\section{Ray geometry for the dimension induction} \label{sec:local-ray-geometry} The construction of zero-supported cocycles requires lifts of derivations that satisfy the Leibniz rule on nonzero ray products. We first state the two assertions to be proved by induction. The rest of this section establishes the local factorization restrictions used in the lifting step: endpoint actions determine which zero products can acquire socle-valued corrections, and factorizations of generating rays determine the equations for the remaining coefficients. \subsection{Derivation lifts and the induction assertions} \label{subsec:table-induction} Since the ray basis $\mathcal P$ of $A_0$ has structure constants zero or one, it defines algebras $A^{(2)}=\F_2[\mathcal P]$ and $A^{(4)}=(\Z/4)[\mathcal P]$. The normalized relative Hochschild cochains of $A^{(2)}$ are finite-dimensional in each cochain degree, and tensoring them with $k$ gives those of $A_0$. Since $-\otimes_{\F_2}k$ is exact, cohomology commutes with this base change, separately in each internal degree and $L$-weight. We may therefore construct the required cocycles over $\F_2$. The following proposition expresses their classes as Bocksteins of derivations; its proof in Section~\ref{sec:integral-deformations} does not use the dimension induction. \begin{proposition} \label{prop:integral-bockstein} The Bockstein connecting homomorphism associated to the short exact coefficient sequence $0\to\F_2\xrightarrow{2}\mathcal O\to\F_2\to0$, where $\mathcal O=\Z/4$, is surjective in every positive internal degree $q>0$: \[ \beta\colon\HH^1(A^{(2)})_q\twoheadrightarrow\HH^2(A^{(2)})_q. \] Explicitly, for any derivation $\bar\delta$ of $A^{(2)}$ and any normalized $\mathcal O$-linear one-cochain $\delta$ of $A^{(4)}$ lifting $\bar\delta$, the class $\beta([\bar\delta])$ is represented by $(d\delta)/2$, where division by~$2$ denotes the inverse of the additive isomorphism $\F_2\xrightarrow{\sim}2\mathcal O$. The same statement holds within each weight component $\chi\in L$ with $\chi(\deg_\JJ)>0$. \end{proposition} For $(d\delta)/2$ to vanish on all nonzero basis products, the lift $\delta$ must satisfy \begin{equation} \delta(rs)=\delta(r)s+r\delta(s)\qquad(rs\ne0). \label{eq:nonzero-leibniz} \end{equation} The following calculation illustrates why the choice of lift matters. \begin{example} \label{ex:bockstein-pair} Consider the multiplication table of $B_0$ from Example~\ref{ex:nonstandard-pair} over $\mathcal O=\Z/4$, and define a normalized one-cochain $\delta$ by \[ \delta(a)=aq,\qquad \delta(b)=qb,\qquad \delta(q)=q^2,\qquad \delta(q^2)=2q^3, \] and $\delta=0$ on the remaining basis elements. The only pairs $(r,s)$ of radical basis elements for which $\delta(r)s+r\delta(s)$ is nonzero are $(a,b)$, $(b,a)$, and $(q,q)$, where $\delta(r)s+r\delta(s)$ takes the values $2w$, $2q^3$, and $2q^3$, respectively. Because $ab=0$ while $ba=q^2$, we obtain \[ (d\delta)(a,b)=2w,\qquad (d\delta)(b,a)=(d\delta)(q,q)=0, \] and $d\delta$ vanishes on all other basis pairs. Thus the reduction $\bar\delta$ modulo $2$ is a derivation in characteristic two, and dividing $d\delta$ by $2$ recovers the cocycle that deforms $ab=0$ to $ab=tw$. Had we instead lifted $\bar\delta$ by setting $\delta(q^2)=0$, its coboundary would take the value $2q^3$ on the nonzero products $(b,a)$ and $(q,q)$; the coefficient $2$ on $q^2$ is therefore essential to concentrate the cocycle on zero products. \end{example} We pair the desired conclusion about $A$ with an assertion about additive weights on multiplication tables. Lifting these weights to integers and reducing their values modulo four will supply coefficients satisfying~\eqref{eq:nonzero-leibniz}. \begin{definition} \label{def:weighted-table} A \emph{weighted multiplication table} $T$ consists of a finite set of objects, identity elements $1_x$, and finitely many nonzero morphism elements, equipped with an associative partial composition under which the product of two composable elements is either an element of $T$ or zero. It also has an additive weight $\ell\colon T\setminus\{0\}\to M$ valued in an abelian group $M$ and a homomorphism $\partial\colon M\to\Z$ such that $\ell(1_x)=0$ and $\partial(\ell(r))>0$ for every nonidentity nonzero element $r$. We include a zero morphism for each ordered pair of objects and omit these zero morphisms from the basis of $k[T]$. Its linearization $k[T]$ is defined as in Definition~\ref{def:ray-linearization}(2), and its weight group is $W(T)$ from Definition~\ref{def:scalar-weight-group}. \end{definition} The table in this definition can be recovered from the rays of its linearization. We establish this before stating the induction assertions. \begin{lemma} \label{lem:table-rays} Let $T$ be a weighted multiplication table such that $k[T]$ is representation-finite. Then $k[T]$ is basic, and $r\mapsto[r]$ identifies $T$ with $\ray(k[T])$ as multiplication tables. \end{lemma} \begin{proof} Because the nonidentity elements of $T$ have positive degree, they span a nilpotent ideal in $B=k[T]$ with quotient $\prod_x k\cdot1_x$, so $B$ is basic. By Lemma~\ref{lem:endpoint-chains}, each local ring $R_x=B(x,x)$ is a truncated polynomial ring and each corner bimodule $B(x,y)$ is cyclic over $R_x$ or $R_y$. Since the nonidentity elements of $T$ in $R_x$ span $\rad R_x$, we may choose for each $R_x\ne k$ an element $q_x\in T$ whose image generates $\rad R_x/(\rad R_x)^2$; its nonzero powers (including $q_x^0=1_x$) then form a basis of $R_x$. Likewise, for each nonzero corner $B(x,y)$, choosing $f\in T$ with nonzero image in the one-dimensional top of $B(x,y)$ generates $B(x,y)$ over a cyclic endpoint by Nakayama's lemma. The nonzero products $q_x^if$ (or $fq_y^i$) lie in $T$, span $B(x,y)$, and have strictly increasing degree, so they are linearly independent and exhaust the elements of $T$ in $B(x,y)$ with pairwise distinct degrees and weights. By Lemma~\ref{lem:endpoint-chains}, the endpoint subbimodules of $B(x,y)$ are the successive spans of these powers, so each nonzero ray of $B$ contains a unique element of $T$. Moreover, whenever two elements $r,s\in T$ have a nonzero product $rs\in T$, its degree $\partial(\ell(rs))=\partial(\ell(r))+\partial(\ell(s))$ is strictly smaller than the degree of any middle-radical insertion $rq_y^ms$ ($m\geq1$); hence $rs\notin M(r,s)$, which means that the ray product in $\ray B$ coincides with the product in $T$. If $rs=0$ in $T$, then $rs\in M(r,s)$, so its ray product is also zero. Thus $T\cong\ray B$. \end{proof} The first induction assertion concerns lifting additive weights. Both propositions below are proved in Subsection~\ref{subsec:main-theorems}. \begin{proposition} \label{prop:table-weight-lifting} Let $T$ be a weighted multiplication table whose linearization $k[T]$ is representation-finite over the algebraically closed field $k$. Every additive weight $T\to\Z/2$ lifts to an additive weight $T\to\Z$. \end{proposition} The second assertion concerns the algebra itself. Its conclusions will be proved together with Proposition~\ref{prop:table-weight-lifting}. \begin{proposition} \label{prop:dimension-induction} Let $A$ be a basic representation-finite algebra over $k$, in any characteristic, and let $A_0$ be the standard form of Construction~\ref{con:projective-algebra}. Then: \begin{enumerate}[label={\upshape(\arabic*)}] \item $A$ admits a normed multiplicative basis consisting of one representative of each nonzero ray; \item the initial-form bijection of Lemma~\ref{lem:ray-degree} identifies the ray multiplication in $\ray A$ with that in $\ray A_0$; \item the weight group $W(\ray A)$ is free abelian. \end{enumerate} \end{proposition} \noindent\textbf{Induction on dimension.} Fix $k$. While Proposition~\ref{prop:integral-ray-weights} already establishes freeness of the weight group for a standard form, the smaller algebra $k[T]$ that arises in the lifting problem is only known a priori to be representation-finite, not yet identified with its standard form. We therefore prove Proposition~\ref{prop:table-weight-lifting} and Proposition~\ref{prop:dimension-induction} together by induction on dimension. For each integer $n\geq1$, the induction consists of two implications: \begin{enumerate}[label={\upshape(\arabic*)}] \item Proposition~\ref{prop:table-weight-lifting} for representation-finite tables with fewer than $n$ nonzero elements implies all three conclusions of Proposition~\ref{prop:dimension-induction} for basic connected algebras $A$ with $\dim_k A=n$ (by Lemma~\ref{lem:recovery-with-lifting} in characteristic two, and by Corollary~\ref{cor:away-from-two-basis} and Proposition~\ref{prop:integral-ray-weights} outside characteristic two). \item Proposition~\ref{prop:dimension-induction} for algebras of dimension at most $n$ implies Proposition~\ref{prop:table-weight-lifting} for any table $T$ with $n$ nonzero elements: when $T$ is connected, applying Proposition~\ref{prop:dimension-induction}(3) to $A=k[T]$ and using Lemma~\ref{lem:table-rays} shows that $W(T)\cong W(\ray A)$ is free abelian, so every homomorphism $W(T)\to\Z/2$ lifts to $\Z$, while a disconnected table reduces immediately to its connected components. \end{enumerate} To establish the characteristic-two implication in (1), we solve \eqref{eq:nonzero-leibniz} separately for each residual character $\chi_i$. In Section~\ref{sec:integral-partial-lifts} we identify a core subsystem of the integer linear equations for a derivation lift with the additive-weight relations of a smaller representation-finite multiplication table, which has an integer solution by the induction hypothesis; the local factorization restrictions proved below then allow us to extend that solution to all rays of $A_0$. Section~\ref{sec:integral-deformations} combines the resulting modulo-four lift with the Bockstein construction, and Section~\ref{sec:binary-families} assembles the polynomial deformation family and completes the induction. \subsection{Rays, endpoint actions and stem products} We begin the local analysis needed for the lifting step by comparing ray products with products of the endpoint subbimodules generated by their factors. These restrictions are independent of the weight $\chi$; Subsection~\ref{sec:primitive-ray-equations} fixes $\chi$ and introduces the endpoint types used to organize its coefficient equations. \begin{setting} \label{set:projective-companion} Throughout Part~\ref{part:characteristic-two} and Appendix~\ref{app:local-rays}, we let $A_0=k[\mathcal P]$ denote the standard form of Construction~\ref{con:projective-algebra}, identified with $k[\mathcal P]$ via Proposition~\ref{prop:scalar-normalization} and equipped with its normed homogeneous ray basis. For each vertex $x$, we write $R_x=A_0(x,x)=k[q_x]/(q_x^{N_x})$ when $R_x\ne k$, where $q_x$ is the chosen homogeneous generator of $\rad R_x$, and we set $q_x=0$ and $N_x=1$ when $R_x=k$. \end{setting} The zero products that can acquire deformation terms are detected by multiplying the endpoint subbimodules generated by their factors. \begin{definition}[{\cite[Section~9.1]{BGRS}}] \label{def:stem-product} For a ray $r\colon x\to y$, let $U_r=R_xrR_y\subseteq A_0(x,y)$ denote the endpoint subbimodule it generates. For composable rays $r\colon x\to y$ and $s\colon y\to z$, their \emph{stem product} $r\circledast s$ is the unique ray $t\colon x\to z$ satisfying $U_t=U_rU_s$ when $U_rU_s\ne0$, and $r\circledast s=0$ when $U_rU_s=0$. Uniqueness is guaranteed by the linear ordering of endpoint subbimodules in Lemma~\ref{lem:endpoint-chains}. \end{definition} Under the bijection between rays and the endpoint subbimodules they generate, the stem product $\circledast$ is composition in Kupisch's stem algebra \cite[Part~I, (1.13)--(1.14) and Theorem~1.2]{Kupisch}; see also \cite[Section~2.3]{BongartzSurvey} for the categorical formulation. Because $U_rU_s=R_xrR_ysR_z$, the stem product captures the entire subspace spanned by products of representatives with arbitrary insertions from the intermediate local ring $R_y$. We will use it to determine which pairs of rays with $rs=0$ can support a nonzero deformation term. We now record the endpoint and factorization conditions used to determine when $rs=0$ but $r\circledast s\ne0$. Ordinary arrows refer to the quiver of $A_0$; arrows of $\Gamma_A$ instead represent irreducible morphisms between all indecomposable $A$-modules. \begin{definition} \label{def:generating-rays} Let $A_0=k[\mathcal P]$ satisfy Setting~\ref{set:projective-companion}. \begin{enumerate}[label={\upshape(\arabic*)}] \item A ray $r\colon x\to y$ is \emph{generating} if it generates the full corner bimodule $A_0(x,y)$ as an $(R_x,R_y)$-bimodule; otherwise $r$ is \emph{proper}. \item Given a generating ray $f\colon x\to y$, we say that the corner bimodule $A_0(x,y)$ (or simply $f$) is \emph{cyclic at $x$} if $A_0(x,y)=R_xf$, and \emph{cyclic at $y$} if $A_0(x,y)=fR_y$. When $A_0(x,y)$ is cyclic at $x$, its rays are the elements $q_x^jf$, and we call the exponent $j$ the \emph{endpoint layer} of $q_x^jf$; the endpoint layer at $y$ is defined symmetrically when $A_0(x,y)$ is cyclic at $y$. \item An \emph{ordinary arrow} is a nonidentity ray that cannot be factored as a product of two nonidentity rays, and an \emph{ordinary path} is a composable sequence of ordinary arrows. Within a corner or quotient category, these terms refer to factorizations inside that corner or quotient. \item Let $x\ne y$, and let $f\colon x\to y$ and $h\colon y\to x$ be rays such that $fh=q_x^b\ne0$ for some integer $b\geq1$. We call $fh$ a \emph{return} at $x$ through $y$. \end{enumerate} The \emph{endpoint socle} of a corner bimodule $A_0(x,y)$ is the subspace of vectors annihilated by both $\rad R_x$ on the left and $\rad R_y$ on the right. By Lemma~\ref{lem:endpoint-chains}, every nonzero corner bimodule $A_0(x,y)$ has a one-dimensional endpoint socle. \end{definition} \subsection{Homogeneous subcategories and a dimension criterion} \label{subsec:homogeneous-local-categories} We realize ray factorizations as morphisms in finite subcategories of a graded covering of $A_0$. These categories are representation-finite, so their projective and injective modules cannot have the configurations excluded by the dimension estimate below. The grading separates rays between the same vertices into different morphism spaces. \begin{definition} \label{def:homogeneous-category} Equip $A_0$ with its weight grading by $L$ (Definition~\ref{def:integer-weights}). A \emph{homogeneous category} is the full subcategory $H$ of its graded covering category (Definition~\ref{def:graded-covering-category}) on a finite set of distinct objects $(x_i,w_i)$, where $w_i\in L$. Thus \[ H(i,j)=A_0(x_i,x_j)_{w_j-w_i}. \] We also use this term for the same construction with the categorical $\Z$-grading $\deg_\JJ$ in place of $L$. \end{definition} By Lemma~\ref{lem:ray-degree}, each nonzero $H(i,j)$ is one-dimensional. Every nonidentity morphism has positive categorical degree $\deg_\JJ(w_j-w_i)$, so $H$ is directed. Moreover, every such $H$ is representation-finite by Lemma~\ref{lem:homogeneous-projection}. We will repeatedly cancel common factors in equal nonzero ray products. \begin{lemma} \label{lem:ray-cancellation} For rays $u,r,s$ with $r$ and $s$ in the same corner, $ur=us\ne0$ implies $r=s$, and $ru=su\ne0$ implies $r=s$. \end{lemma} \begin{proof} Additivity of weights gives $\ell(u)+\ell(r)=\ell(u)+\ell(s)$ in the first case, hence $\ell(r)=\ell(s)$. Rays of the same corner and weight coincide by Lemma~\ref{lem:ray-degree}, since their categorical degrees coincide. The right-hand version is identical. \end{proof} We obtain representation-infinite configurations from the faithful support of a thin projective or injective $H$-module, defined below. Taking the quotient by the annihilator of the module in the next definition is essential, because a morphism that is nonzero in $H$ may still act as zero on the chosen module. \begin{definition} \label{def:faithful-support} Let $H$ be a homogeneous category (with respect to either the weight lattice $L$ or the categorical grading $\deg_\JJ$). Fix an object $X\in H$, set $V=H(X,-)$, and let $\ann V$ be the ideal of morphisms in $H$ that act as zero on $V$. The \emph{faithful projective support} at $X$ is the poset consisting of the objects $Y\in H$ with $V(Y)\ne0$, ordered by $Y\leq Z$ whenever $(H/\ann V)(Y,Z)\ne0$. Equivalently, if we label the points of this poset by the nonzero rays $r$ starting at $X$, then $r\leq s$ if and only if $rv=s$ for some ray $v$ in $H$. The \emph{faithful injective support} at $X$ is defined dually via the thin module $D_k H(-,X)$; in terms of nonzero rays ending at $X$, its order is given by $r\leq s$ if and only if $vr=s$ for some ray $v$ in $H$. \end{definition} When restricted to the objects on which $V$ is nonzero, the quotient $H/\ann V$ is isomorphic to the incidence category of the faithful projective support at $X$ by Lemma~\ref{lem:thin-module-incidence} (positivity of the grading rules out nonidentity isomorphisms). By the same lemma, this incidence category is representation-finite, and hence so is the incidence category of every full subposet (which is an idempotent corner of $H/\ann V$). The dual statements hold for the faithful injective support. The following dimension estimate rules out the small quivers and posets that arise in our local arguments. \begin{lemma} \label{lem:dimension-infinity} Let $B=kQ/(\rho_1,\ldots,\rho_t)$ be a finite-dimensional bound quiver algebra, where each relation $\rho_i$ is a linear combination of radical paths sharing a common source and target. For a nonzero dimension vector $d=(d_v)$, set \[ D=\sum_v d_v^2,\quad N=\sum_{a:v\to w} d_v d_w,\quad F=\sum_{i=1}^t d_{\mathrm{source}(\rho_i)}d_{\mathrm{target}(\rho_i)}. \] If $N-F\geq D$, then $B$ is representation-infinite. \end{lemma} \begin{proof} The affine variety of $B$-modules of dimension vector $d$ is cut out inside $\mathbb{A}^N$ by at most $F$ polynomial equations, and it is nonempty because it contains the zero representation. Every irreducible component of this variety therefore has dimension at least $N-F$. On the other hand, the algebraic group $\prod_v\operatorname{GL}_{d_v}(k)$ has dimension $D$, and the diagonally embedded subgroup $k^*$ acts trivially on every representation, so every isomorphism orbit has dimension at most $D-1$. Since $N-F>D-1$, the representation variety cannot be covered by finitely many orbits. By the Krull--Schmidt theorem, this contradicts finite representation type. \end{proof} A \emph{four-point crown} is the poset with two incomparable minimal elements and two incomparable maximal elements, with every minimal element below both maximal elements. Its Hasse diagram is a four-cycle. For the relation-free quivers ($F=0$) used repeatedly below, the critical dimension vectors achieving $N=D$ are listed in the following table: \begin{center} \small \begin{tabular}{@{}lll@{}} \toprule Underlying graph & Dimension vector $d$ & $N=D$ \\\midrule Four-point crown & $1$ at every vertex & $4$ \\ Four-leaf star ($\widetilde D_4$) & $2$ at center, $1$ at each leaf & $8$ \\ Tree $\widetilde D_5$ (two trivalent vertices) & $2$ at trivalent vertices, $1$ at leaves & $12$ \\ Tree $\widetilde E_7$ (arms $3,3,1$) & $4$ at center; $(3,2,1)$, $(3,2,1)$, $(2)$ on arms & $48$ \\ \bottomrule \end{tabular} \end{center} Because $N$ and $D$ depend only on the underlying undirected graph when $F=0$, these counts apply to any orientation of the listed trees. In particular, neither a four-point crown nor a four-leaf star can appear as a full subposet of any faithful projective or injective support. To rule out more general configurations with relations, we pull representations back along a split bimodule inclusion using the following observation. \begin{lemma} \label{lem:representation-transfer} Let $F\colon \mathcal U\to B$ be a functor between finite categories, and let $B_F$ be the category having the same object set as $\mathcal U$ and morphism spaces $B_F(x,y)=B(Fx,Fy)$. Suppose that the induced functor $\mathcal U\to B_F$ is injective on morphism spaces and admits a retraction $\pi\colon B_F\to \mathcal U$ as a $\mathcal U$-bimodule morphism. If $B$ is representation-finite, then so is $\mathcal U$. \end{lemma} \begin{proof} When several objects of $\mathcal U$ map to the same object of $B$, the category $B_F$ is a matrix enlargement of an idempotent corner of $B$, and is therefore representation-finite by Lemma~\ref{lem:endpoint-chains}(1). Applying Lemma~\ref{lem:split-representation-transfer} to the split inclusion $\mathcal U\hookrightarrow B_F$ with $C=D=\mathcal U$ then shows that $\mathcal U$ is representation-finite. \end{proof} To transfer finite representation type from a finite homogeneous category $H$ to the incidence algebra $\operatorname{Inc}_k(Q)$ of a finite poset $Q$, we use a general criterion for when a family of selected rays in $H$ admits an $\operatorname{Inc}_k(Q)$-bimodule retraction, even when $H$ contains additional morphisms outside $\operatorname{Inc}_k(Q)$ that do not form an ideal. Write $a\lessdot b$ when $aj$, since the required morphism in $A_0(x,y)$ would have degree $e+(j-i)d=Latex,every node/.style={font=\small}] \node (u0) at (0,0) {$u_0$}; \node (u1) at (1,1) {$u_1$}; \node (u2) at (1,2) {$u_2$}; \node (u3) at (2,3) {$u_3$}; \node (v2) at (0,3) {$v_2$}; \node (v1) at (2,2) {$v_1$}; \draw[->] (u0)--(u1); \draw[->] (u1)--(u2); \draw[->] (u2)--(u3); \draw[->] (u2)--(v2); \draw[->] (u1)--(v1); \end{tikzpicture} \] which is an orientation of the affine Dynkin tree $\widetilde D_5$. Assigning dimension~$2$ to the trivalent vertices $u_1,u_2$ and dimension~$1$ to the four leaves gives $N=D=12$ and $F=0$, contradicting finite representation type by Lemma~\ref{lem:dimension-infinity}. We conclude that $l=a+1$. Consequently $q_x(fq_y)=q_x^{a+1}f=0$ and $(fq_y)q_y=q_x^{2a}f=0$ (since $2a\geq a+1=l$), so $fq_y$ is annihilated by both endpoint radicals and therefore spans the endpoint socle of $A_0(x,y)$. \end{proof} \begin{lemma} \label{lem:opposite-controls} In Setting~\ref{set:projective-companion}, let $f\colon x\to y$ and $g\colon y\to z$ be generating rays whose corner bimodules $A_0(x,y)$ and $A_0(y,z)$ are cyclic at $y$ but not at $x$ and $z$, respectively, and let $q=q_y$ generate $\rad R_y$. \begin{enumerate}[label={\upshape(\arabic*)}] \item Either $fq^2=0$ or $q^2g=0$. \item The product subspace $fR_yg\subseteq A_0(x,z)$ has dimension at most one and is annihilated by $\rad R_x$ on the left and by $\rad R_z$ on the right. \item If $fq\ne0$, $qg\ne0$, and $q^2\ne0$, then \begin{equation} fg=0. \label{eq:strong-star-vanishing} \end{equation} \end{enumerate} \end{lemma} \begin{proof} Because the local ring $R_y=A_0(y,y)$ is commutative and hence cyclic on both sides, the noncyclicity assumptions at $x$ and $z$ imply that $x\ne y$ and $z\ne y$ (though $x$ and $z$ may coincide). Moreover, a one-dimensional corner bimodule is automatically cyclic over both endpoint rings, so $\dim A_0(x,y)\geq2$ and $\dim A_0(y,z)\geq2$. Writing $A_0(x,y)=fR_y\cong k[q]/(q^n)$ with $n\geq2$, take any $t\in\rad R_x$ and write $tf=f\varphi(q)$ for a polynomial $\varphi(q)\in k[q]$. Nilpotence of $t$ implies $\varphi(0)=0$, and if the linear coefficient of $\varphi$ were nonzero, then the powers of $\varphi(q)$ would generate $k[q]/(q^n)$, making $A_0(x,y)=R_xf$ cyclic at $x$, contrary to assumption. Thus $(\rad R_x)f\subseteq fq^2R_y$. Furthermore, for any morphism $h\in A_0(y,x)$, the composition $fh$ lies in $\rad R_x$ because $x\ne y$; hence $f(hf)=(fh)f\in fq^2R_y$, and since $\ann_{R_y}(f)=(q^n)\subseteq(q^2)$, it follows that $hf\in(q^2)$. Applying the dual argument to $g$ yields the four inclusions \begin{equation} \begin{gathered} (\rad R_x)f\subseteq fq^2R_y,\qquad g(\rad R_z)\subseteq R_yq^2g,\\ A_0(y,x)f\subseteq(q^2),\qquad gA_0(z,y)\subseteq(q^2). \end{gathered} \label{eq:opposite-control-inclusions} \end{equation} We prove (3) first, then (1) and (2). For (3), suppose that $fq,qg,q^2$ and $fg$ are all nonzero. Let $\mathcal U$ be the path category of the four-leaf star quiver \[ \begin{tikzcd}[row sep=small,column sep=small] & u\arrow[d,"q_1"] & \\ l\arrow[r,"f"] & y\arrow[r,"g"]\arrow[d,"q_2"] & r \\ & d & \end{tikzcd} \] in which the two paths $l\xrightarrow{f}y\xrightarrow{g}r$ and $u\xrightarrow{q_1}y\xrightarrow{q_2}d$ cross at $y$ with no relations. We map the five vertices $l,y,r,u,d$ of $\mathcal U$ to the objects of the $\Z$-graded homogeneous category $H$ of $A_0$ having underlying vertices $x,y,z,y,y$ and grading coordinates \[ -\deg_\JJ(f),\quad 0,\quad \deg_\JJ(g),\quad -\deg_\JJ(q),\quad \deg_\JJ(q), \] respectively, sending the arrows $f,g,q_1,q_2$ to the rays $f,g,q,q$. All four length-two paths in $\mathcal U$ then map to the nonzero rays $fg,fq,qg,q^2$ in $H$, so the functor $\mathcal U\to H$ identifies each nonzero morphism space of $\mathcal U$ with the corresponding one-dimensional morphism space of $H$. Let $Q$ be the reachability poset of this star, and apply Lemma~\ref{lem:configuration-retraction} with $r_{ij}$ the image in $H$ of the unique path from $i$ to $j$ in $\mathcal U$. Because all paths in $\mathcal U$ have length at most two, any factor in $H$ sharing the endpoints of a tree path already coincides with its image by thinness, while any factor reversing a tree arrow would have negative degree. Thus the first cover implication of Lemma~\ref{lem:configuration-retraction} needs to be checked only when the product lands on the other outgoing branch $\{r,d\}$, and the second only when it starts on the other incoming branch $\{l,u\}$. These four possibilities between the two branches are ruled out as follows: \[ \begin{array}{c|c|l} \text{branch comparison}&\text{required equality}&\text{exclusion}\\\hline r\to d&gh=q&gA_0(z,y)\subseteq(q^2)\\ d\to r&qh=g&g\notin(\rad R_y)A_0(y,z)\\ u\to l&hf=q&A_0(y,x)f\subseteq(q^2)\\ l\to u&hq=f&f\notin A_0(x,y)(\rad R_y). \end{array} \] Specifically, a morphism $h:z\to y$ in the first row would give $q=gh\in(q^2)$ by~\eqref{eq:opposite-control-inclusions}, which is impossible for the radical generator $q$, while $qh=g$ in the second row would place the generator $g$ inside the radical of $A_0(y,z)=R_yg$; the third and fourth rows are dual for $f$. The projection $H\to\mathcal U$ is therefore a bimodule retraction, contradicting Lemma~\ref{lem:configuration-retraction} because the four-leaf star $\mathcal U$ is representation-infinite by Lemma~\ref{lem:dimension-infinity}. \smallskip For (1), suppose for a contradiction that $fq^2\ne0$ and $q^2g\ne0$. Then $fq,qg,q^2$ are nonzero, so (3) gives $fg=0$. In this case, the last two inclusions in~\eqref{eq:opposite-control-inclusions} sharpen to \[ A_0(y,x)f\subseteq(q^3),\qquad gA_0(z,y)\subseteq(q^3), \] because if $hf$ had a nonzero $q^2$-term, multiplying on the right by $g$ would give $(hf)g=h(fg)=0$ while $q^2g\ne0$, and dually for $gh$. We now let $\mathcal U$ be the bound quiver category on seven vertices with diagram and single relation \[ \begin{tikzcd}[row sep=large,column sep=large] u' & u\arrow[l,"g_1"']\arrow[d,"q_1"] & \\ l\arrow[r,"f_1"] & y\arrow[r,"g_2"]\arrow[d,"q_2"] & r \\ l_d\arrow[r,"f_2"] & d & \end{tikzcd} \qquad f_1g_2=0. \] We map $l,y,r,u,d$ to the same graded objects as in the star construction above, and we send $l_d$ and $u'$ to $(x,\deg_\JJ(q)-\deg_\JJ(f))$ and $(z,\deg_\JJ(g)-\deg_\JJ(q))$, respectively, mapping $f_1,f_2$ to $f$, $g_1,g_2$ to $g$, and $q_1,q_2$ to $q$. The relation $f_1g_2=0$ holds in $H$ because $fg=0$, and the three remaining length-two paths map to the nonzero rays $q^2,fq,qg$. Defining the projection $\pi\colon H\to \mathcal U$ as before, we verify that no basis morphism of $H$ outside the image of $\mathcal U$ can compose with an arrow image to give a nonzero path image. The possible unwanted products and the reasons they vanish are summarized below: \[ \begin{array}{c|c|l} \text{target ray in }\mathcal U & \text{product in }H & \text{reason for exclusion} \\ \hline q\text{ or }q^2 & gh & gA_0(z,y)\subseteq(q^3) \\ qg & gv\ (v\in\rad R_z) & \text{inclusion~\eqref{eq:opposite-control-inclusions} and }q^3g\notin k\cdot qg \\ g & qh & g\text{ is a generating ray} \\ q\text{ or }q^2 & hf & A_0(y,x)f\subseteq(q^3) \\ fq & vf\ (v\in\rad R_x) & \text{inclusion~\eqref{eq:opposite-control-inclusions} and }fq^3\notin k\cdot fq \\ f & hq & f\text{ is a generating ray}. \end{array} \] At $u$, the first three rows rule out factorizations between $g_1\colon u\to u'$ and the paths through $y$; at $y$, they rule out factorizations between the two outgoing branches $g_2$ and $q_2$. The last three rows provide the dual incoming exclusions at $y$ and $d$, while the remaining vertices have at most one incident branch in the relevant direction. Thus $\pi$ is a bimodule retraction. Assigning dimension~$3$ to $y$, dimension~$2$ to $u$ and $d$, and dimension~$1$ to the four outer vertices $u',l,r,l_d$ gives $N=22$, $D=21$, and $F=1$ from the single relation $f_1g_2=0$. Since $N-F=D$, Lemma~\ref{lem:dimension-infinity} and Lemma~\ref{lem:representation-transfer} again yield a contradiction, completing the proof of~(1). \smallskip For~(2), consider the action of $\rad R_x$ on the left or $\rad R_z$ on the right of $fR_yg$. By~\eqref{eq:opposite-control-inclusions}, any such product is a linear combination of terms $fq^mg$ with $m\geq2$. Since either $fq^2=0$ or $q^2g=0$ by~(1), every such term vanishes. Thus $fR_yg$ is annihilated by both $\rad R_x$ and $\rad R_z$, so it lies inside the one-dimensional endpoint socle of $A_0(x,z)$. \end{proof} \subsection{Stem products and generating paths} We next show that the stem product of Definition~\ref{def:stem-product} is determined purely by the ray category $\mathcal P$ via an explicit local formula, and can be nonzero even when the ordinary ray product vanishes. \begin{proposition}[{\cite[Section~9.1]{BGRS}}] \label{prop:intrinsic-stem-product} In Setting~\ref{set:projective-companion}, the stem product on $A_0=k[\mathcal P]$ is determined by the ray category $\mathcal P$: for any composable rays $r\colon x\to y$ and $s\colon y\to z$, \[ r\circledast s=\begin{cases} rq_y^ns, & \text{where }n\geq0\text{ is the least integer such that }rq_y^ns\ne0,\\ 0, & \text{if }rq_y^ns=0\text{ for all }n\geq0, \end{cases} \] with $q_y^0=1_y$ and $q_y=0$ when $R_y=k$. The operation $\circledast$ is associative and agrees with the ray product $rs$ whenever $rs\ne0$. Moreover, the same formula computes the product of endpoint subbimodules in any algebra having ray category $\mathcal P$ and satisfying parts (2) and (3) of Lemma~\ref{lem:endpoint-chains}. \end{proposition} \begin{proof} Let $r\colon x\to y$ and $s\colon y\to z$ be composable rays in $\mathcal P$, choose representatives $f\in r$ and $h\in s$ in an algebra with ray category $\mathcal P$, and let $u$ generate $\rad R_y$ (with $u=0$ if $R_y=k$). For each $j\geq0$, set $V_j=R_xfu^jR_yhR_z$. We have $fu^j\ne0$ if and only if the ray product $rq_y^j$ is nonzero: indeed, if $fu^j\ne0$, then $fu^j$ generates the subbimodule $U(fu^j)$ and therefore lies outside its radical $\rad U(fu^j)\supseteq M(f,u^j)$, which means $rq_y^j\ne0$ by~\eqref{eq:ray-product}; conversely, if $fu^j=0$, then $fu^j\in M(f,u^j)$ and $rq_y^j=0$. When $rq_y^j\ne0$, its ray product with $s$ vanishes if and only if $fu^jh\in V_{j+1}$, or equivalently $V_j=V_{j+1}$. Consequently, if $n\geq0$ is the smallest integer for which $rq_y^ns\ne0$, then $U_rU_s=V_0=V_n$, and hence \begin{equation} U_rU_s=U_{rq_y^ns}. \label{eq:intrinsic-stem-product} \end{equation} If no such $n$ exists, the chain of equalities $V_j=V_{j+1}$ and the nilpotence of $u$ force $U_rU_s=0$. Thus $r\circledast s$ is given by the stated formula in $\mathcal P$. Associativity of $\circledast$ follows immediately from the associativity of subbimodule multiplication, and when $rs\ne0$ the least index is $n=0$, giving $r\circledast s=rs$. \end{proof} Example~\ref{ex:zero-ray-span} illustrates the difference between the ordinary ray product and the stem product: there $[\alpha][\beta]=0$ in the ray category, whereas $U_\alpha U_\beta=ku$ is nonzero because inserting the middle radical generator $q$ yields $\alpha q\beta=u\ne0$, so that $[\alpha]\circledast [\beta]=[u]$. \begin{corollary} \label{cor:active-zero-pairs} In Setting~\ref{set:projective-companion}, let $r\colon x\to y$ and $s\colon y\to z$ be composable rays such that $rs=0$ and $r\circledast s\ne0$. Then $r$ and $s$ are generating rays, $r\circledast s=rq_ys$, and $U_{r\circledast s}$ is the one-dimensional endpoint socle of $A_0(x,z)$. \end{corollary} \begin{proof} If the corner bimodule $A_0(x,y)$ containing $r$ were cyclic at $x$, then $rq_y\in\rad A_0(x,y)$ would equal $q_x^ar$ for some $a\geq1$. For every $n\geq1$ we would then have $rq_y^ns=q_x^{an}(rs)=0$, giving $r\circledast s=0$ by Proposition~\ref{prop:intrinsic-stem-product}, a contradiction. By the dual argument for $s$, neither $A_0(x,y)$ nor $A_0(y,z)$ is cyclic at its outer endpoint, so both are cyclic only at $y$. Write $r=fq_y^i$ and $s=q_y^jg$ where $f$ and $g$ are the generating rays of $A_0(x,y)$ and $A_0(y,z)$. By Lemma~\ref{lem:opposite-controls}, $fq_y^mg=0$ for all $m\geq2$, and the entire subspace $fR_yg$ lies in the endpoint socle of $A_0(x,z)$. Since $rs=fq_y^{i+j}g=0$ while $rq_y^ns=fq_y^{i+j+n}g\ne0$ for some $n\geq1$, the only possibility is $i=j=0$ and $n=1$. Thus $r=f$ and $s=g$ are generating rays, $r\circledast s=rq_ys$, and $U_{r\circledast s}$ is the endpoint socle of $A_0(x,z)$. \end{proof} \begin{lemma} \label{lem:generating-paths} In Setting~\ref{set:projective-companion}, let $f\colon x\to z$ be a generating ray whose corner bimodule $A_0(x,z)$ has dimension at least two. Then $f$ admits no factorization of the form $f=aq_yb$ through a local radical generator $q_y$. In particular, every ordinary path representing $f$ is vertex-simple (visits no vertex twice), and every nonempty proper subsegment of such a path represents a generating ray. \end{lemma} \begin{proof} Suppose for a contradiction that $f=aq_yb$ for some rays $a\colon x\to y$ and $b\colon y\to z$. If $a=1_x$, then $f=q_xb\in(\rad R_x)A_0(x,z)$ lies in the radical of $A_0(x,z)$, contradicting the assumption that $f$ generates $A_0(x,z)$; the case $b=1_z$ is ruled out similarly. If the corner bimodule $A_0(x,y)$ is cyclic at $x$, then $aq_y\in(\rad R_x)A_0(x,y)$, which again places $f=aq_yb$ inside $(\rad R_x)A_0(x,z)$, and symmetrically if $A_0(y,z)$ is cyclic at $z$. In the remaining case, both $A_0(x,y)$ and $A_0(y,z)$ are cyclic only at $y$. Writing $a=a'q_y^i$ and $b=q_y^jb'$ with $a',b'$ generating, we find $f\in a'R_yb'$. By Lemma~\ref{lem:opposite-controls}(2), $a'R_yb'$ is contained in the one-dimensional endpoint socle of $A_0(x,z)$, whereas the generating ray $f$ of a corner bimodule of dimension at least two cannot lie in its endpoint socle. This contradiction rules out any factorization $f=aq_yb$. Finally, any repeated vertex along an ordinary path representing $f$, or any proper (non-generating) subsegment of that path, would produce a factorization of $f$ through a local radical generator $q_y$. \end{proof}