\documentclass[12pt,a4paper,english,reqno]{amsart} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{amsmath,amsthm,amssymb,amscd,mathtools} \usepackage{booktabs,array,longtable,enumitem} \usepackage[colorlinks=true,allcolors=blue]{hyperref} \usepackage[expansion=false]{microtype} \setlength{\textwidth}{142mm} \setlength{\textheight}{219mm} \setlength{\oddsidemargin}{9mm} \setlength{\evensidemargin}{9mm} \setlength{\headheight}{14pt} \allowdisplaybreaks[2] \emergencystretch=3em \setcounter{tocdepth}{2} \setcounter{MaxMatrixCols}{11} \usepackage{needspace} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{claim}[theorem]{Claim} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \newtheorem*{maintheorem}{Theorem A} \numberwithin{equation}{section} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\Val}{Val} \DeclareMathOperator{\ad}{ad} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\Span}{span} \DeclareMathOperator{\Int}{Int} \DeclareMathOperator{\Id}{Id} \newcommand{\slm}{\mathfrak{sl}} \newcommand{\Mat}{\operatorname{M}} \newcommand{\UT}{\operatorname{UT}} \newcommand{\HH}{\mathcal H} \title[Multilinear values in degrees four to six]{The L'vov--Kaplansky conjecture for multilinear polynomials of degrees four to six} \author{Shengyong Pan} \address{School of Mathematics and Statistics, Beijing Jiaotong University, Beijing, China} \email{shypan@bjtu.edu.cn} \date{} \subjclass[2020]{16R30, 15A24, 14A10} \keywords{L'vov--Kaplansky conjecture, multilinear polynomial, matrix algebra, polynomial identity, flat morphism} \hypersetup{pdftitle={The L'vov-Kaplansky conjecture for multilinear polynomials of degrees four to six},pdfauthor={Shengyong Pan}} \begin{document} \begin{abstract} Let $K$ be an algebraically closed field of characteristic zero. We determine the images of ordinary multilinear polynomials of degrees four, five and six on matrix algebras. For $n\geq3$, a nonidentity has image $\mathfrak{sl}_n(K)$ when its cyclic class is zero and $M_n(K)$ otherwise. In these degrees the only nonzero identity exception is the standard degree-six line on $M_3(K)$. The proof starts with proper sextics and treats arbitrary sextics by unit restrictions and local trace lifting. An image-preserving product substitution then gives degrees four and five as consequences. We also give cyclic and matrix lifting results applicable in arbitrary degree. The finite calculations are given by explicit matrix-unit identities. \end{abstract} \maketitle \tableofcontents \section*{Introduction} Let $K$ be an algebraically closed field of characteristic zero. For a noncommutative polynomial $f$, write $\Val_n(f)$ for its set of values on $M_n(K)$. The L'vov--Kaplansky conjecture asks whether this set is a vector space whenever $f$ is multilinear. The question appears under I.~V.~L'vov's name in \cite[Problem 1.98, p.~12]{Dniester1993}; a survey of matrix polynomial images is \cite{Survey2020}. For $2\times2$ matrices, Kanel--Belov, Malev and Rowen \cite[Theorem 2]{KMR} proved the four possible images $\{0\}$, $KI_2$, $\mathfrak{sl}_2(K)$ and $M_2(K)$. The degree-three case was proved by Vitas \cite[Theorem B]{Vitas2026}. In degree four, traceless containment for nonzero polynomials on $M_n(K)$, $n\geq3$, was established by Buzinski and Winstanley \cite{BuzinskiWinstanley2013}, with the correction in \cite[Theorem 4.1]{FagundesCorrection2022}. Tang's preprint \cite[Theorem 1.1]{Tang2026} gives a proof of the degree-four conjecture under the same field assumptions. Its quartic image classification agrees with the degree-four part of Theorem A below. The methods share trace-pairing tests, scalar values from clock-and-shift matrices, frozen-input linear maps, and cycle and block constructions with exact minors. Tang's main route uses traceful linear slices of codimension at most two to meet nonscalar conjugacy classes \cite[Corollary 2.5]{Tang2026}, followed by quartic coefficient reductions and transfer from sizes two and three. We first treat proper sextics by studying local fibers and corrected trace terms at rank-one square-zero values. Unit deletions give the ordinary sextic case, from which degrees four and five follow by product substitution. The proof of the proper-sextic theorem is independent of lower-degree image classifications, including Tang's theorem. Let $P_d$ be the ordinary multilinear component: every word contains each of $x_1,\ldots,x_d$ once, with coefficients in $K$. Let $\pi:P_d\to C_d$ identify words under cyclic rotation. Its coordinates are the sums of coefficients in the cyclic orbits. Put \[ s_d(x_1,\ldots,x_d)=\sum_{\sigma\in S_d}\operatorname{sgn}(\sigma) x_{\sigma(1)}\cdots x_{\sigma(d)}. \] Our main result is the following. \begin{maintheorem}\label{thm:main-intro}\label{thm:main} Let $d\in\{4,5,6\}$, $f\in P_d$, and $n\geq3$. Define \[ I_{d,n}=\begin{cases}Ks_6,&(d,n)=(6,3),\\\{0\},&\text{otherwise}.\end{cases} \] Then \[ \Val_n(f)= \begin{cases} \{0\},&f\in I_{d,n},\\ \mathfrak{sl}_n(K),&f\notin I_{d,n}\text{ and }\pi f=0,\\ M_n(K),&\pi f\ne0. \end{cases} \] For $n=2$, the four image alternatives are determined by actual identity, centrality and trace conditions, as specified in Corollary~\ref{cor:small-sizes}. \end{maintheorem} Here the image is the set of single evaluations, not its linear span or Zariski closure. In particular, dominance alone does not prove the statement. We construct an open neighborhood of a nonzero rank-one square-zero value and use conjugation and scalar multiplication to recover every nonscalar target. Scalar targets in the traceful case are obtained independently from a homogeneous matrix basis. The paper is organized as follows. In Section~\ref{sec:preliminaries}, we fix notation and collect results on cyclic words, proper components and local fibers. A polynomial is proper if every single unit deletion annihilates it. The first two upper-triangular symbols divide the proper sextics into a one-slot case, products of three commutators, and a mixed two-slot case. Sections~\ref{sec:one-slot}--\ref{sec:proper-mixed} treat these cases and prove Theorem~\ref{thm:proper-sextic-classification}. In Section~\ref{sec:nonproper-complete}, we use this theorem and image-preserving Lie substitutions to determine the values of proper polynomials of lower degree. A maximal nonzero unit restriction of a nonproper sextic is proper. Local trace lifting in the original inputs then gives the ordinary sextic theorem. Replacing one input by a product of fresh variables preserves the value set, so the degree-four and degree-five classifications follow in Corollary~\ref{cor:degrees-four-five}. This product substitution is distinct from the Lie substitutions used for proper polynomials. In Section~\ref{sec:short-tools}, we prove further lifting results applicable in arbitrary degree. In Section~\ref{sec:checks}, we describe the finite calculations as matrix-unit identities and parameter factorizations. The arguments for arbitrary matrix size are given in the preceding sections. \section*{Acknowledgments} The author used ChatGPT, to help draft and revise the manuscript. The author is responsible for the mathematical content of the paper. This work is supported by the Beijing Natural Science Foundation (Grant Nos. 1262017, 1252011). \section{Preliminaries}\label{sec:preliminaries} Throughout the paper, $K$ is algebraically closed of characteristic zero. All varieties and tangent spaces are over $K$. Put $\mathcal H_n=\{A\in M_n(K):A_{ii}=0\}$ and $\slm_n=\{A\in M_n(K):\tr A=0\}$. We use $[A,B]=AB-BA$ and $\ad_A(B)=[A,B]$. The space $P_d$ has basis the $d!$ words $x_{\sigma(1)}\cdots x_{\sigma(d)}$, $\sigma\in S_d$. For $f\in P_d$, set \[ \Val_n(f)=\{f(A_1,\ldots,A_d):A_i\in M_n(K)\}. \] The cyclic projection $\pi:P_d\to C_d$ identifies words under cyclic rotation; its coordinates are the sums of the coefficients in the cyclic orbits. A polynomial is \emph{traceful} on $M_n$ if its trace function is not identically zero. A \emph{proper} multilinear polynomial is killed, in the free algebra, when any one input is replaced by $1$. We denote the proper subspace of $P_d$ by $\Gamma_d$. Properness implies invariance under adding any scalar matrix to an individual input. This definition agrees with the usual multilinear proper component described by products of Lie commutators through the Poincar\'e--Birkhoff--Witt decomposition \cite{Reutenauer}. \subsection{Unit deletions and proper components} For $d\geq2$, deletion of $x_i$ induces $\partial_i:C_d\to C_{d-1}^{(i)}$, where the superscript means that $x_i$ is omitted. For $f_i=f|_{x_i=1}$ one has $\pi(f_i)=\partial_i\pi(f)$. The cyclic deletion kernel is \[ \mathcal K_d=\bigcap_{i=1}^d\ker\partial_i\subseteq C_d. \] It is distinct from the proper word space $\Gamma_d=\bigcap_i\ker(f\mapsto f_i)\subseteq P_d$. \begin{proposition}\label{prop:proper-pbw} For a nonempty subset $B\subseteq[d]$, let $\mathcal L(B)$ denote the multilinear free Lie component on the variables indexed by $B$. Let $\Pi_{\ge2}(d)$ be the set partitions of $[d]$ without singleton blocks. Symmetrized multiplication gives a natural $S_d$-module isomorphism \[ \Gamma_d\cong \bigoplus_{\pi\in\Pi_{\ge2}(d)}\ \bigotimes_{B\in\pi}\mathcal L(B). \] For $\pi=\{B_1,\ldots,B_r\}$, this map sends a tensor of Lie elements to $\frac1{r!}\sum_{\sigma\in S_r}\ell_{\sigma(1)}\cdots\ell_{\sigma(r)}$; it is independent of the ordering of the blocks. Consequently, with $b_0=1$, \[ b_d:=\dim\Gamma_d =\sum_{\pi\in\Pi_{\ge2}(d)}\prod_{B\in\pi}(|B|-1)! =d!\sum_{j=0}^d\frac{(-1)^j}{j!}. \] \end{proposition} \begin{proof} The free associative algebra is the enveloping algebra of the free Lie algebra $\mathcal L$. In characteristic zero, PBW symmetrization is a natural multigraded vector-space isomorphism $\operatorname{Sym}(\mathcal L)\cong U(\mathcal L)$; see \cite[Theorems 0.2 and 0.5; Section 0.4.3]{Reutenauer}. Its multilinear part is the direct sum over all set partitions, including singleton blocks. Deleting $x_i$ kills every Lie factor of length at least two containing $x_i$. On a symmetrized product with singleton factor $x_i$, deletion removes that factor: its $r+1$ possible positions change the coefficient $1/(r+1)!$ into $1/r!$. Distinct remaining partition components are independent by PBW. Hence the kernel of this deletion is precisely the sum of components without singleton block $\{i\}$. Intersecting over $i$ gives the asserted decomposition. Symmetrization commutes with relabeling, proving equivariance. For $|B|=r$, the multilinear Lie component has dimension $(r-1)!$. Indeed, fix one variable innermost on the right. The corresponding right-normed brackets span by Jacobi; their associative expansions have distinct words ending in that fixed variable, proving independence. This gives the partition sum. Separating singleton supports in the full multilinear PBW decomposition gives $d!=\sum_{j=0}^d\binom dj b_{d-j}$. Binomial inversion yields the second formula. \end{proof} In degrees four and five the dimensions are $9$ and $44$. In degree six, the block types $(6),(4,2),(3,3),(2,2,2)$ contribute respectively $120,90,40,15$, giving $b_6=265$. These are coefficient-space decompositions; they do not permit addition of independently attained polynomial values. \subsection{Detection by pairwise collisions} \label{sec:collision-detection} For $d\ge3$ and $i1$, forces all diagonal entries of $L$ to agree. The first column, using $B_{j1}\ne0$, then forces all entries of $M$ to agree. Scalars are zero in the dual of $\slm_n$, so $d\kappa$ is onto. Thus the preceding differential assertion holds at a realization of the pair by common commutators. For $S=0$ the local characterization in $\slm_n$ gives the whole trace-zero space. For $S\ne0$ it gives every nonscalar matrix. To include scalars, take $P=\operatorname{diag}(1,\zeta,\ldots,\zeta^{n-1})$ and $Q=P^{-2}$, where $\zeta$ is a primitive $n$th root. Both are traceless for $n\ge3$ and $F(P,Q)=SI$. The same Fourier basis makes both hollow, so they are common commutators with a regular diagonal $D$. Scaling $B$ gives all scalars. \end{proof} \begin{lemma} \label{w6draft:star-seed} For every $n\ge3$ the polynomial \[ V(D,E,A,B)=s_3([D,E],[D,A],[D,B]) \] has an actual nonzero rank-one square-zero value with differential rank $n^2$. \end{lemma} \begin{proof} Put $k=n-1$. In the lower $k\times k$ corner take \[ Y=\sum_{i=1}^{k-1}E_{i,i+1},\qquad Z=\sum_{i=1}^{k-1}z_iE_{i+1,i},\qquad H=[Y,Z]=\operatorname{diag}(h_i),\quad h_i=z_i-z_{i-1}, \] where $z_0=z_k=0$. Embed them in $M_n$ with first row and column zero, and put $X=\sum_{j=2}^nE_{1j}$. Choose the $z_i$ generically so every $h_i\ne0$. Then $s_3(X,Y,Z)=XH=N$ is rank-one square-zero with every entry of its first row outside the diagonal nonzero. A diagonal $D$ with distinct entries realizes $X,Y,Z$ independently as $[D,E]$, $[D,A]$ and $[D,B]$. For a hollow variation $U$ of $X$ the derivative is \[ L(U)=UH+HU+ZUY-YUZ. \] Its first row and column are arbitrary. On the $r$th upper diagonal in the lower corner its matrix is tridiagonal, of size $k-r$, with \[ (L_r)_{ii}=h_i+h_{i+r},\qquad (L_r)_{i,i-1}=z_{i-1},\qquad (L_r)_{i,i+1}=-z_{i+r}. \] For $r\ge2$ this matrix is generically invertible. Indeed set $z_i=t^i$. The diagonal entry has leading degree $i+r$ except at the last position, where it has leading degree $k-1$ and negative leading coefficient. In the tridiagonal determinant every adjacent transposition has strictly smaller degree than the two corresponding diagonal terms. Thus the diagonal product is the unique highest-degree term. For $r=1$ the row sums vanish, while its upper-left minor of size $k-2$ has a nonzero leading diagonal product. Hence its rank is $k-2$. The nonzero upper and lower off-diagonal entries force both endpoints of a left null vector to be nonzero. The lower diagonals have the same ranks: the weighted transpose interchanging $Y$ and $Z$ intertwines $L$. It remains to fill one cokernel direction on each first off-diagonal of the lower corner. A variation of $Y$ by the last matrix unit in the first column gives $-e_k xZ$ in that corner, where $x=(1,\ldots,1)$. It has only a last row, zero diagonal, and nonzero last subdiagonal entry. A variation of $Z$ by the first matrix unit in the first column gives $e_1xY$, with only a first row, zero diagonal, and nonzero first superdiagonal entry. These fill the two cokernels. All hollow matrices are therefore in the differential image. Conjugation of $N$ supplies all traceless diagonal directions. Finally, varying $Z$ by the second matrix unit in the first column gives trace $3x_1=3\ne0$. All these are permitted common-commutator variations or simultaneous conjugation directions in the original inputs, so $dV$ is onto $M_n$. \end{proof} \begin{proof}[Completion of Theorem~\ref{w6draft:cubic}: shared-factor cubics] Write $S=A+B$, with $A,B$ the two cyclic coefficient sums of $q$. If all three pair identifications in $q$ vanish as polynomials, polarization makes $q$ antisymmetric in every pair of variables. Thus $q$ is a multiple of $s_3$. For a nonzero multiple, Lemma~\ref{w6draft:star-seed} gives the asserted full-rank nilpotent point. Otherwise relabel the cubic slots so that \[ J(P,Q)=q(P,P,Q)=a'P^2Q+b'PQP+c'QP^2\ne0. \] The sum $a'+b'+c'$ is $S$. Lemma~\ref{w6draft:common-pair} supplies an actual common-commutator point $P=[D,E]$, $Q=[D,B_0]$, with nilpotent value $N=J(P,Q)$, and the required differential if either $S\ne0$ or $A=B=0$. Its tangent directions are also tangent directions of $H_q$ on the subspace where the first two commutator inputs agree. The only remaining case is $S=0$ and $A=-B\ne0$. Retain the smooth common-commutator realization from Lemma~\ref{w6draft:common-pair}. Before its change of basis, $P$ is diagonal with distinct entries and $Q=e_1(0,v_2,\ldots,v_n)$, all $v_j\ne0$. Therefore $[P,Q]$ is nonzero rank-one nilpotent. The common factor $D$ is regular semisimple, so \[ K=[[P,Q],D]\ne0: \] a matrix commuting with $D$ is diagonal in its eigenbasis and cannot be nonzero nilpotent. At the actual shared-factor input $(D,X,Y,Z)=(D,E,E,B_0)$ the value is already $N$. The restricted map $J$ gives all $\slm_n$ tangent directions there. Independently vary the first input $X$ by $L$. Cyclicity gives \[ \tr q([D,L],P,Q) =A\tr([D,L][P,Q])=A\tr(LK). \] This is a nonzero linear functional of $L$. Hence the differential of $H_q$ at this same actual point is onto $M_n$. No separate realizations of its trace and traceless parts have been combined. The image of $H_q$ is conjugation invariant and conical, by scaling one of $X,Y,Z$. The degeneration in Lemma~\ref{w6draft:geometry-local}, applied to the obtained open neighborhood of $N$, therefore gives every nonscalar target in $\slm_n$ or $M_n$, respectively. For completeness all scalars in the traceful case have a common-factor realization as well. Take a primitive $n$th root $\zeta$, the cyclic shift $T$, and a diagonal $\Lambda$ with $T\Lambda=\zeta\Lambda T$. Set \[ U=T,\qquad V=\Lambda T,\qquad W=(UV)^{-1}. \] Their cyclic offsets are $1,1,-2$, all nonzero modulo $n\ge3$. Thus all three are hollow in this same basis. The cyclic cubic words are $I$, and the reversed words are $\zeta^{-1}I$, giving \[ q(U,V,W)=(A+\zeta^{-1}B)I. \] At least one of the two distinct primitive roots $\zeta,\zeta^{-1}$ makes the coefficient nonzero when $(A,B)\ne(0,0)$. For any regular diagonal $D$, solve $[D,X]=U$, $[D,Y]=V$ and $[D,Z]=W$ entrywise. Scaling $X$ gives all scalars in one shared-factor evaluation. Together with the trace-zero identity when $A=B=0$, this completes the common-$D$ classification in every size $n\ge3$. \end{proof} \subsection{Values on hollow matrices in one basis} \begin{corollary} \label{w6draft:merge145:hollow:thm:cubic} Let $n\ge3$ and $V_n=\{A\in M_n:A_{ii}=0\text{ for every }i\}$. For a nonzero multilinear cubic $g$, the similarity saturation of $g(V_n^3)$ is $\slm_n$ if $g$ is identically traceless, and $M_n$ otherwise. In particular every traceless target is similar to one actual value on three matrices hollow in the same basis. \end{corollary} \begin{proof} Apply the shared-factor construction in Theorem~\ref{w6draft:cubic}. Its proof gives a smooth nonzero rank-one nilpotent value with the common input $D$ regular semisimple: this is explicit in Lemmas~\ref{w6draft:common-pair} and~\ref{w6draft:star-seed}. Restricting the input space to the open locus where $D$ is regular semisimple preserves smoothness and hence the image neighborhood. This restricted image remains conjugation invariant and conical, with scaling performed in $X$, not in $D$. The local degeneration therefore gives every nonscalar target in the appropriate space. The scalar construction in the proof of Theorem~\ref{w6draft:cubic} also uses regular diagonal $D$. Diagonalizing $D$ makes each of $[D,X],[D,Y],[D,Z]$ hollow in one common basis. Conversely, any three hollow matrices have these three commutator realizations for any fixed regular diagonal $D$. This proves precisely the stated saturated image, including the scalar targets in the traceful case. There is no reliance on an ordinary unrestricted cubic image theorem. \end{proof} \begin{corollary}\label{w6draft:merge145:hollow:cor:delta3} For any fixed regular diagonal $D$ and $n\ge3$, the similarity saturation of $g([D,A],[D,B],[D,C])$ has the same full-target classification. \end{corollary} \begin{proof} For fixed regular diagonal $D$, the map $\ad_D:V_n\to V_n$ is an isomorphism. Apply the preceding corollary; simultaneous similarity conjugates all inputs, including $D$. \end{proof} \section{Double collisions and products of three commutators} \subsection{Repeated variables and coefficient detection} \begin{proposition}\label{w6draft:repetition:dominance} Let $W$ be an $S_m$-submodule of the multilinear free associative component or its cyclic-word quotient, and write \[ W=\bigoplus_{\lambda\vdash m}S^\lambda\otimes M_\lambda. \] For a partition $\mu\vdash m$ and a set partition $\mathcal P$ of type $\mu$, let $T_{\mathcal P}$ identify the variables in each block of $\mathcal P$. Then \[ \bigcap_{\mathcal P\text{ of type }\mu}\ker T_{\mathcal P} =\bigoplus_{\lambda\not\unrhd\mu}S^\lambda\otimes M_\lambda, \] where $\lambda\unrhd\mu$ denotes dominance of partitions. \end{proposition} \begin{proof} Let $H_{\mathcal P}$ permute the variables within the blocks of $\mathcal P$, and let $e_{\mathcal P}=|H_{\mathcal P}|^{-1}\sum_{h\in H_{\mathcal P}}h$. Complete polarization, separately in the repeated letters, gives \[ T_{\mathcal P}(w)=0\quad\Longleftrightarrow\quad e_{\mathcal P}w=0. \] Substitution and polarization descend to cyclic words, so this argument applies to both stated spaces. The intersection over all conjugates is an $S_m$-submodule. On a simple module $S^\lambda$ it is the whole module precisely when $(S^\lambda)^{H_{\mathcal P}}=0$, and is zero otherwise. By Young's rule this invariant space has dimension $K_{\lambda\mu}$, which is nonzero precisely when $\lambda\unrhd\mu$; see \cite{merge:Fulton}. The group algebra acts trivially on the multiplicity factor. Equivalently, its two-sided ideal generated by $e_{\mathcal P}$ is the whole matrix algebra on each detected simple factor. The same conclusion therefore holds with arbitrary multiplicities. \end{proof} \subsection{Arbitrary mixtures of three pair commutators} \label{w6draft:sec:W6} Write $W_6$ for the span of the ninety ordered products of three disjoint pair commutators. The result of this section applies to one arbitrary linear combination of all ninety products. Define \[ \mathcal K=\bigcap_{\sigma}\ker(R_\sigma|_{W_6}), \qquad R_\sigma P=P(D,D,E,E,A,B), \] where $\sigma$ runs over the forty-five choices of two disjoint pairs, with the remaining variables relabelled $A,B$. \begin{theorem} \label{w6draft:W6-mixed} For every $P\in W_6$ and $n\ge3$, the actual image is \[ \Val_n(P)=\begin{cases} \{0\},&P=0\text{ or }(n=3\text{ and }P\in Ks_6),\\ \slm_n(K),&\pi P=0\text{ and }P\text{ is not an identity of }M_n,\\ M_n(K),&\pi P\ne0. \end{cases} \] All coefficients across the fifteen pairings and six factor orders are arbitrary. The double-collision kernel satisfies \[ \dim\mathcal K=11,\quad\pi\mathcal K=0,\qquad \mathcal K\simeq 2S^{(2,1,1,1,1)}\oplus S^{(1,1,1,1,1,1)}. \] Its last summand is $Ks_6$: it is an identity for $n=3$, and every nonzero member has image $\slm_n$ for $n\ge4$. \end{theorem} The proof uses Lemmas~\ref{w6draft:common-pair} and~\ref{w6draft:star-seed}, together with the collision constructions below. No image assertion is inferred by adding separately obtained polynomial values. We repeatedly use the following parameter observation. Suppose $F_t$ is polynomial in $t$ and its matrix inputs, $F_0(a)=N$, and $d(F_0)_a$ is onto a target vector space. The map $(t,x)\mapsto(t,F_t(x))$ is smooth at $(0,a)$. Restrict its source to the open set of surjective matrix Jacobians. Its image contains an open neighborhood of $(0,N)$, hence contains $(t,N)$ for some $t\ne0$. Thus $F_t$ has an actual full-rank point over $N$. The original map therefore has such a point if, for $t\ne0$, $F_t$ is obtained from it by invertible input transformations and invertible output transformations fixing $N$. \begin{lemma} \label{w6draft:fifteen} Every double collision of a member of $W_6$ lies in the span \[ \mathcal W=\langle P^2Z,PZP,ZP^2,\ \operatorname{perm}(P,Q,R),\ \operatorname{perm}(P,S,T)\rangle, \] with $P,Q,R,S,T,Z$ as in Theorem~\ref{w6draft:four:classification}. Every traceful member of $\mathcal W$ has an open image neighborhood of a rank-one square-zero matrix. Every nonzero trace-zero member has image $\slm_n$. Both assertions hold for all $n\ge3$. \end{lemma} \begin{proof} The displayed spanning property follows by listing the three possible pairing types after the collision: two $P$ factors and $Z$, or one each of $P,Q,R$, or one each of $P,S,T$. This is a fifteen-dimensional space. For precision, write its coefficients as $a_0,a_1,a_2$ and $b_\tau,c_\tau$ for $\tau=123,132,213,231,312,321$. The four substitutions $A=D,E$ and $B=D,E$ are common-factor cubics from Lemma~\ref{w6draft:common-pair}. Their joint polynomial kernel is exactly \[ H=\alpha PL+\beta LP+\gamma C+\delta\mathcal T+\eta\mathcal U, \qquad \mathcal U=s_3(P,Q,R)-s_3(P,S,T). \] This assertion requires only coefficient elimination. The three coefficients of each substituted cubic, with an irrelevant overall sign suppressed in the third row, are \[ \begin{array}{c|ccc} A=D&a_0-c_{132}-c_{312}&a_1-c_{123}-c_{321}&a_2-c_{213}-c_{231}\\ A=E&a_0+b_{123}+b_{213}&a_1+b_{132}+b_{231}&a_2+b_{312}+b_{321}\\ B=D&a_0+b_{132}+b_{312}&a_1+b_{123}+b_{321}&a_2+b_{213}+b_{231}\\ B=E&-a_0+c_{123}+c_{213}&-a_1+c_{132}+c_{231}&-a_2+c_{312}+c_{321} \end{array} \] Setting these entries to zero gives $\alpha=(a_0+a_1-a_2)/2$, $\beta=(a_1+a_2-a_0)/2$, $\gamma=(a_0+a_2-a_1)/2$, and two free alternating coefficients $\delta+\eta$, $\delta-\eta$. This proves the stated kernel expression directly. It is five-dimensional: the three trace functions of $PL,\mathcal T,\mathcal U$ are independent (as verified by the trace-coordinate matrix below), while the coefficients of $DEDEAB$ and $DEDAEB$ separate $PL-LP$ and $C-PL$ by the matrix $\left(\begin{smallmatrix}1&0\\-1&1\end{smallmatrix}\right)$. The same argument also verifies independence of the fifteen spanning products. If all four polynomial substitutions vanish, use this kernel expression. The first four terms are invariant under $E\mapsto E+tD$, while \[ \mathcal U(D,E+tD,A,B)=\mathcal U(D,E,A,B)+2tV(D,E,A,B). \] For $\eta\ne0$, Lemma~\ref{w6draft:star-seed} and the parameter observation applied to $\varepsilon H(D,E+\varepsilon^{-1}D,A,B)$ give a full-rank nilpotent point for $H$. For nonzero $\varepsilon$ the factor $\varepsilon$ is absorbed by the invertible substitution $B\mapsto\varepsilon B$. For $\eta=0$, Theorem~\ref{w6draft:four:classification} applies. The trace-zero kernel members have $\eta=\delta=0$ and $\alpha+\beta+\gamma=0$, so that theorem also proves their assertion. Suppose now that one substituted polynomial is nonzero. If $H$ is trace zero, the corresponding common-factor cubic is nonzero and trace zero; Lemma~\ref{w6draft:common-pair} supplies every traceless target in the image of $H$. If $H$ is traceful and one substitution is traceful, the same lemma supplies a full-rank nilpotent point. It remains to consider traceful $H$ for which all four substitutions are trace zero and at least one is a nonzero polynomial. Relabel so that \[ J(D,E,B)=H(D,E,D,B)\ne0. \] This is a trace-zero common-factor cubic. The five independent trace coordinates of $\mathcal W$ are \[ \tau_0=\tr P^2Z,\quad\tau_1=\tr PQR,\quad\tau_2=\tr PRQ, \quad\tau_3=\tr PST,\quad\tau_4=\tr PTS. \] Their independence can be checked by rotating $A$ to the front and reading the coefficients of the remaining words $DDEEB$, $DEDEB$, $DEDBE$, $DEBDE$, $DBDEE$. Their coefficient matrix is \[ \begin{pmatrix} 0&0&-1&0&0\\ -1&0&1&0&0\\ 0&0&-1&1&0\\ 0&-1&0&-1&0\\ 0&0&0&0&1 \end{pmatrix},\qquad\det=1. \] Write $h=\tr H=\sum k_i\tau_i$. The four trace-zero substitutions give \[ k_0+k_1+k_2=0,\qquad k_0-k_3-k_4=0. \] Let $K$ be the cyclic derivative with respect to $A$, so $h=\tr(AK)$. Direct cyclic rotation gives \[ K=k_0[B,P^2]+k_1[RP,D]+k_2[PR,D]+k_3[PS,E]+k_4[SP,E]. \] Choose the nilpotent seed for $J$ from Lemma~\ref{w6draft:common-pair}, retaining its smooth lift, a hollow dense $P$, and a regular diagonal $D$ in the common basis. Here $S=[D,B]$ is the rank-one matrix denoted $Q$ in that lemma. The eigenvalues of $D$ remain freely selectable. Put $\theta=k_0+k_1-k_3$. Replacing $E$ by $E+\lambda D$ leaves $P,S$ and the map $J(D,E,B)$ unchanged and changes $K$ by \[ \lambda\theta[[S,P],D]. \] If $\theta\ne0$, this coefficient is nonzero: $[S,P]$ is a nonzero rank-one nilpotent, and cannot commute with the regular semisimple $D$. Thus one can arrange $K\ne0$ without changing the nilpotent value of $J$. If $\theta=0$, the trace is a linear combination $h=a\tr(PL)+b\tr\mathcal T$, with $(a,b)\ne(0,0)$. Replacing $B$ by $B+\mu D^2$ still leaves $S$, hence $J$, unchanged. The coefficient added to $K$ is \[ \mu\{a([D^2,P^2]+2[PDP,D])+3b[D,[D,P^2]]\}. \] It is nonzero for a generic choice of the diagonal entries $d_i$ of $D$. Indeed the $(i,j)$ entry of the first summand, without $a$, is \[ (d_i-d_j)\sum_\ell(d_i+d_j-2d_\ell)P_{i\ell}P_{\ell j}. \] For $a\ne0$, choose distinct $i,j,\ell$; dense hollowness of $P$ makes the coefficient of $d_\ell$ nonzero, while the second summand depends only on $d_i,d_j$. For $a=0$, the simple spectrum of $P^2$ prevents it from commuting with $D$: otherwise $P$, a polynomial in $P^2$, would be both diagonal and hollow. Thus the second summand is nonzero. In either case choose $\mu$ so that $K\ne0$. These diagonal shifts preserve the smooth-lift first-row and first-column conditions of Lemma~\ref{w6draft:common-pair}. Return directly to the actual point $(D,E,A,B)=(D,E,D,B)$. Its value is $J(D,E,B)=N$. Varying $D,E,B$ with $A=D$ gives all $\slm_n$ tangent directions from $J$. An independent $A$ variation $L$ has output trace $\tr(LK)$, which is nonzero for some $L$. Consequently $dH$ at this same point is onto $M_n$. This proves the remaining case. \end{proof} \begin{proof}[Proof of Theorem~\ref{w6draft:W6-mixed}] The ninety products form a basis: a word has a unique decomposition into its ordered adjacent pairs, and each product has eight disjoint word terms. As an $S_6$-module this is the multilinear part of $(\bigwedge^2 V)^{\otimes3}$. The Pieri rule gives \[ W_6\simeq S^{(3,3)}\oplus2S^{(3,2,1)}\oplus3S^{(2,2,1,1)} \oplus S^{(3,1,1,1)}\oplus S^{(2,2,2)} \oplus2S^{(2,1,1,1,1)}\oplus S^{(1^6)}. \] Apply Proposition~\ref{w6draft:repetition:dominance} with pattern $(2,2,1,1)$. In the displayed decomposition only $(2,1^4)$ and $(1^6)$ fail to dominate this pattern. Thus the joint kernel consists exactly of these summands. Their dimensions are $5$ and $1$, so $\dim\mathcal K=2\cdot5+1=11$. For completeness, its trace vanishing also has a direct cyclic-word explanation. Modulo cyclic permutation of the three factors there are thirty formal products. Each cyclic word of length six appears in exactly two products, obtained by the two possible parities of adjacent pair cuts. Their coefficient matrix is a signed incidence matrix on thirty vertices. The graph is connected: the pair swaps and a one-place cyclic shift generate $S_6$. Its kernel is therefore at most one-dimensional; the alternating $s_6$ supplies that kernel. Equivalently, the trace quotient is the cyclic-factor coinvariant space modulo $Ks_6$. The elementary symmetric/exterior Pieri decompositions \[ \operatorname{Sym}^3(\bigwedge^2 V) =S_{(3,3)}V\oplus S_{(2,2,1,1)}V\oplus\bigwedge^6V, \qquad \bigwedge^3(\bigwedge^2 V) =S_{(3,1,1,1)}V\oplus S_{(2,2,2)}V \] then identify that quotient as $S^{(3,3)}\oplus S^{(2,2,1,1)}\oplus S^{(3,1,1,1)}\oplus S^{(2,2,2)}$. In particular $\pi\mathcal K=0$ and the cyclic rank is $5+9+10+5=29$. If $\pi P\ne0$, some double collision $H$ has nonzero cyclic projection: otherwise polarization in the trace quotient and the preceding Young rule would put its cyclic class in a zero joint kernel. Such an $H$ has an open image neighborhood at a rank-one nilpotent by Lemma~\ref{w6draft:fifteen}. Its image is contained in that of the original multilinear $P$. The local characterization, including its multilinear scalar lemma, gives $\Val_n(P)=M_n$. If $\pi P=0$ and $P\notin\mathcal K$, some double collision is nonzero and trace zero. Lemma~\ref{w6draft:fifteen} gives $\slm_n$ in the image of $P$; trace vanishing gives equality. Finally $S^{(1^6)}=Ks_6$ is an identity of $M_3$ by Amitsur--Levitzki. The following subsections complete $\mathcal K$ in size three and in every larger size. \end{proof} \subsubsection{Completion of the residual module in size three} \label{w6draft:sec:residual3} In this subsection only, index the variables by $0,\ldots,5$. Let $\operatorname{Alt}_5$ mean the unnormalized signed sum over permutations of $x_1,\ldots,x_5$, leaving $x_0$ fixed, and put \begin{align*} F_1&=\operatorname{Alt}_5[x_0,x_1][x_2,x_3][x_4,x_5],\\ F_2&=\operatorname{Alt}_5[x_2,x_3][x_0,x_1][x_4,x_5],\\ F_3&=\operatorname{Alt}_5[x_2,x_3][x_4,x_5][x_0,x_1],\\ A&=F_1-F_3,\qquad B=2F_2-F_1-F_3. \end{align*} Set $A_0=A$, $B_0=B$, and $A_s=-(0\ s)A$, $B_s=-(0\ s)B$ for $s>0$. These elements have \[ \sum_{s=0}^5A_s=\sum_{s=0}^5B_s=0,\qquad \mathcal K=\langle A_s,B_s:0\le s\le5\rangle\oplus Ks_6. \] To prove this basis description, observe that alternation in five variables kills every double collision. The orbit of each alternant is a quotient of the sign-twisted six-point permutation module. The full alternating projections of $F_1,F_2,F_3$ agree, so $A,B$ have no sign component; the displayed sums vanish. Each nonzero orbit therefore affords $S^{(2,1^4)}$. The two $S_5$-alternating vectors $A,B$ are linearly independent: their coefficients at the words $012345$ and $120345$ are respectively $(4,-4)$ and $(0,8)$. They therefore generate two different copies of that irreducible module. Together with $s_6$ their dimension is eleven, equal to $\dim\mathcal K$. Since $s_6$ vanishes on $M_3$, it remains to consider \[ P_{a,b}=\sum_{s=0}^5a_sA_s+\sum_{s=0}^5b_sB_s. \] Adding a constant to all the $a_s$, or to all the $b_s$, does not change this polynomial. Variable permutations act by the same permutation of the coefficient indices and an overall sign. First fix the five inputs in slots $1,\ldots,5$ to be \[ H,\ K,\ E_{12},\ E_{23},\ E_{31}, \qquad H,K\text{ arbitrary traceless diagonal matrices}. \tag{C} \] The linear map in slot zero preserves every root space $KE_{ij}$. Write $m_{ij}(H,K)$ for its multiplier. Expansion of the three commutators, followed by alternation, gives the following exact conditions for a multiplier to vanish identically in $H,K$: \[ \begin{array}{c|l} ij& m_{ij}\equiv0\text{ if and only if}\\\hline 12&a_0=a_3\\ 23&a_0=a_4\\ 31&a_0=a_5\\ 13&a_1=a_2,\ b_1=b_2,\ 2a_0-a_2-a_3-a_4+a_5+3b_3-3b_4=0\\ 21&a_1=a_2,\ b_1=b_2,\ 2a_0-a_2+a_3-a_4-a_5+3b_4-3b_5=0\\ 32&a_1=a_2,\ b_1=b_2,\ 2a_0-a_2-a_3+a_4-a_5-3b_3+3b_5=0 \end{array} \tag{C0} \] These are linear coefficient identities, not checks at a single choice of $H,K$. For example, expand $H=h_1\operatorname{diag}(1,-1,0) +h_2\operatorname{diag}(0,1,-1)$ and similarly $K$; each condition is obtained by setting the four coefficients of $h_i k_j$ to zero. Suppose the $a_s$ are not all equal. Select an $a$-value with minimum multiplicity $r\le3$, put one of its indices in slot zero, and assign the remaining indices to the five fixed slots as follows. If $r=2$, put the other occurrence and an index with a different $a$-value in slots $1,2$. All three remaining $a$-values differ from $a_0$, and $a_1\ne a_2$; thus none of (C0) holds. If $r=1$ and the five remaining pairs $(a_s,b_s)$ are not all equal, put two different such pairs in slots $1,2$. Again no multiplier vanishes. If those five pairs are equal, the last three equations in (C0) reduce to $2(a_0-a_1)\ne0$. If $r=3$, the $a$-values split into two triples. If $b$ is nonconstant on either triple, choose the active index in that triple so that its other two $b$-values differ, and put these in slots $1,2$. Otherwise $b$ is constant on each triple; use the other two members of the active triple in slots $1,2$. The last three equations in (C0) then reduce to the difference of the two $a$-values, which is nonzero. This covers every nonconstant $a$, with arbitrary $b$. For constant $a$, remove it using $\sum A_s=0$ and fix instead \[ H,\ E_{12},\ E_{21},\ E_{23},\ E_{32} \quad\text{in slots }1,\ldots,5. \tag{P} \] For pure $B$ the four multipliers on $E_{12},E_{21},E_{23},E_{32}$ vanish identically precisely when, respectively, $b_2=b_0$, $b_3=b_0$, $b_4=b_0$, $b_5=b_0$. With $t=(b_0+b_1)/2$, the other two vanish precisely as follows: \[ \begin{array}{c|l} 13&b_2+3b_3=4t,\quad b_4+3b_5=4t,\\ 31&3b_2+b_3=4t,\quad3b_4+b_5=4t. \end{array} \tag{P0} \] Assume some $b$-value has multiplicity at most two and place it in slot zero. If its multiplicity is two, put its other occurrence in slot one; then $b_2,\ldots,b_5$ all differ from $b_0=t$. If its multiplicity is one, choose slot one so that the other four values are not all $t$: if an initial choice fails, interchange that slot with one of the four equal values. In both cases the four remaining values differ from $b_0$ and are not all $t$. Among four numbers not all $t$, there are two, say $u,v$, with $u+3v\ne4t$ and $3u+v\ne4t$. Indeed, subtract $t$. A zero and a nonzero offset give such a pair. If all offsets are nonzero and every pair failed, their ratios would all belong to $\{-3,-1/3\}$, which is impossible already for three offsets in characteristic zero. Put $u,v$ in slots $2,3$. All six multipliers are now nonzero polynomials by (P0). In either successful construction, choose diagonal parameters outside the union of the six multiplier zero sets. Varying the single active slot then realizes every hollow $3\times3$ matrix in one evaluation. Every traceless matrix is similar to a hollow matrix: for a nonscalar matrix choose independent vectors $v,Tv$ to make its first diagonal entry zero, then continue on the remaining trace-zero compression; a scalar trace-zero compression is zero. Conjugation equivariance therefore gives every member of $\slm_3$ as an actual value of $P_{a,b}$. The only coefficients left are constant $a$ and two different constant $b$-values on complementary triples. Modulo constants and scaling, these give ten lines, one for each unordered $3+3$ partition. All are variable permutations, up to sign, of \[ G=\frac1{12}(B_1+B_2+B_3). \] They are settled by one explicit smooth point. At \[ (x_0,x_1,x_2,x_3,x_4,x_5) =(E_{31},E_{13},E_{21},E_{12}, \operatorname{diag}(0,1,-1),E_{23}), \] one has $G=-5E_{23}$. Its differential has the following eight values; the unlisted slots of each tangent vector are zero: \[ \begin{array}{c|c|c} \text{slot}&\text{variation}&dG\\\hline 0&E_{31}&-5E_{23}\\ 0&E_{32}&-6E_{13}\\ 1&\operatorname{diag}(1,-1,0)&-5E_{21}\\ 1&E_{32}&-7E_{31}\\ 2&E_{31}&\operatorname{diag}(4,-3,-1)\\ 2&E_{32}&E_{12}\\ 4&E_{32}&\operatorname{diag}(-2,4,-2)\\ 5&E_{32}&7E_{32} \end{array} \tag{J} \] The six off-diagonal matrix units and two independent traceless diagonals span $\slm_3$. Thus $dG$ has rank eight over a rank-one square-zero value. Scaling and conjugation normalize that value to $E_{12}$, and Corollary~\ref{w6draft:geometry-cert} completes these ten lines. If both $a$ and $b$ are constant the polynomial is zero modulo $Ks_6$. Together with the preceding proof outside $\mathcal K$, this proves the size-three classification in Theorem~\ref{w6draft:W6-mixed}. \subsubsection{A single collision in arbitrary size} \label{w6draft:sec:allsizeK} Continue the variable numbering $0,\ldots,5$. Let $\rho$ identify slots zero and one, and retain the notation $A_s,B_s$ of Section~\ref{w6draft:sec:residual3}. Alternation immediately gives \[ \rho P_{a,b}=(a_0-a_1)\rho A+(b_0-b_1)\rho B. \] The terms with distinguished index at least two vanish, and $\rho A_1=-\rho A_0$, $\rho B_1=-\rho B_0$. After a variable permutation some such collision is nonzero unless both coefficient vectors are constant. A multiple of $s_6$ is killed by every collision. For $U=[D,X]$, $V=[D,Y]$, $W=[D,Z]$, direct alternation gives \[ (aA+bB)(D,D,D^2,X,Y,Z)=-4Q_{a+3b,-a+3b}(D;U,V,W), \] where \[ Q_{r,s}(D;U,V,W)=\operatorname{Alt}_{U,V,W} \bigl(rDUVW+sUDVW-rUVDW-sUVWD\bigr). \tag{Q} \] One way to verify the identity without a long word expansion is to follow a path through four vertices with diagonal entries $d_i,d_j,d_k,d_l$. The coefficients of $A$ and $B$, after removing the three factors $(d_i-d_j)(d_j-d_k)(d_k-d_l)$, are respectively $-4(d_i-d_j-d_k+d_l)$ and $-12(d_i+d_j-d_k-d_l)$. The change of coefficients $(a,b)\mapsto(r,s)=(a+3b,-a+3b)$ is invertible. \begin{lemma}\label{w6draft:weighted-cubic} For every $(r,s)\ne(0,0)$ and $n\ge4$, the actual values of (Q), with $D$ regular diagonal and $U,V,W$ hollow, after simultaneous conjugation, fill $\slm_n(K)$. \end{lemma} \begin{proof} Every value is traceless. Indeed \[ Q_{1,0}=\sum_{\rm cyc}[D,[U,V]]W,\qquad Q_{0,1}=\sum_{\rm cyc}U[D,[V,W]], \] and cyclicity of trace reduces both traces to the Jacobi identity. First suppose $rs\ne0$. Write $n=1+k$, index the lower block by $1,\ldots,k$, and take \[ D=\operatorname{diag}(d_0,\ldots,d_k),\quad U=\sum_{j=1}^kE_{0j},\quad V=\sum_{i=1}^{k-1}E_{i,i+1},\quad W=\sum_{i=1}^{k-1}z_iE_{i+1,i}. \] Use $z_0=z_k=0$. The output has only its first row nonzero, with entries \[ N_{0j}=r\{z_j(d_0-d_{j+1})-z_{j-1}(d_0-d_{j-1})\}. \] An absent boundary term is omitted. All these entries are nonzero for generic parameters, so $N$ is a rank-one square-zero matrix. The $U$ derivative covers the first row and column: their respective multipliers are the preceding expression and \[ s\{z_i(d_{i+1}-d_0)-z_{i-1}(d_{i-1}-d_0)\}. \] On the lower off-diagonal band $j-i=h$, the same derivative is tridiagonal. Its diagonal, forward and backward coefficients are \begin{align*} C_{ij}&=r\{z_j(d_i-d_{j+1})-z_{j-1}(d_i-d_{j-1})\} +s\{z_i(d_{i+1}-d_j)-z_{i-1}(d_{i-1}-d_j)\},\\ A_{ij}&=-z_j\{r(d_i-d_{j+1})+s(d_{i+1}-d_j)\},\\ B_{ij}&=z_{i-1}\{r(d_i-d_{j-1})+s(d_{i-1}-d_j)\}. \end{align*} For $h\ge2$, specialize $z_i=t^i$. Every diagonal coefficient except the last has unique leading term $rz_{i+h}(d_i-d_{i+h+1})$; the last has leading term $-rz_{k-1}(d_{k-h}-d_{k-1})$. In a tridiagonal determinant every adjacent transposition has strictly smaller $t$-degree than the corresponding diagonal pair. Thus the leading diagonal product is unique and nonzero. Negative bands follow by transposition and diagonal rescaling of the shifts, interchanging $r,s$. On the first upper band the rows sum to zero, and its upper-left $(k-2)$-minor is generically nonzero by the same leading-term argument. Its rank is $k-2$. Both off-diagonals are nonzero, so both endpoints of a left null vector are nonzero. The first lower band has the same properties. Varying $V$ by $E_{k0}$ gives a hollow lower-block output supported in its last row, whose $(k,k-1)$ entry is $-s z_{k-1}(d_0-d_{k-1})\ne0$. Varying $W$ by $E_{10}$ gives a hollow output in its first row, whose $(1,2)$ entry is $s(d_0-d_2)\ne0$. These fill the two remaining band cokernels. All hollow output directions are therefore covered. Conjugation of $N$ supplies every traceless diagonal direction modulo the hollow matrices. The saturated derivative is onto $\slm_n$ over $N$, and Corollary~\ref{w6draft:geometry-cert} applies. It remains to treat $rs=0$. Transposition interchanges $r,s$, and nonzero scaling is harmless, so take $(r,s)=(1,0)$. Here a regular \emph{zero} is available. Set \[ D=\operatorname{diag}(0,1,\ldots,k),\quad U=\sum_{i=1}^kE_{i0},\quad V=\sum_{i=1}^{k-1}E_{i,i+1},\quad W=V^{\mathsf T}. \tag{Z} \] Since $[V,W]$ is diagonal, (Q) is zero at this point. We show that its derivative in hollow $U,V,W$ alone is onto $\slm_n$. The $U$ derivative covers the entire first row: its multipliers are $-2$ for $j0$. If $c=2$ and $q|_{\ker T}=0$, restrict to $X_0\oplus Y_0=\ker p_u\oplus\ker s_v$. The restrictions $h_0,k_0$ are both nonzero, and the pure bilinear quadratic $q_0$ vanishes on $h_0+k_0=0$. Hence $q_0=(h_0+k_0)(a(x)+b(y))$. Its pure bidegree $(2,0)$ part $h_0a$ and pure bidegree $(0,2)$ part $k_0b$ vanish. Therefore $a=b=0$ and $q_0=0$. A bilinear form vanishing on $\ker p_u\times\ker s_v$ has the form \[ q=p_u B+C s_v, \] so its matrix rank is at most two, a contradiction. This proves the assertion for $r\ge3$. Now let $r=2$ and assume, seeking a contradiction, that the quadratic vanishes on the \emph{entire} full differential kernel generically. In particular it vanishes on $\ker T$. If the corner matrix $M$ is zero, then $T=0$ and $q\ne0$, a contradiction. For $M\ne0$ the endpoint rows $p_u,s_v$ are nonzero. If $c=2$, the same bidegree argument shows \begin{equation} q|_{\ker p_u\times\ker s_v}=0. \label{m3draft:bilinear-vanishing} \end{equation} If $c=1$, the middle rows are multiples of the endpoint rows and $\ker T=\ker p_u\times\ker s_v$, so the same conclusion holds. A fixed bilinear form of matrix rank two cannot satisfy \eqref{m3draft:bilinear-vanishing} for independent generic $u,v$ unless the spans of the $p_u$ and the $s_v$ each have dimension one. Indeed, fix generic $v$. For $y\in\ker s_v$, the functional $q(\,\cdot\,,y)$ belongs to the line $Fp_u$ for every generic $u$. If those lines span dimension at least two, their intersection is zero, forcing $q(\,\cdot\,,\ker s_v)=0$ and matrix rank at most one. The symmetric argument applies to $s_v$. This reasoning can be made over independent transcendental parameter copies, so it is a generic identity over $F$, not a choice of incompatible specializations. It follows that $\operatorname{rank}M=1$. Write $M=ab^t$, with disjoint nonempty supports $I=\operatorname{supp}a$, $J=\operatorname{supp}b$, since $M_{ii}=0$. The other corner row is obtained from $\sigma(M)^t$, so its fixed support is $I$ while that of the first row is $J$. Thus both corner jets actually have rank two. We may write \begin{equation} p_u=\alpha(u)p,\qquad s_v=\delta(v)s,\qquad q(x,y)=p(x)B(y)+C(x)s(y), \label{m3draft:rank-two-decomposition} \end{equation} where $p,s,B,C$ are defined over $F$, $p$ has support $J$, $s$ has support $I$, and $\alpha,\delta$ are nonzero. Since $q$ has matrix rank two, $B$ is not proportional to $s$ and $C$ is not proportional to $p$. The row $h_u$ is independent of $p$, and $k_v$ of $s$. Consequently $T$ has rank three; its kernel projects onto all of $\ker p$ in $X$ and all of $\ker s$ in $Y$. Introduce the $E_{31}$ source amplitudes $z$. The two negative rows on them are \[ \beta_v(z)=v^tA(D^{(2,3,1)})z,\qquad \gamma_u(z)=z^tA(D^{(3,1,2)})u. \] They are independent, since the full image at rank eight is precisely the hyperplane with $31$ coordinate zero. Set $Z=\ker\beta_v\cap\ker\gamma_u$, a codimension-two space over $F(u,v)$, independent of $w$. The following mixed equivariance identities hold for every full source vector $\xi$: \begin{equation} (d^2f_a(\rho_{21},\xi))_{31}=-(df_a\xi)_{32},\qquad (d^2f_a(\rho_{32},\xi))_{31}=(df_a\xi)_{21}. \label{m3draft:mixed-equivariance} \end{equation} They follow by differentiating covariance with $E_{21}$ or $E_{32}$ and taking output $31$. The extra differential term has zero $31$ coordinate; the upper part of the conjugation tangent contributes nothing by \eqref{m3draft:constant-normal}. Thus both $\rho$ directions are radical for the normal Hessian restricted to $W_0\oplus Z$. They can remove upper controls without altering any pairing used below. Subsequent upper source corrections also alter no such pairing. For $z\in Z$, the diagonal part of $df_a(zE_{31})$ is affine-linear in $w$. Its homogeneous $w$ part occurs only at endpoints $11,33$ and is the $13$ two-cycle jet. Write its vertex-relabeled rank-one matrix as $M^{13}=a^{13}(b^{13})^t$ and let $\sigma_{13}$ swap labels $1,3$. These two endpoint terms are \begin{equation} \begin{split} d_1(w,z)&=(w^ta^{13})((b^{13})^tz),\\ d_3(w,z)&=(w^t\sigma_{13}(b^{13})) (\sigma_{13}(a^{13})^tz). \end{split} \label{m3draft:endpoint-jets} \end{equation} Their two $w$ forms have disjoint supports $I,J$ and are independent. Their corresponding $z$ forms have supports $J,I$. Choose a right inverse of the surjective map $T$ over $F(u,v)$, independent of $w$, and solve \[ T(x(w),y(w))=-\operatorname{diag}(df_a(zE_{31})). \] The solutions are affine-linear in $w$. Their homogeneous parts $x_1,y_1$ satisfy \[ \alpha p(x_1)=-d_1(w,z),\qquad \delta s(y_1)=-d_3(w,z),\qquad h_u(x_1)+k_v(y_1)=0. \] After the control adjustments justified by \eqref{m3draft:mixed-equivariance}, the direction $zE_{31}+x(w)+y(w)$ is in the full kernel. Likewise each $\kappa=(\kappa_x,\kappa_y)\in\ker T$, chosen independently of $w$, has a full-kernel lift preserving the pairings. By our contradiction assumption their polarized normal pairing is zero. The raw pairing between $zE_{31}$ and $\kappa$ contains no $w$: an $E_{31},E_{21}$ pair requires an $E_{12}$ background edge to end at $31$, and an $E_{31},E_{32}$ pair requires an $E_{23}$. Taking the homogeneous $w$ coefficient in the pairing and using \eqref{m3draft:rank-two-decomposition} therefore gives \begin{equation} 0=p(x_1)B(\kappa_y)+C(\kappa_x)s(y_1) =-\frac{d_1(w,z)}\alpha B(\kappa_y) -\frac{d_3(w,z)}\delta C(\kappa_x). \label{m3draft:w-pairing} \end{equation} Here $p(\kappa_x)=s(\kappa_y)=0$ explains the absent two polarized terms. The control corrections vanish by \eqref{m3draft:mixed-equivariance} and \eqref{m3draft:constant-normal}. Independence of the $w$ forms in \eqref{m3draft:endpoint-jets}, the surjective projections of $\ker T$, and the nonproportionality of $B,s$ and $C,p$ show that both $b^{13}$ and $\sigma_{13}(a^{13})$ annihilate $Z$. They are independent; hence \begin{equation} \operatorname{span}(\beta_v,\gamma_u) =R:=\operatorname{span}(b^{13},\sigma_{13}(a^{13})), \label{m3draft:fixed-plane} \end{equation} a fixed plane over $F$ with one nonzero $J$-supported vector and one nonzero $I$-supported vector. Pass to the quotient by this fixed plane. Since $\beta_v$ and $\gamma_u$ depend linearly on independent variables, each coefficient vanishes in the quotient. Their coefficient matrices have rank two, so their row spaces equal $R$. Their path formulas, followed by the inverse diagonal-label field automorphisms, show that both row and column supports of $A$ lie in $I\cup J$, and its row space contains both a nonzero $I$-supported vector and a nonzero $J$-supported vector. Only preservation of zero coordinates under field automorphisms is used; no invariance of the plane $R$ is assumed. On the other hand reversing the path in \eqref{m3draft:rank-two-decomposition} yields \[ A=\eta V^t+U\theta^t, \] where $\eta$ has precisely support $I$ and $\theta$ precisely support $J$. The zero diagonal forces $V_i=0$ for $i\in I$ and $U_j=0$ for $j\in J$. If $L$ is the complement of $I\cup J$, absence of $L$-columns and $L$-rows forces $V_\ell=U_\ell=0$ for $\ell\in L$ (use any nonzero entry of $\eta$ or $\theta$). Thus $A$ is supported only in $I\times J$. Its row space is wholly $J$-supported, contrary to \eqref{m3draft:fixed-plane}. This contradiction proves the nonvanishing of the full-kernel quadratic when $r=2$. Finally choose the eight independent non-$31$ output controls. Their local fixed-value fiber is smooth by the Jacobian criterion~\cite{smooth} of dimension $54-8=46$. The remaining corrected equation has nonzero quadratic initial term, so it is a nonzero nonunit in a regular local domain and cuts dimension by one. The full fiber has the expected dimension $45$; miracle flatness applies. All generic conditions hold on a common nonempty open set, providing an actual triangular $K$-point as claimed. \end{proof} \subsection{Original slots for rank-one symbols} \label{m3draft:rank-one-symbol-section} \begin{proposition}\label{m3draft:rank-one-symbol} If the mixed two-step symbol $A$ has rank one, one actual original full slot has rank at least six. This conclusion is independent of the generic triangular differential rank and of the two-cycle jets. \end{proposition} \begin{proof} Write $A=\alpha\beta^t$ over $F$. Its zero diagonal forces disjoint nonempty supports $I=\operatorname{supp}\alpha$ and $J=\operatorname{supp}\beta$. Choose $i\in I,j\in J$. Then the polynomial $A_{ij}$ is nonzero and $A_{ji}=0$ identically. Keep precisely the original slots $x_i=X,x_j=Y$ free and fix the other four to diagonal matrices. The polynomial $A_{ij}$ is independent of diagonal coordinates in slots $i,j$. Thus all six vertex-permuted coefficients \[ c_{rst}=A_{ij}(D^{(r,s,t)}),\qquad r,s,t\text{ distinct}, \] are nonzero in the same polynomial ring of the other four diagonal inputs. Their product is nonzero. Choose one actual background where all six are nonzero simultaneously. On hollow $X,Y$, a word path with off-diagonal output $(r,t)$ must use the third vertex $s$ and the two free edges. Its reverse-order coefficient is the corresponding permutation of $A_{ji}$, which is zero. Hence the exact off-diagonal output is \[ f(\ldots,X,\ldots,Y,\ldots)_{rt} =c_{rst}X_{rs}Y_{st}. \] Closed two-vertex paths contribute only to diagonal outputs. Freeze this actual $Y$ to have all six hollow entries equal to one and diagonal zero. On the six hollow matrix units of the original $X$ slot, the hollow projection is \[ \begin{array}{lll} E_{12}\mapsto c_{123}E_{13},&E_{13}\mapsto c_{132}E_{12},& E_{21}\mapsto c_{213}E_{23},\\ E_{23}\mapsto c_{231}E_{21},&E_{31}\mapsto c_{312}E_{32},& E_{32}\mapsto c_{321}E_{31}. \end{array} \] In the order $(12,13,21,23,31,32)$ its determinant is \[ -c_{123}c_{132}c_{213}c_{231}c_{312}c_{321}\ne0. \] Therefore the \emph{full} original-slot map has rank at least six. The hollow projection only certifies its rank; it is not substituted for the full-slot map in Theorem~\ref{m3draft:rank-six}. Tracefulness and that theorem give full actual image. Nothing here asserts triangular flatness, which need not hold for rank-one symbols. \end{proof} \subsection{Proof of the image theorem in size three} \label{m3draft:assembly-section} \begin{lemma} \label{m3draft:flat-to-image} Let $h:M_3(K)^m\to M_3(K)$ be ordinary multilinear and traceful. If it is flat at a point with value $\lambda E_{13}\ne0$, then its actual image is all of $M_3(K)$. \end{lemma} \begin{proof} The flat locus is open~\cite{flatlocus}, so restrict to a flat open source neighborhood. A flat morphism of finite presentation is open~\cite{flatopen}, and its target-open image contains $\lambda E_{13}$. Every $K$-point of this open image has a nonempty finite-type fiber, hence a $K$-point by algebraic closedness. Thus it is an actual-image neighborhood. The image is invariant under scalar multiplication (scale one original input) and conjugation (conjugate all inputs). Conjugating the open set, assume it contains $E_{13}$. For a nonscalar target $B_0$, choose $v$ with $v,B_0v$ independent; otherwise every vector is an eigenvector and $B_0$ is scalar. Use $B_0v$ as first basis vector and $v$ as third. The conjugate $C$ has $C_{13}=1$. For $t\ne0$, put \[ \Gamma(t)=t^2\operatorname{diag}(t^{-1},1,t) C\operatorname{diag}(t,1,t^{-1}). \] Its entry exponents are $2+e_i-e_j$ with $(e_1,e_2,e_3)=(-1,0,1)$; they are nonnegative and only the $13$ exponent is zero. This polynomial curve therefore has $\Gamma(0)=E_{13}$. Its inverse image of the open neighborhood contains some nonzero $t_0$. Undoing the scalar multiplication and conjugation gives $C$ and then $B_0$ as actual values. This treats every nonscalar Jordan type, not just cyclic matrices. The traceful clock--shift argument in the proof of Theorem~\ref{m3draft:rank-six} supplies all scalars. \end{proof} \begin{proof}[Proof of Theorem~\ref{m3draft:main}] If $c(f)\ne0$, choose an original slot $s$ with $c_s\ne0$ and an actual regular diagonal $D$. The slot $c_s\operatorname{ad}_D^5:M_3\to M_3$ has exactly the six-dimensional hollow image. Theorem~\ref{m3draft:rank-six} applies. If $c(f)=0$ and $f\in W_6$, apply the arbitrary-$W_6$ traceful classification, Theorem~\ref{w6draft:W6-mixed}. Its hypothesis of nonzero cyclic class follows from $\operatorname{tr}f\ne0$. This invokes the theorem for arbitrary simultaneous coefficients across all ninety ordered products, not a sum of independently attained values or a single-pairing special case. Otherwise Lemma~\ref{m3draft:filtration} supplies a nonzero symbol $A$. Proposition~\ref{m3draft:strong-seed} puts the generic full triangular rank $R$ in $\{7,8\}$. If $r=1$, use Proposition~\ref{m3draft:rank-one-symbol}, regardless of $R$. For $R=7$ and $M=0$, Proposition~\ref{m3draft:rank-seven-zero} and Lemma~\ref{m3draft:flat-to-image} apply. For $R=7$ and $M\ne0$, Lemma~\ref{m3draft:symmetric-corner} and the exact two-free-slot restriction yield Proposition~\ref{m3draft:weighted-slot}. Finally, for $R=8$ and $r\ge2$, Proposition~\ref{m3draft:rank-eight} and Lemma~\ref{m3draft:flat-to-image} apply. These alternatives exhaust all possibilities. All conclusions concern actual $K$-valued original tuples, with every extension-field target descended through its full affine fiber. \end{proof} \section{Two free original inputs and the mixed proper-sextic layer}\label{sec:proper-mixed} \label{mx:section} Throughout this section $K$ is algebraically closed of characteristic zero, $f\in K\langle x_1,\ldots,x_6\rangle$ is ordinary multilinear and proper, and $f\in\operatorname{Id}(UT_2)$ but $f\notin\operatorname{Id}(UT_3)$. Proper means that every original unit substitution is the zero free polynomial. The standard triangular-identity filtration identifies this layer with $c(f)=0$ and $f\notin W_6$. Write $\mathcal H_n$ for the vector space of hollow $n\times n$ matrices. All statements about images concern single actual evaluations over $K$. \subsection{Local images and two-input restrictions} \begin{lemma}\label{mx:actual-image} Let $p$ be an ordinary multilinear polynomial. If its actual image contains an open neighborhood of a nonzero rank-one square-zero matrix in $M_n$, and $\operatorname{tr}p$ is not identically zero, then its image is $M_n$. If $p$ is identically traceless and its image contains such a relative open neighborhood in $\mathfrak{sl}_n$, its image is $\mathfrak{sl}_n$. A smooth evaluation chart, or a flat point of the full evaluation map, over that matrix supplies the required actual open neighborhood. \end{lemma} \begin{proof} Conjugate and scale the neighborhood to contain $E_{12}$. For a nonscalar matrix $A$, choose $v$ with $v,Av$ independent and a basis making the $(1,2)$ entry of a conjugate $B$ equal to one. The curve \[ t^2\operatorname{diag}(t^{-1},t,1,\ldots,1)B \operatorname{diag}(t,t^{-1},1,\ldots,1) \] extends to $t=0$ with value $E_{12}$: all entry exponents are nonnegative and only position $(1,2)$ has exponent zero. The inverse image of the actual open neighborhood contains some nonzero $t$. Undoing conjugation and absorbing the scalar into one original multilinear input gives $A$. In the traceless case every nonzero target is nonscalar, and zero is obtained by setting one original input to zero. In the traceful case use the clock--shift homogeneous basis $Z^aX^b$, $(a,b)\in(\mathbb Z/n)^2$. Expand a nonzero-trace evaluation multilinearly in this basis. Some basis tuple has nonzero-trace output. Every word has the same total grade, and only the degree-zero component $KI$ has nonzero trace. Thus $p$ attains a nonzero scalar, and original-slot scaling gives every scalar. Smooth maps and flat locally finitely presented maps are open near the specified point. A nonempty finite-type fiber over a $K$-point has a $K$-point, so this openness concerns actual values, not only geometric image closures. \end{proof} The triangular restriction has the form \[ f(D_l+u_lE_{12}+v_lE_{23}+w_lE_{13}) =u^{\mathsf T}A(D)vE_{13},\qquad A\ne0, \] and $A_{ii}=0$ by original multilinearity. Choose original slots $i,j$, leave their inputs $X,Y$ free, and make the other four inputs diagonal. \begin{lemma}\label{mx:chart} The resulting actual map $F_D(X,Y)$ ignores the diagonal parts of $X,Y$. For distinct matrix vertices its coordinates are \begin{align} (F_D)_{rt}&=\sum_{s\ne r,t} \bigl(a_{rst}X_{rs}Y_{st}+b_{rst}Y_{rs}X_{st}\bigr),\label{mx:offdiag}\\ (F_D)_{rr}&=\sum_{s\ne r}q_{rs} \bigl(\alpha X_{rs}Y_{sr}+\beta Y_{rs}X_{sr}\bigr),\label{mx:diag}\\ q_{rs}&=\prod_{l\notin\{i,j\}}(d_{lr}-d_{ls})=q_{sr}. \end{align} Here $a_{rst},b_{rst}$ are vertex relabelings of $A_{ij},A_{ji}$. Their closed-path coefficients are respectively $\alpha q_{rs},\beta q_{rs}$. Each nonzero symbol gives nonzero polynomials for all its distinct-vertex relabelings, and these may be made nonzero simultaneously at one actual diagonal background. \end{lemma} \begin{proof} If one free input is diagonal and the other is a matrix unit, every surviving word visits at most two vertices. After relabeling this is a $UT_2$ evaluation, hence zero. Linearity proves the first assertion. Two off-diagonal factors can reach $(r,t)$ only along a path $r\to s\to t$, in one of the two orders of the original free slots; closed paths give the diagonal formula. There is no third off-diagonal factor. Properness permits an independent scalar translation in each passive diagonal input. Its closed coefficient is linear in that input's two vertex entries and translation invariant, hence contains their difference. Multiplying the four differences leaves constants $\alpha,\beta$. Their number is even, giving $q_{rs}=q_{sr}$. A distinct-vertex relabeling is an injective renaming of independent coordinates. The finite product of the nonzero weights is nonzero in one polynomial ring, so a common actual background exists. \end{proof} \begin{lemma}\label{mx:derangement} An $m\times m$ matrix with zero diagonal and off-diagonal entries $c_{rs}y_{rs}$, where every $c_{rs}\ne0$ and $m\ge2$, has nonzero determinant as a polynomial in the independent $y_{rs}$. \end{lemma} \begin{proof} A derangement exists. Its determinant term has nonzero coefficient, and different permutations have different directed-edge monomials. Thus that term cannot cancel. \end{proof} \subsection{Leading trace tensors and two low-degree detectors} \begin{lemma}\label{mx:trace-fiber} Let $h=\operatorname{tr}f$ and $g=f-hI/n$. Suppose an actual tuple $a$ has $f(a)=N$ nonzero rank-one square-zero. Suppose a source control space $W$ maps onto $\mathfrak{sl}_n$ under $dg_a$. If, for some $r\ge1$, all ambient derivatives $d^qh_a$ vanish for $q0$ we have $\ell_1>0>\ell_2$, whereas for $h<0$ the signs reverse. This proves nonvanishing in $\mathbb Q$, hence in every characteristic-zero field. The reduced controls therefore have rank ten for every integer $h$. Together with the six first-row and first-column controls, they give rank sixteen for the full original evaluation map at the same rank-one square-zero value. These are variations of two original slots together with simultaneous conjugation of the original inputs. Apply Lemma~\ref{mx:quotient-induction} starting at size four. For \eqref{mx:nine-types} keep $D_n=\operatorname{diag}(0,1,\ldots,n-1)$; all its distinct-vertex and closed weights are nonzero products of powers of distinct differences. This supplies actual smooth rank-one charts for every $n\ge4$. Lemma~\ref{mx:actual-image} proves the result. Tracefulness is automatic here from the nonzero symmetric closed pair. \end{proof} \subsection{Images in the mixed layer} \begin{theorem} \label{mx:traceful-complete} Let $f$ be ordinary multilinear proper of degree six, an identity of $UT_2$ but not of $UT_3$, with nonzero cyclic projection. Then $f(M_n^6)=M_n$ for every $n\ge4$. \end{theorem} \begin{proof} The nonzero cyclic derivative has degree five, below $2n$, so $f$ is traceful in each stated size. If every closed coefficient is zero, choose a nonzero $A_{ij}$ and use Theorem~\ref{mx:zero-cycle}, including its directional subcase. Otherwise choose $M_{ij}\ne0$. If $M_{ij}\ne M_{ji}$ use Theorem~\ref{mx:nonsymmetric}; if they are equal use Theorem~\ref{mx:symmetric}. The closed coefficients have the common four-difference form, so these are exactly the alternatives $\alpha\ne\beta$ and $\alpha=\beta\ne0$. This proves the assertion in all cases. \end{proof} \begin{theorem} \label{mx:traceless-complete} Under the same proper and triangular hypotheses, if the cyclic projection is zero, then $f(M_n^6)=\mathfrak{sl}_n$ for every $n\ge3$. \end{theorem} \begin{proof} The whole polynomial is trace zero. Equation~\eqref{mx:diag} with $X=E_{rs},Y=E_{sr}$ gives $\beta=-\alpha$ for every pair. If $A_{ij}\ne0,A_{ji}=0$, then $\alpha=\beta=0$. Fix $X$ generically. For each target column the map in the hollow entries of $Y$ has the weighted hollow coefficient matrix $(a_{rst}X_{rs})_{r,s\ne t}$. By derangements all these matrices are invertible simultaneously at one actual background. Thus every hollow target is one chart value, and every traceless target is its conjugate. If both directions of a selected nonzero pair survive and its closed coefficients vanish, the hollow-control proof of Theorem~\ref{mx:zero-cycle} applies already in size three: Lemma~\ref{mx:weighted-hollow} has base $m=n-1=2$. The two original slots give all hollow directions at a rank-one row seed; conjugation gives all traceless diagonals. No trace detector is needed because the whole $f$ is traceless. If any closed coefficient is nonzero, the skew branch of Theorem~\ref{mx:nonsymmetric} gives the same smooth $\mathfrak{sl}_n$ conclusion, with its size-three base and all-size induction. These cases exhaust the nonzero symbol. Lemma~\ref{mx:actual-image}, applied relatively in $\mathfrak{sl}_n$, proves equality. \end{proof} \subsection{Completion of the proper classification} \begin{theorem}\label{thm:proper-sextic-classification} Let $f\in\Gamma_6$ and $n\geq3$. If $f$ is not an identity of $M_n$, then \[ \Val_n(f)=\begin{cases}\slm_n,&\pi f=0,\\M_n,&\pi f\ne0.\end{cases} \] The identity case is $f=0$ when $n\geq4$, and $f\in Ks_6$ when $n=3$. \end{theorem} \begin{proof} The identity statements and the trace test are Proposition~\ref{trace-test}. If $c(f)\ne0$, apply Theorem~\ref{higher-uniform}. If $c(f)=0$ and $f\in W_6$, apply Theorem~\ref{w6draft:W6-mixed}, including its explicit $M_3$ identity exception. Otherwise Proposition~\ref{filtration} puts $f$ in the nonzero mixed layer. In the trace-zero case, Theorem~\ref{mx:traceless-complete} applies for every $n\geq3$. In the traceful case use Theorem~\ref{m3draft:main} for $n=3$ and Theorem~\ref{mx:traceful-complete} for $n\geq4$. These are all cases of the proper filtration. None of these results uses a classification of nonproper sextics or of lower-degree images. \end{proof} \section{Nonproper sextics: unit restrictions and trace lifting} \label{sec:nonproper-complete} Throughout this section $n\geq3$. We use the notation and properness convention of Section~\ref{sec:preliminaries}. We now apply Theorem~\ref{thm:proper-sextic-classification}, whose proof is independent of this section and of lower-degree image classifications. We use the standard minimum-degree facts that a nonzero polynomial identity of $M_n$ has degree at least $2n$, and a nonzero polynomial identity of $\operatorname{UT}_3$ has degree at least six. We also use that the centralizer of a free generator $D$ in a free associative algebra is $K[D]$. The geometric steps below use miracle flatness and openness of flat morphisms locally of finite presentation. The only special $M_3$ image input is the preceding original-slot theorem~\ref{thm:original-slot-rank-six}. \subsection{Image-preserving Lie substitutions} \begin{lemma}\label{np:padding} Let $q$ be a nonzero proper multilinear polynomial of degree $d$, where $2\leq d\leq5$. There is a proper ordinary multilinear sextic $Q$ such that \[ \operatorname{Val}_n(Q)=\operatorname{Val}_n(q)\qquad(n\geq2). \] Consequently, by Theorem~\ref{thm:proper-sextic-classification}, for $n\geq3$ this common image is $M_n$ if $q$ is traceful and is $\mathfrak{sl}_n$ if $q$ has identically zero trace. \end{lemma} \begin{proof} For $r\geq2$ use the explicit right-normed polynomial \[ L_r(z_1,\ldots,z_r) =[z_1,[z_2,\ldots,[z_{r-1},z_r]\ldots]]. \] Its image is exactly $\mathfrak{sl}_n$. Indeed, every value is traceless. Conversely, Lemma~\ref{lem:hollow} and $L_r(D,\ldots,D,Z)=\ad_D^{r-1}Z$ realize every hollow target and then every traceless target by simultaneous conjugation. Set $r=7-d$ and replace, for example, the first slot of $q$ by $L_r$ on fresh disjoint variables: \[ Q(z_1,\ldots,z_r,x_2,\ldots,x_d) =q(L_r(z_1,\ldots,z_r),x_2,\ldots,x_d). \] It is ordinary multilinear of degree $r+d-1=6$ and proper: a unit in a new leaf kills the Lie word, and a unit in an untouched slot kills $q$. One image inclusion is immediate from substitution. Conversely, for every actual input tuple, \[ q(A_1,A_2,\ldots,A_d) =q\left(A_1-\frac{\operatorname{tr}A_1}{n}I,A_2,\ldots,A_d\right). \] Realize the displayed traceless first input as one $L_r$-value. This realizes the original $q$-value as one $Q$-value, proving equality of actual sets. For $n\geq3$, the free nonzero $q$ is an actual nonidentity since $d\leq5<2n$. Image equality makes $Q$ an actual nonidentity too. In particular, when $n=3$, $Q$ cannot fall into the degree-six identity line. Tracefulness or trace-zero image is preserved. Applying the already proved proper-sextic theorem to this nonidentity $Q$ gives the asserted alternatives. No classification through degree five is being assumed here. \end{proof} \subsection{Local trace lifting} \begin{lemma}\label{np:trace-lift} Let $f:(M_n)^6\to M_n$ be ordinary multilinear and traceful, put $h=\operatorname{tr}f$ and $g=f-(h/n)I$. Suppose that at an actual input $a$ the value $N=f(a)$ is nonzero rank-one square-zero. Suppose also that there is a linear control space $W$ in the source tangent space such that $dg_a(W)=\mathfrak{sl}_n$, and an integer $r\geq1$ satisfying \[ d^s h_a=0\ (s3$, use the same matrix units on three vertices, with zero arbitrary inputs outside that corner. Every word contains arbitrary edges, so other vertices contribute nothing. Thus the very same trace-coefficient detection works in every $n\geq3$. This is a direct evaluation of the trace expression in size $n$, not an inference from an $M_3$ image theorem. \end{proof} \paragraph{Restriction matrices.} The restrictions in the preceding proof can also be written as matrices. Index the coefficients of $P_\sigma$ by gap assignments $(t_1,\ldots,t_j)\in\{0,\ldots,k-1\}^j$. For every listed partition $a$ and output monomial $v\in\{0,1,2\}^j$, the restriction matrix entry is \[ R_{(a,v),(t_1,\ldots,t_j)} =\prod_{\ell=1}^j{\bf1}_{a_{t_\ell}=v_\ell}. \] It has column rank $k^j$. At maximal $j$, actual square-minor determinants are $1,1$ for $(k,j)=(3,3)$, $1,-1,1$ for $(4,2)$, $1,1,-1,-1$ for $(5,1)$, and five copies of $1$ for $(6,0)$. The thirty cases, including smaller $j$, comprise eight, nine, eight and five cases for $k=3,4,5,6$. These integer minors express the independence of the class sums and the unique labeled Euler orders used in the proof. The passage to arbitrary $n$ is the explicit three-vertex construction in the proof, and the passage from the detector to a full image uses the separate controls and trace lift below. \subsection{The two-free-slot trace-zero charts} Let $q$ be a nonzero proper trace-zero unit restriction of $f$, of degree $d\leq5$. If all one-edge coefficients of $q$ vanish, then $q\in\operatorname{Id}(\operatorname{UT}_2)$. It is not an identity of $\operatorname{UT}_3$, by the degree bound, so it has a nonzero two-step symbol. Choose two original surviving slots $X,Y$ with directional coefficient $C\ne0$, and denote the opposite coefficient by $E$. Freeze all other surviving inputs at independent diagonal matrices and retain the deleted original inputs at $I$. The resulting actual chart ignores the diagonal parts of $X,Y$. Indeed a term involving only one arbitrary off-diagonal edge is an upper-triangular two-vertex evaluation and vanishes by the $\operatorname{UT}_2$ identity. For distinct vertices, its exact off-diagonal formula is \begin{equation}\label{np:two-path} F(X,Y)_{rt}=\sum_{s\ne r,t} \bigl(a_{rst}X_{rs}Y_{st}+b_{rst}Y_{rs}X_{st}\bigr). \end{equation} Every $a_{rst}$ is a relabeling of the same nonzero polynomial $C$ in the remaining diagonal entries; the $b_{rst}$ are relabelings of $E$. Because $q$ is trace zero, the closed contribution of $X_{rs}Y_{sr}$ is \begin{equation}\label{np:closed-pair} k_{rs}(E_{rr}-E_{ss}). \end{equation} This follows by taking the trace of that single matrix-unit evaluation. It holds in every degree: no even-degree sign convention is imposed. Either all $k_{rs}$ vanish polynomially, or all are nonzero relabelings. A finite product in the common diagonal-background polynomial ring shows that all nonzero path coefficients can be made nonzero simultaneously at one actual background. If $E=0$, trace zero forces all $k_{rs}=0$. \begin{lemma}\label{np:weighted-hollow} Let $B$ have size $m\geq2$, with all displayed coefficients nonzero. For hollow $W,H$ and vectors $x,z$ on $B$, put \[ T_W(H)_{rt}=\sum_{s\ne r,t} (a_{rst}H_{rs}W_{st}+b_{rst}W_{rs}H_{st}),\qquad R_x(z)_{rt}=c_{rt}z_rx_t\quad(r\ne t). \] Generically in $(W,x)$, $(H,z)\mapsto T_W(H)+R_x(z)$ is onto $\mathcal H_m$. \end{lemma} \begin{proof} For $m=2$, $R_x$ alone is onto if both entries of $x$ are nonzero. For induction write $B=B_0\cup\{v\}$ and specialize $W=W_0\oplus0$. The old outputs are the smaller map. The new row is controlled by $H_{vs}$ through the weighted hollow matrix with entries $a_{vst}W_{st}$, and the new column by $H_{sv}$ through the matrix with entries $b_{rsv}W_{rs}$. Each determinant is nonzero: a derangement exists on $B_0$, and different permutations give different monomials in the independent entries of $W_0$. Their opens meet the old surjectivity open in the same irreducible space. Prescribe old outputs first using old $H,z$, then correct the new row and column independently by these cross $H$ entries. Such corrections do not change old outputs. This gives one surjective specialization; a nonzero maximal minor gives the asserted open for unrestricted $W$. \end{proof} \begin{lemma}\label{np:bidirectional-control} Suppose both $C$ and $E$ are nonzero in \eqref{np:two-path}, with the trace-zero closed terms \eqref{np:closed-pair}. There is a nonempty open of actual row seeds \[ X=\sum_{t\in B}x_tE_{1t},\qquad Y=0\oplus W, \qquad B=\{2,\ldots,n\},\quad W\in\mathcal H_{n-1}, \] at which $N=F(X,Y)$ has every first-row off-diagonal entry nonzero, and the two original slot spaces together with simultaneous conjugation map onto $\mathfrak{sl}_n$. The assertion holds in either orientation of the two selected slots. \end{lemma} \begin{proof} The value is a single hollow row: \[ N_{1t}=\sum_{s\in B\setminus\{t\}}a_{1st}x_sW_{st}. \] The matrix taking $x$ to this row is generically invertible by the derangement argument. The first-row and first-column variations of $X$ likewise give all first-row and first-column outputs on a common nonempty open; the latter matrix has entries $b_{rs1}W_{rs}$. If every $k_{rs}=0$, lower-block variations $H$ of $X$ and first-column variations $z_r=\delta Y_{r1}$ give the hollow lower block \[ T_W(H)_{rt}+b_{r1t}z_rx_t. \] Lemma~\ref{np:weighted-hollow} makes this onto. These controls do not create first-row or first-column output. Thus the two slots give all hollow directions. The conjugation directions $[E_{t1},N]$ have diagonal parts $N_{1t}(E_{tt}-E_{11})$, supplying the missing traceless diagonals. Now suppose the $k_{rs}$ are nonzero. Work modulo the already controlled first row and column. The target consists of trace-zero diagonals and hollow lower-block entries. The controls are lower hollow $H$, first-column $z$, and the conjugation directions $E_{r1}$. Start with $n=3$, writing \[ \begin{gathered} X=xE_{12}+yE_{13},\qquad W=uE_{23}+vE_{32},\\ N=pE_{12}+q_0E_{13},\quad p=a_{132}yv,\quad q_0=a_{123}xu. \end{gathered} \] The lower $H$ controls give the line $K(E_{22}-E_{33})$. Modulo this line, $z_2$ and the $E_{21}$ conjugation direction have coordinates on $(11,23)$ respectively \[ (k_{12}x,b_{213}y),\qquad(-p,q_0). \] Their determinant is \[ k_{12}a_{123}x^2u+a_{132}b_{213}y^2v\ne0 \] as a polynomial, since the monomials are distinct and both coefficients are nonzero. The $z_3$ direction then supplies coordinate $32$ through $b_{312}x\ne0$, its other coordinates being removable by the preceding controls. This proves the reduced surjectivity in size three. For induction add a lower vertex $v$ and specialize $W=W_0\oplus0$. The old controls supply the old reduced target. Cross $H$ entries give the new hollow row and column through the two invertible weighted hollow matrices from the preceding induction; they change neither old outputs nor diagonals. The control $z_v$ contributes $k_{1v}x_v(E_{11}-E_{vv})$, nonzero modulo the old diagonal space. Absorb its remaining old trace-zero diagonal part with old controls and its hollow part with cross controls. This adds exactly the required new diagonal direction. The reduced map is therefore onto at one specialization and on a nonempty open for unrestricted $W$. Finally intersect this open with the first-row/column determinant opens and with the open where all entries of $N$ in its row are nonzero. All are nonempty opens in the same irreducible parameter space. Since both directional coefficients are nonzero, the argument is unchanged on interchanging the original slots. \end{proof} \subsection{The original-input trace lift} \begin{proposition}\label{np:unit-trace-lifting} Let $f$ be an ordinary multilinear traceful sextic on $M_n$, $n\geq3$. Suppose it has a nonzero proper trace-zero unit restriction $q$ of degree $2\leq d\leq5$. Then $\operatorname{Val}_n(f)=M_n$. \end{proposition} \begin{proof} Every unit slot below remains an original input, and all its tangent and higher-jet directions are retained. \paragraph{A nonzero one-edge coefficient.} Properness gives the exact two-vertex formula \[ q(D_\ell+u_\ell E_{12}) =\sum_s c_su_s\prod_{\ell\ne s}(d_{\ell1}-d_{\ell2})E_{12}. \] Indeed each passive diagonal variable enters linearly, and its unit substitution kills the coefficient; hence its scalar-coordinate polynomial is divisible by that variable's two-vertex difference. Suppose $c_s\ne0$. Keep its original slot $X$ free, take every other surviving input equal to a common regular diagonal $D$, and retain the deleted inputs at $I$. The exact original $X$-slot is, up to a fixed harmless sign, $c_s\operatorname{ad}_D^{d-1}$, an isomorphism on $\mathcal H_n$. For $n=3$ this is an original full-slot minor of rank six, so Theorem~\ref{thm:original-slot-rank-six} finishes the proof. For $n\geq4$ write $h=\operatorname{tr}f=\operatorname{tr}(XG)$, with $G$ multilinear of degree five in the other original inputs. Expand all five of those inputs about their $D$ or $I$ base values as base $+tZ$. The constant coefficient $G_0$ is zero, since $\operatorname{tr}q=0$ for every full $X$. The top coefficient is the original nonzero cyclic gradient, so let $r\geq1$ be the least nonzero coefficient $G_r$. It has an off-diagonal value for some regular diagonal $D$. Otherwise $[D,G_r]=0$ for diagonal $D$, hence for every diagonalizable $D$ by covariance and for every $D$ by density. This is a polynomial identity of degree at most six, less than $2n$, and is therefore zero freely. But the centralizer of $D$ is $K[D]$, whereas every component of $G_r$ has positive degree in the independent $Z$ variables. This forces $G_r=0$, a contradiction. Components with different perturbed-slot subsets can be separated by their $Z$ multidegrees. Choose one actual regular $D$ and perturbation tuple for this nonvanishing, and relabel so that $(G_r)_{b1}\ne0$ for some $b\ne1$. Choose a hollow first-row $X$ avoiding the trace hyperplane $\operatorname{tr}(XG_r)=0$ and the finitely many hyperplanes on which one output row entry vanishes. Then $N=f(a)$ is a full nonzero hollow row. The original $X$-slot gives all hollow controls, and conjugation gives all traceless diagonal controls. The minimal-jet argument preceding Lemma~\ref{np:diagonal-output} supplies exactly the tensor hypotheses of Lemma~\ref{np:trace-lift}, proving full image. \paragraph{No one-edge coefficient: a one-direction pair.} Then $q\in\operatorname{Id}(\operatorname{UT}_2)$ but not $\operatorname{Id}(\operatorname{UT}_3)$, so the two-step chart described above exists. Suppose first that $C\ne0$ and $E=0$. All closed terms vanish by trace zero. For fixed hollow $Y$, the original hollow $X$-slot acts row by row through matrices \[ B_r=(a_{rst}Y_{st})_{s,t\ne r},\qquad (B_r)_{ss}=0. \] Each determinant is nonzero as a polynomial in $Y$: derangements exist since $n-1\geq2$, and distinct permutations give distinct monomials. The product of all these determinants is nonzero, so one actual $Y$ makes the slot an isomorphism of $\mathcal H_n$ onto itself. When $n=3$ this gives an original rank-six minor, and the rank-six theorem applies. For $n\geq4$, again write $h=\operatorname{tr}(XG)$, now retaining $Y$ arbitrary and expanding the other four original slots, including the unit slots, about their independent diagonal or unit backgrounds. Let $r\geq1$ be the first nonzero coefficient $G_r$. Each component has at least the two arbitrary letters $Y$ and one perturbed slot, and after deleting fixed unit letters has total degree at most five. Lemma~\ref{np:diagonal-output} implies that $G_r$ has a nonzero off-diagonal value. This nonvanishing open intersects all the determinant opens in the same irreducible $(D,Y,Z)$ space. Choose a row $X$ with $\operatorname{tr}(XG_r)\ne0$ and every output row entry nonzero. The original slot and orbit again give $\mathfrak{sl}_n$ controls. Minimality kills all lower ambient trace jets, including those varying $Y$ or an original unit input, and makes the first tensor annihilate the control space. Lemma~\ref{np:trace-lift} applies. \paragraph{A bidirectional pair.} Finally suppose $C,E\ne0$. Use the row family from Lemma~\ref{np:bidirectional-control}. On the full two-free-slot chart, $h(X,Y;D,I)=\operatorname{tr}q=0$ for all full $X,Y$. Expand \emph{all four} passive original inputs, including the unit inputs. Let $r\geq1$ be the least nonzero trace coefficient $H_r$. It exists because the fourth coefficient is the original trace with all passive slots arbitrary. Take a nonzero perturbed-slot component, and delete its unperturbed unit letters. It has $k=r+2\geq3$ arbitrary letters, including $X,Y$, and $j\leq6-k$ actual diagonal letters. Lemma~\ref{np:three-vertex-detector} shows that $H_r$ is nonzero on one of the two row orientations. Both orientations have the smooth-control open of Lemma~\ref{np:bidirectional-control}. On that same irreducible parameter family, intersect the nonzero trace coefficient, the control minor, and the conditions that all output row entries are nonzero. This gives one actual tuple $a$ satisfying every requirement simultaneously. Let $W$ be the sum of the two active-slot tangent spaces and the orbit space. Its image under $dg_a$ is $\mathfrak{sl}_n$. Minimality of $r$ gives every lower ambient trace derivative zero and a nonzero order-$r$ tensor annihilating both original active slots and every orbit tangent. In particular this is a statement on all $6n^2$ source coordinates, including the inputs whose base value was $I$. Apply Lemma~\ref{np:trace-lift}. This exhausts the cases. \end{proof} \subsection{Completion of the nonproper classification} \begin{theorem}\label{np:classification} Let $f$ be a nonproper ordinary multilinear sextic over $K$. For every $n\geq3$, \[ \operatorname{Val}_n(f)= \begin{cases} M_n,&\operatorname{tr}f\not\equiv0,\\ \mathfrak{sl}_n,&\operatorname{tr}f\equiv0. \end{cases} \] \end{theorem} \begin{proof} Choose an inclusion-maximal set $S$ of original slots whose unit deletion leaves a nonzero \emph{free} polynomial $q$. Nonproperness provides a nonempty such set. If all six slots are deleted, the coefficient sum $c$ is nonzero; setting all but one original input to $I$ gives $cX$, so the image is full. Otherwise maximality makes $q$ proper of positive degree at most five. Degree one is impossible, since a nonzero linear polynomial has nonzero unit deletion. Thus $2\leq d\leq5$. The minimum matrix-identity degree separately shows that this free nonzero $q$ is an actual nonidentity of every $M_n$, $n\geq3$. If $q$ is traceful, Lemma~\ref{np:padding}, applied only after the proper sextic theorem, gives $\operatorname{Val}_n(q)=M_n$. Every such value is an actual value of $f$ by the chosen unit substitution. If $f$ is trace zero, then $q$ is trace zero, and the same lemma gives \[ \mathfrak{sl}_n=\operatorname{Val}_n(q) \subseteq\operatorname{Val}_n(f)\subseteq\mathfrak{sl}_n. \] These are inclusions of actual values, not spans. The only remaining case is traceful $f$ with trace-zero $q$, exactly Proposition~\ref{np:unit-trace-lifting}. \end{proof} \subsection{The ordinary sextic theorem and small sizes} \begin{theorem}\label{thm:ordinary-sextic} The degree-six classification in Theorem A holds for every $n\geq3$. \end{theorem} \begin{proof} For proper polynomials this is Theorem~\ref{thm:proper-sextic-classification}. Every nonproper polynomial is covered by Theorem~\ref{np:classification}. Proposition~\ref{trace-test} identifies the trace condition with the cyclic projection and specifies the identity exception in each size. \end{proof} \begin{lemma}\label{lem:product-padding} Let $1\leq d\leq e$ and $f\in P_d$. Set \[ F(x_1,\ldots,x_e) =f(x_1,\ldots,x_{d-1},x_d x_{d+1}\cdots x_e). \] Then $F\in P_e$ and $\Val_n(F)=\Val_n(f)$ for every $n\geq1$. Moreover, $F=0$ as a free polynomial if and only if $f=0$. \end{lemma} \begin{proof} Every value of $F$ is a value of $f$, by multiplying its last $e-d+1$ inputs. Conversely set $x_{d+1},\ldots,x_e$ equal to the identity. This gives every value of $f$ as one value of $F$. Expanding a word of $f$ replaces its unique $x_d$ by the fixed ordered block $x_d\cdots x_e$. Different words give different expanded words, so the coefficient map is injective. This proves the final assertion. \end{proof} \begin{corollary}\label{cor:degrees-four-five} Let $d\in\{4,5\}$ and $f\in P_d$. For every $n\geq3$, \[ \Val_n(f)=\begin{cases} \{0\},&f=0,\\ \mathfrak{sl}_n(K),&f\ne0\text{ and }\pi f=0,\\ M_n(K),&\pi f\ne0. \end{cases} \] \end{corollary} \begin{proof} Apply Lemma~\ref{lem:product-padding} with $e=6$, after Theorem~\ref{thm:ordinary-sextic}. If $f\ne0$, then $f$ is not an identity on $M_n$, because $d<2n$. Equality of actual value sets therefore makes $F$ a nonidentity too; in particular $F\notin Ks_6$ when $n=3$. Tracefulness is preserved by the same equality. Proposition~\ref{trace-test} identifies it with the nonvanishing of the cyclic class in each degree. Apply the ordinary sextic classification to $F$ and transfer its value set back to $f$. \end{proof} \begin{corollary}\label{cor:small-sizes} Let $d\in\{4,5,6\}$ and $f=\sum_{\sigma\in S_d}a_\sigma x_{\sigma(1)}\cdots x_{\sigma(d)}$. For $n=1$ put $c=\sum_{\sigma\in S_d}a_\sigma$, the total word coefficient of $f$. Its image is $\{0\}$ when $c=0$ and is $K$ otherwise. For $n=2$, the image is $\{0\}$ for an identity, $KI_2$ for a nonidentity central polynomial, $\slm_2$ for a noncentral trace-zero polynomial, and $M_2$ for a noncentral traceful polynomial. \end{corollary} \begin{proof} The size-one evaluation is $c x_1\cdots x_d$. The size-two assertions are the published ordinary multilinear image theorem of Kanel-Belov, Malev and Rowen~\cite[Theorem 2]{KMR}, whose field hypotheses are satisfied here. Its alternatives are distinguished by the actual identity, centrality and trace conditions. The free cyclic projection alone cannot replace those conditions on $M_2$: for example, $s_4(x_1,x_2,x_3,x_4)x_5x_6$ is an identity of $M_2$ although its cyclic class is nonzero. In degree four the equivalence $\tr f\equiv0\Longleftrightarrow\pi f=0$ still holds on $M_2$: in Proposition~\ref{prop:trace-all-degrees}, $h_f$ has degree three, below the identity bound four. This does not decide identity or centrality. In degrees five and six even the trace equivalence fails; $x_5s_4$ gives the corresponding degree-five example. \end{proof} \section{Further lifting results}\label{sec:short-tools} In this section, we give several lifting results applicable in arbitrary degree. We use the notation of Section~\ref{sec:preliminaries}. These results are independent of the proof of Theorem~\ref{thm:proper-sextic-classification}. \subsection{Root-space lifting and cyclic compatibility}\label{sec:general-lifting} \begin{proposition} Let $\Phi(D,A,Z)$ be a word polynomial linear in $A$. For diagonal $D,C$, write $\Phi(D,E_{ij},C)=m(d_i,d_j;c_i,c_j)E_{ij}$, $i\ne j$. If $m$ is a nonzero polynomial, then some nonempty open set of regular diagonal pairs makes this map an automorphism of $\mathcal H_n$. \end{proposition} \begin{proof} The product of the finitely many multipliers and the eigenvalue differences is a nonzero polynomial in the diagonal coordinates. Its nonvanishing gives the assertion. This only controls root-space coordinates; diagonal equations must be solved separately. \end{proof} \begin{proposition} \label{prop:cycle-monodromy} Let $r\ge2$, with indices modulo $r$, and let $(Lz)_i=\alpha_i z_i+\beta_i z_{i-1}$, where all $\alpha_i,\beta_i$ are nonzero. Put $h=\prod_i(-\beta_i/\alpha_i)$. If $h\ne1$, then $L$ is invertible. If $h=1$, then $\rank L=r-1$, and $Lz=t$ is solvable exactly when \[ \sum_i\ell_i t_i=0, \qquad \ell_{i+1}=-\frac{\alpha_i}{\beta_{i+1}}\ell_i, \] where $\ell$ is any nonzero solution of this recurrence. In the constant-coefficient case $L=pI+qS$, with $(Sz)_i=z_{i-1}$, its eigenvalues are $p+q\omega$ for $\omega^r=1$. Thus it is invertible exactly when $p^r\ne(-q)^r$, including cases in which one coefficient is zero. \end{proposition} \begin{proof} The homogeneous equations give $z_i=(-\beta_i/\alpha_i)z_{i-1}$. Going once around the cycle gives $z_0=hz_0$. If $h\ne1$, the kernel is zero. If $h=1$, one nonzero initial coordinate determines a nonzero kernel vector, so the kernel is one-dimensional. The equations for the left kernel are $\ell_i\alpha_i+\ell_{i+1}\beta_{i+1}=0$. They likewise have a one-dimensional solution space, and orthogonality to this space is necessary and sufficient for membership in $\im L$. Finally, $S$ has the $r$ distinct $r$th roots of unity as eigenvalues in characteristic zero. Alternatively, the identity \[ (pI+qS)\sum_{k=0}^{r-1}p^{r-1-k}(-q)^k S^k =(p^r-(-q)^r)I \] gives its inverse whenever the scalar is nonzero. \end{proof} \begin{lemma} \label{lem:adjoint-transversality} Let $F:M_n(K)^r\to M_n(K)$ be a polynomial map equivariant for simultaneous conjugation, and put $T=F(a)$. With orthogonal complements taken for the trace pairing, one has \[ [M_n,T]\subseteq\im dF_a,\qquad dF_a\text{ is surjective}\ \Longleftrightarrow\ Z(T)\cap(\im dF_a)^\perp=0. \] If $T$ is regular, meaning its minimal polynomial has degree $n$, then surjectivity is equivalent to the existence of tangent vectors $v_1,\ldots,v_n$ such that \[ \det\bigl(\tr(T^{i-1}dF_a(v_j))\bigr)_{1\le i,j\le n}\ne0. \] No semisimplicity assumption on $T$ is required. Surjectivity implies that the image of $F$ contains a Zariski-open neighborhood of $T$. \end{lemma} \begin{proof} Differentiating conjugation equivariance gives \[ dF_a([X,a_1],\ldots,[X,a_r])=[X,T]. \] Moreover, $\tr(Y[X,T])=\tr((TY-YT)X)$, so nondegeneracy of trace gives $[M_n,T]^\perp=Z(T)$. If $W=\im dF_a$, then $W^\perp\subseteq Z(T)$, and the first criterion follows. For regular $T$, choose a cyclic vector $v$. Every commuting matrix is determined by its value on $v$, which is uniquely $p(T)v$ with $\deg p