\documentclass[11pt]{article} \usepackage[margin=0.9in]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{array} \usepackage{longtable} \usepackage{needspace} \usepackage[T1]{fontenc} \usepackage[colorlinks=true,linkcolor=blue]{hyperref} \newcommand{\g}{\mathfrak g} \newcommand{\s}{\mathfrak s} \newcommand{\sderived}{\mathfrak s_{\mathrm{ss}}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\End}{\operatorname{End}} \newcommand{\Aut}{\operatorname{Aut}} \newcommand{\Stab}{\operatorname{Stab}} \newcommand{\Sym}{\operatorname{Sym}} \newcommand{\Gr}{\operatorname{Gr}} \newcommand{\id}{\operatorname{id}} \newcommand{\Lam}{\bigwedge} \newcommand\mono{\hookrightarrow} \newcommand\sminus{\smallsetminus} \newcommand\st{{\textrm{ s.t.\ }}} \newcommand\ket[1]{\mid #1 \rangle} \newcommand\bra[1]{\langle #1 \mid} \newcommand\setof[1]{\{ #1 \}} \newcommand\lt{<} \newcommand\abs[1]{ \mid #1 \mid } \newtheorem{theorem}{Theorem} \newtheorem{thm}{Theorem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{prop}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{cor}[theorem]{Corollary} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{conj}[theorem]{Conjecture} \newtheorem{remark}[theorem]{Remark} \newtheorem{defn}[theorem]{Definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{example}[theorem]{Example} \title{Class-two Lie algebras and quadratic-dual current constructions} \author{Ammar Husain} \date{\today} \begin{document} \maketitle \begin{abstract} For a reductive Lie algebra $\s$, a finite-dimensional $\s$-module $R$ and an $\s$-submodule $K\subseteq\bigwedge^2R$, the class-two radical $R\oplus(\bigwedge^2R/K)$ carries an $\s$-action. For semisimple $\s$ the semidirect products $\s\ltimes(R\oplus(\bigwedge^2R/K))$ are, up to isomorphism, all Lie algebras whose radical has class at most two. We study the isomorphisms that fix the acting algebra pointwise, including their central mixing and shear terms. On the locus where central mixing vanishes, the resulting groupoid admits a description by invariant Grassmannians and equivariant linear groups. We express the wild matrix-pair construction as a functor into class-two Lie algebras and give its equivariant square in $\s$-modules. We work out single-isotypic slices, signed multiplicity-free tensor squares, and explicit low-dimensional multiplicity-space examples. The acting algebra may be zero and its central action need not be semisimple. Finally, we study pairs of Lie brackets on one object whose mixed composites vanish. Tensoring them with algebras carrying two independent commutative associative products, for instance pointwise multiplication and convolution exchanged by Fourier transform, produces further Lie algebras. For these pairs, in representations of $\s$, we define the groupoid of objects that are isomorphic to their bracket exchange $\sigma X$; we give the structure of the switches $X\to\sigma X$, the obstruction to an involutive switch, and its indecomposable objects in small dimension for semisimple $\s$. \end{abstract} Let $k$ be an algebraically closed field of characteristic zero, and let all representations below be finite-dimensional unless otherwise noted. \section{Introduction}\label{sec:intro} In characteristic zero the radical of a finite-dimensional perfect Lie algebra is nilpotent; among the nonsemisimple perfect complex Lie algebras of dimension at most nine, every radical but one is abelian or of class two (Theorem~\ref{thm:small-perfect}). A Lie algebra with semisimple Levi factor $\s$ and radical of class at most two is determined by the $\s$-module $R$, the radical modulo its commutator, and the kernel $K\subseteq\bigwedge^2R$ of its bracket. This reduction is standard and is recalled in Section~\ref{sec:perfect}. Generator-and-relation approaches to two-step nilpotent Lie algebras go back to Gauger~\cite{Gauger1973}; see also Scheuneman~\cite{Scheuneman1967}. Unrestricted classification already contains wild matrix problems when the central commutator has dimension three as $\dim R$ grows \cite{BelitskiiLipyanskiSergeichuk2005}. That problem treats $R$ only as a vector space. Here $R\oplus(\bigwedge^2R/K)$ is a Lie algebra object in $\s\text{-}\mathrm{Rep}$, and isomorphisms are required to fix the acting algebra pointwise. For a nontrivial $\s$, the corresponding orbit problems often live in much smaller multiplicity spaces, although wild behavior can return as those multiplicities grow. The original motivating question came from the standard quadratic-duality construction of a Lie algebra on the tensor product of a $\mathcal P$-algebra and a $\mathcal P^!$-algebra applied with $\mathcal P=\mathrm{Com}\amalg\mathrm{Com}$. Those are two independent commutative associative products, such as pointwise multiplication and convolution. Its quadratic dual governs two Lie brackets whose mixed composites vanish individually. Class-two brackets on a common $R\oplus W$ provide a concrete family satisfying these relations. \subsection*{Organization} Section~\ref{sec:perfect} collects the background: small classifications, the construction and its converse. Section~\ref{sec:isomorphisms} types every block of an isomorphism fixing the acting algebra, describes the locus where central mixing vanishes (Section~\ref{sec:vanishing}), and introduces the wedge multiplicity spaces (Section~\ref{sec:wedge}). Section~\ref{sec:moduli} identifies the moduli on that locus, for a fixed representation and over all representations, as groupoids of invariant subspaces (Sections~\ref{sec:fixed-moduli} and~\ref{sec:global-moduli}). It then expresses the wild matrix-pair construction as a functor and lifts it to $\s$-modules (Section~\ref{sec:wild}). Section~\ref{sec:slices} computes these groupoids in explicit cases. It specializes them to a single semisimple isotypic component (Section~\ref{sec:isotypic}), compares compatible multiplicity-free tensor-square data across semisimple types (Section~\ref{sec:change-ss}), and calculates explicit small binary multiplicity-space slices (Sections~\ref{sec:sl2} and~\ref{sec:binary}). It also treats the first ternary multiplicity space (Section~\ref{sec:ternary}) and lists the components of total dimension at most six for each semisimple acting algebra, including $\s=0$ (Section~\ref{sec:lowdim}). Neither a polarization of $R$ nor a trivialization of a central line is assumed. Section~\ref{sec:pairs} turns from one bracket to two. A pair of Lie brackets $\beta_1,\beta_2$ on one object is \emph{mutually annihilating} when $\beta_1(\beta_2(x,y),z)=0=\beta_2(\beta_1(x,y),z)$; this is stronger than the compatibility of~\cite{DotsenkoKhoroshkin}, which asks only that the cyclic sum of the mixed terms vanish. Such pairs are the algebras over the quadratic dual of $\mathrm{Com}\amalg\mathrm{Com}$, so they tensor with algebras carrying two unrelated commutative associative products to give Lie algebras (Section~\ref{sec:quadratic-dual}). Exchanging the two brackets is an involution $\sigma$ of the groupoid of pairs in $\s\text{-}\mathrm{Rep}$ or $\s\text{-}\mathrm{sRep}$. Section~\ref{sec:switch} defines this groupoid and its full subgroupoid on the objects $X$ that admit an arrow $X\to\sigma X$, with no such arrow chosen. The examples follow the definitions. Section~\ref{sec:pair-examples} gives pairs of class-two brackets on a common $R\oplus W$, in the notation of the earlier sections, and semidirect pairs over a shared abelian algebra. Section~\ref{sec:switch-examples} lists the indecomposable objects of the switchable subgroupoid of dimension at most four in $\mathrm{Vect}_k$ and in $\s\text{-}\mathrm{Rep}$ for semisimple $\s$, with their automorphism groups and switches. Section~\ref{sec:fourier} gives coefficient algebras, including pointwise multiplication and convolution exchanged by Fourier transform, and Section~\ref{sec:transport} the resulting Lie algebras. \section{From perfect Lie algebras to the class-two family}\label{sec:perfect} The radical of a finite-dimensional perfect Lie algebra in characteristic zero is nilpotent (Lemma~\ref{lem:nilradical} below). The following complete small-dimensional list shows how often it has class two. Let $V_j$ be the irreducible $\mathfrak{sl}_2(\mathbb C)$-module of highest weight $j$, let $\mathfrak h_d$ be the $d$-dimensional Heisenberg algebra, and let $\mathfrak f_{r,c}$ be the free nilpotent Lie algebra on $r$ generators of class $c$. In direct sums inside a semidirect product, the brackets between the displayed summands vanish. In each occurrence of $\mathfrak h_3$, the specified $\mathfrak{sl}_2$-action induces $V_1$ on $\mathfrak h_3/Z(\mathfrak h_3)$ and is trivial on its center; its bracket is the invariant alternating pairing into that center. On $\mathfrak f_{2,3}$ and $\mathfrak f_{3,2}$ the action extends those on their generating spaces $V_1$ and $V_2$, respectively. There are two indicated actions on $\mathfrak h_5$, denoted $\mathfrak h_5^{(1,1)}$ and $\mathfrak h_5^{(3)}$: their symplectic quotients are $V_1\oplus V_1$ and $V_3$, respectively, their centers are trivial $\mathfrak{sl}_2$-modules, and their brackets are the corresponding nondegenerate invariant alternating pairings. Finally, $\mathfrak a_{6,4}$ has nonzero brackets $[e_1,e_4]=e_5$, $[e_2,e_3]=-e_5$, and $[e_3,e_4]=e_6$; $\langle e_1,e_2\rangle$ and $\langle e_3,e_4\rangle$ are standard $\mathfrak{sl}_2$-modules, while $e_5,e_6$ are fixed. \begin{theorem}[Small complex perfect Lie algebras] \label{thm:small-perfect} Up to isomorphism, the nonsemisimple perfect complex Lie algebras of dimension at most nine are exactly the entries in the following table. The unique entry whose radical has nilpotency class three is $\mathfrak{sl}_2\ltimes\mathfrak f_{2,3}$; every other radical is abelian or has class two. \begin{center}\small \begin{tabular}{@{}c p{0.79\linewidth}@{}} \hline $\dim\g$ & Isomorphism classes\\ \hline 5 & $\mathfrak{sl}_2\ltimes V_1$\\ 6 & $\mathfrak{sl}_2\ltimes V_2$, $\mathfrak{sl}_2\ltimes\mathfrak h_3$\\ 7 & $\mathfrak{sl}_2\ltimes V_3$, $\mathfrak{sl}_2\ltimes(V_1\oplus V_1)$\\ 8 & $\mathfrak{sl}_2\oplus(\mathfrak{sl}_2\ltimes V_1)$, $\mathfrak{sl}_2\ltimes V_4$, $\mathfrak{sl}_2\ltimes(V_1\oplus V_2)$, $\mathfrak{sl}_2\ltimes(V_1\oplus\mathfrak h_3)$, $\mathfrak{sl}_2\ltimes\mathfrak f_{2,3}$, $\mathfrak{sl}_2\ltimes\mathfrak h_5^{(1,1)}$, $\mathfrak{sl}_2\ltimes\mathfrak h_5^{(3)}$\\ 9 & $\mathfrak{sl}_2\oplus(\mathfrak{sl}_2\ltimes V_2)$, $\mathfrak{sl}_2\oplus(\mathfrak{sl}_2\ltimes\mathfrak h_3)$, $\mathfrak{sl}_2\ltimes V_5$, $\mathfrak{sl}_2\ltimes(V_1\oplus V_3)$, $\mathfrak{sl}_2\ltimes(V_2\oplus V_2)$, $\mathfrak{sl}_2\ltimes(V_1\oplus V_1\oplus V_1)$, $\mathfrak{sl}_2\ltimes(V_2\oplus\mathfrak h_3)$, $\mathfrak{sl}_2\ltimes(\mathfrak h_3\oplus\mathfrak h_3)$, $\mathfrak{sl}_2\ltimes\mathfrak f_{3,2}$, $\mathfrak{sl}_2\ltimes\mathfrak a_{6,4}$\\ \hline \end{tabular} \end{center} The semisimple perfect algebras in this range are $\mathfrak{sl}_2$, $\mathfrak{sl}_2^{\oplus2}$, $\mathfrak{sl}_3(\mathbb C)$, and $\mathfrak{sl}_2^{\oplus3}$ (and the zero algebra, if dimension zero is included). \end{theorem} \begin{proof} The nonsemisimple classification is \cite[Proposition~5.3]{BurdeWagemannPerfect}; the semisimple entries follow from the classification of complex simple Lie algebras. \end{proof} For comparison, the class-two nilpotent Lie algebras themselves can be listed in the same quotient-kernel language. For a vector space $R$ and $K\subseteq\bigwedge^2R$, write $\mathfrak n(R,K)=R\oplus(\bigwedge^2R/K)$, with $[r,r']=r\wedge r'\pmod K$ and the second summand central. Let $R_q$ be a $q$-dimensional space with basis $u_1,\ldots,u_q$ and dual basis $u^1,\ldots,u^q$. The expression $ij$ in the following table abbreviates the functional $u^i\wedge u^j:\bigwedge^2R_q\to k$; in particular it is not a bracket. For forms $\omega_1,\ldots,\omega_d$ put \[ K_q(\omega_1,\ldots,\omega_d) =\bigcap_{a=1}^d\ker\bigl(\omega_a:\bigwedge^2R_q\to k\bigr), \qquad K_q()=\bigwedge^2R_q. \] \begin{theorem}[Class-two nilpotent algebras through dimension six] \label{thm:small-two-step} Over an algebraically closed field of characteristic zero, the following are all the nilpotent Lie algebras of class at most two and dimension $1\le n\le6$, without repetitions. Each displayed pair $(R,K)$ denotes $\mathfrak n(R,K)$, and semicolons separate distinct isomorphism classes. The pair $(R_n,K_n())$ is abelian; omitting it from each row gives exactly the algebras of class two. \begin{center}\small \begin{tabular}{@{}c >{\raggedright\arraybackslash}p{0.79\linewidth}@{}} \hline $n$ & Isomorphism classes as $(R,K)$ pairs\\ \hline 1 & $(R_1,K_1())$\\ 2 & $(R_2,K_2())$\\ 3 & $(R_3,K_3())$\hspace{0.9em};\hspace{0.9em} $(R_2,K_2(12))$\\ 4 & $(R_4,K_4())$\hspace{0.9em};\hspace{0.9em} $(R_3,K_3(12))$\\ 5 & $(R_5,K_5())$\hspace{0.9em};\hspace{0.9em} $(R_4,K_4(12))$\hspace{0.9em};\hspace{0.9em} $(R_4,K_4(12+34))$\hspace{0.9em};\hspace{0.9em} $(R_3,K_3(12,13))$\\ 6 & $(R_6,K_6())$\hspace{0.9em};\hspace{0.9em} $(R_5,K_5(12))$\hspace{0.9em};\hspace{0.9em} $(R_5,K_5(12+34))$\hspace{0.9em};\hspace{0.9em} $(R_4,K_4(12,13))$\hspace{0.9em};\hspace{0.9em} $(R_4,K_4(12,34))$\hspace{0.9em};\hspace{0.9em} $(R_4,K_4(12,13+24))$\hspace{0.9em};\hspace{0.9em} $(R_3,K_3(12,13,23))$\\ \hline \end{tabular} \end{center} \end{theorem} \begin{proof} This is the class-at-most-two part of the classification in \cite[Table~1]{BazzoniMunoz2012}, with the lower-dimensional entries also obtained by deleting abelian direct summands. \end{proof} The complete eight-dimensional class-two list has 35 isomorphism classes~\cite[Theorem~1 and Corollary~2]{BazzoniRojo2025}; the table above is complete through dimension six, the range used in Section~\ref{sec:lowdim}. For space reasons, we do not reproduce them all in this introduction. Let $\g=\sderived\ltimes\mathfrak r$ be a Levi decomposition, with $\sderived$ semisimple and $\mathfrak r$ the solvable radical. Recall that $\g$ is \emph{perfect} when $[\g,\g]=\g$. \begin{lemma}\label{lem:nilradical} For every finite-dimensional Lie algebra in characteristic zero, $[\g,\operatorname{rad}\g]$ is a nilpotent ideal. Consequently the solvable radical of a perfect Lie algebra is nilpotent \cite[Lemma~2.2]{BurdeWagemannPerfect}. \end{lemma} \begin{proof} Write $\mathfrak r=\operatorname{rad}\g$. By Lie's theorem, the operators $\operatorname{ad}(\mathfrak r)$ on $\g$ are upper triangular; let $\chi_i$ be their diagonal characters. For $x\in\g$ and $z\in\mathfrak r$, conjugation by $e^{t\operatorname{ad}x}$ preserves $\operatorname{ad}(\mathfrak r)$. The operator $\operatorname{ad}(e^{t\operatorname{ad}x}z)=e^{t\operatorname{ad}x}\operatorname{ad}(z)e^{-t\operatorname{ad}x}$ is conjugate to $\operatorname{ad}z$ and has the same characteristic polynomial, whose coefficients do not depend on $t$. Its diagonal entries $\chi_i(e^{t\operatorname{ad}x}z)$, extended $k[[t]]$-linearly, are roots of that polynomial, so each is one of the finitely many constant roots $\chi_j(z)$ in $k[[t]]$. Its constant term is $\chi_i(z)$; therefore $\chi_i([x,z])=0$. Thus $\operatorname{ad}([\g,\mathfrak r])$ is strictly upper triangular, and Engel's theorem makes $[\g,\mathfrak r]$ nilpotent. If $\g$ is perfect, its Levi decomposition gives $\mathfrak r=[\sderived,\mathfrak r]+[\mathfrak r,\mathfrak r] =[\g,\mathfrak r]$. \end{proof} \begin{proposition}[Perfectness criterion]\label{prop:perfect} For $\g=\sderived\ltimes\mathfrak r$ with $\sderived$ semisimple, put $R=\mathfrak r/[\mathfrak r,\mathfrak r]$. Then $\g$ is perfect if and only if $R^{\sderived}=0$. \end{proposition} \begin{proof} Since $[\sderived,\sderived]=\sderived$, the equality $[\g,\g]=\g$ is equivalent to $\mathfrak r=[\sderived,\mathfrak r]+[\mathfrak r,\mathfrak r]$. Passing to $R$, this says $\sderived R=R$. Complete reducibility splits $R$ into its trivial isotypic part $R^{\sderived}$ and the sum of its nontrivial irreducible summands. The latter, and only the latter, lie in $\sderived R$, giving the criterion. \end{proof} The following equivariant quotient construction refines the usual generator-and-relation description of two-step nilpotent Lie algebras~\cite{Gauger1973,Scheuneman1967}. \begin{proposition}[Construction]\label{prop:construction} Let $\s$ be any Lie algebra. For an $\s$-module $R$ and an $\s$-submodule $K\subseteq\Lam^2R$, put $W=\Lam^2R/K$. The vector space $\mathfrak r_K=R\oplus W$, with bracket $[r_1,r_2]=r_1\wedge r_2\pmod K$ and $[R,W]=[W,W]=0$, is a class-at-most-two nilpotent Lie algebra acted on by $\s$ through derivations. \end{proposition} \begin{proof} Every triple bracket in $\mathfrak r_K$ vanishes, so Jacobi holds. The $\s$-action on $\Lam^2R$ is $x(r_1\wedge r_2)=(xr_1)\wedge r_2+r_1\wedge(xr_2)$. Since $K$ is stable, the projection to $W$ intertwines this action; this is precisely the derivation identity for the bracket. \end{proof} \begin{corollary}[Radical and perfectness of the construction] \label{cor:construction-perfect} If $\s=\sderived$ is semisimple in Proposition~\ref{prop:construction}, the Lie algebra $\sderived\ltimes\mathfrak r_K$ has radical $\mathfrak r_K$ and is perfect exactly when $R^{\sderived}=0$. \end{corollary} \begin{proof} The quotient by $\mathfrak r_K$ is semisimple, so the nilpotent ideal $\mathfrak r_K$ is the radical. Its derived ideal is $W$; Proposition~\ref{prop:perfect} gives the perfectness criterion. \end{proof} \begin{theorem}[Converse for a class-two radical]\label{thm:construction} Let $\sderived$ be semisimple. Every $\sderived\ltimes\mathfrak r$ whose radical $\mathfrak r$ has class at most two is isomorphic to one of these. The module $R$ is $\mathfrak r/[\mathfrak r,\mathfrak r]$ and $K$ is the kernel of the induced surjection $\Lam^2R\twoheadrightarrow[\mathfrak r,\mathfrak r]$. \end{theorem} \begin{proof} $W=[\mathfrak r,\mathfrak r]$ is central in $\mathfrak r$. Complete reducibility supplies an equivariant linear section of $\mathfrak r\twoheadrightarrow R$ and hence an equivariant vector-space identification $\mathfrak r\cong R\oplus W$. Because $W$ is central, the bracket of lifts depends only on their images in $R$ and defines an equivariant surjection $b:\Lam^2R\twoheadrightarrow W$. With $K=\ker b$, the induced $\Lam^2R/K\cong W$ identifies both brackets, giving the claimed Lie isomorphism. \end{proof} We study the resulting $(R,K)$ systematically, and also allow a reductive algebra \[ \s=\sderived\oplus\bigoplus_{j=1}^{m}kt_j, \] allowing $\sderived=0$, $m=0$, and $\s=0$. For $m>0$ this $\s$ is a specified acting subalgebra, while $\sderived$ is the Levi factor of the resulting Lie algebra. Let $R$ be an $\s$-representation, with $D_j^R=\rho_R(t_j)$. Choose an $\s$-subrepresentation $K\subseteq\Lam^2R$, put $W_K=\Lam^2R/K$, and denote its quotient map by $b_K:\Lam^2R\twoheadrightarrow W_K$. On the vector space \[ \g_K=\s\oplus R\oplus W_K \] use the given $\s$-actions and the bracket \[ [r_1,r_2]=b_K(r_1 \wedge r_2),\qquad [R,W_K]=[W_K,W_K]=0. \] Equivariance of $b_K$ gives the Jacobi identity involving $\s$; the other Jacobi identities follow because $W_K$ is central in $R\oplus W_K$. This construction does not require $\s$ to be semisimple. \paragraph{A central line and the Heisenberg object.} Suppose $K$ is such that $W_K$ is isomorphic in $\s\text{-}\mathrm{Rep}$ to its monoidal unit $\mathbf1$ (the one-dimensional trivial module). The projection $b_K:\Lam^2R\twoheadrightarrow W_K$ is then an invariant, line-valued alternating form, and $R\oplus W_K$ is a two-step central Lie algebra object in $\s\text{-}\mathrm{Rep}$. If the intrinsic map \[ b_K^\sharp:R\longrightarrow\Hom_k(R,W_K), \qquad b_K^\sharp(r)(v)=b_K(r\wedge v), \] is an isomorphism, it is the Heisenberg Lie algebra object of the symplectic object $(R,b_K)$, with its central line kept explicit. The condition $W_K\cong\mathbf1$ alone permits a degenerate form: $\ker b_K^\sharp$ is central and splits off as an abelian factor of the underlying Lie algebra. Such a splitting in $\s\text{-}\mathrm{Rep}$ requires an equivariant complement. A presentation $R\cong X\oplus X^*$ with the evaluation pairing is a further $\s$-equivariant polarization, not part of the construction. \section{General isomorphisms fixing the acting algebra}\label{sec:isomorphisms} Write $\g_K=\g(\s,R,K)$ and $\g_{K'}=\g(\s,R',K')$ for two constructions, with $W=W_K$, $W'=W_{K'}$, and quotient maps $b,b'$. An isomorphism is required to fix the displayed $\s$ pointwise; no preservation of $R$, $W$, or $R\oplus W$ is assumed. The central component below arises from an arbitrary such isomorphism. \begin{theorem}[General isomorphism criterion]\label{thm:general-iso} Every isomorphism $\Phi:\g_K\to\g_{K'}$ fixing $\s$ has the form \begin{equation}\label{eq:general-Phi} \Phi(x,r,w)=\bigl(x+c(r),\;a(r)+p(w),\;h(r)+q(w)\bigr), \end{equation} where the $\s$-intertwiners have types \[ c:R\to Z(\s),\quad a:R\to R',\quad p:W\to R',\quad h:R\to W',\quad q:W\to W'. \] Here $Z(\s)$ has its trivial adjoint $\s$-action. Conversely these maps define an isomorphism if and only if the linear block map \begin{equation}\label{eq:block-iso} T=\begin{pmatrix}a&p\\h&q\end{pmatrix}:R\oplus W \xrightarrow{\sim}R'\oplus W' \end{equation} is invertible and, for $r,v\in R$ and $w,u\in W$, one has \begin{align} p\,b(r\wedge v) &=c(r)\cdot a(v)-c(v)\cdot a(r),\label{eq:general-rR}\\ q\,b(r\wedge v) &=b'(a(r)\wedge a(v)) +c(r)\cdot h(v)-c(v)\cdot h(r),\label{eq:general-rW}\\ 0&=c(r)\cdot p(w),\label{eq:general-RW-R}\\ 0&=b'(a(r)\wedge p(w))+c(r)\cdot q(w),\label{eq:general-RW-W}\\ 0&=b'(p(w)\wedge p(u)).\label{eq:general-WW} \end{align} \end{theorem} \begin{proof} The radical is $Z(\s)\oplus R\oplus W$, since its quotient is $\sderived$. Its characteristic property restricts the $\s$-part of $\Phi(R\oplus W)$ to $Z(\s)$. This part vanishes on $W=[R,R]$, as the bracket of two elements with central $\s$-parts has zero $\s$-part. Thus it is the map $c$ in \eqref{eq:general-Phi}. Bracketing with $\s$, which $\Phi$ fixes, shows that all five maps are intertwiners. The triangular map $\Phi$ is invertible exactly when $T$ is. Comparing respectively the $R'$- and $W'$-components of $[r,v]$, $[r,w]$, and $[w,u]$ gives \eqref{eq:general-rR}--\eqref{eq:general-WW}. Conversely these identities check every bracket, including those with $\s$. \end{proof} \subsection{The locus where central mixing vanishes}\label{sec:vanishing} First decompose $R$ under $\sderived$ and write the action of $t_j$ on each multiplicity space as $D_{j,\lambda}$: \[ R=\bigoplus_{\lambda}V_\lambda\otimes M_\lambda, \qquad D_j^R|_{V_\lambda\otimes M_\lambda} =\id_{V_\lambda}\otimes D_{j,\lambda}. \] Here $\lambda$ runs through the finitely many occurring highest weights, $V_\lambda$ is the corresponding simple $\sderived$-module, and $V_0=k$ is the trivial module. The operators $D_{j,\lambda}$ commute because the $t_j$ do. Next take the joint generalized weight decomposition for the commuting $D_{j,\lambda}$: \[ M_\lambda=\bigoplus_{\mu\in k^m}M_{\lambda,\mu}, \] where $\mu=(\mu_1,\ldots,\mu_m)$ and $M_{\lambda,\mu}$ consists of the vectors killed by sufficiently high powers of every $D_{j,\lambda}-\mu_j\id$. Only finitely many summands are nonzero. Write $\mathbf0_m=(0,\ldots,0)$; if $m=0$, this is the empty tuple and $M_{\lambda,()}=M_\lambda$. Set $R_{\mathbf0}=V_0\otimes M_{0,\mathbf0_m}$ and let $R_{\ne}$ be the sum of the other summands, so $R=R_{\mathbf0}\oplus R_{\ne}$. These are generalized joint weight spaces; the $D_j^R$ need not be diagonalizable. In particular, the $D_{j,0}$ may each have a nonzero generalized nullspace on $M_0$ while $M_{0,\mathbf0_m}=0$. For instance, on $M_0=k^2$, take \[ D_{1,0}=\begin{pmatrix}0&0\\0&1\end{pmatrix},\qquad D_{2,0}=\begin{pmatrix}1&0\\0&0\end{pmatrix}. \] The condition $M_{0,\mathbf0_m}=0$ constrains only the trivial $\sderived$-isotypic part; other types and multiplicities are arbitrary. An $\s$-invariant linear map $c:R\to Z(\s)=\bigoplus_jkt_j$ vanishes on $R_{\ne}$ and factors through \[ R_{\mathbf0}\Big/\sum_jD_j^RR_{\mathbf0}\longrightarrow Z(\s). \] Indeed it vanishes on $[\sderived,R]$ and on $D_j^RR$; on any nonzero joint generalized weight, at least one $D_j^R$ is invertible. Consequently, \begin{equation}\label{eq:no-c} M_{0,\mathbf0_m}=0\quad\Longrightarrow\quad \Hom_{\s}(R,Z(\s))=0. \end{equation} \begin{proposition}[Sufficient, not necessary, vanishing] \label{prop:c-vanish} The condition $M_{0,\mathbf0_m}=0$ forces $c=0$ for every isomorphism fixing $\s$, but it is not necessary. More generally, put \[ Z_{\mathrm{nil}}(\s;R')= \{z\in Z(\s):\rho_{R'}(z)\text{ is nilpotent}\}. \] Every $c$ arising in Theorem~\ref{thm:general-iso} has image in $Z_{\mathrm{nil}}(\s;R')$. Thus $\Hom_{\s}(R,Z_{\mathrm{nil}}(\s;R'))=0$ also forces $c=0$. \end{proposition} \begin{proof} The first assertion follows from \eqref{eq:no-c}. Since $R\oplus W$ is a nilpotent ideal, its image lies in the nilradical of $\g_{K'}$. Every element of the nilradical acts nilpotently by adjoint on $\g_{K'}$: its first bracket lands in that ideal, and subsequent brackets follow its lower central series. The action of $\operatorname{ad}\Phi(r)$ on $(R'\oplus W')/W'\cong R'$ is $\rho_{R'}(c(r))$, which must therefore be nilpotent. For nonnecessity take $\s=kt$, $R=ke_0\oplus ke_1$ with $t e_0=0$, $t e_1=e_1$, and $K=\Lam^2R$. Here $M_{0,(0)}=ke_0\ne0$, whereas the only scalar multiple of $\rho_R(t)=\operatorname{diag}(0,1)$ that is nilpotent is zero. Hence $c=0$ for every automorphism of $\g_K$ fixing $t$. \end{proof} Let $\mathcal C_{\s}$ be the groupoid whose objects are pairs $(R,K)$ as above and whose arrows are \emph{all} Lie algebra isomorphisms $\g(\s,R,K)\to\g(\s,R',K')$ fixing $\s$ pointwise. Define $\mathcal C_{\s}^{0}$ to be its full subgroupoid on objects satisfying $M_{0,\mathbf0_m}(R)=0$. This tests the representation $R$ alone, independently of $K$ and of any prospective arrow. Fullness means that no further restriction is placed on the isomorphisms between selected objects. Proposition~\ref{prop:c-vanish} explains why this object condition simplifies every arrow in the subgroupoid. Let $\mathcal B_{\s}^{0}$ have the same objects, with only the arrows $a_*(x,r,w)=(x,ar,\bar a w)$ induced by $\s$-module isomorphisms $a:R\to R'$ carrying $K$ onto $K'$; here $\bar a$ is the induced map of exterior-square quotients. \begin{theorem}\label{thm:groupoid} For $(R,K),(R',K')\in\mathcal C_{\s}^{0}$, an arrow $\g(\s,R,K)\to\g(\s,R',K')$ exists if and only if there is an $\s$-module isomorphism $a:R\xrightarrow{\sim}R'$ with \begin{equation}\label{eq:orbit} (\Lam^2a)K=K'. \end{equation} For each such $a$, let $\bar a:W_K\to W_{K'}$ be the uniquely induced isomorphism, characterized by $\bar a\,b_K=b_{K'}\Lam^2a$. All the isomorphisms above $a$ are \begin{equation}\label{eq:arrow} \Phi_{a,H}(x,r,w) =\bigl(x,ar,\bar a(w)+H(ar)\bigr), \qquad H\in\Hom_{\s}(R',W_{K'}). \end{equation} \end{theorem} \begin{proof} Begin with an arbitrary isomorphism $\Phi$ fixing $\s$ and apply Theorem~\ref{thm:general-iso}. Equation~\eqref{eq:no-c} forces $c=0$, so \eqref{eq:general-rR} and surjectivity of $b_K$ force $p=0$. Hence $T$ is triangular, with invertible diagonal maps $a$ and $q$; equation \eqref{eq:general-rW} becomes \[ q\,b_K=b_{K'}\Lam^2a. \] This implies \eqref{eq:orbit} and $q=\bar a$. Finally $h=Ha$ for a unique $H\in\Hom_{\s}(R',W_{K'})$. Conversely \eqref{eq:arrow} preserves every bracket and is invertible. \end{proof} \begin{proposition}[Composition and self-loops]\label{prop:groupoid-composition} For composable arrows $(a,H):(R,K)\to(R',K')$ and $(d,J):(R',K')\to(R'',K'')$, composition is \begin{equation}\label{eq:groupoid-composition} (d,J)\circ(a,H) =\bigl(da,\;J+\bar d\,H\,d^{-1}\bigr), \end{equation} where the second component belongs to $\Hom_{\s}(R'',W_{K''})$. Thus at \emph{every} object $(R,K)$ the group of all self-loops is \begin{equation}\label{eq:aut} \Aut_{\mathcal C_{\s}^{0}}(R,K)=\Aut_{\s}(\g_K) \cong\Hom_{\s}(R,W_K)\rtimes \Aut_{\mathcal B_{\s}^{0}}(R,K), \end{equation} where $\Aut_{\s}(\g_K)$ denotes the Lie algebra automorphisms of $\g_K$ fixing $\s$ pointwise, and $a$ acts on a shear $H$ by $\bar a H a^{-1}$. \end{proposition} \begin{proof} For $H\in\Hom_{\s}(R,W_K)$ write $S_H(x,r,w)=(x,r,w+H(r))$, so that $\Phi_{a,H}=S_{H}\circ a_*$ with $S_H$ taken on the target. The identity $d_*S_Hd_*^{-1}=S_{\bar d H d^{-1}}$ gives \eqref{eq:groupoid-composition}; restricting it to self-loops gives \eqref{eq:aut}. \end{proof} \begin{corollary}[Closure of the vanishing locus] \label{cor:vanishing-closure} The full subgroupoid $\mathcal C_{\s}^{0}$ is closed under isomorphisms in $\mathcal C_{\s}$. \end{corollary} \begin{proof} An isomorphism whose source satisfies $M_{0,\mathbf0_m}(R)=0$ has $c=0$ by Proposition~\ref{prop:c-vanish}. The general bracket equations in Theorem~\ref{thm:general-iso} then force $p=0$, as $b_K$ is surjective. The resulting triangular isomorphism has an invertible $\s$-intertwiner $a:R\to R'$ on its diagonal. Hence $R'\cong R$ as $\s$-modules and the target satisfies the same condition. \end{proof} Equivalently, at each object $X=(R,K)$ put $\mathcal H_X=\Hom_{\s}(R,W_K)$. Transport along $a:X\to X'$ is $H\mapsto\bar a H a^{-1}$. Formula \eqref{eq:groupoid-composition} identifies $\mathcal C_{\s}^{0}$ with the split semidirect product groupoid $\mathcal H\rtimes\mathcal B_{\s}^{0}$. The $\mathcal H_X$ consist of the shear loops \[ S_H(x,r,w)=(x,r,w+H(r)) \] and every arrow factors uniquely as $\Phi_{a,H}=S_H\circ a_*$. For a fixed $R$, the restriction of $\mathcal B_{\s}^{0}$ to objects $(R,K)$ is the action groupoid of $\Aut_{\s}(R)$ on the $\s$-subrepresentations $K\subseteq\Lam^2R$. The corresponding restriction of $\mathcal C_{\s}^{0}$ adds the shear loops without changing its isomorphism classes. \subsection{The wedge multiplicity spaces}\label{sec:wedge} Fix an object $(R,K)$ of $\mathcal C_{\s}^{0}$ and keep its $\s$-module $R$ fixed while varying $K$. Write $R=\bigoplus_\lambda V_\lambda\otimes M_\lambda$, and fix a total order on the finitely many occurring highest weights. Define $N_\nu=\Hom_{\sderived}(V_\nu,\Lam^2R)$, so that $\Lam^2R=\bigoplus_\nu V_\nu\otimes N_\nu$. For two distinct $\sderived$-types $\lambda<\kappa$, the cross term is governed by the usual Clebsch--Gordan space $\Hom_{\sderived}(V_\nu,V_\lambda\otimes V_\kappa)$. Within one type, the canonical wedge identity gives the two spaces \[ A_{\nu,\lambda}=\Hom_{\sderived}(V_\nu,\Lam^2V_\lambda),\qquad B_{\nu,\lambda}=\Hom_{\sderived}(V_\nu,\Sym^2V_\lambda). \] Consequently \begin{align}\label{eq:general-N} N_\nu\cong{}&\bigoplus_\lambda \bigl(A_{\nu,\lambda}\otimes\Sym^2M_\lambda \oplus B_{\nu,\lambda}\otimes\Lam^2M_\lambda\bigr)\\ &\quad\oplus\bigoplus_{\lambda<\kappa} \Hom_{\sderived}(V_\nu,V_\lambda\otimes V_\kappa) \otimes M_\lambda\otimes M_\kappa.\nonumber \end{align} This decomposition uses the entire $M_\lambda$, including all its joint generalized central weights. \section{Moduli problems}\label{sec:moduli} Section~\ref{sec:isomorphisms} describes the arrows between two objects. This section assembles the objects and arrows of $\mathcal C_{\s}^{0}$ (Section~\ref{sec:vanishing}) into moduli groupoids, first with the $\s$-module $R$ fixed (Section~\ref{sec:fixed-moduli}) and then over all $R$ (Section~\ref{sec:global-moduli}). Section~\ref{sec:wild} embeds the matrix-pair problem into the categories of class-two Lie algebra objects and compares it with these equivariant moduli. \subsection{Moduli for an arbitrary fixed representation}\label{sec:fixed-moduli} For each $j$ and $\lambda$, set $D_{j,\lambda}^{\{2\}}=D_{j,\lambda}\otimes\id_{M_\lambda} +\id_{M_\lambda}\otimes D_{j,\lambda}$, and write $C_{\nu,\lambda\kappa} =\Hom_{\sderived}(V_\nu,V_\lambda\otimes V_\kappa)$. Under \eqref{eq:general-N}, the $t_j$-action on $N_\nu$ which is denoted $E_{j,\nu}$ has blocks \begin{align}\label{eq:general-E} E_{j,\nu}|_{A_{\nu,\lambda}\otimes\Sym^2M_\lambda} &=\id_{A_{\nu,\lambda}}\otimes D_{j,\lambda}^{\{2\}}|_{\Sym^2M_\lambda},\\ E_{j,\nu}|_{B_{\nu,\lambda}\otimes\Lam^2M_\lambda} &=\id_{B_{\nu,\lambda}}\otimes D_{j,\lambda}^{\{2\}}|_{\Lam^2M_\lambda},\nonumber\\ E_{j,\nu}|_{C_{\nu,\lambda\kappa}\otimes M_\lambda\otimes M_\kappa} &=\id_{C_{\nu,\lambda\kappa}}\otimes (D_{j,\lambda}\otimes\id_{M_\kappa} +\id_{M_\lambda}\otimes D_{j,\kappa}) \quad(\lambda<\kappa).\nonumber \end{align} Put $G_R=\Aut_{\s}(R)$, the group acting in this fixed-$R$ slice of $\mathcal B_{\s}^{0}$. Its elements are $a_g=\bigoplus_\lambda\id_{V_\lambda}\otimes g_\lambda$, where \begin{equation}\label{eq:general-G} G_R=\prod_\lambda G_\lambda,\qquad G_\lambda=\{g_\lambda\in GL(M_\lambda): g_\lambda D_{j,\lambda}=D_{j,\lambda}g_\lambda \text{ for all }j\}. \end{equation} The action induced by $\Lam^2a_g$ on $N_\nu$ has blocks \begin{equation}\label{eq:general-g} \id_{A_{\nu,\lambda}}\otimes\Sym^2g_\lambda,\qquad \id_{B_{\nu,\lambda}}\otimes\Lam^2g_\lambda,\qquad \id_{C_{\nu,\lambda\kappa}}\otimes g_\lambda\otimes g_\kappa. \end{equation} Denote this entire operator by $g_\nu$. It commutes with all $E_{j,\nu}$; the $g_\lambda$ act simultaneously across every $\nu$. \begin{lemma}[Invariant-subspace parameter space] \label{lem:invariant-subspaces} Fix any $\s$-module $R$ occurring in an object of $\mathcal C_{\s}^{0}$. In its decomposition $R=\bigoplus_\lambda V_\lambda\otimes M_\lambda$, the only condition is $M_{0,\mathbf0_m}=0$; in particular $R$ may have several isotypic types and several joint weights. With $N_\nu$, $E_{j,\nu}$, and $G_R$ as above, the possible $\s$-subrepresentations $K\subseteq\Lam^2R$ are precisely \begin{equation}\label{eq:K-L} K=\bigoplus_\nu V_\nu\otimes L_\nu, \qquad L_\nu\subseteq N_\nu,\quad E_{j,\nu}L_\nu\subseteq L_\nu \quad\text{for all }j. \end{equation} For $\mathbf d=(\dim L_\nu)_\nu$, set \begin{equation}\label{eq:X} X_{\mathbf d}(R)= \prod_\nu\bigl\{L\in\Gr_{d_\nu}(N_\nu): E_{j,\nu}L\subseteq L\text{ for all }j\bigr\}. \end{equation} \end{lemma} \begin{proof} Semisimple complete reducibility identifies a submodule of $\Lam^2R=\bigoplus_\nu V_\nu\otimes N_\nu$ with its collection of multiplicity subspaces $L_\nu$. Stability under each $t_j$ is exactly $E_{j,\nu}L_\nu\subseteq L_\nu$. This proves \eqref{eq:K-L} and \eqref{eq:X}. \end{proof} \begin{theorem}[Fixed-representation moduli] \label{thm:moduli} Under the assumptions of Lemma~\ref{lem:invariant-subspaces}, the isomorphism classes with this fixed $R$ form the orbit set \begin{equation}\label{eq:moduli} \coprod_{\mathbf d}X_{\mathbf d}(R)/G_R. \end{equation} \end{theorem} \begin{proof} The commutant of $\sderived$ on $R$ consists of the $g_\lambda$; commuting also with every $t_j$ gives \eqref{eq:general-G}. Taking exterior squares gives the three blocks in \eqref{eq:general-g}. Theorem~\ref{thm:groupoid} now identifies isomorphism classes with the $G_R$-orbits. \end{proof} \begin{corollary}[Stabilizers and self-loops at fixed $R$] \label{cor:fixed-loops} At $X=(R,K)$ the stabilizer is exactly the self-loop group in the base groupoid $\mathcal B_{\s}^{0}$, and the full self-loop group in $\mathcal C_{\s}^{0}$ is its split shear extension: \begin{equation}\label{eq:loops} G_K=\Aut_{\mathcal B_{\s}^{0}}(R,K) =\{g\in G_R:g_\nu L_\nu=L_\nu\text{ for all }\nu\},\qquad \Aut_{\mathcal C_{\s}^{0}}(R,K) \cong\Hom_{\s}(R,W_K)\rtimes G_K. \end{equation} For $g\in G_K$, the induced map $q_g:W_K\to W_K$ is determined by $q_gb_K=b_K\Lam^2a_g$. \end{corollary} \begin{proof} The stabilizer condition is exactly preservation of every $L_\nu$ under the action in \eqref{eq:general-g}. Theorem~\ref{thm:groupoid} and Proposition~\ref{prop:groupoid-composition} identify the full self-loop group as its shear extension. The equation for $q_g$ is the universal property of the quotient by $K$. \end{proof} The quotient in \eqref{eq:moduli} denotes an orbit set; no existence claim for a geometric quotient is needed. A stabilizer preserves $K$ as a subspace, not necessarily pointwise. If $W_K$ is a line, then $q_g=\chi(g)\id_{W_K}$ for an induced scalar $\chi(g)$; this does not require an identification $W_K\cong k$. \subsection{The moduli of all objects}\label{sec:global-moduli} The fixed-$R$ loci assemble by first choosing the isomorphism class of the underlying $\s$-representation. This choice involves the full commuting actions on the multiplicity spaces, including their nilpotent parts. For a set $X$ acted on by a group $G$, write $X//G$ for its \emph{action groupoid}: its objects are points of $X$ and an arrow $x\to x'$ is an element $g\in G$ with $gx=x'$. \begin{lemma}[Intertwiners between representation fibers] \label{lem:module-iso} For decompositions \[ R=\bigoplus_\lambda V_\lambda\otimes M_\lambda, \qquad R'=\bigoplus_\lambda V_\lambda\otimes M'_\lambda, \] the representations $R,R'$ are isomorphic precisely when, for every $\lambda$, there is a linear isomorphism $f_\lambda:M_\lambda\xrightarrow{\sim}M'_\lambda$ such that \begin{equation}\label{eq:module-iso} f_\lambda D_{j,\lambda}=D'_{j,\lambda}f_\lambda \qquad\text{for every }j. \end{equation} Equivalently, their multiplicity spaces are isomorphic as modules for the commuting central operators. If this fails for any $\lambda$, there are no arrows in $\mathcal C_{\s}^{0}$ from an object over $R$ to an object over $R'$, regardless of $K,K'$. If it holds, transporting $K$ along $\Lam^2(\bigoplus_\lambda \id_{V_\lambda}\otimes f_\lambda)$ reduces every arrow to the fixed-$R'$ description. \end{lemma} \begin{proof} Any $\sderived$-intertwiner $R\to R'$ acts independently on each $V_\lambda$-isotypic part and has the form $\bigoplus_\lambda\id_{V_\lambda}\otimes f_\lambda$. It intertwines the $t_j$ exactly when \eqref{eq:module-iso} holds, and is invertible exactly when every $f_\lambda$ is. Theorem~\ref{thm:groupoid} gives the claim about arrows and transport of $K$. \end{proof} \begin{theorem}[Global groupoid decomposition]\label{thm:global-moduli} Let $\mathcal R_{\s}^{0}$ be the set of isomorphism classes of finite-dimensional $\s$-modules $R$ satisfying $M_{0,\mathbf0_m}(R)=0$, and choose one representative of each class. The groupoid $\mathcal C_{\s}^{0}$ is equivalent to the disjoint union, over these representatives $R$, of the split shear groupoids $\mathcal H_R\rtimes(\mathcal K(R)//G_R)$, where $\mathcal K(R)$ is the set of $\s$-subrepresentations of $\Lam^2R$ and $\mathcal H_R$ is the restriction of the shear system $\mathcal H$ to the objects over $R$, with fiber $\Hom_{\s}(R,W_K)$ at $K$. \end{theorem} \begin{proof} Lemma~\ref{lem:module-iso} separates the representation classes. Within each class, Theorem~\ref{thm:groupoid} gives precisely the base arrows and shears. Proposition~\ref{prop:groupoid-composition} gives their composition. Choosing one $R$ in each class produces the asserted equivalence. \end{proof} \begin{corollary}[Global orbit set]\label{cor:global-orbits} With $X_{\mathbf d}(R)$ from \eqref{eq:X} and $G_R$ from \eqref{eq:general-G}, the set of isomorphism classes of all objects of $\mathcal C_{\s}^{0}$ is \begin{equation}\label{eq:global-moduli} \pi_0(\mathcal C_{\s}^{0})\cong \coprod_{[R]\in\mathcal R_{\s}^{0}} \ \coprod_{\mathbf d}X_{\mathbf d}(R)/G_R. \end{equation} \end{corollary} \begin{proof} Take isomorphism classes in Theorem~\ref{thm:global-moduli} and apply Theorem~\ref{thm:moduli} in each representation fiber. \end{proof} Joint weight multiplicities alone need not establish \eqref{eq:module-iso}: commuting operators can have different nilpotent parts on generalized weight spaces of the same dimension. When the $D_{j,\lambda}$ and the $D'_{j,\lambda}$ are each jointly diagonalizable, the dimensions of the ordinary joint weight spaces do suffice. When $m=0$, the condition reduces to $\dim M_\lambda=\dim M'_\lambda$ for each $\lambda$. \subsection{Wildness in class-two Lie algebra objects}\label{sec:wild} For any Lie algebra $\s$, let $\operatorname{LieAlg}_{\leq2}(\s\text{-}\mathrm{Rep})$ be the category of Lie algebra objects of nilpotence class at most two in $\s\text{-}\mathrm{Rep}$, with equivariant Lie homomorphisms. $\s=0$ is allowed. The following construction exhibits a matrix-pair problem inside these categories and compares it with the equivariant moduli above. The word \emph{wild} in this subsection refers to an explicit family of moduli problems, not to a fixed dimension. Put $\mathcal A=k\langle x,y\rangle\text{-}\mathrm{mod}_{\mathrm{fd}}$; an object is a triple $(V,T_1,T_2)$ with $T_i\in\End_k(V)$. Write $\operatorname{Core}(\mathcal A)$ for its isomorphism groupoid. The target category has all Lie homomorphisms as arrows. We give the block construction from \cite[Lemma~7 and Theorem~4]{BelitskiiLipyanskiSergeichuk2005} in functorial form. \Needspace{10\baselineskip} \begin{theorem}[The matrix-pair functor] \label{thm:wild-functor} There is a functor \[ \mathcal F_0:\operatorname{Core}(\mathcal A) \longrightarrow\operatorname{LieAlg}_{\leq2}(\mathrm{Vect}_k) \] such that for every $N>0$ and $\dim V=N$ the Lie algebra $\mathcal F_0(V,T_1,T_2)$ has dimension $350N+3$ and derived algebra of dimension three. \end{theorem} \begin{proof} The entire object construction below is the alternating-form construction of \cite[Lemma~7, formulas (22)--(24), and Lemma~6] {BelitskiiLipyanskiSergeichuk2005}, expressed without a choice of basis; its fixed blocks are part of that construction. We supply the arrow map to make its functoriality explicit. On $V$ set \[ E_V=V^{\oplus100}\oplus V^{\oplus50}\oplus V^{\oplus20}\oplus V\oplus V^{\oplus4},\qquad M_V=E_V\oplus E_V^*. \] Each $C_i\in\End_k(E_V)$ has five diagonal blocks in the displayed order of the summands of $E_V$; a zero denotes the zero endomorphism of its corresponding summand: \[ \begin{aligned} C_1&=\operatorname{diag}\bigl(\id_{V^{\oplus100}},0,0, \id_V,\id_{V^{\oplus4}}\bigr),\\ C_2&=\operatorname{diag}\bigl(0,\id_{V^{\oplus50}},0, \id_V,J_4(0)\otimes\id_V\bigr),\\ C_3&=\operatorname{diag}\bigl(0,0,\id_{V^{\oplus20}}, \id_V,\operatorname{diag}(\id_V,T_1,T_2,0)\bigr). \end{aligned} \] For $e,e'\in E_V$ and $\alpha,\alpha'\in E_V^*$ put \[ h_i\bigl((e,\alpha),(e',\alpha')\bigr) =\alpha'(C_i e)-\alpha(C_i e'),\qquad q_V=(h_1,h_2,h_3):\Lam^2M_V\longrightarrow k^3. \] The first three fixed blocks make $q_V$ surjective for $V\ne0$. Let $K_{T_1,T_2}=\ker q_V$ and give $\mathcal F_0(V,T_1,T_2)=M_V\oplus k^3$ the bracket $q_V$, with $k^3$ central. This has the asserted dimensions. Define $\mathcal F_0(0,0,0)=0$. For an intertwining isomorphism $S:V\to V'$, let $E(S):E_V\to E_{V'}$ be the direct sum of its indicated copies and put $M(S)=E(S)\oplus(E(S)^{-1})^*:M_V\to M_{V'}$. The identities $E(S)C_i=C'_iE(S)$ give $q_{V'}\circ\Lam^2M(S)=q_V$. Hence $M(S)\oplus\id_{k^3}$ is a Lie isomorphism; inverse duals and direct sums respect composition. This proves that $\mathcal F_0$ is a functor on the core. The inverse dual is the reason this arrow formula is stated only for isomorphisms of input triples. \end{proof} \begin{theorem}[Recovering a matrix pair from an output isomorphism] \label{thm:wild-recovery} For every Lie isomorphism $\phi:\mathcal F_0(V,T_1,T_2)\xrightarrow{\sim} \mathcal F_0(V',T'_1,T'_2)$ there exists an isomorphism of input modules $g_\phi:(V,T_1,T_2)\xrightarrow{\sim}(V',T'_1,T'_2)$. No equality $\mathcal F_0(g_\phi)=\phi$ or compatibility of $g_\phi$ with composition is asserted. Consequently $\mathcal F_0$ induces an injection on isomorphism classes of matrix pairs of every dimension. \end{theorem} \begin{proof} For nonzero inputs, $\phi$ induces an isomorphism $P:M_V\to M_{V'}$ on the quotients by their derived algebras and an isomorphism $\Gamma:k^3\to k^3$ on those derived algebras, with $q_{V'}\circ\Lam^2P=\Gamma\circ q_V$. Thus the three alternating forms above are related by simultaneous congruence and an arbitrary invertible substitution among the three forms. The fixed blocks of sizes $100N,50N,20N$ distinguish the three coordinate lines of the form space by rank. Indeed, the corresponding endomorphisms $C_1,C_2,C_3$ have ranks $105N$, $54N$, and $22N+\operatorname{rank}T_1+\operatorname{rank}T_2\leq24N$; mixing two or three coordinates puts the rank in a disjoint range. Thus the substitution is diagonal. The common $N$-block $(\id_V,\id_V,\id_V)$ then forces its three diagonal scalars to agree by the cancellation argument. After absorbing their common scalar, the remaining block $(I_{4N},J_4(0_N),\operatorname{diag}(I_N,T_1,T_2,0_N))$ recovers a simultaneous intertwiner $S:V\to V'$. These are the block comparison and cancellation steps of \cite[Theorem~1(a) and Lemma~7]{BelitskiiLipyanskiSergeichuk2005}; in particular they handle the freedom to change the basis of $k^3$. Taking $g_\phi=S$ proves the assertion. Different positive input dimensions have different output dimensions, while the zero input maps to zero, so the assertion also holds across dimensions. \end{proof} \Needspace{14\baselineskip} \begin{corollary}[The equivariant square] \label{thm:wild-equivariant-square} Let $\s$ be any finite-dimensional Lie algebra. Let $\mathcal A_{\s}$ have as objects triples $(V,T_1,T_2)$ with $V\in\s\text{-}\mathrm{Rep}$ and $T_i\in\End_{\s}(V)$, and as morphisms the $\s$-linear maps intertwining both $T_i$. The construction of Theorem~\ref{thm:wild-functor} defines a functor $\mathcal F_{\s}:\operatorname{Core}(\mathcal A_{\s})\to \operatorname{LieAlg}_{\leq2}(\s\text{-}\mathrm{Rep})$. With the vertical arrows forgetting the $\s$-action, the square \[ \begin{array}{ccc} \operatorname{Core}(\mathcal A_{\s}) &\xrightarrow{\quad\mathcal F_{\s}\quad}& \operatorname{LieAlg}_{\leq2}(\s\text{-}\mathrm{Rep})\\[4pt] \big\downarrow\scriptstyle\mathrm{forget}&& \big\downarrow\scriptstyle\mathrm{forget}\\[4pt] \operatorname{Core}(\mathcal A) &\xrightarrow{\quad\mathcal F_0\quad}& \operatorname{LieAlg}_{\leq2}(\mathrm{Vect}_k) \end{array} \] commutes up to the canonical identification of the underlying vector spaces and brackets. \end{corollary} \begin{proof} All the summands in $E_V$ inherit the $\s$-action from $V$, its dual $E_V^*$ has the dual action, and $k^3$ has the trivial action. Because $T_1,T_2$ intertwine $\s$, each $C_i$ does too. Evaluation $E_V^*\otimes E_V\to k$ is equivariant, so $q_V$ is an $\s$-morphism and its kernel is an $\s$-submodule. The bracket is therefore equivariant. If $S$ is $\s$-linear, so are $E(S)$ and its inverse dual, proving functoriality and the square. \end{proof} \begin{lemma}[Equivariant separation of the coding block] \label{lem:equivariant-coding-block} Let $\mathcal D_{\s}$ be the category of diagrams consisting of two finite-dimensional $\s$-modules and three $\s$-linear maps between them; morphisms are pairs of $\s$-linear maps commuting with the three arrows. For $T_1,T_2\in\End_{\s}(V)$ put \[ \mathcal Q(V,T_1,T_2) =\bigl(\id_{V^{\oplus4}},J_4(0)\otimes\id_V, \operatorname{diag}(\id_V,T_1,T_2,0)\bigr). \] The diagram obtained from the three alternating forms defining $\mathcal F_{\s}(V,T_1,T_2)$, viewed as maps $M_V\to M_V^*$, is isomorphic in $\mathcal D_{\s}$ to a direct sum of $\mathcal Q(V,T_1,T_2)$ and a diagram with no nonzero direct summand isomorphic to a direct summand of any $\mathcal Q(V',T'_1,T'_2)$. Consequently, an isomorphism between two such diagrams induces an isomorphism between their $\mathcal Q$ summands. \end{lemma} \begin{proof} The category $\mathcal D_{\s}$ is abelian, its objects have finite length, and its Hom spaces are finite-dimensional over $k$. Idempotents split, and the Fitting decomposition of a diagram endomorphism is a decomposition by $\s$-submodules. Hence Krull--Schmidt applies in this category; no semisimplicity of the $\s$-action is required. Reordering the target of $M_V\to M_V^*$ and changing the sign on one target summand identifies the forms with the direct sum of the triples $(C_1,C_2,C_3)$ on $E_V$ and $(C_1^*,C_2^*,C_3^*)$ on $E_V^*$. These are equivariant identifications, where $*$ denotes the linear dual with its contragredient action. The four fixed blocks, on copies of $V$ or $V^*$, have arrow triples \[ (\id,0,0),\quad(0,\id,0),\quad(0,0,\id),\quad (\id,\id,\id). \] The remaining blocks are $\mathcal Q(V,T_1,T_2)$ and, after reversing the four dual copies, \[ \mathcal Q^\vee(V,T_1,T_2) =\bigl(\id_{(V^*)^{\oplus4}},J_4(0)\otimes\id_{V^*}, \operatorname{diag}(0,T_2^*,T_1^*,\id_{V^*})\bigr). \] On a diagram whose first arrow is invertible, use it to identify the two vertex spaces and denote the other arrows by $J,D$. Every nonzero direct summand of $\mathcal Q$ has only size-four Jordan blocks for $J$: its underlying $J$-module is a direct summand of a sum of size-four nilpotent blocks. This separates it from the fixed blocks with first arrow invertible, where $J=0$ or $J=\id$. The other fixed blocks have first arrow zero and cannot share a nonzero summand with it. For both $\mathcal Q$ and $\mathcal Q^\vee$, $D$ preserves $\operatorname{im}J$. Its induced action on $\operatorname{coker}J$ is respectively zero and the identity. Every nonzero direct summand has nonzero $\operatorname{coker}J$, since $J$ is nilpotent. Thus these two kinds of coding blocks cannot share a nonzero summand either. All these obstructions already hold after forgetting the $\s$-action and hence also hold in $\mathcal D_{\s}$. Krull--Schmidt now separates the $\mathcal Q$ summands in any isomorphism of the full diagrams. \end{proof} \begin{corollary}[Equivariant recovery from an isomorphism] \label{cor:wild-equivariant-recovery} Under the assumptions of Corollary~\ref{thm:wild-equivariant-square}, if $\phi:\mathcal F_{\s}(V,T_1,T_2)\xrightarrow{\sim} \mathcal F_{\s}(V',T'_1,T'_2)$ is an $\s$-equivariant Lie isomorphism, the block recovery produces an $\s$-linear input isomorphism $g_\phi:V\xrightarrow{\sim}V'$ intertwining $T_1,T_2$. \end{corollary} \begin{proof} The zero-input case is immediate, so suppose $V,V'\ne0$. The induced maps $P:M_V\to M_{V'}$ and $\Gamma:k^3\to k^3$ satisfy $q_{V'}\Lam^2P=\Gamma q_V$, with $P$ equivariant. The ordinary rank and scalar-comparison argument in Theorem~\ref{thm:wild-recovery}, equivalently \cite[Lemma~7]{BelitskiiLipyanskiSergeichuk2005}, gives $\Gamma=\gamma\id$ for some $\gamma\ne0$. This conclusion concerns the original $\Gamma$; it does not require an equivariant choice of the changes of coordinates used to prove it. Writing $h_i^\sharp:M_V\to M_V^*$ for the three forms, we have \[ P^*h_i'^\sharp P=\gamma h_i^\sharp. \] Thus $P$ and $\gamma(P^*)^{-1}$ give an isomorphism of the three-arrow diagrams in $\mathcal D_{\s}$. Lemma~\ref{lem:equivariant-coding-block} gives an equivariant diagram isomorphism $\mathcal Q(V,T_1,T_2)\cong\mathcal Q(V',T'_1,T'_2)$. Its first-arrow equation identifies its two vertex maps with one invertible map $S:V^{\oplus4}\to(V')^{\oplus4}$. The equation $SJ=J'S$ makes it upper triangular Toeplitz: \[ S=\begin{pmatrix} S_0&S_1&S_2&S_3\\ 0&S_0&S_1&S_2\\ 0&0&S_0&S_1\\ 0&0&0&S_0 \end{pmatrix}. \] Each $S_i$ is $\s$-linear and $S_0$ is invertible. Comparing the second and third diagonal blocks in $SD=D'S$ gives $S_0T_i=T'_iS_0$ for $i=1,2$. Take $g_\phi=S_0$. \end{proof} \begin{proposition}[Orbit criterion for trivial-action objects] \label{prop:wild-multiplicity} Let $\s$ be any finite-dimensional Lie algebra. Give a finite-dimensional vector space $M$ the trivial $\s$-action, and let $K\subseteq\Lam^2M$ have codimension three. The object $M\oplus(\Lam^2M/K)$ of $\operatorname{LieAlg}_{\leq2}(\s\text{-}\mathrm{Rep})$ is determined up to isomorphism by the $\mathrm{GL}(M)$-orbit of $K$: for a second such pair $(M',K')$, the objects $M\oplus(\Lam^2M/K)$ and $M'\oplus(\Lam^2M'/K')$ are isomorphic if and only if some linear isomorphism $P:M\to M'$ satisfies $(\Lam^2P)K=K'$. \end{proposition} \begin{proof} The derived object is $\Lam^2M/K$ and its quotient in the class-two Lie algebra object is $M$. An isomorphism therefore induces $P:M\to M'$ with $(\Lam^2P)(K)=K'$, and such a $P$ induces an isomorphism. Since all the $\s$-actions are trivial, the same condition applies in $\s\text{-}\mathrm{Rep}$. \end{proof} \begin{corollary}[Matrix pairs in every equivariant category] \label{cor:wild-trivial-action} Let $\s$ be any finite-dimensional Lie algebra and let $(V,T_1,T_2)$ be a finite-dimensional $k\langle x,y\rangle$-module, where $T_i\in\End_k(V)$ are the actions of $x$ and $y$. Endowing $V$ with the trivial $\s$-action makes $(V,T_1,T_2)$ an object of $\mathcal A_{\s}$ in Corollary~\ref{thm:wild-equivariant-square}. For $V\ne0$, its image $\mathcal F_{\s}(V,T_1,T_2)$ has the form of Proposition~\ref{prop:wild-multiplicity} with $M=M_V$ and $K=K_{T_1,T_2}=\ker q_V$. These objects realize every matrix-pair isomorphism class of positive dimension, and distinct classes give nonisomorphic objects. \end{corollary} \begin{proof} The first claim is the object formula of Theorem~\ref{thm:wild-functor} applied through the equivariant square. Corollary~\ref{cor:wild-equivariant-recovery} lifts any isomorphism of outputs to one of input pairs. \end{proof} \begin{proposition}[Symmetric-module amplification] \label{prop:symmetric-amplification} Let $\s$ be any finite-dimensional Lie algebra and let $U$ be an $\s$-module with an invariant symmetric bilinear form $\beta:U\otimes U\to k$. For a class-two Lie algebra object $L$ in $\s\text{-}\mathrm{Rep}$, write $Q(L)=L/[L,L]$, $D(L)=[L,L]$, and let $b_L:\Lam^2Q(L)\twoheadrightarrow D(L)$ be its induced bracket. The formulas \[ \mathcal P_U(L)=(U\otimes Q(L))\oplus D(L),\qquad [u\otimes q,v\otimes q']_{\mathcal P_U(L)} =\beta(u,v)b_L(q\wedge q'), \] with $D(L)$ central, and $\mathcal P_U(f)=(\id_U\otimes Q(f))\oplus D(f)$ on a Lie homomorphism $f$, define an endofunctor $\mathcal P_U$ of $\operatorname{LieAlg}_{\leq2}(\s\text{-}\mathrm{Rep})$. \end{proposition} \begin{proof} The induced bracket $b_L$ is well-defined because $D(L)$ is central. Symmetry of $\beta$ makes the displayed bracket alternating, and invariance of $\beta$ and equivariance of $b_L$ make it an $\s$-morphism. Its image is central, so Jacobi holds. Every Lie homomorphism $f$ induces $Q(f)$ and $D(f)$ with $D(f)b_L=b_{L'}\Lam^2Q(f)$; multiplying this identity by $\beta$ proves preservation of the bracket. These induced maps respect identities and composition, proving functoriality. \end{proof} \begin{corollary}[Isomorphism classes under amplification] \label{cor:amplification-injective} In Proposition~\ref{prop:symmetric-amplification}, suppose $\beta\ne0$ and $U$ is absolutely irreducible. For objects $L,L'$ with trivial $\s$-action, every isomorphism $\phi:\mathcal P_U(L)\xrightarrow{\sim}\mathcal P_U(L')$ yields an isomorphism $g_\phi:L\xrightarrow{\sim}L'$. No equality $\mathcal P_U(g_\phi)=\phi$ is asserted. Thus $\mathcal P_U$ induces an injection on isomorphism classes of objects with trivial $\s$-action. \end{corollary} \begin{proof} Suppose $L,L'$ have trivial $\s$-action and let $\phi:\mathcal P_U(L)\xrightarrow{\sim}\mathcal P_U(L')$ be arbitrary. Since $\beta\ne0$, their derived objects are $D(L)$ and $D(L')$. The induced map $U\otimes Q(L)\to U\otimes Q(L')$ is $\id_U\otimes P$ for a linear isomorphism $P$, by absolute irreducibility. The induced map $H:D(L)\to D(L')$ then satisfies $H b_L=b_{L'}\Lam^2P$, by evaluating the bracket on a pair $u,v$ with $\beta(u,v)\ne0$. Linear splittings of the quotient maps $L\to Q(L)$ and $L'\to Q(L')$ identify each object with its quotient-plus-derived presentation, so $(P,H)$ gives an isomorphism $g_\phi:L\xrightarrow{\sim}L'$. Its image under $\mathcal P_U$ need not be $\phi$. The forward implication on isomorphism classes follows from functoriality. \end{proof} \begin{corollary}[A matrix-pair family of perfect Lie algebras] \label{cor:wild-perfect-family} Let $\s$ be semisimple, and let $V_\lambda$ be a nontrivial irreducible $\s$-module with a nondegenerate invariant symmetric form $\beta$. Let the subscript $\mathrm{triv}$ denote the full subgroupoid or subcategory on objects with trivial $\s$-action. The trivial-action functors in the first row of vertical arrows are equivalences, and the second row consists of inclusions: \[ \begin{array}{ccc} \operatorname{Core}(\mathcal A) &\xrightarrow{\quad\mathcal F_0\quad}& \operatorname{LieAlg}_{\leq2}(\mathrm{Vect}_k)\\[5pt] \big\downarrow\scriptstyle\mathrm{triv}&& \big\downarrow\scriptstyle\mathrm{triv}\\[5pt] \operatorname{Core}(\mathcal A_{\s})_{\mathrm{triv}} &\xrightarrow{\quad\mathcal F_{\s}|_{\mathrm{triv}}\quad}& \operatorname{LieAlg}_{\leq2}(\s\text{-}\mathrm{Rep})_{\mathrm{triv}}\\[5pt] \big\downarrow\scriptstyle\mathrm{incl}&& \big\downarrow\scriptstyle\mathrm{incl}\\[5pt] \operatorname{Core}(\mathcal A_{\s}) &\xrightarrow{\quad\mathcal F_{\s}\quad}& \operatorname{LieAlg}_{\leq2}(\s\text{-}\mathrm{Rep}). \end{array} \] Both squares commute. For every $k\langle x,y\rangle$-module $(V,T_1,T_2)$, follow the left vertical arrows to give $V$ the trivial $\s$-action, apply $\mathcal F_{\s}$ and then $\mathcal P_{V_\lambda}$, and put \[ \g_{V,T_1,T_2} =\s\ltimes\mathcal P_{V_\lambda}(\mathcal F_{\s}(V,T_1,T_2)). \] These Lie algebras are perfect. Two of them are isomorphic by an isomorphism fixing $\s$ pointwise if and only if their $k\langle x,y\rangle$-modules are isomorphic. For simple $\s$, its adjoint module with the Killing form is one possible $V_\lambda$. \end{corollary} \begin{proof} Giving a vector space the trivial $\s$-action identifies the top row with the middle row, and restricting $\mathcal F_{\s}$ to trivial-action inputs gives the indicated trivial-action outputs. Thus the diagram commutes. For any $L$ with trivial $\s$-action, $(V_\lambda\otimes Q(L))^{\s}=0$. Proposition~\ref{prop:perfect} shows that $\s\ltimes\mathcal P_{V_\lambda}(L)$ is perfect. Its radical is $\mathcal P_{V_\lambda}(L)$; hence an isomorphism fixing $\s$ restricts to an isomorphism of these Lie algebra objects. Corollary~\ref{cor:amplification-injective} and Corollary~\ref{cor:wild-equivariant-recovery} recover an isomorphism of the input matrix-pair modules. Conversely, an input isomorphism induces one of the displayed semidirect products. \end{proof} \section{Explicit slices and low-dimensional components}\label{sec:slices} This section computes the groupoids of Sections~\ref{sec:isomorphisms} and~\ref{sec:moduli} in explicit cases: a single isotypic component (Sections~\ref{sec:isotypic}--\ref{sec:binary}), the first ternary multiplicity space (Section~\ref{sec:ternary}), and all objects of total dimension at most six for semisimple $\s$, including $\s=0$ (Section~\ref{sec:lowdim}). Throughout this section $V_\lambda$ denotes the irreducible module of highest weight $\lambda$ for the displayed semisimple algebra. For a named simple factor, $\omega_i$ is its $i$th fundamental weight in the usual Dynkin numbering. An external tensor product $\boxtimes$ records the specified order of simple factors; their permutations are not arrows fixing the acting algebra pointwise. \subsection{One isotypic component with arbitrary multiplicity}\label{sec:isotypic} The reduction to a single isotypic component is independent of $\mathfrak{sl}_2$. The following statement records exactly what that reduction gives for an arbitrary irreducible type. \begin{theorem}[Parameter spaces for one isotypic component] \label{thm:single-type} Let $\s=\sderived\oplus kt_1\oplus\cdots\oplus kt_m$, and let $V=V_\lambda$ be an irreducible $\sderived$-module. Put $R=V\otimes M$, where the $t_j$ act by commuting $D_j$ on $M$. Assume $R$ belongs to $\mathcal C_\s^0$ (automatically so when $V$ is nontrivial). For each irreducible $V_\nu$ set \[ A_\nu=\Hom_{\sderived}(V_\nu,\Lam^2V),\quad B_\nu=\Hom_{\sderived}(V_\nu,\Sym^2V),\quad N_\nu=(A_\nu\otimes\Sym^2M)\oplus (B_\nu\otimes\Lam^2M). \] Let $E_{j,\nu}$ act as the restriction of $D_j\otimes\id+\id\otimes D_j$ on each multiplicity factor. The possible $K$ are exactly $K_L=\bigoplus_\nu V_\nu\otimes L_\nu$, with $L_\nu\subseteq N_\nu$ invariant under every $E_{j,\nu}$. Their fixed-$R$ base groupoid is the action groupoid of \[ G_R=\{g\in GL(M):gD_j=D_jg\text{ for every }j\} \] on these tuples, with action $\id_{A_\nu}\otimes\Sym^2g\oplus \id_{B_\nu}\otimes\Lam^2g$ on $N_\nu$. \end{theorem} \begin{proof} The exterior-square identity gives $\Lam^2(V\otimes M)=(\Lam^2V\otimes\Sym^2M) \oplus(\Sym^2V\otimes\Lam^2M)$. Decomposing its two $\sderived$-factors yields the $N_\nu$; stability under $t_j$ is precisely stability under $E_{j,\nu}$. Schur's lemma gives $\Aut_\s(R)=G_R$ and the displayed action on $N_\nu$. Theorem~\ref{thm:groupoid} supplies the action groupoid. \end{proof} \begin{corollary}[Self-loops for one isotypic component] \label{cor:single-type-loops} Under the assumptions and notation of Theorem~\ref{thm:single-type}, put $U_L=N_\lambda/L_\lambda$ (zero if $N_\lambda=0$), with $t_j$ acting by the induced $E_{j,\lambda}$. The full self-loop group is \[ \Hom_{k[t_1,\ldots,t_m]}(M,U_L)\rtimes G_{K_L}, \qquad G_{K_L}=\{g\in G_R:gL_\nu=L_\nu\text{ for all }\nu\}. \] \end{corollary} \begin{proof} Only the $V_\lambda$-isotypic quotient of $W_L=\Lam^2R/K_L$ receives an $\sderived$-map from $R$. Intertwining the $t_j$ then identifies the shear maps $R\to W_L$ with $\Hom_{k[t_1,\ldots,t_m]}(M,U_L)$. Proposition~\ref{prop:groupoid-composition} gives the split self-loop group. \end{proof} \subsection{Changing the semisimple algebra in a multiplicity-free slice}\label{sec:change-ss} The labels of the simple summands in $V\otimes V$ belong to the representation theory of $\sderived$. An equivalence with a slice for a different semisimple algebra therefore needs a \emph{chosen} matching of those labels. The numbers of summands of each flip parity determine whether such a matching exists, but do not specify one. \begin{lemma}[The signed tensor-square data] \label{lem:signed-square} Let $V=V_\lambda$ be a simple module for a semisimple algebra $\mathfrak a$, and suppose $V\otimes V$ is multiplicity-free. Denote its set of constituent highest weights by $I_\lambda$. The tensor flip acts on the constituent $V_\alpha$ by a sign $\epsilon_\alpha\in\{+,-\}$. Put \[ I_\lambda^+=\{\alpha:\epsilon_\alpha=+\},\quad I_\lambda^-=\{\alpha:\epsilon_\alpha=-\},\qquad r_+(\lambda)=|I_\lambda^+|,\quad r_-(\lambda)=|I_\lambda^-|. \] For a fixed multiplicity module $(M,D_1,\ldots,D_m)$, the subrepresentation parameters in $\Lam^2(V\otimes M)$ are one $D_j^{\{2\}}$-invariant subspace of $\Lam^2M$ for every $\alpha\in I_\lambda^+$ and one invariant subspace of $\Sym^2M$ for every $\alpha\in I_\lambda^-$. The same group $G=\{g\in GL(M):gD_j=D_jg\ \forall j\}$ acts diagonally on all these choices, by $\Lam^2g$ and $\Sym^2g$, respectively. In particular, writing $\mathcal I(T)$ for the set of subspaces of $T$ invariant under all the restricted $D_j^{\{2\}}$, the fixed-type base groupoid is \begin{equation}\label{eq:signed-standard-model} \left(\prod_{\alpha\in I_\lambda^-}\mathcal I(\Sym^2M) \times\prod_{\alpha\in I_\lambda^+}\mathcal I(\Lam^2M)\right) //G. \end{equation} \end{lemma} \begin{proof} The flip commutes with $\mathfrak a$, so Schur's lemma makes it a scalar on each multiplicity-one constituent; its square is one. The Cauchy identity $\Lam^2(V\otimes M)=(\Lam^2V\otimes\Sym^2M) \oplus(\Sym^2V\otimes\Lam^2M)$ proves the assertion, including the interchange of symmetric and exterior powers of $M$. The central and $G$ actions are those of Theorem~\ref{thm:single-type}. \end{proof} \begin{theorem}[Equivalence across semisimple types] \label{thm:signed-groupoid-equivalence} Let $\s=\mathfrak a\oplus\mathfrak z$ and $\s'=\mathfrak a'\oplus\mathfrak z'$ be reductive, with a chosen identification of their $m$-dimensional centers. Let $V_{\lambda,\mathfrak a}$ and $V_{\mu,\mathfrak a'}$ be simple modules whose tensor squares are each multiplicity-free in their respective representation categories. Let $\mathcal C^0_{\s,\lambda}$ be the full subgroupoid of $\mathcal C^0_{\s}$ on objects $(R,K)$ for which $R\cong_{\s}V_{\lambda,\mathfrak a}\otimes M$ for some finite-dimensional $k[t_1,\ldots,t_m]$-module $M$, subject to the object condition $\lambda\ne0$ or $M_{\mathbf0}=0$, where \[ M_{\mathbf0}=\{u\in M:\text{there is an }e\geq1\text{ such that } t_j^e u=0\text{ for every }j\}. \] This is the space $M_{0,\mathbf0_m}$ of Section~\ref{sec:vanishing} when $\lambda=0$. Define $\mathcal C^0_{\s',\mu}$ likewise, using $\mu$ and $\mathfrak a'$. Choose two bijections of constituent highest weights \begin{equation}\label{eq:signed-bijection} \theta_+:I_\lambda^+\xrightarrow{\sim}I_\mu^+, \qquad \theta_-:I_\lambda^-\xrightarrow{\sim}I_\mu^-, \end{equation} and require the pointed-subset equations \begin{equation}\label{eq:signed-pointed} \theta_\pm(I_\lambda^\pm\cap\{\lambda\}) =I_\mu^\pm\cap\{\mu\}. \end{equation} Then reindexing the $K$-components by $\theta_+$ and $\theta_-$ gives an equivalence of the \emph{full} groupoids \[ \mathcal C^0_{\s,\lambda} \simeq\mathcal C^0_{\s',\mu}. \] The equivalence retains the multiplicity module $M$ on objects and the intertwining isomorphism $f:M\to M'$ on arrows; both modules vary over the allowed representations. Its construction uses the two chosen bijections. The numerical equalities \begin{equation}\label{eq:signed-counts} (r_+(\lambda),r_-(\lambda))=(r_+(\mu),r_-(\mu)) \end{equation} say exactly when bijections \eqref{eq:signed-bijection} exist; \eqref{eq:signed-pointed} is the additional condition on the chosen bijections for this full-groupoid construction. \end{theorem} \begin{proof} Since $\mathfrak a$ is semisimple, $\lambda=0$ exactly when $V_{\lambda,\mathfrak a}$ is one-dimensional, equivalently $I_\lambda^-=\varnothing$; the same holds for $\mu$. The bijection $\theta_-$ therefore implies $\lambda=0$ if and only if $\mu=0$. Thus the two full subgroupoids admit precisely the same multiplicity modules: all finite-dimensional modules when the weights are nonzero, and those with $M_{\mathbf0}=0$ when they are zero. For any object, the multiplicity module $M=\Hom_{\mathfrak a}(V_{\lambda,\mathfrak a},R)$ has its induced $k[t_1,\ldots,t_m]$-action, and evaluation identifies $R$ with $V_{\lambda,\mathfrak a}\otimes M$. Multiplicity-freeness makes every constituent multiplicity space in $V_{\lambda,\mathfrak a}^{\otimes2}$ and $V_{\mu,\mathfrak a'}^{\otimes2}$ a line. Choose an isomorphism of each such line with $k$. For each allowed $M$ this presents the $K$-parameters as in \eqref{eq:signed-standard-model}. A plus slot uses $\Lam^2M$, and a minus slot uses $\Sym^2M$. The two bijections reindex these slots for every $M$. An arrow between objects over $M$ and $M'$ has an underlying $k[t_1,\ldots,t_m]$-module isomorphism $f:M\to M'$. The maps $\Lam^2f$ and $\Sym^2f$ act in the corresponding slots on both sides, so the reindexing carries base arrows bijectively to base arrows and respects their composition. If the input weight $\lambda$ belongs to a sign set, the shear fiber at $(M,K)$ is \begin{equation}\label{eq:signed-shears} \Hom_{k[t_1,\ldots,t_m]}(M,T_\pm/L_\lambda), \qquad T_+=\Lam^2M,\quad T_-=\Sym^2M, \end{equation} with the sign of that set; otherwise the fiber is zero. Equation~\eqref{eq:signed-pointed} makes the corresponding fiber on the $\mu$ side the same quotient construction, including the induced $t_j$ actions. It also matches transport of a shear along $f$, namely $H\mapsto\bar fHf^{-1}$. The composition law \eqref{eq:groupoid-composition} therefore agrees for every arrow, giving a fully faithful, essentially surjective functor of the full groupoids. The line identifications specify this functor. \end{proof} \begin{proposition}[A rectangular $\mathfrak{sl}_2$ comparison] \label{prop:rectangular-comparison} Let $V_a\boxtimes V_b$ be the simple module for $\mathfrak{sl}_2\oplus\mathfrak{sl}_2$, where $a,b\geq0$, and put $h=(a+1)(b+1)-1$. The simple constituents of its tensor square are indexed by $(i,j)$ with $0\leq i\leq a$, $0\leq j\leq b$: \[ V_{2a-2i}\boxtimes V_{2b-2j},\qquad \epsilon_{i,j}=(-1)^{i+j}. \] Those of $V_h\otimes V_h$ for one $\mathfrak{sl}_2$ are $V_{2h-2r}$, $0\leq r\leq h$, with flip sign $(-1)^r$. The \emph{chosen} bijection \begin{equation}\label{eq:serpentine-bijection} r(i,j)=i(b+1)+ \begin{cases}j,&i\text{ even},\\ b-j,&i\text{ odd} \end{cases} \end{equation} preserves these signs. It also matches the input constituents: they are absent on both sides unless $a,b$ are both even; in that case $(a/2,b/2)$ maps to $r=h/2$. Thus \eqref{eq:serpentine-bijection} yields an equivalence of the \emph{full} groupoids $\mathcal C^0_{\s,(a,b)}$ and $\mathcal C^0_{\s',h}$ of Theorem~\ref{thm:signed-groupoid-equivalence}, where $\s=(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)\oplus\mathfrak z$ and $\s'=\mathfrak{sl}_2\oplus\mathfrak z'$ have identified centers. \end{proposition} \begin{proof} Clebsch--Gordan gives the displayed rectangular index set. Interchanging the two copies of $V_a\boxtimes V_b$ interchanges the two $V_a$ factors and the two $V_b$ factors, so the signs on the two Clebsch--Gordan summands are $(-1)^i$ and $(-1)^j$, respectively, so their product is $(-1)^{i+j}$. Indeed the summand indexed by $i$ is obtained by $i$ contractions with the alternating form on the standard two-dimensional module; exchanging the tensor factors changes the sign of each contraction. Equation~\eqref{eq:serpentine-bijection} lists the $b+1$ entries in each successive row alternately forward and backward. Its values are exactly $0,\ldots,(a+1)(b+1)-1$, without repetition; in either kind of row $r(i,j)\equiv i+j\pmod2$. An input constituent requires $2a-2i=a$ and $2b-2j=b$, so it occurs exactly when both $a,b$ are even, at $(i,j)=(a/2,b/2)$. Then $h$ is even and direct substitution gives $r(a/2,b/2)=h/2$, the input index on the other side. If at least one of $a,b$ is odd, $h$ is odd and neither input type occurs. Theorem~\ref{thm:signed-groupoid-equivalence} now applies, including its shear identification. \end{proof} For instance $a=b=1$ gives $h=3$ and the ordered indices $(0,0),(0,1),(1,1),(1,0)$ with signs $+,-,+,-$. For $a=b=2$ one has $h=8$ and the input index $(1,1)\mapsto4$, illustrating the pointed condition when shears can occur. \begin{remark}[Counts can miss the shear constituent] \label{rem:signed-counts-shears} Take no central factors and $M=k$. For $\mathfrak{sl}_2$, the module $V_3$ has $\Sym^2V_3=V_6\oplus V_2$ and $\Lam^2V_3=V_4\oplus V_0$; $V_3$ is absent. For the seven-dimensional simple $G_2$-module $V_{\omega_1}$, one has $\Sym^2V_{\omega_1}=V_{2\omega_1}\oplus V_0$ and $\Lam^2V_{\omega_1}=V_{\omega_2}\oplus V_{\omega_1}$. Both signed counts are $(2,2)$, so their base groupoids are equivalent after a choice of signed bijection. At $K=0$, however, the first full self-loop group is $k^\times$, whereas the second contains the nonzero shear group $\Hom_{\mathfrak g_2}(V_{\omega_1},V_{\omega_1})=k$ and is $k\rtimes k^\times$. In particular these full groupoids are not equivalent: all self-loop groups in the first slice are abelian, while the second slice has a nonabelian one. \end{remark} \begin{theorem}[Stembridge: the tensor-square condition] \label{thm:stembridge-squares} Let $\mathfrak a$ be simple and $\lambda$ a dominant weight. If $V_\lambda\otimes V_\lambda$ is multiplicity-free, then $\lambda=0$ or $\lambda=d\omega_i$ for one fundamental weight $\omega_i$ and an integer $d\geq1$. In type $A_n$ this condition is also sufficient: the complete list is $\lambda=0$ and $\lambda=d\omega_i$, where $1\leq i\leq n$ and $d\geq1$. For the remaining simple types the admissible pairs $(i,d)$ are subject to the type and node restrictions in Stembridge's classification; the condition $\lambda=d\omega_i$ alone is not sufficient in general. \end{theorem} \begin{proof} This is the tensor-square specialization of \cite[Theorem~1.1]{Stembridge2003}. \end{proof} \begin{corollary}[Semisimple tensor squares] \label{cor:semisimple-squares} Let $\mathfrak a=\bigoplus_{h=1}^q\mathfrak a_h$ be semisimple, where the $\mathfrak a_h$ are simple, and write $V_\lambda=\boxtimes_{h=1}^q V_{\lambda_h}$. Then $V_\lambda\otimes V_\lambda$ is multiplicity-free if and only if every $V_{\lambda_h}\otimes V_{\lambda_h}$ is multiplicity-free. This includes $q=0$, for which $V_\lambda=k$. \end{corollary} \begin{proof} There is an isomorphism \( V_\lambda\otimes V_\lambda \cong\boxtimes_{h=1}^q (V_{\lambda_h}\otimes V_{\lambda_h}). \) The simple summands of an external tensor product are indexed by tuples of simple summands of its factors, and their multiplicities are the products of the corresponding multiplicities. Each factor has at least one summand, so a multiplicity exceeds one in the product exactly when one does in a factor. The empty product is the one-dimensional trivial module. \end{proof} \subsection{The explicit $\mathfrak{sl}_2$ tensor squares}\label{sec:sl2} Let $\sderived=\mathfrak{sl}_2$ and retain the $m$ central generators $t_j$. Fix $n\geq1$ and an $\ell$-dimensional multiplicity space $M$ carrying arbitrary commuting operators $D_1,\ldots,D_m$; thus $R=V_n\otimes M$, with $V_n=\Sym^n V_1$ for the standard two-dimensional $\mathfrak{sl}_2$-module $V_1$. The integer $n$ denotes the highest weight, leaving $m$ as the number of central generators. This is a single $\mathfrak{sl}_2$-isotypic component, with no restriction on $\ell$ or on the Jordan parts of the $D_j$. Since $n\geq1$, it satisfies $M_{0,\mathbf0_m}=0$. Write $C_\nu=\Hom_{\mathfrak{sl}_2}(V_\nu,V_n\otimes V_n)$. \begin{lemma}[Clebsch--Gordan reduction for $V_n\otimes M$] \label{lem:sl2-fiber} For $0\leq i\leq n$ put $\nu_i=2n-2i$. The complete list of Clebsch--Gordan multiplicities and flip eigenspaces is \begin{equation}\label{eq:sl2-CAB} \dim C_{\nu_i}=1,\qquad (\dim A_{\nu_i,n},\dim B_{\nu_i,n})= \begin{cases}(1,0)&i\text{ odd},\\(0,1)&i\text{ even},\end{cases} \end{equation} with all three spaces zero for other $\nu$. In particular, \begin{equation}\label{eq:sl2-N} N_{\nu_i}=\begin{cases} \Sym^2M,&i\text{ odd},\\ \Lam^2M,&i\text{ even}, \end{cases} \qquad \dim N_{\nu_i}=\begin{cases} \binom{\ell+1}{2},&i\text{ odd},\\ \binom{\ell}{2},&i\text{ even}. \end{cases} \end{equation} \end{lemma} \begin{proof} Realize $V_n\otimes V_n$ as polynomials homogeneous of degree $n$ in each of $(x_1,y_1)$ and $(x_2,y_2)$. The vectors \[ (x_1y_2-y_1x_2)^i x_1^{n-i}x_2^{n-i} \qquad(0\leq i\leq n) \] are highest-weight vectors of distinct weights $2n-2i$. Their irreducible summands have total dimension $\sum_{i=0}^n(2n-2i+1)=(n+1)^2$, so they exhaust $V_n\otimes V_n$. The tensor flip sends the $i$th vector to $(-1)^i$ times itself, proving \eqref{eq:sl2-CAB}. The wedge identity then gives \eqref{eq:sl2-N}. \end{proof} \begin{theorem}[Fixed-$R$ groupoid for one $\mathfrak{sl}_2$ type] \label{thm:sl2-groupoid} Write $D_j^{\{2\}}=D_j\otimes\id_M+\id_M\otimes D_j$. Let $\mathcal P(R)$ be the set of all tuples $(Q_0,\ldots,Q_n)$, where $Q_i$ is a $D_j^{\{2\}}$-invariant subspace of the space $N_{\nu_i}$ in \eqref{eq:sl2-N} for every $j$. Equivalently, $\mathcal P(R)$ is the product, over $i$, of the unions of the corresponding invariant Grassmannians of all dimensions; by \eqref{eq:sl2-K} below it is identified with the set $\mathcal K(R)$ of Theorem~\ref{thm:global-moduli}. Then every $K$ is uniquely \begin{equation}\label{eq:sl2-K} K_Q=\bigoplus_{i=0}^{n}V_{\nu_i}\otimes Q_i, \qquad W_Q=\bigoplus_{i=0}^{n}V_{\nu_i}\otimes(N_{\nu_i}/Q_i). \end{equation} The group and its action on these tuples are completely explicit: \begin{equation}\label{eq:sl2-G} G_R=\{g\in GL(M):gD_j=D_jg\text{ for all }j\},\qquad gQ_i=\begin{cases} (\Sym^2g)Q_i,&i\text{ odd},\\ (\Lam^2g)Q_i,&i\text{ even}. \end{cases} \end{equation} Thus the full fixed-$R$ base groupoid is $\mathcal P(R)//G_R$, its orbit set is $\mathcal P(R)/G_R$, and $G_{K_Q}=\{g\in G_R:gQ_i=Q_i\text{ for every }i\}$. \end{theorem} \begin{proof} The $t_j$ act on the spaces in \eqref{eq:sl2-N} by the restrictions of $D_j^{\{2\}}$. Since the $V_{\nu_i}$ are pairwise distinct, their submodules are precisely \eqref{eq:sl2-K} with the stated invariance conditions. The commutant of the $\mathfrak{sl}_2$-action on $R$ is $GL(M)$; requiring commutation with all $t_j$ gives \eqref{eq:sl2-G}. Theorem~\ref{thm:groupoid} identifies the base arrows and orbits. \end{proof} \begin{proposition}[Shears for one $\mathfrak{sl}_2$ type] \label{prop:sl2-shears} The shear group is also determined by these data. If $n$ is odd, $\Hom_{\s}(R,W_Q)=0$. If $n$ is even, set $i_*=n/2$ and let $U_Q=N_{\nu_{i_*}}/Q_{i_*}$, equipped with the induced $D_j^{\{2\}}$-actions. Then \begin{equation}\label{eq:sl2-H} \Hom_{\s}(R,W_Q) \cong\Hom_{k[t_1,\ldots,t_m]}(M,U_Q), \end{equation} where $t_j$ acts by $D_j$ on $M$ and by the induced $D_j^{\{2\}}$ on $U_Q$. At each tuple $Q$ the full self-loop group is the semidirect product of this additive shear group with $G_{K_Q}$; $g$ acts on a shear $H$ by $H\mapsto\bar g H g^{-1}$, where $\bar g$ is induced by $\Sym^2g$ or $\Lam^2g$ on $U_Q$. In particular the shear dimension is the nullity of the explicit linear map \[ \Hom_k(M,U_Q)\longrightarrow \bigoplus_{j=1}^{m}\Hom_k(M,U_Q),\qquad F\longmapsto(FD_j-\widehat D_jF)_j, \] where $\widehat D_j$ is the induced action on $U_Q$. These data present the local groupoid as $\mathcal H_R\rtimes(\mathcal P(R)//G_R)$ for every $n,\ell$ and every commuting tuple $(D_j)$; a concrete orbit calculation requires further analysis of this action. \end{proposition} \begin{proof} A $V_n$ constituent occurs in $W_Q$ only if $2n-2i=n$, or $i=n/2$, which requires $n$ even. Its multiplicity module is $U_Q$, and the remaining intertwining condition is exactly \eqref{eq:sl2-H}. The action on shears is that of Theorem~\ref{thm:groupoid}, completing the description. \end{proof} \subsection{Two-dimensional multiplicity spaces}\label{sec:binary} Throughout this subsection keep the setting of Section~\ref{sec:sl2} with $\ell=2$, and write $M_2$ for the two-dimensional multiplicity space, so that $R=V_n\otimes M_2$. The subscript records the dimension, not a highest weight as in $M_\lambda$. \begin{lemma}[Binary exterior-square choices]\label{lem:binary-choices} Suppose all $D_j=\mu_j\id_{M_2}$. For $R=V_n\otimes M_2$, each even index $i\in\{0,\ldots,n\}$ has $N_{\nu_i}=\Lam^2M_2$, a one-dimensional space. Its subspace $Q_i$ is independently either $0$ or $\Lam^2M_2$. Thus every orbit configuration of the odd-indexed $Q_i$ occurs in $2^{\lfloor n/2\rfloor+1}$ distinct copies. These choices do not affect its stabilizer in $G_R=GL(M_2)$. For $n=1,2,3$, they do not affect the full self-loop group either. \end{lemma} \begin{proof} Lemma~\ref{lem:sl2-fiber} gives one $\Lam^2M_2$ at each even index. Both its subspaces are stable under the scalar $D_j^{\{2\}}$ and fixed by every $g\in GL(M_2)$. There are $\lfloor n/2\rfloor+1$ even indices. For $n=1,3$ the shear group vanishes; for $n=2$ it depends only on $N_2/Q_1=\Sym^2M_2/Q_1$. Proposition~\ref{prop:groupoid-composition}, with the shear groups of Proposition~\ref{prop:sl2-shears}, gives the claim for full loops. \end{proof} \begin{lemma}[The intrinsic pole of a plane] \label{lem:intrinsic-pole} For the two-dimensional space $M_2$, the discriminant is a canonical quadratic map $\Delta:\Sym^2M_2\longrightarrow(\det M_2)^{\otimes2}$. Use the polarization \[ \Delta(p,q)=\Delta(p+q)-\Delta(p)-\Delta(q), \qquad \Delta(p,p)=2\Delta(p). \] It induces an isomorphism \[ \psi:\Sym^2M_2\xrightarrow{\sim} (\Sym^2M_2)^*\otimes(\det M_2)^{\otimes2}, \qquad \psi(p)(q)=\Delta(p,q). \] For a plane $U=\ker f\subset\Sym^2M_2$, where $0\ne f\in(\Sym^2M_2)^*$, define its \emph{pole line} intrinsically by \[ \varpi_U=\psi^{-1}\bigl(kf\otimes(\det M_2)^{\otimes2}\bigr) \subset\Sym^2M_2. \] It is independent of the choice of $f$ defining $U$. If $0\ne p_f\in\varpi_U$, then $U=\{q:\Delta(p_f,q)=0\}$, and $\varpi_{gU}=g\varpi_U$ for every $g\in GL(M_2)$. Neither an inner product nor a trivialization of $\det M_2$ is involved. \end{lemma} \begin{proof} For $p\in\Sym^2M_2$ let $\widetilde p:M_2^*\to M_2$ be its associated symmetric linear map. The determinant of this map belongs canonically to $\Hom(\det M_2^*,\det M_2)=(\det M_2)^{\otimes2}$; put $\Delta(p)=-4\det\widetilde p$. Choose any basis $x,y$ of $M_2$ with dual $e_1,e_2$. If $p=ax^2+bxy+cy^2$ and $\delta=(x\wedge y)^2$, then \[ \Delta(p)=(b^2-4ac)\delta,\qquad \psi(p)=(-4c e_1^2+4b e_1e_2-4a e_2^2)\otimes\delta. \] Here the natural symmetric-tensor pairing satisfies $e_1e_2(xy)=1/2$. The displayed formula shows that $\psi$ is invertible. Since $(\det M_2)^{\otimes2}$ is one-dimensional, $kf\otimes(\det M_2)^{\otimes2}$ is a line; its inverse image is $\varpi_U$. The equation $\psi(p_f)\in kf\otimes(\det M_2)^{\otimes2}$ gives $\ker f=\ker \Delta(p_f,-)$ without identifying the target line with $k$. Finally $\Delta(gp)=(\det g)^2\Delta(p)$, so its polarization and the construction of $\varpi_U$ are $GL(M_2)$-equivariant. \end{proof} \begin{lemma}[Divisors and the discriminant parameter] \label{lem:divisor-parameter} Each line $kp\subseteq\Sym^2M_2$ has the effective zero divisor of $p$ on $\mathbb P(M_2^*)$. A plane $U$ has the zero divisor of any $p\in\varpi_U\setminus\{0\}$, with $\varpi_U$ as in Lemma~\ref{lem:intrinsic-pole}. Thus both kinds of subspaces give degree-two divisors on the same projective line. For two divisors with distinct roots, represented by nonzero $p,q\in\Sym^2M_2$, the scalar \begin{equation}\label{eq:family-J} \eta([p],[q])=\frac{\Delta(p,q)^2}{\Delta(p)\Delta(q)}\in k \end{equation} does not depend on their representatives or on an ordering of their roots. Equal divisors and divisors sharing exactly one root both have $\eta=4$; they are separate incidence types. The divisors are disjoint exactly when $\eta\ne4$. On the disjoint locus, $\eta$ is a complete invariant of the ordered pair under $PGL(M_2)$; $\eta=0$ is the harmonic locus. \end{lemma} \begin{proof} The pole construction is equivariant by Lemma~\ref{lem:intrinsic-pole}. Scaling $p$ and $q$ cancels from the ratio, and numerator and denominator both lie in $(\det M_2)^{\otimes4}$, so their quotient is a scalar without any trivialization. The expression is symmetric in $p,q$. For the incidence claims, use an arbitrary basis to put $p=xy$, $q=ax^2+bxy+cy^2$. Then, suppressing the common determinant factors only in this calculation, \[ \Delta(p)=1,\quad \Delta(p,q)=2b,\quad \Delta(q)=b^2-4ac, \quad 4\Delta(p)\Delta(q)-\Delta(p,q)^2=-16ac. \] The roots of $p$ are $[x=0]$ and $[y=0]$; $q$ vanishes at one of them exactly when $ac=0$. For distinct-root $q$, equality of divisors is the case $a=c=0$. When $ac\ne0$, the span of $p,q$ is nondegenerate for $\Delta$. Two indexed pairs with the same $\eta$ can be rescaled to have the same Gram matrix, and an isometry between their spans extends to $\Sym^2M_2$; a reflection on the orthogonal complement adjusts its determinant. The resulting projective orthogonal transformation comes from $PGL(M_2)$. Over algebraically closed $k$, varying $4b^2/(b^2-4ac)$ with $ac\ne0$ attains every value except $4$. This proves completeness and the harmonic assertion. \end{proof} \begin{lemma}[The four typed formulas for $\eta$] \label{lem:eta-typed} Put $V=\Sym^2M_2$ and $H=(\det M_2)^{\otimes2}$. The inverse of $\psi\otimes\id_{H^{-1}}:V\otimes H^{-1}\to V^*$ is a map $\gamma:V^*\to V\otimes H^{-1}$. Define \emph{separate} maps on this twisted space by \[ \begin{aligned} \Delta_{\mathrm{tw}}:V\otimes H^{-1}&\longrightarrow H^{-1}, &\Delta_{\mathrm{tw}}(p\otimes\lambda) &=\Delta(p)\otimes\lambda^{\otimes2},\\ \Delta_{\mathrm{tw}}:(V\otimes H^{-1})^2&\longrightarrow H^{-1}, &\Delta_{\mathrm{tw}}(p\otimes\lambda,q\otimes\mu) &=\Delta(p,q)\otimes\lambda\otimes\mu. \end{aligned} \] The targets use the canonical contraction $H\otimes H^{-2}\cong H^{-1}$. The first map is quadratic and the second is bilinear; since $H^{-1}$ is a line, every element of $V\otimes H^{-1}$ is a pure tensor, and the formulas are unchanged by rescaling its two factors inversely. Now define the dual-domain maps \[ \Delta^\vee=\Delta_{\mathrm{tw}}\circ\gamma:V^*\to H^{-1}, \qquad \Delta^\vee(g,h)=\Delta_{\mathrm{tw}}(\gamma(g),\gamma(h)) \quad(g,h\in V^*). \] Let $0\ne f_1,f_2\in V$ and $0\ne g_1,g_2\in V^*$ represent lines $\mathsf L(f_i)$ and planes $\mathsf D(g_i)$. Whenever both associated divisors have distinct roots, the parameter of Lemma~\ref{lem:divisor-parameter} is \[ \begin{aligned} \eta(\mathsf L(f_1),\mathsf L(f_2)) &=\frac{\Delta(f_1,f_2)^2}{\Delta(f_1)\Delta(f_2)},\\ \eta(\mathsf D(g_1),\mathsf L(f_2)) &=\frac{g_1(f_2)^2}{\Delta^\vee(g_1)\Delta(f_2)},\\ \eta(\mathsf L(f_1),\mathsf D(g_2)) &=\frac{g_2(f_1)^2}{\Delta(f_1)\Delta^\vee(g_2)},\\ \eta(\mathsf D(g_1),\mathsf D(g_2)) &=\frac{\Delta^\vee(g_1,g_2)^2} {\Delta^\vee(g_1)\Delta^\vee(g_2)}. \end{aligned} \] Every quotient is a scalar: the mixed denominators lie in $H^{-1}\otimes H=k$, and the two other ratios cancel their respective determinant twists. In particular $g_i(f_j)$ is the natural evaluation $V^*\times V\to k$. \end{lemma} \begin{proof} For $g\in V^*$ choose a nonzero $d\in H$ and write $\gamma(g)=p\otimes d^{-1}$. Then $\psi(p)=g\otimes d$, so the pole line of $\mathsf D(g)$ is $kp$. Moreover $\Delta^\vee(g)=\Delta(p)d^{-2}$ and $g(f)=\Delta(p,f)d^{-1}$. Substitution in the line--line formula cancels $d$ and gives both mixed formulas. With two functionals, write $\gamma(g_i)=p_i\otimes d_i^{-1}$; then $\Delta^\vee(g_1,g_2)=\Delta(p_1,p_2)d_1^{-1}d_2^{-1}$, which gives the last formula. The calculation is independent of the chosen $d_i$ and of scaling the representatives. \end{proof} \begin{lemma}[Stabilizers of ordered divisor pairs] \label{lem:divisor-stabilizers} For two doubled divisors, equal roots have projective point stabilizer and distinct roots have two-point stabilizer. For a doubled divisor and a two-root divisor, the stabilizer is a two-point stabilizer when the double root is on the two-root divisor, and $C_2$ when it is off it. For two two-root divisors, the stabilizer is an unordered-pair stabilizer when they are the same divisor, trivial when they share one root, $C_2$ when disjoint with $\eta\ne0$, and $V_4$ when disjoint with $\eta=0$. The same list applies whichever slot is a line or a plane, using the pole line of a plane. \end{lemma} \begin{proof} The projective action on degree-two divisors is the standard $PGL(M_2)$-action on unordered pairs of points of $\mathbb P(M_2^*)$. Fixing one or two points gives the first claims. A point off an unordered pair leaves its transposition, while three distinct marked points have trivial stabilizer. For disjoint two-root divisors, their nondegenerate span in $\Sym^2M_2$ has projective stabilizer $C_2$ if $\Delta(p,q)\ne0$: the normalized quadratics have the same sign. If $\Delta(p,q)=0$, their signs vary independently, giving $V_4$. Lemma~\ref{lem:divisor-parameter} identifies the latter case with $\eta=0$. The polarity is equivariant, so a plane slot has the same stabilizer as its pole divisor. \end{proof} \paragraph{Notation for representatives.} For the following tables alone, choose a basis $e_1,e_2$ of $M_2^*$ and write $x=e_1^*$, $y=e_2^*$ for its dual basis in $M_2$. Thus $x^2=e_1^*e_1^*$ and $xy=e_1^*e_2^*$ are elements of $\Sym^2M_2$. For a nonzero quadratic $p\in\Sym^2M_2$ and a nonzero $f\in\Sym^2M_2^*$, write \[ \mathsf L(p)=kp,\qquad \mathsf D(f)=(k\langle f\rangle)^\perp\subseteq\Sym^2M_2. \] The perpendicular is the annihilator under the natural $\Sym^2M_2^*$--$\Sym^2M_2$ pairing, so each $\mathsf D(f)$ is a plane dual to the displayed line $k\langle f\rangle$. For example, $\mathsf D(e_2^2)=\langle x^2,xy\rangle$ and $\mathsf D(e_1e_2)=\langle x^2,y^2\rangle$. These basis-dependent representatives do not enter the classification or its invariants. \begin{theorem}[The discrete $V_1\otimes M_2$ slice] \label{thm:small-n-one} Let every central generator act on $M_2$ as $D_j=\mu_j\id_{M_2}$, with arbitrary $\mu_j\in k$. For $R=V_1\otimes M_2$, every $K$ is uniquely \[ K=(V_2\otimes L)\oplus(V_0\otimes U),\qquad L\in\{0,\Lam^2M_2\},\quad U\subseteq\Sym^2M_2. \] The following table gives all six $GL(M_2)$-orbits of $U$. Each row represents $2^1$ distinct fixed-$R$ isomorphism classes as $L$ varies, yielding twelve classes in total. The indicated stabilizer is the full self-loop group for every scalar tuple $(\mu_j)$. Write $B\subseteq GL(M_2)$ for the stabilizer of a line in $M_2$, and $N(T)$ for the stabilizer of an unordered pair of distinct lines in $M_2$, where $T$ stabilizes each line separately. The discriminant polarity identifies a plane $U$ with its pole in $\mathbb P(\Sym^2M_2)$. \[ \begin{array}{c|l|c|c} \dim U&\text{orbit of }U&\text{representative }U &\Aut_{\s}(\g_K)\\\hline 0&U=0&0&GL(M_2)\\ 1&\text{double-root line}&\mathsf L(x^2)&B\\ 1&\text{distinct-root line}&\mathsf L(xy)&N(T)\\ 2&\text{tangent plane (double-root pole)}&\mathsf D(e_2^2)&B\\ 2&\text{secant plane (distinct-root pole)}&\mathsf D(e_1e_2)&N(T)\\ 3&U=\Sym^2M_2&\Sym^2M_2&GL(M_2) \end{array} \] \end{theorem} \begin{proof} Lemma~\ref{lem:sl2-fiber} gives $\Lam^2R\cong(V_2\otimes\Lam^2M_2)\oplus (V_0\otimes\Sym^2M_2)$. The scalar actions induce $2\mu_j\id$ on both multiplicity spaces, so all subspaces are stable and $G_R=GL(M_2)$. The line $\Lam^2M_2$ has exactly two subspaces, each fixed by $GL(M_2)$. The discriminant polarity identifies lines and planes in $\Sym^2M_2$ equivariantly; each has two orbits, according to whether its quadratic has a double root or two distinct roots. The stabilizers are respectively $B$ and $N(T)$. The zero and full subspaces are fixed by $GL(M_2)$. Proposition~\ref{prop:sl2-shears} gives no shear since $n=1$ is odd; Theorem~\ref{thm:groupoid} makes the listed groups full. \end{proof} \begin{proposition}[Shears for scalar central actions on $V_2\otimes M_2$] \label{prop:small-two-general-shifts} Let $R=V_2\otimes M_2$, and let the central generators act on $M_2$ by $D_j=\mu_j\id_{M_2}$, with arbitrary scalars $\mu_j\in k$. For \[ Q=(Q_0,Q_1,Q_2),\qquad Q_0,Q_2\subseteq\Lam^2M_2,\quad Q_1\subseteq\Sym^2M_2 \] put $U_Q=\Sym^2M_2/Q_1$. Regard $M_2$ and $U_Q$ as modules over $A=k[t_1,\ldots,t_m]$, with $t_j$ acting on them by $\mu_j\id_{M_2}$ and $2\mu_j\id_{U_Q}$, respectively. Then the additive group of all shears at $(R,K_Q)$ is \begin{equation}\label{eq:small-two-module-hom} \mathcal H_Q=\Hom_A(M_2,U_Q) =\{F\in\Hom_k(M_2,U_Q): F(\mu_j u)=2\mu_jF(u)\text{ for every }j,u\}. \end{equation} The full self-loop group is $\mathcal H_Q\rtimes G_{K_Q}$, with $g\in G_{K_Q}$ acting by $F\mapsto\bar g Fg^{-1}$, where $\bar g$ is induced by $\Sym^2g$ on $U_Q$. Using the $V_2$-isotypic summand $V_2\otimes U_Q\subseteq W_Q$ from Lemma~\ref{lem:sl2-fiber}, each $F\in\mathcal H_Q$ is the automorphism \begin{equation}\label{eq:small-two-shift} S_F(x,v\otimes u,w) =\bigl(x,v\otimes u,w+\iota_Q(v\otimes F(u))\bigr) \end{equation} on decomposable vectors, extended linearly. Here $\iota_Q$ includes that summand in $W_Q$. The shears satisfy $S_F^{-1}=S_{-F}$ and $S_FS_{F'}=S_{F+F'}$. \end{proposition} \begin{proof} The $V_2$-isotypic component of $W_Q$ is $V_2\otimes U_Q$, since the other two components have types $V_4$ and $V_0$. An $\mathfrak{sl}_2$-intertwiner $R\to W_Q$ is therefore uniquely given by a linear map $F:M_2\to U_Q$ under the fixed isotypic identification. It intertwines every $t_j$ exactly when $F(\mu_j u)=2\mu_jF(u)$, proving \eqref{eq:small-two-module-hom}. Because $W_Q$ is central in $R\oplus W_Q$, formula \eqref{eq:small-two-shift} preserves the bracket of two elements of $R$; the module condition preserves brackets with $\s$. Its inverse and composition follow by addition. The factorization and conjugation action are Proposition~\ref{prop:groupoid-composition}. \end{proof} \begin{theorem}[The twenty-four discrete $V_2\otimes M_2$ classes] \label{thm:small-n-two} Keep $D_j=\mu_j\id_{M_2}$ for arbitrary $\mu_j\in k$. For $R=V_2\otimes M_2$, write \[ K=(V_4\otimes L_0)\oplus(V_2\otimes U) \oplus(V_0\otimes L_2),\quad L_0,L_2\in\{0,\Lam^2M_2\},\quad U\subseteq\Sym^2M_2. \] The two $\Lam^2M_2$ summands contribute $2^2$ distinct choices of $(L_0,L_2)$ for every orbit type of $U$. They do not change the stabilizer or the shear group. The table lists the six orbit types of $U$ once each; each row therefore represents four distinct isomorphism classes, giving twenty-four classes in total. The labels $\mathrm d,\mathrm e,\mathrm t,\mathrm s$ denote a double-root line, a distinct-root line, a tangent plane, and a secant plane in $\Sym^2M_2$, respectively. The column $h$ is the dimension of the shear group when all $\mu_j=0$. \[ \begin{array}{c|c|c|c|c} U&\dim U&\text{representative }U&G_K&h\\\hline 0&0&0&GL(M_2)&6\\ \mathrm d&1&\mathsf L(x^2)&B&4\\ \mathrm e&1&\mathsf L(xy)&N(T)&4\\ \mathrm t&2&\mathsf D(e_2^2)&B&2\\ \mathrm s&2&\mathsf D(e_1e_2)&N(T)&2\\ \Sym^2M_2&3&\Sym^2M_2&GL(M_2)&0 \end{array} \] For at least one $\mu_j\ne0$, the full self-loop group in every row is $G_K$. \end{theorem} \begin{proof} Lemma~\ref{lem:sl2-fiber} gives the displayed decomposition; again $G_R=GL(M_2)$ and all its multiplicity subspaces are invariant under the scalar $D_j^{\{2\}}=2\mu_j\id$. The two $\Lam^2M_2$ summands each contribute a fixed binary choice. The six orbits of $U$ and their stabilizers were proved in Theorem~\ref{thm:small-n-one}. This proves the $2^2\cdot6=24$ distinct classes. Proposition~\ref{prop:small-two-general-shifts} identifies the shears with $\Hom_{k[t_1,\ldots,t_m]}(M_2,\Sym^2M_2/U)$. If some $\mu_j\ne0$, its intertwining equation $\mu_jF=2\mu_jF$ forces $F=0$. \end{proof} \begin{corollary}[Scalar weights and the shears in the table] \label{cor:small-two-zero} For the arbitrary scalar tuple $(\mu_j)$ of Theorem~\ref{thm:small-n-two}, the shear group in each row is the polynomial-module Hom space in \eqref{eq:small-two-module-hom}, with $U_Q=\Sym^2M_2/U$. If at least one $\mu_j\ne0$, it is zero. If every $\mu_j=0$, the full self-loop group is \begin{equation}\label{eq:small-two-aut} \Hom_k(M_2,\Sym^2M_2/U)\rtimes G_K, \end{equation} and its additive factor has dimension $h=2(3-\dim U)$, as recorded in the table. In particular, the rows with $U=0$ have a six-dimensional group of nonidentity shears, explicitly given by \eqref{eq:small-two-shift}. \end{corollary} \begin{proof} The module-intertwining equation in \eqref{eq:small-two-module-hom} is $\mu_jF=2\mu_jF$ for every $j$. A nonzero $\mu_j$ forces $F=0$; when all are zero, every linear $F:M_2\to\Sym^2M_2/U$ intertwines. Its dimension is $2(3-\dim U)$, and Proposition~\ref{prop:small-two-general-shifts} supplies the stated action and explicit automorphisms. \end{proof} \begin{proposition}[Parameter space for $V_3\otimes M_2$] \label{prop:family-parameter} Take $n=3$ and $D_j=\mu_j\id_{M_2}$ for all central generators, where $\mu_j\in k$ are arbitrary, including the case $m=0$. Write \[ \mathcal S(M_2)=\{0\}\sqcup\mathbb P(\Sym^2M_2) \sqcup\mathbb P((\Sym^2M_2)^*)\sqcup\{\Sym^2M_2\}. \] The two projective planes parameterize respectively lines and planes $U\subseteq\Sym^2M_2$. The complete parameter space of objects with this fixed $R$ is the disjoint union of standard algebraic varieties \begin{equation}\label{eq:family-varieties} \mathcal P(R)=\{0,\Lam^2M_2\}^{2}\times\mathcal S(M_2)^2. \end{equation} Its points $(L_0,U_1,L_2,U_3)$ give exactly \begin{equation}\label{eq:family-K} K=(V_6\otimes L_0)\oplus(V_4\otimes U_1) \oplus(V_2\otimes L_2)\oplus(V_0\otimes U_3). \end{equation} Every orbit of the pair $(U_1,U_3)$ occurs in $2^2$ distinct choices of $(L_0,L_2)$, all fixed separately by $GL(M_2)$. There is no shear at any object, so the full group of self-loops is the stabilizer of the ordered pair $(U_1,U_3)$ in $GL(M_2)$. \end{proposition} \begin{proof} Lemma~\ref{lem:sl2-fiber} gives $N_6=N_2=\Lam^2M_2$ and $N_4=N_0=\Sym^2M_2$. These have dimensions one and three, respectively. All $D_j$ are scalar, so $D_j^{\{2\}}=2\mu_j\id$ on every $N_{\nu_i}$; every subspace is invariant and $G_R=GL(M_2)$. Theorem~\ref{thm:sl2-groupoid} gives \eqref{eq:family-varieties} and \eqref{eq:family-K}. Since $n=3$ is odd, Proposition~\ref{prop:sl2-shears} gives zero shear. The groupoid factorization gives the claimed full self-loop group. \end{proof} \begin{proposition}[Extreme slots in the $V_3\otimes M_2$ slice] \label{prop:family-extremes} Under the assumptions of Proposition~\ref{prop:family-parameter}, fix a choice of $(L_0,L_2)$. If both $U_1,U_3$ belong to $\{0,\Sym^2M_2\}$, their four ordered choices are distinct orbits, each with full self-loop group $GL(M_2)$. If exactly one belongs to $\{0,\Sym^2M_2\}$, its slot and its value are fixed. The other slot has four orbits: the line or plane associated to a doubled divisor, each with self-loop $B$, and the line or plane associated to a two-root divisor, each with self-loop $N(T)$. Thus these cases have twenty orbits. Representatives for the nonextreme slot are $\mathsf L(x^2)$, $\mathsf D(e_2^2)$, $\mathsf L(xy)$, and $\mathsf D(e_1e_2)$, respectively. Here $B\subseteq GL(M_2)$ is a point stabilizer, $T$ fixes two distinct points separately, and $N(T)$ fixes their unordered pair. \end{proposition} \begin{proof} The extreme subspaces are fixed by all of $GL(M_2)$. Lemma~\ref{lem:divisor-parameter} and Lemma~\ref{lem:divisor-stabilizers} give the two single-divisor orbits and their stabilizers for each of the four choices of extreme slot and value. Proposition~\ref{prop:sl2-shears} gives zero shears for $n=3$; hence the listed stabilizers are the full self-loop groups. \end{proof} \begin{theorem}[The line--plane orbit tables for $V_3\otimes M_2$] \label{thm:family-moduli} Under the assumptions of Proposition~\ref{prop:family-parameter}, fix any of the $2^2$ choices of $(L_0,L_2)$. When both $U_1,U_3$ are lines or planes, their fixed-$R$ isomorphism classes, representatives, orbit parameters, and full self-loop groups are exactly the forty rows of the four tables below. Each row occurs for every choice of $(L_0,L_2)$, without changing its self-loop group. Together with Proposition~\ref{prop:family-extremes}, these are all the objects and arrows in this fixed-$R$ groupoid. \end{theorem} In each table, a line has the divisor of its quadratic, and a plane has the divisor of its pole from Lemma~\ref{lem:intrinsic-pole}. Both divisors lie on $\mathbb P(M_2^*)$, as in Lemma~\ref{lem:divisor-parameter}. ``Same divisor'' means equality as effective divisors; ``one common root'' and ``disjoint'' compare their supports. Under $\pi:GL(M_2)\twoheadrightarrow PGL(M_2)$, write $\widetilde C_2$ and $\widetilde V_4$ for the inverse images of the indicated projective stabilizers. Each group in the last column is the full self-loop group. For the generic rows set $q_\eta=4\eta x^2+4\eta xy+(\eta-4)y^2$ with $\eta\notin\{0,4\}$, and for the harmonic rows set $q_0=x^2+y^2$. In the examples using planes, put $f_\eta=(4-\eta)e_1^2+4\eta e_1e_2-4\eta e_2^2$. It is the discriminant polar of $q_\eta$, up to a nonzero scalar, so $\mathsf D(f_\eta)$ has pole $[q_\eta]$. The parameter $\eta$ was defined in \eqref{eq:family-J}. Lemma~\ref{lem:eta-typed} gives its formula for each of the four line--plane slot types. \Needspace{15\baselineskip} \noindent\textbf{$U_1$ a line; $U_3$ a line}\par \begin{center} \small\begin{tabular}{c|l|c|c} representative $(U_1,U_3)$&relation of divisors&orbit parameter&$\Aut_{\mathcal C_{\s}^{0}}(R,K)$\\\hline $(\mathsf L(x^2),\mathsf L(x^2))$ & same doubled divisor & point & $B$\\ $(\mathsf L(x^2),\mathsf L(y^2))$ & distinct doubled divisors & point & $T$\\ $(\mathsf L(x^2),\mathsf L(xy))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection nonempty} & point & $T$\\ $(\mathsf L(x^2),\mathsf L(y(x+y)))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection empty} & point & $\widetilde C_2$\\ $(\mathsf L(xy),\mathsf L(x^2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection nonempty} & point & $T$\\ $(\mathsf L(xy),\mathsf L((x+y)^2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection empty} & point & $\widetilde C_2$\\ $(\mathsf L(xy),\mathsf L(xy))$ & same divisor & point & $N(T)$\\ $(\mathsf L(xy),\mathsf L(x(x+y)))$ & one common root & point & $k^\times$\\ $(\mathsf L(xy),\mathsf L(q_\eta))$ & disjoint, generic & $\eta\in\mathbb A^1\setminus\{0,4\}$ & $\widetilde C_2$\\ $(\mathsf L(xy),\mathsf L(q_0))$ & disjoint, harmonic & $\eta=0$ & $\widetilde V_4$\\ \end{tabular} \end{center} \Needspace{15\baselineskip} \noindent\textbf{$U_1$ a line; $U_3$ a plane}\par \begin{center} \small\begin{tabular}{c|l|c|c} representative $(U_1,U_3)$&relation of divisors&orbit parameter&$\Aut_{\mathcal C_{\s}^{0}}(R,K)$\\\hline $(\mathsf L(x^2),\mathsf D(e_2^2))$ & same doubled divisor & point & $B$\\ $(\mathsf L(x^2),\mathsf D(e_1^2))$ & distinct doubled divisors & point & $T$\\ $(\mathsf L(x^2),\mathsf D(e_1e_2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection nonempty} & point & $T$\\ $(\mathsf L(x^2),\mathsf D(e_1e_2-e_1^2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection empty} & point & $\widetilde C_2$\\ $(\mathsf L(xy),\mathsf D(e_2^2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection nonempty} & point & $T$\\ $(\mathsf L(xy),\mathsf D(e_1^2-2e_1e_2+e_2^2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection empty} & point & $\widetilde C_2$\\ $(\mathsf L(xy),\mathsf D(e_1e_2))$ & same divisor & point & $N(T)$\\ $(\mathsf L(xy),\mathsf D(e_1e_2-e_2^2))$ & one common root & point & $k^\times$\\ $(\mathsf L(xy),\mathsf D(f_\eta))$ & disjoint, generic & $\eta\in\mathbb A^1\setminus\{0,4\}$ & $\widetilde C_2$\\ $(\mathsf L(xy),\mathsf D(e_1^2+e_2^2))$ & disjoint, harmonic & $\eta=0$ & $\widetilde V_4$\\ \end{tabular} \end{center} \Needspace{15\baselineskip} \noindent\textbf{$U_1$ a plane; $U_3$ a line}\par \begin{center} \small\begin{tabular}{c|l|c|c} representative $(U_1,U_3)$&relation of divisors&orbit parameter&$\Aut_{\mathcal C_{\s}^{0}}(R,K)$\\\hline $(\mathsf D(e_2^2),\mathsf L(x^2))$ & same doubled divisor & point & $B$\\ $(\mathsf D(e_2^2),\mathsf L(y^2))$ & distinct doubled divisors & point & $T$\\ $(\mathsf D(e_2^2),\mathsf L(xy))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection nonempty} & point & $T$\\ $(\mathsf D(e_2^2),\mathsf L(y(x+y)))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection empty} & point & $\widetilde C_2$\\ $(\mathsf D(e_1e_2),\mathsf L(x^2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection nonempty} & point & $T$\\ $(\mathsf D(e_1e_2),\mathsf L((x+y)^2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection empty} & point & $\widetilde C_2$\\ $(\mathsf D(e_1e_2),\mathsf L(xy))$ & same divisor & point & $N(T)$\\ $(\mathsf D(e_1e_2),\mathsf L(x(x+y)))$ & one common root & point & $k^\times$\\ $(\mathsf D(e_1e_2),\mathsf L(q_\eta))$ & disjoint, generic & $\eta\in\mathbb A^1\setminus\{0,4\}$ & $\widetilde C_2$\\ $(\mathsf D(e_1e_2),\mathsf L(q_0))$ & disjoint, harmonic & $\eta=0$ & $\widetilde V_4$\\ \end{tabular} \end{center} \Needspace{15\baselineskip} \noindent\textbf{$U_1$ a plane; $U_3$ a plane}\par \begin{center} \small\begin{tabular}{c|l|c|c} representative $(U_1,U_3)$&relation of divisors&orbit parameter&$\Aut_{\mathcal C_{\s}^{0}}(R,K)$\\\hline $(\mathsf D(e_2^2),\mathsf D(e_2^2))$ & same doubled divisor & point & $B$\\ $(\mathsf D(e_2^2),\mathsf D(e_1^2))$ & distinct doubled divisors & point & $T$\\ $(\mathsf D(e_2^2),\mathsf D(e_1e_2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection nonempty} & point & $T$\\ $(\mathsf D(e_2^2),\mathsf D(e_1e_2-e_1^2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection empty} & point & $\widetilde C_2$\\ $(\mathsf D(e_1e_2),\mathsf D(e_2^2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection nonempty} & point & $T$\\ $(\mathsf D(e_1e_2),\mathsf D(e_1^2-2e_1e_2+e_2^2))$ & \shortstack[l]{exactly one divisor doubled;\\support intersection empty} & point & $\widetilde C_2$\\ $(\mathsf D(e_1e_2),\mathsf D(e_1e_2))$ & same divisor & point & $N(T)$\\ $(\mathsf D(e_1e_2),\mathsf D(e_1e_2-e_2^2))$ & one common root & point & $k^\times$\\ $(\mathsf D(e_1e_2),\mathsf D(f_\eta))$ & disjoint, generic & $\eta\in\mathbb A^1\setminus\{0,4\}$ & $\widetilde C_2$\\ $(\mathsf D(e_1e_2),\mathsf D(e_1^2+e_2^2))$ & disjoint, harmonic & $\eta=0$ & $\widetilde V_4$\\ \end{tabular} \end{center} The orbit-parameter column uses the intrinsic invariant \eqref{eq:family-J}. The generic and harmonic rows partition the disjoint locus $\mathbb A^1\setminus\{4\}$; the table separates coincident divisors from those sharing one root even though both have $\eta=4$. No arrow joins different rows, different values of $\eta$, or different $(L_0,L_2)$. \begin{proof}[Proof of Theorem~\ref{thm:family-moduli}] By Lemma~\ref{lem:intrinsic-pole}, a line and a plane each determine a projective quadratic, equivariantly for $GL(M_2)$. For two doubled divisors and for a doubled divisor paired with a two-root divisor, their equality or incidence gives precisely the first six rows in each table. Two two-root divisors can be the same, share exactly one root, or be disjoint. Lemma~\ref{lem:divisor-parameter} classifies the disjoint pairs by $\eta$ and separates the other two incidence types. In the chosen basis, $\Delta(q_\eta)=64\eta(x\wedge y)^2$ and $\Delta(xy,q_\eta)=8\eta(x\wedge y)^2$, so the displayed $q_\eta$ realizes its stated value of $\eta$. The formula for $\psi$ in Lemma~\ref{lem:intrinsic-pole} checks the displayed plane annihilators; in particular $\psi(q_\eta)$ is proportional to $f_\eta$. Lemma~\ref{lem:divisor-stabilizers} gives the projective stabilizers in each row. Their inverse images in $GL(M_2)$ are the groups displayed, with $k^\times$ for the trivial projective stabilizer. Proposition~\ref{prop:sl2-shears} gives zero shears, and Lemma~\ref{lem:binary-choices} shows that the binary $L_i$ do not change these groups. Theorem~\ref{thm:sl2-groupoid} identifies the rows and parameters as all the arrows and self-loops. \end{proof} \begin{corollary}[Trivial central action] \label{cor:zero-mu} If every $\mu_j=0$ in Proposition~\ref{prop:family-parameter}, then the $t_j$ act trivially on $R$ and $W_K$, and \[ \g(\s,V_3\otimes M_2,K) \cong \bigl(\mathfrak{sl}_2\ltimes((V_3\otimes M_2)\oplus W_K)\bigr) \oplus\bigoplus_{j=1}^m kt_j. \] Its fixed-$\s$ isomorphism classes and full self-loop groups are precisely those of Proposition~\ref{prop:family-extremes} and Theorem~\ref{thm:family-moduli}. \end{corollary} \begin{proof} The induced action on $\Lam^2R$ is $D_j^{\{2\}}=2\mu_j\id=0$, and so the action on $W_K$ vanishes. The stated direct sum follows from the brackets. Proposition~\ref{prop:family-parameter} and Theorem~\ref{thm:family-moduli} apply at these scalar values. \end{proof} \subsection{A ternary multiplicity space: conics with a marked point or line}\label{sec:ternary} The binary tables above do not describe the first ternary multiplicity space. Here take $R=V_1\otimes M_3$, where $M_3$ is a three-dimensional multiplicity space; the subscript records the dimension, not a highest weight as in $M_\lambda$. Suppose that $D_j=\mu_j\id_{M_3}$ for every central generator $t_j$. Put $P=\mathbb P(M_3^*)$, so an element of $\Sym^2M_3$ is a conic equation on $P$. The scalar-action assumption makes every subspace below an $\s$-subrepresentation, for arbitrary $\mu_j\in k$. \begin{lemma}[The two independent pieces]\label{lem:ternary-pieces} Every allowed $K$ has a unique expression \[ K=(V_2\otimes L)\oplus(V_0\otimes U),\qquad L\subseteq\Lam^2M_3,\quad U\subseteq\Sym^2M_3. \] Write $a=\dim L$ and $d=\dim U$. Under the canonical, determinant-twisted isomorphism $\Lam^2M_3\simeq M_3^*\otimes\det M_3$, the choices $a=0,1,2,3$ correspond respectively to no mark, a point $p\in P$, a line $\ell\subset P$, and the whole plane. An isomorphism in the fixed-$\s$ groupoid exists precisely when one $g\in GL(M_3)$ carries both $L$ to $L'$ and $U$ to $U'$. Its self-loop group is the common stabilizer of $L$ and $U$ in $GL(M_3)$. \end{lemma} \begin{proof} The Cauchy decomposition is \[ \Lam^2(V_1\otimes M_3) = (\Sym^2 V_1\otimes\Lam^2M_3) \oplus(\Lam^2V_1\otimes\Sym^2M_3) = (V_2\otimes\Lam^2M_3)\oplus(V_0\otimes\Sym^2M_3). \] The two simple $\mathfrak{sl}_2$ factors are inequivalent, and the central action on both multiplicity spaces is scalar, proving the first assertion. The determinant-twisted identification sends a decomposable $u\wedge v$ to the covector $w\mapsto u\wedge v\wedge w$. It sends a one-dimensional subspace to a point of $P$ and a two-dimensional subspace to a projective line of $P$, equivariantly. By Theorem~\ref{thm:sl2-groupoid}, arrows are exactly the indicated simultaneous changes of basis. Proposition~\ref{prop:sl2-shears} eliminates shears because $V_1$ does not occur in $\Lam^2R$. \end{proof} The following table covers all possible dimensions, including a single conic. Its entries specify which geometric data have to be classified \emph{together}. The complete orbit test immediately after the table also handles continuous marked families. In the last three rows $U^\perp$ means the annihilator under the contraction pairing $\Sym^2M_3\times\Sym^2M_3^*\to k$. Its conics live on the \emph{dual} plane $P^*=\mathbb P(M_3)$; the point $p\in P$ becomes a line of $P^*$, and a line $\ell\subset P$ becomes a point of $P^*$. \begin{center} \small \begin{tabular}{c|l|l|l} $d$ & linear-system invariant & $a=1$ & $a=2$\\\hline $0$ & empty system & point of $P$ & line of $P$\\ $1$ & one conic, rank $1,2,$ or $3$ & conic with point & conic with line\\ $2$ & pencil of conics & pencil with point & pencil with line\\ $3$ & net of conics & net with point & net with line\\ $4$ & dual pencil $U^\perp\subset\Sym^2M_3^*$ & dual pencil with line & dual pencil with point\\ $5$ & dual single conic $U^\perp$ & dual conic with line & dual conic with point\\ $6$ & complete system & point of $P$ & line of $P$ \end{tabular} \end{center} For $a=0$ and $a=3$ the mark is unique, so the invariant is just the system in the second column. In particular, the $d=0,6$ rows each have exactly one class for every $a$. \begin{theorem}[The complete single-conic incidence table] \label{thm:ternary-single-conic} For $d=1$, the following are all classes for $a=1,2$. Coordinates $[X:Y:Z]$ are on $P$. For $d=5$ the same table applies to the conic $U^\perp$ on $P^*$, with the $a=1$ and $a=2$ columns interchanged. Thus it gives seven point classes and seven line classes for each of $d=1$ and $d=5$. \begin{center} \small \begin{tabular}{c|l|p{2.15in}|p{2.25in}} rank & representative & point classes & line classes\\\hline $1$ & $X^2$ & on $X=0$; off $X=0$ & $X=0$; any other line\\ $2$ & $XY$ & node; on one component away from node; off both & a component; through node but not a component; avoids node\\ $3$ & $XZ-Y^2$ & on conic; off conic & tangent; secant \end{tabular} \end{center} The self-loop group of any entry is the inverse image in $GL(M_3)$ of the projective stabilizer of the displayed conic and indicated point or line. For $a=0,3$ there are exactly the three rank classes. \end{theorem} \begin{proof} Every ternary quadratic has one of the displayed ranks. A change of coordinates preserving $X^2$ is transitive on and off its support line and on the lines other than that support. A change preserving $XY$ preserves or exchanges the two component lines: it is transitive separately on the node, their smooth points, and the complement. Dually, lines are either a component, pass through the node without being a component, or avoid the node, and the same coordinate transformations act transitively in each case. The stabilizer of a smooth conic is $PGL_2$ acting on its Veronese plane; it is transitive on the conic and its complement and, on dual lines, on tangents and secants. These incidence conditions are mutually exclusive and invariant. The contraction pairing is $GL(M_3)$ equivariant, so annihilating $U$ converts $d=5$ to the same classification in $P^*$; the point/line interchange follows from projective duality. Lemma~\ref{lem:ternary-pieces} gives the loop groups. \end{proof} For pencils and nets, incidence is not exhausted by the simple predicates in Theorem~\ref{thm:ternary-single-conic}. Here is a finite, explicit equivalence test for \emph{every} row of the dimension table. It is useful when the stabilizer of a pencil or net has a continuous family of marked orbits. Fix a basis $(x,y,z)$ of $M_3$ and the ordered quadratic basis $\mathcal B=(x^2,y^2,z^2,xy,xz,yz)$. Let $B$ be the $6\times d$ full-rank matrix of a basis of $U$ in $\mathcal B$, and let $H$ be the $3\times a$ full-rank matrix of a basis of $L$ in $(x\wedge y,x\wedge z,y\wedge z)$. Empty matrices are allowed. Write $S_2(G)$ and $E_2(G)$ for the matrices of $\Sym^2G$ and $\Lam^2G$ in these ordered bases; their entries are explicit quadratic polynomials in the nine entries of $G$. \begin{theorem}[Polynomial orbit criterion] \label{thm:ternary-matrix-test} For two entries $(B,H)$ and $(B',H')$ with the same $(d,a)$, form the polynomial system in $G=(g_{ij})$, an auxiliary variable $u$, and square matrices $C$ of size $d$ and $J$ of size $a$: \begin{equation}\label{eq:ternary-orbit-test} S_2(G)B=B'C,\qquad E_2(G)H=H'J,\qquad u\det(G)\det(C)\det(J)=1. \end{equation} For a matrix of size zero its determinant is $1$. The two Lie algebra objects are isomorphic in the fixed-$\s$ groupoid if and only if \eqref{eq:ternary-orbit-test} has a solution over $k$. Equivalently, its generated ideal is proper. For a self-loop, the solutions projected to $G$ give the \emph{full} loop group $\Stab_{GL(M_3)}(L,U)$. This is a necessary-and-sufficient isomorphism criterion for all pencils, nets, and their marked incidences, including nonclosed orbits. If the input coefficients lie in a specified computable subfield $k_0\subseteq k$, with effective field operations and equality testing, it is a decision procedure: a Gr\"obner basis over $k_0$ tests whether the generated ideal contains $1$. \end{theorem} \begin{proof} The first two equations say exactly that $G$ maps the two subspaces onto their primed counterparts; the invertibility equation makes $G,C,J$ isomorphisms. Conversely, every simultaneous isomorphism gives unique $C,J$ relative to the chosen column bases and a suitable $u$. Lemma~\ref{lem:ternary-pieces} identifies these simultaneous isomorphisms with all arrows. Since $k$ is algebraically closed, the weak Nullstellensatz says the finite polynomial system has a solution exactly when its ideal is proper. For coefficients in $k_0$, extension from $k_0$ to $k$ is faithfully flat, so the ideal contains $1$ over $k$ exactly when it does over $k_0$. A Gr\"obner basis computes this membership using effective arithmetic in $k_0$. Substituting the same input twice gives the stated self-loop equations. \end{proof} \begin{proposition}[The eight unmarked pencil types] \label{prop:ternary-pencils} When $d=2$, the underlying pencil $U$ belongs to exactly one row of the following table. The partition records multiplicities in the base scheme when the pencil contains a smooth conic. In the last three rows every member is singular. In each row the \emph{marked} classes are its points or lines modulo the stabilizer of the displayed pencil, tested exactly by \eqref{eq:ternary-orbit-test} above. \begin{center} \small \begin{tabular}{c|l|l} type & pencil $U\subset\Sym^2M_3$ & intrinsic description\\\hline $A$ & $\langle x^2-z^2,y^2-z^2\rangle$ & four simple base points\\ $B$ & $\langle x^2-z^2,yz\rangle$ & base multiplicities $2,1,1$\\ $C$ & $\langle xy,z^2\rangle$ & base multiplicities $2,2$\\ $D$ & $\langle xy,xz-y^2\rangle$ & base multiplicities $3,1$\\ $E$ & $\langle x^2,xz-y^2\rangle$ & one base point of length $4$\\ $F$ & $\langle xy,xz\rangle$ & common line, no common singular point\\ $G$ & $\langle x^2,xy\rangle$ & common line and singular point\\ $H$ & $\langle x^2,y^2\rangle$ & common singular point, no common line \end{tabular} \end{center} For $d=4$, take annihilators of these eight pencils in $\Sym^2M_3$ after placing their displayed equations in $\Sym^2M_3^*$: point and line marks exchange roles in the dual plane. \end{proposition} \begin{proof} Suppose a smooth member $F$ exists and let $G$ be an independent member. Regard their polarizations as symmetric bilinear forms on $M_3^*$. Then $F$ is nondegenerate and $T=F^{-1}G$ is $F$-self-adjoint. Simultaneous congruence of $(F,G)$ is classified by the Jordan blocks of $T$: on a block of size $r$ and eigenvalue $\lambda$ the pair is $(\mathsf E_r,\mathsf E_rJ_r(\lambda))$, where $\mathsf E_r$ has ones on its antidiagonal and $J_r(\lambda)$ has ones on its superdiagonal \cite{Thompson1991}. This form is valid over our field $k$. Indeed, generalized eigenspaces of a self-adjoint operator are orthogonal. On one of them write $T=\lambda\id+N$ and choose a maximal chain generated by $v$ with $F(v,N^{r-1}v)\ne0$; such a vector exists because $F(-,N^{r-1}-)$ is a nonzero symmetric form. The cyclic span is nondegenerate. Replacing $v$ by an invertible polynomial in $N$ applied to $v$ normalizes its Gram matrix to $\mathsf E_r$: the required square root of a unit exists in $k[t]/(t^r)$ since $k$ is algebraically closed and $2$ is invertible. Its orthogonal complement is $N$-stable, so induction gives the asserted blocks. A change of the two pencil generators replaces $T$ by \[ (a\id+bT)^{-1}(c\id+dT),\qquad ad-bc\ne0, \] provided the new first form is nondegenerate. This changes eigenvalues by a fractional linear transformation and preserves the sizes of their Jordan blocks. There are at most three distinct eigenvalues, so projective transformations normalize their positions without a parameter. The complete list of block patterns is therefore \[ \begin{array}{c|l} A&J_1(\lambda)\oplus J_1(\mu)\oplus J_1(\nu), \quad\lambda,\mu,\nu\text{ distinct}\\ B&J_2(\lambda)\oplus J_1(\mu),\quad\lambda\ne\mu\\ C&J_1(\lambda)^{\oplus2}\oplus J_1(\mu),\quad\lambda\ne\mu\\ D&J_3(\lambda)\\ E&J_2(\lambda)\oplus J_1(\lambda). \end{array} \] The omitted scalar operator would make $F,G$ dependent. Computing these patterns for the displayed representatives identifies them with $A$--$E$. Their base schemes have, respectively, local lengths $(1,1,1,1)$, $(2,1,1)$, $(2,2)$, $(3,1)$ and $(4)$, as substitution in their two equations shows. This argument uses changes of the smooth member as well as projective coordinates; it does not fix a conic and discard the cross-ratio of four points on it. Now suppose every member is singular. Some member has rank two: two independent rank-one quadratics are squares of independent linear forms and their sum has rank two. Choose coordinates in which this member is $xy$, and subtract a multiple of it from the other generator to write \[ G=ax^2+cy^2+dxz+eyz+fz^2. \] The coefficient of $t$ in the determinant of the symmetric matrix of $xy+tG$ forces $f=0$. With $f=0$ that determinant is \[ \tfrac14\bigl(t^2de-t^3(ae^2+cd^2)\bigr). \] Thus $de=0$ and $ae^2+cd^2=0$. If $d\ne0$, then $e=c=0$ and the pencil is $\langle xy,x(ax+dz)\rangle$; replacing the third coordinate gives $\langle xy,xz\rangle$. The case $e\ne0$ is symmetric. This is type $F$, with a common line and no common singular point. If $d=e=0$, both quadratics depend only on $x,y$ and have the common singular point $[0:0:1]$. A plane in the three-dimensional space of binary quadratics has two orbits, by the discriminant polarity of Lemma~\ref{lem:intrinsic-pole}: it is equivalent to $\langle x^2,xy\rangle$ or $\langle x^2,y^2\rangle$. These are $G$ and $H$, distinguished by the presence or absence of a common line. This proves both exhaustion and separation in the singular case. Contraction is equivariant, proving the $d=4$ statement. The marked assertion follows from Lemma~\ref{lem:ternary-pieces} and Theorem~\ref{thm:ternary-matrix-test}. \end{proof} \begin{lemma}[Why marked-pencil and marked-net families remain] \label{lem:ternary-marked-families} Let $U$ be a pencil through four points in general position. Its projective stabilizer is the finite group $S_4$ permuting those four points. Consequently the point-marked classes above this one pencil are the points of $P/S_4$, and the line-marked classes are those of $P^*/S_4$; both families have dimension two. Likewise, any net with finite projective stabilizer has two-dimensional families of point and line marks. For a fixed system $U$ in any row, the precise marked moduli are the orbits of $\Stab_{PGL(M_3)}(U)$ on $P$ or $P^*$, and \eqref{eq:ternary-orbit-test} is an explicit test for equality in those families. \end{lemma} \begin{proof} The base scheme of the generic pencil consists of four points in general position. A projectivity preserving the pencil permutes the base scheme. Conversely any permutation of a projective frame of four points is realized by a unique projectivity, and it preserves the two-dimensional space of conics through them. Thus its stabilizer is $S_4$. A finite group has finite orbits, so quotienting a plane of possible marks leaves dimension two. The same argument applies to the point and dual-line planes for any net with finite stabilizer. For any $U$, fixing a representative of its orbit leaves precisely its stabilizer acting on the mark; this is also immediate by specializing \eqref{eq:ternary-orbit-test} to $B=B'$. \end{proof} \begin{proposition}[A net with genuinely moving marks] \label{prop:ternary-net-marks} Set \[ U_{\mathrm{He}}=\langle x^2-3yz,\ y^2-3xz,\ z^2-3xy\rangle \subset\Sym^2M_3. \] Its projective stabilizer $\Gamma\subset PGL(M_3)$ is finite. Hence the marked point and line classes above this one net are, respectively, $P/\Gamma$ and $P^*/\Gamma$, each a two-dimensional family. These families occur for every choice of the central weights $\mu_j$. For any net $U$, the canonical evaluation map $\operatorname{ev}_p:U\to\mathcal O_P(2)|_p$ takes values in the fiber of the quadratic line bundle at $p$. It is zero exactly when the point mark lies in the base scheme; no trivialization of that fiber is needed. The restriction map $U\to H^0(\ell,\mathcal O_\ell(2))$ records the conics containing a line mark. The full marked position, subject to the stabilizer action, is retained by \eqref{eq:ternary-orbit-test}. \end{proposition} \begin{proof} An infinitesimal projectivity preserving $U_{\mathrm{He}}$ is induced by $T\in\mathfrak{gl}(M_3)$ for which $Tq\in U_{\mathrm{He}}$ for its three displayed generators. Expanding coefficients of $x^2,y^2,z^2,xy,xz,yz$ gives eight independent linear constraints on the nine entries of $T$: its off-diagonal entries are zero and its three diagonal entries are equal. Thus the projective stabilizer has zero Lie algebra, and is finite in characteristic zero. Its orbits on $P$ and $P^*$ are finite, leaving two-dimensional marked families. Evaluation and restriction commute with projective change of coordinates, and their stated kernels have the claimed geometric meanings. The scalar $t_j$-actions add no restriction to $L$ or $U$. \end{proof} \subsection{Low-dimensional components of the isomorphism groupoids}\label{sec:lowdim} In this subsection $\s$ is semisimple, including $\s=0$; in particular there are no $kt_j$ factors. Write $\mathcal C_{\s}^{\leq6}$ for the full subgroupoid of $\mathcal C_{\s}$ on objects with $\dim\mathfrak n(R,K)=\dim R+\dim(\bigwedge^2R/K)\leq6$. No condition on $M_{0,\mathbf0_0}(R)$ is imposed, so the zero-action objects are included. \begin{theorem}[Bounded isomorphism components] \label{thm:bounded-pi0} Choose an $\s$-module $R$ in each isomorphism class $[R]$. Then \begin{equation}\label{eq:bounded-pi0} \pi_0(\mathcal C_{\s}^{\leq6})\ \cong\ \coprod_{\substack{[R]\\\dim R\leq6}}\quad \coprod_{d=0}^{6-\dim R} \Gr_{\binom{\dim R}{2}-d}^{\s}(\bigwedge^2R) \big/\Aut_{\s}(R). \end{equation} Here $\Gr_q^{\s}(\bigwedge^2R)$ is the set of invariant $q$-planes, with an empty Grassmannian interpreted as an empty summand. The quotient is the orbit set for the induced exterior-square action. \end{theorem} \begin{proof} The total dimension is invariant under isomorphism. Since $Z(\s)=0$, the central mixing map $c:R\to Z(\s)$ in Theorem~\ref{thm:general-iso} vanishes for every arrow in $\mathcal C_{\s}$, including when $\s=0$. The bracket identities and surjectivity of the canonical projection $\bigwedge^2R\to\bigwedge^2R/K$ then give $p=0$ and show that an arrow exists precisely when an $\s$-module isomorphism $a:R\to R'$ satisfies $(\bigwedge^2a)K=K'$. Conversely this $a$ induces an arrow. Partitioning the objects by $[R]$ and $d=\dim(\bigwedge^2R/K)$ gives the displayed orbit sets. \end{proof} Theorem~\ref{thm:bounded-pi0} computes components separately for each fixed acting algebra $\s$. The table uses patterns for that algebra: a label $\mathfrak a+\mathrm{junk}$ means $\s=\mathfrak a\oplus\mathfrak j$ for any chosen semisimple $\mathfrak j$, which acts trivially on $R$ and $\bigwedge^2R/K$. Substituting a particular $\mathfrak j$ makes the row a point of $\pi_0(\mathcal C_{\mathfrak a\oplus\mathfrak j}^{\leq6})$. The same row and orbit condition work for \emph{every} such $\mathfrak j$: its trivial action imposes no additional condition on $K$ or on an intertwiner of $R$. In each application the chosen $\s$ stays fixed during the isomorphism test. The plus sign belongs only to the \emph{acting-algebra label}; it does not adjoin $\mathfrak s$ to $\mathfrak n(R,K)$. \Needspace{8\baselineskip} \begin{theorem}[Small-dimensional components for specified actions] \label{thm:small-equivariant-table} Let $k$ be algebraically closed of characteristic zero and let $\dim\mathfrak n(R,K)\leq6$. The table lists every component in this dimension range for each active-algebra pattern, with the acting algebra fixed pointwise in its isomorphisms. For each row labelled $\mathfrak a+\mathrm{junk}$ and each chosen semisimple $\mathfrak j$, the row gives a point of $\pi_0(\mathcal C_{\s}^{\leq6})$ for the fixed algebra $\s=\mathfrak a\oplus\mathfrak j$. Taking $\mathfrak j=0$ in the rows labelled $0+\mathrm{junk}$ exhausts $\pi_0(\mathcal C_0^{\leq6})$. The two binary-form ranks for $R\cong2V_1$ are distinct for each fixed $\s=\mathfrak{sl}_2\oplus\mathfrak j$. Each row specifies $R$, the quotient dimension $d$, its orbit condition, and a representative kernel $K$. The acting algebra is never counted in the first column's dimension. Within each dimension the rows are ordered by $\dim\mathfrak a$ from smallest to largest; the wildcard $\mathfrak j$ does not enter this ordering. The proof gives the case reduction for all representation types of total dimension at most six. Here $\mathrm{triv}$ is the one-dimensional trivial module. For $\mathfrak{sl}_2$, $V_n$ is its irreducible module of highest weight $n$ (dimension $n+1$). For every other simple factor, $\omega_i$ denotes its $i$th fundamental weight in the usual Dynkin numbering, and $V_\lambda$ is the irreducible module of highest weight $\lambda$; thus $V_{2\omega_1}$ has highest weight twice the first fundamental weight. In particular, $V_{\omega_1}$ is the standard module of each displayed $\mathfrak{sl}_n$ ($n\ge3$) or $\mathfrak{sp}_{2n}$. The acting algebra in each row specifies which module is meant; $V_{\omega_2}\subset\bigwedge^2V_{\omega_1}$ in the $\mathfrak{sp}_4$ rows. The symbol $\boxtimes$ denotes an external tensor product over the displayed simple factors. For each isotypic summand $V_a\otimes M_a$ of $R$, choose a basis $e_{a,1},\ldots,e_{a,m_a}$ of the multiplicity space $M_a$. Write $e_{a,ij}=e_{a,i}\wedge e_{a,j}$ in $\bigwedge^2M_a$, and write $x_{a,i}$ for the same basis vector when it occurs in $\operatorname{Sym}M_a$; thus $x_{a,i}x_{a,j}$ belongs to $\operatorname{Sym}^2M_a$. The label $a=0$ denotes the trivial representation; in a row with one nontrivial constituent we label it $a$, and for two constituents we use $a,b$ in displayed order. A cross term $V_a\otimes V_b\otimes M_a\otimes M_b$ uses $e_{a,i}\otimes e_{b,j}$ in its multiplicity factors. Consequently every example in the ``Example $K$'' column is a submodule of the stated decomposition of $\bigwedge^2R$, rather than a span requiring a basis of $R$. In particular the same-isotypic term is \[ \bigwedge^2(V_a\otimes M_a) =\bigl(\operatorname{Sym}^2V_a\otimes\bigwedge^2M_a\bigr) \oplus\bigl(\bigwedge^2V_a\otimes\operatorname{Sym}^2M_a\bigr). \] For $K=0$ or $K=\bigwedge^2R$ the ``Example $K$'' column says precisely that. For a one-dimensional quotient, the table writes $K=\ker\omega_{\mathbf q}$. List the nontrivial symplectic irreducible types in $R$ in the order displayed, and let $\mathbf q=(q_0,q_1,\ldots,q_t)$: $q_0$ is the even rank of an alternating form on the trivial multiplicity space, while $q_i$ is the rank of a symmetric form on the multiplicity space of the $i$th symplectic type. Zeros are retained, including $q_0=0$ if there is no trivial summand. Choose standard forms of these ranks and fixed nonzero invariant alternating forms on the symplectic irreducibles. Their direct sum $\omega_{\mathbf q}$ is an invariant two-form on $R$; $\ker\omega_{\mathbf q}$ is its kernel as a map $\bigwedge^2R\to k$. Over $k$, these ranks determine the orbit under $\Aut_{\s}(R)$. Explicitly, on the trivial multiplicity space choose the form $\sum_{h=1}^{q_0/2}e_{0,2h-1}^*\wedge e_{0,2h}^*$. On the $i$th symplectic isotypic summand $V_{\lambda_i}\otimes M_i$, use $\beta_i\otimes\sum_{h=1}^{q_i}(x_{i,h}^*)^2$, where $(x^*)^2(u,v)=x^*(u)x^*(v)$; all cross terms vanish. When $t=0$, the rank tuple is written simply $(q_0)$. For the three trivial-action rows with $\dim R=4$ and $\dim K=4$, the \emph{annihilator pencil} is $\mathbb P(K^\perp)\subset\mathbb P(\bigwedge^2R^*)$. The Klein quadric $Q\subset\mathbb P(\bigwedge^2R^*)$ consists of decomposable two-forms. Thus ``contained,'' ``secant,'' and ``tangent'' below describe how this annihilator pencil meets $Q$. For $R=V_1\otimes M$ with $\dim M=2$, invariant alternating quotient forms correspond to $\operatorname{Sym}^2M^*$; the rank of the annihilator line in the two dimension-five $2V_1$ rows is the rank of its binary quadratic form. The rank-one form gives a three-dimensional Heisenberg summand and two central generators; the rank-two form is the evaluation pairing after choosing an identification $V_1\cong V_1^*$ between two copies of $V_1$. Thus $\operatorname{Heisen}_{\omega}(V_1)$ has $R=V_1$ and bracket given by its invariant alternating form, and $\operatorname{Heisen}(V_1)$ has $R=2V_1$ with bracket $b((u,\alpha),(v,\beta))=\beta(u)-\alpha(v)$. For the hyperbolic form $x_{a,1}^*x_{a,2}^*$, which over $k$ is equivalent to the standard form above, the kernel of $\operatorname{Heisen}(V_1)$ is $(V_2\otimes e_{a,12})\oplus\langle x_{a,1}^2,x_{a,2}^2\rangle$. In the two rows with $R=2V_1$ and $d=2$, $M=M_a$ is a two-dimensional multiplicity space and $U\subset\operatorname{Sym}^2M$ is the kernel line, as in Theorem~\ref{thm:small-n-one}; $\bigwedge^2(V_1\otimes M)=V_2\otimes\bigwedge^2M\oplus \mathrm{triv}\otimes\operatorname{Sym}^2M$. A rank-one line in $\operatorname{Sym}^2M$ is generated by a square $m^2$ for some $0\ne m\in M$. Apart from the zero-dimensional row, the $0+\mathrm{junk}$ rows in the final column use the $(R_q,K_q(\cdots))$ labels of Theorem~\ref{thm:small-two-step}, including its abelian rows. For the other rows, ``abelian'' means $K=\bigwedge^2R$. Write $\operatorname{Heisen}_{\omega}(S)=S\oplus k$ when $S$ has a nondegenerate invariant alternating form $\omega$, with $[u,v]=\omega(u,v)$ in its trivial central line; write $\operatorname{Heisen}(V)=V\oplus V^*\oplus k$ for the evaluation bracket. The Heisenberg labels in the final column refer to rows in which $\omega_{\mathbf q}$ is nondegenerate on all of $R$. A dash leaves that column unfilled; it does not assert that no decomposition or other description exists. \begingroup\normalfont\scriptsize\setlength{\tabcolsep}{0.35pt} \renewcommand{\arraystretch}{1.14} \begin{longtable}{@{}>{\raggedright\arraybackslash}p{0.18in} >{\raggedright\arraybackslash}p{1.07in} >{\raggedright\arraybackslash}p{1.40in} >{\raggedright\arraybackslash}p{0.18in} >{\raggedright\arraybackslash}p{1.15in} >{\raggedright\arraybackslash}p{1.64in} >{\raggedright\arraybackslash}p{1.00in}@{}} \hline $n$ & $\mathfrak a+\mathrm{junk}$ & $R$ & $d$ & $W$ and orbit & Example $K$ & Special form\\ \hline \endfirsthead \hline $n$ & $\mathfrak a+\mathrm{junk}$ & $R$ & $d$ & $W$ and orbit & Example $K$ & Special form\\ \hline \endhead 0 & $0+\mathrm{junk}$ & $0$ & 0 & $W=0$ & $0$ & zero algebra\\ \noalign{\hrule height 1.2pt} 1 & $0+\mathrm{junk}$ & $\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & $(R_1,K_1())$\\ \noalign{\hrule height 1.2pt} 2 & $0+\mathrm{junk}$ & $2\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & $(R_2,K_2())$\\ \hline 2 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \noalign{\hrule height 1.2pt} 3 & $0+\mathrm{junk}$ & $2\mathrm{triv}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2)$ & $\ker\omega_{(2)}$ & $(R_2,K_2(12))$\\ \hline 3 & $0+\mathrm{junk}$ & $3\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & $(R_3,K_3())$\\ \hline 3 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & $\operatorname{Heisen}_{\omega}(V_1)$\\ \hline 3 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 3 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 3 & $\mathfrak{sl}_3+\mathrm{junk}$ & $V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 3 & $\mathfrak{sl}_3+\mathrm{junk}$ & $V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \noalign{\hrule height 1.2pt} 4 & $0+\mathrm{junk}$ & $3\mathrm{triv}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2)$ & $\ker\omega_{(2)}$ & $(R_3,K_3(12))$\\ \hline 4 & $0+\mathrm{junk}$ & $4\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & $(R_4,K_4())$\\ \hline 4 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 4 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 4 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_3$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 4 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 4 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 4 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 4 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 4 & $\mathfrak{sl}_3+\mathrm{junk}$ & $\mathrm{triv}\oplus V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 4 & $\mathfrak{sl}_3+\mathrm{junk}$ & $\mathrm{triv}\oplus V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 4 & $\mathfrak{sp}_4+\mathrm{junk}$ & $V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 4 & $\mathfrak{sl}_4+\mathrm{junk}$ & $V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 4 & $\mathfrak{sl}_4+\mathrm{junk}$ & $V_{\omega_3}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \noalign{\hrule height 1.2pt} 5 & $0+\mathrm{junk}$ & $3\mathrm{triv}$ & 2 & $W=2\mathrm{triv}$; $\dim K=1$ & $\langle e_{0,23}\rangle$ & $(R_3,K_3(12,13))$\\ \hline 5 & $0+\mathrm{junk}$ & $4\mathrm{triv}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2)$ & $\ker\omega_{(2)}$ & $(R_4,K_4(12))$\\ \hline 5 & $0+\mathrm{junk}$ & $4\mathrm{triv}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(4)$ & $\ker\omega_{(4)}$ & $(R_4,K_4(12+34))$\\ \hline 5 & $0+\mathrm{junk}$ & $5\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & $(R_5,K_5())$\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,2)$ & $\ker\omega_{(0,2)}$ & $\operatorname{Heisen}(V_1)$\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2,0)$ & $\ker\omega_{(2,0)}$ & \textemdash\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2,1)$ & $\ker\omega_{(2,1)}$ & $\operatorname{Heisen}_{\omega}(R)$\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $3\mathrm{triv}\oplus V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_1\oplus V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_3$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & $\operatorname{Heisen}_{\omega}(R)$\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_4$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus2V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus V_1$ & 2 & $W=V_1$ & $\bigwedge^2V_1$ & \textemdash\\ \hline 5 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus V_3$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,0,1)$ & $\ker\omega_{(0,0,1)}$ & \textemdash\\ \hline 5 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1,0)$ & $\ker\omega_{(0,1,0)}$ & \textemdash\\ \hline 5 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1,1)$ & $\ker\omega_{(0,1,1)}$ & $\operatorname{Heisen}_{\omega}(R)$\\ \hline 5 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_2\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes\mathrm{triv}\oplus V_1\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes\mathrm{triv}\oplus V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_3+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_3+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sp}_4+\mathrm{junk}$ & $V_{\omega_1}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & $\operatorname{Heisen}_{\omega}(R)$\\ \hline 5 & $\mathfrak{sp}_4+\mathrm{junk}$ & $V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sp}_4+\mathrm{junk}$ & $\mathrm{triv}\oplus V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_4+\mathrm{junk}$ & $\mathrm{triv}\oplus V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_4+\mathrm{junk}$ & $\mathrm{triv}\oplus V_{\omega_3}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_5+\mathrm{junk}$ & $V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 5 & $\mathfrak{sl}_5+\mathrm{junk}$ & $V_{\omega_4}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \noalign{\hrule height 1.2pt} 6 & $0+\mathrm{junk}$ & $3\mathrm{triv}$ & 3 & $W=\bigwedge^2R$ & $0$ & $(R_3,K_3(12,13,23))$\\ \hline 6 & $0+\mathrm{junk}$ & $4\mathrm{triv}$ & 2 & $W=2\mathrm{triv}$; pencil contained in $Q$ & $\langle e_{0,14},e_{0,23},e_{0,24},e_{0,34}\rangle$ & $(R_4,K_4(12,13))$\\ \hline 6 & $0+\mathrm{junk}$ & $4\mathrm{triv}$ & 2 & $W=2\mathrm{triv}$; pencil secant to $Q$ & $\langle e_{0,13},e_{0,14},e_{0,23},e_{0,24}\rangle$ & $(R_4,K_4(12,34))$\\ \hline 6 & $0+\mathrm{junk}$ & $4\mathrm{triv}$ & 2 & $W=2\mathrm{triv}$; pencil tangent to $Q$ & $\langle e_{0,14},e_{0,23},e_{0,34},e_{0,13}-e_{0,24}\rangle$ & $(R_4,K_4(12,13+24))$\\ \hline 6 & $0+\mathrm{junk}$ & $5\mathrm{triv}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2)$ & $\ker\omega_{(2)}$ & $(R_5,K_5(12))$\\ \hline 6 & $0+\mathrm{junk}$ & $5\mathrm{triv}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(4)$ & $\ker\omega_{(4)}$ & $(R_5,K_5(12+34))$\\ \hline 6 & $0+\mathrm{junk}$ & $6\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & $(R_6,K_6())$\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2V_1$ & 2 & $W=2\mathrm{triv}$; $\operatorname{rk}U=1$ & $(V_2\otimes e_{a,12})\oplus\langle x_{a,1}^2\rangle$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2V_1$ & 2 & $W=2\mathrm{triv}$; $\operatorname{rk}U=2$ & $(V_2\otimes e_{a,12})\oplus\langle x_{a,1}x_{a,2}\rangle$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2\mathrm{triv}\oplus2V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_1$ & 2 & $W=2\mathrm{triv}$ & $V_1\otimes M_a\otimes M_0$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_1$ & 2 & $W=V_1$ & $(V_1\otimes e_{a,1}\otimes e_{0,2})\oplus(\bigwedge^2V_1\otimes x_{a,1}^2)\oplus\langle e_{0,12}\rangle$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_2$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2)$ & $\ker\omega_{(2)}$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_3$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $3V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $3\mathrm{triv}\oplus V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $3\mathrm{triv}\oplus V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2,0)$ & $\ker\omega_{(2,0)}$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $3\mathrm{triv}\oplus V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2,1)$ & $\ker\omega_{(2,1)}$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $3\mathrm{triv}\oplus V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $4\mathrm{triv}\oplus V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_1\oplus V_2$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_1\oplus V_3$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_2$ & 3 & $W=\bigwedge^2R$ & $0$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $V_5$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus2V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus2V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,2)$ & $\ker\omega_{(0,2)}$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus V_1$ & 3 & $W=\bigwedge^2R$ & $0$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus V_1\oplus V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus V_3$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_2+\mathrm{junk}$ & $\mathrm{triv}\oplus V_4$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $2(V_1\boxtimes\mathrm{triv})\oplus\mathrm{triv}\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $2(\mathrm{triv}\boxtimes\mathrm{triv})\oplus V_1\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $2(\mathrm{triv}\boxtimes\mathrm{triv})\oplus V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes V_1\oplus V_1\boxtimes\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes V_1\oplus\mathrm{triv}\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus2(\mathrm{triv}\boxtimes V_1)$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 2 & $W=2\mathrm{triv}$ & $V_1\boxtimes V_1$ & \textemdash\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_2$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_3$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_2\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_2\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_2\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_3\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes\mathrm{triv}\oplus V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,0,1)$ & $\ker\omega_{(0,0,1)}$ & \textemdash\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes\mathrm{triv}\oplus V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1,0)$ & $\ker\omega_{(0,1,0)}$ & \textemdash\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes\mathrm{triv}\oplus V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1,1)$ & $\ker\omega_{(0,1,1)}$ & \textemdash\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes\mathrm{triv}\oplus V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_2$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes\mathrm{triv}\oplus V_2\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $2V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $2V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_{\omega_1}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2)$ & $\ker\omega_{(2)}$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_{\omega_2}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(2)$ & $\ker\omega_{(2)}$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $3\mathrm{triv}\oplus V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $3\mathrm{triv}\oplus V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $V_{2\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $V_{2\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $V_{\omega_1}$ & 3 & $W=\bigwedge^2R$ & $0$ & \textemdash\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $V_{\omega_1}\oplus V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_3+\mathrm{junk}$ & $V_{\omega_2}$ & 3 & $W=\bigwedge^2R$ & $0$ & \textemdash\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes\mathrm{triv}\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\boxtimes V_1\oplus\mathrm{triv}\boxtimes V_1\boxtimes\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_2\oplus\mathfrak{sl}_2)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes\mathrm{triv}\boxtimes V_1$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sp}_4+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sp}_4+\mathrm{junk}$ & $\mathrm{triv}\oplus V_{\omega_1}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 6 & $\mathfrak{sp}_4+\mathrm{junk}$ & $\mathrm{triv}\oplus V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $V_1\boxtimes V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $V_1\boxtimes V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_1}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_2}$ & 1 & $W=\mathrm{triv}$; $\mathbf q=(0,1)$ & $\ker\omega_{(0,1)}$ & \textemdash\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $V_2\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $V_2\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes\mathrm{triv}\oplus V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes\mathrm{triv}\oplus V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sp}_4)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_4+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_4+\mathrm{junk}$ & $2\mathrm{triv}\oplus V_{\omega_3}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_4+\mathrm{junk}$ & $V_{\omega_2}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_3\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes V_{\omega_1}\oplus V_{\omega_1}\boxtimes\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_3\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes V_{\omega_1}\oplus V_{\omega_2}\boxtimes\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_3\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes V_{\omega_2}\oplus V_{\omega_1}\boxtimes\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_3\oplus\mathfrak{sl}_3)+\mathrm{junk}$ & $\mathrm{triv}\boxtimes V_{\omega_2}\oplus V_{\omega_2}\boxtimes\mathrm{triv}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_4)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $(\mathfrak{sl}_2\oplus\mathfrak{sl}_4)+\mathrm{junk}$ & $V_1\boxtimes\mathrm{triv}\oplus\mathrm{triv}\boxtimes V_{\omega_3}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sp}_6+\mathrm{junk}$ & $V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_5+\mathrm{junk}$ & $\mathrm{triv}\oplus V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_5+\mathrm{junk}$ & $\mathrm{triv}\oplus V_{\omega_4}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_6+\mathrm{junk}$ & $V_{\omega_1}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \hline 6 & $\mathfrak{sl}_6+\mathrm{junk}$ & $V_{\omega_5}$ & 0 & $W=0$ & $\bigwedge^2R$ & abelian\\ \noalign{\hrule height 1.2pt} \end{longtable} \endgroup \end{theorem} \begin{proof} Fix the acting algebra $\s$ throughout an orbit calculation. If $R=0$, then $K=0$ and $\mathfrak n(R,K)=0$, giving the first row for every $\s$. Henceforth suppose $R\ne0$. If $R$ is a sum of trivial $\s$-modules, the action on $\bigwedge^2R/K$ is also trivial. These are exactly the cases represented by the $0+\mathrm{junk}$ rows, taking $\mathfrak j=\s$; Theorem~\ref{thm:small-two-step} gives their complete kernel-orbit list through dimension six. Thus an additional row for a nonzero displayed $\mathfrak a$ must have at least one nontrivial irreducible constituent in $R$. Remove from $\s$ the direct sum $\mathfrak j$ of simple ideals acting trivially on $R$. The remaining algebra $\mathfrak a$ acts faithfully; this is the active part in the row label. Here is the dimension reduction behind the possible active parts. Every nontrivial irreducible constituent of $R$ has dimension at most six. For a simple factor the possibilities in this range are $V_1,\ldots,V_5$ for $\mathfrak{sl}_2$; $V_{\omega_1},V_{\omega_2},V_{2\omega_1},V_{2\omega_2}$ for $\mathfrak{sl}_3$ (dimensions $3,3,6,6$); $V_{\omega_1},V_{\omega_3},V_{\omega_2}$ for $\mathfrak{sl}_4$ (dimensions $4,4,6$); $V_{\omega_1},V_{\omega_4}$ for $\mathfrak{sl}_5$; $V_{\omega_1},V_{\omega_5}$ for $\mathfrak{sl}_6$; $V_{\omega_1},V_{\omega_2}$ for $\mathfrak{sp}_4$ (dimensions $4,5$); and $V_{\omega_1}$ for $\mathfrak{sp}_6$. For an irreducible on more than one factor, the only possibilities of dimension at most six are $V_1\boxtimes V_1$, $V_1\boxtimes V_2$, and $V_2\boxtimes V_1$ for two $\mathfrak{sl}_2$ factors, and $V_1\boxtimes V_{\omega_1}$ or $V_1\boxtimes V_{\omega_2}$ for $\mathfrak{sl}_2\oplus\mathfrak{sl}_3$. This follows from multiplying the dimensions of the nontrivial factors and the Weyl dimension formula for the simple types; in particular all other simple types have least nontrivial irreducible dimension above six. Take direct sums of these constituents and trivial modules with total dimension at most six, retaining every active factor. This bounds all possible pairs $(\mathfrak a,R)$, even when a displayed row abbreviates a family of choices of $\mathfrak j$. Put $r=\dim R$ and $d=\dim(\bigwedge^2R/K)$. Since $r+d\leq6$, the remaining orbit calculations break down as follows. For $d=0$, $K=\bigwedge^2R$: every allowed module $R$ gives its own abelian component, with no extra kernel choice. For $d=1$, the quotient is trivial as an $\s$-module, and $K$ is the kernel of a nonzero invariant alternating form, considered up to scalar and $\Aut_{\s}(R)$. Its radical $T$ and a nondegenerate equivariant complement $S$ give $R\cong S\oplus T$. For $r\leq5$ its rank is two or four. In rank two, $S$ is either $2\mathrm{triv}$ or the standard $V_1$ of an $\mathfrak{sl}_2$ factor; $T$ is any complementary module of dimension $r-2$. In rank four, $S$ is one of $4\mathrm{triv}$, $V_1\oplus2\mathrm{triv}$, $2V_1$ for one $\mathfrak{sl}_2$, $V_1\boxtimes\mathrm{triv}\oplus \mathrm{triv}\boxtimes V_1$ for two $\mathfrak{sl}_2$ factors, $V_3$ for $\mathfrak{sl}_2$, or $V_{\omega_1}$ for $\mathfrak{sp}_4$; here $T$ has dimension $r-4$. In each case the nondegenerate invariant form is unique up to an $\s$-module automorphism and scalar. Keeping the isomorphism types of $S$ and $T$, with the same fixed $\s$, is therefore the complete rank-one-quotient orbit test. For $r=5$ only $d=0,1$ occur, so this finishes that dimension. For $r=4$, the only further value is $d=2$. Complete reducibility makes the two-dimensional quotient either $2\mathrm{triv}$ or a standard $V_1$ for an $\mathfrak{sl}_2$ factor. A quotient $V_1$ occurs precisely for $R=V_1\oplus2\mathrm{triv}$ and uses one of the two cross-term copies of $V_1$; $\operatorname{GL}_2$ on the trivial multiplicity space is transitive on its nonzero quotient lines. A quotient $2\mathrm{triv}$ requires at least two independent invariant alternating forms. Apart from the purely trivial case, this leaves $R=2V_1$ for one factor, $R=V_1\oplus2\mathrm{triv}$, or the sum of the two standards for two distinct $\mathfrak{sl}_2$ factors. For $2V_1$ the kernel choice is a line in $\operatorname{Sym}^2M$, $\dim M=2$, with the two rank orbits described above; in the other active cases the two trivial summands of $\bigwedge^2R$ give a unique quotient. For $R=4\mathrm{triv}$, the annihilator pencil is contained in, secant to, or tangent to the Klein quadric; these are the three classes $(R_4,K_4(12,13))$, $(R_4,K_4(12,34))$ and $(R_4,K_4(12,13+24))$ of Theorem~\ref{thm:small-two-step}. For $r=3$, a nontrivial $R$ is $V_2$ or $V_1\oplus\mathrm{triv}$ for $\mathfrak{sl}_2$, or a standard module or its dual for $\mathfrak{sl}_3$. Their exterior squares are respectively $V_2$, $V_1\oplus\mathrm{triv}$, and the opposite standard. The first and last admit only $K=0$ or the full exterior square; the middle admits in addition its two individual summands as kernels. For $r=2$, the only active $R$ is $V_1$ for $\mathfrak{sl}_2$, with $K=0$ or the full exterior square. There is no active $R$ of smaller dimension. Finally $r=6$ forces $d=0$, already covered. This exhausts the dimension budget and gives every row of the displayed table. For each displayed row, direct exterior-square decomposition and substitution of its example $K$ verify the stated orbit condition. The equality $\Gr_q^{\mathfrak a\oplus\mathfrak j}(\bigwedge^2R) =\Gr_q^{\mathfrak a}(\bigwedge^2R)$ and $\Aut_{\mathfrak a\oplus\mathfrak j}(R)=\Aut_{\mathfrak a}(R)$ when $\mathfrak j$ acts trivially makes each such calculation uniform in the wildcard $\mathfrak j$. \end{proof} \begin{remark}[No continuous moduli below the cutoff] \label{rem:eta-cutoff} The cutoff $6$ is small enough that there are no continuous moduli: by Theorem~\ref{thm:small-equivariant-table}, each $\pi_0(\mathcal C_{\s}^{\leq6})$ is finite. Larger cutoffs bring in continuous moduli, for instance the invariant $\eta$ of \eqref{eq:family-J}. For semisimple $\s$, the slice of Theorem~\ref{thm:family-moduli} is $\s=\mathfrak{sl}_2$ with $R=V_3\otimes M_2$, the case $m=0$ of Proposition~\ref{prop:family-parameter}; already $\dim R=8$. By \eqref{eq:family-K}, \[ \dim\bigl(\textstyle\bigwedge^2R/K\bigr) =7(1-\dim L_0)+5(3-\dim U_1)+3(1-\dim L_2)+(3-\dim U_3). \] The rows of Theorem~\ref{thm:family-moduli} with parameter $\eta$ have $U_1$ and $U_3$ each a line or a plane, so their total dimension is at least $8+5+1=14$. This bound is attained, with $U_1,U_3$ planes and $L_0=L_2=\bigwedge^2M_2$, for every $\eta\ne4$. Within the cutoff, $V_3$ occurs only with a one-dimensional multiplicity space. \end{remark} \section{Mutually annihilating pairs of Lie brackets}\label{sec:pairs} The class-two objects above supply one source of pairs of Lie brackets on the same object. Pairing these brackets with two separately retained commutative associative products produces a further Lie algebra (Section~\ref{sec:quadratic-dual}). The rest of this section studies the pairs themselves: their groupoid and the exchange of the two brackets, the assembly of pairs in vector spaces from ordinary Lie algebras, and examples. The operadic statements below are over the characteristic-zero field $k$. The algebraic tensor construction allows infinite-dimensional objects, so the finite-dimensional convention is relaxed in Section~\ref{sec:quadratic-dual}; it is restored from Section~\ref{sec:switch} to the end of this section. Operations in the super setting are even, and $|x|$ denotes the parity of a homogeneous element. \subsection{The quadratic-dual pair and its tensor construction}\label{sec:quadratic-dual} \begin{definition}[Two independent commutative products]\label{def:commcomm} Write $\mathrm{CommComm}=\mathrm{Com}\amalg\mathrm{Com}$ for the coproduct of the two reduced operads governing commutative associative algebras without a specified unit. An algebra over it has products $\mu_1,\mu_2$ satisfying, separately for $i=1,2$, \[ \mu_i(a,b)=\mu_i(b,a),\qquad \mu_i(\mu_i(a,b),c)=\mu_i(a,\mu_i(b,c)). \] There are no relations involving both products. In super vector spaces, commutativity means $\mu_i(a,b)=(-1)^{|a||b|}\mu_i(b,a)$. Units may exist in particular examples, but are not part of the operadic data. \end{definition} \begin{definition}[Mutually annihilating brackets]\label{def:lielie-zero} Write $\mathrm{LieLieZero}$ for the operad with two skew-symmetric binary operations $\beta_1,\beta_2$, each satisfying Jacobi, and with \begin{equation}\label{eq:mixed-zero} \beta_1(\beta_2(x,y),z)=0,\qquad \beta_2(\beta_1(x,y),z)=0. \end{equation} In super vector spaces, skew-symmetry and Jacobi mean \[ \beta_i(x,y)=-(-1)^{|x||y|}\beta_i(y,x),\qquad \sum_{\mathrm{cyc}}(-1)^{|x||z|}\beta_i(x,\beta_i(y,z))=0 . \] \end{definition} \begin{proposition}[Quadratic duality]\label{prop:quadratic-duality} With the sign-twisted dual convention for binary quadratic operads, \[ (\mathrm{Com}\amalg\mathrm{Com})^!\cong\mathrm{LieLieZero}. \] \end{proposition} \begin{proof} The binary generator space consists of two symmetric generators. Its sign-twisted dual consists of two skew-symmetric generators. In arity three, split the free-operad space according to the two vertex labels. On each monochromatic summand, orthogonality of relations gives the usual $\mathrm{Com}^!=\mathrm{Lie}$ calculation. The original relation space has zero component in the mixed summands, so its orthogonal contains those entire summands. Every mixed composite is therefore zero in the dual operad. This is the quadratic-dual construction of \cite[Theorem~7.6.5]{LodayVallette}. \end{proof} The tensor construction uses only the quadratic dual $\mathcal P^!$. For a binary quadratic operad $\mathcal P$ with finite-dimensional generator space, the canonical morphism \[ \mathrm{Lie}\longrightarrow\mathcal P\otimes_{\mathrm H}\mathcal P^! \] uses the coevaluation of the generator space and its sign-twisted dual; $\otimes_{\mathrm H}$ denotes the aritywise tensor product of operads. It gives a Lie algebra on the tensor product of a $\mathcal P$-algebra and a $\mathcal P^!$-algebra \cite[Corollary~2.2.9(b)]{GinzburgKapranov}, \cite[Proposition~7.6.10]{LodayVallette}. For Proposition~\ref{prop:quadratic-duality} this becomes the following explicit formula. \begin{theorem}[The paired tensor bracket]\label{thm:paired-tensor} Let $(A,\mu_1,\mu_2)$ be a $\mathrm{CommComm}$-algebra and $(B,\beta_1,\beta_2)$ a $\mathrm{LieLieZero}$-algebra. The formula \begin{equation}\label{eq:paired-bracket} [a\otimes x,b\otimes y]=\mu_1(a,b)\otimes\beta_1(x,y)+\mu_2(a,b)\otimes\beta_2(x,y) \end{equation} defines a Lie algebra $L(A,B)$ on $A\otimes B$. In super vector spaces the right-hand side is multiplied by $(-1)^{|x||b|}$, the sign for interchanging the middle tensor factors. Morphisms preserving the two operations on each factor induce Lie algebra morphisms on their tensor products. No scalar weights are needed: rescaling $\beta_1$ or $\beta_2$ by a scalar gives another $\mathrm{LieLieZero}$-algebra, and likewise for $\mu_1,\mu_2$. \end{theorem} \begin{proof} Commutativity and skew-symmetry give skew-symmetry of the tensor bracket. In the ordinary case, its Jacobiator on pure tensors is the sum, over $i,j\in\{1,2\}$, of \[ \sum_{\mathrm{cyc}}\mu_j(\mu_i(a,b),c)\otimes\beta_j(\beta_i(x,y),z). \] If $i\neq j$, each summand vanishes by \eqref{eq:mixed-zero}. If $i=j$, associativity and commutativity give a common coefficient multiplying the Jacobiator of $\beta_i$. In the super case the symmetry morphisms give the displayed Koszul sign and the same argument uses super Jacobi. \end{proof} \begin{remark}[Compatibility is weaker]\label{rem:compatibility-weaker} The operad $\mathrm{Lie}_2$ of \cite{DotsenkoKhoroshkin} requires each bracket to be Lie and only the mixed cyclic identity \[ \sum_{\mathrm{cyc}}\big(\beta_1(\beta_2(x,y),z)+\beta_2(\beta_1(x,y),z)\big)=0 \] in the ordinary case. It guarantees that every scalar linear combination is Lie. In \eqref{eq:paired-bracket}, however, mixed nested brackets carry different coefficients from the two products on $A$. Their cyclic cancellation alone does not justify the construction with unrestricted $\mu_1,\mu_2$. There is a quotient $\mathrm{Lie}_2\twoheadrightarrow\mathrm{LieLieZero}$; the latter imposes the individual mixed vanishings. In order to have the capability to do two different multiplications with no law between them, the bracket side becomes stricter as $\mathrm{LieLieZero}$ vs just $\mathrm{Lie}_2$. \end{remark} \subsection{The groupoid of pairs and the bracket exchange}\label{sec:switch} From here to the end of Section~\ref{sec:pairs} we return to the global conventions of the earlier sections: $k$ is algebraically closed of characteristic zero, all representations are finite-dimensional, and $\s$ is reductive, possibly zero, with a central action that need not be semisimple. Let $\mathcal M$ be $\s\text{-}\mathrm{Rep}$ or its super analogue $\s\text{-}\mathrm{sRep}$; the case $\s=0$ gives $\mathrm{Vect}_k$ and $\mathrm{sVect}_k$, which are included as this case. Skew-symmetry and Jacobi use the symmetry of $\mathcal M$; in particular an odd element need not have zero self-bracket. All subobjects, centers and isomorphisms are taken in $\mathcal M$, so arrows are even and $\s$-equivariant. Since the central action may be nonsemisimple, equivariant splittings are not automatic; none of the arguments below uses one. \begin{definition}[The pair groupoid]\label{def:pair-groupoid} Let $\mathcal G_{\mathcal M}=\operatorname{Core}\mathrm{Alg}_{\mathrm{LieLieZero}}(\mathcal M)$ be the groupoid of $\mathrm{LieLieZero}$-algebras in $\mathcal M$ (Definition~\ref{def:lielie-zero}) and their isomorphisms. Its objects are triples $X=(B,\beta_1,\beta_2)$ with $B\in\mathcal M$, each $\beta_i\colon B\otimes B\to B$ a Lie bracket in $\mathcal M$, and \[ \beta_1\circ(\beta_2\otimes\mathrm{id})=0,\qquad \beta_2\circ(\beta_1\otimes\mathrm{id})=0 . \] The case $\beta_1=\beta_2$ is allowed, in which case it satisfies the mixed identities exactly when the common bracket has nilpotency class at most two. An arrow $X\to X'$ is an isomorphism $g\colon B\to B'$ in $\mathcal M$ with $g\,\beta_i=\beta_i'\,(g\otimes g)$ for $i=1,2$; the bracket labels are preserved. If both brackets have class at most two, each $(B,\beta_i)$ is an object of $\operatorname{LieAlg}_{\leq2}(\mathcal M)$, defined as in Section~\ref{sec:wild}, with quotient $B/\operatorname{im}\beta_i$ and derived object $\operatorname{im}\beta_i$ playing the roles of $R$ and $\bigwedge^2R/K$ in Proposition~\ref{prop:construction}. \end{definition} The mixed identities are equivalent to $\operatorname{im}\beta_1\subseteq Z(\beta_2)$ and $\operatorname{im}\beta_2\subseteq Z(\beta_1)$. \begin{definition}[Bracket exchange and underlying object]\label{def:sigma} Let $\sigma\colon\mathcal G_{\mathcal M}\to\mathcal G_{\mathcal M}$ be the strict involution given by $\sigma(B,\beta_1,\beta_2)=(B,\beta_2,\beta_1)$ on objects and by the same underlying isomorphism on arrows. Let $\operatorname{Fgt}$ send an object to its underlying vector space, or super vector space, and an arrow to its underlying linear map, so that $\operatorname{Fgt}\circ\sigma=\operatorname{Fgt}$. Every $\operatorname{Hom}$-set of $\mathcal G_{\mathcal M}$ is thereby a subset of the linear maps between underlying spaces, cut out by $\s$-equivariance and the bracket equations. In particular $\sigma$ restricts to a group isomorphism $\operatorname{Aut}(X)\to\operatorname{Aut}(\sigma X)$, $g\mapsto\sigma(g)$, and $\operatorname{Fgt}(\sigma(g))=\operatorname{Fgt}(g)$. \end{definition} \begin{definition}[Switch; the switchable groupoid]\label{def:switchable} A \emph{switch} of $X$ is an arrow $F\colon X\to\sigma X$ of $\mathcal G_{\mathcal M}$. Its underlying linear isomorphism $\operatorname{Fgt}(F)\colon B\to B$ is $\s$-equivariant and satisfies \[ \operatorname{Fgt}(F)\beta_1=\beta_2\,(\operatorname{Fgt}(F)\otimes \operatorname{Fgt}(F)),\qquad \operatorname{Fgt}(F)\beta_2=\beta_1\,(\operatorname{Fgt}(F)\otimes \operatorname{Fgt}(F)). \] Let $\mathcal G^{\mathrm{sw}}_{\mathcal M}$ be the full subgroupoid of $\mathcal G_{\mathcal M}$ on the objects $X$ with $\operatorname{Hom}(X,\sigma X)\neq\varnothing$. No switch is chosen as part of an object. This subgroupoid is replete, and $\sigma$ maps it to itself. At each object $X$ we record the two hom-sets \[ \operatorname{Hom}(X,X)=\operatorname{Aut}(X)\qquad\text{and}\qquad\operatorname{Hom}(X,\sigma X). \] An arrow $h\colon X\to Y$ transports them to $Y$ by $g\mapsto hgh^{-1}$ and $F\mapsto\sigma(h)\,F\,h^{-1}$. \end{definition} \begin{proposition}[The set of switches]\label{prop:switch-set} Let $X\in\mathcal G^{\mathrm{sw}}_{\mathcal M}$ and $F_0\in\operatorname{Hom}(X,\sigma X)$. \begin{enumerate} \item Precomposition $g\mapsto F_0\circ g$ is a bijection $\operatorname{Aut}(X)\to\operatorname{Hom}(X,\sigma X)$, and postcomposition $h\mapsto h\circ F_0$ is a bijection $\operatorname{Aut}(\sigma X)\to\operatorname{Hom}(X,\sigma X)$. Thus $\operatorname{Hom}(X,\sigma X)$ is a right $\operatorname{Aut}(X)$-torsor and a left $\operatorname{Aut}(\sigma X)$-torsor. \item The composite $c=\sigma(F_0)\circ F_0\colon X\to\sigma X\to X$ lies in $\operatorname{Aut}(X)$. \item The map $\theta(g)=F_0^{-1}\circ\sigma(g)\circ F_0$ is an automorphism of the group $\operatorname{Aut}(X)$ with $\theta^2(g)=c^{-1}gc$. Replacing $F_0$ by $F_0\circ h$ replaces $\theta$ by $h^{-1}\theta(-)h$. Hence the class $[\theta]\in\operatorname{Out}(\operatorname{Aut}X)$ depends only on $X$, and $[\theta]^2=1$. \end{enumerate} \end{proposition} \begin{proof} For switches $F,F'\colon X\to\sigma X$, the composite $F^{-1}\circ F'$ is an arrow $X\to X$ and $F'\circ F^{-1}$ is an arrow $\sigma X\to\sigma X$, which gives (1). Since $\sigma$ is a strict involution, $\sigma(F_0)\colon\sigma X\to X$, which gives (2). For (3), $\sigma(g)\in\operatorname{Aut}(\sigma X)$, so $\theta(g)\in\operatorname{Aut}(X)$ and $\theta$ is multiplicative. Using $\sigma(F_0)=c\,F_0^{-1}$ and $\sigma^2=\mathrm{id}$, \[ \theta^2(g)=F_0^{-1}\,\sigma(F_0)^{-1}\,g\,\sigma(F_0)\,F_0=c^{-1}gc . \] Finally $(F_0h)^{-1}\sigma(g)(F_0h)=h^{-1}\theta(g)h$. \end{proof} \begin{definition}[Homotopy fixed points]\label{def:hfp} Let $\mathcal G^{h\mathbb Z/2}_{\mathcal M}$ have objects $(X,F)$ with $F\colon X\to\sigma X$ and $\sigma(F)\circ F=\mathrm{id}_X$; such an $F$ is an \emph{involutive switch}. An arrow $a\colon(X,F)\to(Y,F')$ is an arrow $a\colon X\to Y$ of $\mathcal G_{\mathcal M}$ with $\sigma(a)\circ F=F'\circ a$ as arrows $X\to\sigma Y$. The forgetful functor $\mathcal G^{h\mathbb Z/2}_{\mathcal M}\to\mathcal G^{\mathrm{sw}}_{\mathcal M}$ is faithful and in general not full, since arrows of $\mathcal G^{h\mathbb Z/2}_{\mathcal M}$ must commute with the chosen switches. An object $X$ lies in its essential image exactly when some $F\in\operatorname{Hom}(X,\sigma X)$ satisfies $\sigma(F)\circ F=\mathrm{id}_X$. \end{definition} \begin{proposition}[Lifting and fibers]\label{prop:lifting} Let $F_0$, $c$ and $\theta$ be as in Proposition~\ref{prop:switch-set}. For $g\in\operatorname{Aut}(X)$, \[ \sigma(F_0g)\circ(F_0g)=c\,\theta(g)\,g , \] so $F_0g$ is involutive exactly when $c\,\theta(g)\,g=1$. An element $a\in\operatorname{Aut}(X)$ is an arrow $(X,F_0g)\to(X,F_0g')$ exactly when $g'=\theta(a)\,g\,a^{-1}$. Hence, if $F_0$ is itself involutive, the objects of $\mathcal G^{h\mathbb Z/2}_{\mathcal M}$ with underlying object $X$, up to the arrows over $\operatorname{Aut}(X)$, are classified by \[ H^1(\mathbb Z/2,\operatorname{Aut}X)=\{g:\ \theta(g)\,g=1\}\big/\big(g\sim\theta(a)\,g\,a^{-1}\big), \] where the generator of $\mathbb Z/2$ acts on $\operatorname{Aut}(X)$ through $\theta$, and $\operatorname{Aut}(X,F_0g)=\{a\in\operatorname{Aut}(X):\ \theta(a)\,g=g\,a\}$. \end{proposition} \begin{proof} $\sigma(F_0g)\,F_0g=\sigma(F_0)\,\sigma(g)\,F_0\,g=c\,F_0^{-1}\sigma(g)F_0\,g=c\,\theta(g)\,g$. The arrow condition $\sigma(a)\,F_0g=F_0g'\,a$ is $g'=F_0^{-1}\sigma(a)F_0\,g\,a^{-1}=\theta(a)\,g\,a^{-1}$. \end{proof} \begin{remark}[Parity in super categories]\label{rem:parity} In $\s\text{-}\mathrm{sRep}$ the parity map $(-1)^{|\cdot|}$ of $B$ commutes with all even maps and preserves both brackets, since they are even. It therefore underlies a central element $\mathsf P_X\in\operatorname{Aut}(X)$, with $\sigma(\mathsf P_X)=\mathsf P_{\sigma X}$. It underlies an arrow $X\to\sigma X$ only when the two brackets of $X$ agree. If $F\colon X\to\sigma X$ is an involutive switch, so is $F\circ\mathsf P_X$. Their classes in $H^1(\mathbb Z/2,\operatorname{Aut}X)$ need not coincide. \end{remark} \begin{definition}[Decomposability]\label{def:decomposable} Direct sums are formed blockwise in $\mathcal G_{\mathcal M}$, and $\sigma(X\oplus X')=\sigma X\oplus\sigma X'$. Hence $\mathcal G^{\mathrm{sw}}_{\mathcal M}$ is closed under them, using $F\oplus F'\colon X\oplus X'\to\sigma X\oplus\sigma X'$. An object $X$ of $\mathcal G^{\mathrm{sw}}_{\mathcal M}$ is \emph{decomposable} if $X\cong X_1\oplus X_2$ with $X_1,X_2$ nonzero objects of $\mathcal G^{\mathrm{sw}}_{\mathcal M}$, so that each summand is itself switchable. No switch is specified. In $\mathcal G^{h\mathbb Z/2}_{\mathcal M}$ the summands must be objects $(X_i,F_i)$ with $F=F_1\oplus F_2$. \end{definition} \begin{proposition}[Symmetrization]\label{prop:symmetrize} For every object $X$ of $\mathcal G_{\mathcal M}$, the object $X\oplus\sigma X$ lies in $\mathcal G^{\mathrm{sw}}_{\mathcal M}$, and the exchange of the two summands underlies an involutive switch $X\oplus\sigma X\to\sigma(X\oplus\sigma X)$. \end{proposition} \begin{proof} Write $X=(B,\beta_1,\beta_2)$. Then $X\oplus\sigma X=(B\oplus B,\ \beta_1\oplus\beta_2,\ \beta_2\oplus\beta_1)$ and $\sigma(X\oplus\sigma X)=(B\oplus B,\ \beta_2\oplus\beta_1,\ \beta_1\oplus\beta_2)$. The exchange $\tau(b,b')=(b',b)$ satisfies $\tau(\beta_1\oplus\beta_2)=(\beta_2\oplus\beta_1)(\tau\otimes\tau)$ and $\tau(\beta_2\oplus\beta_1)=(\beta_1\oplus\beta_2)(\tau\otimes\tau)$, and $\tau^2=\mathrm{id}$. \end{proof} When $X$ itself is not switchable, $X\oplus\sigma X$ is a direct sum of two nonzero objects of $\mathcal G_{\mathcal M}$ but need not be decomposable in $\mathcal G^{\mathrm{sw}}_{\mathcal M}$; the object $\mathsf S$ of Proposition~\ref{prop:vect4} is an example. \subsection{Assembly from ordinary Lie algebras}\label{sec:assembly} The following structural description is for ordinary vector spaces over $k$. Its construction is meaningful in other abelian symmetric monoidal categories, but the converse below uses a linear splitting. \begin{theorem}[Fiber product and central identification]\label{thm:fiber-product} Take ordinary Lie algebras $L_1,L_2$, surjections $p_i\colon L_i\to H$ onto an abelian Lie algebra, and injections \[ j_i\colon C\hookrightarrow Z(L_i)\cap\ker p_i . \] The vector space \begin{equation}\label{eq:fiber-product} B=(L_1\times_HL_2)/\Delta^-_C,\qquad \Delta^-_C:=\{(j_1c,-j_2c):c\in C\}, \end{equation} has mutually annihilating brackets induced by \[ \beta_1((x_1,x_2),(y_1,y_2))=([x_1,y_1],0),\qquad \beta_2((x_1,x_2),(y_1,y_2))=(0,[x_2,y_2]). \] Every $\mathrm{LieLieZero}$-algebra in vector spaces admits such a presentation with $\ker p_i=[L_i,L_i]$. In these normalized presentations, isomorphism classes of pairs are in bijection with isomorphism classes of the diagrams $(L_1,L_2,H,C,p_1,p_2,j_1,j_2)$. The two bracket labels are preserved. \end{theorem} \begin{proof} Since $H$ is abelian, the displayed brackets take values in the fiber product. The subspace $\Delta^-_C$ is central for both, so they descend. Jacobi holds in each coordinate and every mixed composition is zero. Conversely, put $I_i=\operatorname{im}\beta_i$, $C=I_1\cap I_2$, and $H=B/(I_1+I_2)$. Mixed vanishing gives $I_1\subseteq Z(B,\beta_2)$ and $I_2\subseteq Z(B,\beta_1)$. Choose a linear section $s\colon H\to B$ and set $L_i=s(H)+I_i$, using $\beta_i$ on $L_i$. Because the other derived subspace is central for that bracket, $[L_i,L_i]=I_i$. The map \[ (s(h)+u,\ s(h)+w)\longmapsto s(h)+u+w \] from $L_1\times_HL_2$ onto $B$ preserves both brackets and has kernel $\{(c,-c):c\in C\}$. This proves the asserted presentation. To check independence of the section, write $s'-s=u+w$ with linear maps $u\colon H\to I_1$, $w\colon H\to I_2$. The map $s(h)+a\mapsto s(h)+w(h)+a$ identifies the first $L_1$ with the one constructed from $s'$: $w(H)$ is central for $\beta_1$. It fixes $I_1$ and induces the identity on $H$. For $L_2$ use the corresponding addition of $u(h)$. Both maps fix $C$. An isomorphism of pairs preserves $I_1,I_2,C,H$ and therefore induces an isomorphism of the resulting diagrams, after these section changes. Conversely, a diagram isomorphism induces an isomorphism of \eqref{eq:fiber-product}. Recovering a diagram from its assembled pair returns an isomorphic diagram: choose sections of both $p_i$ and use their paired section. \end{proof} \subsection{Examples of pairs}\label{sec:pair-examples} \begin{example}[Two class-two brackets with common quotient and center]\label{ex:common-quotient} Let $R$ and $W$ be $\s$-modules and let $b_1,b_2\colon\bigwedge^2R\to W$ be $\s$-module maps. On $B=R\oplus W$ put \[ \beta_i((r,w),(r',w'))=(0,b_i(r\wedge r')). \] Both brackets have central image, so all same-label and mixed nested brackets vanish, and $X=(B,\beta_1,\beta_2)$ is an object of $\mathcal G_{\s\text{-}\mathrm{Rep}}$. No surjectivity or nondegeneracy is required. If $b_i$ is surjective with kernel $K_i$, then $(B,\beta_i)$ is the class-two radical $R\oplus(\bigwedge^2R/K_i)$ of Proposition~\ref{prop:construction}, with $W$ identified with $W_{K_i}=\bigwedge^2R/K_i$ through $b_i$. Conversely, start from two objects $(R_1,K_1)$ and $(R_2,K_2)$ of the earlier sections. Choose $\s$-module isomorphisms $\phi_R\colon R_1\to R_2$ and $\phi_W\colon W_{K_1}\to W_{K_2}$. The second bracket is transported to $R_1\oplus W_{K_1}$ by \[ \widetilde b_2(r\wedge r')=\phi_W^{-1}\,b_{K_2}(\phi_Rr\wedge\phi_Rr'). \] These identifications are data; the two brackets are subsequently classified under a simultaneous change of coordinates, that is, by arrows of $\mathcal G_{\s\text{-}\mathrm{Rep}}$. Two unrelated class-two brackets on the same space do not automatically have zero mixed compositions: their central subspaces must interact as required. \end{example} \begin{proposition}[Arrows between common-quotient pairs]\label{prop:common-quotient-switch} In Example~\ref{ex:common-quotient}, put $\operatorname{rad}b_i=\{r\in R:\ b_i(r\wedge R)=0\}$ and suppose $\operatorname{rad}b_1\cap\operatorname{rad}b_2=0$. Then $\operatorname{Fgt}$ identifies $\operatorname{Hom}(X,\sigma X)$ with the set of maps \[ (r,w)\longmapsto\big(a(r),\ h(r)+q(w)\big),\qquad a\in\operatorname{Aut}_{\s}(R),\ q\in\operatorname{Aut}_{\s}(W),\ h\in\operatorname{Hom}_{\s}(R,W), \] subject to \[ q\,b_1=b_2\circ\textstyle\bigwedge^2a,\qquad q\,b_2=b_1\circ\textstyle\bigwedge^2a , \] with $h$ unrestricted. The same holds for $\operatorname{Hom}(X,X)$ with the conditions $q\,b_i=b_i\circ\bigwedge^2a$. In particular $X\in\mathcal G^{\mathrm{sw}}_{\s\text{-}\mathrm{Rep}}$ exactly when such a pair $(a,q)$ exists, and a switch is involutive exactly when $a^2=\mathrm{id}_R$, $q^2=\mathrm{id}_W$ and $ha+qh=0$; for instance when $h=0$. When $b_1$ and $b_2$ are surjective with kernels $K_1,K_2$, the conditions imply $(\bigwedge^2a)K_1=K_2$ and $(\bigwedge^2a)K_2=K_1$. \end{proposition} \begin{proof} An element $(r,w)$ is central for $\beta_i$ exactly when $r\in\operatorname{rad}b_i$, so the common center $Z(\beta_1)\cap Z(\beta_2)$ is $(\operatorname{rad}b_1\cap\operatorname{rad}b_2)\oplus W=W$. An arrow $X\to\sigma X$ carries $Z(\beta_1)$ to $Z(\beta_2)$ and $Z(\beta_2)$ to $Z(\beta_1)$, hence preserves $W$. It therefore has the displayed form, where $a$ is the induced automorphism of $B/W=R$ and $q$ its restriction to $W$, both $\s$-equivariant. Both brackets depend only on the $R$-components and take values in $W$, so the arrow conditions $g\beta_1=\beta_2(g\otimes g)$ and $g\beta_2=\beta_1(g\otimes g)$ reduce to the two displayed equations, and impose nothing on $h$. Conversely every such map satisfies them. The case of $\operatorname{Hom}(X,X)$ is identical, and the square of the displayed map is $(r,w)\mapsto\big(a^2r,\ (ha+qh)(r)+q^2w\big)$. The statement about kernels follows by applying the equations to $K_i=\ker b_i$. \end{proof} In the notation of Theorem~\ref{thm:general-iso}, $p=0$ and the maps $a$, $h$, $q$ are the blocks of the arrow; the shear is $H=ha^{-1}$ as in Proposition~\ref{prop:groupoid-composition}, and the condition on kernels is the condition of Theorem~\ref{thm:groupoid} with the two kernels exchanged. \begin{example}[Shared abelian acting algebra]\label{ex:shared-abelian} Let $H$ be an abelian Lie algebra, allowed to be zero, and let $U_1,U_2$ be arbitrary Lie algebras. Choose Lie algebra homomorphisms \[ \delta_1\colon H\to\operatorname{Der}(U_1),\qquad \delta_2\colon H\to\operatorname{Der}(U_2). \] Thus the derivations in each action commute. On $B=H\oplus U_1\oplus U_2$ set \begin{align*} \beta_1((h,u,w),(h',u',w'))&=(0,\,[u,u']_{U_1}+\delta_1(h)u'-\delta_1(h')u,\,0),\\ \beta_2((h,u,w),(h',u',w'))&=(0,\,0,\,[w,w']_{U_2}+\delta_2(h)w'-\delta_2(h')w). \end{align*} Each bracket is a semidirect-product bracket with an additional central summand. The first has image in $U_1$, which is central for the second; the second has image in $U_2$, which is central for the first. Internal brackets on $U_1,U_2$ are retained, and neither algebra need be nilpotent. The choice $H=0$ includes brackets supported on the separate summands of $U_1\oplus U_2$. In the presentation of Theorem~\ref{thm:fiber-product}, this example uses the split extensions $H\ltimes U_1$ and $H\ltimes U_2$ over the displayed $H$, without requiring these quotient maps to be their full abelianizations. \end{example} \begin{proposition}[Switches of semidirect pairs]\label{prop:shared-abelian-switch} In Example~\ref{ex:shared-abelian}, let $\phi\colon U_1\to U_2$ be a Lie algebra isomorphism and $\alpha$ a linear automorphism of $H$ with \[ \phi\,\delta_1(h)\,\phi^{-1}=\delta_2(\alpha h),\qquad \phi\,\delta_1(\alpha h)\,\phi^{-1}=\delta_2(h)\qquad(h\in H). \] Then $(h,u,w)\mapsto(\alpha h,\ \phi^{-1}w,\ \phi u)$ is $\operatorname{Fgt}$ of an arrow $X\to\sigma X$. Its square is $\alpha^2\oplus\mathrm{id}$, so it is an involutive switch when $\alpha^2=\mathrm{id}_H$; in particular when $\alpha=\mathrm{id}_H$ and $\phi$ intertwines $\delta_1$ with $\delta_2$. \end{proposition} \begin{proof} Write $g$ for the displayed map. Applied to $\beta_1$, \[ g\beta_1((h,u,w),(h',u',w'))=\big(0,0,\ \phi[u,u']_{U_1}+\phi\delta_1(h)u'-\phi\delta_1(h')u\big), \] and the first condition rewrites the last entry as $[\phi u,\phi u']_{U_2}+\delta_2(\alpha h)\phi u'-\delta_2(\alpha h')\phi u$, which is the $U_2$-entry of $\beta_2(g(h,u,w),g(h',u',w'))$. Applied to $\beta_2$, the second condition in the form $\phi^{-1}\delta_2(h)\phi=\delta_1(\alpha h)$ gives $g\beta_2=\beta_1(g\otimes g)$ in the same way. The square is $(h,u,w)\mapsto(\alpha^2h,u,w)$. \end{proof} \subsection{Small examples and equivariant switches}\label{sec:switch-examples} As in Definition~\ref{def:sigma}, every $\operatorname{Hom}$-set below is described as a subset of the linear maps of $\operatorname{Fgt}(X)$. \begin{proposition}[Total dimension at most two]\label{prop:dim2} Let $\s=0$. \begin{enumerate} \item In superdimensions $1\,|\,0$, $0\,|\,1$ and $0\,|\,2$ both brackets vanish. \item In superdimension $2\,|\,0$ every class of $\mathcal G_{\mathcal M}$ is represented by exactly one of \[ (\beta_1,\beta_2)=(0,0),\qquad \big([x,y]_1=y,\ 0\big),\qquad \big(0,\ [x,y]_2=y\big). \] Only $(0,0)$ lies in $\mathcal G^{\mathrm{sw}}_{\mathcal M}$. \item In superdimension $1\,|\,1$, with $x$ even and $\vartheta$ odd, every class of $\mathcal G_{\mathcal M}$ is represented by exactly one of the zero pair, \[ \begin{gathered} \big([x,\vartheta]_1=\vartheta,\ 0\big),\qquad \big(0,\ [x,\vartheta]_2=\vartheta\big),\qquad \big(0,\ [\vartheta,\vartheta]_2=x\big),\\ O_\gamma=\big([\vartheta,\vartheta]_1=x,\ [\vartheta,\vartheta]_2=\gamma x\big)\qquad(\gamma\in k). \end{gathered} \] The objects of $\mathcal G^{\mathrm{sw}}_{\mathcal M}$ among them are the zero pair and the objects $O_{\pm1}$ with $\gamma=\pm1$. The identity map underlies an involutive switch $O_{+1}\to\sigma O_{+1}$, and $x\mapsto -x$, $\vartheta\mapsto\vartheta$ underlies one for $O_{-1}$. \end{enumerate} \end{proposition} \begin{proof} (1) An even bracket on $1\,|\,0$ vanishes by skew-symmetry. On $0\,|\,1$ and $0\,|\,2$ every bracket of two odd vectors lies in the even part, which is zero. (2) A nonabelian Lie bracket on $k^2$ has zero center and image of dimension one. If $\beta_1\neq0$, then $\operatorname{im}\beta_1\subseteq Z(\beta_2)$ forces $\beta_2=0$, and symmetrically. If $\beta_2=0$, every element of $GL_2$ preserves $\beta_2$, so the arrows act on $\beta_1$ alone and bring it to $[x,y]_1=y$; likewise with the labels exchanged. The two nonzero representatives are not isomorphic, since arrows preserve labels. A switch carries $\beta_1$ to $\beta_2$, so a switchable object has both brackets zero or both nonzero; only $(0,0)$ remains. (3) Write $[x,\vartheta]_i=\alpha_i\vartheta$ and $[\vartheta,\vartheta]_i=\gamma_ix$. Jacobi for each bracket is $\alpha_i\gamma_i=0$, and mutual annihilation is $\alpha_1\alpha_2=\alpha_1\gamma_2=\alpha_2\gamma_1=0$. The even linear automorphisms are $x\mapsto ax$, $\vartheta\mapsto b\vartheta$, and they act on both brackets at once by $(\alpha_i,\gamma_i)\mapsto(a\alpha_i,\ a^{-1}b^2\gamma_i)$ with the same $a,b$. If some $\alpha_i\neq0$, the equations force the other three coefficients to vanish, and $a$ normalizes $\alpha_i$ to $1$. If $\alpha_1=\alpha_2=0$, both $\gamma_i$ are rescaled by the single factor $a^{-1}b^2$, so the ratio $[\gamma_1:\gamma_2]\in\mathbb P^1(k)$ is a complete invariant, and the displayed representatives realize each value once. The exchange $\sigma$ sends $[\gamma_1:\gamma_2]$ to $[\gamma_2:\gamma_1]$ and exchanges the two semidirect classes, so the switchable classes are the zero pair and $\gamma^2=1$. The displayed maps satisfy $g\beta_1=\beta_2(g\otimes g)$ and $g\beta_2=\beta_1(g\otimes g)$ and square to the identity. \end{proof} Write $\mathfrak r_2$ for the nonabelian two-dimensional Lie algebra, $[e_1,e_2]=e_2$. For the next proposition, recall Theorem~\ref{thm:small-two-step} and its notation $ij=u^i\wedge u^j$. The objects $\mathsf A_\epsilon$, $\mathsf N$, $\mathsf D$ below are instances of Example~\ref{ex:common-quotient}, with $R$ spanned by the noncentral basis vectors and $W$ by the central ones: $\mathsf A_\epsilon$ has $R=R_2$, $W=k$, $b_1=12$, $b_2=\epsilon\cdot12$; $\mathsf N$ has $R=R_3$, $W=k$, $b_1=23$, $b_2=13$; and $\mathsf D$ has $R=R_2$, $W=k^2$, $b_1=(12,0)$, $b_2=(0,12)$. The objects $\mathsf B_\epsilon$ and $\mathsf S$ are instances of Example~\ref{ex:shared-abelian}: for $\mathsf B_\epsilon$ take $H=ke_1$, $U_1=ke_2$, $U_2=ke_3$ abelian, $\delta_1(e_1)=1$, $\delta_2(e_1)=\epsilon$; for $\mathsf S$ take $H=0$ and $U_1=\langle e_1,e_2\rangle\cong U_2=\langle e_3,e_4\rangle\cong\mathfrak r_2$. Equivalently, $\mathsf S=X\oplus\sigma X$ for $X=(\mathfrak r_2,[\ ,\ ],0)$, as in Proposition~\ref{prop:symmetrize}. \begin{proposition}[Switchable pairs in $\mathrm{Vect}_k$ through dimension four]\label{prop:vect4} Let $\s=0$, so that $\mathcal M=\mathrm{Vect}_k$. Every object of $\mathcal G^{\mathrm{sw}}_{\mathcal M}$ of dimension at most four is isomorphic to a direct sum of the following objects, which are indecomposable in $\mathcal G^{\mathrm{sw}}_{\mathcal M}$ and pairwise nonisomorphic. All unlisted brackets are zero, $\epsilon=\pm1$, and $F_0$ gives $\operatorname{Fgt}$ of an involutive switch $F_0\colon X\to\sigma X$, fixing unlisted basis vectors. \begin{center} \begin{tabular}{llll} & $\beta_1$ & $\beta_2$ & $F_0$\\\hline $\mathbf 1$ & $0$ & $0$ & $\mathrm{id}$\\ $\mathsf A_\epsilon$ & $[e_1,e_2]_1=e_3$ & $[e_1,e_2]_2=\epsilon e_3$ & $e_3\mapsto\epsilon e_3$\\ $\mathsf B_\epsilon$ & $[e_1,e_2]_1=e_2$ & $[e_1,e_3]_2=\epsilon e_3$ & $e_1\mapsto\epsilon e_1,\ e_2\leftrightarrow e_3$\\ $\mathsf N$ & $[e_2,e_3]_1=e_4$ & $[e_1,e_3]_2=e_4$ & $e_1\leftrightarrow e_2$\\ $\mathsf D$ & $[e_1,e_2]_1=e_3$ & $[e_1,e_2]_2=e_4$ & $e_3\leftrightarrow e_4$\\ $\mathsf S$ & $[e_1,e_2]_1=e_2$ & $[e_3,e_4]_2=e_4$ & $e_1\leftrightarrow e_3,\ e_2\leftrightarrow e_4$ \end{tabular} \end{center} In every row $\operatorname{Hom}(X,\sigma X)=F_0\circ\operatorname{Hom}(X,X)$, and $\operatorname{Fgt}(\operatorname{Hom}(X,X))\subseteq GL(\operatorname{Fgt}X)$ consists of the following maps: \begin{itemize} \item $\mathsf A_\epsilon$: $\begin{pmatrix}M&0\\ w^{T}&\det M\end{pmatrix}$ with $M\in GL_2$ and $w\in k^2$; \item $\mathsf B_\epsilon$: $e_1\mapsto e_1+pe_2+qe_3$, $e_2\mapsto be_2$, $e_3\mapsto de_3$; \item $\mathsf N$: $e_1\mapsto ae_1+xe_4$, $e_2\mapsto ae_2+ye_4$, $e_3\mapsto de_3+pe_1+qe_2+re_4$, $e_4\mapsto ade_4$; \item $\mathsf D$: $\begin{pmatrix}M&0\\ N&\det M\cdot I_2\end{pmatrix}$ with $M\in GL_2$; \item $\mathsf S$: $\operatorname{Aut}(\mathfrak r_2)\times\operatorname{Aut}(\mathfrak r_2)$ acting blockwise. \end{itemize} The decomposable objects of dimension at most four are $\mathbf 1^{\oplus n}$ with $n\ge2$, $\mathsf A_\epsilon\oplus\mathbf 1$ and $\mathsf B_\epsilon\oplus\mathbf 1$. Every object in this range has an involutive switch, so here $\mathcal G^{\mathrm{sw}}_{\mathcal M}$ coincides with the essential image of $\mathcal G^{h\mathbb Z/2}_{\mathcal M}$. \end{proposition} \begin{proof} Let $X=(B,\beta_1,\beta_2)$ be switchable with $n=\dim B\le4$, and put $I_i=\operatorname{im}\beta_i$. A switch carries $\beta_1$ to $\beta_2$, so either both brackets vanish, giving $X\cong\mathbf 1^{\oplus n}$, or both are nonzero; assume the latter. A switch carries $I_1$ onto $I_2$, so $d=\dim I_1=\dim I_2\ge1$. Mutual annihilation gives $I_j\subseteq Z(\beta_i)$ for $j\ne i$, so $\beta_i$ factors through $\bigwedge^2(B/I_j)$ and $d\le\binom{n-d}{2}$. For $n\le4$ this forces $n\ge3$ and $d=1$. Write $\beta_i=\omega_i\otimes v_i$ with $I_i=kv_i$ and $\omega_i$ a nonzero two-form vanishing on $I_j$. Since $\dim B/I_j\le3$, the form $\omega_i$ is decomposable, so $Z_i=\operatorname{rad}\omega_i=Z(\beta_i)$ has codimension two, and mutual annihilation reads $I_2\subseteq Z_1$, $I_1\subseteq Z_2$. Every decomposable $\omega\otimes v$ is a Lie bracket, since the Jacobiator is $v\otimes(\omega\wedge\iota_v\omega)$ and $\omega\wedge\iota_v\omega=0$ for decomposable $\omega$. Call $\beta_i$ of central type if $v_i\in Z_i$ and of noncentral type otherwise. A switch carries $(I_1,Z_1)$ to $(I_2,Z_2)$, so both brackets have the same type. \emph{Central type, $I_1=I_2=I$, $Z_1=Z_2=Z$.} Two decomposable forms with the same radical are proportional, so $\beta_2=t\beta_1$ for a scalar $t\neq0$ that every arrow preserves; $\sigma$ replaces $t$ by $t^{-1}$, so switchability forces $t=\pm1$. Choosing $e_1,e_2$ spanning a complement of $Z$ and $e_3=v_1$ gives $\mathsf A_t$ when $n=3$. When $n=4$, a vector $e_4\in Z\setminus I$ is central for both brackets and outside their images, and $X\cong\mathsf A_t\oplus\mathbf 1$. \emph{Central type, $I_1=I_2=I$, $Z_1\neq Z_2$.} Since $n=3$ would force $Z_1=Z_2=I$, we have $n=4$ and $Z_1\cap Z_2=I$. Take $e_4=v_1$, $e_1\in Z_1\setminus I$, $e_2\in Z_2\setminus I$ and $e_3$ completing a basis. Then $\omega_1$ is a multiple of $e_2^*\wedge e_3^*$ and $\omega_2$ of $e_1^*\wedge e_3^*$, and rescaling $e_1,e_2$ gives $\mathsf N$. \emph{Central type, $I_1\neq I_2$.} Then $I_1+I_2\subseteq Z_1\cap Z_2$, so $n=4$ and $Z_1=Z_2=I_1+I_2=Z$. Both forms lie in the line $\bigwedge^2(B/Z)^*$, and rescaling $v_2$ gives $\beta_i=\omega\otimes v_i$, which is $\mathsf D$ with $e_3=v_1$, $e_4=v_2$. \emph{Noncentral type.} Here $v_i\notin Z_i$, so $I_1\neq I_2$ and $Z_1\neq Z_2$. Define linear forms $c_i$ by $\beta_i(x,v_i)=c_i(x)v_i$; then $\ker c_i=Z_i+I_i$. An arrow $g\colon X\to X'$ carries $v_i$ into $I_i'$, so $c_i'\circ g=c_i$, and $\sigma$ exchanges $c_1$ and $c_2$. If $n=3$, then $Z_1=I_2$ and $Z_2=I_1$, so $\ker c_1=\ker c_2=I_1+I_2$ and $c_2=\lambda c_1$. The scalar $\lambda$ is preserved by arrows and inverted by $\sigma$, so $\lambda=\pm1$. With $e_2=v_1$, $e_3=v_2$ and $e_1$ normalized by $c_1(e_1)=1$, the object is $\mathsf B_\lambda$. If $n=4$, then $\dim(Z_1\cap Z_2)\in\{0,1\}$. If it is one, write $Z_1\cap Z_2=ke_4$; the same normalization gives $\mathsf B_\lambda\oplus\mathbf 1$ with $\lambda=\pm1$, since $e_4$ is central for both brackets and $\ker c_1=\ker c_2=\langle e_2,e_3,e_4\rangle$. If it is zero, then $B=Z_1\oplus Z_2$, $\beta_1$ is a nonabelian bracket on $Z_2$ vanishing when one argument lies in $Z_1$, and symmetrically, so $X\cong\mathsf S$. These cases exhaust the switchable objects. Each listed row is switchable by the displayed map, which one checks satisfies $f\beta_1=\beta_2(f\otimes f)$ and $f\beta_2=\beta_1(f\otimes f)$ for $f=\operatorname{Fgt}(F_0)$, together with $\operatorname{Fgt}(\sigma(F_0)\circ F_0)=f^2=\mathrm{id}$. The rows are pairwise nonisomorphic: $n$, the type, $\dim(I_1+I_2)$ and the invariants $t$ and $\lambda$ separate them. For indecomposability, a nonzero switchable summand of dimension at most two has both brackets zero by Proposition~\ref{prop:dim2}, so it lies in $Z(\beta_1)\cap Z(\beta_2)$ and meets $I_1+I_2$ trivially. In every listed row with nonzero brackets the common center is contained in $I_1+I_2$, so no such summand exists, and for $n\le4$ every decomposition has a summand of dimension at most two. Finally, an arrow preserves $I_i$ and $Z_i$; for each row, solving $g\beta_i=\beta_i(g\otimes g)$ on these flags gives the displayed groups. For example, for $\mathsf N$ one finds $g(e_1)\in Z_1$, $g(e_2)\in Z_2$, and the two bracket equations give $g(e_4)=a_1de_4=a_2de_4$, so the scalars on $e_1$ and $e_2$ agree. By Proposition~\ref{prop:switch-set}, $\operatorname{Hom}(X,\sigma X)=F_0\circ\operatorname{Hom}(X,X)$. \end{proof} \begin{proposition}[Supertranslation pairs in dimension $1\,|\,N$]\label{prop:supertranslation} Let $\s=0$ and $B=kH\oplus\Pi E$ in $\mathrm{sVect}_k$, with $H$ even. \begin{enumerate} \item For symmetric forms $b_1,b_2\colon\operatorname{Sym}^2E\to k$, the brackets $[H,x]_i=0$ and $[x,y]_i=b_i(x,y)H$ define an object $X(b_1,b_2)$ of $\mathcal G_{\mathcal M}$. \item $\operatorname{Fgt}$ identifies the arrows $X(b_1,b_2)\to X(b_1',b_2')$ with the pairs $(a,S)$ with $a\in k^\times$ and $S\in GL(E)$, acting by $H\mapsto aH$, $x\mapsto Sx$, such that $b_i'(Sx,Sy)=a\,b_i(x,y)$ for $i=1,2$. \item If $b_1$ is nondegenerate, put $T=b_1^{-1}b_2$. The classes are represented by \[ b_1=\bigoplus_j\mathsf E_{r_j},\qquad b_2=\bigoplus_j\mathsf E_{r_j}J_{r_j}(\lambda_j), \] where $\mathsf E_r$ has ones on its antidiagonal, $J_r(\lambda)$ is the Jordan block with ones on its superdiagonal, the $J_{r_j}(\lambda_j)$ are the Jordan blocks of $T$, and the blocks are unordered. \item $X(b_1,b_2)$ lies in $\mathcal G^{\mathrm{sw}}_{\mathcal M}$ exactly when the elementary divisors of the pencil $(b_1,b_2)$, regarded as points of $\mathbb P^1$ with their block sizes, form a multiset invariant under $\lambda\mapsto\lambda^{-1}$. \item For a general pair in dimension $1\,|\,N$ with $[H,x]_i=A_ix$ and $[x,y]_i=b_i(x,y)H$, the Jacobi identities are \[ b_i(A_ix,y)+b_i(x,A_iy)=0,\qquad b_i(x,y)A_iz+b_i(y,z)A_ix+b_i(z,x)A_iy=0, \] and for $i\neq j$ mutual annihilation is \[ A_jA_i=0,\qquad b_j(A_ix,y)=0,\qquad b_i(x,y)A_jz=0 . \] \end{enumerate} \end{proposition} \begin{proof} (1) All brackets land in $kH$, which is central for both, so every nested bracket of either label vanishes. (2) Even maps of $1\,|\,N$ are block diagonal, and the arrow condition is the displayed equation. (3) The shared factor $a$ is absorbed by replacing $S$ with $a^{-1/2}S$, so the classes are those of pairs of symmetric forms under simultaneous congruence; for $b_1$ nondegenerate these are given by the stated canonical form~\cite{Thompson1991}. Thompson works over $\mathbb C$; the canonical forms have rational entries and the invariants (elementary divisors and minimal indices) are defined algebraically, so the same statements hold over every algebraically closed field of characteristic zero by the Lefschetz principle. (4) The exchange $\sigma$ sends $(b_1,b_2)$ to $(b_2,b_1)$, which changes the pencil parameter by $\lambda\mapsto\lambda^{-1}$. Congruence classes of symmetric pencils are determined by the elementary divisors and the minimal indices~\cite{Thompson1991}, and minimal indices do not depend on the choice of generators of the pencil. (5) Expand the Jacobi identity and the mixed compositions on $H$ and on odd vectors. \end{proof} \begin{proposition}[Supertranslation pairs with an involutive switch]\label{prop:supertranslation-involutive} Let $\s=0$ and $N=2$ in Proposition~\ref{prop:supertranslation}, with even generator $H$ and odd basis $\vartheta_1,\vartheta_2$, and write forms as Gram matrices in this basis. \begin{enumerate} \item For $\mu\in k^\times$ with $\mu^2\ne1$, take $r_1=r_2=1$, $\lambda_1=\mu^2$ and $\lambda_2=\mu^{-2}$: \[ b_1=\mathsf E_1\oplus\mathsf E_1=\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad b_2=\mathsf E_1J_1(\mu^2)\oplus\mathsf E_1J_1(\mu^{-2})=\begin{pmatrix}\mu^2&0\\0&\mu^{-2}\end{pmatrix}. \] Then $H\mapsto H$, $\vartheta_1\mapsto\mu\vartheta_2$, $\vartheta_2\mapsto\mu^{-1}\vartheta_1$ is $\operatorname{Fgt}$ of an involutive switch $X\to\sigma X$. \item Take one block with $r_1=2$ and $\lambda_1=1$: \[ b_1=\mathsf E_2=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad b_2=\mathsf E_2J_2(1)=\begin{pmatrix}0&1\\1&1\end{pmatrix}. \] Then $H\mapsto -H$, $\vartheta_1\mapsto\vartheta_1$, $\vartheta_2\mapsto\tfrac12\vartheta_1-\vartheta_2$ is $\operatorname{Fgt}$ of an involutive switch $X\to\sigma X$. \end{enumerate} In both cases $b_1\ne b_2$. The elementary divisors are the points $\mu^2$ and $\mu^{-2}$ of $\mathbb P^1$, each with block size one, in (1), and the point $1$ with block size two in (2); each multiset is invariant under $\lambda\mapsto\lambda^{-1}$, as Proposition~\ref{prop:supertranslation}(4) requires. \end{proposition} \begin{proof} By Proposition~\ref{prop:supertranslation}(2), applied with target $\sigma X=X(b_2,b_1)$, a pair $(a,S)$ underlies an arrow $F\colon X\to\sigma X$ exactly when \[ b_2(Sx,Sy)=a\,b_1(x,y),\qquad b_1(Sx,Sy)=a\,b_2(x,y)\qquad(x,y\in E). \] Since $\operatorname{Fgt}\circ\sigma=\operatorname{Fgt}$ and $\operatorname{Fgt}$ is faithful, $\sigma(F)\circ F=\mathrm{id}_X$ exactly when $\operatorname{Fgt}(F)^2=\mathrm{id}$, that is, $a^2=1$ and $S^2=\mathrm{id}_E$. (1) Here $a=1$. Since $S$ exchanges the two $b_1$- and $b_2$-orthogonal lines $k\vartheta_1$ and $k\vartheta_2$, all mixed values vanish on both sides. On the diagonal, \[ b_1(S\vartheta_1,S\vartheta_1)=\mu^2=b_2(\vartheta_1,\vartheta_1),\qquad b_1(S\vartheta_2,S\vartheta_2)=\mu^{-2}=b_2(\vartheta_2,\vartheta_2), \] \[ b_2(S\vartheta_1,S\vartheta_1)=\mu^2\mu^{-2}=b_1(\vartheta_1,\vartheta_1),\qquad b_2(S\vartheta_2,S\vartheta_2)=\mu^{-2}\mu^2=b_1(\vartheta_2,\vartheta_2). \] Finally $S^2\vartheta_1=\mu\mu^{-1}\vartheta_1$ and $S^2\vartheta_2=\mu^{-1}\mu\vartheta_2$. (2) Here $a=-1$. With $S\vartheta_1=\vartheta_1$ and $S\vartheta_2=\tfrac12\vartheta_1-\vartheta_2$, \[ \begin{aligned} b_1(S\vartheta_1,S\vartheta_1)&=0=-b_2(\vartheta_1,\vartheta_1), & b_2(S\vartheta_1,S\vartheta_1)&=0=-b_1(\vartheta_1,\vartheta_1),\\ b_1(S\vartheta_1,S\vartheta_2)&=-1=-b_2(\vartheta_1,\vartheta_2), & b_2(S\vartheta_1,S\vartheta_2)&=-1=-b_1(\vartheta_1,\vartheta_2),\\ b_1(S\vartheta_2,S\vartheta_2)&=-1=-b_2(\vartheta_2,\vartheta_2), & b_2(S\vartheta_2,S\vartheta_2)&=-1+1=0=-b_1(\vartheta_2,\vartheta_2). \end{aligned} \] Moreover $S^2\vartheta_2=\tfrac12\vartheta_1-(\tfrac12\vartheta_1-\vartheta_2)=\vartheta_2$, and $a^2=1$. \end{proof} By contrast, the pair of Proposition~\ref{prop:obstruction} also has an inversion-invariant multiset of elementary divisors, but none of its switches is involutive. \begin{proposition}[Switchable without an involutive switch]\label{prop:obstruction} Let $\s=0$ and $N=2$ in Proposition~\ref{prop:supertranslation}, with even generator $H$, odd basis $\vartheta_1,\vartheta_2$, $b_1=\operatorname{diag}(1,1)$, $b_2=\operatorname{diag}(1,-1)$, that is, \[ \beta_1:\ [\vartheta_1,\vartheta_1]=H,\ [\vartheta_2,\vartheta_2]=H;\qquad \beta_2:\ [\vartheta_1,\vartheta_1]=H,\ [\vartheta_2,\vartheta_2]=-H . \] Then $X$ lies in $\mathcal G^{\mathrm{sw}}_{\mathcal M}$ but not in the essential image of $\mathcal G^{h\mathbb Z/2}_{\mathcal M}$. \end{proposition} \begin{proof} In the canonical form of Proposition~\ref{prop:supertranslation}(3), $b_1=\mathsf E_1\oplus\mathsf E_1$ and $b_2=\mathsf E_1J_1(1)\oplus\mathsf E_1J_1(-1)$, so $T=b_1^{-1}b_2=\operatorname{diag}(1,-1)$. The elementary divisors of the pencil $(b_1,b_2)$ are therefore $\lambda-1$ and $\lambda+1$, that is, the points $1$ and $-1$ of $\mathbb P^1$, each with block size one. Inversion $\lambda\mapsto\lambda^{-1}$ fixes each of these points, so it fixes the multiset, and Proposition~\ref{prop:supertranslation}(4) puts $X$ in $\mathcal G^{\mathrm{sw}}_{\mathcal M}$. Explicitly, the map $H\mapsto H$, $\vartheta_1\mapsto\vartheta_1$, $\vartheta_2\mapsto -i\vartheta_2$ satisfies the switch equations. Every arrow $X\to X$ or $X\to\sigma X$ conjugates $T=b_1^{-1}b_2=\operatorname{diag}(1,-1)$ to $T$ or to $T^{-1}=T$, so it preserves the eigenlines $k\vartheta_1$ and $k\vartheta_2$. Hence $\operatorname{Fgt}(\operatorname{Hom}(X,X))$ consists of $H\mapsto cH$, $\vartheta_1\mapsto h_1\vartheta_1$, $\vartheta_2\mapsto h_2\vartheta_2$ with $h_1^2=h_2^2=c$, and $\operatorname{Fgt}(\operatorname{Hom}(X,\sigma X))$ of $H\mapsto cH$, $\vartheta_1\mapsto h_1\vartheta_1$, $\vartheta_2\mapsto -ih_2\vartheta_2$. For such a switch $F$, $\sigma(F)\circ F$ acts by $c^2$, $c$ and $-c$ on $H$, $\vartheta_1$ and $\vartheta_2$, so it is never $\mathrm{id}_X$. \end{proof} \begin{proposition}[Equivariant switches with a symplectic module]\label{prop:symplectic} Let $V$ be a nontrivial irreducible $\s$-module with a nonzero invariant alternating form $\omega$, and let $z,z_1,z_2$ span trivial modules. Let $X_\epsilon=(V\oplus kz,\ \omega\otimes z,\ \epsilon\,\omega\otimes z)$ for $\epsilon=\pm1$ and $Y=(V\oplus kz_1\oplus kz_2,\ \omega\otimes z_1,\ \omega\otimes z_2)$. Then: \begin{enumerate} \item $\operatorname{Fgt}(\operatorname{Hom}(X_\epsilon,X_\epsilon))$ consists of $v\mapsto mv$, $z\mapsto m^2z$, and $\operatorname{Fgt}(\operatorname{Hom}(X_\epsilon,\sigma X_\epsilon))$ of $v\mapsto mv$, $z\mapsto\epsilon m^2z$, with $m\in k^\times$; \item $\operatorname{Fgt}(\operatorname{Hom}(Y,Y))$ consists of $v\mapsto mv$, $z_i\mapsto m^2z_i$, and $\operatorname{Fgt}(\operatorname{Hom}(Y,\sigma Y))$ of $v\mapsto mv$, $z_1\mapsto m^2z_2$, $z_2\mapsto m^2z_1$. \end{enumerate} In both cases $m=1$ gives an involutive switch. \end{proposition} \begin{proof} An arrow is $\s$-equivariant, so it preserves the $V$-isotypic and trivial isotypic parts. By Schur's lemma it acts by a scalar $m$ on $V$ and by some $C$ on the trivial part. The bracket equations read $C(\omega(v,v')z)=m^2\omega(v,v')z$ for $X_\epsilon\to X_\epsilon$ and $C(\omega(v,v')z)=\epsilon m^2\omega(v,v')z$ for $X_\epsilon\to\sigma X_\epsilon$, and similarly with $z_1,z_2$ for $Y$. Since $\omega\neq0$, they determine $C$ as stated. \end{proof} \begin{proposition}[Semisimple acting algebra through dimension four]\label{prop:ss4} Let $\s$ be semisimple and $\mathcal M=\s\text{-}\mathrm{Rep}$. The indecomposable objects of $\mathcal G^{\mathrm{sw}}_{\mathcal M}$ of dimension at most four are the objects of Proposition~\ref{prop:vect4} with trivial $\s$-action, the nontrivial simple modules with both brackets zero, and the objects $X_\epsilon$ and $Y$ of Proposition~\ref{prop:symplectic} with $V$ the standard module of a simple ideal of $\s$ isomorphic to $\mathfrak{sl}_2$. \end{proposition} \begin{proof} Restricting along $0\to\s$ gives a faithful functor $\mathcal G^{\mathrm{sw}}_{\mathcal M}\to\mathcal G^{\mathrm{sw}}_{\mathrm{Vect}_k}$, so the restriction $X_0$ of $X$ is one of the objects of Proposition~\ref{prop:vect4}, and $X$ is determined by a homomorphism $\rho\colon\s\to\mathfrak d=\operatorname{Der}(\beta_1)\cap\operatorname{Der}(\beta_2)$, the Lie algebra of $G=\operatorname{Aut}(X_0)$, up to conjugation by $G$. By the theorems of Levi--Malcev and Malcev--Harish-Chandra~\cite[Ch.~I, \S6, no.~8]{BourbakiLie}, $\rho(\s)$ is contained in a Levi subalgebra of $\mathfrak d$, and any two Levi subalgebras are conjugate by $\exp(\operatorname{ad}z)$ with $z\in[\mathfrak d,\operatorname{rad}\mathfrak d]$. The argument of Lemma~\ref{lem:nilradical}, applied to the representation of $\mathfrak d$ on $\operatorname{Fgt}X_0$ instead of the adjoint representation, shows that such $z$ acts nilpotently, so $\exp(z)$ lies in $G$. From the groups in Proposition~\ref{prop:vect4}, together with those of $\mathsf A_\epsilon\oplus\mathbf 1$ and $\mathsf B_\epsilon\oplus\mathbf 1$ (which add a solvable factor), a Levi subalgebra of $\mathfrak d$ is $\mathfrak{sl}_n$ for $\mathbf 1^{\oplus n}$, $\mathfrak{sl}_2$ acting on $\langle e_1,e_2\rangle$ for $\mathsf A_\epsilon$, $\mathsf A_\epsilon\oplus\mathbf 1$ and $\mathsf D$, and zero for $\mathsf B_\epsilon$, $\mathsf B_\epsilon\oplus\mathbf 1$, $\mathsf N$ and $\mathsf S$. In the last cases $\s$ acts trivially. For $\mathbf 1^{\oplus n}$, $B$ is a semisimple $\s$-module with zero brackets, and it is indecomposable exactly when it is simple. For the remaining cases, a nonzero $\rho$ maps onto $\mathfrak{sl}_2$, so $\langle e_1,e_2\rangle$ is the standard module of a simple ideal isomorphic to $\mathfrak{sl}_2$ and the central vectors are trivial. This gives $X_\epsilon$ and $Y$, and $X_\epsilon\oplus\mathbf 1$, which is decomposable. Finally, $X_\epsilon$ and $Y$ are indecomposable because their restrictions $\mathsf A_\epsilon$ and $\mathsf D$ are. \end{proof} The restriction to $\mathfrak{sl}_2$ comes from the dimension bound alone. The constructions of Proposition~\ref{prop:symplectic} apply to every nontrivial irreducible symplectic module, and multiplicity spaces permit the two brackets to use distinct invariant forms. There just isn't enough room yet with the cutoff imposed. \section{Paired current algebras}\label{sec:currents} This section applies Theorem~\ref{thm:paired-tensor} to particular coefficient algebras, among them those whose two products are pointwise multiplication and convolution, exchanged by Fourier transform. Most of these coefficient algebras are infinite-dimensional, so the finite-dimensional convention is relaxed throughout this section. For the analytic examples take $k=\mathbb C$. All tensor products are algebraic unless a topology is explicitly specified. If we do not demand the existence of such an exchange, then there are many more examples. In addition we want the two commutative algebras not to be algebras for $\mathrm{Lie}_2^!$ which would have some relation between the two products rather than them being completely independently specified (at least a priori). In the former case, we would have just done the Lie algebra with $\mathrm{Lie}_2^! \otimes_H \mathrm{Lie}_2$. \subsection{Coefficient algebras supplied by Fourier analysis}\label{sec:fourier} \begin{example}[Rescaled products]\label{ex:rescaled} For a commutative associative algebra $(A,\mu_1)$ and an element $c\in A$, define $\mu_2(a,b)=c\,\mu_1(a,b)$. In the super case take $c$ even. Associativity and commutativity of the second product follow from those of the first. This is an example of $\mathrm{CommComm}$, even though this particular algebra satisfies additional identities. A scalar rescaling $\mu_2=r\mu_1$ makes \eqref{eq:paired-bracket} the single-product current bracket for $\beta_1+r\beta_2$. An algebra element $c$ need not be a scalar. This reduces to the very well studied cases. \end{example} \begin{example}[Schwartz functions on the line]\label{ex:schwartz} Take $A=\mathcal S(\mathbb R,\mathbb C)$, defined by the seminorms \[ p_{m,n}(f)=\sup_{x\in\mathbb R}|x^mf^{(n)}(x)|<\infty\qquad(m,n\ge0). \] With Lebesgue measure fixed, put \[ \mu_1(f,g)(x)=f(x)g(x),\qquad \mu_2(f,g)(x)=(f*g)(x)=\int_{\mathbb R}f(y)g(x-y)\,dy . \] Both products are jointly continuous and preserve $\mathcal S$. Leibniz's rule proves the pointwise assertion; for convolution one differentiates under the integral and expands $x^m=(y+(x-y))^m$, bounding the resulting convolutions by an $L^1$ norm times a supremum norm. Associativity of convolution follows from Fubini. Neither product has a unit in this space: the pointwise unit would be $1$, and the convolution unit would be the Dirac distribution. Standard Fourier transform with usual normalizations interchanges the products and squares to reflection. \end{example} \begin{example}[Feichtinger algebra]\label{ex:feichtinger} Fix $g(t)=e^{-\pi t^2}$ and set \[ V_gf(x,\xi)=\int_{\mathbb R}f(t)\,g(t-x)\,e^{-2\pi i\xi t}\,dt,\qquad S_0(\mathbb R)=\{f\in L^2(\mathbb R):V_gf\in L^1(\mathbb R^2)\}. \] The norm $\|V_gf\|_1$ makes $S_0$ a Banach space, continuously embedded in $L^1\cap C_0$. It is closed under both pointwise multiplication and convolution, with continuous product maps, and Fourier transform exchanges them. Replacing $g$ by any nonzero window in $S_0$ gives an equivalent norm; this assertion is not for an arbitrary $L^2$ window. The same two-product construction is available on $S_0(G)$ for locally compact abelian groups, with Haar measure and the dual Haar measure fixed. See \cite{Jakobsen2018} for these analytic facts. \end{example} \begin{example}[Laurent polynomials and smooth loops]\label{ex:laurent} Let $\mathbb T_\ell=\mathbb R/\ell\mathbb Z$, $\ell>0$, with normalized Haar measure $dx/\ell$. Put $e_n(x)=e^{2\pi inx/\ell}$. On the trigonometric polynomials, identified with $\mathbb C[t,t^{-1}]$ by $t^n\mapsto e_n$, the products are \[ t^m\cdot t^n=t^{m+n},\qquad t^m\circ t^n=\delta_{mn}t^n . \] The latter is circular convolution with the normalized Haar measure. Thus both products close on the Laurent polynomials, which are dense in $C^\infty(\mathbb T_\ell)$ in its smooth topology. Fourier series identify the latter with rapidly decreasing sequences on $\mathbb Z$: pointwise multiplication of functions becomes Cauchy convolution of sequences, and circular convolution becomes componentwise multiplication. However, this example has no map to itself exchanging its two products (Proposition~\ref{prop:laurent-no-exchange}) unlike the previous. \end{example} \begin{proposition}[No product exchange on the Laurent model]\label{prop:laurent-no-exchange} There is no linear isomorphism $\Psi\colon k[t,t^{-1}]\to k[t,t^{-1}]$ with \[ \Psi(a\cdot b)=\Psi(a)\circ\Psi(b),\qquad \Psi(a\circ b)=\Psi(a)\cdot\Psi(b). \] \end{proposition} \begin{proof} The first identity makes $\Psi$ an isomorphism of commutative algebras from $\cdot$ to $\circ$. Since $\Psi$ is surjective, $\Psi(1)\circ\Psi(b)=\Psi(1\cdot b)=\Psi(b)$ shows that $\Psi(1)$ is a unit for $\circ$. If $u=\sum_nu_nt^n$ were such a unit, then $u\circ t^n=u_nt^n=t^n$ would give $u_n=1$ for every $n\in\mathbb Z$, which is impossible for a Laurent polynomial. \end{proof} \subsection{The resulting Lie algebras and Fourier transport}\label{sec:transport} For the Schwartz or Feichtinger coefficients, write $L=L(A,B)$. Explicitly, \begin{equation}\label{eq:L-fourier} [f\otimes x,g\otimes y]=fg\otimes\beta_1(x,y)+(f*g)\otimes\beta_2(x,y). \end{equation} The two coefficient products remain separate structures. For finite-dimensional $B$ these are continuous brackets on the corresponding spaces of $B$-valued functions. With infinite-dimensional $B$, Theorem~\ref{thm:paired-tensor} supplies the algebraic tensor product; a completed tensor product needs a specified topology. When $\beta_2=0$ this is the usual current bracket for $(A,\mu_1)$ and $(B,\beta_1)$, including nonunital $A$. When $\beta_1=0$, only the second current bracket remains. For the Laurent model, \begin{equation}\label{eq:laurent-bracket} [t^m\otimes x,t^n\otimes y]=t^{m+n}\otimes\beta_1(x,y)+\delta_{mn}t^n\otimes\beta_2(x,y). \end{equation} The first term is the Laurent current algebra; the second separately is the direct sum of the Lie algebras $(B,\beta_2)$ over the modes. Suppose $\Psi\colon A\to A'$ exchanges the two products, and $F\colon X\to\sigma X'$ is an arrow from $X=(B,\beta_1,\beta_2)$ to the exchange of $X'=(B',\beta_1',\beta_2')$, so that $\operatorname{Fgt}(F)$ carries $\beta_1$ to $\beta_2'$ and $\beta_2$ to $\beta_1'$. Then the tensor formula gives \[ \Psi\otimes \operatorname{Fgt}(F)\colon L(A,B)\xrightarrow{\ \sim\ }L(A',B'). \] This applies with $A=A'$ equal to $\mathcal S(\mathbb R^n)$ or $S_0(\mathbb R^n)$ and $\Psi$ equal to $\mathrm{FT}$ followed by the coordinate identification of the dual space through $x\cdot\xi$, the formulas of Example~\ref{ex:schwartz} written on $\mathbb R^n$; for $X'=X$ it gives an automorphism of $L(A,B)$. An involutive switch $F\colon X\to\sigma X$, so that $\operatorname{Fgt}(F)^2=\operatorname{Fgt}(\sigma(F)\circ F)=\mathrm{id}_B$, gives $(\mathrm{FT}\otimes \operatorname{Fgt}(F))^2=J\otimes\mathrm{id}$, where $Jf(x)=f(-x)$. Consequently the combined automorphism has fourth power the identity; an involution on the bracket factor does not remove Fourier reflection. \begin{proposition}[A simultaneous modified Yang--Baxter operator]\label{prop:mcybe} In the Laurent model \eqref{eq:laurent-bracket}, the subspaces $L_+=k[t]\otimes B$ and $L_-=t^{-1}k[t^{-1}]\otimes B$ are subalgebras. Let $P_\pm$ be the projections of $L=L_+\oplus L_-$ and $\mathcal R=P_+-P_-$. Then \[ [\mathcal R\xi,\mathcal R\zeta]-\mathcal R\big([\mathcal R\xi,\zeta]+[\xi,\mathcal R\zeta]\big)=-[\xi,\zeta] , \] so $\mathcal R$ solves the modified classical Yang--Baxter equation. The associated bracket \[ [\xi,\zeta]_{\mathcal R}=[\mathcal R\xi,\zeta]+[\xi,\mathcal R\zeta] =2\big([\xi_+,\zeta_+]-[\xi_-,\zeta_-]\big) \] is a Lie bracket on the same space: it is the bracket of $L_+\times L_-$ with the bracket of the second factor negated, rescaled by $2$. \end{proposition} \begin{proof} When $\beta_2=0$, the algebra $L(k[t,t^{-1}],B)$ is the loop algebra $B[t,t^{-1}]$ of the Lie algebra $(B,\beta_1)$, and $\mathcal R$ is its splitting into $B[t]$ and $t^{-1}B[t^{-1}]$. This is the situation of \cite[Proposition~5]{SemenovTianShansky1983}: for a Lie algebra that is a direct sum of two subalgebras, $\mathcal R=P_+-P_-$ satisfies the modified classical Yang--Baxter equation \cite[Section~3, equation~(12)]{SemenovTianShansky1983}, and the $\mathcal R$-bracket \cite[Section~1, equation~(1)]{SemenovTianShansky1983} equals $2[x_+,y_+]-2[x_-,y_-]$. The loop-algebra case appears in the example following that proposition, for the associative algebra $\mathrm{Mat}(n,\mathbb C)\otimes\mathbb C[\lambda,\lambda^{-1}]$ with the operator \cite[equation~(13)]{SemenovTianShansky1983}. This follows the same argument. \iffalse \begin{align*} [\mathcal R\xi,\zeta] &= [\xi_+,\zeta_+]+[\xi_+,\zeta_-]-[\xi_-,\zeta_+]-[\xi_-,\zeta_-]\\ [\xi,\mathcal R\zeta] &= [\xi_+,\zeta_+]-[\xi_+,\zeta_-]+[\xi_-,\zeta_+]-[\xi_-,\zeta_-]\\ [\xi,\zeta]_{\mathcal R} = [\mathcal R\xi,\zeta]+[\xi,\mathcal R\zeta] &= 2[\xi_+,\zeta_+]-2[\xi_-,\zeta_-]\\ \mathcal R\big([\mathcal R\xi,\zeta]+[\xi,\mathcal R\zeta]\big) &= 2[\xi_+,\zeta_+]+2[\xi_-,\zeta_-]\\ [\mathcal R\xi,\mathcal R\zeta] &= [\xi_+,\zeta_+]-[\xi_+,\zeta_-]-[\xi_-,\zeta_+]+[\xi_-,\zeta_-]\\ [\mathcal R\xi,\mathcal R\zeta]-\mathcal R\big([\mathcal R\xi,\zeta]+[\xi,\mathcal R\zeta]\big) &= -[\xi_+,\zeta_+]-[\xi_+,\zeta_-]-[\xi_-,\zeta_+]-[\xi_-,\zeta_-]=-[\xi,\zeta]\\ \end{align*} \fi \end{proof} For a single pair, Golubchik and Sokolov~\cite{GolubchikSokolov2006} construct a nonconstant solution of the classical Yang--Baxter equation from any pair of compatible Lie brackets that share a nondegenerate invariant symmetric form. Mutually annihilating pairs are compatible, so their construction applies to them. The next proposition shows how restrictive a common invariant form is in the mutually annihilating case. An invariant symmetric bilinear form $(\ ,\ )$ on $B$ is \emph{$\beta$-invariant} if $(\beta(x,y),z)=(x,\beta(y,z))$ for all $x,y,z$. \begin{proposition}[Common invariant forms]\label{prop:common-form} Let $\beta_1,\beta_2$ be Lie brackets on a finite-dimensional vector space $B$, and let $(\ ,\ )$ be a nondegenerate symmetric form that is $\beta_1$- and $\beta_2$-invariant. Put $I_i=\operatorname{im}\beta_i$. \begin{enumerate} \item $Z(\beta_i)=I_i^{\perp}$ for $i=1,2$. \item The pair is mutually annihilating if and only if $I_1\perp I_2$. \item If the pair is mutually annihilating and each $\beta_i$ has class at most two, then $I_1+I_2$ is isotropic. In particular $\dim(I_1+I_2)\le\tfrac12\dim B$. \item If the pair is mutually annihilating and $\beta_1+\beta_2$ is perfect, then $B=I_1\oplus I_2$ is an orthogonal decomposition, $\beta_1$ vanishes on $I_2$ and $\beta_2$ vanishes on $I_1$. Thus $(B,\beta_1,\beta_2)\cong(\mathfrak g_1,0)\oplus(0,\mathfrak g_2)$ with $\mathfrak g_i=(I_i,\beta_i)$ perfect. \end{enumerate} \end{proposition} \begin{proof} (1) Fix $i$ and $z\in B$. By definition $z\in Z(\beta_i)$ means $\beta_i(z,y)=0$ for all $y\in B$. Since the form is nondegenerate, this holds exactly when $(\beta_i(z,y),w)=0$ for all $y,w\in B$. By invariance $(\beta_i(z,y),w)=(z,\beta_i(y,w))$, so the condition becomes $(z,\beta_i(y,w))=0$ for all $y,w$. The elements $\beta_i(y,w)$ span $I_i$, so this says $z\in I_i^{\perp}$. (2) The identity $\beta_1(\beta_2(x,y),z)=0$ for all $x,y,z$ says that every element of $I_2$ is central for $\beta_1$, that is, $I_2\subseteq Z(\beta_1)$. Likewise $\beta_2(\beta_1(x,y),z)=0$ for all $x,y,z$ says $I_1\subseteq Z(\beta_2)$. By (1), the first inclusion is $I_2\subseteq I_1^{\perp}$ and the second is $I_1\subseteq I_2^{\perp}$. Since the form is symmetric, both inclusions say $(a,b)=0$ for all $a\in I_1$ and $b\in I_2$, that is, $I_1\perp I_2$. So the two mixed identities hold together exactly when $I_1\perp I_2$. (3) Class at most two for $\beta_i$ means $\beta_i(\beta_i(x,y),z)=0$ for all $x,y,z$, that is, $I_i\subseteq Z(\beta_i)$. By (1) this is $I_i\subseteq I_i^{\perp}$, so $(a,a')=0$ for all $a,a'\in I_i$. Now let $a_1+a_2$ and $b_1+b_2$ lie in $I_1+I_2$, with $a_i,b_i\in I_i$. Then \[ (a_1+a_2,\ b_1+b_2)=(a_1,b_1)+(a_1,b_2)+(a_2,b_1)+(a_2,b_2). \] The first and last terms vanish because each $I_i$ is isotropic, and the middle terms vanish by (2). So $V=I_1+I_2$ is isotropic, that is, $V\subseteq V^{\perp}$. Nondegeneracy gives $\dim V^{\perp}=\dim B-\dim V$, hence $\dim V\le\dim B-\dim V$, which is the stated bound. (4) Every value $(\beta_1+\beta_2)(x,y)=\beta_1(x,y)+\beta_2(x,y)$ lies in $I_1+I_2$, so perfectness of $\beta_1+\beta_2$ gives $B=I_1+I_2$. Let $c\in I_1\cap I_2$. For $a\in I_1$ we have $(c,a)=0$ because $c\in I_2$ and $I_2\perp I_1$; for $b\in I_2$ we have $(c,b)=0$ because $c\in I_1$. Hence $c\perp I_1+I_2=B$, and $c=0$ by nondegeneracy. So $B=I_1\oplus I_2$, and the summands are orthogonal by (2); in particular the form restricts to a nondegenerate form on each $I_i$. Write $x=x_1+x_2$ and $y=y_1+y_2$ with $x_i,y_i\in I_i$. Since $I_2\subseteq Z(\beta_1)$, every term of $\beta_1(x,y)$ involving $x_2$ or $y_2$ vanishes, so $\beta_1(x,y)=\beta_1(x_1,y_1)$, which lies in $I_1$. Thus $\beta_1$ restricts to a Lie bracket on $I_1$ and is zero on every pair with an argument in $I_2$. Symmetrically, $\beta_2(x,y)=\beta_2(x_2,y_2)\in I_2$. With $\mathfrak g_i=(I_i,\beta_i|_{I_i})$, the identity map of $B$ is therefore an isomorphism $(B,\beta_1,\beta_2)\cong(\mathfrak g_1,0)\oplus(0,\mathfrak g_2)$ in $\mathcal G_{\mathcal M}$. Finally $(\beta_1+\beta_2)(x,y)=\beta_1(x_1,y_1)+\beta_2(x_2,y_2)$, so $B=\operatorname{im}(\beta_1+\beta_2)=\beta_1(I_1,I_1)\oplus\beta_2(I_2,I_2)$. Comparing with $B=I_1\oplus I_2$ gives $\beta_i(I_i,I_i)=I_i$, so each $\mathfrak g_i$ is perfect. \end{proof} \section*{Declaration of generative AI and AI-assisted technologies\\in the manuscript preparation process} During preparation of this manuscript, the author used OpenAI ChatGPT-6 Astra and Anthropic Claude Opus 5.5 to assist with drafting and revising exposition, exploring calculations, and preparing the \LaTeX{} source and PDF. The choices of definitions, which morphisms to allow in which categories, and which special cases to consider were specified in the author's prompts during this session. The author reviewed and edited the generated material. The author remains responsible for the mathematical statements and the final contents of the released report. \nocite{RenZhu2017,GalitskiTimashev1999,Eberlein2003,GrunewaldOHalloran1988} \begin{thebibliography}{99} \bibitem{BurdeWagemannPerfect} D.~Burde and F.~Wagemann, ``Cohomology of perfect Lie algebras,'' preprint, arXiv:2411.14952 (2024). \href{https://arxiv.org/abs/2411.14952}{arXiv:2411.14952}. \bibitem{BazzoniMunoz2012} G.~Bazzoni and V.~Mu\~noz, ``Classification of minimal algebras over any field up to dimension 6,'' \emph{Transactions of the American Mathematical Society} \textbf{364} (2012), 1007--1028. \href{https://doi.org/10.1090/S0002-9947-2011-05471-1} {doi:10.1090/S0002-9947-2011-05471-1}. \bibitem{BazzoniRojo2025} G.~Bazzoni and J.~Rojo, ``8-dimensional 2-step nilpotent Lie algebras over algebraically closed fields of characteristic 0,'' arXiv:2507.14907 (2025). \href{https://arxiv.org/abs/2507.14907}{arXiv:2507.14907}. \bibitem{Gauger1973} M.~A. Gauger, ``On the classification of metabelian Lie algebras,'' \emph{Transactions of the American Mathematical Society} \textbf{179} (1973), 293--329. \href{https://doi.org/10.1090/S0002-9947-1973-0325719-0} {doi:10.1090/S0002-9947-1973-0325719-0}. \bibitem{Scheuneman1967} J.~Scheuneman, ``Two-step nilpotent Lie algebras,'' \emph{Journal of Algebra} \textbf{7} (1967), 152--159. \bibitem{BelitskiiLipyanskiSergeichuk2005} G.~Belitskii, R.~Lipyanski, and V.~V. Sergeichuk, ``Problems of classifying associative or Lie algebras and triples of symmetric or skew-symmetric matrices are wild,'' \emph{Linear Algebra and its Applications} \textbf{407} (2005), 249--262. \href{https://doi.org/10.1016/j.laa.2005.05.007} {doi:10.1016/j.laa.2005.05.007}. \bibitem{Stembridge2003} J.~R. Stembridge, ``Multiplicity-free products and restrictions of Weyl characters,'' \emph{Representation Theory} \textbf{7} (2003), 404--439. \href{https://doi.org/10.1090/S1088-4165-03-00150-X} {doi:10.1090/S1088-4165-03-00150-X}. \bibitem{RenZhu2017} B.~Ren and L.~S. Zhu, ``Classification of 2-step nilpotent Lie algebras of dimension 9 with 2-dimensional center,'' \emph{Czechoslovak Mathematical Journal} \textbf{67} (2017), 953--965. \href{https://doi.org/10.21136/CMJ.2017.0253-16} {doi:10.21136/CMJ.2017.0253-16}. \bibitem{GalitskiTimashev1999} L.~Yu. Galitski and D.~A. Timashev, ``On the classification of metabelian Lie algebras,'' \emph{Journal of Lie Theory} \textbf{9} (1999), 125--156. \href{https://jolt.centre-mersenne.org/articles/10.5802/jolt.163/} {Journal of Lie Theory archive}. \bibitem{Eberlein2003} P.~Eberlein, ``The moduli space of 2-step nilpotent Lie algebras of type $(p,q)$,'' in \emph{Explorations in Complex and Riemannian Geometry}, Contemporary Mathematics \textbf{332}, American Mathematical Society (2003), 37--72. \href{https://doi.org/10.1090/conm/332/05929} {doi:10.1090/conm/332/05929}. \bibitem{GrunewaldOHalloran1988} F.~Grunewald and J.~O'Halloran, ``Varieties of nilpotent Lie algebras of dimension less than six,'' \emph{Journal of Algebra} \textbf{112} (1988), 315--325. \href{https://doi.org/10.1016/0021-8693(88)90093-2} {doi:10.1016/0021-8693(88)90093-2}. \bibitem{LodayVallette} Jean-Louis Loday and Bruno Vallette, \textit{Algebraic Operads}, Grundlehren der mathematischen Wissenschaften \textbf{346}, Springer, Heidelberg, 2012. \href{https://www.math.univ-paris13.fr/~vallette/Operads.pdf}{PDF}. \bibitem{DotsenkoKhoroshkin} Vladimir Dotsenko and Anton Khoroshkin, ``Character formulas for the operad of two compatible brackets and for the bihamiltonian operad,'' \textit{Functional Analysis and Its Applications} \textbf{41} (2007), no.~1, 1--17. \href{https://doi.org/10.1007/s10688-007-0001-0}{doi:10.1007/s10688-007-0001-0}. \bibitem{GinzburgKapranov} V.~Ginzburg and M.~Kapranov, ``Koszul duality for operads,'' \emph{Duke Mathematical Journal} \textbf{76} (1994), no.~1, 203--272. \href{https://arxiv.org/abs/0709.1228}{arXiv:0709.1228}. \bibitem{Jakobsen2018} M.~S. Jakobsen, ``On a (no longer) new Segal algebra: a review of the Feichtinger algebra,'' \emph{Journal of Fourier Analysis and Applications} \textbf{24} (2018), 1579--1660. \href{https://doi.org/10.1007/s00041-018-9596-4}{doi:10.1007/s00041-018-9596-4}. \bibitem{GolubchikSokolov2006} I.~Z. Golubchik and V.~V. Sokolov, ``Compatible Lie brackets and the Yang--Baxter equation,'' \emph{Theoretical and Mathematical Physics} \textbf{146} (2006), no.~2, 159--169. \bibitem{Thompson1991} R.~C. Thompson, ``Pencils of complex and real symmetric and skew matrices,'' \emph{Linear Algebra and its Applications} \textbf{147} (1991), 323--371. \bibitem{SemenovTianShansky1983} M.~A. Semenov-Tian-Shansky, ``What is a classical $r$-matrix?,'' \emph{Funktsional'nyi Analiz i ego Prilozheniya} \textbf{17} (1983), no.~4, 17--33; English translation: \emph{Functional Analysis and Its Applications} \textbf{17} (1983), no.~4, 259--272. \href{https://doi.org/10.1007/BF01076717}{doi:10.1007/BF01076717}. \bibitem{BourbakiLie} N.~Bourbaki, \emph{Lie Groups and Lie Algebras, Chapters 1--3}, Elements of Mathematics, Springer, Berlin, 1989. \end{thebibliography} \end{document}