\section{Nonvanishing of proper theta series}\label{sec:proper} For a support in $\LL$, a weight has \emph{finite sublevel sets} if, for every real $R$, only finitely many full exponents have weight at most $R$. We use the word proper for this property. We retain the original charge filtration even when discussing exponents modulo a primitive direction. \begin{proposition}\label{prop:proper} Let a consistent positive rational conical scattering diagram be finite at each charge order, over a field $\kk$ of characteristic zero, with injective charge map $P$ and wall functions $(1+z^{Pn})^d$, $d\in\ZZ_{>0}$. Suppose its supports are cylindrical along a primitive $b\in\LL$, its normals annihilate $b$, and no wall correction is parallel to $b$. Fix a simultaneously general endpoint as in Section~\ref{sec:setup}. If a theta function has proper support for every weight in a nonempty relatively open subset of $b^\perp$, then its specialization at $z^b=c$ is nonzero for every $c\in\kk^*$. \end{proposition} \begin{proof} Write $H=z^b$. Properness implies finite fibers modulo $\ZZ b$, since every such fiber has constant weight. Thus specialization is defined coefficientwise in projected characters. First, positive formal multiplication holds for this diagram. Adjoin $t_i$, $i\in J$, and give a correction $Pn$ the monomial $t^n$. At each endpoint, \[ \tht_{u,q}=z^u+\sum_{n\ne0}c_{u,n}(q)z^{u+Pn}t^n, \qquad c_{u,n}(q)\in\ZZ_{\ge0}. \] Triangular elimination in total $t$-degree gives, for any fixed inputs $u_1,\ldots,u_k$, \[ \prod_{j=1}^k\tht_{u_j} =\sum_{n\in\NN^J}A_n t^n\tht_{u_1+\cdots+u_k+Pn},\qquad A_0=1. \] At a fixed charge there are finitely many decompositions, so elimination and the displayed sum are defined. Consistent theta transport shows that $A_n$ is independent of the general endpoint. For any fixed product and output label $a$, choose the endpoint close enough to $a$ that $\nu(q)\nu(a)>0$ for every normal in the required finite truncation with $\nu(a)\ne0$. A final backward ray $q+\RR_{\ge0}a$ cannot cross one of these hyperplanes. When $\nu(a)=0$, the general ray is parallel and disjoint. There can be no final nontrivial bend for output exponent $a$, so coefficient extraction gives the theta coefficient as the corresponding positive character coefficient. In particular, for every positive integer $K$, $\tht_u^K-\tht_{Ku}$ is a theta-positive formal sum, with nonnegative auxiliary degrees. Fix a weight $w$ for which the support is proper. If $\tht_u$ has minimum $w$-weight $\mu$, its $K$th power has all weights at least $K\mu$. Positivity implies the same lower bound for $\tht_{Ku}$. At a crossing with incoming exponent $e$, orient the primitive normal $\nu$ so that $\kappa=\nu(e)>0$. Index the binomial factors at this crossing by $a$, writing them as $(1+z^{v_a})^{d_a}$ with $d_a\in\ZZ_{>0}$. A selected correction has a finite witness \begin{equation}\label{eq:witness} \delta=\sum_a\lambda_a v_a,\qquad 0\le\lambda_a\le\kappa d_a, \end{equation} where $v_a$ are their correction directions and the $\lambda_a$ are integers with finite support. A rational broken line allows rational exponents and rational $\lambda_a$ with the same inequalities. Clearing all denominators by $K$ and replacing $\gamma(t)$ by $\gamma(Kt)$ gives an ordinary line of input $Ku$ and final exponent multiplied by $K$. Therefore every rational line of input $u$ has weight at least $\mu$. This uses only finitely many factors even at a crossing with infinitely many coplanar walls. Consider a line with finitely many recorded events at $t_1<\cdots0,\qquad a_i=\delta_i/\kappa_i,\qquad S_i=I+a_i\nu_i. \] The product $a_i\nu_i$ denotes the rank-one map $x\mapsto a_i\nu_i(x)$. Since $\nu_i(\delta_i)=0$, its square is zero. Thus $S_i^{-1}=I-a_i\nu_i$, and $e_i=S_i e_{i-1}$. These shears record the change of exponent and the reverse reconstruction of the path. For a final weight $w$, the weight transported back to event $i$ is \[ w_i=w\circ S_s\cdots S_{i+1}. \] Suppose an admissible rational correction $\delta_i'$ with finite witness satisfies $w_i(\delta_i')0$ ensure that all new pairings remain positive. Hence \eqref{eq:witness} continues to hold at every later event. The endpoint remains exactly $q$ under the following reverse reconstruction: \[ C_j(\varepsilon)=S_{j+1}(\varepsilon)^{-1}\cdots S_s(\varepsilon)^{-1},\quad \gamma_j^\varepsilon(t)=C_j(\varepsilon)q-te_j(\varepsilon),\quad t_i(\varepsilon)=\frac{\nu_i(C_i(\varepsilon)q)}{\nu_i(e_i(\varepsilon))}. \] Take $t_0(\varepsilon)=-\infty$, $t_{s+1}(\varepsilon)=0$, and use segment $j$ on $[t_j(\varepsilon),t_{j+1}(\varepsilon)]$ for $0\le j\le s$, with the first interval open at $-\infty$. Empty products of shears are the identity. At each crossing, \[ \begin{aligned} &C_{i-1}(\varepsilon)q-t_i(\varepsilon)e_{i-1}(\varepsilon)\\ &\quad=C_i(\varepsilon)q-a_i(\varepsilon)\nu_i(C_i(\varepsilon)q) -t_i(\varepsilon)e_{i-1}(\varepsilon)\\ &\quad=C_i(\varepsilon)q-t_i(\varepsilon) \bigl(\kappa_i(\varepsilon)a_i(\varepsilon)+e_{i-1}(\varepsilon)\bigr)\\ &\quad=C_i(\varepsilon)q-t_i(\varepsilon)e_i(\varepsilon). \end{aligned} \] Thus the adjacent segments coincide, the last segment ends at $q$, and the first retains input $u$. The denominators are nonzero, so strict time order persists. Take the finite union of the witnesses of all original corrections and of $\delta_i'$. At the original crossing each used wall is met in its relative interior, since the original line avoids joints. Small perturbation preserves all these finitely many support conditions. The fixed endpoint's simultaneous generality makes the new rational record avoid every joint of the full diagram; unrecorded crossings take constant terms. Hence the perturbed path is legal. Its final exponent changes by \[ e_s(\varepsilon)-e_s(0) =\varepsilon S_s\cdots S_{i+1}(\delta_i'-\delta_i), \] which strictly lowers its weight. It follows that any ordinary line realizing $\mu$ minimizes the transported weight at \emph{every} crossing, including constant-term crossings. This is the tautness property used below. Choose $w$ in the given proper open region, avoiding the following countable family of hyperplanes in $b^\perp$. For every finite rational record, every suffix shear product $T$, and every wall direction $v_a$, exclude $w(Tv_a)=0$. Each shear fixes $b$, so $T$ is invertible, fixes $b$, and cannot send $v_a\notin\RR b$ into $\RR b$. These are proper hyperplanes. Also exclude $w(a)=0$ for all $a\in\LL\setminus\ZZ b$. Primitivity of $b$ makes these proper as well. Then distinct projected lattice exponents have different weights, and all transported weights are nonzero on all wall directions. Properness ensures that a minimum is attained: take any occurring exponent and minimize over the finite nonempty sublevel set below its weight. Positivity realizes a minimizing exponent by an ordinary broken line. At any crossing of such a line, tautness forces the unique correction \[ \delta_{\min}=\kappa\sum_{a:w_i(v_a)<0}d_a v_a. \] Indeed, any positive-weight factor with nonzero choice can be removed to lower the weight, and any negative-weight factor not at its maximum can be increased. Each change uses only one factor. Since the minimizing correction has a finite witness, there can be only finitely many negative-weight factors. The formula therefore defines a finite and unique complete correction. Take two minimizing lines. Their final exponents differ by $jb$, $j\in\ZZ$, because their projected weights are equal. Reverse them from their common endpoint. Before a nontrivial bend, their positions differ by $-tjb$, so their projected paths coincide. Use the finite union of the nontrivial bend times of these two existing lines. Inductively, after all later corrections have been shown equal, the positions at the next such time differ by $-tjb$ and their exponents by $jb$. Cylindrical supports give the same wall factors; normals annihilating $b$ give the same crossing powers. Their suffix shears, and hence transported weights, are equal. The unique minimizing correction above is therefore the same for both lines, including when one had chosen the constant term. Reversing it preserves the exponent difference $jb$ and equality of suffix shears. At the initial segments both inputs are $u$, so $j=0$. If a final exponent is parallel to $b$, all normal pairings vanish, and the line must have been straight; this case gives the same conclusion. Thus the lowest projected coefficient has exactly one complete exponent. It is $aH^r$ with $a$ a positive integer. Its value $ac^r$ is nonzero in characteristic zero. Higher projected terms cannot cancel it, proving the proposition. \end{proof}