot
1=n^{2}+1.$

Similar computations show that $N_{2}=2n^{2}-6n+5,N_{3}=n^{2}+1$ and
$N_{4}=n^{2}/2+3n/2+2$. Having computed $N_{1},\dots,N_{d}$, we can then
compute the size of the 2-cell $W_{2}$ as the weighted sum
$\sum_{i=1}^{d}n_{i}N_{i}$ by 4.15.(4). Doing so yields $\lvert
W_{2}\rvert=n^{4}-6n^{3}+20n^{2}-21n+10$. ∎

##  5\. Representative 0-cells

We assume $(W,S)$ is a nontrivially $\mathbf{a}(2)$-finite Coxeter system and
prove 4.17 in this section. We complete the proof in § 5.2 after preparing a
series of technical definitions and lemmas in § 5.1.

###  5.1. Lemmas on heaps

Let $G$ denote the Coxeter diagram of $(W,S)$. Then $G$ has no cycles,
therefore for any two vertices $u,v$ in $G$ there is a unique path from $u$ to
$v$ (see Remark 2.14). For pairwise distinct vertices $u,v,t$, we say that $u$
and $v$ _lie on different sides of_ a vertex $t$ if the path connecting $u$ to
$v$ passes through $t$. We call two vertices _neighbors_ of each other if they
are adjacent in $G$, define the _degree_ of a vertex $v$ in $G$ to be the
number of its neighbors, and call a vertex an _end vertex_ if it has degree 1.
For each element $w\in\operatorname{\mathrm{FC}}(W)$ and any two elements
$i,j$ in the heap $H(w)$, we denote the label of $i$ by $e(i)$, write
$(i,j):=\\{h\in H(w):i\precneqq h\precneqq j\\}$, and write $[i,j]:=\\{h\in
H(w):i\preceq h\preceq j\\}$.

The following two lemmas allow us to deduce the existence of additional
elements and structure in $H(w)$ from certain local configurations. The vertex
whose existence is asserted in the lemmas allows us to expand the lattice
embedding of the heap vertically and horizontally, which is why we name the
lemmas vertical and horizontal completion.

######  Lemma 5.1 (Vertical Completion).

Let $a$ be an end vertex in $G$ and let $b$ be its only neighbor. Suppose
$b\in\mathcal{L}(w)$ and $e(i)=a$ for an element $i\in H(w)$. Then $H(w)$
contains an element that is covered by $i$ and has label $b$.

###### Proof.

Since $b\in\mathcal{L}(w)$, some minimal element $j\in H(w)$ has label $b$ by
Remark 2.13. The elements $i$ and $j$ are comparable since $a$ and $b$ are
adjacent in $G$. Moreover, we must have $j\leq i$ since $j$ is minimal, so
there is a chain $j=j_{1}\prec\dots j_{p-1}\prec j_{p}=i$ of coverings
connecting $j$ to $i$ in $H(w)$. Since $b$ is the only neighbor of $a$, we
must have $e(j_{p-1})=b$. It follows that $j_{p-1}$ has the desired
properties. ∎

######  Lemma 5.2 (Horizontal Completion).

Let $a$ be an end vertex in $G$ and let $b$ be its only neighbor. Suppose that
$C=(j\prec i\prec j^{\prime})$ is a chain of coverings in $H(w)$ with $e(i)=a$
and $e(j)=e(j^{\prime})=b$. Then $i,j,j^{\prime}$ account for all elements in
$[j,j^{\prime}]$ that are labeled by $a$ or $b$. Moreover, if $C$ is not
convex, then $j$ and $j^{\prime}$ are connected by a chain $D=(j=j_{1}\prec
j_{2}\prec\dots\prec j_{q}=j^{\prime})$ of coverings where $q\geq 3$. In this
case, at least one element in $D$ is labeled by a neighbor of $b$ in $G$ that
is distinct from $a$.

###### Proof.

An element in $(j,j^{\prime})$ with label $b$ would be comparable to $i$ and
thus contradict the fact that $i$ covers $j$ and $j^{\prime}$ covers $i$, so
$j$ and $j^{\prime}$ are the only elements with label $b$ in $[j,j^{\prime}]$.
Since $b$ is the only neighbor of $a$, it further follows that
$(j,j^{\prime})$ contains no element with label $a$ other than $i$, because
otherwise such an element would cover or be covered by $i$, violating
Condition (1) of 2.12. We have proved the first claim.

If $C$ is not convex, then $j$ and $j^{\prime}$ must both be comparable to
some element $k\in(j,j^{\prime})$ that is not in $C$. Concatenating chains of
coverings $j\prec\dots\prec k$ and $k\prec\dots\prec j^{\prime}$ connecting
$j$ to $k$ and $k$ to $j^{\prime}$ yields a chain $D$ in $[j,j^{\prime}]$ that
connects $j$ to $j^{\prime}$ and has at least three elements, as desired. The
chain $D$ corresponds to a walk of length at least 3 from $b$ to $b$ in $G$ by
Remark 2.14. Such a walk must involve a neighbor of $b$, and such a neighbor
cannot be $a$ by the first claim since $D$ lies in the interval
$[j,j^{\prime}]$. The proof is complete. ∎

The next result proves that certain subsets of heaps are antichains.

######  Lemma 5.3.

Let $x,y,z$ be elements of $S$ such that $x$ and $z$ lie on different sides of
$y$. Suppose that $H(w)$ contains two elements $i$ and $k$ labeled by $x$ and
$z$, respectively. Then the set $\\{i,k\\}$ is an antichain in $H(w)$ whenever
one of the following conditions holds.

  1. (1)

The heap $H(w)$ contains two elements $j,j^{\prime}$ labeled by $y$ such that
$j\prec i\prec j^{\prime}$ and $k\in(j,j^{\prime})$.

  2. (2)

The heap $H(w)$ contains an element $j$ labeled by $y$ and the sets
$\\{i,j\\}$ and $\\{j,k\\}$ are antichains in $H(w)$.

###### Proof.

We prove the lemma by contradiction. Suppose $i$ and $k$ are comparable and
assume, without loss of generality, that $i\preceq k$ in $H(w)$. Then by
Remark 2.14, there is a chain of coverings $i=i_{1}\prec i_{2}\prec\dots\prec
i_{l}=k$ in $H(w)$ whose elements’ labels form a walk on $G$ from $x$ to $z$.
The walk must pass through $y$, so $e(i_{p})=y$ for some $1\leq p\leq l$.
Elements sharing a label are comparable in a heap, so $i_{p}$ is comparable to
$j$ and $j^{\prime}$ under Condition (1) and is comparable to $j$ under
Condition (2).

Suppose Condition (1) holds. Then the interval $(j,j^{\prime})$ contains no
element with label $y$ by Lemma 5.2, so we have either $i_{p}\preceq j$ or
$j^{\prime}\preceq i_{p}$ in $H(w)$. But in these cases we would have
$i\preceq i_{p}\preceq j$ or $j^{\prime}\preceq i_{p}\preceq k$, respectively,
contradicting the fact that $k$ and $i$ lie in $(j,j^{\prime})$.

Under Condition (2), we have either $i_{p}\prec j$ or $j\prec i_{p}$, which
would imply that $i\preceq i_{p}\preceq j$ or $j\preceq i_{p}\preceq k$ and
violate the assumption that $\\{i,j\\}$ and $\\{j,k\\}$ are antichains. ∎

Lemma 5.3 will be used in tandem with Lemma 5.2 in Section 5.2. It also helps
establish our next proposition, 5.7, which concerns the following notions:

######  Definition 5.4.

We define a _special quadruple in $S$_ to be an ordered tuple $(a,b,c,d)$ of
four distinct elements in $S$ with the following properties.

  1. (1)

We have $m(a,c)=m(a,d)=m(b,d)=2$, $m(a,b)\geq 3,m(b,c)\geq 3$ and $m(c,d)=3$.
In particular, the generators $a,b,c,d$ induce a subgraph of the form
$a$–$b$–$c$–$d$ in $G$ if we ignore edge weights.

  2. (2)

The vertices $a$ and $c$ have degrees 1 and 2 in $G$, respectively. In other
words, the vertex $b$ is the only neighbor of $a$, and the vertices $b,d$ are
the only neighbors of $c$.

######  Definition 5.5.

Let $(a,b,c,d)$ be a special quadruple in $S$. We call an antichain
$\\{i,j\\}$ of size 2 in $H(w)$ a _trapped antichain relative to $(a,b,c,d)$_
if the following conditions hold.

  1. (1)

Neither $i$ nor $j$ is maximal or minimal in $H(w)$.

  2. (2)

The label of $i$ is $a$ and the label of $j$ is $c$.

######  Example 5.6.

In each Coxeter system $(W,S)$ of type $\tilde{C}_{n-1}$ where $n\geq 5$, the
tuple $(1,2,3,4)$ is a special quadruple. In the heap of the FC element
$w=2413524\in W$, the elements with labels 1 and 3 form a trapped antichain
relative to $(1,2,3,4)$. Note that $n(w)=3$, hence $\mathbf{a}(w)=n(w)>2$ by
2.15. The next proposition shows that the fact that $\mathbf{a}(w)\neq 2$ is
to be expected.

######  Proposition 5.7.

Let $(W,S)$ be a nontrivially $\mathbf{a}(2)$-finite Coxeter system, let $w\in
W_{2}$, and suppose $\lvert\mathcal{L}(w)\rvert=\lvert\mathcal{R}(w)\rvert=2$.
Then the heap $H(w)$ cannot contain any trapped antichain relative to a
special quadruple in $S$.

###### Proof.

Suppose that $H(w)$ contains a trapped antichain $A=\\{i,j\\}$ relative to a
special quadruple $(a,b,c,d)$ in $S$, with $e(i)=a$ and $e(j)=c$. We will
derive a contradiction. To start, note that $A$ is a maximal antichain in
$H(w)$ by Corollary 2.16. By Lemma 2.17 and Definition 4.1.(2), it follows
that $w$ can be written as a product $w=x*y$ such that $H(x)$ and $H(y)$
coincide with the ideal $\mathcal{I}_{A}$ and filter $\mathcal{F}_{A}$ when
viewed naturally as subsets of $H(w)$. Lemma 3.18 then implies that
$\mathcal{L}(x)=\mathcal{L}(w)$, so $x$ has two left descents. On the other
hand, Condition (1) of Definition 5.5 implies that $\mathcal{I}_{A}$ properly
contains $A$, so $x\neq ac$ and $l(x)>2$. It follows that
$x\notin\mathcal{S}(W)$, because no stub in $\mathcal{S}(W)$ with length at
least 3 has two left descents by Lemma 3.9.(1).

Since $i$ is not minimal, it must cover an element $g\in H(w)$. The label of
$g$ must be a neighbor of $e(i)=a$, which has to be $b$ by Condition (2) of
Definition 5.4. Since $b$ is also adjacent to $c$ in $G$, the element $g$ is
comparable to $j$ as well. If $j\preceq g$ then $j\preceq g\prec i$ and $A$
would not be antichain, so $g\preceq j$. It follows that neither the set
$H(x)\setminus\\{i\\}$ nor the set $H(x)\setminus\\{j\\}$ contains a maximal
element labeled by $b$, therefore $x$ admits no right lower star operation
with respect to $\\{a,b\\}$ or $\\{c,b\\}$ by 2.22. On the other hand, since
$x$ is not a stub it must admit a right lower star operation with respect to
some pair $\\{s,t\\}$ of noncommutating generators, and one of $s,t$ has to
come from the set $\mathcal{R}(x)=\\{a,c\\}$. Condition (2) of Definition 5.4
now forces $\\{s,t\\}=\\{c,d\\}$. It follows that $j$ covers an element $h\in
H(x)\subseteq H(w)$ labeled by $d$. A similar argument shows that $H(w)$
contains an element $h^{\prime}$ that lies in $\mathcal{F}_{A}$, covers $j$,
and has label $d$.

Since $m(c,d)=3$ by Definition 5.4, the chain $h\preceq j\preceq h^{\prime}$
cannot be convex by 2.12, so Lemma 5.2 implies that $h$ and $h^{\prime}$ are
connected by a chain of coverings containing an element $k$ labeled by some
neighbor $e$ of $d$ distinct from $c$. The vertices $a,b,c,d,e$ induce a
subgraph of $G$ of the form $a$–$b$–$c$–$d$–$e$, so we may now apply Lemma 5.3
as follows: first, since $c,e$ lie on different sides of $d$, Part (1) of the
lemma implies that $\\{j,k\\}$ forms an antichain; next, since $a,e$ lie on
different sides of $c$ and $\\{i,j\\},\\{j,k\\}$ are both antichains, Part (2)
of the lemma implies that $\\{i,k\\}$ forms an antichain. It follows that
$\\{i,j,k\\}$ is an antichain in $H(w)$. 2.15 then implies that
$\mathbf{a}(w)\geq n(w)\geq 3$, contradicting the assumption that
$\mathbf{a}(w)=2$. ∎

###  5.2. Proof of 4.17

Let $(W,S)$ be a nontrivially $\mathbf{a}(2)$-finite Coxeter system, and
recall that each part of 4.17 claims that if $(W,S)$ has a certain type $X$,
then a representative 0-cell $I(x,y):=I(x,y)\subseteq W$ where
$x,y\in\mathcal{S}(W)$ equals a certain set. Denoting this set by $K(x,y)$, we
first note that $K(x,y)\subseteq I(x,y)$ regardless of what $X$ is:

######  Proposition 5.8.

We have $K(x,y)\subseteq I(x,y)$ for every Coxeter system $(W,S)$ in 4.17.

###### Proof.

Let $w\in K(x,y)$. Then $w$ has reduced factorizations $w=x\cdot w^{\prime}$
and $w=w^{\prime\prime}\cdot y^{-1}$ by inspection, so we can conclude that
$w\in I(x,y)$ by 3.17 once we can show $\mathbf{a}(w)=2$. If $w$ is a product
$st$ of two commuting generators $s,t\in S$, then $\mathbf{a}(w)=2$ by
2.3.(2). Otherwise, we observe that $w$ can be obtained from such a product
$24$ via a sequence of right upper star operations, therefore
$\mathbf{a}(w)=\mathbf{a}(24)=2$ by Corollary 2.21. Specifically, the element
$w=243524$ in Part (7) can be built from $24$ by applying right upper star
operations with respect to $\\{4,5\\},\\{2,3\\},\\{2,3\\}$ and $\\{4,5\\}$
successively, and a suitable sequence of star operations taking $24$ to $w$ is
straightforward to find in all other cases. ∎

To prove 4.17 it remains to prove that $I(x,y)\subseteq K(x,y)$ for each group
$W$ in the theorem. We will do so by treating one Coxeter type at a time
below, and the following notions will be used frequently. For every element
$w\in\operatorname{\mathrm{FC}}(W)$ and any generator $p\in S$, we denote the
number of elements labeled by $p$ in the heap by $o_{w}(p)$, where we drop the
subscript $w$ if there is no danger of confusion. We define a _$p$ -interval_
in $H(w)$ to be an interval of the form $[j,j^{\prime}]\subseteq H(w)$ where
$j,j^{\prime}$ are consecutive elements labeled by $p$ in the sense that
$j\preceq j^{\prime}$, $j\neq j^{\prime}$, and $j,j^{\prime}$ are the only
elements with label $p$ in the interval $[j,j^{\prime}]$. Finally, when we
wish to emphasize the ambient group $W$ that a 0-cell $I(x,y)$ lies in, we
will write $I_{W}(x,y)$ for $I(x,y)$.

The following facts will also be used frequently without further comment.
First, if $x,y$ are short stubs, then every element $w\in I(x,y)$ has left
descent set $\mathcal{L}(w)=\mathcal{L}(x)=\operatorname{\mathrm{Supp}}(x)$
and right descent set
$\mathcal{R}(w)=\mathcal{R}(y^{-1})=\mathcal{R}(y)=\operatorname{\mathrm{Supp}}(y)$
by Lemma 3.9.(1). Second, recall from Remark 2.13 that the left and right
descents of an FC elements are precisely the labels of the minimal and maximal
elements of its heap, respectively. Third, consider the situation where an
$\mathbf{a}(2)$-finite Coxeter system $(W,S)$ is isomorphic to a parabolic
subgroup $W^{\prime}_{J}$ of another $\mathbf{a}(2)$-finite Coxeter system
$(W^{\prime},S^{\prime})$ for some subset $J$ of $S^{\prime}$ via a group
isomorphism $\phi:W\rightarrow W^{\prime}_{J}$ induced by a bijection
$f:S\rightarrow J$ with the property that $m(f(s),f(t))=m(s,t)$ for all
$s,t\in S$. (By “induced” we mean the restriction of $\phi$ to $S$ coincides
with $f$.) Then the next lemma holds for any two stubs $x,y\in\mathcal{S}(W)$,
where $f(Z)=\\{f(z):z\in Z\\}$ for every subset $Z$ of a Coxeter group and any
map $f$ whose domain contains $Z$. The lemma allows us to deduce 0-cells of
$\mathbf{a}$-value 2 in $W$ from such cells in $W^{\prime}$ by “parabolic
restriction”.

######  Lemma 5.9.

In the above setting, let $x^{\prime}=\phi(x),y^{\prime}=\phi(y)$, denote the
0-cell $I_{W}(x,y)$ in $W$ by $I_{1}$, denote the 0-cell
$I_{W^{\prime}_{J}}(x^{\prime},y^{\prime})$ in $W^{\prime}_{J}$ by $I_{2}$,
and denote the 0-cell $I_{W^{\prime}}(x^{\prime},y^{\prime})$ in $W^{\prime}$
by $I_{3}$. Then we have $I_{2}=\phi(I_{1})$ and $I_{2}=\\{z\in
I_{3}:\operatorname{\mathrm{Supp}}(z)\subseteq J\\}$. In particular, we have

$I_{W}(x,y)=\\{\phi^{-1}(z):z\in
I_{W^{\prime}}(x^{\prime},y^{\prime}),\operatorname{\mathrm{Supp}}(z)\subseteq
J\\}.$

###### Proof.

We first sketch why $I_{2}=\phi(I_{1})$: the fact that $\phi$ is induced by
the bijection $f:S\rightarrow J$ implies that $\phi$ is in fact an isomorphism
of Coxeter systems, so that $\phi$ in turn induces an isomorphism from the
Hecke algebra of $(W,S)$ to the Hecke algebra of $(W^{\prime}_{J},J)$ that
sends the Kazhdan–Lusztig basis element $C_{w}$ to $C_{\phi(w)}$ for all $w\in
W$. It follows that $I_{2}=\phi(I_{1})$.

Next, let $\mathbf{a}_{S^{\prime}}$ and $\mathbf{a}_{J}$ denote the
$\mathbf{a}$-functions associated to the Coxeter system
$(W^{\prime},S^{\prime})$ and $(W^{\prime}_{J},J)$, respectively. By 2.3.(7),
the system $(W^{\prime}_{J},J)$ is $\mathbf{a}(2)$-finite since
$(W^{\prime},S^{\prime})$ is, so 3.17 and its analog for left cells imply that

$I_{2}=\\{z\in W^{\prime}_{J}:\mathbf{a}_{J}(z)=2,x\leq^{R}z,y\leq^{L}z\\}$

and

$I_{3}=\\{z\in
W^{\prime}:\mathbf{a}_{S^{\prime}}(z)=2,x\leq^{R}z,y\leq^{L}z\\}.$

An element $z\in W^{\prime}$ is in the parabolic subgroup $z\in
W^{\prime}_{J}$ if and only if $\operatorname{\mathrm{Supp}}(w)\subseteq J$,
and for such an element we have $\mathbf{a}_{S^{\prime}}(z)=\mathbf{a}_{J}(z)$
by 2.3.(8). It follows that $I_{2}=\\{z\in
I_{3}:\operatorname{\mathrm{Supp}}(z)\subseteq J\\}$. ∎

####  5.2.1. Type $\tilde{C}_{n-1}$

Suppose $X=\tilde{C}_{n-1}(n\geq 5)$. We complete the proof of 4.17.(4) by
proving the following:

######  Proposition 5.10.

If $X=\tilde{C}_{n-1}(n\geq 5)$, then we have

$I(24,24)\subseteq\\{24,2124,2\cdot z,212\cdot z\\}$

where $z=45\dots(n-1)n(n-1)\dots 54$.

To prove the proposition, let

$z_{p}=p(p-1)\dots 212\dots(p-1)p,$
$z_{p}^{\prime}=p(p+1)\dots(n-1)n(n-1)\dots(p+1)p$

for all $1<p<n$. For example, the elements $212$ and $z$ from the proposition
equal $z_{2}$ and $z_{4}^{\prime}$, respectively. We will take advantage of
the following result of Ernst:

######  Lemma 5.11 ([13]).

Let $n\geq 4$ and let $w\in\operatorname{\mathrm{FC}}(\tilde{C}_{n-1})$. Let
$1<p<n$ and suppose $[j,j^{\prime}]$ is a $p$-interval in $H(w)$ for two
elements $j\preceq j^{\prime}$ in $H(w)$.

  1. (1)

If the interval $[j,j^{\prime}]$ contains no element labeled by $(p+1)$, then
either $\mathbf{a}(w)=1$ or $w$ has a reduced factorization $w=u\cdot
z_{p}\cdot u^{\prime}$ where
$\operatorname{\mathrm{Supp}}(u)\cup\operatorname{\mathrm{Supp}}(u^{\prime})\subseteq\\{p+1,p+2\dots,n\\}$.

  2. (2)

If the interval $[j,j^{\prime}]$ contains no element labeled by $(p-1)$, then
either $\mathbf{a}(w)=1$ or $w$ has a reduced factorization $w=u\cdot
z^{\prime}_{p}\cdot u^{\prime}$ where
$\operatorname{\mathrm{Supp}}(u)\cup\operatorname{\mathrm{Supp}}(u^{\prime})\subseteq\\{1,2\dots,p-1\\}$.

###### Proof.

Part (1) is a restatement of Lemma 3.6 of [13]. (We note that each element “of
Type I” in [13] has a unique reduced word by §3.1 of the paper and thus has
$\mathbf{a}$-value 1 by 2.5.(2).) Part (2) follows from Part (1) by the
symmetry in the Coxeter diagram of $\tilde{C}_{n-1}$. ∎

###### Proof of 5.10.

Let $w\in I(24,24)$. Then $2$ and $4$ each labels both a maximal and a minimal
element in $H(w)$. It follows that $o(2),o(4)\geq 1$, and that if
$3\in\operatorname{\mathrm{Supp}}(w)$ then every element labeled by 3 in
$H(w)$ has to lie in a 2-interval. Note that every 2-interval in $H(w)$ must
contain an element labeled by $1$: otherwise, Lemma 5.11.(2) implies that
$w=u\cdot z^{\prime}_{2}\cdot u^{\prime}$ where $4$ labels no maximal element
in $z_{2}^{\prime}$ and does not appear in $u$ or $u^{\prime}$, so $4$ labels
no maximal element in $H(w)$, a contradiction. On the other hand, since
$(1,2,3,4)$ is a special quadruple in $S$, no 2-interval in $H(w)$ can contain
both an element labeled by 1 and an element labeled by 3 by 5.7. It follows
that $3\notin\operatorname{\mathrm{Supp}}(w)$. For $i\in\\{2,4\\}$, let
$a_{i}$ and $b_{i}$ stand for the minimal and maximal element in $H(w)$
labeled by $i$, respectively. Since $3\notin\operatorname{\mathrm{Supp}}(w)$,
Remark 2.14 implies that no two elements in $H(w)$ labeled by two generators
in $S$ that lie on different sides of 3 are comparable. Since every element in
$H(w)$ must be comparable to a maximal and to a minimal element, it follows
that $H(w)$ equals the disjoint union of the two convex intervals
$I_{2}:=[a_{2},b_{2}]$ and $I_{4}:=[a_{4},b_{4}]$ as sets, where each element
in $I_{2}$ and $I_{4}$ carries a label smaller and larger than 3,
respectively. It follows that $w=x\cdot y$ for FC elements $x,y$ such that
$H(x)=I_{1}$ and $H(y)=I_{2}$. Applying Lemma 5.11.(1) with $p=2$ and Lemma
5.11.(2) with $p=4$ reveals that $x\in\\{2,212\\}$ and
$y\in\\{4,z^{\prime}_{4}=z\\}$, respectively, therefore we have
$w\in\\{24,2124,2z,212z\\}$. We conclude that
$I(24,24)\subseteq\\{24,2124,2z,212z\\}$. ∎

####  5.2.2. Type $B_{n}$

We complete the proof of 4.17.(2)–(3) by deducing the following proposition
from 5.10.

######  Proposition 5.12.

If $X=B_{n}(n\geq 3)$, then $I(13,13)\subseteq\\{13\\}$ if $n=3$ and
$I(24,24)\subseteq\\{24,2124\\}$ if $n>3$.

###### Proof.

Let $k=n+2\geq 5$, consider the Coxeter system of type
$(W^{\prime},S^{\prime})$ of type $\tilde{C}_{k-1}$, and let
$J=\\{1,2,\dots,n\\}\subseteq S^{\prime}$. The group $B_{n}$ is naturally
isomorphic to parabolic subgroup $W^{\prime}_{J}$ of $W^{\prime}$ via the
isomorphism $\phi:W\rightarrow W^{\prime}_{J}$ with $\phi(i)=i$ for each
generator $i\in S$. Thus, Lemma 5.9 implies that if $n>3$ then
$I(24,24)=I_{W}(24,24)\subseteq\\{\phi^{-1}(w^{\prime}):w^{\prime}\in
I_{W^{\prime}}(24,24),\operatorname{\mathrm{Supp}}(w^{\prime})\subseteq J\\}$,
where the last set equals $\\{\phi^{-1}(24),\phi^{-1}(2124)\\}=\\{24,2124\\}$.
It follows that $I(24,24)\subseteq\\{24,2124\\}$. Similarly, if $n=3$, then
using the methods from Remark 4.13 we can see that
$I_{W^{\prime}}(13,13)=\\{13,134543,132413,13245413\\}$, whence Lemma 5.9
implies that $I(13,13)\subseteq\\{13\\}$ since $4\notin J$. ∎

####  5.2.3. Type $A_{n}$

We complete the proof of 4.17.(1) by deducing the following proposition from
5.12.

######  Proposition 5.13.

If $X=A_{n}(n\geq 3)$, then $I(13,13)\subseteq\\{13\\}$.

###### Proof.

We use parabolic restriction as in the proof of 5.12. More precisely, let
$(W,S)$ and $(W^{\prime},S^{\prime})$ be the Coxeter systems of type $A_{n}$
and $B_{n+1}$, respectively, let $J=\\{2,3,\dots,n+1\\}$, and consider the
isomorphism $\phi:W\rightarrow W^{\prime}_{J}$ with $\phi(i)=i+1$ for all
$i\in S$. We have $I_{W^{\prime}}(24,24)\subseteq\\{24,2124\\}$ by 5.12 and
$I(13,13)=I_{W}(13,13)\subseteq\\{\phi^{-1}(z):z\in
I_{W^{\prime}}(24,24),\operatorname{\mathrm{Supp}}(z)\in J\\}$ by Lemma 5.9,
therefore $I(13,13)\subseteq\\{\phi^{-1}(24)\\}=\\{13\\}$. ∎

####  5.2.4. Type $E_{q,r}$

We complete the proof of 4.17.(5) by deducing the following result from
4.17.(1):

######  Proposition 5.14.

If $X=E_{q,r}$ where $r\geq q\geq 1$, then $I(x,x)\subseteq\\{x\\}$ for all
$x\in\\{(-1)v,1v,(-1)1\\}$.

###### Proof.

Let $x\in\\{(-1)v,1v,(-1)1\\}$. To prove the proposition we may assume that
$r\geq q>1$ in the Coxeter system $W$ of type $E_{q,r}$: any system
$(W^{\prime},S^{\prime})$ of type $E_{1,r}$ lies in $W$ as a parabolic
subgroup, and Lemma 5.9 implies that if $I_{W}(x,x)\subseteq\\{x\\}$ then
$I_{W^{\prime}}(x,x)\subseteq\\{x\\}$. Under the assumption that $r\geq q>1$,
it further suffices to show that $I((-1)1,(-1)1)\subseteq\\{(-1)1\\}$: since
$\mathcal{S}(W)$ contains a single $\sim$-class by 3.30.(1), if
$I((-1)1,(-1)1)\subseteq\\{(-1)1\\}$ then all 0-cells have size 1 in $W_{2}$
by 4.12, which in turn implies that the 0-cell $I(y,y)$ has to contain only
the element $y$ that is obviously in it for all short stubs
$y\in\mathcal{S}(W)$.

Let $x=(-1)1$ and let $w\in I(x,x)$. Suppose
$v\in\operatorname{\mathrm{Supp}}(w)$, so that some element $i$ in heap $H(w)$
has label $v$. Since $\mathcal{R}(w)=\mathcal{L}(w)=\\{-1,1\\}$, the element
$i$ is neither minimal nor maximal in $H(w)$, so it must cover some element
$j$ and be covered by some element $j^{\prime}$ in $H(w)$ by Lemma 5.1. The
labels of $j$ and $j^{\prime}$ must both be $0$, the only neighbor of $v$ in
$S$. Since $m(v,0)=3$, the chain $j\prec i\prec j^{\prime}$ cannot be convex
by 2.12, so Lemma 5.2 implies that $j$ and $j^{\prime}$ are connected by a
chain in $H(w)$ that contains an element $k$ labeled by either $-1$ or $1$.
Lemma 5.3 then implies that the set $\\{i,k\\}$ forms a trapped antichain
relative to either the special quadruple $(v,0,-1,-2)$ or the special
quadruple $(v,0,1,2)$. This contradicts 5.7, so
$v\notin\operatorname{\mathrm{Supp}}(w)$. But then $w$ lies in the parabolic
subgroup of $W$ of type $A_{q+r+1}$ generated by the set
$\\{-q,\cdots,-1,0,1,\cdots,r\\}$. It follows from 4.17.(1), Lemma 5.9 and
4.12 that $I(x,x)$ is the singleton $\\{x\\}$. The proof is complete. ∎

####  5.2.5. Type $F_{n}$

We complete the proof of 4.17.(6)–(7) by deducing the following proposition
from 5.12.

######  Proposition 5.15.

If $X=F_{n}(n\geq 3)$, then $I(24,24)\subseteq\\{24\\}$ if $n=4$ and
$I(24,24)\subseteq\\{24,243524\\}$ if $n>4$.

###### Proof.

Consider the parabolic group $W_{J}$ of $W$ generated by the set
$J=\\{2,3,\cdots,n\\}$. It is a Coxeter group of type $B_{n-1}$, therefore
$I_{W_{J}}(24,24)=\\{24\\}$ if $n-1=3$ and $I_{W_{J}}(35,35)=\\{35,3235\\}$ if
$n-1>3$ by 4.17.(2)–(3). In the latter case, computing the 0-cell
$I_{W_{J}}(24,24)$ from $I_{W_{J}}(35,35)$ using the methods of Remark 4.13
gives $I_{W_{J}}(24,24)=\\{24,243524\\}$.

Let $w\in I(24,24)$. We claim that $1\notin\operatorname{\mathrm{Supp}}(w)$,
so that $w$ lies in the parabolic subgroup of type $B_{n-1}$ generated by the
set $J=\\{2,3,\dots,n\\}$. It follows that $I(24,24)=I_{W}(24,24)\subseteq
I_{W_{J}}(24,24)$, so $I(24,24)\subseteq\\{24\\}$ if $n=4$ and
$I(24,24)\subseteq\\{24,243524\\}$ if $n>4$ by the last paragraph. To prove
the claim, suppose that $1\in\operatorname{\mathrm{Supp}}(w)$, so that some
element $i$ in heap $H(w)$ has label $1$. This leads to a contradiction to 5.7
in the same way the element $i$ does in the proof of 5.14: Lemmas 5.1, 5.2 and
5.3 force $i$ to be part of an antichain $\\{i,k\\}$ contained in an
$2$-interval $[j,j^{\prime}]\subseteq H(w)$ where $j\prec i\prec j^{\prime}$
and $e(k)=3$, and this antichain is a trapped antichain relative to the
special quadruple $(1,2,3,4)$. It follows that
$1\notin\operatorname{\mathrm{Supp}}(w)$, as claimed. ∎

####  5.2.6. Type $H_{n}$

We complete the proof of 4.17.(8)–(9) by deducing the following proposition
from 5.12.

######  Proposition 5.16.

If $X=H_{n}(n\geq 3)$, then $I(13,13)\subseteq\\{13\\}$ if $n=3$ and
$I(24,24)\subseteq\\{24,2124\\}$ if $n>3$.

###### Proof.

If we can show $I(24,24)=\\{24,2124\\}$ whenever $n>3$, then we can use Remark
4.13 to obtain $I(13,13)=\\{13,132413\\}$ when $n=4$. Viewing $H_{3}$ as the
parabolic subgroup of $H_{4}$ generated by $\\{1,2,3\\}$ naturally, we can
then use Lemma 5.9 to deduce that $I(13,13)=\\{13\\}$ in $H_{3}$. Thus, it
suffices to prove that $I(24,24)\subseteq\\{24,2124\\}$ in $H_{n}$ whenever
$n>3$.

Assume $n>3$ and let $w\in I(24,24)$. Let $\underline{w}$ be a reduced word of
$w$. We claim that the heap $H(w)=H(\underline{w})$ does not contain a convex
chain of the form $i\prec j\prec i^{\prime}\prec j^{\prime}$ where
$e(i)=e(i^{\prime})=1,e(j)=e(j^{\prime})=2$ or where
$e(i)=e(i^{\prime})=2,e(j)=e(j^{\prime})=1$. By 2.12, the claim implies that
$\underline{w}$ is the reduced word of an element $w^{\prime}$ in the Coxeter
group of type $B_{n}$. Moreover, we have
$\mathbf{a}(w^{\prime})=n(w^{\prime})=n(w)=2$ by 3.11.(3), whence 3.17.(3)
implies that $w^{\prime}\in I_{B_{n}}(24,24)=\\{24,2124\\}$ by reduced word
considerations. It follows that $w\in\\{24,2124\\}$ in $H_{n}$, so that
$I(24,24)\subseteq\\{24,2124\\}$ in $H_{n}$, as desired.

It remains to prove the claim, which we do by contradiction. Suppose
$e(i)=e(i^{\prime})=1$ and $e(j)=e(j^{\prime})=2$ in the chain $C$. Since
$1\notin\mathcal{L}(w)$, the element $i$ must cover an element $h$ with label
$2$ in $H(w)$ by Lemma 5.1. Let $C=\\{h,i,j,i^{\prime},j^{\prime}\\}$ and let
$k\in[h,j^{\prime}]$. Applying Lemma 5.2 to the chains $h\prec i\prec j$ and
$j\prec i^{\prime}\prec j^{\prime}$, we note that if $e(k)\in\\{1,2\\}$ then
$k\in C$. Also observe that $e(k)\neq 3$, for otherwise $\\{i,k\\}$ is a
trapped antichain in $H(w)$ with relative to the special quadruple
$(1,2,3,4)$, contradicting 5.7. Finally, we cannot have $e(k)>3$ either,
because otherwise $e(k)$ and $2$ lie on different sides of the vertex 3 in
$G$, so $h$ must be connected to $k$ by a chain in $[h,j^{\prime}]$ passing
through an element labeled by 3, contradicting the last observation. It
follows that $[h,j^{\prime}]=C$ as sets. But then the chain $h\prec i\prec
j\prec i^{\prime}\prec j^{\prime}$ is convex, which contradicts the fact that
$w\in\operatorname{\mathrm{FC}}(H_{n})$ by 2.12. A similar contradiction can
be derived if $e(i)=e(i^{\prime})=2$ and $e(j)=e(j^{\prime})=1$, and the proof
is complete. ∎

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