% Provenance: corrected synthesis of the chiral-trace, Yang--Mills, infinite-colour, and chamber no-go results developed in stages 14--16.
\section[Trace and Chamber Obstructions]
{Trace Functionals, Infinite Colors, and Chamber Obstructions}
\label{p4:retained:ch:trace-and-chamber-obstructions}

At a root of unity, finite-dimensionality turns a surface skein into an
endomorphism and turns the Yang--Mills functional into a matrix trace.  It
is tempting to carry that formula unchanged to generic parameter.  This
section shows exactly where the temptation fails.  The generic
identity-double functional does not annihilate an arbitrary filling ideal,
the formal infinite color is not a projector in the naive disk topology,
and opposite twists cannot preserve one color-split disk lattice.  These are
obstructions to a particular scalarization architecture, not obstructions to
the full-slide quotient of Section~\ref{p4:retained:ch:arbitrary-genus-slide-quotient}.
Mapping-torus trace constructions provide useful comparison data
\cite{MappingTori}; a Heegaard filling amplitude is nevertheless a mixed
vacuum coefficient rather than that trace.

\subsection{Finite-root trace architecture}

Let $V$ be a finite-dimensional reduced handlebody state space, let
$v\in V$ be the normalized empty vector, and let $\lambda\in V^*$ satisfy
$\lambda(v)=1$.  Put
\[
 P=v\lambda,
\]
so $P^2=P$ and $P$ has rank one.  Suppose the reduced surface skein algebra
$\mathcal A$ acts through an isomorphism
\[
 \Phi:\mathcal A\xrightarrow{\sim}\End(V),
\]
and let $e_H=\Phi^{-1}(P)$.  In a fixed reduced convention, the Yang--Mills
functional has the form
\[
 \operatorname{YM}_r(x)=c_r\tr(\Phi(x)),\qquad c_r\ne0.
\]
This is the finite-root trace formula for the skein-theoretic Yang--Mills
measure \cite{YangMills}.

\begin{theorem}[Doubled rank-one overlap]
\label{p4:retained:thm:c09-doubled-overlap}
For every $U\in\operatorname{GL}(V)$,
\begin{align*}
 \operatorname{YM}_r\bigl(e_H\Phi^{-1}(UPU^{-1})\bigr)
   &=c_r\tr(PUPU^{-1}),\\
 \tr(PUPU^{-1})
   &=(\lambda Uv)(\lambda U^{-1}v).
\end{align*}
Consequently
\[
 \frac{\operatorname{YM}_r(e_H\Phi^{-1}(UPU^{-1}))}
      {\operatorname{YM}_r(e_H)}
 = (\lambda Uv)(\lambda U^{-1}v).
\]
This ratio is unchanged by $U\mapsto zU$ and by $U\mapsto U^{-1}$.
In a unitary convention it is $|\langle v,Uv\rangle|^2$.
\end{theorem}

\begin{proof}
 At a finite root, the handlebody vacuum and cotest turn the filling data into
the rank-one operator $P$.  Multiplying by the transported rank-one projector
and taking the trace produces the two vacuum coefficients; the paired factors
explain projective invariance, while a linear lift is needed to retain the
chiral phase.  The later no-go results identify which parts of this finite
calculation fail when the color set or filling ideal becomes generic.

The first equality is the definition of $e_H$ and the trace functional.
For $w\in V$,
\[
 PUPU^{-1}w
  =v\,\lambda(Uv)\,\lambda(U^{-1}w).
\]
The trace of the rank-one map $w\mapsto a\,\mu(w)$ is $\mu(a)$, giving the
second equality.  Since $\tr(P)=1$, one has
$\operatorname{YM}_r(e_H)=c_r$.  Scalar rescaling cancels between $U$ and
$U^{-1}$, and the two factors are exchanged by inversion.  In the unitary
case they are complex conjugates.
\end{proof}

The doubled expression is projectively safe but loses the oriented phase.  A
linear lift of the mapping-class action gives the normalized one-sided vacuum
matrix coefficient.

\begin{theorem}[Normalized one-sided vacuum matrix coefficient at a root]
\label{p4:retained:thm:c09-chiral-overlap}
Fix a Maslov or framing central extension and a linear representation
$\widetilde\rho$ on $V$.  For
$U_{\widetilde\phi}=\widetilde\rho(\widetilde\phi)$ and
$u_{\widetilde\phi}=\Phi^{-1}(U_{\widetilde\phi})$,
\[
 \boxed{
 \frac{\operatorname{YM}_r(e_Hu_{\widetilde\phi})}
      {\operatorname{YM}_r(e_H)}
 =\lambda(U_{\widetilde\phi}v).}
\]
If the central generator acts by $\kappa$, changing the lift by its $m$-th
power multiplies the coefficient by $\kappa^m$.
\end{theorem}

\begin{proof}
Since $\Phi(e_Hu_{\widetilde\phi})=PU_{\widetilde\phi}$,
\[
 \tr(PU_{\widetilde\phi})
 =\tr\bigl(v(\lambda U_{\widetilde\phi})\bigr)
 =\lambda(U_{\widetilde\phi}v).
\]
Divide by $\operatorname{YM}_r(e_H)=c_r$.  The central statement follows
from linearity of the chosen lift.
\end{proof}

For a Heegaard double-coset invariant, conjugation invariance of $P$ is not
enough.  If $h_+$ and $h_-$ extend over the right and left handlebodies, one
needs the one-sided identities
\[
 U_{h_+}v=v,
 \qquad
 \lambda U_{h_-}=\lambda,
\]
together with the declared anomaly characters.  This distinction between a
simultaneously transported trace and a mixed chiral coefficient will remain
important at generic parameter.

\subsection{The generic Yang--Mills row does not descend}

Let $H=S^1\times D^2$, let $s_0=\varnothing$, and let $z=s_1$ be the
zero-framed core.  The generic coefficient functional arising from the
identity double $S^1\times S^2$ is
\[
 \lambda_H(s_n)=\delta_{n0}.
\]
It is the generic handlebody coefficient functional on its declared test
domain \cite{YangMills}.

\begin{theorem}[Genus-one no-descent]
\label{p4:retained:thm:c09-yang-mills-no-descent}
Attach a two-handle to $H$ along a longitude with the standard product
framing and cap the resulting sphere.  The filled manifold is $S^3$, and
its full slide kernel contains
\[
 z+(A^2+A^{-2})\varnothing.
\]
Nevertheless
\[
 \lambda_H\bigl(z+(A^2+A^{-2})\varnothing\bigr)=A^2+A^{-2}\ne0
\]
at generic $A$.  Hence $\lambda_H$ does not descend through this filling
quotient.  The same local obstruction occurs after any number of standard
stabilizations.
\end{theorem}

\begin{proof}
After filling, the core $z$ becomes a trivial zero-framed unknot in $S^3$.
The trivial-circle relation therefore identifies it with
$-(A^2+A^{-2})\varnothing$, proving that the displayed element belongs to
the complete kernel.  Its value under $\lambda_H$ is immediate from
$\lambda_H(z)=0$ and $\lambda_H(\varnothing)=1$.  Stabilization can be
performed away from the solid-torus summand and does not change the local
relation.
\end{proof}

There is no contradiction with the ordinary bracket on $S^3$.  The bracket
is the correct covector for the longitude filling; $\lambda_H$ is the
covector for the identity double.  Different fillings generally require
different annihilator lines.

\subsection{The infinite color is not a naive projector}

Write the formal Kirby or Yang--Mills color as
\[
 \Omega_\infty=\sum_{n\geq0}\Delta_ns_n,
 \qquad
 \Delta_n=(-1)^n[n+1],
\]
and let $\Omega_N$ be its cutoff at $n=N$.  Fusion contains the vacuum
$s_0$ in $s_is_j$ precisely when $i=j$, with coefficient one.

\begin{proposition}[Vacuum-square obstruction]
\label{p4:retained:prop:c09-vacuum-square}
The vacuum coefficient of the cutoff square is
\[
 [s_0](\Omega_N^2)=\sum_{i=0}^{N}\Delta_i^2.
\]
With
\[
 [n+1]=A^{-2n}R_n(A),
 \qquad R_n(A)=1+A^4+\cdots+A^{4n},
\]
the next increment is
\[
 \Delta_{N+1}^2=A^{-4(N+1)}R_{N+1}(A)^2.
\]
Its $A$-adic valuation tends to $-\infty$, and its magnitude diverges for
$|A|<1$.  Thus $\Omega_\infty^2$ is not defined in the naive disk
completion.  At $A=1$, the increments are $(N+2)^2$, so the same cutoffs
also fail first-stage cyclotomic Cauchy convergence.
\end{proposition}

\begin{proof}
 The infinite-color obstruction is read from the vacuum coefficient of a
finite cutoff square. Fusion selects equal colors, the balanced
quantum-integer factorization gives the exact negative disk valuation of the
next increment, and specializing at the cyclotomic point turns the same
increment into quadratic growth. The two convergence failures therefore
come from the same cutoff viewed in the two topologies.

The fusion rule gives the vacuum coefficient.  The displayed factorization
of the balanced quantum integer gives the exact valuation
$-4(N+1)$ because $R_{N+1}(0)=1$.  Substitution $A=1$ gives
$\Delta_n=(-1)^n(n+1)$.
\end{proof}

The infinite color is rigorously useful as a weak capping distribution on
a declared test domain.  The proposition only rules out treating a fixed
unnormalized color cutoff as an idempotent in the naive coefficient
topology.

\subsection{Scalar extension erases the simultaneous disk-adic/cyclotomic filtration}

Let
\[
 I_N=q^N(q;q)_N\Lambda
\]
in a non-Laurent polynomial ring $\Lambda$.  The two factors encode
different smallness conditions.

\begin{proposition}[Scalar-extension collapse]
\label{p4:retained:prop:c09-scalar-collapse}
For every $N$:
\begin{enumerate}
\item $q^N(q;q)_N$ is a unit in an algebraic Laurent-series field
      $\C((q))^{\mathrm{alg}}$;
\item in $\C[[q]]$, $(q;q)_N$ is a unit and
      $I_N\C[[q]]=q^N\C[[q]]$;
\item if $q$ is a Laurent monomial in $A$, then localization at $A$ makes
      $q^N$ a unit and retains only the cyclotomic factor.
\end{enumerate}
\end{proposition}

\begin{proof}
 Scalar extension is tested factor by factor. The cyclotomic factors are units
in the disk power-series ring, while the disk factor becomes a unit after
passing to a Laurent field; localizing in the surgery variable removes the
disk filtration as well. Each allowed extension therefore forgets one of the
two smallness conditions.

Every factor $1-q^k$ has constant coefficient one and is invertible in
$\C[[q]]$.  The Laurent-series field also contains $q^{-N}$.  If $q$ is a
Laurent monomial in $A$, it is invertible after localizing $A$.  The three
claims follow.
\end{proof}

A vector space over a Laurent field can host generic recoupling operators,
but it cannot remember simultaneous disk-adic and cyclotomic divisibility.  The
non-Laurent lattice must be chosen before scalar extension.

\subsection{Opposite twists force different disk chambers}

The local twist obstruction is already visible on one color line.

\begin{theorem}[Two-sided color-line obstruction]
\label{p4:retained:thm:c09-two-sided-twist}
Let $\Lambda_y$ be a discrete valuation ring with uniformizer $y$ and
fraction field $F$.  On $Fe_n$, let
\[
 Te_n=(-1)^ny^{n(n+2)}e_n.
\]
For $n>0$, no rank-one fractional $\Lambda_y$-lattice spanning $Fe_n$ is
preserved by both $T$ and $T^{-1}$.
\end{theorem}

\begin{proof}
 The two-sided twist obstruction reduces every rank-one fractional lattice to
a single valuation shift. That shift cancels from preservation by the twist
and its inverse, whose valuations have opposite signs for every nonzero
color. Hence one fixed diagonal lattice cannot serve both directions; the
positive repair is to let arrows move between declared disk chambers.

Every fractional ideal in a discrete valuation ring is principal, so such a
lattice is $y^w\Lambda_ye_n$ up to a unit.  The basis weight $w$ cancels
from the preservation condition.  The valuations of $T$ and $T^{-1}$ are
$n(n+2)$ and $-n(n+2)$, and both are nonnegative only for $n=0$.
\end{proof}

The theorem rules out one color-split lattice and every diagonal monomial
renormalization.  It does not rule out a non-diagonal enlargement.  Its
positive conclusion is architectural: use a chamber family with arrows
\[
 U_{\gamma,c}:L_c\longrightarrow L_{\gamma c}
\]
rather than demand that every generator preserve one disk lattice.

\subsection{Exact disk damping and its sign wall}

Set $Q=A^4$ for this calculation.  For genus $g>1$, the empty-channel term
of the generic Yang--Mills distribution is
\[
 Y_{g,n}=[n+1]^{2-2g}
 =A^{4n(g-1)}R_n(A)^{2-2g}
 =Q^{n(g-1)}
   \left(\frac{1-Q}{1-Q^{n+1}}\right)^{2g-2}.
\]
It has exact $A$-valuation $4n(g-1)$, so the untwisted tail is
coefficientwise disk-coercive.  For
\[
 \theta_n=(-1)^nA^{n(n+2)},
\]
one instead obtains
\begin{align*}
 \ord_A(Y_{g,n}\theta_n)&=n^2+(4g-2)n,\\
 \ord_A(Y_{g,n}\theta_n^{-1})&=-n^2+(4g-6)n.
\end{align*}
The first is proper and the second is unbounded below.  Replacing $A$ by
$A^{-1}$ exchanges the two sectors.

A related local genus-two channel makes the wall especially transparent.
Let $c=m+n-1$ and let $\kappa\in\{0,1,2\}$ encode the raw-row,
metric-symmetric, and raw-column gauges.  For
\[
 \mathsf K_{m,n,\kappa}(t)=[t+1]_y^\kappa y^{c\,t(t+2)},
\]
the inside and outside degrees are
\[
 v_{\mathrm{in}}\mathsf K=c\,t(t+2)-2\kappa t,
 \qquad
 v_{\mathrm{out}}\mathsf K=-c\,t(t+2)-2\kappa t.
\]
Hence the inside chamber is coercive when $m+n>1$, the outside chamber when
$m+n<1$, and the wall $m+n=1$ is noncoercive in every displayed gauge.
Gauge changes alter the linear term, not the quadratic wall.

\begin{computation}[Escaping ray in the literal Poincar\'e kernel]
\label{p4:retained:comp:c09-poincare-ray}
For every even $t\geq0$, one fully resolved path in the marked genus-two
Poincar\'e word has
\[
 v_{\mathrm{in}}(\operatorname{path}_t)=-3t(7t+16),
 \qquad
 v_{\mathrm{out}}(\operatorname{path}_t)=3t(11t+4).
\]
The formulas follow by summing the exact twist and recoupling degrees along
that admissible ray and were checked symbolically over the integers.  The
negative first formula disproves termwise coercivity of the literal inside
expansion.  The positive second formula is evidence only: one ray cannot
prove global coercivity of the outside expansion.
\end{computation}

\subsection{What a chamber trace must contain}

A generic chiral construction therefore needs, for each genus and chamber,
a completed multiplier algebra $\widehat{\mathcal A}_{g,c}$, a completed
handlebody module $\widehat{\mathcal M}_{g,c}$, a right vacuum $v^+_{H,c}$,
a left cotest $\lambda^-_{H,c}$, lifted implementers
$u_{\widetilde\phi,c}$, and a trace or pairing on an explicitly declared
product domain.  The required identities include
\begin{align*}
 u_{\widetilde\phi}xu_{\widetilde\phi}^{-1}&=\alpha_\phi(x),&
 u_{\widetilde\phi\widetilde\psi}
   &=u_{\widetilde\phi}u_{\widetilde\psi},\\
 u_{h_+}v^+_H&=v^+_H,&
 \lambda^-_Hu_{h_-}&=\lambda^-_H,
\end{align*}
as well as slide annihilation, anomaly normalization, sector-transition cocycles, and
stabilization.

The finite-root trace theorem supplies the model for these identities.  The
no-descent and two-sided-twist theorems prove that the generic objects cannot
be obtained by simply reusing one identity-double row and one color lattice
for every filling and both twist directions.