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\title[Lusztig's sharp bound]{Lusztig's sharp bound for Hecke algebras}
\author{Qiyue Tang}
\address{Zhili College, Tsinghua University, Beijing 100084, China}
\email{tangqy24@mails.tsinghua.edu.cn}
\date{November 8, 2026}
\subjclass[2020]{Primary 20C08; Secondary 20F55}
\keywords{Hecke algebra, Coxeter group, unequal parameters, cone type,
  parabolic subgroup.}
\hypersetup{pdftitle={Lusztig's sharp bound for Hecke algebras},
  pdfauthor={Qiyue Tang},
  pdfsubject={Lusztig's Conjecture 13.4 for positive weight functions},
  pdfkeywords={Hecke algebra, Coxeter group, unequal parameters, sharp bound}}

\begin{document}

\begin{abstract}
We prove Lusztig's sharp boundedness conjecture for Hecke algebras of
finite-rank Coxeter systems with positive integer-valued weight functions.
\end{abstract}

\maketitle

\section{Introduction}

\subsection{The boundedness conjecture}

Kazhdan and Lusztig introduced their basis of a Hecke algebra and the
associated cell relations in \cite{KazhdanLusztig1979}. Lusztig later
introduced the $a$-function in his study of cells in affine Weyl groups
\cite{Lusztig1985}. Its value at a group element is the supremum of the
degrees of the corresponding Kazhdan--Lusztig structure constants.
A uniform bound on these degrees makes the $a$-function
finite-valued. The boundedness problem can also be expressed in the
standard basis, where the multiplication rule gives a direct path
interpretation of degree. We determine the optimal uniform bound in
this form.

Let $(W,S)$ be a Coxeter system with $S$ finite. Write $e$ for its
identity, $\ell$ for its length function, $\leq$ for Bruhat order, and
$m_{st}$ for the order of $st$ for $s,t\in S$. A weight function is a
map $L:W\longrightarrow\Z$ such that
$L(xy)=L(x)+L(y)$ whenever $\ell(xy)=\ell(x)+\ell(y)$. We assume
$L(s)>0$ for every $s\in S$. Let $v$ be an indeterminate and put
$\mathcal A=\Z[v,v^{-1}]$. The Hecke algebra $\HH$ has standard basis
$(T_w)_{w\in W}$, with $T_e=1$ and relations
\[
 T_xT_y=T_{xy}\quad\text{if }\ell(xy)=\ell(x)+\ell(y),
 \qquad
 T_s^2=1+(v^{L(s)}-v^{-L(s)})T_s\quad(s\in S).
\]
Define its standard structure constants by
\[
 T_xT_y=\sum_{z\in W}f_{x,y,z}T_z.
\]
These sums have finite support. For $X\in\HH$, write $[T_z]X$ for
its $T_z$-coefficient. The degree of a nonzero Laurent polynomial is
its largest exponent of $v$, and $\deg0=-\infty$.

A subset $I\subseteq S$ is \emph{spherical} if the standard parabolic
subgroup $W_I=\langle I\rangle$ is finite. For such $I$, let $w_I$ be
the longest element of $W_I$, and set
\begin{equation}\label{eq:capacity}
 N_L(W)=\max_{I\subseteq S,\;W_I\text{ finite}}L(w_I).
\end{equation}
The empty subset is included, with $w_{\varnothing}=e$. The maximum
is finite because $S$ is finite. Lusztig conjectured that
\[
 \deg f_{x,y,z}\leq N_L(W)\qquad(x,y,z\in W)
\]
in \cite[\S13.4]{Lusztig}. In equal parameters, normalized by
$L=\ell$, the prediction is
related to Xi's earlier Question~1.13(iv) in \cite{Xi1994}, as Xi
explains in \cite[Introduction]{Xi2012}.
Lusztig's comparison of the two bases shows that the same uniform
bound then holds for the Kazhdan--Lusztig structure constants
\cite[Lemma 13.5]{Lusztig}.

\subsection{Previous results}

For a finite Coxeter group, the bound is $L(w_S)$
\cite[Lemma 13.3]{Lusztig}. Lusztig proved the equal-parameter affine
case in his work on affine cells \cite[\S7]{Lusztig1985}; in
\cite[\S13.4]{Lusztig}, he observes that the proof also applies to
positive weight functions. These cases suggested that finite
parabolic subgroups should control the bound even when $W$ is infinite.

Several families of infinite Coxeter groups were subsequently treated.
In equal parameters, Belolipetsky proved the bound for right-angled
groups, where $m_{st}\in\{2,\infty\}$ for distinct $s,t$
\cite[\S4.2]{Belolipetsky2004}. Xi proved it for complete Coxeter
graphs, where $m_{st}\ne2$ for distinct $s,t$
\cite[Theorem 2.1]{Xi2012}, and Zhou established the rank-three case
\cite{Zhou2013}.

For positive weights, Shi and Yang computed the $a$-function of
universal Coxeter groups, where $m_{st}=\infty$ for all distinct
$s,t$, and obtained the expected bound
\cite[Theorem 3.7]{ShiYang2015}. They then proved the bound for
complete Coxeter graphs \cite[Theorem 3.2]{ShiYang2016}.
Li and Shi treated Coxeter systems with $m_{st}\ne3$ for all
distinct $s,t$ \cite[Theorem 6.2]{LiShi2019}. Their hypothesis allows
commuting generators and groups of arbitrary finite rank.
Gao proved the bound for every weighted Coxeter group of rank three
\cite[Theorem 2.1]{Gao2022}.

In his 2020 list of open problems, Lusztig asked whether the standard
structure constants admit some uniform finite degree bound
\cite[\S2(b)]{Lusztig2020}. Chen and Hu proved this for every
finite-rank Coxeter system with a positive integer-valued weight
function \cite[Theorem 1.3]{ChenHu}.
The bound they obtain can exceed $N_L(W)$, even in equal parameters;
see \cite[Example 5.3]{ChenHu}. Their theorem supplies the general
finiteness result used here. We identify the resulting maximum with
the value in \eqref{eq:capacity}.

\subsection{Main result}

\begin{theorem}\label{thm:main}
Let $(W,S)$ be a Coxeter system of finite rank, and let $L$ be a positive
integer-valued weight function. Then
\[
 \max_{x,y,z\in W}\deg f_{x,y,z}=N_L(W).
\]
\end{theorem}

The bound is attained inside a finite standard parabolic subgroup.
Indeed, for every spherical $I\subseteq S$,
\[
 \deg f_{w_I,w_I,w_I}=L(w_I);
\]
we prove this in Lemma~\ref{alg:finite-profile}. Taking $I$ that
attains \eqref{eq:capacity} proves the lower bound in the theorem.
The main argument establishes the upper bound.

\subsection{Outline of the proof}

By the theorem of Chen and Hu, the degrees of the standard structure
constants have an attained global maximum $D$. Trace symmetry produces
a coefficient $f_{a,y,y}$ of degree $D$. We choose $y$ of minimum
length among all such inputs and follow a path of fold weight $D$ in
the standard multiplication expansion of $T_aT_y$.

The first step is geometric. Every state of this path has the same
continuation cone, consisting of the right factors that extend it
with additive length. We use the boundary description of these cones
from Parkinson and Yau \cite{ParkinsonYau}. Every boundary root must
occur at a fold: a missing boundary root would give a common right
reflection that preserves the path and shortens $y$. Dyer's criterion
\cite{Dyer} then identifies the boundary as the canonical simple
system of a finite reflection subgroup. An accepted continuation
places it in a finite standard parabolic component. Two common-suffix
lemmas show that the cone is the continuation cone of $w_J$ for a
spherical $J$, and give
\[
 y=uw_J,\qquad
 \ell(uw_Ju^{-1})=2\ell(u)+\ell(w_J),
\]
where $u$ is the minimal representative of its right coset $uW_J$.

The second step takes place in the Hecke algebra. Expanding
$T_yT_{y^{-1}}$ bounds the degree of every coefficient of
$T_uT_hT_{u^{-1}}$ by $D-L(h)$ for $h\in W_J$. We compress left
multiplication by $T_a$ to the standard coordinates indexed by
$uW_J$. Minimality of $y$ forces a coefficient indexed by an element
whose support is all of $J$ to attain its degree bound. A finite
power of that element reaches $w_J$ under the Demazure rule,
which crosses at an ascent and stays at a descent. Positivity of the
fold expansions compares powers of the compressed
operator with the original powers. For a positive integer $K$, this
gives
\[
 (K-1)D\ \geq\ \deg\bigl(\coeff{uw_Ju^{-1}}{T_a^K}\bigr)
 \ \geq\ KD-L(w_J).
\]
Hence $D\leq L(w_J)\leq N_L(W)$, which proves the upper bound.

\section{Standard multiplication and a maximum-degree closed path}

For $s\in S$, write $\xi_s=v^{L(s)}-v^{-L(s)}$. The standard
multiplication rules are
\begin{equation}\label{eq:simple-multiplication}
 T_sT_w=
 \begin{cases}
   T_{sw},&\ell(sw)=\ell(w)+1,\\
   T_{sw}+\xi_sT_w,&\ell(sw)=\ell(w)-1,
 \end{cases}
\end{equation}
and the analogous formulas hold for right multiplication. To compute
$T_xT_y$ using a fixed reduced expression
$x=s_m\cdots s_1$, start at $y$ and apply $s_1,\ldots,s_m$ on the left.
At each step a \emph{crossing} replaces the current state $w$ by $s_iw$
and contributes $1$. If $s_iw<w$, a \emph{fold} may instead leave the
state unchanged and contributes $\xi_{s_i}$. We refer to the sequence
of these choices as a path. Its fold weight is the sum of $L(s_i)$ over
its folds. A path is closed if its initial and final states agree.

Each coefficient $f_{x,y,z}$ is the sum of the fold monomials of the
paths from $y$ to $z$. The same statement holds for a finite product of
standard basis elements, by concatenating their reduced words.
A \emph{positive fold polynomial} is an element of $\mathcal A$
admitting a polynomial expression in the $\xi_s$ with nonnegative
integer coefficients; zero is allowed.

\begin{lemma}\label{lem:fold-positive}
A nonzero polynomial with nonnegative integer coefficients in the
elements $\xi_s$ has positive leading Laurent coefficient. Its degree
is the largest weighted degree of its displayed monomials. Consequently,
in a finite sum of such nonzero polynomials the degree is the largest
degree of a summand.
\end{lemma}

\begin{proof}
The monomial $\prod_s\xi_s^{n_s}$ is monic of degree
$\sum_s n_sL(s)$. All its other exponents are strictly smaller because
each $L(s)$ is positive. The terms of largest displayed degree therefore
have positive leading coefficients and cannot cancel.
\end{proof}

Define the $\mathcal A$-linear map $\tau:\HH\longrightarrow\mathcal A$
by $\tau(T_w)=\delta_{w,e}$, where $\delta$ is the Kronecker delta.
Define the $\mathcal A$-bilinear form
$\langle T_x,T_y\rangle=\delta_{x,y}$ on $\HH$.
For $sx>x$, left multiplication by $T_s$ on the ordered pair
$(T_x,T_{sx})$ has the symmetric matrix
\[
 \begin{pmatrix}0&1\\1&\xi_s\end{pmatrix}.
\]
Thus $\langle T_sX,Y\rangle=\langle X,T_sY\rangle$ for all
$X,Y\in\HH$. Moving the simple factors of a reduced expression
of $x$ from the first argument to the second gives
\[
 \delta_{x,y}=\langle T_x,T_y\rangle
 =\langle T_e,T_{x^{-1}}T_y\rangle
 =\tau(T_{x^{-1}}T_y).
\]
Consequently $\tau(T_xT_y)=\delta_{x^{-1},y}=\tau(T_yT_x)$.
By bilinearity, we obtain the trace identities
\begin{equation}\label{eq:trace}
 \tau(XY)=\tau(YX),\qquad
 \tau(T_{x^{-1}}T_y)=\delta_{x,y},\qquad
 \coeff{z}{X}=\tau(T_{z^{-1}}X).
\end{equation}
These are the identities of \cite[Lemma 10.4(a), (b)]{Lusztig}.
Reversing the defining relations gives the $\mathcal A$-linear
anti-involution $T_w\mapsto T_{w^{-1}}$, which preserves $\tau$.
It follows that
\begin{equation}\label{eq:transpose}
 f_{x,y,z}=f_{x^{-1},z,y}.
\end{equation}
Indeed, the two sides are respectively
$\tau(T_{z^{-1}}T_xT_y)$ and
$\tau(T_{y^{-1}}T_{x^{-1}}T_z)$, which agree by the
anti-involution and cyclicity of $\tau$.

By \cite[Theorem 1.3]{ChenHu}, there is a finite bound for all
$\deg f_{x,y,z}$. The degrees of the nonzero structure constants
are integers, and $f_{e,e,e}=1$. Thus the maximum
\begin{equation}\label{eq:D}
 D=\max_{x,y,z\in W}\deg f_{x,y,z}
\end{equation}
exists and is attained, with $D\geq0$.

\begin{lemma}\label{lem:closed}
If $\deg f_{x,y,z}=D$ for some $x,y,z$, then
$\deg f_{a,y,y}=D$ for some $a\in W$.
\end{lemma}

\begin{proof}
Extracting the $T_y$ coefficient from
$T_{x^{-1}}(T_xT_y)=(T_{x^{-1}}T_x)T_y$ and using
\eqref{eq:transpose} gives
\begin{equation}\label{eq:square-trace}
 \sum_z f_{x,y,z}^{2}
   =\sum_a f_{x^{-1},x,a}f_{a,y,y}.
\end{equation}
The sums are finite. The left side has degree $2D$ and positive leading
coefficient. On the right, both factors in each summand have degree at
most $D$. Lemma~\ref{lem:fold-positive} shows that at least one summand
has degree $2D$, so both its factors have degree $D$.
\end{proof}

Fix an element $y$ of minimum ordinary Coxeter length
among all inputs for which $\deg f_{a,y,y}=D$ for some $a\in W$.
Fix such an $a$, a reduced expression $a=s_m\cdots s_1$, and a closed
path from $y$ to $y$ of fold weight $D$ for that expression. Such a path
exists by Lemma~\ref{lem:fold-positive}.
If $S=\varnothing$, then $W=\{e\}$ and $\HH=\mathcal A$, so both
sides of Theorem~\ref{thm:main} are zero.

\section{Coxeter conventions and cone boundaries}
\label{sec:coxeter}

We use the real geometric representation of $(W,S)$ on the vector
space with basis $\Pi=\{\alpha_s:s\in S\}$ and symmetric bilinear form
\[
 (\alpha_s,\alpha_t)=
 \begin{cases}
  -\cos(\pi/m_{st}),&m_{st}<\infty,\\
  -1,&m_{st}=\infty.
 \end{cases}
\]
For a vector $\lambda$, the action of $s\in S$ is
$s\lambda=\lambda-2(\lambda,\alpha_s)\alpha_s$.
This representation is faithful and preserves the form.
The root system $\Phi=W\Pi$ is the disjoint union
$\Phi^+\sqcup(-\Phi^+)$, where
\[
 \Phi^+=\Phi\cap\sum_{s\in S}\R_{\geq0}\alpha_s.
\]
We write $\beta>0$ for $\beta\in\Phi^+$ and $\beta<0$ for
$\beta\in-\Phi^+$. Every root has squared norm one. For a root
$\beta=w\alpha_s$, its reflection is $r_\beta=wsw^{-1}\in W$,
acting by $r_\beta(\lambda)=\lambda-2(\lambda,\beta)\beta$.
For multiplication by a simple reflection, strict Bruhat
comparisons are exactly the corresponding length comparisons.
We use the two inversion conventions
\[
 N(w)=\{\beta\in\Phi^+:w\beta\in-\Phi^+\},
 \qquad \Inv(w)=N(w^{-1}).
\]
Both inversion sets are finite and have cardinality $\ell(w)$.
For $\beta\in\Phi^+$, reflection descent is characterized by
$\ell(wr_\beta)<\ell(w)$ if and only if $\beta\in N(w)$.
If $sw<w$, then
\begin{equation}\label{eq:left-root-deletion}
  N(sw)=N(w)\setminus\{-w^{-1}\alpha_s\}.
\end{equation}
For these root-sign and inversion facts, see \cite[Section 4.4]{BB}.
An inversion set and its complement in $\Phi^+$ are closed under
positive linear combinations which are roots. For example, if two roots
are sent to positive roots by $w$, then any positive linear combination
of them which is a root is also sent to a positive root. This proves
the assertion for the complement; the other assertion is identical.

For $J\subseteq S$, put $\Pi_J=\{\alpha_s:s\in J\}$ and
$\Phi_J=W_J\Pi_J$, with $\Phi_J^+=\Phi_J\cap\Phi^+$. The set
\[
 W^J=\{q\in W:\ell(qs)>\ell(q)\text{ for all }s\in J\}
\]
consists of the minimal representatives of the cosets $wW_J$.
Every $w$ has a unique factorization $w=qp$, with $q\in W^J$,
$p\in W_J$, and $\ell(w)=\ell(q)+\ell(p)$. Moreover,
$\ell(qk)=\ell(q)+\ell(k)$ for every $k\in W_J$.
For $q\in W^J$, the roots $q\alpha_s$, $s\in J$, are positive;
their nonnegative linear combinations give
$q\Phi_J^+\subseteq\Phi^+$.
We write $D_L(w)=\{s\in S:sw<w\}$ and
$D_R(w)=\{s\in S:ws<w\}$.
These sets are spherical. If $J$ is spherical, $w_J$ permutes
$\Pi_J$ after a change of sign and
$\ell(w_Jp)=\ell(pw_J)=\ell(w_J)-\ell(p)$ for $p\in W_J$.
We use the Bruhat subword property and the existence of joins of
bounded subsets in weak order. These standard facts hold for arbitrary
Coxeter systems; see \cite[Chapters 2--3]{BB}.
The support $\supp(w)$ is the set of simple reflections occurring in
any reduced expression for $w$; it is independent of the expression.

We record the simple-coset rule in a form needed later. If $q\in W^J$
and $s\in S$, either $sq\in W^J$, or $sq=qt$ for a simple
reflection $t\in J$. To see the second assertion, if $sq$ is not
$J$-minimal, choose $t\in J$ with $sq\alpha_t<0$. Since
$q\alpha_t>0$ and a simple reflection changes the sign of only its own
positive simple root, $q\alpha_t=\alpha_s$, whence $sq=qt$.
It follows that a simple left ascent on $qp$ either changes the minimal
representative $q$ by an ascent and leaves $p$ fixed, or leaves $q$
fixed and makes a simple left ascent on $p$. The alternative
$sq\in W^J$ with $sq<q$ is a descent on $qp$ for every $p\in W_J$.

Two reflection-subgroup facts will also be used. Every reflection
subgroup has its canonical simple system and the positive roots induced
from $\Phi^+$. Dyer's criterion says that a subset $B\subseteq\Phi^+$
is such a canonical simple system if the inner products of
distinct members belong to
\begin{equation}\label{eq:dyer}
 \{-\cos(\pi/m):m\in\Z,\ m\ge2\}\ \cup\ (-\infty,-1].
\end{equation}
Equivalently, it is enough that each pair be the canonical simple pair
of its two-generated reflection subgroup
\cite[Theorem 3.9 and Corollary 3.10]{Dyer}. All positive roots of the
generated subgroup are then nonnegative linear combinations of $B$.
Finally, every finite subgroup of $W$ is contained in a conjugate of a
finite standard parabolic subgroup \cite[Proposition 29]{Casselman}.

Define the continuation cone of $w$ by
\[
 \mathcal C(w)=\{c\in W:\ell(wc)=\ell(w)+\ell(c)\}.
\]
The basic inversion criterion is
\begin{equation}\label{eq:cone-inversions}
 c\in\mathcal C(w)
 \quad\Longleftrightarrow\quad
 N(w)\cap\Inv(c)=\varnothing.
\end{equation}
For a cone $C$, define its boundary $\partial C$ to be the set of
positive roots $\gamma$ for which there exist $c\notin C$ and $s\in S$
such that $cs\in C$ and $r_\gamma c=cs$. This definition depends only
on $C$. The following boundary facts are due to Parkinson and Yau
\cite[Theorems 2.6 and 2.8, Corollary 2.7]{ParkinsonYau}; we include
the elementary proofs in our inversion conventions.

\Needspace{6\baselineskip}
\begin{lemma}\label{lem:boundary}
Let $C$ be a continuation cone and put $B=\partial C$.
\begin{enumerate}
\item For every $\gamma\in B$, there exists a single $c_\gamma\in W$
such that
\[
 N(w)\cap\Inv(c_\gamma)=\{\gamma\}
 \quad\text{for every }w\text{ with }\mathcal C(w)=C.
\]
In particular, $B\subseteq N(w)$ for every such $w$, and $B$ is finite.
\item For every $\gamma\in B$ and every representative $w$ of $C$,
$\ell(wr_\gamma)=\ell(w)-1$.
\item The boundary gives the exact half-space description
\begin{equation}\label{eq:boundary-halfspaces}
 C=\{c\in W:c^{-1}\gamma>0\text{ for every }\gamma\in B\}.
\end{equation}
No boundary root can be omitted from this description.
\end{enumerate}
\end{lemma}

\begin{proof}
Choose $c,s$ in the definition of a boundary root. Necessarily $cs<c$.
Indeed, if $cs>c$, for every representative $w$ of $C$ the membership
$cs\in C$ and nonmembership $c\notin C$ would give
\[
 \ell(w)+\ell(c)+1=\ell(wcs)
 \le\ell(wc)+1<\ell(w)+\ell(c)+1.
\]
Thus $\gamma=-c\alpha_s>0$ and
$\Inv(cs)=\Inv(c)\setminus\{\gamma\}$.
Applying \eqref{eq:cone-inversions} to $c$ and $cs$ proves
$N(w)\cap\Inv(c)=\{\gamma\}$ simultaneously for every representative.
This proves (i).

For (ii), use the same pair $c,s$. Since $cs=r_\gamma c\in C$,
\[
 \ell(w)+\ell(c)-1
   =\ell(wr_\gamma c)
   \le\ell(wr_\gamma)+\ell(c).
\]
On the other hand, $\gamma\in N(w)$, so reflection descent gives
$\ell(wr_\gamma)<\ell(w)$. These two inequalities give (ii).

If $c\in C$, its inversion set avoids $N(w)$ and hence avoids $B$,
which proves one inclusion in (iii). Conversely, take $c\notin C$
and a reduced word for $c$. Let $b$ be the first prefix which does not
belong to $C$, and write $b=b's$, where $b'\in C$ is the preceding
prefix. Then $\gamma=-b\alpha_s$ lies in $B$, and
$\gamma\in\Inv(b)\subseteq\Inv(c)$. Hence $c^{-1}\gamma<0$.
This proves \eqref{eq:boundary-halfspaces}, including the case $b'=e$.
Finally, the witness from (i) has inversion set meeting $B$ only in
$\gamma$. It violates precisely that boundary inequality, proving
irredundancy.
\end{proof}

For later use, if $c\in\mathcal C(w)$, then the cone
$\mathcal C(wc)$ is determined by $\mathcal C(w)$ and $c$. Indeed,
$d\in\mathcal C(wc)$ is equivalent to
$\ell(cd)=\ell(c)+\ell(d)$ and $cd\in\mathcal C(w)$.
Thus the same accepted continuation applied to elements with equal
cones again gives elements with equal cones.



\section{Maximal paths and finite reflection subgroups}
\label{geom:sec-paths}

We call a coefficient \emph{maximal} if its degree is the global
maximum $D$ in \eqref{eq:D}. In this section, a path for a fixed reduced
expression of a single left input is called maximal if its fold weight
is $D$. Throughout this section, contributions to products are compared
using Lemma~\ref{lem:fold-positive}.

\begin{lemma}\label{geom:endpoints}
If $\deg f_{x,b,c}=D$, then $\mathcal C(b)=\mathcal C(c)$.
\end{lemma}

\begin{proof}
Suppose that $bs>b$ and $cs<c$ for a simple reflection $s$. In
\[
 T_xT_{bs}=(T_xT_b)T_s,
\]
the coefficient of $T_c$ has the positive fold contribution
$\xi_s f_{x,b,c}$. Its degree is at least $D+L(s)>D$, a contradiction.
Applying the same argument to
$f_{x^{-1},c,b}=f_{x,b,c}$, by \eqref{eq:transpose}, excludes the
opposite discrepancy. Thus $b$ and $c$ have the same simple right
ascents.

If $s$ is such a common ascent, the crossing from $c$ to $cs$ in the
same identity shows that $f_{x,bs,cs}$ contains the contribution
$f_{x,b,c}$. Therefore $\deg f_{x,bs,cs}=D$. This argument can be
iterated along any common accepted prefix. If the two continuation
cones differed, take a reduced expression for an element belonging to
one cone but not the other. At its first unaccepted letter, the states
reached after the preceding common prefix would have different simple
right ascents, contrary to what was just proved. Hence the cones agree.
\end{proof}

\begin{lemma}\label{geom:cuts}
Suppose that $x=hk$ is reduced and that
\begin{equation}\label{geom:eq-cut}
 \deg f_{k,b,d}+\deg f_{h,d,c}=D.
\end{equation}
Then
\[
 \mathcal C(b)=\mathcal C(d)=\mathcal C(c).
\]
Consequently, every state of a path of fold weight $D$ for a reduced
expression of $x$ has the same continuation cone as its endpoints.
\end{lemma}

\begin{proof}
Associativity and positivity imply $\deg f_{x,b,c}=D$, so
Lemma~\ref{geom:endpoints} gives $\mathcal C(b)=\mathcal C(c)$.
We first compare the simple right ascents of $b$ and $d$.
If $bs>b$ and $ds<d$, expand
\[
 T_xT_{bs}=T_h\bigl((T_kT_b)T_s\bigr).
\]
The contribution through $T_d$, followed by the fold at $s$, and then
through $T_c$ under left multiplication by $T_h$, is
$\xi_s f_{k,b,d}f_{h,d,c}$. By \eqref{geom:eq-cut}, this gives degree
at least $D+L(s)$ in one ordinary two-factor coefficient, which is
impossible.

For the other discrepancy, suppose that $bs<b$ and $ds>d$.
The endpoint equality gives $cs<c$. The right multiplication rule
implies the exact identities
\begin{equation}\label{geom:eq-discrepancy}
 f_{k,b,d}=f_{k,bs,ds},\qquad
 f_{h,ds,c}=f_{h,d,cs}+\xi_s f_{h,d,c}.
\end{equation}
Indeed, the first identity follows by taking the $T_d$ coefficient in
$T_kT_b=(T_kT_{bs})T_s$; a fold at $d$ is unavailable because $ds>d$.
For the second, take the $T_c$ coefficient in
$T_hT_{ds}=(T_hT_d)T_s$, using $cs<c$. The contribution through $ds$
in $T_h(T_kT_{bs})=T_xT_{bs}$ now has degree at least $D+L(s)$.
This is again a contradiction. Hence $b$ and $d$ have the same simple
right ascents.

Let $s$ be a common right ascent of $b,d,c$. Crossing at the output
in each of the two cut coefficients gives contributions
\[
 f_{k,b,d}\quad\text{in }f_{k,bs,ds},\qquad
 f_{h,d,c}\quad\text{in }f_{h,ds,cs}.
\]
The new degrees have sum at least $D$. Their product is a positive
summand of $f_{x,bs,cs}$, so the global bound makes the sum exactly
$D$. Thus \eqref{geom:eq-cut} remains valid after any common accepted
prefix. The preceding comparison of simple ascents can be repeated
at each prefix. The first-discrepancy argument from
Lemma~\ref{geom:endpoints} proves equality of the full cones.

Finally, cut a maximal path at any one of its states $d$. Its two
subpaths have fold weights with sum $D$, so the two corresponding
coefficients have degrees with sum at least $D$. Since $x=hk$ is
reduced, their product is a positive summand of $f_{x,b,c}$, and the
sum cannot exceed $D$. The cut therefore satisfies
\eqref{geom:eq-cut}, proving the last assertion.
\end{proof}

An accepted continuation preserves a maximal path itself. More
precisely, let $c$ belong to the common cone of all its states and
multiply each state $w$ on the right by $c$. A crossing remains a
crossing. If a step is a fold at $sw<w$, then
\[
 \ell(swc)\leq\ell(sw)+\ell(c)
      =\ell(wc)-1,
\]
so it is still a fold at a descent. All fold labels and weights are
unchanged. The resulting path is maximal, and its states again have
equal cones, either by Lemma~\ref{geom:cuts} or by the accepted
continuation observation following Lemma~\ref{lem:boundary}.

Return to the fixed closed maximal path for the reduced expression
$a=s_m\cdots s_1$, with initial state $y$ chosen above. Set
\[
 C=\mathcal C(y),\qquad B=\partial C.
\]
At a fold with letter $s$ and state $w$, call
\[
 \eta=-w^{-1}\alpha_s\in\Phi^+
\]
its \emph{fold root}. Thus $\eta\in N(w)$ and
$N(sw)=N(w)\setminus\{\eta\}$.

\begin{lemma}\label{geom:no-blocker}
Let $\gamma\in B$, and let $\eta$ be any fold root of the fixed path.
If $\eta\ne\gamma$, then $r_\gamma\eta\in\Phi^+$.
\end{lemma}

\begin{proof}
Let $w$ be the fold state, with letter $s$, and use the common witness
$c_\gamma$ from Lemma~\ref{lem:boundary}. Both inversion sets
\[
 N(sw),\qquad \Inv(c_\gamma)
\]
contain $\gamma$ and omit $\eta$. For the first set this follows from
the single-root deletion formula and $\gamma\ne\eta$; for the second
it follows from
$N(w)\cap\Inv(c_\gamma)=\{\gamma\}$.

Suppose instead that $r_\gamma\eta<0$, and put
$\delta=-r_\gamma\eta>0$. The reflection formula gives
\[
 \eta+\delta=\lambda\gamma
 \quad\text{for some }\lambda>0.
\]
The coefficient is positive because both sides lie in the pointed
positive cone spanned by $\Pi$. If $\delta$ were omitted by either of
the displayed inversion sets, closure of its complement under positive
root combinations would imply that it also omits
$\gamma=\lambda^{-1}(\eta+\delta)$. Thus $\delta$ belongs to both
sets. Moreover $\delta\ne\gamma$: equality would imply
$r_\gamma\eta=-\gamma$ and hence $\eta=\gamma$.
Since $N(sw)\subseteq N(w)$, this contradicts the singleton witness
intersection. Therefore $r_\gamma\eta>0$.
\end{proof}

\begin{lemma}\label{geom:all-boundaries}
Every root in the complete boundary $B$ occurs as a fold root of the
fixed closed maximal path.
\end{lemma}

\begin{proof}
If $\gamma\in B$ never occurs, multiply every path state on the right
by $r_\gamma$. Crossings remain crossings. At a fold with root
$\eta\ne\gamma$, its new descent test is
\[
 (wr_\gamma)^{-1}\alpha_s=-r_\gamma\eta<0
\]
by Lemma~\ref{geom:no-blocker}. Thus all folds survive, with their
original weights. The transformed path is closed, has the same reduced
left expression $a$, and has fold weight $D$. It follows that
$\deg f_{a,yr_\gamma,yr_\gamma}=D$.
But Lemma~\ref{lem:boundary} gives
$\ell(yr_\gamma)=\ell(y)-1$, contrary to the minimality of $y$.
\end{proof}

For a reflection subgroup $R$ of $W$, put
\[
 \Psi_R=\{\beta\in\Phi:r_\beta\in R\},\qquad
 \Psi_R^+=\Psi_R\cap\Phi^+.
\]
These are its induced root system and positive roots. For $h\in R$,
the inversion set in its canonical Coxeter system is
\[
 N_R(h)=\{\beta\in\Psi_R^+:h\beta<0\}.
\]

\begin{lemma}\label{geom:restriction}
For every $w\in W$, the intersection $N(w)\cap\Psi_R^+$ is an
inversion set of $R$.
\end{lemma}

\begin{proof}
Choose $q$ of minimum ambient Coxeter length in the coset $wR$,
and write $w=qh$ with $h\in R$. For $\beta\in\Psi_R^+$, the
reflection $r_\beta$ belongs to $R$. Minimality gives
$\ell(qr_\beta)\geq\ell(q)$, and equality is impossible because a
reflection changes length parity. Thus $\ell(qr_\beta)>\ell(q)$,
so $q\beta\in\Phi^+$ by the reflection descent criterion.
It follows also that $q$ sends every negative
$R$-root to an ambient negative root. Since $h$ permutes $\Psi_R$,
for every $\beta\in\Psi_R^+$ we have
\[
 w\beta<0\quad\Longleftrightarrow\quad h\beta<0.
\]
Consequently $N(w)\cap\Psi_R^+=N_R(h)$.
\end{proof}

\begin{proposition}\label{geom:finite-canonical}
The set $B$ is the canonical simple system of a finite reflection
subgroup
\[
 H=\langle r_\gamma:\gamma\in B\rangle.
\]
\end{proposition}

\begin{proof}
First take distinct $\gamma,\eta\in B$, and let
$R=\langle r_\gamma,r_\eta\rangle$. Its canonical Coxeter system is
dihedral \cite[Corollary 3.16 and the following discussion]{Dyer}.
By Lemma~\ref{geom:all-boundaries}, there is a path state
at which $\gamma$ is a fold root. Restrict the inversion sets before
and after its single-root deletion to $\Psi_R^+$, using
Lemma~\ref{geom:restriction}. We obtain an inversion set $A$ of $R$
containing both $\gamma$ and $\eta$, such that
$A\setminus\{\gamma\}$ is also an inversion set. Here both roots
belong to the original inversion set because every path state has
cone $C$. Similarly, another fold gives an inversion set $A'$ which
contains both roots and from which $\eta$ alone can be removed.

Recall explicitly the inversion sets of a dihedral system. In infinite
dihedral type, the two alternating reduced words give two disjoint
strings of positive roots, each starting at one canonical simple root.
Every nonempty inversion set is a finite initial segment of one string;
the root which can be removed while leaving an inversion set is its
terminal root. These descriptions follow directly by adding the one
new inversion root at each letter of an alternating reduced word.
Thus $A$ would place $\eta$ strictly before $\gamma$ in one string,
whereas $A'$ would place $\gamma$ strictly before $\eta$ in that same
string. This is impossible. Hence $R$ is finite.

In finite dihedral type of order $2m$, write its positive roots in
angular order as $\rho_1,\ldots,\rho_m$, with the two canonical
simple roots at the ends. The same alternating-word calculation says
that its inversion sets are the initial and final segments of this
ordered list, including the empty and full sets. Suppose that
$\gamma=\rho_i$ and $\eta=\rho_j$, with $i<j$. Because $A$
contains both and permits removal of $\rho_i$, it must be the final
segment $\{\rho_i,\ldots,\rho_m\}$. Likewise,
$A'=\{\rho_1,\ldots,\rho_j\}$. In particular,
$A\cup A'=\Psi_R^+$.

The common witness for $\gamma$ meets the ambient inversion set at
each of these two states exactly in $\gamma$. Thus the restriction
of $\Inv(c_\gamma)=N(c_\gamma^{-1})$ meets $A$ and $A'$ exactly
in $\gamma$. Since their union is all of $\Psi_R^+$, that restriction
is the singleton $\{\gamma\}$. By Lemma~\ref{geom:restriction} it
is an inversion set of $R$. A singleton inversion set is the root of
a canonical simple generator. The same reasoning applies to $\eta$.
Therefore $\gamma,\eta$ are precisely the canonical simple pair
of $R$.

Dyer's criterion \eqref{eq:dyer} now shows that the entire set $B$ is
the canonical simple system of $H$; the assertion is immediate when
$B$ has at most one member. Every positive $H$-root is a nonnegative
linear combination of $B$. Since $B\subseteq N(y)$, its image under
$y$ is a nonzero nonpositive combination of ambient simple roots.
It is therefore a negative root. Hence
\[
 \Psi_H^+\subseteq N(y).
\]
The set on the right is finite. The canonical length of any $h\in H$
is the cardinality of its $H$-inversion set and is consequently bounded
by $|N(y)|$. Since $B$ is finite, there are only finitely many words of
that bounded length in the canonical generators. Thus $H$ is finite.
\end{proof}

\section{Parabolic boundaries and common suffixes}
\label{geom:sec-parabolic}

\begin{proposition}\label{geom:parabolic-boundary}
There exist a spherical subset $J\subseteq S$ and an element $d\in W$
such that $B=d\Pi_J$.
\end{proposition}

\begin{proof}
By finite-subgroup containment \cite[Proposition 29]{Casselman},
choose a spherical $I\subseteq S$ and $g\in W$ with
$H\subseteq gW_Ig^{-1}$. Then $g^{-1}B\subseteq\Phi_I$.
Indeed, the reflection in $g^{-1}\beta$ belongs to $W_I$ for each
$\beta\in B$. The positive choice of this root is inverted by its
own reflection, and the inversions of an element of $W_I$ all lie
in $\Phi_I$, as follows from a reduced expression in $I$.

Choose a linear functional $\delta$ strictly positive on every
ambient simple root. The functional
$f=\delta\circ g$ on $\operatorname{span}(\Phi_I)$ satisfies
$f(g^{-1}\beta)>0$ for all $\beta\in B$. Perturb $f$ to a regular
functional while preserving these finitely many strict inequalities.
The chambers of the finite reflection arrangement of $W_I$ are its
translates of the standard chamber. Thus there is $k\in W_I$ such
that the roots on which this functional is positive are exactly
$k\Phi_I^+$. Put $c=gk$. We have
\begin{equation}\label{geom:eq-accepted-residue}
 c^{-1}B\subseteq\Phi_I^+,
 \qquad c\in C,
\end{equation}
where the second assertion follows from
\eqref{eq:boundary-halfspaces}. If $B=\varnothing$, one may take
$I=\varnothing$ and $c=e$ instead.

Extend every state of the fixed path by $c$. All folds survive and all
extended states have the same cone. Their right $W_I$-components are
therefore equal. To verify this last assertion, write an extended state
as $w=qb$, with $q\in W^I$ and $b\in W_I$. For $r\in W_I$,
parabolic length additivity gives
\[
 r\in\mathcal C(w)\quad\Longleftrightarrow\quad
 \ell(br)=\ell(b)+\ell(r).
\]
Within $W_I$, this accepted set has the unique element of greatest
length $b^{-1}w_I$. Its length is $\ell(w_I)-\ell(b)$, and any
accepted $r$ of that length satisfies $br=w_I$. Thus
$\mathcal C(w)\cap W_I$ determines $b$. Equality of the extended
cones proves that the same $b$ occurs at every extended state.

Suppose that an extended fold has letter $t\in S$ and root
$\beta'=-b^{-1}q^{-1}\alpha_t$ lying in $\Phi_I^+$. Then
$q^{-1}\alpha_t\in\Phi_I$. This root is positive: if it were
negative, the fact that $q\in W^I$ sends $\Phi_I^+$ to positive
roots would imply $\alpha_t<0$. Let
$r=r_{q^{-1}\alpha_t}\in W_I$. We have $tq=qr$ and
$\ell(tq)=\ell(q)+1$. On the other hand, parabolic additivity gives
$\ell(qr)=\ell(q)+\ell(r)$. Hence $\ell(r)=1$, so $r=s$ for
some $s\in I$. The fold condition $tqb<qb$ becomes $sb<b$, and
its root is
\[
 \beta'=-b^{-1}\alpha_s,\qquad s\in D_L(b).
\]

Each original boundary root occurs at a fold, and its extended fold
root is its image under $c^{-1}$. By
\eqref{geom:eq-accepted-residue}, the preceding argument therefore
gives a subset $K\subseteq D_L(b)\subseteq I$ such that
\[
 c^{-1}B=\{-b^{-1}\alpha_s:s\in K\}.
\]
The subset $K$ is spherical. Its longest element satisfies
$w_K\Pi_K=-\Pi_K$, so
\[
 B=(cb^{-1}w_K)\Pi_K,
\]
which proves the assertion.
\end{proof}

For the remaining argument use left weak order in its right-suffix
form: write $d\preccurlyeq x$ if $x=hd$ and
$\ell(x)=\ell(h)+\ell(d)$. We call $d$ an \emph{additive right
suffix} of $x$. In particular, $N(d)\subseteq N(x)$. We will use
bounded joins for this order, obtained by inversion from the usual
right weak order \cite[Theorem 3.2.1 and the following discussion]{BB}.

\begin{lemma}\label{geom:top-covers}
Let $J$ be spherical and $q\in W^J$. If $d$ is an additive right
suffix of $qw_J$ and of every $qsw_J$, $s\in J$, then it is an
additive right suffix of every element of $qW_J$.
\end{lemma}

\begin{proof}
For each $t\in J$, opposition gives $s=w_Jtw_J\in J$ and
$qsw_J=qw_Jt$. Since $w_J\alpha_t\in-\Phi_J^+$ and
$q\Phi_J^+\subseteq\Phi^+$, we have
\[
 (qw_Jt)\alpha_t=-qw_J\alpha_t\in\Phi^+.
\]
Thus $qw_Jt$ does not invert $\alpha_t$. Since $d$ is a right
suffix of this element, $\alpha_t\notin N(d)$. It follows that
$d\in W^J$.

Choose a reduced left-ascent gallery from $d$ to $qw_J$: such a
gallery exists because $d\preccurlyeq qw_J$. Write each state in
its right parabolic decomposition $rp$, with $r\in W^J$ and
$p\in W_J$. The initial decomposition is $d\cdot e$, and the final
one is $q w_J$. The simple-coset rule in Section~\ref{sec:coxeter} shows that a step
with letter $s$ has exactly one of the following forms:
\begin{enumerate}
\item $sr\in W^J$, $\ell(sr)=\ell(r)+1$, and
$rp\longmapsto(sr)p$;
\item $sr=rt$ for $t\in J$, and
$rp\longmapsto r(tp)$ with $\ell(tp)=\ell(p)+1$.
\end{enumerate}
The possibility $sr\in W^J$ with $sr<r$ is excluded because it
would be a descent on $rp$ for every $p\in W_J$.

Record the labels of the steps of the second type in chronological
order as $t_1,\ldots,t_n$. They give the reduced expression
\[
 w_J=t_n\cdots t_1.
\]
For an arbitrary $k\in W_J$, choose a reduced subword representing
$k$ in this expression, using the Bruhat subword property. Retain
every step of the first type and precisely those steps of the second
type selected by this subword.

This selection still gives a left-ascent gallery. Indeed, discarding
a step of the second type leaves the minimal representative $r$
unchanged. The sequence of minimal representatives at every retained
position is consequently the original sequence. The simple-coset
alternative at a position, and the label $t=r^{-1}sr$ in the second
alternative, depend only on $s$ and $r$, so these labels remain
unchanged. As the retained finite steps are executed chronologically,
their successive products are suffixes of the selected reduced
expression for $k$. Each is therefore a left ascent. Every step of
the first type is an ascent independently of the current finite
component, by parabolic length additivity. The new gallery starts at
$d$, ends at $qk$, and is reduced. Hence $d\preccurlyeq qk$ for
every $k\in W_J$.
\end{proof}

\begin{lemma}\label{geom:ribbons}
Let $J$ be spherical and $q\in W^J$. There exist a spherical subset
$K$, an element $p\in W^K$, and $z\in W$ such that
\begin{equation}\label{geom:eq-ribbon}
 z\Pi_J=\Pi_K,\qquad
 qk=p(zkz^{-1})z\quad(k\in W_J),
\end{equation}
where the outer factorizations are length additive:
\[
 \ell(qk)=\ell(p)+\ell(zkz^{-1})+\ell(z)
        =\ell(p)+\ell(k)+\ell(z).
\]
The element $z$ is the greatest common additive right suffix of
$qW_J$. In particular,
if $d$ is a common right suffix and $z=z_0d$ is reduced, then
\begin{equation}\label{geom:eq-ribbon-cancel}
 qw_Jd^{-1}=pw_Kz_0\quad\text{is reduced},\qquad
 d\Pi_J=z_0^{-1}\Pi_K.
\end{equation}
\end{lemma}

\begin{proof}
We describe one shortening step. Suppose that every element of
$qW_J$ has a common simple right descent $t$. Since $q\in qW_J$
and $q\in W^J$, we have $t\notin J$. The right descent set of
$qw_J$ contains $I=J\cup\{t\}$, so $I$ is spherical. Its right
parabolic decomposition has the form
\[
 qw_J=rw_I,\qquad r\in W^I.
\]
Put $u=w_Iw_J$ and $J'=w_IJw_I\subseteq I$. Finite opposition
gives $u\Pi_J=\Pi_{J'}$. In particular, $u\in W^J$, and
conjugation by $u$ identifies the simple generators of $W_J$ and
$W_{J'}$, preserving their lengths. Since
$\ell(u)=\ell(w_I)-\ell(w_J)$, cancellation in the preceding
reduced equality gives $q=ru$ reduced.

For every $k\in W_J$, we have
\begin{equation}\label{geom:eq-one-ribbon}
 qk=r(uku^{-1})u.
\end{equation}
This expression is reduced. Indeed, $u\in W^J$ gives
$\ell(uk)=\ell(u)+\ell(k)$, while
$\ell(uku^{-1})=\ell(k)$. Thus the last two factors in
\eqref{geom:eq-one-ribbon} are length additive. Their product belongs
to $W_I$, and $r\in W^I$ makes the whole product additive. The new
minimal representative $r$ lies in $W^{J'}$ and is strictly shorter
than $q$, because $J\subsetneq I$ implies $\ell(u)>0$.

Repeat this step for the new coset whenever it has a common simple
right descent. The length of the minimal representative strictly
decreases, so the procedure terminates.

To verify that the factors compose with additive lengths, consider
two successive steps. The first gives \eqref{geom:eq-one-ribbon};
write the second as
\[
 rj=p(u'j(u')^{-1})u'\quad(j\in W_{J'}),
 \qquad u'\Pi_{J'}=\Pi_{J''},
\]
with the displayed product reduced and $p\in W^{J''}$.
Substitute $j=uku^{-1}\in W_{J'}$, whose length is $\ell(k)$.
Then
\begin{align*}
 qk&=p\bigl((u'u)k(u'u)^{-1}\bigr)u'u,\\
 \ell(qk)&=\ell(rj)+\ell(u)\\
        &=\ell(p)+\ell(k)+\ell(u')+\ell(u).
\end{align*}
For $k=e$, this gives $q=pu'u$ with
$\ell(q)=\ell(p)+\ell(u')+\ell(u)$. The length inequalities
\[
 \ell(q)\leq\ell(p)+\ell(u'u)
 \leq\ell(p)+\ell(u')+\ell(u)
\]
are therefore equalities. Thus $u'u$ is reduced, and
$(u'u)\Pi_J=\Pi_{J''}$. Induction gives $K,p,z$ satisfying
\eqref{geom:eq-ribbon}, with all concatenations reduced.
At the terminal stage $pW_K$ has no common simple right descent.
It therefore has no nonidentity common additive right suffix:
the final simple letter of such a suffix would be a common right descent.

The element $z$ in \eqref{geom:eq-ribbon} is a common suffix. Let
$d$ be any other common suffix. The pair $d,z$ is bounded above in
left weak order, for example by $q$, so its join $b=d\vee z$
exists. Every element of $qW_J$ is an upper bound for this pair;
therefore $b$ is also a common suffix. As $z\preccurlyeq b$, write
$b=cz$ reduced. For every $k\in W_J$, the comparisons
$b\preccurlyeq qk$ and \eqref{geom:eq-ribbon} give
\[
 p(zkz^{-1})z=h_kcz
\]
with both expressions reduced. Cancelling $z$ preserves length
additivity and yields
\[
 p(zkz^{-1})=h_kc,\qquad
 \ell\bigl(p(zkz^{-1})\bigr)=\ell(h_k)+\ell(c).
\]
Conjugation by $z$ maps $W_J$ onto $W_K$, so $c$ is a common right
suffix of $pW_K$. The terminal property forces $c=e$. Hence $b=z$
and $d\preccurlyeq z$, proving that $z$ is greatest.

Finally, conjugation by $z$ takes $w_J$ to $w_K$. Substituting
$z=z_0d$ in the reduced equality for $qw_J$ and cancelling $d$
gives the first assertion of \eqref{geom:eq-ribbon-cancel}.
The equality $z\Pi_J=\Pi_K$ gives the second.
\end{proof}

\begin{theorem}\label{geom:standard-boundary}
Let $C$ be a continuation cone. Suppose that its complete boundary
has the form $\partial C=d\Pi_J$ for a spherical subset $J$ and
some $d\in W$. Then there is a spherical subset $K\subseteq S$
such that
\[
 \partial C=\Pi_K,\qquad C=\mathcal C(w_K).
\]
Every representative $g$ of $C$ has a reduced factorization $g=pw_K$
with $p\in W^K$.
\end{theorem}

\begin{proof}
Put $B=\partial C=d\Pi_J$. Positivity of $B$ gives $d\in W^J$,
and \eqref{eq:boundary-halfspaces} gives $d\in C$, since
$d^{-1}B=\Pi_J$. Choose a representative $g$ of $C$. The product
$gd$ is reduced and sends every simple $J$-root to a negative root,
because $B\subseteq N(g)$. Its right parabolic component is thus
$w_J$, and we may write
\[
 gd=qw_J,\qquad q\in W^J.
\]
For $s\in J$, put $\gamma=d\alpha_s$. Boundary shortness gives
$\ell(gr_\gamma)=\ell(g)-1$, whereas
\[
 gr_\gamma d=gds=qw_Js,
 \qquad
 \ell(qw_Js)=\ell(g)+\ell(d)-1.
\]
It follows that $gr_\gamma d$ is reduced. Thus $d$ is a common
right suffix of $qw_J$ and every $qw_Js$. Opposition permutes $J$,
so these are also the elements $qsw_J$ used in
Lemma~\ref{geom:top-covers}. That lemma makes $d$ a common right
suffix of the whole coset $qW_J$.

Apply Lemma~\ref{geom:ribbons}, and write its greatest common suffix
as $z=z_0d$ reduced. Formula \eqref{geom:eq-ribbon-cancel} becomes
\begin{equation}\label{geom:eq-standardization}
 g=pw_Kz_0\quad\text{reduced},\qquad
 B=z_0^{-1}\Pi_K.
\end{equation}
The full boundary formula now puts $z_0^{-1}$ in $\mathcal C(g)$,
since $z_0B=\Pi_K\subseteq\Phi^+$. Hence
\[
 \ell(gz_0^{-1})=\ell(g)+\ell(z_0).
\]
But the reduced factorization in \eqref{geom:eq-standardization}
gives $\ell(gz_0^{-1})=\ell(g)-\ell(z_0)$. Therefore $z_0=e$.
This proves $B=\Pi_K$ and the asserted factorization of $g$.

Finally, \eqref{eq:boundary-halfspaces} says that $C$ consists
exactly of the elements $c$ for which $c^{-1}\Pi_K$ is positive.
For each such $c$, the inverse $c^{-1}$ sends every positive root of
$\Phi_K$ to a positive root, since those roots are nonnegative
combinations of $\Pi_K$.
As $N(w_K)=\Phi_K^+$, the inversion criterion
\eqref{eq:cone-inversions} identifies this set with
$\mathcal C(w_K)$. The proof also includes $J=\varnothing$;
equivalently, an empty boundary gives $C=W$, whose only representative
is $e$.
\end{proof}

\begin{proposition}\label{prop:geometry}
Let $y$ have minimum length among all inputs supporting a closed
coefficient of the global maximum degree $D$, as chosen after
Lemma~\ref{lem:closed}. Then there exist a spherical subset
$J\subseteq S$ and $u\in W^J$ such that
\[
 y=uw_J,\qquad \mathcal C(y)=\mathcal C(w_J),\qquad
 \ell(uw_Ju^{-1})=2\ell(u)+\ell(w_J).
\]
\end{proposition}

\begin{proof}
Propositions~\ref{geom:finite-canonical} and
\ref{geom:parabolic-boundary}, followed by
Theorem~\ref{geom:standard-boundary}, give a spherical $J$ with
$\mathcal C(y)=\mathcal C(w_J)$ and $y=uw_J$ reduced,
$u\in W^J$. Inverting this factorization shows that
$w_Ju^{-1}$ is reduced. Thus
$u^{-1}\in\mathcal C(w_J)=\mathcal C(y)$, and consequently
\[
 \ell(uw_Ju^{-1})=\ell(y)+\ell(u)
                =2\ell(u)+\ell(w_J).
\]
\end{proof}


\section{Parabolic compression and the sharp bound}
\label{alg:section}

Apply Proposition~\ref{prop:geometry} to the minimum-length input $y$
supporting a closed path of degree $D$, chosen above. Thus there are
a spherical subset $J\subseteq S$ and an element $u\in W^J$ such that
\[
 y=uw_J,
 \qquad
 \ell(uw_Ju^{-1})=2\ell(u)+\ell(w_J).
\]
Retain the fixed $a$ with $\deg f_{a,y,y}=D$, and put
\[
 w=w_J,\qquad g=uwu^{-1}.
\]
We shall prove $D\leq L(w)$.

\begin{lemma}[Finite longest-element coefficients]
\label{alg:finite-profile}
Let $I\subseteq S$ be spherical, and let $b\in W_I$. The coefficients
$f_{b,w_I,w_I}$ and $f_{w_I,w_I,b}$ are monic Laurent polynomials of
degree $L(b)$.
\end{lemma}

\begin{proof}
Choose a reduced expression for $b$ and expand $T_bT_{w_I}$ by successive
left multiplication. Every simple reflection in $I$ is a left descent of
$w_I$. Consequently, the path which folds at every letter is legal,
remains at $w_I$, and contributes a product of fold factors of total
weight $L(b)$. Its leading coefficient is one. Every other path uses at
least one crossing, and hence omits at least one strictly positive letter
weight from its fold weight. Thus the all-fold path is the unique path
of weight $L(b)$, proving the assertion for $f_{b,w_I,w_I}$.

Since $w_I^{-1}=w_I$, the trace pairing and cyclicity give
\[
 f_{w_I,w_I,b}
 =\tau(T_{b^{-1}}T_{w_I}T_{w_I})
 =\tau(T_{w_I}T_{b^{-1}}T_{w_I})
 =f_{b^{-1},w_I,w_I}.
\]
The first assertion, applied to $b^{-1}$, completes the proof because
$L(b^{-1})=L(b)$.
\end{proof}

\begin{lemma}[A degree estimate from a square]
\label{alg:charged}
For $h\in W_J$ and $z\in W$, one has
\begin{equation}
 \deg\coeff{z}{T_uT_hT_{u^{-1}}}\leq D-L(h).
 \label{alg:charged-equation}
\end{equation}
\end{lemma}

\begin{proof}
The expressions $uw$ and $wu^{-1}$ are reduced. Expanding their
product gives
\begin{equation}
 T_yT_{y^{-1}}
 =T_uT_w^2T_{u^{-1}}
 =\sum_{h\in W_J} f_{w,w,h}\,T_uT_hT_{u^{-1}}.
 \label{alg:square}
\end{equation}
For a fixed $z$, each summand in its standard coordinate has a positive
fold expansion. By Lemma~\ref{alg:finite-profile}, the factor
$f_{w,w,h}$ has degree $L(h)$. If the coordinate of the remaining triple
product is nonzero, their product therefore has degree
\[
 L(h)+\deg\coeff{z}{T_uT_hT_{u^{-1}}}.
\]
Lemma~\ref{lem:fold-positive} shows that this leading degree cannot be
cancelled by the other summands in \eqref{alg:square}. The coordinate on
the left is $f_{y,y^{-1},z}$ and has degree at most $D$. This proves
\eqref{alg:charged-equation}; the zero coordinate satisfies it under the
convention $\deg 0=-\infty$.
\end{proof}

Let $\HH_J=\bigoplus_{h\in W_J}\mathcal A T_h$ be the parabolic
Hecke subalgebra, and let $E_J:\HH\longrightarrow\HH_J$ be the
standard-coordinate projection:
\[
 E_J(T_x)=
 \begin{cases}
 T_x,&x\in W_J,\\
 0,&x\notin W_J.
 \end{cases}
\]
Define $\varphi:\HH\longrightarrow\HH_J$ by
\begin{equation}
 \varphi(X)=E_J(T_{u^{-1}}XT_u).
 \label{alg:compression-definition}
\end{equation}
Let $\mathcal A^{(W)}$ and $\mathcal A^{(W_J)}$ be the free
$\mathcal A$-modules with coordinate bases $(\delta_x)_{x\in W}$
and $(\delta_h)_{h\in W_J}$, respectively. We write $\lambda_W$
and $\lambda_J$ for the left regular representations in the
standard bases, identified with these coordinate bases. Thus
\[
 [\lambda_W(X)]_{x,z}=\coeff{x}{XT_z}.
\]
These matrices are column-finite. Define $\mathcal A$-linear maps
$V:\mathcal A^{(W_J)}\longrightarrow\mathcal A^{(W)}$ and
$V^*:\mathcal A^{(W)}\longrightarrow\mathcal A^{(W_J)}$ by
\[
 V\delta_h=\delta_{uh}\quad(h\in W_J),
 \qquad
 V^*\delta_x=
 \begin{cases}
 \delta_h,&x=uh\text{ for }h\in W_J,\\
 0,&x\notin uW_J.
 \end{cases}
\]
In particular, $V^*V=1$.

\begin{proposition}[Compression to the parabolic algebra]
\label{alg:compression}
For every $X\in\HH$,
\begin{equation}
 \lambda_J(\varphi(X))=V^*\lambda_W(X)V.
 \label{alg:compression-equation}
\end{equation}
If
\[
 P=\varphi(T_a)=\sum_{h\in W_J}p_hT_h,
\]
then every $p_h$ has a positive fold expansion, and
\begin{equation}
 p_h=f_{a,u,uh}
 =\coeff{a^{-1}}{T_uT_{h^{-1}}T_{u^{-1}}},
 \qquad
 \deg p_h\leq D-L(h).
 \label{alg:p-coefficients}
\end{equation}
\end{proposition}

\begin{proof}
First, $E_J$ is an $\HH_J$-bimodule map. Indeed, if $x\notin W_J$ and
$s\in J$, then $sx$ and $xs$ also lie outside $W_J$. The multiplication
rules therefore show that left or right multiplication by $T_s$ takes
the span of the outside standard basis elements into itself. On
$\HH_J$, the projection is the identity. The asserted bimodule property
follows by linearity and multiplication of the generators.

For $h,k\in W_J$, the trace pairing now yields
\begin{align*}
 [\lambda_J(\varphi(X))]_{h,k}
 &=\tau\bigl(T_{h^{-1}}E_J(T_{u^{-1}}XT_u)T_k\bigr)\\
 &=\tau\bigl(T_{h^{-1}}T_{u^{-1}}XT_uT_k\bigr)\\
 &=\tau\bigl(T_{(uh)^{-1}}XT_{uk}\bigr)\\
 &=\coeff{uh}{XT_{uk}}.
\end{align*}
For the second equality, use the bimodule property and
$\tau\circ E_J=\tau$. For the third, use $u\in W^J$, which gives
$T_uT_k=T_{uk}$ and
$T_{h^{-1}}T_{u^{-1}}=T_{(uh)^{-1}}$. This proves
\eqref{alg:compression-equation} with the stated row and column indices.

Taking the $(h,e)$ entry with $X=T_a$ gives $p_h=f_{a,u,uh}$, so $p_h$
has a positive fold expansion. Alternatively, cyclicity of the trace
gives
\begin{align*}
 p_h
 &=\tau(T_{h^{-1}}T_{u^{-1}}T_aT_u)\\
 &=\tau(T_uT_{h^{-1}}T_{u^{-1}}T_a)\\
 &=\coeff{a^{-1}}{T_uT_{h^{-1}}T_{u^{-1}}}.
\end{align*}
Applying Lemma~\ref{alg:charged} to $h^{-1}$ proves the degree estimate,
since $L(h^{-1})=L(h)$.
\end{proof}

\begin{lemma}[A coefficient of full support]
\label{alg:full-support}
There exists $h\in W_J$ such that
\begin{equation}
 \deg p_h=D-L(h).
 \label{alg:saturation}
\end{equation}
Every such $h$ satisfies $\supp(h)=J$.
\end{lemma}

\begin{proof}
The diagonal entry of \eqref{alg:compression-equation} indexed by
$k\in W_J$ reads
\begin{equation}
 \coeff{k}{PT_k}=f_{a,uk,uk}.
 \label{alg:diagonal}
\end{equation}
At $k=w$, this becomes
\[
 f_{a,y,y}=\sum_{h\in W_J}p_h f_{h,w,w}.
\]
Every nonzero summand has a positive fold expansion, and
Lemma~\ref{alg:finite-profile} gives
$\deg f_{h,w,w}=L(h)$. By \eqref{alg:p-coefficients}, the degree of
each summand is at most $D$. Since the left side has degree $D$, at
least one summand attains that degree. This is precisely
\eqref{alg:saturation}.

Now let $h$ satisfy \eqref{alg:saturation}, and put $I=\supp(h)\subseteq J$.
Suppose $I\subsetneq J$. Then $W_I$ is finite and $h\in W_I$.
Lemma~\ref{alg:finite-profile} shows that
$p_h f_{h,w_I,w_I}$ has degree $D$. It is a positive fold contribution
to $\coeff{w_I}{PT_{w_I}}$, so that coordinate has degree at least $D$.
By \eqref{alg:diagonal}, it is $f_{a,uw_I,uw_I}$, whose degree is at
most $D$ by the definition of $D$. Thus $uw_I$ supports a closed
path of degree $D$. Since $u\in W^J$ and $I\subsetneq J$,
\[
 \ell(uw_I)=\ell(u)+\ell(w_I)
 <\ell(u)+\ell(w_J)=\ell(y).
\]
This contradicts the global minimality of $y$ and proves the claim.
\end{proof}

\begin{lemma}[A finite power reaches the longest element]
\label{alg:demazure}
Let $h\in W_J$ have support $J$. There is an integer $K\geq1$ such that
\begin{equation}
 \coeff{w}{T_h^K}
 \text{ is monic of degree }KL(h)-L(w).
 \label{alg:demazure-degree}
\end{equation}
One may take $K\leq\ell(w)-\ell(h)+1$.
\end{lemma}

\begin{proof}
Fix a reduced expression for $h$. Starting at $e$, read successive
copies of this expression by the right Demazure rule: cross at an
ascent, and remain at the current element at a descent. The first
copy ends at $h$. Length never decreases. If a complete subsequent
copy causes no increase, the state stays fixed throughout that copy.
Every letter in the expression is then a right descent of this state.
The support assumption says that all of $J$ are right descents.
This state lies in $W_J$ and sends $\Pi_J$, hence $\Phi_J^+$,
to negative roots. Its length is therefore $|\Phi_J^+|=\ell(w)$,
so it equals $w$. Hence, until $w$ is reached,
each complete copy increases length by at least one. It follows that
some $K\leq\ell(w)-\ell(h)+1$ copies end at $w$.

Expand $T_h^K$ using the concatenation of these $K$ reduced words,
starting at $T_e$ and multiplying on the right. For any path ending at
$w$, let $U_+$, $U_-$, and $F$ denote the total letter weights of its
ascending crossings, descending crossings, and folds, respectively.
Each letter belongs to exactly one of these three classes. Moreover,
ascending and descending crossings change the weight of the state by
plus and minus the corresponding letter weight. Therefore
\[
 KL(h)=U_++U_-+F,
 \qquad
 L(w)=U_+-U_-,
 \qquad
 F=KL(h)-L(w)-2U_-.
\]
There is exactly one path with no descending crossing: it crosses at
every ascent and folds at every descent. This is the Demazure path
just constructed, and it ends at $w$. It has fold weight $KL(h)-L(w)$.
Every other path ending at $w$ has $U_->0$, because the simple weights
are strictly positive, and consequently has smaller fold weight.
The unique path of greatest fold weight contributes a fold monomial
with leading coefficient one. This proves \eqref{alg:demazure-degree}.

If $J=\varnothing$, then $h=w=e$ and the same conclusion holds with
$K=1$.
\end{proof}

Choose $h$ as in Lemma~\ref{alg:full-support}, and choose $K$ as in
Lemma~\ref{alg:demazure}. Selecting $p_hT_h$ in every factor of $P^K$
gives the positive fold contribution $p_h^KT_h^K$. Its $T_w$
coordinate has degree
\[
 K\bigl(D-L(h)\bigr)+KL(h)-L(w)=KD-L(w).
\]
All other contributions also have positive fold expansions. Thus
\begin{equation}
 \deg\coeff{w}{P^K}\geq KD-L(w).
 \label{alg:power-lower-bound}
\end{equation}

\begin{proposition}[Comparison of powers]
\label{alg:positive-defect}
For every integer $m\geq1$, each matrix entry of
\[
 \lambda_J\bigl(\varphi(T_a^m)-P^m\bigr)
\]
is a positive fold polynomial, possibly zero. In particular, for the
chosen $K$,
\begin{equation}
 \deg\coeff{w}{\varphi(T_a^K)}\geq KD-L(w).
 \label{alg:compressed-lower-bound}
\end{equation}
\end{proposition}

\begin{proof}
Put $M=\lambda_W(T_a)$ and $Q=VV^*$. The latter is the coordinate
projection onto $uW_J$. Equation~\eqref{alg:compression-equation} and
multiplicativity of the regular representations give
\[
 \lambda_J(\varphi(T_a^m))=V^*M^mV,
 \qquad
 \lambda_J(P^m)=V^*M Q M\cdots Q M V,
\]
where the second product contains $m$ copies of $M$ and $m-1$ copies
of $Q$. For $h,k\in W_J$, set $x_0=uk$ and $x_m=uh$. The difference
of the two $(h,k)$ entries is exactly
\begin{equation}
 \sum_{\substack{x_1,\ldots,x_{m-1}\in W\\
                  x_i\notin uW_J\text{ for some }i}}
       \prod_{i=1}^m f_{a,x_{i-1},x_i}.
 \label{alg:defect-paths}
\end{equation}
Indeed, $V^*M^mV$ allows all intermediate states, whereas each inserted
$Q$ requires the corresponding state to belong to $uW_J$. Every
factor in \eqref{alg:defect-paths} has a positive fold expansion.
The sum is finite: every standard multiplication has finite support,
and only finitely many successive multiplications are involved.
For $m=1$ the difference is zero.

Take the $(w,e)$ entry and apply Lemma~\ref{lem:fold-positive}.
The coefficient $\coeff{w}{\varphi(T_a^K)}$ is the sum of
$\coeff{w}{P^K}$ and a positive fold polynomial. Its degree is
therefore at least that in \eqref{alg:power-lower-bound}, proving
\eqref{alg:compressed-lower-bound}.
\end{proof}

\begin{lemma}[An involution coefficient]
\label{alg:involution-coordinate}
For every $X\in\HH$,
\begin{equation}
 \coeff{w}{\varphi(X)}=\coeff{g}{X}.
 \label{alg:involution-equation}
\end{equation}
\end{lemma}

\begin{proof}
The expression $g=uwu^{-1}$ is reduced by
Proposition~\ref{prop:geometry}, and $g^{-1}=g$ because $w^{-1}=w$.
Using the bimodule property of $E_J$, cyclicity, and the trace pairing,
we obtain
\begin{align*}
 \coeff{w}{\varphi(X)}
 &=\tau(T_wT_{u^{-1}}XT_u)\\
 &=\tau(T_uT_wT_{u^{-1}}X)\\
 &=\tau(T_gX)\\
 &=\coeff{g}{X}.
\end{align*}
The reducedness assertion is what gives
$T_uT_wT_{u^{-1}}=T_g$ in the third line.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
For each integer $m\geq1$, a product of $m$ standard basis elements
has every coefficient of degree at most $(m-1)D$. To see this, the
assertion for $m=1$ follows
from its single coefficient $1$. If
$T_{x_1}\cdots T_{x_m}=\sum_z c_zT_z$ satisfies the asserted bound,
then the coefficient of $T_t$ after multiplication by $T_{x_{m+1}}$
is
\[
 \sum_z c_z f_{z,x_{m+1},t}.
\]
Each summand has degree at most $(m-1)D+D=mD$, proving the induction.

Apply this upper bound to $T_a^K$, and combine
\eqref{alg:compressed-lower-bound} with
\eqref{alg:involution-equation}. It follows that
\[
 (K-1)D
 \geq\deg\coeff{g}{T_a^K}
 =\deg\coeff{w}{\varphi(T_a^K)}
 \geq KD-L(w).
\]
Subtracting $(K-1)D$ gives
\[
 D\leq L(w_J)\leq N_L(W).
\]
Conversely, for every spherical $I\subseteq S$,
Lemma~\ref{alg:finite-profile} gives
\[
 \deg f_{w_I,w_I,w_I}=L(w_I).
\]
The definition of $D$ therefore implies $D\geq N_L(W)$. Hence
$D=N_L(W)$, proving the theorem.
\end{proof}


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\end{document}