\section{Transport to the tangent diagram}\label{sec:star}
Assume the divisor in Lemma~\ref{lem:valuation} exists and retain its notation. The construction of $n_0$ in Lemma~\ref{lem:valuation} gives $\Lambda b=\rho^{-1}a_0$ for a positive rational number $\rho$. Since $\Lambda$ is alternating and $\Lambda P=R$,
\[
 \pair{a_0}{Pn}=\rho(Pn)^{\mathsf T}\Lambda b
              =-\rho b^{\mathsf T}\Lambda Pn.
\]
Thus
\begin{equation}\label{eq:rho}
 \pair{a_0}{Pn}=-\rho\pair{Rn}{b}\qquad(n\in\QQ^J).
\end{equation}
Let $\D_b$ be the tangent diagram: retain walls whose closed support $W$ contains $b$ and replace $W$ by $W+\RR b$. Its normals annihilate $b$. For a retained wall, write its supporting inequalities as $\eta_j(x)\ge0$ within its hyperplane. Those satisfying $\eta_j(b)=0$ define $W+\RR b$; those with $\eta_j(b)>0$ hold on a fixed compact set after translation by a sufficiently large multiple of $b$. The finitely many walls not containing $b$ miss that translated set. Thus at every finite charge order a compact path in this diagram translates sufficiently far in the positive $b$ direction to a path in $\D$ with the same wall product. Applying this to a closed path proves that its product is the identity. Thus $\D_b$ is consistent and positive. Its charge filtration is the original pointed filtration given by $\NN^J$.

Choose a positive-chamber point $q_0$ simultaneously general for $\D_b$, for $\D$ at all endpoints $q_0+lb$ with integers $l\ge0$, and for the transport line $q_0+\RR b$. The countable exceptional-set construction in Section~\ref{sec:setup} applies, adding the proper subspaces $U+\RR b$ for joints and hyperplanes parallel to $b$. Choose $m_*\in\LL$ with $\pair{a_0}{m_*}<0$, and set
\[
 m_l=m_*+lb,\qquad q_l=q_0+lb.
\]
Since $g$ is a maximum of finitely many linear functions, one linear piece $h$ is active at every sufficiently large $q_l$, with $h(b)=g(b)>0$. There is also a constant $c_*$ such that $g(m_l)=lg(b)+c_*$ for all sufficiently large $l$.

\begin{lemma}\label{lem:uniform}
For every broken line of input $m_l$ ending at $q_l$, its total correction $Pn$ satisfies
\begin{equation}\label{eq:uniform}
 h(Pn)\le c_*-h(m_*).
\end{equation}
There is a nonempty relatively open set of weights in $b^\perp\subset\LL_\RR^*$ for which these supports, after shifting by $-lb$, have uniformly finite sublevel sets.
\end{lemma}
\begin{proof}
Let $e=m_l+Pn$ be the final exponent. The initial asymptotic derivative of $g\circ\gamma$ is $-g(m_l)$. Its left derivative at zero is $-\max_{h'\text{ active at }q_l}h'(e)$. Convexity implies $h(e)\le g(m_l)$ and gives \eqref{eq:uniform}.

Put $w=\ell-\ell(b)h/h(b)$. Then $w(b)=0$, $\ell(b)>0$, and
\[
 w(Pn)\ge |n|_1-\frac{\ell(b)}{h(b)}\bigl(c_*-h(m_*)\bigr).
\]
A sufficiently small perturbation of $w$ in $b^\perp$ retains a positive linear coefficient in front of $|n|_1$, because $\|Pn\|\le\|P\||n|_1$. Each bounded sublevel therefore permits only finitely many charges, with a bound independent of $l$. In particular, a fixed exponent class modulo $\ZZ b$ has only a fixed finite set of possible full exponents after the shift.
\end{proof}

\begin{lemma}\label{lem:transport}
For fixed $l>0$, the wall product along $q_0+tb$, $0\le t\le l$, and its inverse converge in the completion of the valuation $v$ and preserve $v$. Their actions on finite Laurent inputs agree with chargewise theta transport.
\end{lemma}
\begin{proof}
If the path crosses a wall of charge $n$, then
$\pair{Rn}{q_0}+t\pair{Rn}{b}=0$ for some $0<t\le l$. Set $\epsilon_0=\min_{i\in J}\pair{Re_i}{q_0}>0$. Then $\pair{Rn}{q_0}\ge\epsilon_0|n|_1$. Formula~\eqref{eq:rho} gives the positive integer
\begin{equation}\label{eq:crossing}
 s(n):=\pair{a_0}{Pn}\ge\frac{\rho\epsilon_0}{l}|n|_1>0.
\end{equation}
Retain $F=H-c$ and $V=Z/F^k$ from Lemma~\ref{lem:valuation}. The rational function field is $\kk(S,V)(F)$, because $H=c+F$ and $Z=VF^k$. Lemma~\ref{lem:valuation} says that $V,S_1,\ldots,S_{d-2}$ have algebraically independent residues. Consequently every nonzero element of $\kk(S,V)$ has valuation zero, and $F$ has valuation one. Completion is therefore the Laurent series field $\kk(S,V)((F))$. Let $A=\kk[H^{\pm1},S_1^{\pm1},\ldots,S_{d-2}^{\pm1}]$. There is an injective map
\begin{equation}\label{eq:common}
 A((Z))\longrightarrow\kk(S,V)((F)),\qquad
 \sum_{r\ge\mu}Z^rf_r(H,S)\longmapsto
 \sum_{r\ge\mu}V^rF^{kr}f_r(c+F,S).
\end{equation}
Here $A((Z))$ consists of Laurent series bounded below in $Z$ with coefficients in $A$. Negative powers of $H$ expand as $(c+F)^{-a}=c^{-a}(1+F/c)^{-a}$ for $a>0$, using the formal binomial series; this is defined because $c\ne0$. Thus the map is defined since each $f_r(c+F,S)$ has nonnegative $F$-order, and any fixed $F$-degree receives only finitely many terms. Its valuation is $\min_r(kr+\operatorname{ord}_{H=c}f_r)$: the lowest coefficient is a nonzero Laurent polynomial in the transcendental $V$. This proves injectivity as well.

In $A((Z))$, the notation $O(Z^j)$ means an element of $Z^jA[[Z]]$. A crossed wall changes $Z$ and each $S_i$ by a unit factor $1+O(Z^{s(n)})$, and sends $F$ to $F+O(Z^{s(n)})$. If $ks(n)>1$, its associated graded action for $v$ is the identity. If $ks(n)=1$, then $k=s(n)=1$ and that action is the invertible shear
\[
 Z\mapsto Z,\qquad F\mapsto F+\alpha(S)Z,
\]
where $\alpha(S)$ is the coefficient obtained by evaluating the linear-in-$Z$ correction to $F$ at $H=c$. On the residue coordinate it sends $V$ to $V/(1+\alpha(S)V)$, whose denominator is nonzero. Thus the wall and its inverse preserve $v$. Their difference from the identity increases relative valuation by at least $ks(n)-1$. By \eqref{eq:crossing} this tends to infinity with charge. Fix a desired precision $F^N$. By \eqref{eq:crossing}, only bounded-charge walls can alter a rational input modulo that precision relative to its valuation. The remaining maps act identically on the required associated graded pieces. Interspersing the finite lower-charge maps preserves this estimate because those maps preserve $v$; equivalently one conjugates the higher-order maps by a finite valuation-preserving product. Thus the truncated path products form a Cauchy system. The reverse path gives a Cauchy system of inverses, and their composites are the identity at every finite precision. The limits are inverse valuation-preserving automorphisms.

For a finite Laurent input of minimum $Z$-degree $\mu$, each output term is an input term plus a sum $PN$ of crossed charges. Formula~\eqref{eq:crossing} gives $\pair{a_0}{PN}\ge(\rho\epsilon_0/l)|N|_1$. Outputs of $Z$-degree at most $R$ therefore use only bounded total charge. Their complete coefficients stabilize in a finite truncation. Consequently the chargewise output belongs to $A((Z))$, with finite Laurent coefficients and a finite negative-$Z$ part. The same finite truncations define the valuation product; the map \eqref{eq:common} sends a high-$Z$ tail to a high-$F$ tail. Their limits agree.
\end{proof}

The initial $\tht_{m_l,+}$ is an element of $C$, so has $v\ge0$. Lemma~\ref{lem:transport} implies this for its expansion at $q_l$. In the common series ring its finite negative-$Z$ part must be regular: its nonnegative-$Z$ tail already has nonnegative valuation. Formula~\eqref{eq:valuation}, or its series version in \eqref{eq:common}, therefore makes the coefficient of each $Z^r$, $r<0$, divisible by $(H-c)^{-kr}$. Write that finite coefficient as $f_r(H,S)=\sum_\beta c_{r,\beta}(H)S^\beta$. For $N\ge1$,
\[
 A/((H-c)^N)=
 \bigl(\kk[H^{\pm1}]/((H-c)^N)\bigr)
 [S_1^{\pm1},\ldots,S_{d-2}^{\pm1}].
\]
The $S^\beta$ are a free basis over the displayed coefficient ring. Taking $N=-kr$ proves that every $c_{r,\beta}(H)$ is divisible by $(H-c)^{-kr}$.

\begin{lemma}\label{lem:stabilization}
For each fixed charge bound $K$, all sufficiently large integers $l$ give a coefficient-preserving bijection between broken lines of input $m_l$ and endpoint $q_l$ in $\D$, and broken lines of input $m_*$ and endpoint $q_0$ in $\D_b$, with total charge at most $K$. It is given by
\[
 \beta(t)=\gamma_l(t)-l(1-t)b.
\]
\end{lemma}
\begin{proof}
Writing the bends as $Pa_j$ at $-u_j$ gives, for all $u\ge0$,
\[
 \beta(-u)=q_0+um_*+P\sum_j\min(u,u_j)a_j,
 \qquad
 \frac{\|\beta(-u)\|}{1+u}\le\|q_0\|+\|m_*\|+\|P\|K.
\]
Only finitely many walls affect this cutoff. For a closed wall cone not containing $b$, its distance from $b$ is positive; the displayed bound shows that $l(1+u)b+\beta(-u)$ misses it for every $u\ge0$ when $l$ is large. For a wall containing $b$, write its face inequalities in its supporting hyperplane as $\eta_i(x)\ge0$. The cylinder $W+\RR b$ is defined by those with $\eta_i(b)=0$. The inequalities with $\eta_i(b)>0$ hold strictly after adding $l(1+u)b$, uniformly in $u$, for large $l$. Thus support membership is exactly the cylindrical support condition.

All retained normals annihilate $b$, so crossing powers and coefficients agree. Removing or adding the drift gives inverse finite bend records. The simultaneous generality of $q_0$ ensures that each reconstructed record avoids joints in the complete diagram, including higher-order walls, where one takes constant terms. This proves the full coefficient assertion.
\end{proof}

Every correction of $\D_b$ has zero $a_0$-pairing by \eqref{eq:rho}; hence all terms of its theta of input $m_*$ have $Z$-degree $\pair{a_0}{m_*}<0$. Fix a class in $\LL/\ZZ b$ and one of the weights from Lemma~\ref{lem:uniform}. Every shifted exponent in this class has the same weight. The coercive lower bound on that weight therefore supplies an integer $K$ bounding all charges occurring in this class, uniformly in $l$. The same bound holds for the star series: each of its individual finite-charge contributions occurs in a sufficiently large $l$ by Lemma~\ref{lem:stabilization}, and hence satisfies the bound. Thus the entire class, on both sides, is decided by a single finite truncation. Lemma~\ref{lem:stabilization} then identifies the entire finite $H$-Laurent coefficient of $H^{-l}\tht_{m_l,q_l}$ with that of $\tht^{\D_b}_{m_*,q_0}$ for sufficiently large $l$. Multiplication by $H^{-l}$ preserves the $(H-c)$ divisibility because $c\ne0$. We obtain
\begin{equation}\label{eq:starzero}
 \left.\tht^{\D_b}_{m_*,q_0}\right|_{H=c}=0
\end{equation}
coefficientwise in projected characters. The weight bound also passes to this star theta, so its support has finite sublevel sets for a relatively open nonempty set of weights in $b^\perp$.

Finally, $\D_b$ has no correction parallel to $b$. Such a correction would have charge parallel to $n_0$, by injectivity of $P$. Its original support contains $b$, so that original wall is incoming. But $n_0$ has at least two mutable indices in its support, whereas every incoming direction is initial. The positive representative with this property is supplied by the algorithm in \cite[Appendix~C]{GHKK}. Proposition~\ref{prop:proper} below contradicts \eqref{eq:starzero}.