\subsubsection{The three-corner and hook coordinates}

We use the following elementary insertion rules in the anchor
calculations. In rank-\(r\) Garfinkle insertion a new largest
positive letter adds a horizontal domino at the end of row one,
and a new largest negative absolute letter adds a vertical
domino at the end of column one. In general one first inserts
into the tableau on smaller absolute labels and reinserts the
larger dominoes in increasing order. A fully covered horizontal
domino moves to the end of the next row. If only its northwest
square is covered, it turns into the vertical domino immediately
to the right of that square. The transposed rules apply to a
vertical domino, and a disjoint domino is unchanged.
These are the rules H1--H3 and V1--V3 in
\cite[\S2.1]{TaskinDomino}. A decreasing positive block of
\(j\) letters all larger than the old labels adds a horizontal
domino to each of the first \(j\) rows. A negative block with
decreasing absolute labels, all larger in absolute value than
the old labels, adds a vertical domino to each of the first
\(j\) columns. Indeed each next letter fully covers the first
block domino and moves the existing block dominoes one row or
column farther. The old smaller tableau is unchanged.

\begin{lemma}\label{fin:rectangleanchor}
The three-corner configurations in the off-wall reverse-insertion
argument, up to transposition, have ordinary bipartitions
\[
 \begin{aligned}
 &((d^{h-1},d-1),(1)),\quad h=d+r,\\
 &((1),(h^{d-1},h-1)),\quad h=d+r+1,
 \end{aligned}
 \qquad d\geq2.
\]
Here \((d^{h-1},d-1)\) has \(h-1\) rows of length \(d\)
and a final row of length \(d-1\). Each shape has a pair of
coordinates in the same leading column, with a common
recording tableau whose last domino is the overlapping
vertical corner, and with insertion last corners in the two
different proper-move components.
\end{lemma}

\begin{proof}
The staircase boundary in the three-corner configuration
is
\[
 (L,L-1,\ldots,t+2,t,t,t-2,t-3,\ldots,1),
 \qquad 2\leq t\leq L-2.
\]
Let \(j=L-t-1\). Its removable corners are the upper
horizontal corner \(\{(j,t+1),(j,t+2)\}\), the vertical
corner \(\{(j+1,t),(j+2,t)\}\), and the lower horizontal
corner \(\{(j+2,t-1),(j+2,t)\}\).
With beta-set length \(L-1\), the even runner divided
by two is \(\{t-1\}\cup\{t+1,\ldots,L-1\}\), while the
odd runner after subtracting one and dividing by two
is \(\{0,\ldots,t-3\}\cup\{t-1\}\).
Their difference in sizes is \(L-2t+1\).
A packed runner pair with size difference \(D\) has
core \(\delta_{D-1}\) for \(D\geq1\) and \(\delta_{-D}\)
for \(D\leq0\), as follows by merging the consecutive
packed beta sets. Thus the core condition gives
\((L,t)=(r+2d,d)\) or \((r+2d+1,d+r+1)\).
The runner partitions, ordered in the fixed-core
convention, are precisely the displayed labels.
Removing the vertical corner packs the runner with
the singleton and gives a pure rectangle with its
southeast box deleted. Removing either horizontal
corner leaves the singleton, and the two mixed
predecessors are different: the lower removal
decreases the even beta rank \(t-1\) by one and
the upper removal decreases the rank \(t+1\) by one.

We now construct an anchor for each configuration.
Write the pure rectangle as \(R=(W^H)\), \(n=WH\),
with \((W,H)=(d,d+r)\) in the first case and
\((d+r+1,d)\) in the second case. Its mixed companion
has one component \(R^-\), obtained by deleting the
southeast box from \(R\), and the other \((1)\).
Put
\[
 p=(W-2)(H-1),\qquad m=(W-1)H,\qquad c=p+H.
\]
Here \(c\) is a letter index in this calculation.
Use consecutive absolute intervals for the blocks below.
In the first case let \(P\) be the concatenation of
the first \(d-2\) blocks in which block \(i\) consists
of \(r+i\) decreasing positive letters followed by
\(i\) negative letters of decreasing absolute value.
In the second case let \(P\) be the concatenation
of the first \(d-1\) blocks in which block \(i\)
consists of \(i-1\) decreasing positive letters
followed by \(r+i\) negative letters of decreasing
absolute value. In both cases \(P\) uses the labels
\(1,\ldots,p\).
Set \(B=(-m,-(m-1),\ldots,-(c+1))\) and
\(C=(n-1,n-2,\ldots,m+1)\). In the first case
with \(d\geq3\) set
\(A=(c,c-1,\ldots,p+2)\) and \(z=-(p+1)\).
In the second case, and in the first case with \(d=2\),
set \(A=(c-1,c-2,\ldots,p+1)\) and \(z=-c\).
The two words are
\[
 w=P A\,(-n)\,B C z,\qquad
 w'=P\,n\,A B C z .
 \tag{1}
\]

Here is the insertion calculation. In the first case
the shape of the first \(i\) pure blocks is
\[
 (r+2i,r+2i-1,\ldots,i+1,i,i,i-1,\ldots,1).
\]
Indeed the positive block adds two boxes to each of
the first \(r+i\) rows, and the negative block then
adds two to each of the first \(i\) columns, whose
heights before this addition are \(r+2i-j\) for
\(1\leq j\leq i\). In the second case the first \(i\)
pure blocks have shape
\[
 (r+2i-1,r+2i-2,\ldots,r+i+1,r+i,r+i,r+i-1,\ldots,1).
\]
The additions are to the first \(i-1\) rows and
then the first \(r+i\) columns. Before the column
additions the latter heights are \(r+2i-1-j\).
These statements start with the core at \(i=0\).

For the first case set \(k=d-1\) and \(h=H-1\).
After \(PA\) the first \(k-1\) column heights are
\(h+k-j\) and column \(k\) has height \(h\).
The letters of \(B\) occupy pairs in column \(j\)
at rows \(h+k-j+1,h+k-j+2\), \(1\leq j<k\).
The largest domino of the first prefix in (1) is
vertical at rows \(h+1,h+2\) of column \(k\).
In the second prefix it is horizontal at columns
\(k,k+1\) of row \(h+1\), since the decreasing
\(A\)-block has carried it through its \(h\) rows.
The \(B\)-pairs are disjoint from this horizontal
pair, and the \(C\)-pairs are in the first \(h\)
rows. When \(z\) is inserted, its initial vertical
pair fully covers the first old \(B\)-pair. Full
covering continues from column to column, so the new
portion with \(n\) deleted is the vertical pair
in column \(k\) at rows \(h+1,h+2\).
This fully covers the old vertical largest domino
in \(w\), and covers only the northwest square
of the old horizontal largest domino in \(w'\).
The two rules give the same final largest domino
at rows \(h+1,h+2\) of column \(k+1\).
When \(d=2\) the \(B\)-block is empty and \(k=1\).
The final letter adds the vertical pair directly
at rows \(h+1,h+2\) in column one, giving the
same conclusion. In all cases the final recording
corner for \(w\) is the largest final vertical
pair. That for \(w'\) is the horizontal pair
in row \(h+2\) at columns \(k,k+1\).

For the second case put \(k=d-1\) and \(t=r+k\).
After \(PA\) the first \(t\) column heights are
\(k+t-j+1\), and column \(t+1\) has height \(k\).
The \(B\)-pairs are at rows \(k+t-j+2,k+t-j+3\)
of columns \(j=1,\ldots,t\).
The largest domino of the first prefix in (1)
is vertical at rows \(k+1,k+2\) of column \(t+1\).
The second is horizontal at columns \(t+1,t+2\)
of row \(k+1\). The \(C\)-block lies in the first
\(k\) rows. The final negative letter begins at
rows \(k+t+1,k+t+2\) of column one. The same
full-cover cascade leaves the vertical pair in
column \(t+1\) at rows \(k+1,k+2\) after erasing
\(n\). The largest domino is then in column
\(t+2\) at rows \(k+1,k+2\) in both tableaux.
It is the final recording corner for \(w\), while
for \(w'\) that corner is horizontal at columns
\(t+1,t+2\) of row \(k+2\). In both configurations
the smaller labels agree by largest-label
deletion, so \(P_r(w)=P_r(w')\) with the claimed
largest domino. The shape follows from the
displayed row and column additions.

It remains to place the inverse coordinates of (1)
in one leading column. They lie in one proper
induced module from \(B_m\times\mathfrak S_H\).
Indeed reversing the last \(H\) entries of the
two inverses to make them increasing gives
minimal coset representatives with respective
last strings
\[
 (-c,m+1,\ldots,n-1),\qquad
 (p+1,m+1,\ldots,n-1).
\]
Their first \(m\) entries are the remaining positive
absolute values in order. The common proper right
factor is the inverse of \(PABz\) in \(B_m\) and
the longest element of \(\mathfrak S_H\).
In the second case \(PABz\) is the next pure block
word. In the first case its tableau is obtained by
deleting the largest \(H\) insertion labels from
the full tableau. The positions just calculated
leave the pure prefix shape at \(i=d-1\).
The runner calculation gives proper ordinary
label \(((d-1)^H,\varnothing)\) in the first
case, and \((\varnothing,(W-1)^H)\) in the
second. For example the first pure prefix shape
at \(i\) has, with beta-set length \(r+2i+1\),
even ranks \(i,\ldots,r+2i\) and odd ranks
\(0,\ldots,i-1\). For the second pure prefix
use length \(r+2i\), giving even ranks
\(r+i,\ldots,r+2i-1\) and odd ranks
\(0,\ldots,r+i-1\). This yields the stated
proper labels with the fixed-core runner order.

The induced module has a coordinate counit
onto the pure rectangle, by the pure-cell
construction: in the positive convention
its type \(A\) rectangle is transposed, and
the proper source removes its last row of
length \(H\). In the first case the target
short eigenvalue is negative and the
criterion is \(b/a>H-W=r\). In the second
it is positive and the criterion is
\(b/a<W-H=r+1\). Both hold.
The part of the induced module with no
negative sign in the coset representative
is the type \(A\) canonical induction from
the corresponding proper \(A\) cell and the
symmetric factor. Indeed the no-sign
representatives lie in \(\mathfrak S_n\).
Their bar transition uses only the type \(A\)
parabolic action. Its image in the pure
target is a nonzero type \(A\) submodule,
because it contains the embedded proper copy.
It is therefore the whole pure target.
The coordinate counit has exactly one
preimage per target coordinate, so all
coset layers with a negative letter are in
its kernel. Thus \(w^{-1}\) is in the kernel.

For \((w')^{-1}\) it suffices to show that
its left \(\mathfrak S_n\) factor has ordinary
insertion shape different from \(R\).
Let \(\rho\) be the list of positions of \(P\) in increasing
signed-value order. If \(P\) has \(i\) blocks, the negative
positions occur in block order \(i,\ldots,1\) and increase
within each block. The positive positions follow in block
order \(1,\ldots,i\) and decrease within each block.
Every position of block \(j\) is smaller than every position
of block \(j+1\), and within a block the positive positions
are smaller than the negative positions.

An increasing subsequence using a negative position of block
\(j\) uses negatives from that block alone and at most one
positive from each later block. A decreasing subsequence whose
last negative position belongs to block \(j\) uses at most
\(i-j+1\) negative positions and can use positives from just one
block of index at most \(j\). For a subsequence using a single
sign, the same block orders apply: an increasing subsequence
uses at most one positive per block, and a decreasing one
uses at most one negative per block or positives from one block.

In the first case the negative and positive blocks of index \(j\)
have lengths \(j\) and \(r+j\). Thus the longest increasing and
decreasing subsequences have lengths at most \(i\) and \(r+i+1\).
There are \(i(r+i+1)\) positions, so Schensted's theorem forces
a rectangle of width \(i\) and height \(r+i+1\). In the second
case the lengths of the negative and positive blocks of index
\(j\) are \(r+j\) and \(j-1\). The same bounds give width at most
\(r+i\) and height at most \(i\), with \(i(r+i)\) positions.
Both rectangles are \(((W-2)^{H-1})\) for the indicated prefix
lengths, with the empty shape when \(P\) is empty.
The two left factors, obtained by listing
positions in signed-value order, are
\[
\begin{split}
 &(c,c+1,\ldots,m,n,\rho,c-1,\ldots,p+1,n-1,\ldots,m+1),\\
 &(c+1,\ldots,m,n,\rho,c,c-1,\ldots,p+2,n-1,\ldots,m+1,p+1).
\end{split}
\]
In the first list every increasing subsequence
has length at most \(W\). It either uses the
initial increasing block, with at most one
subsequent large letter if it omits \(n\),
or uses at most \(W-2\) from \(\rho\) and at
most one from each of the remaining two
decreasing blocks. A decreasing subsequence
uses at most one initial letter and at most
\(H-1\) from one of the three later bands,
which are ordered by increasing values.
Thus the first shape is \(R\), since there
are \(WH\) letters. The second list has the
decreasing subsequence
\(n,n-1,\ldots,m+1,p+1\) of length \(H+1\).
Thus the ordinary shape of the left factor
of \((w')^{-1}\) differs from \(R\). Its minimal
representative for \(B_m\times\mathfrak S_H\)
has only positive entries. In the type \(A\)
induction the corresponding coordinate belongs
to a cell of a different character from the pure
target, so the coordinate kernel contains it.

The two \(B_{n-1}\) predecessors of \(w^{-1}\)
are both the pure \(R^-\), by the recording
corner computation. Among pure children of
this label only \(R\) has \(\mathfrak S_n\)
type \(R\). The sole mixed child is the
claimed companion. Its leading label is
therefore that companion, since the pure
rectangle has multiplicity zero in the
kernel. The two predecessors of \((w')^{-1}\)
differ, forcing the same companion label.
It occurs with multiplicity one in the
proper induction by Pieri, hence once in
the kernel. The coordinate quotient
criterion now puts both inverse indices
in one leading column. Their recording
tableau ends vertically and their two
insertion last corners are the lower
horizontal and vertical corners.
Transposition follows from sign reversal
and the longest-element leading duality,
as explained below.
\end{proof}

We will also use longest-element duality at the
level of leading rows and columns. For completeness,
write \(D_w\) for the trace-dual element specified by
\(\tau(C_xD_w)=\delta_{x,w}\) for every \(x\).
The scaled coordinates of \(D_w\) are trace dual to those
of \(C_w\), since every scaled Schur trace weight tends
to one.
The limiting matrix coordinate map for the \(C_w\)
is invertible. Thus the scaled matrices of \(D_w\)
have limits, and the supported block is the transpose
of the corresponding signed matrix unit.
In the identity
\((D_w N_{w_0}^{-1})^\dagger
 =(-1)^{\ell(w^{-1}w_0)}C_{w^{-1}w_0}\),
the central factor \(N_{w_0}^{-1}\) is scalar in
each simple block. The determinant of \(N_{w_0}\)
on \(E\), computed along a reduced expression
using the two generator eigenvalues, shows that
this scalar has degree \(A(E\otimes\operatorname{sgn})-A(E)\).
Indeed \(w_0\) contains \(n\) short and \(n(n-1)\)
long reflections. In the determinant each
generator's exponent is its weighted character
value at that reflection divided by \(\dim E\)
when recovering the central scalar. This yields
the weighted reflection character scalar, equal
to \(A(E)-A(E\otimes\operatorname{sgn})\) by
the Schur degree-plus-valuation formula in
Lemma~\ref{fin:orientation}. Its leading scalar
is a sign, since its absolute value is determined
by the determinant, a signed monomial. Realize the
sign-twisted block by composing its matrices with
\(\dagger\). This is again a positive model, since
\(\dagger\) commutes with the adjoint involution.
The trace-dual identity sends the transposed unit
to the leading unit of \(C_{w^{-1}w_0}\) in the
sign-twisted block. Inversion transposes the two
matrix indices again. Thus right translation by
\(w_0\) takes rows to rows and columns to columns
with the sign-twisted label. In a signed permutation \(w\), right
translation by \(w_0\) reverses all letter signs
and therefore transposes the two domino tableaux.

\begin{lemma}\label{fin:hookanchor}
The two-corner configurations left by the
noncritical moves have \(n=r+2\) and ordinary
labels \(((1^p),(k))\), where \(p,k\geq1\) and
\(p+k=n\). For each such label there are two
full leading anchor columns and two actual
irreducible anchor cells, one with each
possible last corner.
\end{lemma}

\begin{proof}
In the overlapping \(2\times2\) configuration
without a larger rectangle or another corner,
the boundary is a staircase outside the plateau.
It has shape
\[
 (n,n-1,\ldots,k+2,k+1,k+1,k-1,\ldots,1),
 \qquad 1\leq k<n.
\]
It is the staircase \(\delta_n\) with one added
box. Its beta set with length \(n\) is
\(\{1,3,\ldots,2n-1\}\) with \(2k-1\) replaced
by \(2k\). The even runner is \(\{k\}\) after
division and the odd ranks are
\(\{0,\ldots,n-1\}\setminus\{k-1\}\).
The core is \(\delta_{n-2}\) and the fixed-core
label is \(((1^p),(k))\), with \(p=n-k\).

The two words
\[
 \begin{aligned}
 w&=(n,p-1,\ldots,1,-(n-1),\ldots,-p),\\
 w'&=(-n,p-1,\ldots,1,-(n-1),\ldots,-p)
 \end{aligned}
 \tag{2}
\]
have the same insertion tableau, with last
insertion domino vertical at
\(\{(p,k+1),(p+1,k+1)\}\).
After deleting \(n\), the \(p-1\) decreasing
positive letters extend the first \(p-1\) rows
of the core and the \(k\) negative letters
extend its first \(k\) columns. In \(w\) the
largest horizontal domino has first moved
through the \(p-1\) rows to
\(\{(p,k),(p,k+1)\}\). The last negative
letter covers its northwest square and
turns it vertically to the right. In \(w'\)
the initially vertical largest domino is
pushed successively into columns \(2,\ldots,k+1\)
by the negative letters. The positive
additions do not meet it. These are
respectively the H3 and full-cover V2
rules. The final recording corner of \(w\)
is the lower horizontal one and that of
\(w'\) is the vertical one, as follows by
comparing the shapes before and after the
last insertion. Deleting the absolute label
\(n\) gives the proper asymptotic label
\(((1^{p-1}),(k))\). Thus the common insertion
tableau ends in the first quotient component.

We show that the inverses of (2) have leading
matrices in one column of the positive label
\(E=((1^k),(p))\). Write \(q=v^a\), \(Q=v^b\).
Use the \(n\)-dimensional orthonormal model
with \(N_{s_0}=\operatorname{diag}(Q,-Q^{-1},
\ldots,-Q^{-1})\). For \(1\leq i<n\), let
\(x_i=Q^2q^{-2(i-1)}\) and on coordinates
\(i,i+1\) let \(N_{s_i}\) have block
\[
 \begin{gathered}
 \begin{pmatrix}
 (q-q^{-1})/(1+x_i)&\beta_i\\
 \beta_i&(q-q^{-1})x_i/(1+x_i)
 \end{pmatrix},\\
 \beta_i=\frac{\sqrt{(1+x_iq^2)(1+x_iq^{-2})}}{1+x_i},
 \end{gathered}
\]
acting by \(q\) elsewhere. Use the positive square
root for \(v>1\). The block has trace \(q-q^{-1}\)
and determinant \(-1\). The adjacent three-term
braid follows by multiplication of the \(3\)-by-\(3\)
blocks with \(x_{i+1}=x_i/q^2\). The only needed
off-diagonal squares are
\(1+(q-q^{-1})^2x_i/(1+x_i)^2\).
At the short edge the product of the two
\(2\)-by-\(2\) blocks has trace zero and determinant
one, so its square is \(-1\) in either order.
Disjoint blocks commute. These checks give a
representation. Take its \(k\)-th exterior
power, scaling the long generators by \(q^{1-k}\)
and the short generator by \((-Q^{-1})^{1-k}\).
The braid relations are preserved, and the two
eigenvalues on the exterior powers after scaling
are the original two Hecke eigenvalues. At \(v=1\)
the model has the sign-subgroup weights indexed by
\(k\)-subsets, with a column on the chosen positions
and a row on the complement. The wreath-product
classification gives the label \(E\).

Put \(m=n-1\), \(S=\{p,\ldots,n-1\}\), and let
\(e_S\) be the associated exterior basis vector.
The first inverse is \(xu\), length-additively,
where
\[
 x=s_1\cdots s_{n-1},\qquad
 u=(p-1,\ldots,1,-m,\ldots,-p)\in B_m .
\]
The other inverse is \(s_0xu\) with length one
larger. For the commuting elements
\(N_{s_{j-1}}\cdots N_{s_1}N_{s_0}
 N_{s_1}\cdots N_{s_{j-1}}\), the eigenvalue
on a subset with \(j\) chosen and \(c\) earlier
chosen positions is \(Qq^{-2c}\).
If \(j\) is unchosen it is
\(-Q^{-1}q^{2(j-1-c)}\).
This follows by induction using the displayed
two-coordinate matrices. The element \(u\)
is the product of the reversal on the first
\(p-1\) positions, the sign changes at
\(p,\ldots,m\), and the reversal of that
negative block. The sign changes are length
additive, whereas the last reversal shortens
length. Thus its standard-basis product uses
the inverse of the last reversal's standard
element. On \(e_S\) these factors give the
scalar
\[
 \xi=(-1)^{\binom{k}{2}}Q^k
 q^{\binom{p-1}{2}-\binom{k}{2}} .
\]
In applying the long generators in \(x\) from
right to left, there is a unique path from
\(S\) to \(\{p+1,\ldots,n\}\). It moves each
chosen position once to the right, and gives
the matrix entry
\(\xi q^{p-1}\prod_{i=p}^{n-1}\beta_i\).
There is also a unique path to
\(\{1,p+1,\ldots,n-1\}\). It keeps \(n-1\)
in the first step, passes the remaining
chosen-chosen pairs, and moves \(p\) to \(1\).
Including the short generator, this entry
in the other inverse is
\[
 \xi Q\frac{q-q^{-1}}{1+x_{n-1}}
   (-q^{-1})^{k-1}\prod_{i=1}^{p-1}\beta_i .
\]
Path uniqueness follows successively from
the required membership of \(n,n-1,\ldots,1\),
as each adjacent position is visited only once.

Put \(\kappa=(n-1)a-b\), so \(0<\kappa<a\).
For \(i\leq n-2\) the factors \(\beta_i\)
tend to one, while the last is asymptotic
to \(v^\kappa\). The last upper diagonal
entry has degree \(a-2(b-(n-2)a)\).
Both displayed entries therefore have
degree
\[
 kb+a\bigl(\tbinom{p}{2}-\tbinom{k}{2}\bigr)
       +\kappa=A(E)
\]
and absolute leading coefficient one.
For the last equality the cross hooks in
the column component are \(2-i\),
\(1\leq i\leq k\), and those in the row
component are \(n-1\) at its first box
and \(p-j\) at the later boxes. In
\cite[Theorem~3.2]{ChlouverakiJacon},
the column cross factors all have positive
degree, and only the first row cross factor
has positive degree. Together with the two
ordinary quantum hook products and the
monomial \(q^{-k(k+1)}\), they give the
displayed value and leading coefficient
one. Now the leading trace identity for
\(N_wN_w^*\) is the sum of squares of
all leading matrix entries in the
orthonormal models, with each block
weighted by its reciprocal leading
Schur coefficient. That sum is one.
The single entry of absolute leading
coefficient one therefore forces each
of these two inverse matrices to be a
signed unit, with the common input
column \(S\). This common-column assertion
is independent of the model: it is
equality of their leading products
\(t_{w^{-1}}^*t_{w^{-1}}\) and
\(t_{(w')^{-1}}^*t_{(w')^{-1}}\).
Together with the proper fixed-corner
moves, this identifies one full leading
column. Exchanging \(p,k\) and applying
longest-element duality supplies the
other last corner.

We give an actual cell anchor independently
of the preceding leading calculation.
On a basis \(f_1,\ldots,f_n\), use the
integral model in which \(N_{s_i}f_i=f_{i+1}\),
\(N_{s_i}f_{i+1}=f_i+(q-q^{-1})f_{i+1}\),
and the other coordinates have eigenvalue \(q\).
Let \(N_{s_0}f_1=Qf_1\) and
\[
 N_{s_0}f_j=-Q^{-1}f_j+Q^{-1}(q-q^{-1})q^{j-2}f_1
 \quad(j\geq2).
\]
These generators obey the quadratic and
braid relations: the long generators are
the type \(A\) permutation model, the upper
row commutes with \(N_{s_i}\) for \(i\geq2\)
because its consecutive coefficients have
ratio \(q\), and the short braid is obtained
on \(f_1,f_2\) by the trace-zero product
and on \(f_j\), \(j\geq3\), by substituting
the coefficient \(Q^{-1}(q-q^{-1})q^{j-2}\)
in both products. This model has the same
specialization at \(v=1\) as the orthogonal
model above. Put
\[
 c_j=f_j+q^{-1}f_{j-1}+\cdots+q^{-(j-1)}f_1,
 \qquad t_j=Qq^{-(j-1)}+Q^{-1}q^{j-1}.
\]
Use \(c_0=0\). Direct substitution gives
\[
 \begin{aligned}
 N_{s_i}c_i&=-q^{-1}c_i+c_{i-1}+c_{i+1},\\
 N_{s_i}c_j&=q c_j\quad(j\ne i),\\
 (N_{s_0}+Q^{-1})c_j&=t_jc_1 .
 \end{aligned}
\]
Since the coefficients of \(C_s\) in
this basis are fixed by bar, declaring the
basis fixed defines a compatible bar.
Replace \(c_n\) by \(c_n/t_n\), retaining
the other \(c_j\). The coefficient into
\(c_n/t_n\) for \(N_{s_{n-1}}\) becomes
\(t_n\), and the returning short coefficient
becomes one, so the new basis still spans
an integral invariant lattice.

To check its form, in the \(f\)-basis set
\(h=(1,q,\ldots,q^{n-1})^{\mathsf t}\) and
\(\eta=(q^2-1)/(q^2Q^2+1)\). The matrix
\(1+\eta hh^{\mathsf t}\) is positive and
makes every generator self-adjoint. The
long generators fix the line of \(h\), and
the short adjoint equation reduces to
\((1+\eta)(q-q^{-1})=\eta q(Q^2+1)\).
In the \(c_j\)-basis,
\(h^{\mathsf t}c_j=q^{j-1}+q^{j-3}+\cdots+q^{1-j}\).
For \(i,j<n\) the rank-one Gram contribution
has degree \((i+j-2)a-2b<0\).
After dividing the last column by \(t_n\)
the mixed contributions still have negative
degree. The final squared norm tends to one,
since its numerator and \(t_n^2\) have the
same leading monomial \(Q^{-2}q^{2(n-1)}\).
The Euclidean part tends to the identity
on the earlier coordinates. Hence the
new Gram tends to the identity.

Take the scaled \(k\)-th exterior power of
this lattice, as before. Its basis is integral,
bar fixed, and has limiting Gram identity.
In this basis the exterior vector on the
coordinates \(1,\ldots,k-1,n\) spans the
one-dimensional proper source of
\(B_{k-1}\times\mathfrak S_{p+1}\).
Its long eigenvalues on the first factor
are \(-q^{-1}\) and its short eigenvalue
is \(Q\), while the symmetric factor is
trivial. Indeed in the displayed formulas
all neighbouring replacements on the first
block duplicate an exterior factor, and
every factor is a \(q\)-eigenvector on the
symmetric block. This is an actual
proper singleton by the asymptotic
classification. Its degree is
\[
 (k-1)b-a\binom{k-1}{2}+a\binom{p+1}{2}
 = A(E),
\]
using \(p+k=n\). The Schur ratio has
leading coefficient one. The coordinate
counit therefore constructs an actual
irreducible cell containing the identity
coset source index
\[
 (-(k-1),\ldots,-1,n,n-1,\ldots,k).
\]
This index is an involution. Deleting
the absolute letter \(n\) gives the
proper asymptotic label
\(((1^p),(k-1))\), so the last domino
is in the second quotient component.
The sign-dual construction after
exchanging \(p,k\) gives an actual
anchor in the first component.
\end{proof}