\subsection{Proper parabolic sources and the finite induction step}\label{fin:propagation}

We use ordinary bipartition labels \((\alpha,\beta)\),
where the first component corresponds to the positive short
eigenvalue. Tensoring with the full sign character sends this label
to \((\beta^{\mathsf t},\alpha^{\mathsf t})\), with \({}^{\mathsf t}\)
denoting partition transposition.

\begin{lemma}\label{fin:orientation}
Let \(n\geq1\). If \(A(E)\leq A(E\otimes\operatorname{sgn})\), there are a proper
standard parabolic subgroup and a generic irreducible \(F\) of that
subgroup such that \(E\) occurs in \(\operatorname{Ind}F\) and
\(A(F)=A(E)\). If \(b/a\) is nonintegral, \(F\) can be chosen so that
\(E\) has multiplicity one and every other constituent has strictly
greater spectral degree.
\end{lemma}

\begin{proof}
Write \(U=E\otimes\operatorname{sgn}\), with ordinary label
\((\alpha,\beta)\). The hypothesis is
\(a_{\mathrm{Sch}}(U)\leq a_{\mathrm{Sch}}(U\otimes\operatorname{sgn})\).
Set \(b/a=r+\delta\) with \(r\) a nonnegative
integer and \(0\leq\delta<1\). For \(N\) at least the lengths of
\(\alpha,\beta\), the charged symbol has upper entries
\(\alpha_i+N+r-i+\delta\) for \(1\leq i\leq N+r\), and lower entries
\(\beta_j+N-j\) for \(1\leq j\leq N\). Partitions are padded by
zeroes. Let \(Z\) be the multiset of these entries, \(Z_\varnothing\)
the corresponding multiset for two empty partitions, and
\(m=2N+r\). The statistic \(n(Z)\) is the sum of the pairwise
minima over unordered pairs of occurrences, since the \(i\)-th
entry in decreasing order is the minimum in exactly \(i-1\)
pairs with an earlier entry. The cancellation-free Schur formula gives
\[
 a_{\mathrm{Sch}}(U)=a\bigl(n(Z)-n(Z_\varnothing)\bigr);
\]
see \cite[Theorem~3.2 and \S4.3]{ChlouverakiJacon}. The same formula,
applied to the minimum and maximum degrees of the Schur element,
gives
\[
 \frac{a_{\mathrm{Sch}}(U)-a_{\mathrm{Sch}}(U\otimes\operatorname{sgn})}{a}
 =
 \sum_{z\in Z} z(m-1+\delta-z)
 -
 \frac{m(m-1)(m-2+3\delta)}{6}.                         \tag{*}
\]
For this identity the degree-plus-valuation of the Schur product,
divided by \(2\), is
\(b(|\alpha|-|\beta|)
 +2a(n(\alpha^{\mathsf t})-n(\alpha)+n(\beta^{\mathsf t})-n(\beta))\).
For the cross factors one uses
\[
 \begin{aligned}
 &\sum_{(i,j)\in\alpha}(\alpha_i-i+\beta_j^{\mathsf t}-j+1)\\
 &\quad+\sum_{(i,j)\in\beta}(\beta_i-i+\alpha_j^{\mathsf t}-j+1)\\
 &=n(\alpha^{\mathsf t})+n(\beta^{\mathsf t})-n(\alpha)-n(\beta)
   +2\sum_j\alpha_j^{\mathsf t}\beta_j^{\mathsf t}.
 \end{aligned}
\]
The cross term cancels against the contribution of the sorted union
of the two partitions in the Schur monomial. In the symbol sum on
the right of \((*)\), adjoining a box \((i,j)\) changes the sum by
\(-2(j-i)-(r+\delta)\) in the upper row and
\(-2(j-i)+(r+\delta)\) in the lower row. The box sums
\(\sum_{(i,j)\in\alpha}(j-i)=n(\alpha^{\mathsf t})-n(\alpha)\) and the
corresponding sum for \(\beta\) give the negative of the
degree-plus-valuation divided by \(2a\). The empty-symbol entries
are \(\delta,\ldots,N+r-1+\delta\) and \(0,\ldots,N-1\).
Summing the quadratic over these two lists gives the subtracted
constant in \((*)\).

The ordinary Pieri rule for \(B_{n-k}\times\mathfrak S_k\)
with sign on \(\mathfrak S_k\) raises \(k\) selected entries of the
source symbol by one, with the condition that each row is a
strictly increasing admissible beta set. This is the two-colour
Pieri rule \cite[Lemma~7.6]{GeckIancuFamilies}.
Raising the \(k\) largest entries increases \(n(Z)\) by
\(\binom{k}{2}\), the sign degree on \(\mathfrak S_k\) divided by
\(a\). For any selection the increase is at least this much,
because a pair of raised entries contributes one and a mixed
pair contributes a nonnegative amount. When all entries are
distinct, equality occurs only for the \(k\) largest: if a raised
entry is below an unraised entry, the mixed pair contributes
strictly positively. The Pieri multiplicities are one, since each
resulting bipartition specifies the vertical strips in both rows.
We will find the source by lowering a nonempty top set of target
entries. Its rank is \(n-k\), so \(B_{n-k}\times\mathfrak S_k\)
is proper, with the trivial-group convention for \(B_0\).

Remove initial pairs \((\delta,0)\) from the two rows and subtract
one from the remaining entries for each removal. These operations
amount to reducing \(N\) without changing the bipartition. If the
minimum of the normalized symbol is at least one, lower all entries
by one. If a successive gap is greater than one, lower all entries
above it by one. Both operations give admissible rows and the
required top set. For integral ratios, the remaining case has
consecutive levels \(0,\ldots,t\), for some \(t\geq0\), with zero
single. If every level is single, both rows occur at positive
rank. Choose adjacent levels in different rows and lower the
suffix above their cut. The only new tie is across rows, so this
again gives an admissible source.

If there are \(k>0\) double levels in the integral case, let \(D\)
be their set, a subset of \(\{1,\ldots,t\}\). Substitution in \((*)\)
gives
\[
 \sum_{i\in D}i(t+k-i)-\sum_{j=0}^{k-1}j(t+k-j)\ \geq\ kt>0.
\]
Indeed the \(k\) smallest values of \(i(t+k-i)\) on
\(\{1,\ldots,t\}\) occur at \(1,\ldots,k\): the expression is
increasing up to \(k\) and is at least \(kt\) from \(k\) to \(t\),
by concavity and the endpoint values. Thus this case cannot
satisfy the inequality in the lemma.

For \(0<\delta<1\), suppose the minimum is less than one and all
successive gaps are at most one. In \((*)\) the difference is affine
as a function of \(\delta\) while the integer indices in the rows
are held fixed. At \(\delta=0\) the integer levels are consecutive
and zero is single. A missing level would give a gap at least
\(2-\delta>1\). Hence the difference at zero is nonnegative, and
is positive if there is a double level. At \(\delta=1\), shift
the upper indices up by one and adjoin an upper zero to obtain
the symbol at ratio \(r+1\). The expression in \((*)\) at
\(\delta=1\) is exactly the expression for this enlarged symbol
at integral ratio. Its levels are consecutive, since a missing
level would give a gap at least \(1+\delta\) in the original
symbol. If zero is double, normalize it once: the original
lower zero must have been followed by a lower one, since the
upper \(\delta\) was absent. The new zero is therefore single.
The difference at one is again nonnegative. If the difference
at zero vanishes, each original index occurs once. The sequence
of rows, from the smallest index upwards, is then an upper
prefix followed by a lower suffix, since a lower-to-upper
transition would give a gap \(1+\delta\). Positive rank forces
both rows to occur. The upper-to-lower transition becomes a
positive double level at the endpoint one, making that endpoint
strictly positive. Affine interpolation now makes the difference
strictly positive throughout the remaining nonintegral case.
This excludes it under the hypothesis of the lemma. At each
cut used in the construction the gap is at least one, so
lowering by one leaves the selected entries a top set in the
source. This gives the asserted equality. Off the walls all
source entries are distinct, giving the strict-minimum and
multiplicity claims. Twisting back proves the statement for
\(A(E)\).
\end{proof}

\begin{lemma}\label{fin:unit}
Let a finite split rational algebra have an integral basis
\(t_w\), an involution permuting it, and a positive symmetric trace
for which \(\langle x,y\rangle=\tau(xy^*)\) makes this basis
orthonormal. Suppose that the algebra over \(\mathbb Z\) contains
its unit, that its irreducible matrix representations can be
chosen rational with positive adjoint forms, and that the trace is
a positive linear combination of their matrix traces. The
nonzero coefficients of the unit in the given basis are signs,
and the corresponding signed basis elements are orthogonal
self-adjoint idempotents. Their left and right multiplications
are coordinate projections.
\end{lemma}

\begin{proof}
For an integral basis element \(t_w\), its matrix in a rational
simple block has integral characteristic polynomial. Its
eigenvalues satisfy the monic integer characteristic polynomial
of left multiplication on the integral algebra. Hence they are
algebraic integers and the block characteristic coefficients,
being rational, are integers. For any such matrix \(M\), computed
with its positive adjoint form, Schur triangularization gives
\(\operatorname{tr}(MM^*)\geq\sum_\lambda|\lambda|^2\), summing
over its eigenvalues with multiplicity. For the roots of a
nonzero irreducible factor of degree \(k\) of its characteristic
polynomial, the product has absolute value at least one.
Their squared moduli therefore sum to at least \(k\), and
Cauchy--Schwarz bounds the modulus of their sum by this sum of
squared moduli. Adding the factors gives
\(\operatorname{tr}(MM^*)\geq|\operatorname{tr}M|\).
Equality forces normality and equality in the root inequalities.
Each nonzero factor then has identical roots of modulus one,
hence its roots are \(1\) or \(-1\) by rationality and
irreducibility. The triangle inequality forces a common sign.

The coefficient of \(t_w\) in the unit is \(\tau(t_w)\), since
\(\tau(t_w^*)=\tau(t_w)\). The weighted trace inequalities give
\[
 1=\tau(t_wt_w^*)\geq|\tau(t_w)|.
\]
An integral nonzero coefficient is therefore a sign, and its
signed basis element is a self-adjoint projection in all blocks.
These projections sum to the identity. For a vector in the range
of one projection, the sum of its squared projection norms is its
squared norm, so all other projections annihilate it.
Multiplication by each signed basis projection is an integral
orthogonal projection in the regular orthonormal basis. Its
diagonal entries lie in \(\{0,1\}\), and each equals the sum of
squares of the entries in its row. This forces all off-diagonal
entries to vanish, on either side.
\end{proof}

\begin{proposition}\label{fin:close}
For finite \(B_n\), suppose the spectral thresholds are
\(C\)-coordinate ideals as in Lemma~\ref{fin:e1}, and no generator
edge at a fixed spectral degree joins different actual two-sided
cells, as in Lemma~\ref{fin:e2}. If the proper standard parabolic
subgroups satisfy \textup{P1--P15}, then so does \(B_n\).
\end{proposition}

\begin{proof}
The threshold assertion is Geck's E1. By
\cite[Proposition~3.6, Lemma~3.7 and Corollary~3.8]{GeckFinite},
it gives P1, P4, P7 and P8, identifies the spectral label with
\(a\), and identifies the leading algebra with the
\(\gamma\)-product. The leading trace \(\tau_J\) takes the value \(n_d\) on
a distinguished element \(d\) and zero elsewhere. The leading
matrix construction \cite[Propositions~2.3 and~3.4--3.6]{GeckLeading}
gives split rational blocks with positive adjoint forms and
positive trace weights. The unit is integral, since its coordinates
are the values of that trace. Apply Lemma~\ref{fin:unit}. The unit
is a sum of signed basis projections \(n_dt_d\) indexed by
distinguished elements. Self-adjointness gives \(d=d^{-1}\).
The coefficient of \(t_d\) in \(t_xt_y\) is
\[
 \gamma_{x,y,d}
 =\tau_J(t_xt_yt_d)
 =n_d\,\delta_{x,y^{-1}}\,
       \mathbf 1_{\{t_y n_dt_d=t_y\}},
\]
where the indicator is one when its condition holds and zero
otherwise. The unique right unit of \(t_y\) gives P2, P3 and P5.
Thus P1--P8 hold. Lusztig's parabolic argument
\cite[\S14.12]{Lusztig} gives P12 from P3, P4 and P8 for the ambient
and source groups.

Fix an exact \(a\)-layer of value \(A\). In its generic
Hecke algebra, the simple blocks have \(A(E)=A\). If a proper
parabolic two-sided cell has value \(A\), its lower boundary has
larger intrinsic \(a\) by P4 and P11 in that subgroup.
By P12 the source has ambient degree \(A\) and its boundary
has ambient degree greater than \(A\). In the ambient quotient
by degrees strictly greater than \(A\), the source cell quotient
therefore embeds on its original coordinates.
The ambient threshold ideal of degrees at least \(A\) therefore
contains every \(hC_pk\) from the source. Its image in the
quotient is the exact \(A\)-layer. The defining bound for the
\(a\)-function gives degree at most \(A\) at all surviving output
coordinates. Apply Lemma~\ref{lem:double-induction} to the quotient
and source. The constant comparison image is
stable under both actions, which commute on it with independent
parameters.

We verify the generic block support of the comparison image and
of the leading ideal generated by the source. The generic image
is the two-sided ideal generated by the source cell quotient:
the two canonical comparison matrices are invertible and
multiplication in the middle source is identified with the
original source action. Thus the generic blocks in the image
are exactly those irreducibles whose restrictions contain a
constituent of that source cell. The leading ideal has the same
block support. To see this, complete at \(v^{-1}=0\) and scale
the source and ambient products by \(v^{-A}\). The resulting
finite free algebras reduce to their leading algebras. A lift of
a unit of the special fibre has invertible left and right
multiplication matrices over the complete ring. If \(u\) is
this lift, solve \(ue=u\) and \(e'u=u\). Associativity and
invertibility imply \(e\) is a left unit and \(e'\) a right
unit, so they agree. On the ambient layer the map in each block
is \(h\mapsto v^{-A}\rho_E(h)\), which is multiplicative for the
scaled product. It is integral by balancedness, and its reduction
is the leading representation. The direct sum of the reduced maps
is an isomorphism by the leading matrix construction, so the direct
sum over the complete ring is an isomorphism as well. The lifted
source unit is therefore an idempotent matrix in each such block,
and its generic and special ranks agree. Its generic rank is
positive exactly when that block sees the source quotient, and
its special rank is positive exactly when the corresponding
leading block sees the source leading unit. Hence the block
supports, and therefore dimensions, agree. The constant image
contains the ordinary leading ideal by stability and inclusion
of the source basis. Equality of dimensions identifies them.

For each actual two-sided cell, the spectral degrees of all
constituents have one common value \(A\). The longest-element
dual cell has constituents obtained by the sign twist, with
common value \(A'\). Suppose \(A\leq A'\).
Lemma~\ref{fin:orientation} gives, for each constituent, a
proper source constituent of the same spectral degree. The
source support just proved puts every corresponding leading
block in a constant image. These images give commuting
independent actions on the whole cell quotient.

For the other inequality use the dual cell. Here are the
duality conventions for the transfer. Identify the coefficient
functional \(D_w(C_x)=\delta_{w,x}\) with a trace-dual element of
the finite Hecke algebra. For the longest element \(w_0\),
\cite[Proposition~11.5]{Lusztig} gives
\[
 (D_wN_{w_0}^{-1})^\dagger
   =(-1)^{\ell(w^{-1}w_0)}C_{w^{-1}w_0}.
\]
The map \(D\mapsto(DN_{w_0}^{-1})^\dagger\) sends \(hDk\) to
\(h^\dagger(DN_{w_0}^{-1})^\dagger k^\dagger\), since \(w_0\)
is central in type \(B\). It gives a signed identification of the
dual cell quotient with the longest-element dual cell quotient,
by the order reversal in \cite[Corollary~11.7]{Lusztig}.
Dualizing interchanges the transpose of a left action with the
right action, and
\(C_s^\dagger=(v^{L(s)}+v^{-L(s)})1-C_s\).
The scalar terms disappear from commutators, so the independent
generator commutators vanish on one cell quotient exactly when
they vanish on its dual. This proves the cell-quotient assertion
in both orientations.

Taking the coefficient of \(u^A\) in the right action at an
independent parameter \(u\) shows that the full left action in
a cell quotient commutes with every right multiplication in its
leading algebra. In particular it preserves the coordinate
projections onto right-unit blocks. P3 and P8 place a basis index
in the same left cell as its right unit, and any left edge inside
that two-sided cell preserves the right unit. This proves P13 and
also proves that a left preorder relation inside a single
two-sided cell is a left cell equivalence. P6 and P13 give P14
by \cite[\S14.14]{Lusztig}. The assumed separation of two-sided
cells at equal degree, combined with the left equivalence just
proved and its inverse, gives P9 and P10. P4 and reversal of
each edge of an equal-degree path give P11. Finally the exact
\(a\)-layer is the direct sum of its cell quotients,
because an equal-degree edge cannot join different cells.
The commuting independent actions on these quotients give P15
on the entire layer.
\end{proof}