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\title[Actual images in the second commutator layer]{Actual images and local fibers\\in the second commutator layer}

\author[S. Pan]{Shengyong Pan}
\address{School of Mathematics and Statistics, Beijing Jiaotong University, Beijing 100044, China}
\email{shypan@bjtu.edu.cn}
\subjclass[2020]{Primary 16R99; Secondary 16S50, 15A30}
\keywords{matrix polynomial images, proper polynomials, local fibers, commutators, L'vov--Kaplansky conjecture}

\date{Working draft, 9 November 2026}
\setcounter{tocdepth}{1}
\hypersetup{pdftitle={Actual images and local fibers in the second commutator layer},
  pdfauthor={Shengyong Pan}}
\begin{document}
\begin{abstract}
Let $K$ be an algebraically closed field of characteristic zero and let
$J$ be the commutator ideal of the free associative algebra. We prove that,
for every proper multilinear polynomial $f\in J^2\setminus J^3$ and every
$n\ge3$, the set of values of $f$ on $M_n(K)$ is $\mathfrak{sl}_n(K)$ if
$\tr f\equiv0$, and $M_n(K)$ otherwise. In the first case the evaluation
map is smooth at an input with nonzero rank-one square-zero value. In the
second case it is flat at such an input, with a local complete-intersection
fiber of the expected dimension. The theorem allows arbitrary proper
multilinear corrections of the same degree in $J^3$. We use restrictions
to two active variables with commuting passive inputs and detect the
first nonzero trace term on the exact fiber of the traceless projection.
\end{abstract}
\maketitle
\tableofcontents

\section{Introduction}\label{paper:introduction}
Throughout, $K$ is algebraically closed of characteristic zero.
We write $M_n=M_n(K)$ and $\slm_n=\mathfrak{sl}_n(K)$. For a polynomial
$f$ in $m$ original variables, let
\[
 \Val_n(f)=\{f(A_1,\ldots,A_m):A_i\in M_n\}
\]
be its set of single values, and write $S_f=\Val_n(f)$. No linear span
or Zariski closure is taken in this notation. An ordinary multilinear polynomial uses each of its
$m$ distinct original variables exactly once in every word. It is
proper if substituting $1$ for any variable gives zero in the free
algebra, equivalently if it is invariant under adding a scalar to any input.
Let $J$ denote the two-sided commutator ideal of the unital free
associative algebra.

The L'vov--Kaplansky conjecture predicts that the actual image of an
ordinary multilinear polynomial on $M_n$ is one of
$\{0\}$, $KI_n$, $\slm_n$, and $M_n$ (see \cite{survey}).
We studied the quartic case by repeated-input restrictions in
\cite{Pan4}. Tang gives a quartic classification in
\cite[Theorem~1.1]{Tang4}. In \cite[Theorem~A]{Pan46}, we classify the
images in degrees four through six, starting with proper sextics.

Here we fix the second commutator layer and allow arbitrary degree.
For $n\ge3$ we determine the images and construct local fibers for
proper multilinear polynomials in $J^2\setminus J^3$.

\Needspace{14\baselineskip}

\begin{theorem}[The proper second layer in every size at least three]
\label{indep2:main}
Let $f$ be an ordinary proper multilinear polynomial in $m$ distinct
original variables, and suppose $f\in J^2\setminus J^3$. For every
fixed $n\ge3$ its actual image is
\[
 \Val_n(f)=
 \begin{cases}
 \slm_n(K),&\tr f\text{ is identically zero on }M_n(K)^m,\\
 M_n(K),&\tr f\text{ is not identically zero on }M_n(K)^m.
 \end{cases}
\]
Put $h=\tr f$ and $g=f-n^{-1}hI_n$. There is an input
$a\in M_n^m$ with $f(a)=N$, $\rank N=1$ and $N^2=0$, such that
\[
 \rank(dg_a)=n^2-1,\qquad
 h|_{g^{-1}(N)}\ne0\ \text{in }\mathcal O_{g^{-1}(N),a}
 \quad\text{if }h\not\equiv0.
\]
At this input the map to $\slm_n$ is smooth with local fiber
dimension $mn^2-n^2+1$ if $h\equiv0$. If $h\not\equiv0$, the map
to $M_n$ is flat with local complete-intersection fiber of
dimension $mn^2-n^2$.

These assertions hold for $f+R$ for every proper ordinary multilinear
$R\in J^3$ in the same variables. The trace and the input $a$ are
taken for the complete polynomial $f+R$.
\end{theorem}

The proof uses complete restrictions to two active variables.
Section~\ref{complete2:section} treats a nonzero restriction with
one common passive matrix. If all such restrictions vanish,
$\operatorname{Id}(UT_3)=J^3$ (see \cite[Theorem~2.1]{UT}) gives a
nonzero pair of symbols with independent diagonal passive matrices.
Section~\ref{indep2:section} treats that pair in an \'etale chart of
the original source. The alternating polynomial $s_4$ belongs to
this second case.

The two local conditions in Theorem~\ref{indep2:main} hold at the
same original input. If the restriction has nonzero trace, its
full differential is surjective at a rank-one square-zero input.
If $h\not\equiv0$ and its restriction is zero, the first nonzero term
in the transverse expansion of $h$ gives a bilinear test on the
two full active matrices. Theorems~\ref{complete2:seed}
and~\ref{indep2:seed} give a nilpotent input at which this test
is nonzero and the traceless differential along the restriction
is surjective. Solving $g=N$ in those tangent directions leaves
the transverse variables unchanged. The selected term therefore
survives on the exact fiber $g^{-1}(N)$ (see
Section~\ref{complete2:original-section} and
Proposition~\ref{indep2:trace-transfer}).
A correction in $J^3$ can change this term even though it vanishes
on the restriction. The input is selected after the correction is
included in $h$. The local-image criterion of
Section~\ref{higher:local} then gives the stated sets of values.

Section~\ref{paper:coordinates} gives coordinates for the proper
second quotient and the vanishing lemma for restrictions with few
active variables. Section~\ref{paper:ordered} constructs the one-order
operators. Section~\ref{m3full:section} gives the size-three
two-symbol construction and its traceful extension to larger sizes.
Sections~\ref{complete2:section} and~\ref{indep2:section} treat the
common-background and independent-diagonal restrictions, respectively.
Section~\ref{paper:further} formulates the remaining local-fiber
questions and relates them to the higher-degree results in
\cite{Pan46,PanHigher,PanMesyanHigher}.

\section{Local fibers and actual images}\label{higher:local}

\subsection{A single local condition for arbitrary degree}
For noncentral-valued $f$ define the target vector space
\[
 V_f=\begin{cases}\slm_n(K),&\tr f\equiv0,\\M_n(K),&\tr f\not\equiv0.\end{cases}
\]
Interior and closure in the next theorem use the Zariski topology on $V_f$.

\begin{theorem}[Local characterization]\label{local}
Let $f$ be noncentral-valued and multilinear. With $N=E_{12}$,
\[
 \boxed{\quad S_f=V_f
 \quad\Longleftrightarrow\quad N\in\Int_{V_f}(S_f)
 \quad\Longleftrightarrow\quad
 N\notin\overline{V_f\setminus S_f}^{\,V_f}.\quad}
\]
Consequently LK, for this field class and all degrees and sizes, is
equivalent to the middle condition for every noncentral-valued $f$.
Moreover, $N\in S_f$ already holds for every such $f$; the issue is coverage
of a neighborhood, not merely the existence of a nilpotent value.
\end{theorem}
\begin{proof}
First note that $S_f$ is invariant under conjugation and multiplication by
nonzero scalars, and contains zero.

If $\tr f\not\equiv0$, then $KI_n\subseteq S_f$. To see this, choose Weyl
matrices $U,V$ with $UV=qVU$, $U^n=V^n=I$, where $q$ is a primitive
$n$th root. The $U^aV^b$ form a basis, and only its degree-zero element has
nonzero trace. Expand a nonzero-trace evaluation multilinearly in that
basis. Some basis tuple has nonzero output trace. All words at this tuple
have the same total $(\mathbb Z/n\mathbb Z)^2$ degree, so their sum is a
multiple of one basis element. It must be a nonzero multiple of $I$.
Scaling one input gives all scalars. If $\tr f\equiv0$, the only scalar in
$V_f$ is zero, already in $S_f$.

Here is the elementary degeneration underlying the result. Every nonscalar
$W$ has a similar representative
$\left(\begin{smallmatrix}u&v\\w&Z\end{smallmatrix}\right)$ with $v\ne0$.
For otherwise every row covector would be a left eigenvector, forcing $W$
to be scalar. The polynomial curve
\[
 R(t)=\begin{pmatrix}tu&v\\t^2w&tZ\end{pmatrix}
 =t\begin{pmatrix}t^{-1}&0\\0&I\end{pmatrix}
 W\begin{pmatrix}t&0\\0&I\end{pmatrix}\quad(t\ne0)
\]
has nonzero rank-one square-zero limit $R(0)$, conjugate to $N$.
If $W\in\slm_n$, the whole curve lies in $\slm_n$.
Suppose an open neighborhood $O$ of $N$ in $V_f$ lies in $S_f$.
After a fixed conjugation the curve passes through $N$. Its inverse image
of $O$ is an open subset of $\mathbb A^1$ containing zero, hence contains
some nonzero $t$. Conjugation and scalar invariance now give $W\in S_f$.
The scalar discussion completes $S_f=V_f$. The reverse implication and
the closure formulation are immediate from the definition of interior.
Central-valued polynomials have image either zero or $KI_n$ by multilinearity,
so the asserted equivalence with LK follows.

Finally, a nonscalar value has, after conjugation, a nonzero off-diagonal
entry. Expand its inputs in matrix units. Some matrix-unit tuple gives an
output with a nonzero off-diagonal entry. Regard these units as directed
edges. Any nonzero word is an Euler trail using the same edges. An open
trail has uniquely determined initial and terminal vertices, so every
nonzero word in this tuple is a multiple of the same off-diagonal unit.
The output is therefore a nonzero multiple of $E_{ij}$, $i\ne j$.
Scaling and conjugation give $N$ itself.
\end{proof}

This argument is independent of degree, properness, and a chosen coefficient
family. We use it as a characterization and make no priority claim for the
local reformulation. The proof also applies to a restricted evaluation
domain whenever its image has the stated conjugation and scalar invariance,
and all the required scalar targets have separately been obtained.

\begin{corollary}[Two sufficient certificates]\label{cert}
Let $d=\dim V_f$ and $q=mn^2$. Either of the following suffices for $S_f=V_f$:
\begin{enumerate}
\item There is $a\in f^{-1}(N)$ with $\rank(df_a:M_n^m\to V_f)=d$.
\item There is $a\in f^{-1}(N)$ with
\[
 \dim\mathcal O_{f^{-1}(N),a}=q-d.
\]
\end{enumerate}
In particular the required fiber dimension is $(m-1)n^2$ in the traceful
case and $(m-1)n^2+1$ in the trace-zero case.
\end{corollary}
\begin{proof}
The first condition makes $f$ smooth at $a$. Smooth morphisms are open
\cite{smooth}; restricting to the smooth locus produces the neighborhood
required in Theorem~\ref{local}.
For the second, the local rings of source and target at these closed points
are regular of dimensions $q,d$. The source is Cohen--Macaulay, so the
displayed dimension equality implies flatness by miracle flatness
\cite{miracle}. The flat locus is open \cite{flatlocus}, and flat morphisms
locally of finite presentation are open \cite{flatopen}. Apply
Theorem~\ref{local} again. Over algebraically closed $K$, a nonempty fiber
of a finite-type scheme over a $K$-point has a $K$-point, so this yields
actual matrix evaluations.
\end{proof}

The first condition implies the second, but neither is claimed necessary
for polynomial surjectivity. In particular, generic full rank only proves
dominance; it does not locate an open image neighborhood at $N$.
The distinction is real even for invariant constructible cones:
$\{0\}\cup\{A:\det A\ne0\}$ is dense, contains every scalar, and omits $N$.
It is an illustration of the logical issue, not a claimed polynomial image.

The rank-one attainment used in Theorem~\ref{local} is already in
Kanel-Belov--Malev--Rowen: see Lemma~5 on p.~6 of
arXiv:1005.0191v3, the author version cited in \cite{BMR2012},
and its explicit restatement for arbitrary $n$ in Lemma~2 on
p.~2 of arXiv:1306.4389v3 \cite{nc:BMR2016}.
These locators refer to the specified preprint versions.

\subsection{Higher-order trace on a smooth traceless fiber}
\begin{proposition}[Exact local trace criterion]\label{higher-trace}
Let $f:M_n^m\to M_n$ be multilinear, let $h=\tr f$ and
$g=f-n^{-1}hI$. Suppose $f(a)=N$ is nonzero rank-one square-zero and
$\rank(dg_a)=n^2-1$. Put $Y=g^{-1}(N)$ and $Q=mn^2$.
The following conditions are equivalent:
\begin{enumerate}
\item $h|_Y\ne0$ in $\mathcal O_{Y,a}$;
\item $\dim\mathcal O_{f^{-1}(N),a}=Q-n^2$;
\item $f$ is flat at $a$.
\end{enumerate}
Each implies $\Val_n(f)=M_n$. In formal local coordinates on $Y$, a
nonzero leading term of any positive weighted degree suffices for (1).
No upper bound on the order of the first nonzero term is asserted.
\end{proposition}
\begin{proof}
The rank assumption makes $Y$ smooth at $a$, with regular local domain
$R=\mathcal O_{Y,a}$ of dimension $Q-n^2+1$. Since $h(a)=0$, its
restriction lies in the maximal ideal. If it is nonzero, it is a
nonzerodivisor and $\dim R/(h)=Q-n^2$. If it is zero, the dimension stays
$Q-n^2+1$. Since $f^{-1}(N)$ is cut out in $Y$ by $h=0$, this proves
(1)$\Leftrightarrow$(2). Miracle flatness proves (2)$\Rightarrow$(3),
and the dimension formula for flat local maps gives the converse.
Corollary~\ref{cert} proves surjectivity. Completion is faithfully flat,
so a nonzero formal leading term detects a nonzero element of $R$.
\end{proof}

This is the local-ring form of the higher-order mechanism. The compatible
algebraic degeneration tools in Glazer--Hendel
\cite[Proposition 2.10 and Corollaries 2.12--2.15]{GH} provide a framework
for proving flatness from weighted initial maps. In the present
one-equation fiber problem, the preceding elementary argument suffices;
no convolution or Waring decomposition is needed.

For clarity, the corrected higher terms can be defined recursively.
Let $K_a=\ker dg_a$ and choose a linear right inverse
$R_a:\slm_n\to T_a(M_n^m)$ of $dg_a$. The formal implicit function
construction gives unique homogeneous maps $v_r:K_a\to\slm_n$ of degree
$r\ge2$ such that
\[
 \phi(u)=a+u+R_a\sum_{r\ge2}v_r(u),\qquad g(\phi(u))=N.
\]
They are obtained by
\[
 v_r(u)=-[\text{degree }r]\left(
 g\left(a+u+R_a\sum_{2\le j<r}v_j(u)\right)-N\right).
\]
The trace germ to test is $H(u)=h(\phi(u))$, not $h(a+u)$.
If $dh_a=0$, its first two possibly nonzero homogeneous terms are
\begin{align*}
 H_2(u)&=\tfrac12d^2h_a(u,u),\\
 H_3(u)&=\tfrac16d^3h_a(u,u,u)
 -\tfrac12d^2h_a\bigl(u,R_a d^2g_a(u,u)\bigr).
\end{align*}
Thus an uncorrected cubic derivative cannot in general certify the trace
on the fiber. For positive weights on $u$, a nonzero initial form of $H$
certifies Proposition~\ref{higher-trace}. If the entire germ is zero, this
particular seed is not flat; this does not rule out another seed or
surjectivity by another construction.

\section{Coordinates in the second layer}\label{paper:coordinates}

\subsection{Coordinates on the proper second layer}
Let $A=K\langle x_1,\ldots,x_m\rangle$ be the unital free associative
algebra, $J$ its two-sided commutator ideal, and $\mathcal P_m$ its
multilinear component. Write $\mathcal P_m^{\rm prop}$ for the proper
subspace, consisting of polynomials killed by every unit deletion.
All quotients in this subsection are free-algebra quotients; no
matrix-identity ideal is imposed. Put
\[
 \mathcal Q_m^{\rm prop}
 =\frac{J^2\cap\mathcal P_m^{\rm prop}}
        {J^3\cap\mathcal P_m^{\rm prop}}.
\]
For a subset $S\subseteq[m]$ of size at least two, set $a=\min S$
and, for $j\in S\setminus\{a\}$, define the left-normed comb
\[
 b_{S,j}=[x_j,x_a,x_{i_1},\ldots,x_{i_{|S|-2}}],
 \qquad i_1<\cdots<i_{|S|-2},\quad
 \{i_1,\ldots,i_{|S|-2}\}=S\setminus\{a,j\}.
\]

\begin{proposition}\label{m3full:second-basis}
For $m\ge4$, the classes
\[
 b_{S,j}b_{S^c,\ell},\qquad
 |S|,|S^c|\ge2,\quad
 j\in S\setminus\{\min S\},\quad
 \ell\in S^c\setminus\{\min S^c\},
\]
form a basis of $\mathcal Q_m^{\rm prop}$. In particular its
ordered-bipartition components are independent, and
\begin{equation}\label{m3full:second-dimension}
 \begin{split}
 D_m:=\dim\mathcal Q_m^{\rm prop}
 &=\sum_{p=2}^{m-2}\binom mp(p-1)(m-p-1)\\
 &=2^{m-2}(m^2-5m+4)+2m-2.
 \end{split}
\end{equation}
\end{proposition}

\begin{proof}
We first compute the dimension without assuming independence of the
proposed products. Use the multigrading by the occurrences of the
original letters, and put
\[
 s=t_1+\cdots+t_m,\qquad C=\prod_{i=1}^m(1-t_i)^{-1},
 \qquad H_A=(1-s)^{-1},\qquad B=1+(s-1)C.
\]
Order the letters. For each word $w$ whose proper prefixes are
nondecreasing but which is not itself nondecreasing, let
$g_w=w-\operatorname{sort}(w)$. Its leading word in degree-lexicographic
order is $w$. These leading words are prefix-free. Reduction at the
first nonordered prefix decreases the order and terminates at the
sorted representative. Therefore the $g_w$ generate $J$ as a right
ideal. Prefix-freeness makes their right multiples have distinct
leading words, proving
\[
 J=\bigoplus_w g_wA,\qquad
 \sum_w\mathbf t^{\deg w}=H_J/H_A=B.
\]
Because $J^{r-1}$ is also a left ideal, the same decomposition gives
$J^r=\bigoplus_w g_wJ^{r-1}$. Consequently
\[
 H_{J^r}=B^rH_A,\qquad H_{J^r/J^{r+1}}=B^rC.
\]
On a nonempty squarefree support of size $p$, the coefficient of $B$
is $p-1$, and its constant coefficient is zero.

The PBW decomposition orders the original generators before a basis
of the derived free Lie algebra. It expresses $A$ as the vector-space
tensor product of sorted commutative monomials and the proper algebra.
This decomposition respects each $J^r$. Indeed, the derivations
$\partial_i(x_j)=\delta_{ij}$ preserve $J$, hence every $J^r$; on a
multilinear component they are unit deletions. Successive coefficient
extraction uses these derivations and the projections
$f\mapsto f-x_i\partial_i f$, as well as multiplication by original
generators, all of which preserve the ideal filtration. Thus removing
the sorted singleton factors divides the squarefree exponential
generating series by $e^x$. Since the squarefree series of $B$ is
\[
 B(x)=\sum_{p\ge2}(p-1)\frac{x^p}{p!}=(x-1)e^x+1,
\]
the series for the proper second layer is $B(x)^2$. Taking its
$x^m/m!$ coefficient gives both expressions in
\eqref{m3full:second-dimension}.

It remains to prove spanning. The proper algebra is generated by
nontrivial free Lie commutators, each a linear combination of combs.
Modulo $J^2$, Jacobi moves the least letter of a comb into its initial
bracket. Later adjoint letters can be reordered: interchanging adjacent
ones has an error of the form $[L,[x,y]]$, a difference of two products
of nontrivial commutators. The resulting one-factor normal forms are
exactly $b_{S,j}$. Their classes are independent modulo $J^2$: in
$UT_2$, set $x_j=E_{12}$ and every other letter of $S$ equal to
$\diag(1,0)$. The selected comb is a nonzero multiple of $E_{12}$,
and every other canonical comb begins with two equal diagonal inputs
and vanishes. Since $J^2$ vanishes on $UT_2$, this isolates every
coefficient.

Now express a proper element of $J^2$ as a sum of products of combs.
After normalizing its one-factor terms, their canonical remainder must
be zero by that independence. Modulo $J^3$, only two-factor terms
remain. Normalize each factor modulo $J^2$; its error multiplied by
the other factor lies in $J^3$. The displayed products therefore
span. Their number is the first expression in
\eqref{m3full:second-dimension}, already proved to be the dimension,
so they are a basis.
\end{proof}

\begin{lemma}[Vanishing on common-background sources]
\label{m3full:passive-vanishing}
If an ordinary multilinear polynomial lies in $J^r$, then its
restriction with fewer than $r$ active original slots and every other
slot equal to one common matrix $D$ is identically zero. The identity
holds for every $D$, not just diagonal matrices, and all derivatives
tangent to that linear source vanish.
\end{lemma}
\begin{proof}
Expand a member of $J^r$ as a sum of products of $r$ commutator-ideal
factors with possible word contexts and take its original
multihomogeneous component. The original variable sets in each term
are disjoint. Fewer than $r$ active variables leave one ideal factor
with only passive inputs. It is evaluated in the commutative algebra
$K[D]$ and is zero. This proves an identity on the entire linear
restriction, and differentiation within that restriction preserves
zero. No assertion is made about transverse derivatives.
\end{proof}


\section{Ordered symbols and smooth nilpotent values}\label{paper:ordered}
Write $\mathcal H_n=\{A\in M_n:A_{ii}=0\text{ for every }i\}$ for the
hollow subspace in the fixed basis. Every traceless matrix is similar
to a hollow matrix, by the prescribed-diagonal theorem of
Fillmore~\cite{Fillmore}.

\subsection{Ordered words and mixed differential depths}

The fixed-depth hypothesis can be removed when the complete
specialization has one order of its two active letters. The following
statement does not require homogeneity.

\begin{theorem}\label{mixed:ordered}
Let $0\ne F(u,v,w)=\sum a_{rst}u^rv^sw^t\in K[u,v,w]$, and put
\[
 H_F(D,A,B)=\sum a_{rst}D^rAD^sBD^t.
\]
For $n\ge3$, let $\mathcal V_F$ be the union of the conjugates of
$H_F(D,A,B)$, where $D$ is regular diagonal and $A,B\in\mathcal H_n$.
Then
\[
 \mathcal V_F=
 \begin{cases}
 M_n(K),&F(u,v,u)\ne0,\\
 \slm_n(K),&F(u,v,u)=0.
 \end{cases}
\]
\end{theorem}
\begin{proof}
Suppose first that $F(u,v,u)\ne0$. Number the indices cyclically and
put $A=\sum_iE_{i,i-1}$ and $D=\diag(d_1,\ldots,d_n)$. Then
\begin{equation}\label{mixed:entry}
 (H_F(D,A,B))_{ij}=F(d_i,d_{i-1},d_j)B_{i-1,j}.
\end{equation}
Each $F(d_i,d_{i-1},d_i)$ is a nonzero polynomial. Each
$F(d_i,d_{i-1},d_{i+1})$ is also nonzero, since $n\ge3$ and its three
indices are distinct. Choose $D$ outside the zero set of the product
of these polynomials and the differences $d_i-d_j$. The source
coordinates $B_{i-1,i}$ and $B_{i-1,i+1}$ are distinct hollow
coordinates. Equation~\eqref{mixed:entry} therefore realizes every
matrix supported on the diagonal and the forward cyclic band, by
one hollow $B$. Set the wrap entry to zero to obtain any upper Jordan
form. This proves $\mathcal V_F=M_n$.

Now suppose that $F(u,v,u)=0$. Write $F=(u-w)G$, where $G\ne0$.
The exact word identity
\[
 H_F(D,A,B)=[D,H_G(D,A,B)]
\]
shows that every value is traceless. For fixed diagonal $D$, this
commutator is hollow. We show that every hollow target occurs.
For a column $j$, the off-diagonal part of $H_G$ is a linear function
of the $n-1$ entries $B_{kj}$, $k\ne j$. Its matrix, with row and
column indices different from $j$, is
\[
 M^{(j)}_{ik}=
 \begin{cases}
 G(d_i,d_k,d_j)A_{ik},&i\ne k,\\
 0,&i=k.
 \end{cases}
\]
For pairwise distinct $i,k,j$, the coefficient $G(d_i,d_k,d_j)$ is
nonzero. A derangement of the $n-1\ge2$ indices contributes a
nonzero monomial to $\det M^{(j)}$. Distinct derangements give
 distinct monomials in the independent entries $A_{ik}$, so they
cannot cancel. Thus every $\det M^{(j)}$ is a nonzero polynomial in
$D$ and hollow $A$. Their product with the differences $d_i-d_j$
is nonzero. Choose one pair $(D,A)$ where this product does not
vanish. All column maps are then invertible simultaneously.
Given $T\in\mathcal H_n$, prescribe the off-diagonal entries of
$H_G$ to be $T_{ij}/(d_i-d_j)$ and solve these column systems for
one hollow $B$. This gives $H_F(D,A,B)=T$. Every traceless matrix
is similar to a hollow matrix, proving the second assertion.
\end{proof}

\begin{corollary}\label{mixed:differential}
Let $0\ne P(s,t)=\sum p_{rs}s^rt^s\in K[s,t]$, and put
\[
 H_P(D,A,B)=\sum p_{rs}(\ad_D^rA)(\ad_D^sB).
\]
For $n\ge3$, the union of the conjugates of its values with $D$
regular diagonal and $A,B\in\mathcal H_n$ is $M_n$ if
$P(s,-s)\ne0$, and is $\slm_n$ if $P(s,-s)=0$.
If $P$ is homogeneous, these conditions are equivalent to
$P(1,-1)\ne0$ and $P(1,-1)=0$, respectively.
\end{corollary}
\begin{proof}
Apply Theorem~\ref{mixed:ordered} to
$F(u,v,w)=P(u-v,v-w)$. This substitution is injective, and
$F(u,v,u)=P(u-v,v-u)$.
\end{proof}

\subsection{An ordered symbol with nonzero trace specialization}

\begin{lemma}[An ordered symbol with nonzero trace specialization]
\label{allocall:ordered-smooth}
Let $h(u,v,w)=\sum_{a,b,d}h_{abd}u^av^bw^d\in K[u,v,w]$ satisfy
$h(u,v,u)\ne0$. For $n\ge3$, the morphism
\[
 H_h:M_n(K)\times\slm_n(K)\times\slm_n(K)\longrightarrow M_n(K),
 \qquad
 H_h(D,A,B)=\sum_{a,b,d}h_{abd}D^aAD^bBD^d,
\]
has a nonzero rank-one square-zero value at a point where its
differential is onto $M_n(K)$.
\end{lemma}

\begin{proof}
For a diagonal $D=\diag(d_1,\ldots,d_n)$, entrywise multiplication gives
\begin{equation}\label{allocall:ordered-entry}
 (H_h(D,A,B))_{i\ell}
 =\sum_{j=1}^n A_{ij}B_{j\ell}h(d_i,d_j,d_\ell).
\end{equation}
Let $\mathcal D\subseteq K^n$ be the open set defined by pairwise
distinct $d_i$ and
\[
 h(d_1,d_j,d_1)\ne0\qquad(2\le j\le n).
\]
It is nonempty: each displayed factor is a nonzero polynomial by
$h(u,v,u)\ne0$, and the product of these factors and the pairwise
differences is nonzero in the polynomial ring $K[d_1,\ldots,d_n]$.
Put $\widehat q=(0,q_2,\ldots,q_n)^{\mathsf t}\ne0$, take $A$ hollow,
and set $B=\widehat q e_1^{\mathsf t}$. Both $A$ and $B$ are traceless,
and \eqref{allocall:ordered-entry} becomes
\begin{equation}\label{allocall:ordered-column}
 H_h(D,A,B)=t e_1^{\mathsf t},\qquad
 t_i=\sum_{j>1}A_{ij}q_jh(d_i,d_j,d_1).
\end{equation}
Consider the parameter space
\[
 \mathcal Z=
 \left\{(d,q,A):
 d\in\mathcal D,\ q\in K^{n-1}\setminus\{0\},\ A_{ii}=0,\
 \sum_{j>1}A_{1j}q_jh(d_1,d_j,d_1)=0\right\}.
\]
The last equation is a nonzero linear functional of the hollow
entries of $A$ at every base point $(d,q)$: some $q_j$ is nonzero,
and all its possible $h(d_1,d_j,d_1)$ factors are nonzero.
Its kernel is therefore a vector bundle of hyperplanes over the
irreducible base $\mathcal D\times(K^{n-1}\setminus\{0\})$.
In particular, $\mathcal Z$ is irreducible. Every point of
$\mathcal Z$ has $t_1=0$, so its output in
\eqref{allocall:ordered-column} is square-zero.

Fix a column $\ell>1$. Varying the $n-1$ hollow entries $B_{j\ell}$,
$j\ne\ell$, gives outputs confined to column $\ell$, with column
vectors equal to columns $j\ne\ell$ of
\[
 (M_\ell)_{ij}=A_{ij}h(d_i,d_j,d_\ell).
\]
These are legitimate variations in the full $B$-space $\slm_n$.
Simultaneous conjugation in all three original inputs is also a
legitimate source direction. For its generator $E_{1\ell}$, the
output derivative is
\begin{equation}\label{allocall:ordered-orbit}
 [E_{1\ell},t e_1^{\mathsf t}]
 =t_\ell E_{11}-t e_\ell^{\mathsf t}.
\end{equation}
Modulo matrices supported in the first column, this supplies $-t$
in column $\ell$. Let $\Delta_\ell$ be the determinant of the matrix
obtained from $M_\ell$ by replacing its $\ell$-th column by $t$.
If $\Delta_\ell\ne0$, these $n$ derivative directions span the
entire $\ell$-th output column modulo the first column.

We show that $\Delta_\ell$ is not identically zero on $\mathcal Z$.
Specialize $\widehat q=e_\ell$. Then the equation defining
$\mathcal Z$ reduces to $A_{1\ell}=0$, and
\[
 t_i=A_{i\ell}h(d_i,d_\ell,d_1).
\]
Choose a fixed-point-free permutation $\sigma$ of $[n]$ with
$\sigma(1)\ne\ell$. Such a permutation exists because $n\ge3$:
choose an $n$-cycle whose successor of $1$ is an index outside
$\{1,\ell\}$. The corresponding determinant term has the monomial
\[
 \prod_{i=1}^n A_{i,\sigma(i)}.
\]
It uses neither a diagonal entry of $A$ nor the entry $A_{1\ell}$.
Its coefficient, up to sign, is a product of nonzero polynomials
in $d$. Indeed, in an ordinary column $j\ne\ell$ the factor is
$h(d_i,d_j,d_\ell)$ with $i\ne j$. If $i=\ell$, this is the
nonzero trace specialization $h(d_\ell,d_j,d_\ell)$. Otherwise
$i,j,\ell$ are distinct, and the factor is a copy of the nonzero
polynomial $h$ in three independent variables. In the replacement
column $\ell$, the row $i$ is neither $\ell$ nor $1$, by the two
conditions on $\sigma$. Its factor $h(d_i,d_\ell,d_1)$ is again a
copy of $h$ in three independent variables. Different permutations
give different monomials in the entries of $A$, so this term cannot
cancel. Restricting a nonzero coefficient polynomial to the nonempty
open set $\mathcal D$ cannot make it identically zero. This proves
the required nonvanishing of $\Delta_\ell$ on $\mathcal Z$.

It remains to control the first output column. At fixed $D,B$, the
hollow $A$-variation $E_{ij}$ with $j>1$ and $i\ne j$ has output
\begin{equation}\label{allocall:ordered-first-column}
 q_jh(d_i,d_j,d_1)E_{i1}.
\end{equation}
For $i=1$ choose $j=2$. For each $i>1$ choose
$j(i)\notin\{1,i\}$, which is possible since $n\ge3$.
Each selected $h$ factor is a nonzero polynomial: for $i=1$ this
is the defining trace specialization, and for $i>1$ its three
coordinates are distinct. Requiring these factors and the finitely
many selected $q_j$ to be nonzero gives a nonempty open subset of
the base $(d,q)$. The hyperplane equation only constrains $A$, so
its inverse image in $\mathcal Z$ is nonempty and open.

This open set and all the nonempty opens $\{\Delta_\ell\ne0\}$,
$2\le\ell\le n$, meet because $\mathcal Z$ is irreducible.
At a point of their intersection, \eqref{allocall:ordered-first-column}
fills the first column, and the $B$-variations together with
\eqref{allocall:ordered-orbit} fill every other column modulo it.
Thus the full differential of $H_h$ is onto $M_n$.
Moreover, $\Delta_\ell\ne0$ implies $t\ne0$, since its replacement
column is $t$. The common value is therefore nonzero of rank one
and square-zero. The variations of $A$ need not preserve the
hyperplane equation defining $\mathcal Z$; they are variations in
the original full source, as required.
\end{proof}

\section{Two-symbol nilpotent seeds}\label{m3full:section}

\subsection{The complete shared-background two-symbol map on \texorpdfstring{$M_3$}{M3}}
\label{m3full:two-symbol}
Let $P,Q\in K[s,t]$ be homogeneous of the same degree $d\ge2$,
each divisible by $st$; either may be zero. Define the actual
three-input polynomial map
\begin{equation}\label{m3full:H}
 H_{P,Q}(D,X,Y)
 =\sum_{r+s=d}P_{rs}(\ad_D^rX)(\ad_D^sY)
  +\sum_{r+s=d}Q_{rs}(\ad_D^rY)(\ad_D^sX).
\end{equation}
Put
\[
 p=P(1,-1),\quad q=Q(1,-1),\quad
 \epsilon=(-1)^d,\quad \tau=p+\epsilon q.
\]
We retain this reverse-symbol convention throughout the proof. The next
statement supplies smooth nilpotent points for every nonzero pair in size three.

Write $E=\slm_3$ if $\tau=0$ and $E=M_3$ if $\tau\ne0$.

\begin{theorem}\label{m3full:complete-two-symbol}
The actual image of \eqref{m3full:H} on $M_3^3$ is
\[
 \begin{cases}
 \{0\},&P=Q=0,\\
 \slm_3,&(P,Q)\ne(0,0),\ \tau=0,\\
 M_3,&\tau\ne0.
 \end{cases}
\]
In either nonzero case, there is an actual original input $(D,X,Y)$
with a nonzero rank-one square-zero value at which the map into its
stated target is smooth. The local fiber is a complete intersection
of dimension $19$ in the trace-zero case and $18$ in the traceful case.
\end{theorem}

\begin{proof}[Proof of Theorem~\ref{m3full:complete-two-symbol}]
For regular diagonal $D=\diag(d_1,d_2,d_3)$, direct multiplication
gives the exact trace identity
\begin{equation}\label{m3full:trace-identity}
 \tr H_{P,Q}(D,X,Y)
 =\tau\sum_{i,j}(d_i-d_j)^d X_{ij}Y_{ji}.
\end{equation}
Equivalently the sum is $\tr((\ad_D^dX)Y)$, so density extends this
identity to all $D$. It is a nonzero polynomial function, as seen at
$D=\diag(1,0,0)$, $X=E_{12}$, $Y=E_{21}$. Moreover a nonzero symbol
is detected on a path through three distinct indices: if $P\ne0$,
use $X=E_{ij}$ and $Y=E_{j\ell}$, so that the reversed product is
zero and the output is
$P(d_i-d_j,d_j-d_\ell)E_{i\ell}$. Its two differences are
independent. The case $Q\ne0$ is obtained by interchanging $X,Y$.
Thus no nonzero pair defines an identity or central-valued map on
$M_3$.

\subsubsection{A column-star chart for the two nonzero symbols}
Assume initially that $P,Q$ are both nonzero. At a regular diagonal
$D$ take
\begin{equation}\label{m3full:star-input}
 X=uE_{23}+vE_{32},\qquad Y=c_2E_{21}+c_3E_{31},
 \qquad uv c_2c_3\ne0.
\end{equation}
Write
\[
 F_{ij\ell}=P(d_i-d_j,d_j-d_\ell),\qquad
 K_{ij\ell}=Q(d_i-d_j,d_j-d_\ell).
\]
The exact value of the complete map is
\begin{equation}\label{m3full:star-value}
 N=n_2E_{21}+n_3E_{31},\qquad
 n_2=uc_3F_{231},\quad n_3=vc_2F_{321}.
\end{equation}
Choose $D$ so that all triple-distinct symbol values needed below
are nonzero. Each is a nonzero polynomial under an invertible linear
change of the two diagonal differences, so these finitely many open
conditions are simultaneously possible. In particular $n_2n_3\ne0$,
and $N$ is nonzero of rank one and square-zero.

Four variations of the original $Y$ slot give pure off-block output
directions:
\[
 \begin{array}{c|cccc}
 \delta Y&E_{12}&E_{13}&E_{21}&E_{31}\\ \hline
 dH(\delta Y)&uK_{123}E_{13}&vK_{132}E_{12}
                 &vF_{321}E_{31}&uF_{231}E_{21}.
 \end{array}
\]
Their multipliers are nonzero. We work modulo these four output
coordinates when adding the remaining directions.

Suppose first that $p^2\ne q^2$. The variations
$\delta Y=E_{23},E_{32}$ give a basis of the two lower diagonal
coordinates: their determinant is
\[
 uv\epsilon(q^2-p^2)(d_2-d_3)^{2d}\ne0.
\]
The variations $\delta X=E_{12},E_{13}$ give respectively
\[
 c_2F_{121}E_{11}+c_2K_{212}E_{22}+c_3K_{312}E_{32},
 \qquad
 c_3F_{131}E_{11}+c_3K_{313}E_{33}+c_2K_{213}E_{23}.
\]
Together with the previous six controls these have rank eight.
With respect to the trace pairing, normalize their one-dimensional
conormal by $C_{11}=1$. Its only other possibly nonzero entries are
\[
 C_{23}=-\frac{c_2F_{121}}{c_3K_{312}},\qquad
 C_{32}=-\frac{c_3F_{131}}{c_2K_{213}}.
\]
Simultaneous conjugation of the full original tuple $(D,X,Y)$ gives
output tangent $[Z,N]$. If every such tangent were annihilated by
$C$, then $[C,N]=0$, which forces
\[
 C_{23}n_3=n_2,\qquad C_{32}n_2=n_3.
\]
These equations cannot hold for every choice of the independent
nonzero parameters $u,v,c_2,c_3$: for fixed diagonal data and $c_2,c_3$,
the ratio $n_2/n_3$ varies with $u/v$. If $p=0$, both displayed
entries of $C$ vanish and the contradiction is immediate. Choose
parameters where one equation fails. An actual orbit tangent then
supplies the ninth direction.

Suppose next that $p^2=q^2$ and $\tau\ne0$. Necessarily
$p\ne0$ and $q=\epsilon p$. Put
\[
 \begin{gathered}
 x=d_2-d_1,\quad y=d_3-d_1,\quad
 f=F_{231},\quad g=F_{321},\\
 h=K_{312}=Q(y,-x),\quad j=K_{213}=Q(x,-y),
 \quad A=(-x)^d,\quad B=(-y)^d.
 \end{gathered}
\]
The lower-diagonal variations now give $E_{22}+E_{33}$. In target
coordinates $(11,22,33,23,32)$, the two $X$ variations, that normalized
lower-diagonal variation, and the simultaneous orbit directions
generated by $E_{12},E_{13}$ have columns
\begin{equation}\label{m3full:resonant-columns}
 \begin{pmatrix}
 pc_2A&pc_3B&0&n_2&n_3\\
 pc_2A&0&1&-n_2&0\\
 0&pc_3B&1&0&-n_3\\
 0&c_2j&0&0&-n_2\\
 c_3h&0&0&-n_3&0
 \end{pmatrix}.
\end{equation}
Its determinant is
\begin{equation}\label{m3full:resonant-determinant}
 2fguv c_2^2c_3^2
       \bigl(p^2x^dy^d-Q(y,-x)Q(x,-y)\bigr).
\end{equation}
If the final polynomial is nonzero, choose the regular diagonal
outside its zero set as well as the preceding symbol zero sets.
This proves rank nine at the same actual nilpotent value. Interchanging
$X,Y$ and $P,Q$ gives the same conclusion when the corresponding
expression for $P$ is nonzero.

If both final polynomials vanish identically, unique factorization in
$K[s,t]$ forces each symbol to be a monomial: a nonmonomial
irreducible factor of either symbol would divide $s^dt^d$.
The endpoint equalities $P(1,-1)=p$ and $q=\epsilon p$ then fix the
coefficients. The pairs not yet covered are exactly
\begin{equation}\label{m3full:residual-pairs}
 P(s,t)=p(-1)^{d-r}s^rt^{d-r},\qquad
 Q(s,t)=p(-1)^\ell s^\ell t^{d-\ell},
 \quad 1\le r,\ell\le d-1,
\end{equation}
where $p\ne0$. Conversely each such pair makes both normal
polynomials zero. We now treat every one of these pairs uniformly.

\subsubsection{All residual monomial pairs}
For \eqref{m3full:residual-pairs}, set
\[
 k=r+\ell-d,\qquad \sigma=(-1)^k,\qquad c_0=p(-1)^{d-r}.
\]
At a fixed regular diagonal $D$, make the invertible linear changes
on the hollow input spaces
$U=\ad_D^rX$ and $V=\ad_D^{d-r}Y$. Then
\begin{equation}\label{m3full:monomial-normalization}
 c_0^{-1}H_{P,Q}=J_k(D,U,V)
      :=UV+\sigma\ad_D^k(V)\ad_D^{-k}(U).
\end{equation}
Any negative exponent in this formula denotes only the inverse of
$\ad_D$ on its hollow root spaces at the chosen regular diagonal.
It is not a negative exponent in the original polynomial, nor a
claim about a rational map at singular $D$.

Take $D=\diag(0,x,y)$ with $xy(y-x)\ne0$, and put
\[
 a=x^k,\qquad b=y^k,\qquad c=(y-x)^k.
\]
These three scalars are nonzero for every integer $k$. Choose
\begin{equation}\label{m3full:monomial-input}
 V=E_{21}+E_{31},\qquad
 U=\begin{pmatrix}0&1&-1\\ S&0&a\\ T&b&0\end{pmatrix}.
\end{equation}
The exact normalized value is independent of $S,T$:
\begin{equation}\label{m3full:monomial-value}
 J_k=N=\begin{pmatrix}0\\a\\b\end{pmatrix}
               \begin{pmatrix}1&1/a&-1/b\end{pmatrix}.
\end{equation}
Its row-column pairing is $1-1=0$, so it is nonzero of rank one and
square-zero. Set $\Delta=Ta-Sb$ and choose $\Delta\ne0$, for
example $S=0,T=1$.

Here are entry formulas giving explicit tangent certificates without
any bound on $k$. For arbitrary hollow $U=(u_{ij})$, $V=(v_{ij})$
at this fixed $D$, the entries of $J_k$ are
\begin{equation}\label{m3full:monomial-entries}
\begin{aligned}
 (J_k)_{11}&=u_{12}v_{21}+u_{13}v_{31}+v_{12}u_{21}+v_{13}u_{31},\\
 (J_k)_{12}&=u_{13}v_{32}+(b/c)v_{13}u_{32},\\
 (J_k)_{13}&=u_{12}v_{23}+\sigma(a/c)v_{12}u_{23},\\
 (J_k)_{21}&=u_{23}v_{31}+(c/b)v_{23}u_{31},\\
 (J_k)_{22}&=u_{21}v_{12}+u_{23}v_{32}+v_{21}u_{12}+v_{23}u_{32},\\
 (J_k)_{23}&=u_{21}v_{13}+(a/b)v_{21}u_{13},\\
 (J_k)_{31}&=u_{32}v_{21}+\sigma(c/a)v_{32}u_{21},\\
 (J_k)_{32}&=u_{31}v_{12}+(b/a)v_{31}u_{12},\\
 (J_k)_{33}&=u_{31}v_{13}+u_{32}v_{23}+v_{31}u_{13}+v_{32}u_{23}.
\end{aligned}
\end{equation}
They follow by multiplying the six root multipliers of $\ad_D^k$,
and also verify \eqref{m3full:monomial-value} directly.

Use row-major target order $(11,12,13,21,22,23,31,32,33)$.
When $\sigma=1$, the nine hollow input columns in the order
\[
 u_{12},u_{13},u_{23},u_{32},v_{12},v_{13},v_{21},v_{23},v_{32}
\]
have determinant $4ab\Delta/c$ at
\eqref{m3full:monomial-input}, by differentiating
\eqref{m3full:monomial-entries}. When $\sigma=-1$, retain the first
eight of those columns and add the output column $[E_{12},N]$.
Their determinant is $2ab\Delta/c$. For an additional direct check
of this last normal direction, a trace-pairing conormal to all hollow
$U,V$ directions is
\[
 C=\begin{pmatrix}
 c\Delta&0&0\\
 -2ac\Delta&-c\Delta-2a^2b^2&2a^3b\\
 2bc\Delta&-2ab^3&-c\Delta+2a^2b^2
 \end{pmatrix}.
\]
The first eight columns have minor $2a^3b^2/c^2$ on target rows
$(11,12,13,21,22,23,31,33)$, and $[C,N]$ has entries
$21=2ac\Delta$, $31=2bc\Delta$, with every other entry zero.
Thus the orbit direction is genuinely outside the hollow tangent
image.

Every hollow $U$ or $V$ direction is realized by an actual $X$ or
$Y$ direction at fixed $D$ through the invertible changes above.
The last column in the odd case is obtained by simultaneously
conjugating the actual original variables $D,X,Y$; equivariance of
\eqref{m3full:H} gives precisely the corresponding commutator output.
No differentiation of inverse powers at a singular diagonal is used.
Restoring $c_0$ rescales all outputs by a nonzero scalar. These
nonzero determinants prove the traceful smooth assertion for all
residual pairs, including negative, zero, and positive $k$.

\subsubsection{The trace-zero branch and the one-order boundary}
Assume $P,Q\ne0$ and $\tau=0$. Use the column-star input
\eqref{m3full:star-input}, with the same nonzero symbol opens and
notation $f,g,h,j,A,B$ as above. The four pure off-block directions
remain available. In the remaining traceless coordinates
$(11,22,23,32)$, with $33$ determined by the trace, the $X$ directions
$E_{12},E_{13}$ and simultaneous orbit directions $E_{12},E_{13}$
give respectively
\[
 \begin{array}{ll}
 R_2=(pc_2A,-pc_2A,0,c_3h),&
 R_3=(pc_3B,0,c_2j,0),\\
 O_2=(n_2,-n_2,0,-n_3),&
 O_3=(n_3,0,-n_2,0).
 \end{array}
\]
Here $q=-\epsilon p$, so the repeated diagonal contribution has
opposite sign from \eqref{m3full:resonant-columns}.
If $p\ne0$, a normalized lower-block $Y$ variation gives
$B_0=(0,1,0,0)$, representing $E_{22}-E_{33}$. Expansion gives
\[
 \det(R_2,R_3,B_0,O_2)
 =jc_2\bigl(Agpv c_2^2+fhu c_3^2\bigr).
\]
It is a nonzero polynomial in the independent nonzero source
parameters. If $p=0$, then $q=0$, and instead
\[
 \det(R_2,R_3,O_2,O_3)=-fghjuv c_2^2c_3^2\ne0.
\]
Choose one common input outside the required zero sets. Together
with the four off-block directions these give rank eight. Since
\eqref{m3full:trace-identity} vanishes identically, the original map
is smooth into $\slm_3$ there.

If exactly one symbol is nonzero and its trace specialization is
nonzero, Lemma~\ref{allocall:ordered-smooth} applies to
$h(u,v,w)=P(u-v,v-w)$, after interchanging $X,Y$ if necessary.
This is an actual smooth point of the same three-input map, since
the hollow inputs used there are allowed here. If its trace
specialization is zero, the proof of Theorem~\ref{mixed:ordered}
fixes one regular diagonal $D$ and one hollow first input for which
the second-input linear map is onto the entire hollow subspace.
Its column derangement determinants are simultaneously nonzero;
this is a fixed-input assertion, not just surjectivity of a union.
Choose the second input giving the connected column-star nilpotent
$E_{21}+E_{31}$. The second-input variations fill all six hollow
coordinates, while simultaneous conjugation gives both traceless
diagonal directions, modulo those hollow coordinates. Hence the
full original differential has rank eight. This proves the smooth
assertion on the one-order trace-zero boundary as well.

\subsubsection{Actual image and the original fibers}
A smooth point gives an open image neighborhood of its value in
the specified affine target. The image of \eqref{m3full:H} is
conjugation invariant and is a cone by scaling $X$. The rank-one
contraction in the proof of Theorem~\ref{local} therefore supplies
every nonscalar target in that target space. If $\tau=0$, this gives
all nonzero traceless matrices; zero is obtained by $X=0$.

For $\tau\ne0$, scalar attainment must also be checked for this
repeated-input map. Take a regular diagonal $D$, index cyclically
modulo three, and put
\[
 X_{i,i-1}=1,\qquad
 Y_{i-1,i}=(d_i-d_{i-1})^{-d},
\]
with all other entries zero. The forward product contributes $p$
to every diagonal entry and the reversed product contributes
$\epsilon q$; all off-diagonal entries vanish. Thus the actual
value is $\tau I_3$. Scaling $X$ supplies every scalar, proving
image $M_3$. This step does not apply ordinary multilinear scalar
extraction to a map with repeated $D$. Finally the Jacobian
criterion gives smooth fibers of dimension $27-\dim E$; in the
regular source local ring their $\dim E$ defining equations form
a regular sequence. This completes the proof of
Theorem~\ref{m3full:complete-two-symbol}.
\end{proof}

\subsection{Traceful pairs in every size}\label{highlayers:section}

Use the map $H_{P,Q}$ in \eqref{m3full:H}, with homogeneous degree
$d\ge2$ symbols divisible by $st$, and put
$p=P(1,-1)$, $q=Q(1,-1)$, $\epsilon=(-1)^d$ and
$\tau=p+\epsilon q$.

\begin{theorem}[Traceful two-symbol pairs]\label{highlayers:all-symbols}
Suppose $\tau\ne0$. For every $n\ge3$, the map
$H_{P,Q}:M_n^3\to M_n$ has actual image $M_n$ and a smooth
original three-input point over a nonzero rank-one square-zero matrix.
Its local fiber there is a complete intersection of dimension $2n^2$.
\end{theorem}

We prove the theorem by producing compatible seeds for an exact-fiber
extension. Surjectivity in a smaller matrix size alone is insufficient.

\begin{lemma}\label{highlayers:singleton}
Suppose $P,Q\ne0$, $\tau\ne0$, and at size $h\ge3$ there is
a regular diagonal $D_0$ and hollow $X_0,Y_0$ with a nonzero
rank-one square-zero value $N_0$. Assume the derivative of
\[
 (D,X,Y,g)\longmapsto gH_{P,Q}(D,X,Y)g^{-1},
 \qquad D\text{ regular diagonal},\quad X,Y\text{ hollow},
 \quad g\in GL_h,
\]
is onto $M_h$ at this input, and
$\det(\ad_{D_0}^jX_0)\ne0$ for $0\le j\le d$.
Then there is a seed with these same properties in size $h+1$.
\end{lemma}
\begin{proof}
We use an exact singleton extension with a rank-one target and allow
one endpoint coefficient to be zero.
Write $N_0=u_0v_0^{\mathsf t}$, where
$v_0^{\mathsf t}u_0=0$. Put
$D=\diag(D_0,t)$, $E=D_0-tI$, and initially
$X=\diag(X_0,0)$. If the upper and lower new $Y$ rectangles
are $v,w$, their output rectangles are
\begin{equation}\label{highlayers:rectangles}
 A_tv,\quad wB_t,\qquad
 A_t=\sum_{i+j=d}P_{ij}(\ad_{D_0}^iX_0)E^j,
 \quad B_t=\sum_{i+j=d}Q_{ij}(-1)^iE^i(\ad_{D_0}^jX_0).
\end{equation}
The highest power of $t$ in each sum has an invertible coefficient:
homogeneity makes its other exponent unique, and the corresponding
symbol coefficient is nonzero. Choose $t$ so that $D$ is regular
and $A_t,B_t$ are invertible.

If $q\ne0$, take the exact target
$N_1=\left(\begin{smallmatrix}N_0&0\\v_0^{\mathsf t}&0\end{smallmatrix}\right)$.
It has rank one and square zero. Solve \eqref{highlayers:rectangles}
with $v=0$ and $w\ne0$. If $q=0$, then $p\ne0$; instead take
$N_1=\left(\begin{smallmatrix}N_0&u_0\\0&0\end{smallmatrix}\right)$,
solving with $v\ne0$ and $w=0$. The old output block is unchanged
and the new bottom-right entry is zero, so these are exact values
at single actual inputs.

For the new upper and lower $X$ components $u,z$, the bottom-right
derivative is
\begin{equation}\label{highlayers:bottom-normal}
 L(u,z)=(-1)^d\bigl(pzE^dv+qwE^du\bigr).
\end{equation}
It is nonzero by the chosen rectangle. Old input and old similarity
variations have zero bottom-right derivative. Their old outputs
are onto $M_h$, and their rectangular contributions can be removed
using \eqref{highlayers:rectangles}. Hence, together with $L$, the
derivative is onto $M_{h+1}$. Every $(u,z)\in\ker L$ extends to
a tangent to the exact target fiber, without changing these raw
cross components of $X$.

The provisional $X$ has a zero last row and column. Along such a
kernel tangent, the first possible determinant coefficient is
\begin{equation}\label{highlayers:Schur}
 (-1)^{j+1}\det(\ad_{D_0}^jX_0)\cdot
 zE^j(\ad_{D_0}^jX_0)^{-1}E^ju
 \quad(0\le j\le d),
\end{equation}
in order two. The last diagonal entry remains identically zero
on the hollow source, so no second-order diagonal correction enters
this coefficient. If $q\ne0$, choose a nonzero $u$ in the
hyperplane $wE^du=0$, then choose $z$ outside the finitely many
annihilating hyperplanes in \eqref{highlayers:Schur}. If $q=0$,
make the dual choice, first fixing $z\ne0$ in $zE^dv=0$.
The hyperplanes have a nonzero vector since $h\ge3$, and all
matrices in these Schur forms are invertible.

The auxiliary source is smooth and its exact target fiber is smooth
at the provisional point. The selected tangent therefore lifts to
a formal arc in that fiber. All the determinant coefficients
\eqref{highlayers:Schur} are nonzero on it; the surjectivity minor
and regular-diagonal conditions remain units. Thus the finite-type
exact fiber localized at these finitely many functions has a
$K((t'))$-point and is nonempty. The Nullstellensatz supplies an
actual $K$-point. This uses nonemptiness of a localized algebraic
fiber, not algebraization of an arbitrary formal arc. Removing the
similarity parameter only conjugates the rank-one target and preserves
the raw determinant conditions. Surjectivity of the auxiliary map
implies surjectivity for the actual three-input map, since all its
directions are induced by actual $D,X,Y$ variations.
\end{proof}

\subsubsection{Nonresonant pairs}
Assume $P,Q\ne0$ and $p^2\ne q^2$. Use the smooth column-star
point \eqref{m3full:star-input} from the proof of
Theorem~\ref{m3full:complete-two-symbol}. Its value is
$N=n_2E_{21}+n_3E_{31}$ with $n_2n_3\ne0$.
The six hollow $Y$ columns span the coordinates
$12,13,21,31,22,33$. In the remaining quotient coordinates
$(11,23,32)$, the two raw upper-row $X$ columns are $R_2,R_3$,
and the actual orbit columns generated by $E_{12},E_{13}$ are
\[
 O_2=(n_2,0,-n_3),\qquad O_3=(n_3,-n_2,0).
\]
The matrix $[R_2\ R_3\ O_2\ O_3]$ has rank three by that
proof. The orbit columns are independent, so a nonzero vector in
its kernel has a nonzero raw upper-row $X$ component. Cancel its
remaining output using the six $Y$ columns. Raw variations in
$X_{21},X_{31}$ have zero output, since they and $Y$ have only a
lower-left column, so the raw lower column can be chosen freely.

At this point every $\ad_D^jX$ has an invertible lower
two-by-two block $A_j$ and zero first row and column. A kernel
tangent with nonzero upper row $u$ and lower column $z$ gives
Schur determinant coefficient
$-\det(A_j)u_jA_j^{-1}z_j$, where $u_j,z_j$ are obtained
from $u,z$ by invertible root multipliers. For fixed $u\ne0$
this is a nonzero linear functional of $z$. Choose $z$ outside
the finitely many kernels for $0\le j\le d$. As in the proof
of Lemma~\ref{highlayers:singleton}, lift in this same smooth
exact target fiber and localize. We obtain a genuine size-three
seed with all determinant conditions, while its rank-one
nilpotent value and surjectivity are retained. The lemma now gives
such a point in every size. This argument establishes determinant
nonvanishing on the actual nilpotent fiber, rather than intersecting
unrelated open sets in the unrestricted source.

\subsubsection{The entire traceful resonance}
Suppose $p^2=q^2$ and $\tau\ne0$. Then $q=\epsilon p\ne0$.
If both symbols are monomials, use
\eqref{m3full:monomial-input} with $S=0,T=1$.
Its smoothness was proved in Section~\ref{m3full:section}.
The first original input $X$ has exactly one nonzero determinant
cycle, $X_{12}X_{23}X_{31}$. Every power $\ad_D^j$ multiplies
this cycle by nonzero root differences, so all determinant conditions
of Lemma~\ref{highlayers:singleton} hold at this same point.
This extends the residual monomial cases without importing any
unproved all-size smoothness implication.

Otherwise interchange the actual $X,Y$ roles, and hence $P,Q$,
if necessary, so that $Q$ is not a monomial. The resonance remains
$q=\epsilon p\ne0$. Define
\begin{align*}
 \Phi_1(x,y,z)&=qx^dQ(y,z)-Q(y,-x)Q(x,z),\\
 \Phi_2(x,y,z)&=qy^dQ(x,z)-Q(x,-y)Q(y,z).
\end{align*}
These are both nonzero. Indeed, if $\Phi_1=0$, setting $x=1$
gives $qQ(y,z)=Q(y,-1)Q(1,z)$. The two factors are nonzero,
and their Cartesian-product monomial support has constant total
degree only when each consists of one monomial. Thus $Q$ itself
is a monomial. Conversely a monomial satisfies that identity.
Finally $\Phi_2(x,y,z)=\Phi_1(y,x,z)$.

Choose $D=\diag(0,x,y,-z)$ regular and set
\[
 a=Q(x,z),\quad b=Q(y,z),\quad
 f=P(x-y,y),\quad g=P(y-x,x),\quad u=a/f,\quad v=b/g.
\]
Require these four symbol values, $\Phi_1,\Phi_2$, and every
$P,Q$ value on a distinct-three-vertex path to be nonzero. For any
such path, its two differences are independent linear forms in
$x,y,z$. Each required factor is therefore a nonzero polynomial.
Their finite product with the regularity factors is nonzero, so
one common choice exists. With a further parameter $\rho\ne0$, put
\begin{equation}\label{highlayers:four-seed}
 X=E_{14}+\rho E_{41}+uE_{23}+vE_{32},\qquad
 Y=E_{21}+E_{31}.
\end{equation}
Direct multiplication of the complete map gives
\[
 H_{P,Q}(D,X,Y)=N=(0,a,b,0)^{\mathsf t}(1,0,0,1).
\]
The forward paths $2\to3\to1$, $3\to2\to1$ and the reverse
paths $2\to1\to4$, $3\to1\to4$ are the only surviving paths.
Thus the value is independent of $\rho$, nonzero of rank one,
and square-zero. The two two-cycles supporting $X$ show that
\[
 \det(\ad_D^jX)=\rho uv
       \bigl((d_1-d_4)(d_2-d_3)\bigr)^{2j}\ne0
       \quad(j\ge0).
\]

It remains to prove smoothness at a single such input. Put
$A=\{1,4\}$, $B=\{2,3\}$, and let $\sigma$ exchange
$1,4$ and $2,3$. Write $h_i=X_{i,\sigma(i)}$.
The eight cross-block hollow $Y$ directions split into four pairs.
For $a'\in A,b'\in B$, the columns
$Y_{a'b'},Y_{\sigma(a')\sigma(b')}$ and the rows
$H_{\sigma(a')b'},H_{a'\sigma(b')}$ give the matrix
\[
 \begin{pmatrix}
 h_{\sigma(a')}P_{\sigma(a'),a',b'}&
 h_{\sigma(b')}Q_{\sigma(a'),\sigma(b'),b'}\\
 h_{b'}Q_{a',b',\sigma(b')}&h_{a'}P_{a',\sigma(a'),\sigma(b')}
 \end{pmatrix},
\]
where $P_{ijk}=P(d_i-d_j,d_j-d_k)$ and similarly for $Q$;
the analogous pairs with $A,B$ exchanged supply the other block.
Each determinant is the difference of a nonzero multiple of
$\rho$ and a nonzero multiple of $uv$. A single $\rho\ne0$
avoids their finitely many roots. The eight cross-block outputs
are then independent. The within-pair $Y$ directions supply
$E_{11}+E_{44}$ and $E_{22}+E_{33}$, by $q=\epsilon p\ne0$.

Let $C$ annihilate these directions and all remaining hollow $X$
and actual orbit directions under the trace pairing. The preceding
ten outputs force
\[
 C_A=\begin{pmatrix}c&e\\h&-c\end{pmatrix},\qquad
 C_B=\begin{pmatrix}k&\ell\\m&-k\end{pmatrix},
\]
with no cross blocks. The raw $X_{42}$ direction gives
$P_{421}E_{41}$, so $e=0$. Actual simultaneous conjugation
of $D,X,Y$ gives $[Z,N]$ and hence $[C,N]=0$. Comparing
the two rank-one factors of $N$ gives
\[
 h=-2c,\qquad \ell=-(c+k)a/b,\qquad m=-(c-k)b/a.
\]
The $X_{12}$ direction has diagonal part
$qx^d(E_{11}+E_{22})$ and remaining entry $Q(y,-x)E_{32}$.
Pairing with $C$ yields $(c+k)\Phi_1/b=0$.
Likewise $X_{13}$ yields $(c-k)\Phi_2/a=0$.
Thus $c=k=0$, and then $C=0$. These actual hollow and orbit
directions span all $M_4$, including its trace direction.
Lemma~\ref{highlayers:singleton} now extends this guarded point
to every $n\ge4$; size three is separately covered by
Theorem~\ref{m3full:complete-two-symbol}.

\subsubsection{Actual images and local fibers}
If one complete symbol is zero, the traceful assertion follows
from Lemma~\ref{allocall:ordered-smooth}. Otherwise the
nonresonant and resonant cases exhaust $\tau\ne0$. At each
smooth point, openness and the cone/conjugation argument in
Theorem~\ref{local} give all nonscalar values. For scalars in
this repeated-input map, choose regular diagonal $D$, use cyclic
indices modulo $n$, and set
\[
 X_{i,i-1}=1,\qquad Y_{i-1,i}=(d_i-d_{i-1})^{-d}.
\]
All other entries are zero. Direct multiplication gives
$H_{P,Q}=\tau I_n$; scaling $X$ gives every scalar.
This is not ordinary-multilinear scalar extraction with the
repeated $D$ treated as independent. The Jacobian criterion gives
the asserted three-input fiber dimension and regular sequence.
This proves Theorem~\ref{highlayers:all-symbols}.

\section{Complete two-active coefficient fibers in every size at least three}
\label{complete2:section}

We retain the standing algebraically closed, characteristic-zero field
$K$. The results concern the unchanged original ordinary multilinear
polynomial. We first treat a nonzero complete two-active restriction
with one common passive matrix. Section~\ref{indep2:section} then treats
the genuine second-layer portion of the common kernel of all these tests.

\begin{theorem}[The complete original coefficient fiber]
\label{complete2:main}
Let $f$ be ordinary multilinear in $m=d+2$ original variables, $d\ge2$.
Suppose that for some two original positions its \emph{complete}
restriction, with $X,Y$ in those positions and $D$ in all others, is
$H_{P,Q}$ of \eqref{m3full:H}, where $P,Q$ are homogeneous of degree $d$,
$st\mid P,Q$, and $(P,Q)\ne(0,0)$. For each $n\ge3$,
\[
 \Val_n(f)=
 \begin{cases}
  \slm_n,&\tr f\text{ is an identity on }M_n,\\
  M_n,&\tr f\text{ is not an identity on }M_n.
 \end{cases}
\]
In the first case $f:M_n^m\to\slm_n$ has a smooth original input over
a nonzero rank-one square-zero matrix; its local fiber dimension is
$mn^2-n^2+1$. In the second case $f:M_n^m\to M_n$ has a flat original
input over such a matrix, and its local fiber is a complete intersection
of dimension $mn^2-n^2$.

The restriction identity may be an actual identity on the fixed $M_n$;
a free identity suffices to apply the statement in every size.
Arbitrary same-degree ordinary corrections with zero complete selected
restriction are permitted, including the entire proper $J^3$ and all
higher powers, with every crossing partition and order. The target is
always chosen from the \emph{actual size-$n$ trace of the corrected
polynomial}, and the nilpotent input may change with that correction.
\end{theorem}

Theorem~\ref{highlayers:all-symbols} already proves the smooth
full-matrix conclusion when the trace of the restriction is nonzero.
The remaining problem is not merely to find a trace-zero restriction
value. One must select a nilpotent input of that \emph{same restriction}
where the first surviving trace term of the complete original map does
not disappear on its exact traceless fiber. The following theorem
provides the needed compatibility.

Put $V_n=\{A\in M_n:A_{ii}=0\}$, the hollow subspace, and keep
$p=P(1,-1)$, $q=Q(1,-1)$, $\epsilon=(-1)^d$ and
$\tau=p+\epsilon q$. At a fixed regular diagonal $D$, write
\[
 H_D(X,Y)=H_{P,Q}(D,X,Y),\qquad
 \Gamma_D(X,Y,g)=gH_D(X,Y)g^{-1},
 \quad X,Y\in V_n,\quad g\in GL_n.
\]
For $\tau=0$, let $\mathcal G_D$ consist of hollow pairs whose raw value
is nonzero rank-one square-zero and for which $d\Gamma_D$ is onto
$\slm_n$ at $(X,Y,I)$. Let $\mathcal G_D^{\mathrm{guard}}$ be the subset
satisfying
\begin{equation}\label{complete2:guards}
 \det(\ad_D^jX)\ne0\qquad(0\le j\le d).
\end{equation}
The simultaneous conjugation in $\Gamma_D$ is realized in actual
inputs by
\[ (D,X,Y)\longmapsto(gDg^{-1},gXg^{-1},gYg^{-1}). \]
In the original polynomial it conjugates \emph{every} passive $D$
slot as well. Raw tests and determinant conditions below refer to
$X,Y$ before this auxiliary conjugation.

\begin{theorem}[Compatible nilpotent bilinear detection]
\label{complete2:seed}
Assume $(P,Q)\ne(0,0)$ and $\tau=0$. For every $n\ge3$ there is a
nonempty Zariski-open set $\Omega_n$ of regular diagonal matrices such
that, for each fixed $D\in\Omega_n$, the tensors
\[
 \{X\otimes Y:(X,Y)\in\mathcal G_D\}
 \quad\text{span }V_n\otimes V_n.
\]
For $n\ge4$ one may require \eqref{complete2:guards}; if $p\ne0$,
these guards can also be required in size three. After allowing
arbitrary diagonal parts in the original $X,Y$, every nonzero bilinear
form on $M_n\times M_n$ is nonzero at one actual smooth nilpotent seed
of this same $H_{P,Q}$ and this same $D$.
\end{theorem}

The span in this statement is a linear-functional detector, not an
assertion that a polynomial image can be added. Given a nonzero
bilinear form, it supplies \emph{one} compatible original input.
The fixed open $\Omega_n$ is independent of that form. In particular,
a form whose coefficients depend on $D$ is first specialized at a
suitable $D\in\Omega_n$ and then detected at that same $D$.

The proof splits exhaustively into $p\ne0$ and $p=q=0$.
In the latter case the cubic cycle norm separates invertible hollow
operators from a two-dimensional cycle kernel. The former case needs
one four-vertex Laurent calculation. Once a guarded spanning set is
available in size four, a tensor uses at most four vertices, and the
remaining size induction is uniform. We first record those common
mechanisms, so that they are not repeated in each coefficient family.

\subsection{Diagonal enrichment and exact guarded extension}

\begin{lemma}\label{complete2:diagonal-enrichment}
For fixed regular diagonal $D$, hollow tensor detection by
$\mathcal G_D$ implies full $M_n\times M_n$ bilinear detection by
actual smooth nilpotent seeds of $H_{P,Q}$.
\end{lemma}
\begin{proof}
All active adjoint exponents are positive. Adding arbitrary diagonal
matrices $D_X,D_Y$ to the original $X,Y$ therefore leaves the value
unchanged. The hollow controls and simultaneous conjugation, including
these added parts, retain the same smooth certificate.
Suppose a bilinear form vanishes on every enriched pair. Its coefficient
bilinear in $D_X,D_Y$ is zero. Its coefficients linear in either added
part vanish on the projections of the hollow seed set. Each projection
spans $V_n$: otherwise a nonzero functional vanishing on that projection,
tensored with any nonzero functional on the other factor, would
contradict the hollow tensor span. Thus both mixed parts vanish.
The remaining hollow-by-hollow part vanishes by hypothesis.
\end{proof}

\begin{lemma}[Trace-zero guarded singleton extension]
\label{complete2:extension}
Suppose $P,Q\ne0$, $\tau=0$, and $(X_0,Y_0)\in
\mathcal G_{D_0}^{\mathrm{guard}}$ in size $h\ge3$. For a generic new
diagonal entry $z$ the same symbol pair has a guarded smooth
rank-one square-zero seed in size $h+1$ at $D=\diag(D_0,z)$.
Moreover, any raw function nonzero at the provisional input below can
be retained while restoring all guards on its exact saturated fiber.
\end{lemma}
\begin{proof}
Write the raw value as $N_0=u_0v_0^{\mathsf t}$, with
$v_0^{\mathsf t}u_0=0$, and set $E=D_0-zI$ and
$X=\diag(X_0,0)$. The full new upper and lower $Y$ rectangle maps are
\begin{equation}\label{complete2:rectangles}
 F(z)y,\quad wG(z),\qquad
 \begin{aligned}
 F(z)&=\sum_{r+s=d}P_{rs}(\ad_{D_0}^{r}X_0)E^s,\\
 G(z)&=\sum_{r+s=d}Q_{rs}(-1)^rE^r(\ad_{D_0}^{s}X_0).
 \end{aligned}
\end{equation}
The largest $s$ in the first sum and largest $r$ in the second have
invertible leading coefficients by \eqref{complete2:guards}.
Homogeneity makes the other exponent unique, so there is no leading
cancellation. Choose one $z$ making $D$ regular and both matrices
invertible. For $\lambda\ne0$ take $y=0$ and
$w=\lambda v_0^{\mathsf t}G(z)^{-1}$. The complete raw value is
\begin{equation}\label{complete2:extended-value}
 N_1=\begin{pmatrix}N_0&0\\\lambda v_0^{\mathsf t}&0\end{pmatrix}
     =\binom{u_0}{\lambda}(v_0^{\mathsf t},0),
 \qquad N_1^2=0,\quad\rank N_1=1.
\end{equation}

Old controls give the old $\slm_h$ block after correcting any new
rectangles; \eqref{complete2:rectangles} fills both rectangles.
A genuine new orbit direction
$Z=\left(\begin{smallmatrix}0&e\\0&0\end{smallmatrix}\right)$ gives
\[
 [Z,N_1]=
 \begin{pmatrix}
  \lambda e v_0^{\mathsf t}&-N_0e\\
  0&-\lambda v_0^{\mathsf t}e
 \end{pmatrix}.
\]
Choose $v_0^{\mathsf t}e\ne0$. Its bottom-right entry supplies the
missing scalar direction; its old-block trace is the opposite scalar,
and the remaining old traceless part and rectangles can be corrected.
Thus $d\Gamma_D$ is onto $\slm_{h+1}$. This is an actual orbit
normal and does not require either endpoint to be nonzero.

Prescribe any raw new $X$ upper column $a$ and lower row $b$.
Their complete derivative is traceless and can be canceled by the
old, $Y$-rectangle and orbit controls just described. These corrections
do not change the specified raw $X$ cross components. Hence $(a,b)$
extends to a kernel tangent of the \emph{exact} $\Gamma_D=N_1$ fiber.
For a formal arc parameter $t$, the first determinant coefficient is
\begin{equation}\label{complete2:Schur}
 \det(\ad_D^jX(t))=
 (-1)^{j+1}t^2\det(\ad_{D_0}^jX_0)
 bE^j(\ad_{D_0}^jX_0)^{-1}E^ja+O(t^3).
\end{equation}
Choose $a\ne0$ and then $b$ outside the finitely many annihilating
hyperplanes. All coefficients in \eqref{complete2:Schur} are nonzero.
The exact fiber is smooth at the provisional point, so the kernel
tangent lifts to a formal arc in it. A surjectivity minor, and any raw
function nonzero at the provisional point, are units on this arc.
The finite-type localization inverting these functions and all guards
is nonempty, and has an actual point over the algebraically closed
base field. This uses nonemptiness of a localization, not an assertion
that an arbitrary formal arc algebraizes. Finally discard the auxiliary
$g$; the raw value is conjugate to $N_1$, and raw guards and tests have
been retained. No global irreducibility of the exact fiber is needed.
\end{proof}

The rectangle and Schur mechanisms also appear in
Lemma~\ref{highlayers:singleton}. Here the trace-zero orbit normal
permits arbitrary raw cross tangents and preserves raw tests in the
same exact nilpotent fiber, including when $p=q=0$.

\begin{lemma}[Four vertices suffice for the size induction]
\label{complete2:induction}
Suppose the same nonzero trace-zero pair has guarded hollow tensor
spanning in size four at generic diagonal parameters. Then it has
that spanning property in every size $n\ge4$.
\end{lemma}
\begin{proof}
Work over the algebraic closure of $K(d_1,\ldots,d_n)$. On an old
$h=n-1$ vertex set choose a finite guarded spanning collection defined
over the algebraic closure of its old diagonal field. The new diagonal
entry is transcendental over these data, so
Lemma~\ref{complete2:extension} applies. Spanning survives scalar
extension to the full diagonal field.

If a bilinear $T$ vanishes at all guarded good size-$n$ pairs, its value
at each provisional extension must be zero: a nonzero value would be
retained by the exact-fiber guard restoration. That value is
\[
 T_h(X_0,Y_0)+\lambda T\!\left(\diag(X_0,0),
       \begin{pmatrix}0&0\\v_0^{\mathsf t}G(z)^{-1}&0\end{pmatrix}\right).
\]
It vanishes for every $\lambda\ne0$, hence its constant term vanishes.
The old spanning collection forces $T_h=0$. This works after removing
any vertex. For $n\ge5$, a pair of hollow matrix units uses at most
four vertices and lies in some $(n-1)$-vertex subset. All coefficients
of $T$ are therefore zero, proving the induction.
\end{proof}

\begin{lemma}[One diagonal open independent of the test]
\label{complete2:descent}
A hollow tensor-spanning assertion proved at the generic diagonal
point by actual smooth nilpotent seeds descends to one nonempty open
$\Omega_n$ of regular diagonal tuples, with the same fixed-$D$
assertion at every $D\in\Omega_n$. The guards may be included.
\end{lemma}
\begin{proof}
Select finitely many generic seed tensors forming a basis. Place $D$,
their raw inputs, auxiliary invertible matrices, raw rank-one
square-zero equations, a nonzero entry in each value, chosen
surjectivity minors, the tensor-basis determinant, and the required
nonzero guard determinants in one finite-type incidence scheme.
Its projection to diagonal $D$ is dominant. By constructibility its
image contains a nonempty open. Every nonempty finite-type fiber over
a $K$-point has an actual $K$-point, since $K$ is algebraically closed.
This gives the same diagonal open for all tests simultaneously.

When a construction uses Laurent series, a nonzero raw test yields
an extension-field point of this incidence, hence its nonemptiness.
One does not descend a fixed fiber whose target itself has Laurent
coordinates. In applications below the Laurent parameter changes raw
$X,Y,g$ while $D$, and any $D$-dependent coefficients of the test,
are fixed.
\end{proof}


\subsection{The nonzero-endpoint trace-zero stratum}
\label{complete2:nonzero-endpoint-section}

Throughout this subsection, the complete symbols satisfy
\begin{equation}\label{complete2:endpoint-hypothesis}
 p=P(1,-1)\ne0,\qquad q=Q(1,-1)=-\epsilon p,
 \qquad \epsilon=(-1)^d.
\end{equation}
In particular, both symbols are nonzero. There is no restriction on
their other coefficients. We write
$\mathcal G_D^{\mathrm{guard}}$ for the hollow pairs $(X,Y)$ such that
$H_D(X,Y)$ is nonzero of rank one and square zero, the differential of
$\Gamma_D$ is onto $\slm_n$ at $(X,Y,I)$, and
\[
 \det(\ad_D^jX)\ne0\qquad(0\le j\le d).
\]
The guards and all tensor evaluations in this subsection concern the
\emph{raw} matrices $X,Y$, before the auxiliary conjugation in
$\Gamma_D$.

\begin{theorem}\label{complete2:nonzero-endpoint}
Assume \eqref{complete2:endpoint-hypothesis}. For every $n\ge3$ there
is a nonempty open set $\Omega_n$ of regular diagonal matrices such
that, at each fixed $D\in\Omega_n$,
\[
 \Span\{X\otimes Y:(X,Y)\in\mathcal G_D^{\mathrm{guard}}\}
       =V_n\otimes V_n.
\]
Consequently, after arbitrary diagonal parts are allowed in the two
original active inputs, the actual smooth nonzero rank-one
square-zero seeds detect every nonzero bilinear form on
$M_n\times M_n$.
\end{theorem}

We prove the tensor assertion first over the algebraic closure of the
field of generic diagonal parameters. This keeps $D$ fixed throughout
each fiber argument. In particular, the coefficients of a bilinear
form under examination are allowed to belong to this field, but they
are held fixed during every source deformation. The final passage to
one common open set of diagonal parameters is
Lemma~\ref{complete2:descent}.

\subsubsection{Guarded detection in size three}

\begin{lemma}\label{complete2:guarded-m3}
Under \eqref{complete2:endpoint-hypothesis}, at a generic regular
diagonal $D$ of size three the tensors of
$\mathcal G_D^{\mathrm{guard}}$ span $V_3\otimes V_3$.
Moreover, the exact local fiber at a generic column-star seed has a
nonempty open subset on which all the raw adjoint-power guards hold.
One may impose simultaneously any prescribed finite collection of
additional guards of the form $\det(W\circ X)\ne0$, where
$W\circ X=(w_{ij}X_{ij})$ and all six off-diagonal $w_{ij}$ are
nonzero constants in the fixed coefficient field.
\end{lemma}

\begin{proof}
Choose $D=\diag(d_1,d_2,d_3)$ with all distinct-three-vertex path
values of $P$ and $Q$ nonzero. These conditions hold at the generic
diagonal: on any such path the two differences are independent
linear forms, and neither symbol is zero. Use the column-star chart
\eqref{m3full:star-input}--\eqref{m3full:star-value}, namely
\begin{equation}\label{complete2:column-star}
 X=uE_{23}+vE_{32},\qquad
 Y=c_2E_{21}+c_3E_{31},\qquad uv c_2c_3\ne0.
\end{equation}
For clarity, put
\begin{gather*}
 f=P(d_2-d_3,d_3-d_1),\qquad
 g=P(d_3-d_2,d_2-d_1),\\
 h=Q(d_3-d_1,d_1-d_2),\qquad
 j=Q(d_2-d_1,d_1-d_3),\\
 A=(d_1-d_2)^d,\qquad B=(d_1-d_3)^d,\qquad
 C=(d_2-d_3)^d,\\
 n_2=uc_3f,\qquad n_3=vc_2g,\qquad \eta=n_2/n_3.
\end{gather*}
The exact value is $N=n_2E_{21}+n_3E_{31}$. The trace-zero
column calculation in the proof of
Theorem~\ref{m3full:complete-two-symbol}, following
\eqref{m3full:resonant-columns}, gives the remaining four-column
minor
\begin{equation}\label{complete2:star-minor}
 jc_2\bigl(Agpv c_2^2+fhu c_3^2\bigr).
\end{equation}
Together with the four pure off-block $Y$ columns already displayed
there, it proves that $\Gamma_D$ is smooth onto $\slm_3$ at a
nonempty open set of these star parameters. All the columns use
hollow input variations or true simultaneous-conjugation directions;
no variation of the diagonal parameters is needed.

We first identify a complete kernel tangent, since both the guard
restoration and the tensor calculation depend on its full output.
The hollow $Y$ tangent image is spanned by the four off-block
coordinates $12,13,21,31$ and the line $E_{22}-E_{33}$.
Modulo this image, use coordinates $(11,23,32)$. The raw $X$
directions $E_{12},E_{13}$ and the orbit directions generated by
$E_{12},E_{13}$ have columns
\[
 R_2=(pAc_2,0,hc_3),\quad R_3=(pBc_3,jc_2,0),\quad
 O_2=(n_2,0,-n_3),\quad O_3=(n_3,-n_2,0).
\]
Set
\begin{equation}\label{complete2:star-L}
 L_2=pAc_2+\eta hc_3,\qquad
 L_3=pBc_3+\eta^{-1}jc_2.
\end{equation}
The following are variations of the actual auxiliary source at
$(X,Y,I)$:
\begin{equation}\label{complete2:complete-kernel}
 \begin{split}
 \delta X&=L_3E_{12}-L_2E_{13},\\
 \delta g&=\frac{hc_3L_3}{n_3}E_{12}
              -\frac{jc_2L_2}{n_2}E_{13},\\
 \delta Y&=\frac{L_2L_3}{pCu}E_{32}.
 \end{split}
\end{equation}
The first two lines have zero output in the displayed quotient.
Their remaining output is
$-L_2L_3(E_{22}-E_{33})$. The last line cancels it, because
$dH_D(E_{32}\text{ in }Y)=pCu(E_{22}-E_{33})$.
Thus \emph{the complete differential} of $\Gamma_D$ on
\eqref{complete2:complete-kernel} is zero. In particular, the
$Y_{32}$ correction cannot be omitted.

The two scalars $L_2,L_3$ are nonzero for generic star parameters.
Independently add $z_2E_{21}+z_3E_{31}$ to $\delta X$; it has
zero output, since every product of this lower column with the
current $Y$ is zero. For a fixed entry-weight matrix $W$ as in the
lemma, the determinant along any arc with this tangent has
coefficient
\begin{equation}\label{complete2:star-guard-coefficient}
 [t^2]\det(W\circ X(t))
   =w_{12}w_{23}w_{31}L_3uz_3
       -w_{13}w_{32}w_{21}L_2vz_2.
\end{equation}
Indeed, a hollow $3\times3$ determinant is the sum of its two
oriented three-cycle products. Each such product here contains two
entries vanishing at $t=0$, so second-order corrections to the arc
cannot affect this coefficient. The right side is a nonzero linear
form in $(z_2,z_3)$. This applies to each
$w_{ab}=(d_a-d_b)^j$, $0\le j\le d$, and to every additional
$W$ under consideration. Choose $(z_2,z_3)$ outside the finitely
many kernels.

Let $\mathcal F$ be the exact fiber
$\Gamma_D^{-1}(N)$ and let $x_*$ be the chosen star point with
auxiliary parameter $I$. The source and the map are smooth at
$x_*$, so $\mathcal F$ is smooth there, its local ring is a regular
local domain, and the selected kernel tangent lifts to a formal arc
in this same fiber. Formula
\eqref{complete2:star-guard-coefficient} makes every required guard
a nonzero element of that local domain. A chosen surjectivity minor
is a unit. Their product therefore defines a nonempty open subset
of the local fiber. More concretely, its formal point proves that
the corresponding finite-type localization is nonempty, and over
an algebraically closed coefficient field it has an actual point.
Discarding the auxiliary $g$ only conjugates the value; it does not
change the raw guards or the saturated differential rank.

This argument also gives the precise continuation principle needed
below. If a raw bilinear form $T$ vanishes on all guarded good pairs,
then its germ on $\mathcal F$ at $x_*$ is zero. Otherwise its
product with the nonzero guard product would be nonzero in the
local domain, and localizing also at $T$ would give a guarded good
pair with $T\ne0$. Consequently both the value of $T$ and every
kernel derivative of $T$ vanish at each generic column star. We do
not assert that the unguarded star itself satisfies the determinant
guards.

The same continuation principle holds at row stars. Write
$R^{\mathrm{op}}(s,t)=R(t,s)$. Direct transposition gives
\begin{equation}\label{complete2:transpose}
 H_{P,Q}(D,X^{\mathsf t},Y^{\mathsf t})
   =\epsilon\,
      H_{Q^{\mathrm{op}},P^{\mathrm{op}}}(D,X,Y)^{\mathsf t}.
\end{equation}
The transformed endpoints are $-p$ and $\epsilon p$, so the
transformed pair again satisfies
\eqref{complete2:endpoint-hypothesis}. Moreover
\[
 \ad_D^j(X^{\mathsf t})=(-1)^j(\ad_D^jX)^{\mathsf t}.
\]
Thus transposition preserves the guards on the \emph{first} raw
slot as well as the rank-one, square-zero and saturated smoothness
conditions. Applying the column argument to the transformed symbols
proves the row assertion for the original symbols.

We now compute the annihilator. Vanishing at the column stars and
row stars, with each isolated vertex in turn, kills every coefficient
$T(E_{ab},E_{cd})$ for which $\{a,b\}\ne\{c,d\}$.
To see the coverage explicitly, two different edges in a
three-vertex set share exactly one endpoint. A column star covers
either orientation of the complementary $X$ edge with a $Y$ edge
ending at the isolated vertex, and a row star covers the orientations
with that $Y$ edge starting there. Independence of $u,v,c_2,c_3$
then kills the individual coefficients.

At \eqref{complete2:column-star}, consider the raw variation
\begin{equation}\label{complete2:parity-shift}
 \delta Y=uE_{23}+\epsilon vE_{32}.
\end{equation}
It preserves the exact value: the new lower diagonal contributions
are multiples of $\epsilon p+q=0$. Its tangent therefore belongs
to the exact local fiber above. Since the germ of $T$ on that fiber
is zero, its derivative on this tangent is zero. Expanding in $u,v$
and repeating with each isolated vertex gives
\[
 T(E_{ab},E_{ab})=0,\qquad
 T(E_{ba},E_{ab})=-\epsilon T(E_{ab},E_{ba})
       \quad(a<b).
\]
Only this derivative of the germ is used; no guard condition on the
points of the shifted-star family is being presumed. The remaining
form has the shape
\begin{equation}\label{complete2:edge-annihilator}
 T(X,Y)=\sum_{a<b}c_{ab}
                  (X_{ab}Y_{ba}-\epsilon X_{ba}Y_{ab}).
\end{equation}

Apply its derivative to the \emph{complete} kernel tangent
\eqref{complete2:complete-kernel}. The result is
\begin{equation}\label{complete2:edge-derivative}
 0=c_{12}c_2L_3-c_{13}c_3L_2
                  +\frac{c_{23}}{pC}L_2L_3.
\end{equation}
For fixed nonzero $c_2,c_3$, varying $u/v$ makes $\eta$ vary
through a nonempty open subset of the multiplicative line; the
smoothness restriction \eqref{complete2:star-minor} removes only a
proper closed subset. Hence \eqref{complete2:edge-derivative} is a
Laurent-polynomial identity in $\eta$. With
$\zeta=c_{23}/(pC)$, its coefficients of $\eta^{-1}$ and $\eta$
respectively give
\begin{equation}\label{complete2:edge-norm}
 c_{12}=-\zeta pA,\qquad c_{13}=\zeta pB,
 \qquad \zeta c_2c_3(hj-p^2AB)=0.
\end{equation}
The final equation is the constant coefficient after the first two
are substituted.

Suppose first that $Q$ is not a monomial. Put
$x=d_2-d_1$, $y=d_3-d_1$. If the norm difference in
\eqref{complete2:edge-norm} vanished identically, it would give
\[
 Q(y,-x)Q(x,-y)=p^2x^dy^d=q^2x^dy^d.
\]
Unique factorization forces both factors on the left to be monomials,
and the invertible substitutions force $Q$ itself to be a monomial,
a contradiction. Thus $hj-p^2AB\ne0$ at the generic diagonal,
so $\zeta=0$ and $T=0$. If $Q$ is a monomial but $P$ is not,
apply the same argument to the transformed pair in
\eqref{complete2:transpose} and the form
$\widetilde T(X,Y)=T(X^{\mathsf t},Y^{\mathsf t})$.
Its reverse symbol $P^{\mathrm{op}}$ is not a monomial. This
transposition, rather than an exchange of the two active inputs,
retains the guard on the required raw $X$ slot.

It remains to treat two monomial symbols. Write
\[
 P=\alpha_0s^rt^{d-r},\qquad
 Q=\beta_0s^\ell t^{d-\ell},\qquad
 k=r+\ell-d,\qquad \sigma=(-1)^k.
\]
The endpoint condition says $\beta_0/\alpha_0=-\sigma$.
As in \eqref{m3full:monomial-normalization}, make the invertible
hollow root-space changes
$U=\ad_D^rX$, $V=\ad_D^{d-r}Y$. The present trace-zero
normalization is
\begin{equation}\label{complete2:minus-monomial}
 \alpha_0^{-1}H_D(X,Y)
   =J^-(U,V):=UV-\sigma\ad_D^k(V)\ad_D^{-k}(U).
\end{equation}
Negative powers here mean only inverse root multipliers at the fixed
regular diagonal. They do not change the original polynomial.
Put $a=x^k$, $b=y^k$, $c=(y-x)^k$, all nonzero, and take the
same support as \eqref{m3full:monomial-input}:
\begin{equation}\label{complete2:minus-monomial-seed}
 U=\begin{pmatrix}0&1&-1\\0&0&a\\\theta&b&0\end{pmatrix},
 \qquad V=E_{21}+E_{31},\qquad \theta\ne0.
\end{equation}
Its exact normalized value is
\[
 J^-(U,V)=
       \begin{pmatrix}0\\a\\b\end{pmatrix}
       \begin{pmatrix}1&-1/a&1/b\end{pmatrix},
\]
which is nonzero of rank one and square zero.

For completeness, the trace-zero tangent minor is recorded explicitly.
Use the eight columns
\[
 u_{12},u_{13},u_{23},u_{32},v_{12},v_{13},v_{21},v_{23}
\]
and output rows $(11,12,13,21,22,23,31,32)$, with the last diagonal
entry determined by trace zero. The entry formulas
\eqref{m3full:monomial-entries} apply with the sign of every
reverse-order summand changed. Their indicated derivative matrix is
\[
 \begin{pmatrix}
 1&1&0&0&0&-\theta&1&0\\
 0&0&0&0&0&-b^2/c&0&0\\
 0&0&0&0&-\sigma a^2/c&0&0&1\\
 0&0&1&0&0&0&0&-c\theta/b\\
 -1&0&0&0&0&0&-1&-b\\
 0&-a/b&0&0&0&0&a/b&0\\
 0&0&0&1&0&0&b&0\\
 -b/a&0&0&0&\theta&0&0&0
 \end{pmatrix}.
\]
Its determinant is
\begin{equation}\label{complete2:minus-monomial-minor}
 \begin{cases}
 -ab(\theta c+2ab^2)/c^2,&\sigma=1,\\
 ab(-\theta c+2ab^2)/c^2,&\sigma=-1.
 \end{cases}
\end{equation}
Choose $\theta\ne0$ away from its one root. These hollow directions
already give the required rank eight. Furthermore, the raw $X$ has
exactly one nonzero determinant cycle, $X_{12}X_{23}X_{31}$.
Every $\ad_D^j$ multiplies its factors by nonzero root differences,
so all the guards hold at this very seed.

Equations \eqref{complete2:edge-norm} leave at most the form
\[
 T=\zeta p\bigl[-A(X_{12}Y_{21}-\epsilon X_{21}Y_{12})
                +B(X_{13}Y_{31}-\epsilon X_{31}Y_{13})
                +C(X_{23}Y_{32}-\epsilon X_{32}Y_{23})\bigr].
\]
At \eqref{complete2:minus-monomial-seed}, undoing the root-space
changes gives
\[
 T(X,Y)=-2\zeta\alpha_0.
\]
Indeed, only $Y_{21},Y_{31}$ are nonzero, and the two contributions
inside the brackets both equal $-(-1)^{d-r}$, while
$p=\alpha_0(-1)^{d-r}$. Thus $\zeta=0$ also in the last
case. This proves the lemma.
\end{proof}

\subsubsection{The four-vertex tensor coefficients}

\begin{proof}[Proof of Theorem~\ref{complete2:nonzero-endpoint}]
Size three is Lemma~\ref{complete2:guarded-m3}. By the guarded
extension and support argument of Lemmas~\ref{complete2:extension}
and \ref{complete2:induction}, size four is the only further tensor
calculation needed. We give that calculation at one fixed generic
$D=\diag(d_1,d_2,d_3,d_4)$.

Let $T$ be a bilinear form on $V_4\times V_4$ vanishing on all
guarded good pairs. Restriction to each three-vertex set is zero.
Here the use of the extension lemma is essential: if
$H_{D_0}(X_0,Y_0)=u_0v_0^{\mathsf t}$ is an old guarded seed,
its provisional extension has new lower $Y$ row
$\lambda v_0^{\mathsf t}G^{-1}$ and $X=\diag(X_0,0)$.
A nonzero value of $T$ there would survive restoration of the guards
in its exact fiber. Thus the provisional value is zero for every
$\lambda\ne0$. Its constant term in $\lambda$ is the value of
the restriction of $T$ at the old seed, so the size-three tensor
span makes that restriction zero. Consequently we may write
\begin{equation}\label{complete2:disjoint-form}
 T(X,Y)=\sum_{\substack{a,b,c,e\text{ pairwise distinct}}}
                 C_{ab,ce}X_{ab}Y_{ce}.
\end{equation}

Designate vertex $4$ as new, and isolate vertex $1$ in the old
three-vertex block. Write $z=d_4$, $D_0=\diag(d_1,d_2,d_3)$.
Start with the generic column star \eqref{complete2:column-star},
whose value is $N_0=ne_1^{\mathsf t}$, where
$n=(0,n_2,n_3)^{\mathsf t}$. The two rectangle matrices of
Lemma~\ref{complete2:extension}, in the convention of
\eqref{highlayers:rectangles}, are
\begin{equation}\label{complete2:weighted-rectangles}
 \begin{split}
 F(z)_{ab}&=P(d_a-d_b,d_b-z)X_{ab},\\
 G(z)_{ab}&=Q(z-d_a,d_a-d_b)X_{ab}
                    \qquad(a\ne b),
 \end{split}
 \qquad F(z)_{aa}=G(z)_{aa}=0.
\end{equation}
At the generic four-tuple each of these six weights is nonzero:
the two displayed arguments are independent diagonal differences.
They are fixed constants during the following deformation.

Use the exact-fiber tangent and formal arc from the proof of
Lemma~\ref{complete2:guarded-m3}, imposing also
$\det F(z)\ne0$ and $\det G(z)\ne0$ as its additional
entry-weight guards. Formula
\eqref{complete2:star-guard-coefficient} permits all these choices
simultaneously. Thus there is an arc
\[
 (X(t),Y(t),g(t))\in\Gamma_{D_0}^{-1}(N_0),
 \qquad g(t)=I+O(t),
\]
on which every required determinant has nonzero quadratic leading
term. Only the raw $X,Y$ and the auxiliary $g$ vary with $t$.
The diagonal entries $d_1,d_2,d_3,z$, and the coefficients of
$T$, do not vary. The surjectivity minor at the initial star remains
a unit. The raw old value factors as
\begin{equation}\label{complete2:moving-factor}
 \begin{split}
 H_{D_0}(X(t),Y(t))&=g(t)^{-1}N_0g(t)=u(t)v(t)^{\mathsf t},\\
 u(t)&=g(t)^{-1}n,\\
 v(t)^{\mathsf t}&=e_1^{\mathsf t}g(t)=e_1^{\mathsf t}+O(t).
 \end{split}
\end{equation}
The change of the right factor will be included in the inverse
calculation; it is not replaced by the constant row $e_1^{\mathsf t}$.

Split $G(z)$ into vertex $1$ and the block $B=\{2,3\}$:
\[
 G(z,t)=\begin{pmatrix}0&\rho(t)\\\kappa(t)&B(t)\end{pmatrix}.
\]
At $t=0$, $B(0)$ is the invertible hollow two-by-two block whose
entries are the nonzero weights in
\eqref{complete2:weighted-rectangles} times $u,v$.
Both $\rho(t)$ and $\kappa(t)$ are $O(t)$. Its scalar Schur
complement satisfies
\begin{equation}\label{complete2:Schur-pole}
 s(t)=-\rho(t)B(t)^{-1}\kappa(t)
            =\gamma t^2+O(t^3),\qquad \gamma\ne0,
\end{equation}
because $\det G(z,t)=\det B(t)\,s(t)$ and its quadratic
coefficient was required to be nonzero. Write
$v(t)^{\mathsf t}=(v_1(t),v_B(t))$.
The block inverse gives exactly
\[
 \begin{split}
 (v(t)^{\mathsf t}G(z,t)^{-1})_1
    &=(v_1-v_BB^{-1}\kappa)s^{-1},\\
 (v(t)^{\mathsf t}G(z,t)^{-1})_B
    &=-(v_1-v_BB^{-1}\kappa)s^{-1}\rho B^{-1}
                         +v_BB^{-1}.
 \end{split}
\]
Here $v_1=1+O(t)$ and $v_B=O(t)$, so the Schur correction
$v_BB^{-1}\kappa$ is $O(t^2)$. In particular,
\begin{equation}\label{complete2:Laurent-row}
 (v(t)^{\mathsf t}G(z,t)^{-1})_1
       =\gamma^{-1}t^{-2}+O(t^{-1}),\qquad
 (v(t)^{\mathsf t}G(z,t)^{-1})_2,
 (v(t)^{\mathsf t}G(z,t)^{-1})_3=O(t^{-1}).
\end{equation}
Thus the moving auxiliary conjugation does not cancel the leading
pole.

Over the Laurent extension field form the provisional size-four
input
\begin{equation}\label{complete2:Laurent-extension}
 \widehat X=\begin{pmatrix}X(t)&0\\0&0\end{pmatrix},\qquad
 \widehat Y=\begin{pmatrix}
       Y(t)&0\\v(t)^{\mathsf t}G(z,t)^{-1}&0
     \end{pmatrix}.
\end{equation}
The complete output, not merely its leading term, is
\[
 H_D(\widehat X,\widehat Y)
 =\begin{pmatrix}u(t)v(t)^{\mathsf t}&0\\v(t)^{\mathsf t}&0\end{pmatrix}
 =\begin{pmatrix}u(t)\\1\end{pmatrix}
       \begin{pmatrix}v(t)^{\mathsf t}&0\end{pmatrix}.
\]
It is nonzero of rank one and square zero. The rectangle matrices
are invertible, so Lemma~\ref{complete2:extension} gives saturated
smoothness at this provisional point as well as restoration of all
size-four raw $X$ guards while retaining any raw function nonzero
there.

By \eqref{complete2:disjoint-form}, terms supported on three old
vertices are absent. The raw $X$ rectangles involving the new vertex
are zero. Equations \eqref{complete2:column-star} and
\eqref{complete2:Laurent-row} therefore give
\begin{equation}\label{complete2:Laurent-isolation}
 [t^{-2}]\,T(\widehat X,\widehat Y)
   =\gamma^{-1}\bigl(C_{23,41}u+C_{32,41}v\bigr).
\end{equation}
Indeed, terms with $Y_{42}$ or $Y_{43}$ must use an old $X$
edge incident to vertex $1$. Such $X$ entries are $O(t)$, whereas
those two $Y$ entries are only $O(t^{-1})$, so they cannot contribute
to the displayed coefficient. The only possible order $t^{-2}$
terms use $Y_{41}$ and $X_{23}=u+O(t)$ or $X_{32}=v+O(t)$.

If either of the two coefficients on the right of
\eqref{complete2:Laurent-isolation} were nonzero, choose the
independent generic star parameters $u,v$ so that their displayed
linear combination is nonzero, retaining the finitely many smoothness
and guard conditions. Then $T$ is nonzero at the provisional
Laurent input. The guard-restoration part of
Lemma~\ref{complete2:extension} supplies, over an algebraically
closed extension field, a guarded good raw pair with $T\ne0$.
To turn this into the required contradiction at the fixed generic
$D$, use the finite-type incidence defined by the conditions
\[
 \begin{gathered}
 \rank H_D(X,Y)=1,\quad H_D(X,Y)^2=0,\quad H_D(X,Y)\ne0,\\
 d\Gamma_D\text{ onto }\slm_4,\quad
 \prod_{j=0}^d\det(\ad_D^jX)\ne0,\quad T(X,Y)\ne0.
 \end{gathered}
\]
Rank one and the differential condition are imposed on the appropriate
open minor charts. The construction gives a point of this incidence
over an extension of the fixed diagonal coefficient field, so it is
nonempty. The algebraically closed coefficient field therefore
supplies an actual point, contradicting the hypothesis on $T$.
It is the incidence of \emph{raw} rank-one square-zero values,
smoothness, guards and $T\ne0$ that is descended here. We do not
attempt to descend an exact target fiber whose target coordinates
have themselves become Laurent series.

It follows that $C_{23,41}=C_{32,41}=0$. The same argument with
each isolated old vertex kills all disjoint-edge coefficients whose
$Y$ edge points from the designated new vertex. Transposition
\eqref{complete2:transpose}, applied to the transformed symbol pair
and transformed bilinear form, gives the identical lower-row argument
for coefficients whose $Y$ edge points to that vertex. It preserves
all the raw $X$ guards and does not invoke the size-four conclusion
being proved. Finally the designation of the new vertex is free:
for any tested $Y$ edge one may designate either of its endpoints
as the new vertex and use the other three vertices as the old block.
This relabeling includes terms whose $X$ edge contains the original
label $4$; permuting only the old three labels would not suffice.
Every coefficient in \eqref{complete2:disjoint-form} is zero.

This proves guarded size-four spanning. Lemma~\ref{complete2:induction}
now proves it in every size $n\ge5$: a pair of hollow matrix units
uses at most four vertices and is contained in an $(n-1)$-vertex
set. The induction is carried out over the generic diagonal field.
For each old vertex subset, its finite spanning collection may first
be chosen over the algebraic closure of its own diagonal field;
the added eigenvalue is transcendental over those old seed data.
The rectangle inverses in Lemma~\ref{complete2:extension} therefore
exist by their invertible leading coefficients. Scalar extension
preserves the old span, even when the coefficients of $T$ depend
on all diagonal entries. Thus no incompatible numerical diagonal
choices are made at different stages. Lemma~\ref{complete2:descent}
then supplies one nonempty open $\Omega_n$ and actual spanning
collections at every $D\in\Omega_n$.

The full-matrix detection statement now follows from
Lemma~\ref{complete2:diagonal-enrichment}. Its hypothesis holds
because $st$ divides both complete symbols, so adding arbitrary
diagonal parts to the two original active inputs does not change
the value or the available hollow and simultaneous-conjugation
smoothness controls.
\end{proof}


\subsection{The complete zero-endpoint seed detector}
\label{complete2:zero-endpoint-section}

We retain the notation $H_D$, $V_n$, and $\Gamma_D$ and the adjoint-power
guards introduced above. Throughout this subsection
\[
 P(1,-1)=Q(1,-1)=0,\qquad (P,Q)\ne(0,0).
\]
Since $st$ divides both symbols, every nonzero symbol is divisible by
$st(s+t)$, and in particular $d\ge3$. For path indices put
\[
 P_{ikj}=P(d_i-d_k,d_k-d_j),\qquad
 Q_{ikj}=Q(d_i-d_k,d_k-d_j).
\]
At diagonal $D$ the complete map, including both orders, satisfies
\begin{equation}\label{complete2:zero-entry}
 (H_D(X,Y))_{ij}
 =\sum_k\bigl(P_{ikj}X_{ik}Y_{kj}+Q_{ikj}Y_{ik}X_{kj}\bigr),
 \qquad (H_D(X,Y))_{ii}=0.
\end{equation}
The terms with $k=i$ or $k=j$ vanish. Thus $H_D$ maps
$V_n\times V_n$ into $V_n$.

\begin{theorem}[Zero-endpoint detection in every size]
\label{complete2:zero-endpoint}
For every $n\ge3$ there is a nonempty Zariski-open set $\Omega_n$ of
regular diagonal matrices such that, at each fixed $D\in\Omega_n$,
the tensors $X\otimes Y$, $(X,Y)\in\mathcal G_D$, span
$V_n\otimes V_n$. Thus these are actual nonzero rank-one square-zero
values, and $d\Gamma_D$ is onto $\slm_n$ at $(X,Y,I)$.
For $n\ge4$ the same spanning assertion holds for
$\mathcal G_D^{\mathrm{guard}}$, with all adjoint-power guards on
the raw first input $X$.
Allowing arbitrary diagonal additions to the original $X$ and $Y$
gives tensors spanning $M_n\otimes M_n$.
\end{theorem}

The one-order fixed-input operators are given in
Theorem~\ref{mixed:ordered} and Corollary~\ref{mixed:differential}.
The size-three star differential is also contained in the proof of
Theorem~\ref{m3full:complete-two-symbol}. We write the cycle operators
in the convention \eqref{m3full:H} and prove the tensor-detection and
exact-fiber assertions below.

\subsubsection{The common-zero-row detector in size three}
\label{complete2:zero-three-section}

\begin{lemma}\label{complete2:zero-three}
The unguarded hollow assertion of Theorem~\ref{complete2:zero-endpoint}
holds in size three, including either one-order boundary.
\end{lemma}
\begin{proof}
Choose one regular diagonal $D$ for which every triple-distinct
evaluation of every nonzero symbol is nonzero. These are finitely
many nonzero polynomial conditions: for a fixed triple the two
successive differences are independent linear coordinates. Their
simultaneous nonvanishing therefore defines a nonempty open set.

Fix an index $i$, and let $R_i\subset V_3$ be the linear space of
hollow matrices with row $i$ zero. For $X,Y\in R_i$ their complementary
two-by-two hollow blocks have products supported on the diagonal.
Those diagonal contributions vanish in \eqref{complete2:zero-entry}.
Consequently $H_D(X,Y)$ is supported on column $i$, off its diagonal.
Every nonzero value on $R_i\times R_i$ is therefore rank one and
square-zero. We show that a nonempty open subset of this entire
irreducible pair space consists of smooth seeds.

Relabel $i=1$. If both symbols are nonzero, take
\[
 X=uE_{23}+vE_{32},\qquad
 Y=c_2E_{21}+c_3E_{31},\qquad uv c_2c_3\ne0.
\]
The value is
\[
 N=n_2E_{21}+n_3E_{31},\qquad
 n_2=P_{231}uc_3,\quad n_3=P_{321}vc_2,
\]
with $n_2n_3\ne0$. Four actual $Y$ variations and two actual $X$
variations give the following pure hollow directions:
\[
\begin{array}{c|cccccc}
 \text{variation}&\delta Y_{31}&\delta Y_{21}&\delta Y_{13}
 &\delta Y_{12}&\delta X_{13}&\delta X_{12}\\ \hline
 \text{output}&P_{231}uE_{21}&P_{321}vE_{31}
 &Q_{132}vE_{12}&Q_{123}uE_{13}
 &Q_{213}c_2E_{23}&Q_{312}c_3E_{32}.
\end{array}
\]
Here each variation denotes a unit change of the indicated raw entry.
All multipliers are nonzero; any potential diagonal contribution is
zero by the endpoint hypothesis. The fixed-$D$ derivative therefore
fills $V_3$. The orbit directions generated by $E_{12},E_{13}$ have
diagonal projections $n_2(E_{11}-E_{22})$ and
$n_3(E_{11}-E_{33})$. Thus $d\Gamma_D$ has rank eight.

If $Q=0$ and $P\ne0$, instead take both $X$ and $Y$ in $R_1$ with all
four entries $21,31,23,32$ nonzero. The output again has both lower
column entries nonzero. The $Y$ variations $E_{21},E_{31}$ give the
two pure lower-column directions, with multipliers
$P_{321}X_{32}$ and $P_{231}X_{23}$. The $X$ variations $E_{12},E_{13}$
give the two pure upper-row directions, with multipliers
$P_{123}Y_{23}$ and $P_{132}Y_{32}$. Finally the $Y$ variations
$E_{12},E_{13}$ give the two lower off-diagonal directions, with
multipliers $P_{312}X_{31}$ and $P_{213}X_{21}$. Thus the fixed-$D$
hollow derivative is again onto, and the same two orbit directions
complete it to $\slm_3$. If $P=0$, exchange the actual input slots.
This proves the required nonempty openness for every $R_i\times R_i$.

Let a bilinear form $T$ vanish on all smooth nilpotent seeds. Its
restriction to $R_i\times R_i$ vanishes on a nonempty open subset,
hence vanishes identically. Given any two hollow units $E_{ab}$ and
$E_{cd}$, there is an index $i$ different from both row indices $a,c$.
Both units belong to $R_i$, so $T(E_{ab},E_{cd})=0$. All hollow tensor
coefficients vanish. This is a size-three argument: the complementary
block has size two, and no common-zero-row assertion in larger size
is being used.
\end{proof}

\subsubsection{A nonzero cubic cycle norm}
\label{complete2:nonzero-cycle-section}

Set $u=-s-t$ and define the homogeneous polynomial
\begin{equation}\label{complete2:cycle-norm}
 \Delta_{P,Q}(s,t)
 =P(s,t)P(t,u)P(u,s)+Q(s,t)Q(t,u)Q(u,s).
\end{equation}
Its coefficients are cubic polynomials in the symbol coefficients.
A nonzero one-order symbol has nonzero cycle norm, since all three
linear substitutions in its product are invertible.

\begin{lemma}\label{complete2:invertible-hollow-operator}
If $\Delta_{P,Q}\ne0$, then for every $n\ge3$ the determinant of
\[
 L_{D,X}:V_n\longrightarrow V_n,\qquad Y\longmapsto H_D(X,Y),
\]
is a nonzero polynomial in diagonal $D$ and hollow $X$. Consequently,
for $D$ in a nonempty regular-diagonal open set there is a dense open
set of hollow $X$ on which $L_{D,X}$ is invertible, even after imposing
all adjoint-power guards on $X$.
\end{lemma}
\begin{proof}
First take $n=3$. Use input order
\[
 (Y_{13},Y_{21},Y_{32};Y_{12},Y_{23},Y_{31})
\]
and output order
\[
 (H_{12},H_{23},H_{31};H_{13},H_{21},H_{32}).
\]
In these orders the full matrix of $L_{D,X}$ is block diagonal with
blocks
\begin{equation}\label{complete2:cycle-blocks}
 \begin{pmatrix}
 Q_{132}X_{32}&0&P_{132}X_{13}\\
 P_{213}X_{21}&Q_{213}X_{13}&0\\
 0&P_{321}X_{32}&Q_{321}X_{21}
 \end{pmatrix},\qquad
 \begin{pmatrix}
 Q_{123}X_{23}&P_{123}X_{12}&0\\
 0&Q_{231}X_{31}&P_{231}X_{23}\\
 P_{312}X_{31}&0&Q_{312}X_{12}
 \end{pmatrix}.
\end{equation}
Their determinants are respectively
\[
 X_{13}X_{21}X_{32}
 (P_{132}P_{213}P_{321}+Q_{132}Q_{213}Q_{321}),
\]
\[
 X_{12}X_{23}X_{31}
 (P_{123}P_{231}P_{312}+Q_{123}Q_{231}Q_{312}).
\]
For $s=d_1-d_2$, $t=d_2-d_3$, and $u=d_3-d_1$, the second parenthesis
is $\Delta_{P,Q}(s,t)$ and the first is
$(-1)^{3d}\Delta_{P,Q}(t,s)$. Hence, with precisely the displayed
input and output orders,
\begin{equation}\label{complete2:cycle-determinant}
 \det L_{D,X}
 =(-1)^{3d}\left(\prod_{i\ne j}X_{ij}\right)
   \Delta_{P,Q}(s,t)\Delta_{P,Q}(t,s).
\end{equation}
A different pair of coordinate orders changes only the overall sign.
When the cycle norm is nonzero, a regular $D$ making both factors
nonzero exists. This gives the size-three assertion.

For the induction with both symbols nonzero, suppose that, at one
fixed regular diagonal $D_0$ of size $h$, the determinant of
$L_{D_0,X}$ is a nonzero polynomial in $X\in V_h$. For each
$0\le j\le d$, the polynomial $\det(\ad_{D_0}^{j}X)$ is also nonzero:
a matrix supported on a Hamiltonian directed cycle is invertible and
remains invertible after its edges are multiplied by the corresponding
nonzero $j$th powers of root differences. The intersection of these
finitely many nonempty opens in the full affine space $V_h$ is
nonempty. Choose $X_0$ in it. This intersection takes place before
any nilpotent-value condition is imposed.

Put
\[
 D=\diag(D_0,z),\quad E=D_0-zI_h,\quad
 X=\begin{pmatrix}X_0&0\\0&0\end{pmatrix},\quad
 Y=\begin{pmatrix}B&v\\w&0\end{pmatrix},\qquad B\in V_h.
\]
The complete hollow operator splits into three independent blocks:
\begin{equation}\label{complete2:cycle-rectangles}
 H_D(X,Y)=
 \begin{pmatrix}
 L_{D_0,X_0}(B)&F(z)v\\
 wG(z)&0
 \end{pmatrix},
 \quad
 \begin{aligned}
 F(z)&=\sum_{r+s=d}P_{rs}(\ad_{D_0}^rX_0)E^s,\\
 G(z)&=\sum_{r+s=d}Q_{rs}(-1)^rE^r(\ad_{D_0}^sX_0).
 \end{aligned}
\end{equation}
In particular these rectangles use the full $P,Q$ coefficients.
Let $s_* =\max\{s:P_{d-s,s}\ne0\}$ and
$r_* =\max\{r:Q_{r,d-r}\ne0\}$. Their leading coefficients as
matrix polynomials in $z$ are
\begin{equation}\label{complete2:cycle-leading-rectangles}
 [z^{s_*}]F=P_{d-s_*,s_*}(-1)^{s_*}\ad_{D_0}^{d-s_*}X_0,
 \qquad
 [z^{r_*}]G=Q_{r_*,d-r_*}\ad_{D_0}^{d-r_*}X_0.
\end{equation}
They are invertible. Homogeneity leaves exactly one symbol monomial
at each stated largest exponent, so there is no possible leading-term
cancellation. Thus both rectangle determinants are nonzero
polynomials in $z$. Choose $z$ avoiding their roots and the eigenvalues
of $D_0$. The complete operator in size $h+1$ is invertible. In the
new full hollow $X$-space its invertibility open again intersects all
adjoint-power opens, preparing the next induction step. No intersection
with a constrained nilpotent fiber is asserted here.

For completeness, the one-order cases do not use an extension with
an absent rectangle. If $Q=0$, choose regular $D$ with all
triple-distinct $P_{ikj}\ne0$. In column $j$ the operator is the
$(n-1)$-square hollow matrix
\[
 M_j(X)_{ik}=P_{ikj}X_{ik}\quad(i,k\ne j,\ i\ne k).
\]
A derangement $\sigma$ of the indices different from $j$ contributes
\[
 \operatorname{sgn}(\sigma)
 \prod_{i\ne j}P_{i,\sigma(i),j}X_{i,\sigma(i)}
\]
to its determinant. This coefficient is nonzero, and distinct
permutations give distinct monomials in the original entries of $X$.
Thus each $\det M_j$ is nonzero and their product is nonzero. This is
the fixed-input operator from Theorem~\ref{mixed:ordered}, not merely
an assertion about a union of images. If $P=0$, write
$Q^{\mathrm{sw}}(s,t)=Q(t,s)$ and use the exact identity
\begin{equation}\label{complete2:one-order-transpose}
 H_{0,Q}(D,X,Y)^t
 =(-1)^d H_{Q^{\mathrm{sw}},0}(D,X^t,Y^t).
\end{equation}
It transfers the preceding invertibility while keeping the first
input fixed; replacing a fixed-$Y$ assertion by a fixed-$X$ assertion
without this transpose would not suffice. The same unconstrained
intersection with the adjoint-power opens applies to both one-order
cases.

Finally, a nonzero coefficient in $X$ of the polynomial
$\det L_{D,X}$ remains nonzero on a nonempty diagonal open set.
Intersecting with the regular locus gives the asserted common open
in $D$; for every $D$ there, the required opens in $X$ are dense.
\end{proof}

\begin{lemma}\label{complete2:invertible-operator-detection}
Under the hypothesis of Lemma~\ref{complete2:invertible-hollow-operator},
the tensors of guarded smooth nilpotent seeds span $V_n\otimes V_n$
for every $n\ge3$.
\end{lemma}
\begin{proof}
Fix an allowed $D$, and let $U_D\subset V_n$ be the dense open set
where $L_{D,X}$ and all the guard matrices are invertible. For
$X\in U_D$, a column index $j$, and a column $w$ with $w_j=0$ and
$w_i\ne0$ for every $i\ne j$, set
\begin{equation}\label{complete2:column-preimage}
 N=we_j^t,\qquad Y=L_{D,X}^{-1}(N).
\end{equation}
This $Y$ is the actual preimage under the complete operator. With both
symbols present it need not itself have column support. The value
$H_D(X,Y)=N$ is nonzero rank one and square-zero. Its actual $Y$
derivative fills $V_n$. Moreover
\[
 \diag[E_{ji},N]=w_i(E_{jj}-E_{ii})\quad(i\ne j),
\]
so the actual orbit directions complete the derivative to $\slm_n$.

Suppose a bilinear form $T$ annihilates every guarded smooth seed.
For fixed $X\in U_D$, the linear functional
\[
 N\longmapsto T\bigl(X,L_{D,X}^{-1}N\bigr)
\]
vanishes on the full-support open subset of each hollow column space.
It therefore vanishes on that entire column space. All these column
spaces span $V_n$, so $T(X,Y)=0$ for every hollow $Y$. This holds for
a dense open set of $X$, and consequently $T=0$. This argument uses
all column-target preimages of one actual operator. It does not add
values of $H_D$ or assume a column shape for its second input.
\end{proof}

\subsubsection{The zero cycle norm: exact lower kernels}
\label{complete2:zero-cycle-section}

It remains to treat $\Delta_{P,Q}=0$. Both symbols are then nonzero,
since a nonzero one-order symbol has nonzero cycle norm. In size four
choose a regular diagonal tuple such that every triple-distinct
$P_{ikj}$ and $Q_{ikj}$ is nonzero. Their finite product is a nonzero
polynomial, so these choices form a nonempty open set. We first use
three lower vertices $a,b,c$, reserving a fourth vertex $0$.

\begin{lemma}\label{complete2:zero-cycle-kernels}
Let $A\in V_3$ have all six entries nonzero. The complete lower
operator $L_A:B\mapsto H_{D_{\mathrm{low}}}(A,B)$ has rank four.
Its kernel has a basis supported on the two opposite directed
three-cycles, with each nonzero coordinate a fixed-$D$ nonzero scalar
times the corresponding entry of $A$. Its conormal for the hollow
trace pairing also has a basis of two invertible matrices supported
on opposite directed three-cycles.
\end{lemma}
\begin{proof}
In input coordinates $(B_{ac},B_{ba},B_{cb})$ and output coordinates
$(H_{ab},H_{bc},H_{ca})$ one block is
\begin{equation}\label{complete2:zero-lower-block}
 \begin{pmatrix}
 Q_{acb}A_{cb}&0&P_{acb}A_{ac}\\
 P_{bac}A_{ba}&Q_{bac}A_{ac}&0\\
 0&P_{cba}A_{cb}&Q_{cba}A_{ba}
 \end{pmatrix}.
\end{equation}
The opposite cycle gives the other block, exactly as in
\eqref{complete2:cycle-blocks}. Each determinant is zero by
\eqref{complete2:cycle-norm}, while its upper-left two-by-two minor is
nonzero. Each block has rank exactly two.

For the displayed block an explicit right-kernel matrix $B^{(1)}$ is
\begin{equation}\label{complete2:zero-right-kernel}
 B^{(1)}_{ac}=A_{ac},\qquad
 B^{(1)}_{ba}=-\frac{P_{bac}}{Q_{bac}}A_{ba},\qquad
 B^{(1)}_{cb}=-\frac{Q_{acb}}{P_{acb}}A_{cb},
\end{equation}
with its other entries zero. The first two equations are immediate;
the third is precisely
$P_{acb}P_{bac}P_{cba}+Q_{acb}Q_{bac}Q_{cba}=0$.
Interchanging $b,c$ gives $B^{(2)}$ on the opposite cycle. In
particular the six entries of $A$ remain independent variables;
the coefficients in these two kernel vectors depend only on $D$.

For an equally explicit conormal, a left null vector of
\eqref{complete2:zero-lower-block} is
\[
 \left(1,
 -\frac{Q_{acb}A_{cb}}{P_{bac}A_{ba}},
 \frac{Q_{acb}Q_{bac}A_{ac}}{P_{bac}P_{cba}A_{ba}}\right).
\]
Its three coordinates are nonzero. Transposing its output positions
for the trace pairing gives
\begin{equation}\label{complete2:zero-left-conormal}
 C_1=E_{ba}
 -\frac{Q_{acb}A_{cb}}{P_{bac}A_{ba}}E_{cb}
 +\frac{Q_{acb}Q_{bac}A_{ac}}{P_{bac}P_{cba}A_{ba}}E_{ac}.
\end{equation}
It satisfies $\tr(C_1L_A(B))=0$ for every hollow $B$, and is an
invertible directed-cycle matrix. Interchanging $b,c$ gives $C_2$.
Their supports are disjoint and opposite. They form a basis of the
two-dimensional conormal of the full rank-four lower operator.
\end{proof}

\subsubsection{A full size-four star differential}
\label{complete2:zero-four-star-section}

\begin{lemma}\label{complete2:zero-four-star}
For every $D$ in the size-four open just chosen, a nonempty open
subset of the entire column-star parameter space
\[
 X=\begin{pmatrix}0&0\\0&A\end{pmatrix},\qquad
 Y=ve_0^t,\qquad (A,v)\in V_3\times K^3,
\]
consists of nonzero rank-one square-zero seeds for which the fixed-$D$
derivative $dH_D$ is onto $V_4$. The same assertion holds for row stars.
The lower block $A$ can simultaneously satisfy all its
adjoint-power guards.
\end{lemma}
\begin{proof}
At the displayed input,
\begin{equation}\label{complete2:zero-four-value}
 H_D(X,Y)=N=we_0^t,\qquad w=Fv,\qquad
 F_{ij}=P_{ij0}A_{ij}\quad(i,j\in\{a,b,c\}),
\end{equation}
where the diagonal of $F$ is zero. The upper-row $Y$ derivative is
$r\mapsto rG$, with
\[
 G_{ij}=Q_{0ij}A_{ij}\quad(i,j\in\{a,b,c\}),\qquad G_{ii}=0.
\]
The determinants of $F$ and $G$ each have the two distinct cycle
monomials $A_{ab}A_{bc}A_{ca}$ and $A_{ac}A_{cb}A_{ba}$ with nonzero
coefficients. Each determinant is therefore a nonzero polynomial in
the same six original entries of $A$. The determinants
$\det(\ad_{D_{\mathrm{low}}}^jA)$ are likewise nonzero polynomials,
as is seen on a single directed-cycle support. Choose one $A$ where
all six entries, both rectangle determinants, and all those lower
guards are nonzero. This is an intersection of opens in the full
irreducible space $V_3$.

The complete fixed-$D$ hollow $Y$ derivative is the direct sum of
these two invertible three-dimensional rectangles and the rank-four
operator $L_A$. Its rank is $3+3+4=10$. Its two-dimensional hollow
conormal is precisely the span of $C_1,C_2$ from
Lemma~\ref{complete2:zero-cycle-kernels}, embedded in the lower block.

Now vary the three entries of the raw upper row of $X$ by
$a=(a_j)_{j\in\{a,b,c\}}$. Its only possibly nonzero output is the
lower hollow block
\begin{equation}\label{complete2:zero-four-upper-variation}
 (dH_D)_{ij}=Q_{i0j}v_i a_j\quad(i\ne j).
\end{equation}
The diagonal entries are zero by the endpoint hypothesis. For a lower
conormal $C$, its pairing with this output is
\[
 \tr(C\,dH_D)=\sum_j a_j(M_Cv)_j,\qquad
 (M_C)_{ji}=C_{ji}Q_{i0j},\quad (M_C)_{jj}=0.
\]
Both $M_{C_1}$ and $M_{C_2}$ are still invertible directed-cycle
matrices: every multiplier on their supports is nonzero. Their
opposite disjoint supports prevent proportionality. It follows that
$M_{C_1}v,M_{C_2}v$ are independent on a nonempty open set of $v$.
Indeed, dependence for every $v$ would make every vector an eigenvector
of $M_{C_1}^{-1}M_{C_2}$, forcing that operator to be scalar, contrary
to the supports.

Choose $v$ in that open with both $v$ and $Fv$ of full support. These
conditions are simultaneously possible because $F$ is invertible
and the vector space is irreducible. The three actual upper-row
$X$ directions then fill the two-dimensional quotient of $V_4$ by
the $Y$ image. Thus the full fixed-$D$ hollow derivative has rank
$12$. The value $N$ in \eqref{complete2:zero-four-value} is nonzero rank
one and square-zero, with all three complementary column entries
nonzero. The orbit directions $E_{0i}$ have diagonal projections
$w_i(E_{00}-E_{ii})$, giving the remaining three independent directions
and rank $15$ for $d\Gamma_D$.

All the conditions used are open in the whole star parameter space,
and we have shown their simultaneous nonemptiness. To obtain row
stars for the same original symbols use
\begin{equation}\label{complete2:two-order-transpose}
 H_{P,Q}(D,X^t,Y^t)
 =(-1)^dH_{Q^{\mathrm{sw}},P^{\mathrm{sw}}}(D,X,Y)^t,
 \qquad R^{\mathrm{sw}}(s,t)=R(t,s).
\end{equation}
The transformed pair has zero endpoints and zero cycle norm, since
its norm is $\Delta_{P,Q}(t,s)$. Its triple-distinct weights are
nonzero under our same conditions on $D$, by reversal of the path
and homogeneity. The preceding column proof therefore applies to it.
Transpose preserves all ranks and the adjoint-power guards, since
$\ad_D^j(X^t)=(-1)^j(\ad_D^jX)^t$.
\end{proof}

\subsubsection{Restoring guards in the exact raw fiber}
\label{complete2:zero-raw-fiber-section}

\begin{lemma}\label{complete2:zero-raw-restoration}
At any column star in the nonempty open constructed in
Lemma~\ref{complete2:zero-four-star}, the exact raw fiber $H_D=N$ is
smooth, and its local ring contains a nonzero product of all raw
$X$ guard determinants. Every function whose germ on this fiber is
nonzero can be retained at an actual guarded smooth nilpotent seed
in this same fiber. The corresponding assertion holds at row stars.
\end{lemma}
\begin{proof}
The surjectivity of the fixed-$D$ hollow differential makes
\[
 \mathcal F_N=\{(X',Y')\in V_4\times V_4:H_D(X',Y')=N\}
\]
smooth at the star $p$. Its local ring $R=\mathcal O_{\mathcal F_N,p}$
is a regular local domain. The map from the three raw upper-row
$X$ variations to the two-dimensional quotient by the $Y$ image has
rank two. Choose a nonzero row $a$ in its one-dimensional kernel.
Its complete output lies in the $Y$ image and can therefore be
cancelled by an actual hollow $Y$ variation. Independently, every raw
lower-column $X$ variation $z$ has zero output: both products with
the current $Y=ve_0^t$ vanish. Consequently, for every $z$, the raw
variation
\[
 \dot X=\begin{pmatrix}0&a\\z&0\end{pmatrix}
\]
can be completed by a hollow $\dot Y$ to a tangent vector of the
exact fiber $\mathcal F_N$. No output conjugation is involved.

Put $E=D_{\mathrm{low}}-d_0I_3$ and
$A_j=\ad_{D_{\mathrm{low}}}^jA$. All $E,A_j$ are invertible.
Smoothness lifts any such tangent to a formal arc in $\mathcal F_N$.
Along it the raw $X$ remains hollow, its upper and lower cross
blocks start with $ta$ and $tz$, and its lower block starts with $A$.
The Schur complement gives, for every guard exponent $j$,
\begin{equation}\label{complete2:zero-raw-guard-coefficient}
 \det(\ad_D^jX(t))
 =(-1)^{j+1}t^2\det(A_j)\,
       aE^jA_j^{-1}E^jz+O(t^3).
\end{equation}
In particular second-order changes in the lower block or the cross
blocks cannot alter the displayed coefficient; the upper-left raw
entry is identically zero. For each $j$ the row
$aE^jA_j^{-1}E^j$ is nonzero. Choose one $z$ outside the finitely many
annihilating hyperplanes. Every determinant in
\eqref{complete2:zero-raw-guard-coefficient} is then nonzero on the
formal arc. Their product is consequently a nonzero element of $R$.
The fixed-$D$ and saturated smoothness minors already nonzero at $p$
are units in $R$.

Let $b\in R$ be any nonzero function germ. The product of $b$, the
guard product, and these minors is nonzero because $R$ is a domain.
On a sufficiently small finite-type neighborhood in the component
through $p$, its nonvanishing defines a nonempty open set. As $K$ is
algebraically closed, this open set has an actual $K$-point. At that
point the raw value is still exactly $N$, all guards hold, the seed
is smooth, and $b$ is nonzero. This proves the stated retention
property, not merely a formal existence claim. The transpose
identity \eqref{complete2:two-order-transpose} transports the entire
raw-fiber argument to row stars.
\end{proof}

\begin{corollary}\label{complete2:zero-raw-germ}
If a bilinear form $T$ on $V_4\times V_4$ vanishes at every guarded
smooth nilpotent seed, its entire germ on the exact raw fiber at
every star in Lemma~\ref{complete2:zero-four-star} is zero.
\end{corollary}
\begin{proof}
A nonzero germ could be retained at a guarded smooth seed in the same
raw fiber by Lemma~\ref{complete2:zero-raw-restoration}, contradicting
the hypothesis on $T$.
\end{proof}

\subsubsection{Six independent monomials and the guarded base}
\label{complete2:zero-four-span-section}

\begin{lemma}\label{complete2:zero-four-span}
For each $D$ in the above size-four diagonal open, the tensors of
guarded smooth nonzero rank-one square-zero seeds span
$V_4\otimes V_4$.
\end{lemma}
\begin{proof}
Suppose a bilinear form $T$ annihilates all such tensors. By
Corollary~\ref{complete2:zero-raw-germ} it vanishes at every generic
column or row star. Polynomial continuation on the whole irreducible
star parameter space makes its restriction identically zero there.

First consider two hollow units $E_{ab},E_{cd}$ with
$\{a,b\}\ne\{c,d\}$. The second edge has an endpoint outside
$\{a,b\}$. If $d$ is such an endpoint, use the column star isolated at
$d$: $E_{ab}$ belongs to its lower block and $E_{cd}$ to its column.
If $c$ is such an endpoint, use the row star isolated at $c$ instead.
Thus
\begin{equation}\label{complete2:zero-different-edges}
 T(E_{ab},E_{cd})=0\qquad\text{if }\{a,b\}\ne\{c,d\}.
\end{equation}
This eliminates four-distinct-vertex coefficients as well, without
an embedding assertion from size three. No relation between the
remaining same-edge and reversed-edge coefficients has been assumed.

Fix now an isolated vertex $0$ and a generic lower block $A$. Choose
a $v$ making its column star one of the smooth points above. For
either of the explicit lower kernel matrices $B^{(\nu)}$ in
\eqref{complete2:zero-right-kernel}, the line
\begin{equation}\label{complete2:zero-kernel-line}
 (X,Y_t)=\left(
 \begin{pmatrix}0&0\\0&A\end{pmatrix},
 ve_0^t+t\begin{pmatrix}0&0\\0&B^{(\nu)}\end{pmatrix}\right)
\end{equation}
lies in the exact same raw fiber: the additional full output is
$tL_A(B^{(\nu)})=0$. The zero germ from
Corollary~\ref{complete2:zero-raw-germ}, restricted to this line, gives
\[
 T\left(\begin{pmatrix}0&0\\0&A\end{pmatrix},
         \begin{pmatrix}0&0\\0&B^{(\nu)}\end{pmatrix}\right)=0.
\]
Let $\mathcal C$ be the directed three-cycle supporting one such
kernel vector, and write $B_e=\gamma_eA_e$ on it. Each
$\gamma_e\in K^*$ depends only on $D$, and the cycle contains exactly
one orientation of each unordered lower edge. Using
\eqref{complete2:zero-different-edges}, the preceding identity is
\begin{equation}\label{complete2:zero-six-monomials}
 \sum_{e\in\mathcal C}\gamma_e
 \left(T(E_e,E_e)A_e^2+
       T(E_{\bar e},E_e)A_{\bar e}A_e\right)=0,
\end{equation}
where $\bar e$ denotes the reverse edge. This identity holds for a
nonempty open set of the six independent entries of $A$, hence is a
polynomial identity in them. Its three square monomials and three
unordered-edge product monomials are six distinct monomials.
Every coefficient displayed in \eqref{complete2:zero-six-monomials}
therefore vanishes. The opposite kernel cycle kills the other three
square coefficients and the opposite three ordered reverse-edge
coefficients on this lower triangle.

Varying the isolated vertex covers every unordered edge of the
four-vertex set. All coefficients left after
\eqref{complete2:zero-different-edges} consequently vanish, and $T=0$.
The two kernel shifts were taken in a fiber of the full operator,
at a common original input, rather than in fibers of its summands.
\end{proof}

\subsubsection{Completion in all sizes and diagonal enrichment}
\label{complete2:zero-completion-section}

\begin{proof}[Proof of Theorem~\ref{complete2:zero-endpoint}]
If the cycle norm is nonzero, Lemmas~\ref{complete2:invertible-hollow-operator}
and \ref{complete2:invertible-operator-detection} give even guarded
hollow detection in every size at least three. If the cycle norm is
zero, Lemma~\ref{complete2:zero-three} gives the required unguarded
size-three result, while Lemma~\ref{complete2:zero-four-span} supplies
the guarded size-four base for both nonzero symbols.

For this latter case apply the exact guarded singleton extension
\ref{complete2:extension} and the support induction
\ref{complete2:induction}. Their hypotheses concern the complete
trace-zero map and guarded smooth nilpotent seeds, all of which have
just been verified; they do not require a nonzero endpoint. They give
full guarded hollow tensor span in every size $n\ge5$. The generic
fixed-diagonal descent in \ref{complete2:descent} then gives a
nonempty numerical diagonal open $\Omega_n$ supporting a full
spanning collection at each one of its points. In particular $D$ is
not selected separately for different tensor coefficients. The
size-three branch needs no claim that its raw first input can satisfy
the guards.

The common diagonal-enrichment lemma
\ref{complete2:diagonal-enrichment} upgrades these hollow assertions
to full $M_n\otimes M_n$ detection. Its hypotheses hold because
$st$ divides both symbols, and the smoothness controls used here are
actual hollow input directions and simultaneous conjugation of the
entire original tuple.
\end{proof}

The conclusion is a detector for the complete zero-endpoint
restriction with arbitrary coefficient dependence between $P$ and
$Q$. Together with the original trace transfer in Theorem~\ref{complete2:main},
it applies to every ordinary coefficient fiber having this
restriction, including corrections whose complete chosen-pair
restriction is zero. No standalone outer-adjoint or corank-one
hypothesis is needed, and no size-two or all-pairs-zero assertion is
part of Theorem~\ref{complete2:zero-endpoint}.


\subsection{The complete original trace on the same nilpotent fiber}
\label{complete2:original-section}

We now prove Theorem~\ref{complete2:main}. This step uses the actual
matrix-entry polynomial $\tr f$ at the specified size, not a formal
free trace coefficient and not a selected low-activity detector.
In particular, all fixed-size polynomial identities are included
before the first nonzero trace term is selected.

\begin{proof}[Proof of Theorems~\ref{complete2:seed} and \ref{complete2:main}]
For the seed theorem, $\tau=0$ implies $q=-\epsilon p$.
The case $p\ne0$ is Theorem~\ref{complete2:nonzero-endpoint}; the case
$p=0$ is Theorem~\ref{complete2:zero-endpoint}. In the latter proof the
cycle norm is either nonzero or identically zero. If only one symbol
is nonzero its norm cannot vanish, since it is a product of three
nonzero polynomials under invertible substitutions. Thus these cases
exhaust every nonzero pair, without imposing a product or fixed-depth
hypothesis. Lemmas~\ref{complete2:descent} and
\ref{complete2:diagonal-enrichment} give the stated common diagonal
open set and the full, rather than hollow, bilinear detector.

For the original map, if $\tau\ne0$, the smooth nilpotent input of
Theorem~\ref{highlayers:all-symbols} is an input of the complete
restriction of $f$. Its tangent directions are original-source
directions, so the full original differential is onto $M_n$.
Writing $h=\tr f$ and $g=f-n^{-1}hI_n$, we obtain
$\rank(dg_a)=n^2-1$ and $dh_a|_{\ker(dg_a)}\ne0$ at this input $a$.
Thus $h|_{g^{-1}(N)}$ has a nonzero linear term at $a$.
Theorem~\ref{local} gives the actual image and smoothness gives the
fiber assertion.
Hence assume $\tau=0$ for the rest of the proof, and put
\[
 \pi(A)=A-\frac{\tr A}{n}I_n,\qquad g=\pi f,\qquad h=\tr f.
\]

Choose one passive original variable as an anchor $D$. In each of the
other $m-3$ passive positions put $D+Z_i$; retain the original active
variables $X,Y$. These are invertible linear coordinates on $M_n^m$.
The linear subspace $S=\{Z_1=\cdots=Z_{m-3}=0\}$ is the complete
restriction source. On it $h=0$ and $g=H_{P,Q}$.
If $h$ is an actual identity, any seed in
Theorem~\ref{complete2:seed} gives an original smooth input into
$\slm_n$. Theorem~\ref{local} then gives $\Val_n(f)=\slm_n$ and the
local fiber dimension $mn^2-n^2+1$.

Suppose instead that $h$ is a nonzero polynomial in actual matrix
entries. Expand it by total degree in the normal variables:
\begin{equation}\label{complete2:normal-expansion}
 h(D,X,Y,Z)=\sum_{j=v}^{m-3}H_j(D,X,Y,Z),\qquad
 H_v\ne0,\quad 1\le v\le m-3.
\end{equation}
Every $H_j$ is homogeneous of degree $j$ in $Z$ and bilinear in the
\emph{full} original matrices $X,Y$. Indeed, every word of $f$ uses
each original slot once. Lower components in
\eqref{complete2:normal-expansion} vanish as actual size-$n$
polynomials, including cancellations arising from trace identities.
Each component is separately invariant under simultaneous
conjugation: apply that invariance to $h(D,X,Y,tZ)$ and compare powers
of the scalar $t$.

The conjugation-saturation of the nonempty diagonal open set
$\Omega_n$ in Theorem~\ref{complete2:seed} is dense in $M_n$.
Indeed, if a polynomial vanishes on that saturation, then for each
fixed conjugating matrix it vanishes on the corresponding conjugate
of the full diagonal space by density of $\Omega_n$ in that space.
It therefore vanishes on the dense regular semisimple locus.
Invariance of $H_v$ now shows that there are $D_0\in\Omega_n$
and fixed normal matrices
$Z^0$ for which
\[
 T(X,Y)=H_v(D_0,X,Y,Z^0)
\]
is a nonzero bilinear form on $M_n\times M_n$.
For if it vanished for every such $D_0,Z^0,X,Y$, conjugation and density
would force $H_v=0$. The seed detector chooses original $X_0,Y_0$
at this \emph{same} $D_0$ such that $T(X_0,Y_0)\ne0$, the value
$N=H_{P,Q}(D_0,X_0,Y_0)$ is nonzero rank-one square-zero, and the
restriction differential onto $\slm_n$ is surjective.
Let $a=(D_0,X_0,Y_0,0)$ in the original linear coordinates.

At $a$ the map $g$ is smooth, already using directions in $T_aS$.
Choose an $(n^2-1)$-dimensional control space $L\subset T_aS$ on
which $dg_a$ is an isomorphism onto $\slm_n$. Complete it to linear
source coordinates $(\ell,w,Z)$ centered at $a$. The formal implicit
function theorem solves the \emph{exact} partial fiber $g=N$ uniquely
as $\ell=\phi(w,Z)$ in a formal neighborhood, with
$\phi(0,0)=0$. Since $L\subset T_aS$, this solution does not change
any normal coordinate $Z_i$.

The total lowest Taylor term of $h$ at $a$ is precisely
\begin{equation}\label{complete2:surviving-leading-term}
 H_v(D_0,X_0,Y_0,Z).
\end{equation}
It is nonzero by the chosen $Z^0$. A variation of any coefficient
variable $D,X,Y$ in $H_v$ adds at least one to its total degree;
all $H_j$ with $j>v$ have larger total degree already. Substitution
of the exact control correction $\phi(w,Z)$ therefore leaves
\eqref{complete2:surviving-leading-term} unchanged. This is a
statement about the complete lowest Taylor term in actual original
coordinates, not just a normal symbol before solving the fiber.

The local ring of $g^{-1}(N)$ at $a$ is regular of dimension
$mn^2-n^2+1$ and is a domain. Its completion is the formal ring just
used, so $h$ is a nonzero element of that local ring. It vanishes at
$a$ and is a non-zero-divisor. Adding the trace equation gives the
full fiber $f^{-1}(N)$ as a local complete intersection of dimension
$mn^2-n^2$. The same local trace-germ and miracle-flatness argument as
Proposition~\ref{higher-trace} proves that $f$ is flat at this original
input. Corollary~\ref{cert} and Theorem~\ref{local}, including their
ordinary multilinear scalar extraction, give the actual image $M_n$.
No scalar value is supplied merely by taking a closure.
If $v=1$, the surviving linear trace term supplies the missing normal
and the point is smooth into $M_n$. If $v\ge2$, its Jacobian rank is
exactly $n^2-1$: it is flat but not smooth. These alternatives are
compatible with the asserted local complete-intersection fiber.
\end{proof}

\begin{corollary}[Full higher-ideal coefficient fibers]
\label{complete2:higher-corrections}
Let $F$ satisfy the complete restriction hypothesis of
Theorem~\ref{complete2:main}. Every same-degree ordinary polynomial
$f=F+R$ with zero complete selected restriction for $R$ has the image
and local-fiber classification in that theorem. In particular, $R$
may range over the entire proper multilinear $J^3$ component and all
higher powers. The target is determined separately by $\tr(F+R)$ on
the fixed $M_n$.
\end{corollary}
\begin{proof}
The complete selected restriction is unchanged. The vanishing of
proper $J^3$ on the whole two-active source is
Lemma~\ref{m3full:passive-vanishing}. We use that vanishing only to
identify the restriction. Transverse derivatives of $R$ are not
set to zero: they are included in $h$ and in every coefficient of
\eqref{complete2:normal-expansion}. The theorem reselects the seed
for the corrected polynomial, rather than asserting that an old
nilpotent input remains suitable.
\end{proof}

\subsection{Coefficient scope, examples, and the remaining boundary}
\label{complete2:scope}

For a fixed ordered pair of original active positions, the free
proper degree-$m$ $J^2$ coefficient space maps onto the full symbol
space
\[
 \{(P,Q):\deg P=\deg Q=m-2,\ st\mid P,Q\},
 \qquad \dim=2(m-3).
\]
To see surjectivity without a dimension heuristic, partition the
$m-2$ passive original leaves into sets of sizes $r,s\ge1$ and form
one iterated commutator on the first active leaf and its $r$ passive
leaves, and another on the second active leaf and its $s$ passive
leaves. Their ordered product restricts, up to a nonzero sign, to
$(\ad_D^rX)(\ad_D^sY)$. Reversing the product gives the other order.
All passive leaves remain distinct. As $r+s=m-2$ varies these are
exactly the two monomial bases. Thus the kernel of this \emph{fixed
pair} map has codimension $2(m-3)$. This is not a codimension formula
for the common kernel over all pairs and is not a quotient by
size-$n$ polynomial identities. Every nonzero fiber of this map is
covered, with the actual trace deciding its target.

\begin{remark}[A high-activity fixed-size trace identity does not obstruct the transfer]
\label{complete2:fourteen-blind}
Take twenty-eight distinct original variables, arranged into fourteen
pairs, and write $A_i=[a_i,a_i']$, $B_i=[b_i,b_i']$, $1\le i\le7$.
Set
\[
 R=\sum_{\sigma\in S_7}\operatorname{sgn}(\sigma)
 A_{\sigma(1)}B_1A_{\sigma(2)}B_2\cdots A_{\sigma(7)}B_7.
\]
This is proper, ordinary multilinear, and belongs to $J^{14}$.
On $M_3$, every common-$D$ restriction with at most fourteen active
original slots is zero. With fewer than fourteen, one basic
commutator has both entries $D$. With fourteen, either that still
happens or each factor has one active entry. In the latter case all
seven $A_i$ belong to $\operatorname{im}\ad_D$. For regular diagonal
$D$ this space has dimension six, so alternation kills the value.
Conjugation and polynomial density extend the identity to every $D$.
This covers every choice of active original slots.

Nevertheless $\tr R$ is not an identity. Use
\[
 (A_1,\ldots,A_7)=(E_{12},E_{13},E_{21},E_{23},
 E_{11}-E_{22},E_{31},E_{32}),
\]
\[
 (B_1,\ldots,B_7)=(E_{21},E_{32},E_{12},E_{31},E_{13},E_{13},E_{21}).
\]
Here is a direct certificate requiring no enumeration of $7!$ terms.
Let $(u_i,v_i)$ be the following ordered coordinate pairs:
\[ (1,2),(1,3),(2,1),(2,3),(1,1),(3,1),(3,2). \]
Then $B_i=E_{v_i,u_{i+1}}$ cyclically and
\[
 \tr(C_1B_1\cdots C_7B_7)=\prod_{i=1}^7(C_i)_{u_i v_i}.
\]
The coordinate-evaluation matrix of the seven displayed $A_i$ is
$I_7$. Hence the alternating trace is its determinant, namely one.
All values have independent original lifts:
$E_{ij}=[E_{ii},E_{ij}]$ and $E_{11}-E_{22}=[E_{12},E_{21}]$.
Embedding these inputs in a three-by-three block proves actual trace
nonvanishing for every $n\ge3$.
With $D=\diag(0,1,2)$, the thirteen off-diagonal factors can be written
$[D,E_{ij}/(d_i-d_j)]$, using one active original slot each;
the diagonal factor uses two. Thus fifteen active slots already
detect the trace. The example is not invisible to every common-$D$
template.

Label the same twenty-eight original leaves $x_1,\ldots,x_{28}$,
and let $C_{14}$ denote a left-normed commutator on fourteen distinct
leaves. Put
\[
 F=[C_{14}(x_1,\ldots,x_{14}),C_{14}(x_{15},\ldots,x_{28})].
\]
The complete restriction with active $x_1,x_{15}$ is
$[\ad_D^{13}X,\ad_D^{13}Y]$; the two signs from the left-normed
combs cancel. Thus $F+R$ has actual image $M_n$ for every $n\ge3$
by Theorem~\ref{complete2:main}. In size three its entire trace is
still zero on every common-$D$ source with at most fourteen active
slots. The theorem uses the first surviving actual normal term,
not the false inference that a low-activity detector must see all
traceful higher-ideal coefficients.
\end{remark}

The common zero restriction region remains outside this theorem.
It contains proper $J^3$ itself and also genuine second-layer
examples such as $s_4$, whose two passive equal inputs make every
chosen-pair restriction zero by alternation. Individual members may
have other classifications. In particular, $\tau=0$ alone is not
the hypothesis: some \emph{complete} pair $(P,Q)$ must be nonzero.
The fixed-size actual trace is still required to determine the target.
Size two is not covered by the present seed theorem. No resolution of
general LK is asserted.

Here the zero-restriction limitation refers to
Theorem~\ref{complete2:main}. The independent-diagonal
Theorem~\ref{indep2:main} covers every genuine proper second-layer
member of that region, including $s_4$, while retaining the exclusions
of pure $J^3$, size two, and general LK.



The actual-image criterion, scalar extraction, local trace-germ
criterion, one-order column operators, cubic cycle determinants,
and highest-power rectangle arguments are established earlier in
this manuscript. The proofs here reuse those mechanisms and supply
the compatible full tensor detectors, exact raw-fiber controls and
whole original trace transfer needed for this complete coefficient
region. They do not claim historical priority for those ingredients.

\section{The entire proper second commutator layer}
\label{indep2:section}

The common-background theorem of Section~\ref{complete2:section}
retains only the sums of the independent passive coefficients at each
fixed depth. We now keep each passive diagonal independently. This
recovers every proper ordinary polynomial in the second commutator
layer, including the kernel of all common-background pair tests. The
field remains algebraically closed of characteristic zero. The ideal
$J$ is the two-sided commutator ideal of the unital free associative
algebra, as in Proposition~\ref{m3full:second-basis}.


The common-background branch is already Theorem~\ref{complete2:main}.
The additional proof has two parts: a compatible nilpotent tensor
span for the complete independent-diagonal restriction, and an
\'etale chart of the entire original source that retains its first
nonzero trace normal. In particular, a triangular value alone is
not used as an ambient flatness certificate.

\subsection{The exact original two-slot restriction}

Choose two original positions $X,Y$ and label the other positions
$D_1,\ldots,D_d$, where $d=m-2\ge2$. For diagonal passive inputs write
$\boldsymbol v_a=((D_i)_{aa})_{i=1}^d$. For a subset $S\subseteq[d]$
and a vector $a=(a_i)$ put $a_S=\prod_{i\in S}a_i$, with empty
product $1$.

\begin{proposition}[Independent passive coefficients]
\label{indep2:normal}
For each original pair of a proper multilinear $f\in J^2$ there are
unique arrays $(c_S),(e_S)$, indexed by
$\varnothing\ne S\subsetneq[d]$, for which
\begin{equation}\label{indep2:symbols}
 P(a,b)=\sum_{\varnothing\ne S\subsetneq[d]}c_Sa_Sb_{S^c},
 \qquad
 Q(a,b)=\sum_{\varnothing\ne S\subsetneq[d]}e_Sa_Sb_{S^c},
\end{equation}
and the complete actual restriction is
\begin{equation}\label{indep2:path}
 \begin{split}
 H_D(X,Y)_{ac}
 &=f(X,Y;D_1,\ldots,D_d)_{ac}\\
 &=\sum_b\bigl[
 P(\boldsymbol v_a-\boldsymbol v_b,
   \boldsymbol v_b-\boldsymbol v_c)X_{ab}Y_{bc}\\
 &\hspace{42mm}+
 Q(\boldsymbol v_a-\boldsymbol v_b,
   \boldsymbol v_b-\boldsymbol v_c)Y_{ab}X_{bc}\bigr].
 \end{split}
\end{equation}
The positions on the first line are the original positions, regardless
of the order used to display their names. At least one pair has
$(P,Q)\ne(0,0)$ when $f\notin J^3$.
\end{proposition}
\begin{proof}
Expand each ordinary word according to the order of $X,Y$ and the
three passive gaps before, between and after them. This gives two
path coefficients separately linear in each triple
$((D_i)_{aa},(D_i)_{bb},(D_i)_{cc})$. Properness is scalar-translation
invariance in every passive variable, so each triple occurs through
one of its two successive differences. Thus each coefficient has a
unique expansion in the $2^d$ products $a_Sb_{S^c}$ before the two
extreme subsets are removed.

A one-active evaluation with diagonal passive inputs vanishes. An
off-diagonal entry is an evaluation of $f$ on a two-dimensional upper
triangular block, where $J^2$ vanishes; diagonal entries also vanish.
Make $X$ diagonal, and let $Y$ have a single off-diagonal entry.
The two independent endpoint entries of $X$ give
$P(0,b)=Q(a,0)=0$. Interchanging the roles of $X,Y$ gives
$P(a,0)=Q(0,b)=0$. Hence both extreme coefficients vanish separately.
This proves \eqref{indep2:symbols} and \eqref{indep2:path} for the
unchanged original polynomial.

There is also exact finite extraction. On three vertices put
\[
 D_i^{(S)}=
 \begin{cases}E_{11},&i\in S,\\-E_{33},&i\notin S.
 \end{cases}
\]
Then
\[
 c_S=f(E_{12},E_{23};D^{(S)})_{13},\qquad
 e_S=f(E_{23},E_{12};D^{(S)})_{13}.
\]
Every $UT_3$ evaluation expands into diagonal and strictly upper
triangular inputs. Terms with at most one radical input vanish as
above, and a word with three radical inputs vanishes because the
radical has cube zero. The remaining terms have exactly two original
active positions and are precisely these path tests. Thus vanishing
of all the pair symbols would make $f$ an identity of $UT_3$.
The classical triangular identity theorem
$\operatorname{Id}(UT_r)=J^r$, used in the commutator-degree
framework of \cite[Theorem 2.1]{UT}, gives the final assertion for $r=3$.
\end{proof}

The last implication uses the triangular identity theorem, not a
claim that a nonzero free polynomial is nonzero on every fixed matrix
algebra. Here the fixed-size issue can be settled directly: for
$n\ge3$ the map
\[
 UT_3\longrightarrow M_n,\qquad
 T\longmapsto\diag(T,t_{11}I_{n-3})
\]
is a unital algebra embedding, since $T\mapsto t_{11}$ is a
character. Consequently
\begin{equation}\label{indep2:fixed-pi}
 \operatorname{Id}(M_n)\subseteq\operatorname{Id}(UT_3)=J^3
 \qquad(n\ge3).
\end{equation}
In particular the theorem's polynomial is not an identity in any of
its asserted sizes. No such inclusion is used for $n=2$.

If some common-$D$ restriction is nonzero, the ordinary homogeneous
symbols obtained by setting $a=s\boldsymbol1$, $b=t\boldsymbol1$
satisfy the hypotheses of Theorem~\ref{complete2:main}. We therefore
assume for the new branch that all original pair restrictions on a
common $D$ vanish, and choose a nonzero independent pair from
Proposition~\ref{indep2:normal}. Then
\[
 \sum_{|S|=r}c_S=\sum_{|S|=r}e_S=0\qquad(1\le r\le d-1).
\]
It follows as a polynomial identity in all $d$ coordinates that
\begin{equation}\label{indep2:endpoints}
 P(a,-a)=a_{[d]}\sum_S(-1)^{d-|S|}c_S=0,
 \qquad Q(a,-a)=0.
\end{equation}
Thus $H_D$ is hollow for every diagonal tuple and all full $X,Y$.
It ignores diagonal parts of the active inputs, since both $S$ and
$S^c$ are nonempty. The seed argument below only requires
\eqref{indep2:symbols}, a nonzero pair, and
\eqref{indep2:endpoints}; it does not require the stronger grouped
coefficient sums once the endpoints have been established.


\subsection{Independent-vector nilpotent tensor detection}

Let $V_n=\{X\in M_n:X_{aa}=0\}$ and, for a fixed diagonal tuple
$D=(D_1,\ldots,D_d)$, define
\[
 \Gamma_D(X,Y,g)=gH_D(X,Y)g^{-1},\qquad g\in GL_n.
\]
Its orbit directions are actual directions of the original polynomial:
one simultaneously conjugates every $D_i,X,Y$. All tests and guards
below concern the raw $X,Y$, before this conjugation. For
$S\subseteq[d]$ define the composition of commuting adjoint operators
\begin{equation}\label{indep2:KS}
 K_S(X)=\left(\prod_{i\in S}\ad_{D_i}\right)(X),\qquad
 K_\varnothing(X)=X.
\end{equation}
Thus $(K_S(X))_{ab}=X_{ab}\prod_{i\in S}((D_i)_{aa}-(D_i)_{bb})$;
\eqref{indep2:KS} is not a product of the matrices
$\ad_{D_i}(X)$.

\begin{theorem}[One fixed diagonal open for all bilinear tests]
\label{indep2:seed}
Assume \eqref{indep2:symbols}, \eqref{indep2:endpoints} and
$(P,Q)\ne(0,0)$. For every $n\ge3$ there is a nonempty open set
$\Omega_n$ of diagonal tuples, with each $D_i$ regular, such that at
every fixed $D\in\Omega_n$ the tensors $X\otimes Y$ of hollow pairs
satisfying
\[
 N=H_D(X,Y)\ne0,\qquad \rank N=1,\qquad N^2=0,\qquad
 \rank d\Gamma_D|_{(X,Y,I)}=n^2-1
\]
span $V_n\otimes V_n$. For $n\ge4$ the same assertion holds after
imposing all $2^d$ guards
\begin{equation}\label{indep2:guards}
 \det K_S(X)\ne0\qquad(S\subseteq[d]),
\end{equation}
including $S=\varnothing$. Allowing arbitrary diagonal parts of $X,Y$
then supplies one such original nilpotent seed where any prescribed
nonzero bilinear form on full $M_n\times M_n$ is nonzero.
\end{theorem}

The span is only a bilinear detector. It selects one input compatible
with a subsequently specified nonzero test and does not add actual
polynomial values. Both the fixed-$D$ quantifier and the full active
matrix spaces will be needed for the trace normal.

For notation write
\[
 P_{ikj}=P(\boldsymbol v_i-\boldsymbol v_k,
                  \boldsymbol v_k-\boldsymbol v_j),\qquad
 Q_{ikj}=Q(\boldsymbol v_i-\boldsymbol v_k,
                  \boldsymbol v_k-\boldsymbol v_j).
\]
For every distinct triple $i,k,j$ each nonzero symbol gives a nonzero
polynomial in the diagonal parameters: the two vector differences
can be prescribed independently. The product of all such nonzero
path weights and all coordinatewise eigenvalue differences is
therefore nonzero. We may impose their simultaneous nonvanishing on
one diagonal open. At any such $D$, each guard determinant is a
nonzero polynomial on the full affine space $V_h$, $h\ge3$: take $X$
supported on a directed Hamiltonian cycle, whose nonzero edges remain
nonzero after every subset rescaling. This observation alone does
not put the guards on a nilpotent fiber; that step is proved below.

\begin{lemma}[The common zero row in size three]
\label{indep2:three}
The unguarded assertion of Theorem~\ref{indep2:seed} holds in size
three on the simultaneous path-weight open just described.
\end{lemma}
\begin{proof}
Let $R_i\subset V_3$ be the matrices with row $i$ zero. For
$X,Y\in R_i$, the complementary two-by-two hollow block can contribute
only diagonal products, and these vanish by
\eqref{indep2:endpoints}. Every nonzero output is consequently a
column star supported in column $i$, of rank one and square zero.

Use $i=1$ first. If both symbols are nonzero, put
\[
 X=uE_{23}+vE_{32},\qquad Y=c_2E_{21}+c_3E_{31},
 \qquad uv c_2c_3\ne0.
\]
The value has entries $N_{21}=P_{231}uc_3$ and
$N_{31}=P_{321}vc_2$, both nonzero. The $Y$ variations $E_{21},E_{31}$
control respectively $H_{31},H_{21}$; the $Y$ variations
$E_{12},E_{13}$ control $H_{13},H_{12}$ through $Q$; the $X$
variations $E_{12},E_{13}$ control $H_{32},H_{23}$ through $Q$.
The indicated weights and parameters are all nonzero, so these are
six independent hollow controls. The actual orbit directions
$E_{12},E_{13}$ have diagonal projections
$N_{21}(E_{11}-E_{22})$ and $N_{31}(E_{11}-E_{33})$.
Thus the saturated differential has rank eight.

If $Q=0$, take every entry $21,31,23,32$ of both $X,Y$ nonzero.
The $Y_{21},Y_{31}$ variations give the two lower-column coordinates;
$X_{12},X_{13}$ give the upper row using $Y_{23},Y_{32}$; and
$Y_{12},Y_{13}$ give the lower off-diagonal coordinates using
$X_{31},X_{21}$. Additional lower-column components, if present,
are removed with the first two controls. Again the hollow rank is
six and the orbit rank is eight. If $P=0$, exchange the actual
active inputs. Relabeling vertices gives the construction on each
$R_i\times R_i$.

The good conditions are a nonempty open in the whole irreducible
space $R_i\times R_i$. A bilinear form vanishing on all good seeds
therefore vanishes identically on each such space. Any two hollow
matrix units omit a common row in size three; hence every tensor
coefficient is zero. This proves the assertion without imposing a
one-variable homogeneity condition or a size-three guard.
\end{proof}

\subsubsection{Cycle determinants with vector differences}

Put $a=\boldsymbol v_1-\boldsymbol v_2$,
$b=\boldsymbol v_2-\boldsymbol v_3$, and $c=-a-b$. Define
\begin{equation}\label{indep2:cycle-norm}
 \Delta(a,b)=P(a,b)P(b,c)P(c,a)+Q(a,b)Q(b,c)Q(c,a).
\end{equation}
The norm is a polynomial in $2d$ independent coordinates. In size
three, use the input and output orders
\[
 (Y_{13},Y_{21},Y_{32};Y_{12},Y_{23},Y_{31}),\qquad
 (H_{12},H_{23},H_{31};H_{13},H_{21},H_{32}).
\]
The operator $L_{D,X}:Y\mapsto H_D(X,Y)$ then has the two cyclic blocks
\begin{equation}\label{indep2:cycle-blocks}
 \begin{pmatrix}
 Q_{132}X_{32}&0&P_{132}X_{13}\\
 P_{213}X_{21}&Q_{213}X_{13}&0\\
 0&P_{321}X_{32}&Q_{321}X_{21}
 \end{pmatrix},\qquad
 \begin{pmatrix}
 Q_{123}X_{23}&P_{123}X_{12}&0\\
 0&Q_{231}X_{31}&P_{231}X_{23}\\
 P_{312}X_{31}&0&Q_{312}X_{12}
 \end{pmatrix}.
\end{equation}
Their determinants are respectively
\[
 (-1)^{3d}X_{13}X_{21}X_{32}\Delta(b,a),\qquad
 X_{12}X_{23}X_{31}\Delta(a,b).
\]
The sign follows from simultaneous reversal of every difference:
$P(-b,-a)=(-1)^dP(b,a)$, and similarly for $Q$. Only total degree
$d$ is used, not homogeneity in one scalar difference. If $\Delta$
is nonzero, both oriented factors are simultaneously nonzero on an
open. If $\Delta=0$, both $P,Q$ are nonzero: when only one is present,
the three invertible linear substitutions in its product cannot
make a nonzero polynomial vanish in a domain.

\begin{lemma}[Independent diagonal rectangle operators]
\label{indep2:rectangles-lemma}
Suppose $P,Q\ne0$ and every $K_S(X_0)$ is invertible in size $h$.
Adjoin new independent diagonal entries $z=(z_1,\ldots,z_d)$ and set
$E_i=D_i^{\rm old}-z_iI_h$, $E_S=\prod_{i\in S}E_i$. At
$X=\diag(X_0,0)$ the complete two new $Y$ rectangles are $F(z)y$
and $wG(z)$, where
\begin{equation}\label{indep2:rectangles}
 F(z)=\sum_S c_SK_S(X_0)E_{S^c},\qquad
 G(z)=\sum_S e_S(-1)^{|S|}E_SK_{S^c}(X_0).
\end{equation}
Both determinant polynomials in the independent $z_i$ are nonzero.
\end{lemma}
\begin{proof}
The block formula follows directly from \eqref{indep2:path}. In
particular the $P$ and $Q$ sums act on separate rectangles. The old
hollow operator is unchanged, and diagonal outputs vanish identically.

Assign weights $w_i=2^{i-1}$ to the $z_i$. All subset weight sums are
distinct, and every proper submonomial of a squarefree subset monomial
has smaller weight. Choose $S_*$ with $c_{S_*}\ne0$ maximizing
$\sum_{i\in S_*^c}w_i$. The unique top-weight term of $F$ is
\[
 (-1)^{|S_*^c|}c_{S_*}K_{S_*}(X_0)z_{S_*^c}.
\]
Every other monomial in every entry has strictly smaller weight.
Thus in the determinant the unique top-weight monomial has coefficient
\[
 \det\bigl((-1)^{|S_*^c|}c_{S_*}K_{S_*}(X_0)\bigr)\ne0
\]
and monomial $z_{S_*^c}^{h}$. Any determinant product using a
lower-weight term has smaller total weight and cannot cancel it.
For $G$ select $S_*$ maximizing its subset weight among $e_S\ne0$.
The signs from $(-1)^{|S|}E_S$ cancel in the top coefficient, which is
$e_{S_*}K_{S_*^c}(X_0)$, again invertible. This proves actual
matrix determinant nonvanishing in every fixed size; it makes no
assumption that a formal noncommutative leading word avoids matrix
identities.
\end{proof}

\begin{lemma}[Invertible hollow operators and bilinear detection]
\label{indep2:invertible}
If $\Delta\ne0$, the guarded hollow assertion of
Theorem~\ref{indep2:seed} holds in every $n\ge3$.
\end{lemma}
\begin{proof}
When both symbols are present, the cyclic determinants give the
invertible $L_{D,X}$ locus in size three. Intersect it with all subset
guards in the full affine hollow $X$-space. Given such a size-$h$
point, the old operator and the two rectangles of
\eqref{indep2:rectangles} are independent blocks of the operator at
$X=\diag(X_0,0)$. Lemma~\ref{indep2:rectangles-lemma} permits one
choice of $z$ avoiding both determinant zero sets and every
coordinatewise eigenvalue collision. The new operator is invertible.
Its nonempty open locus in the entire new $V_{h+1}$ intersects every
new subset guard, preparing the next step. No constrained nilpotent
fiber is used in this open intersection.

If $Q=0$, a different proof avoids an absent rectangle. For each
output column $j$, the operator on vertices other than $j$ is the
hollow $(n-1)$-square matrix
\[
 M_j(X)_{ik}=P_{ikj}X_{ik}\quad(i,k\ne j,\ i\ne k).
\]
A derangement of these $n-1\ge2$ vertices contributes a determinant
monomial with nonzero path-weight coefficient. Distinct permutations
give distinct monomials in the independent entries of $X$, so
$\det M_j$ is nonzero. The product over $j$, and with all the subset
guards, is nonzero. For $P=0$ the row version follows by transposition
of the full map, with the two difference arguments interchanged.
Thus it also produces an invertible fixed-$X$ operator, rather than
merely an invertible operator with the other input fixed.

A nonzero coefficient of $\det L_{D,X}$ remains nonzero on a
nonempty diagonal open. At each such fixed $D$ let $U_D\subset V_n$
be its dense invertibility-and-guards open. For $X\in U_D$, choose
any full-support hollow column star $N=we_j^t$ and set
$Y=L_{D,X}^{-1}N$. This is an actual input; with both orders present
$Y$ need not itself be a column star. The $Y$ derivative fills $V_n$,
and the orbit directions have diagonal projections
$w_i(E_{jj}-E_{ii})$, $i\ne j$. Hence this is a smooth nilpotent seed.

If $T$ vanishes at all such seeds, for fixed $X\in U_D$ the linear
functional $N\mapsto T(X,L_{D,X}^{-1}N)$ vanishes on a dense open
of every hollow column space. It vanishes on all those spaces and
therefore on their span $V_n$. Thus $T(X,Y)=0$ for all hollow $Y$;
density in $X$ gives $T=0$. No addition of actual polynomial values
occurs in this linear-functional argument.
\end{proof}


\subsubsection{The zero cycle norm on four vertices}
\label{indep2:zero-cycle-section}

It remains to treat $\Delta=0$. Both symbols are then nonzero,
since a nonzero one-order symbol has nonzero cycle norm. In size four
choose a diagonal tuple with every $D_i$ regular such that every triple-distinct
$P_{ikj}$ and $Q_{ikj}$ is nonzero. Their finite product is a nonzero
polynomial, so these choices form a nonempty open set. We first use
three lower vertices $a,b,c$, reserving a fourth vertex $0$.

\begin{lemma}\label{indep2:zero-cycle-kernels}
Let $A\in V_3$ have all six entries nonzero. The complete lower
operator $L_A:B\mapsto H_{D_{\mathrm{low}}}(A,B)$ has rank four.
Its kernel has a basis supported on the two opposite directed
three-cycles, with each nonzero coordinate a fixed-$D$ nonzero scalar
times the corresponding entry of $A$. Its conormal for the hollow
trace pairing also has a basis of two invertible matrices supported
on opposite directed three-cycles.
\end{lemma}
\begin{proof}
In input coordinates $(B_{ac},B_{ba},B_{cb})$ and output coordinates
$(H_{ab},H_{bc},H_{ca})$ one block is
\begin{equation}\label{indep2:zero-lower-block}
 \begin{pmatrix}
 Q_{acb}A_{cb}&0&P_{acb}A_{ac}\\
 P_{bac}A_{ba}&Q_{bac}A_{ac}&0\\
 0&P_{cba}A_{cb}&Q_{cba}A_{ba}
 \end{pmatrix}.
\end{equation}
The opposite cycle gives the other block, exactly as in
\eqref{indep2:cycle-blocks}. Each determinant is zero by
\eqref{indep2:cycle-norm}, while its upper-left two-by-two minor is
nonzero. Each block has rank exactly two.

For the displayed block an explicit right-kernel matrix $B^{(1)}$ is
\begin{equation}\label{indep2:zero-right-kernel}
 B^{(1)}_{ac}=A_{ac},\qquad
 B^{(1)}_{ba}=-\frac{P_{bac}}{Q_{bac}}A_{ba},\qquad
 B^{(1)}_{cb}=-\frac{Q_{acb}}{P_{acb}}A_{cb},
\end{equation}
with its other entries zero. The first two equations are immediate;
the third is precisely
$P_{acb}P_{bac}P_{cba}+Q_{acb}Q_{bac}Q_{cba}=0$.
Interchanging $b,c$ gives $B^{(2)}$ on the opposite cycle. In
particular the six entries of $A$ remain independent variables;
the coefficients in these two kernel vectors depend only on $D$.

For an equally explicit conormal, a left null vector of
\eqref{indep2:zero-lower-block} is
\[
 \left(1,
 -\frac{Q_{acb}A_{cb}}{P_{bac}A_{ba}},
 \frac{Q_{acb}Q_{bac}A_{ac}}{P_{bac}P_{cba}A_{ba}}\right).
\]
Its three coordinates are nonzero. Transposing its output positions
for the trace pairing gives
\begin{equation}\label{indep2:zero-left-conormal}
 C_1=E_{ba}
 -\frac{Q_{acb}A_{cb}}{P_{bac}A_{ba}}E_{cb}
 +\frac{Q_{acb}Q_{bac}A_{ac}}{P_{bac}P_{cba}A_{ba}}E_{ac}.
\end{equation}
It satisfies $\tr(C_1L_A(B))=0$ for every hollow $B$, and is an
invertible directed-cycle matrix. Interchanging $b,c$ gives $C_2$.
Their supports are disjoint and opposite. They form a basis of the
two-dimensional conormal of the full rank-four lower operator.
\end{proof}

\subsubsection{A full size-four star differential}
\label{indep2:zero-four-star-section}

\begin{lemma}\label{indep2:zero-four-star}
For every $D$ in the size-four open just chosen, a nonempty open
subset of the entire column-star parameter space
\[
 X=\begin{pmatrix}0&0\\0&A\end{pmatrix},\qquad
 Y=ve_0^t,\qquad (A,v)\in V_3\times K^3,
\]
consists of nonzero rank-one square-zero seeds for which the fixed-$D$
derivative $dH_D$ is onto $V_4$. The same assertion holds for row stars.
The lower block $A$ can simultaneously satisfy all its
subset adjoint guards.
\end{lemma}
\begin{proof}
At the displayed input,
\begin{equation}\label{indep2:zero-four-value}
 H_D(X,Y)=N=we_0^t,\qquad w=Fv,\qquad
 F_{ij}=P_{ij0}A_{ij}\quad(i,j\in\{a,b,c\}),
\end{equation}
where the diagonal of $F$ is zero. The upper-row $Y$ derivative is
$r\mapsto rG$, with
\[
 G_{ij}=Q_{0ij}A_{ij}\quad(i,j\in\{a,b,c\}),\qquad G_{ii}=0.
\]
The determinants of $F$ and $G$ each have the two distinct cycle
monomials $A_{ab}A_{bc}A_{ca}$ and $A_{ac}A_{cb}A_{ba}$ with nonzero
coefficients. Each determinant is therefore a nonzero polynomial in
the same six original entries of $A$. The determinants
$\det K_S(A)$ for all $S\subseteq[d]$ are likewise nonzero polynomials,
as is seen on a single directed-cycle support. Choose one $A$ where
all six entries, both rectangle determinants, and all those lower
guards are nonzero. This is an intersection of opens in the full
irreducible space $V_3$.

The complete fixed-$D$ hollow $Y$ derivative is the direct sum of
these two invertible three-dimensional rectangles and the rank-four
operator $L_A$. Its rank is $3+3+4=10$. Its two-dimensional hollow
conormal is precisely the span of $C_1,C_2$ from
Lemma~\ref{indep2:zero-cycle-kernels}, embedded in the lower block.

Now vary the three entries of the raw upper row of $X$ by
$a=(a_j)_{j\in\{a,b,c\}}$. Its only possibly nonzero output is the
lower hollow block
\begin{equation}\label{indep2:zero-four-upper-variation}
 (dH_D)_{ij}=Q_{i0j}v_i a_j\quad(i\ne j).
\end{equation}
The diagonal entries are zero by the endpoint hypothesis. For a lower
conormal $C$, its pairing with this output is
\[
 \tr(C\,dH_D)=\sum_j a_j(M_Cv)_j,\qquad
 (M_C)_{ji}=C_{ji}Q_{i0j},\quad (M_C)_{jj}=0.
\]
Both $M_{C_1}$ and $M_{C_2}$ are still invertible directed-cycle
matrices: every multiplier on their supports is nonzero. Their
opposite disjoint supports prevent proportionality. It follows that
$M_{C_1}v,M_{C_2}v$ are independent on a nonempty open set of $v$.
Indeed, dependence for every $v$ would make every vector an eigenvector
of $M_{C_1}^{-1}M_{C_2}$, forcing that operator to be scalar, contrary
to the supports.

Choose $v$ in that open with both $v$ and $Fv$ of full support. These
conditions are simultaneously possible because $F$ is invertible
and the vector space is irreducible. The three actual upper-row
$X$ directions then fill the two-dimensional quotient of $V_4$ by
the $Y$ image. Thus the full fixed-$D$ hollow derivative has rank
$12$. The value $N$ in \eqref{indep2:zero-four-value} is nonzero rank
one and square-zero, with all three complementary column entries
nonzero. The orbit directions $E_{0i}$ have diagonal projections
$w_i(E_{00}-E_{ii})$, giving the remaining three independent directions
and rank $15$ for $d\Gamma_D$.

All the conditions used are open in the whole star parameter space,
and we have shown their simultaneous nonemptiness. To obtain row
stars for the same original symbols use
\begin{equation}\label{indep2:two-order-transpose}
 H_{P,Q}(D,X^t,Y^t)
 =(-1)^dH_{Q^{\mathrm{sw}},P^{\mathrm{sw}}}(D,X,Y)^t,
 \qquad R^{\mathrm{sw}}(a,b)=R(b,a).
\end{equation}
The transformed pair has zero endpoints and zero cycle norm, since
its norm is $\Delta(b,a)$. Its triple-distinct weights are
nonzero under our same conditions on $D$, by reversal of the path
and homogeneity. The preceding column proof therefore applies to it.
Transpose preserves all ranks and the subset adjoint guards, since
$K_S(X^t)=(-1)^{|S|}K_S(X)^t$.
\end{proof}

\subsubsection{Restoring guards in the exact raw fiber}
\label{indep2:zero-raw-fiber-section}

\begin{lemma}\label{indep2:zero-raw-restoration}
At any column star in the nonempty open constructed in
Lemma~\ref{indep2:zero-four-star}, the exact raw fiber $H_D=N$ is
smooth, and its local ring contains a nonzero product of all raw
$X$ guard determinants. Every function whose germ on this fiber is
nonzero can be retained at an actual guarded smooth nilpotent seed
in this same fiber. The corresponding assertion holds at row stars.
\end{lemma}
\begin{proof}
The surjectivity of the fixed-$D$ hollow differential makes
\[
 \mathcal F_N=\{(X',Y')\in V_4\times V_4:H_D(X',Y')=N\}
\]
smooth at the star $p$. Its local ring $R=\mathcal O_{\mathcal F_N,p}$
is a regular local domain. The map from the three raw upper-row
$X$ variations to the two-dimensional quotient by the $Y$ image has
rank two. Choose a nonzero row $a$ in its one-dimensional kernel.
Its complete output lies in the $Y$ image and can therefore be
cancelled by an actual hollow $Y$ variation. Independently, every raw
lower-column $X$ variation $z$ has zero output: both products with
the current $Y=ve_0^t$ vanish. Consequently, for every $z$, the raw
variation
\[
 \dot X=\begin{pmatrix}0&a\\z&0\end{pmatrix}
\]
can be completed by a hollow $\dot Y$ to a tangent vector of the
exact fiber $\mathcal F_N$. No output conjugation is involved.

Put $E_i=D_i^{\mathrm{low}}-(D_i)_{00}I_3$,
$E_S=\prod_{i\in S}E_i$ and $A_S=K_S(A)$, with the latter
computed using the lower diagonal tuple. All $E_i,A_S$ are invertible.
Smoothness lifts any such tangent to a formal arc in $\mathcal F_N$.
Along it the raw $X$ remains hollow, its upper and lower cross
blocks start with $ta$ and $tz$, and its lower block starts with $A$.
The Schur complement gives, for every subset $S\subseteq[d]$,
\begin{equation}\label{indep2:zero-raw-guard-coefficient}
 \det K_S(X(t))
 =(-1)^{|S|+1}t^2\det(A_S)\,
       aE_SA_S^{-1}E_Sz+O(t^3).
\end{equation}
In particular second-order changes in the lower block or the cross
blocks cannot alter the displayed coefficient; the upper-left raw
entry is identically zero, even for $S=\varnothing$. For each $S$ the row
$aE_SA_S^{-1}E_S$ is nonzero. Choose one $z$ outside the finitely many
annihilating hyperplanes. Every determinant in
\eqref{indep2:zero-raw-guard-coefficient} is then nonzero on the
formal arc. Their product is consequently a nonzero element of $R$.
The fixed-$D$ and saturated smoothness minors already nonzero at $p$
are units in $R$.

Let $b\in R$ be any nonzero function germ. The product of $b$, the
guard product, and these minors is nonzero because $R$ is a domain.
On a sufficiently small finite-type neighborhood in the component
through $p$, its nonvanishing defines a nonempty open set. As $K$ is
algebraically closed, this open set has an actual $K$-point. At that
point the raw value is still exactly $N$, all guards hold, the seed
is smooth, and $b$ is nonzero. This proves the stated retention
property, not merely a formal existence claim. The transpose
identity \eqref{indep2:two-order-transpose} transports the entire
raw-fiber argument to row stars.
\end{proof}

\begin{corollary}\label{indep2:zero-raw-germ}
If a bilinear form $T$ on $V_4\times V_4$ vanishes at every guarded
smooth nilpotent seed, its entire germ on the exact raw fiber at
every star in Lemma~\ref{indep2:zero-four-star} is zero.
\end{corollary}
\begin{proof}
A nonzero germ could be retained at a guarded smooth seed in the same
raw fiber by Lemma~\ref{indep2:zero-raw-restoration}, contradicting
the hypothesis on $T$.
\end{proof}

\subsubsection{Six independent monomials and the guarded base}
\label{indep2:zero-four-span-section}

\begin{lemma}\label{indep2:zero-four-span}
For each $D$ in the above size-four diagonal open, the tensors of
guarded smooth nonzero rank-one square-zero seeds span
$V_4\otimes V_4$.
\end{lemma}
\begin{proof}
Suppose a bilinear form $T$ annihilates all such tensors. By
Corollary~\ref{indep2:zero-raw-germ} it vanishes at every generic
column or row star. Polynomial continuation on the whole irreducible
star parameter space makes its restriction identically zero there.

First consider two hollow units $E_{ab},E_{cd}$ with
$\{a,b\}\ne\{c,d\}$. The second edge has an endpoint outside
$\{a,b\}$. If $d$ is such an endpoint, use the column star isolated at
$d$: $E_{ab}$ belongs to its lower block and $E_{cd}$ to its column.
If $c$ is such an endpoint, use the row star isolated at $c$ instead.
Thus
\begin{equation}\label{indep2:zero-different-edges}
 T(E_{ab},E_{cd})=0\qquad\text{if }\{a,b\}\ne\{c,d\}.
\end{equation}
This eliminates four-distinct-vertex coefficients as well, without
an embedding assertion from size three. No relation between the
remaining same-edge and reversed-edge coefficients has been assumed.

Fix now an isolated vertex $0$ and a generic lower block $A$. Choose
a $v$ making its column star one of the smooth points above. For
either of the explicit lower kernel matrices $B^{(\nu)}$ in
\eqref{indep2:zero-right-kernel}, the line
\begin{equation}\label{indep2:zero-kernel-line}
 (X,Y_t)=\left(
 \begin{pmatrix}0&0\\0&A\end{pmatrix},
 ve_0^t+t\begin{pmatrix}0&0\\0&B^{(\nu)}\end{pmatrix}\right)
\end{equation}
lies in the exact same raw fiber: the additional full output is
$tL_A(B^{(\nu)})=0$. The zero germ from
Corollary~\ref{indep2:zero-raw-germ}, restricted to this line, gives
\[
 T\left(\begin{pmatrix}0&0\\0&A\end{pmatrix},
         \begin{pmatrix}0&0\\0&B^{(\nu)}\end{pmatrix}\right)=0.
\]
Let $\mathcal C$ be the directed three-cycle supporting one such
kernel vector, and write $B_e=\gamma_eA_e$ on it. Each
$\gamma_e\in K^*$ depends only on $D$, and the cycle contains exactly
one orientation of each unordered lower edge. Using
\eqref{indep2:zero-different-edges}, the preceding identity is
\begin{equation}\label{indep2:zero-six-monomials}
 \sum_{e\in\mathcal C}\gamma_e
 \left(T(E_e,E_e)A_e^2+
       T(E_{\bar e},E_e)A_{\bar e}A_e\right)=0,
\end{equation}
where $\bar e$ denotes the reverse edge. This identity holds for a
nonempty open set of the six independent entries of $A$, hence is a
polynomial identity in them. Its three square monomials and three
unordered-edge product monomials are six distinct monomials.
Every coefficient displayed in \eqref{indep2:zero-six-monomials}
therefore vanishes. The opposite kernel cycle kills the other three
square coefficients and the opposite three ordered reverse-edge
coefficients on this lower triangle.

Varying the isolated vertex covers every unordered edge of the
four-vertex set. All coefficients left after
\eqref{indep2:zero-different-edges} consequently vanish, and $T=0$.
The two kernel shifts were taken in a fiber of the full operator,
at a common original input, rather than in fibers of its summands.
\end{proof}



\subsection{Exact guarded extension and one common diagonal open}

\begin{lemma}[Singleton extension in the exact saturated fiber]
\label{indep2:extension}
Suppose $P,Q\ne0$ and a hollow smooth nilpotent seed in size $h\ge4$
satisfies every subset guard. For a generic new diagonal vector
$z=(z_1,\ldots,z_d)$ it extends to a guarded smooth nilpotent seed
in size $h+1$. Any raw function nonzero at the provisional input in
the proof can be retained while restoring the guards.
\end{lemma}
\begin{proof}
Write $H_{D^{\rm old}}(X_0,Y_0)=N_0=u_0v_0^t$, with
$v_0^tu_0=0$, and provisionally take $X=\diag(X_0,0)$. By
Lemma~\ref{indep2:rectangles-lemma}, both $F(z),G(z)$ are invertible
for a generic $z$, simultaneously with every $E_i$ invertible and
all diagonal entries distinct in each passive coordinate. Put
\begin{equation}\label{indep2:provisional-input}
 Y=\begin{pmatrix}Y_0&0\\\lambda v_0^tG(z)^{-1}&0\end{pmatrix},
 \qquad\lambda\ne0.
\end{equation}
The complete raw value, with no discarded cross terms, is
\begin{equation}\label{indep2:extended-value}
 N_1=\begin{pmatrix}u_0v_0^t&0\\\lambda v_0^t&0\end{pmatrix}
     =\binom{u_0}{\lambda}(v_0^t,0),
 \qquad \rank N_1=1,\quad N_1^2=0.
\end{equation}
The old saturated controls produce the old $\slm_h$ block. Any
extra cross components they create are corrected with the two
invertible $Y$ rectangles, without changing that old block. These
rectangles themselves supply the two full cross blocks. A genuine
orbit direction $Z=\left(\begin{smallmatrix}0&e\\0&0\end{smallmatrix}\right)$
has
\[
 [Z,N_1]=\begin{pmatrix}
 \lambda e v_0^t&-N_0e\\0&-\lambda v_0^te
 \end{pmatrix}.
\]
Choosing $v_0^te\ne0$ gives the last trace-balanced diagonal
direction. The saturated derivative is therefore onto
$\slm_{h+1}$.

These controls can be taken in the old raw variables, the new $Y$
rectangles and the auxiliary orbit variables. None changes the new
raw cross blocks of $X$. Prescribe an arbitrary column $a$ and row
$b$ for those two cross variations. Their output can be cancelled
by the preceding controls, producing a tangent of the exact
saturated fiber $\Gamma_D=N_1$ with these prescribed raw $X$
components.

That fiber is smooth at the provisional point. Lift the tangent to
a formal arc and keep raw $X$ hollow. The new diagonal raw entry is
identically zero, including for the empty-subset guard. With
$A_S=K_S(X_0)$ and $E_S=\prod_{i\in S}(D_i^{\rm old}-z_iI_h)$,
the Schur complement gives
\begin{equation}\label{indep2:extension-Schur}
 \det K_S(X(t))
 =(-1)^{|S|+1}t^2\det(A_S)\,
       bE_SA_S^{-1}E_Sa+O(t^3).
\end{equation}
The coefficient is a nonzero bilinear form in $(a,b)$, since
$E_S,A_S$ are invertible. Fix any $a\ne0$ and choose $b$ outside
the finitely many annihilating hyperplanes. All coefficients are
then nonzero. Unknown second-order corrections cannot cancel them:
the upper-left old block begins at $A_S$, both cross blocks begin
in order one, and the new scalar diagonal is identically zero.

The smooth local fiber ring is a regular local domain. The arc
shows that its product of guards is nonzero. Multiply it by a
surjectivity minor and any raw function nonzero at the provisional
point; those latter functions are units locally. A finite-type
localization of the component through the point is nonempty and
has an actual point over the algebraically closed field. This is
not an assertion that a chosen formal arc algebraizes. Discarding
the auxiliary conjugator leaves a raw value conjugate to $N_1$,
so rank one and square zero persist, while the raw guards and test
have been retained. In contrast to the size-four restoration, this
argument fixes the saturated value, not necessarily the raw value.
\end{proof}

\begin{lemma}[Tensor induction and fixed-diagonal descent]
\label{indep2:induction}
Guarded hollow tensor spanning in size four at generic independent
diagonal parameters implies the same assertion in every $n\ge4$,
on one nonempty open of diagonal tuples independent of the test.
\end{lemma}
\begin{proof}
Work first over an algebraic closure of the field of all independent
diagonal entries. On an old $h=n-1$ vertex set choose finitely many
guarded seed tensors spanning $V_h\otimes V_h$ over the algebraic
closure of the old diagonal function field. The new $d$ entries
$z_i$ remain algebraically independent over that field and over the
finitely many algebraic coordinates of these seeds. Hence the two
nonzero determinant polynomials from
Lemma~\ref{indep2:rectangles-lemma} stay nonzero, and
Lemma~\ref{indep2:extension} applies to each seed. The spanning
property is preserved by scalar extension.

Let $T$ vanish on all guarded good pairs in the new size. It must
vanish at each provisional input: otherwise exact-fiber guard
restoration retains a nonzero value of $T$. At
\eqref{indep2:provisional-input} its value is
\[
 T_h(X_0,Y_0)+\lambda T\!\left(\diag(X_0,0),
 \begin{pmatrix}0&0\\v_0^tG(z)^{-1}&0\end{pmatrix}\right).
\]
It vanishes for every nonzero $\lambda$, so its constant term
vanishes. The old spanning collection gives $T_h=0$. This works on
every $(n-1)$-vertex subset. For $n\ge5$, any two hollow matrix
units use at most four vertices and fit in such a subset. Every
coefficient of $T$ is therefore zero, closing the induction from
the guarded size-four base.

To obtain the claimed numerical open, choose finitely many generic
seed tensors forming a basis and put their coordinates into one
finite-type incidence scheme. Its conditions consist of the raw
rank-one square-zero equations, a nonzero raw entry for each value,
the chosen saturated surjectivity minors, the tensor-basis
determinant, and all subset guards. The projection to the
independent diagonal tuples is dominant, because the construction
has a point over the algebraic closure of their function field.
Its constructible image contains a nonempty open. Intersect with
the coordinatewise regular-diagonal open. Every nonempty finite-type
fiber over a $K$-point has a $K$-point, since $K$ is algebraically
closed. Thus every $D$ in this one open supports a spanning
collection simultaneously. The diagonal tuple is not reselected
separately for different bilinear forms or tensor coordinates.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{indep2:seed}]
Lemma~\ref{indep2:three} handles size three. If $\Delta\ne0$,
Lemma~\ref{indep2:invertible} gives all sizes, including the guards.
If $\Delta=0$, both symbols are present. The full four-vertex
construction in Lemma~\ref{indep2:zero-four-span} gives the guarded
base, and Lemma~\ref{indep2:induction} gives every larger size and
the one common diagonal open. These cases exhaust every nonzero
symbol pair satisfying the zero-endpoint identities.

It remains to pass from hollow to full bilinear tests. Arbitrary
independent diagonal parts can be added to $X,Y$ without changing
$H_D$, by the nonempty proper subset conditions in
\eqref{indep2:symbols}. The hollow differential and its actual
simultaneous-conjugation controls persist; the added diagonal parts
are conjugated along with the other original inputs.
Suppose a bilinear form on $M_n\times M_n$ vanishes at every such
enriched seed. Expanding in the two independent added diagonal
matrices first kills its diagonal-by-diagonal part. Its two mixed
parts vanish on the projections of the hollow seed set. Each
projection spans $V_n$: otherwise a nonzero linear functional
annihilating that projection, tensored with any nonzero functional
on the other factor, would contradict hollow tensor spanning.
Thus both mixed parts vanish. The hollow-by-hollow part vanishes
by the already established tensor span. The full form is zero,
proving the final assertion.
\end{proof}

If one also requires the full enriched input to retain the guards,
the nonempty-subset guards are unchanged. The empty-subset guard
$\det X$ may change after a diagonal addition; restrict those additions
to the open set where it stays nonzero. This open contains the zero
addition at a guarded hollow seed, so the same polynomial-continuation
argument applies. The trace transfer below only needs the enriched
smooth nilpotent seed, not its guards.




\subsection{An original-input chart along the commuting-passive locus}
\label{indep2:chart-section}

The independent passive matrices cannot be replaced by the linear
common-$D$ coordinates used in Theorem~\ref{complete2:main}. In
particular, a nonzero polynomial on the original source need not
remain nonzero on the locus of commuting passive tuples. We retain
all transverse directions by the following chart. Write
$\mathcal D_n$ for the vector space of diagonal matrices and
$V_n$ for the hollow matrices. As above, $d=m-2$.

\begin{lemma}[A full original-input chart]
\label{indep2:chart}
Let $D=(D_1,\ldots,D_d)\in\mathcal D_n^d$, with $D_1$ having
pairwise distinct diagonal entries. Take
\[
 A\in V_n,\qquad Z_i\in V_n\quad(2\le i\le d),
 \qquad X,Y\in M_n,
\]
and restrict to the open set where $g=I_n+A$ is invertible. The map
\begin{equation}\label{indep2:chart-map}
 \begin{split}
 \Psi(D,A,X,Y,Z)=\big(&gD_1g^{-1},\,
              g(D_2+Z_2)g^{-1},\ldots,g(D_d+Z_d)g^{-1},\\
              &gXg^{-1},\,gYg^{-1}\big)
 \end{split}
\end{equation}
puts these entries in their specified original positions. Its source
has dimension $mn^2$, and $\Psi$ is \'etale at every point
with $A=0$ and $Z=0$. In particular, it is dominant onto $M_n^m$.

The subspace $S=\{Z=0\}$ gives a local \'etale chart on the
simultaneously conjugated commuting-passive locus with regular
semisimple first passive input. Its dimension and codimension are
\[
 \dim S=3n^2+(d-1)n,
 \qquad \operatorname{codim}S=(d-1)(n^2-n).
\]
If $d\Gamma_D$ is onto $\slm_n$ at $(X,Y,I_n)$, the restriction of
$d(f\circ\Psi)$ to $T_{(D,0,X,Y,0)}S$ is also onto $\slm_n$.
\end{lemma}
\begin{proof}
The number of coordinates in the source of \eqref{indep2:chart-map}
is
\[
 dn+(n^2-n)+(d-1)(n^2-n)+2n^2=(d+2)n^2=mn^2.
\]
At a point $(D,0,X,Y,0)$ its derivative is
\begin{equation}\label{indep2:chart-differential}
 \begin{split}
 \delta B_1&=\delta D_1+[\delta A,D_1],\\
 \delta B_i&=\delta D_i+\delta Z_i+[\delta A,D_i]
                       \quad(2\le i\le d),\\
 \delta B_X&=\delta X+[\delta A,X],\qquad
 \delta B_Y=\delta Y+[\delta A,Y].
 \end{split}
\end{equation}
Given an arbitrary tangent vector $(\delta B_1,\ldots,
\delta B_d,\delta B_X,\delta B_Y)$ to the original source, the
diagonal of $\delta B_1$ determines $\delta D_1$. For $a\ne b$,
\[
 [\delta A,D_1]_{ab}
   =\big((D_1)_{bb}-(D_1)_{aa}\big)\delta A_{ab}.
\]
The nonzero eigenvalue differences therefore determine the hollow
matrix $\delta A$ uniquely. For each $i\ge2$, subtract
$[\delta A,D_i]$ from $\delta B_i$ and take its diagonal and hollow
parts; these determine $\delta D_i$ and $\delta Z_i$ uniquely.
Finally, the last line of \eqref{indep2:chart-differential}
determines the unrestricted $\delta X,\delta Y$. Thus the derivative
is an isomorphism between smooth spaces of equal dimension, proving
\'etaleness at the stated point.

The source is a nonempty open subset of an affine space, hence is
irreducible. The \'etale locus is open and contains the stated
points, and an \'etale map is open. Consequently the image of
$\Psi$ contains a nonempty open subset of the irreducible original
source. This proves dominance, and in particular injectivity of
pullback on its ring of polynomial functions.

For completeness, the commuting-passive locus used here is smooth.
A regular semisimple matrix $B_1$ has centralizer $K[B_1]$, of
dimension $n$. Every matrix commuting with it has a unique expression
\[
 B_i=\sum_{r=0}^{n-1}b_{ir}B_1^r.
\]
These expressions identify its centralizer bundle with the trivial
rank-$n$ bundle over the regular semisimple locus: the displayed
powers are linearly independent in every fiber, and the inverse
coefficients are regular locally, by an invertible matrix-entry
minor. Choosing $B_2,\ldots,B_d$ in this bundle already makes them
commute with one another. The passive locus thus has dimension
$n^2+(d-1)n$; the two full active matrices add $2n^2$ dimensions.
The image of $S$ lies in this locus. Its tangent map is injective by
\eqref{indep2:chart-differential}, and its dimension equals that of
the locus. This proves the asserted local chart statement and the
codimension formula.

It remains to compare the allowed conjugation directions. By the
actual restriction formula \eqref{indep2:path}, on $S$ one has
\[
 (f\circ\Psi)(D,A,X,Y,0)=gH_D(X,Y)g^{-1}.
\]
The coordinate $A$ provides every hollow infinitesimal conjugator.
If $B$ is diagonal, then $[B,D_i]=0$ for every $i$, and simultaneous
conjugation by $I_n+tB$ to first order has exactly the same input
tangent as
\[
 \delta X=[B,X],\qquad \delta Y=[B,Y],
 \qquad \delta A=0,\quad\delta D_i=0,\quad\delta Z_i=0.
\]
Both active variations are allowed because $X,Y$ are full matrices.
Equivariance gives the output derivative $[B,H_D(X,Y)]$.
Decomposing an arbitrary infinitesimal conjugator into its hollow
and diagonal parts proves that every direction supplied by
$d\Gamma_D$ is supplied in $T S$. No diagonal-torus direction has
been lost by taking $A$ hollow.
\end{proof}

\subsection{The first actual trace term on the same exact fiber}
\label{indep2:trace-section}

The next argument uses the complete matrix-entry trace polynomial
at the fixed size $n$. It does not substitute a free cyclic class
for that polynomial, and does not require a trace to appear at a
prescribed small order.

\begin{proposition}[Trace transfer in the independent chart]
\label{indep2:trace-transfer}
Suppose the complete selected restriction of the original ordinary
multilinear polynomial $f$ is \eqref{indep2:path}, and its nonzero
symbol pair satisfies the zero-endpoint hypotheses of
Theorem~\ref{indep2:seed}. Fix $n\ge3$ and put $h=\tr f$ on $M_n^m$.
If $h=0$, the original map into $\slm_n$ has a smooth input over a
nonzero rank-one square-zero matrix, with local fiber dimension
$mn^2-n^2+1$, and its actual image is $\slm_n$.

If $h\ne0$, the original map into $M_n$ has a flat input over such
a matrix, with local complete-intersection fiber of dimension
$mn^2-n^2$, and its actual image is $M_n$. More precisely, the
construction has a first nonzero normal trace order
$1\le v\le d-1$. Its differential has rank $n^2$ when $v=1$ and
rank exactly $n^2-1$ when $v\ge2$. Thus in the latter case the
constructed point is flat but not smooth.
\end{proposition}
\begin{proof}
Let $\Omega_n\subset\mathcal D_n^d$ be the single nonempty open
set supplied by Theorem~\ref{indep2:seed}; it already requires
regular diagonal passive matrices. Put
\[
 \pi(B)=B-\frac{\tr B}{n}I_n,\qquad
 F=f\circ\Psi,\qquad G=\pi\circ F.
\]
Trace invariance cancels $g$ from the pulled-back trace:
\begin{equation}\label{indep2:trace-pullback}
 (h\circ\Psi)(D,A,X,Y,Z)
   =\widetilde h(D,X,Y,Z)
   =\tr f(D_1,D_2+Z_2,\ldots,D_d+Z_d,X,Y),
\end{equation}
with the same original slot order as in
\eqref{indep2:chart-map}. In particular, this is a polynomial in
$D,X,Y,Z$ and is independent of $A$. The zero endpoints in
\eqref{indep2:path} make $H_D(X,Y)$ hollow for all full $X,Y$.
Consequently
\[
 \widetilde h(D,X,Y,0)=0,
 \qquad F|_S=gH_D(X,Y)g^{-1}.
\]

If $h=0$ on the original source, choose any full seed from
Theorem~\ref{indep2:seed}. For example, its assertion applied to
any nonzero bilinear form ensures that the seed set is nonempty.
At the corresponding point $p=(D,0,X,Y,0)$, the value $N=F(p)$ is
nonzero rank-one square-zero. Lemma~\ref{indep2:chart} makes
$d(F|_S)_p$ onto $\slm_n$. Since $\Psi$ is \'etale at $p$, the
original differential is onto $\slm_n$ as well. The original map is
therefore smooth there, and its fiber has dimension
$mn^2-(n^2-1)$. Smoothness gives an open image neighborhood of $N$;
Theorem~\ref{local} gives the asserted actual image.

Now assume that the actual polynomial $h$ is nonzero. Dominance of
$\Psi$, proved in Lemma~\ref{indep2:chart}, implies
$\widetilde h\ne0$. Expand by total degree in the independent hollow
coordinates $Z_2,\ldots,Z_d$:
\begin{equation}\label{indep2:normal-expansion}
 \widetilde h(D,X,Y,Z)
    =\sum_{j=v}^{d-1}H_j(D,X,Y,Z),
 \qquad H_v\ne0,\qquad 1\le v\le d-1.
\end{equation}
Every $H_j$ is homogeneous of degree $j$ in the $Z$ coordinates
and bilinear in the \emph{full} active matrices $X,Y$.
Indeed, each original passive position other than the first
contributes either its $D_i$ or its $Z_i$, at most once, and each
active position occurs exactly once. The term of degree zero
vanishes by the preceding hollow restriction. The bound is
$d-1=m-3$ because there are precisely that many transverse passive
positions. All coefficients below order $v$ vanish as polynomials
in actual size-$n$ entries. This includes every fixed-size trace
identity and every cancellation involving higher-ideal terms of
$f$.

There are $D^*\in\Omega_n$ and fixed hollow matrices
$Z^0=(Z_2^0,\ldots,Z_d^0)$ such that
\begin{equation}\label{indep2:trace-detector}
 T(X,Y)=H_v(D^*,X,Y,Z^0)
\end{equation}
is a nonzero bilinear form on $M_n\times M_n$.
Here is a coefficient-level justification of this simultaneous
choice. Expand $H_v$ in the matrix-entry variables of $X,Y,Z$.
At least one coefficient is a nonzero polynomial in the independent
diagonal coordinates of $D$. Its nonvanishing open set meets the
nonempty open $\Omega_n$ in the irreducible affine space
$\mathcal D_n^d$. At such a $D^*$ the polynomial in $X,Y,Z$ remains
nonzero. Some coefficient of an $X$-entry times a $Y$-entry is then
a nonzero polynomial in $Z$. Since $K$ is infinite, choose $Z^0$
where this coefficient does not vanish. This proves the assertion
without a density claim for commuting tuples in the original
source.

Apply Theorem~\ref{indep2:seed} to this $T$ at this \emph{same}
$D^*$. It gives full original matrices $X_0,Y_0$ for which
\[
 T(X_0,Y_0)\ne0,
 \qquad N=H_{D^*}(X_0,Y_0),
 \qquad \rank N=1,
 \qquad N^2=0,
\]
and the saturated differential is onto $\slm_n$.
Set $p=(D^*,0,X_0,Y_0,0)$ in the chart and $a=\Psi(p)$ in the
original source. The two properties required for the trace
argument now hold at one input: $G(p)=N$, and
\[
 d(G|_S)_p:T_pS\longrightarrow\slm_n
 \quad\hbox{is onto},\qquad
 H_v(D^*,X_0,Y_0,Z)\ne0.
\]
The second assertion follows by evaluation at $Z^0$; the first is
Lemma~\ref{indep2:chart}. Nothing here adds values from different
input tuples.

We next solve the exact traceless equations without changing the
normal coordinates. Write $q=mn^2$ and $r=n^2-1$.
Choose an $r$-dimensional linear subspace $L\subset T_pS$ on which
$dG_p$ is an isomorphism onto $\slm_n$. In the affine coordinates
of the chart centered at $p$, choose coordinates $\ell$ along
$L$ and complementary coordinates $w$ within $S$, leaving every
coordinate of $Z$ untouched. Thus
\[
 (\ell,w,Z)\quad\hbox{are local coordinates},\qquad
 \#\ell=r,
 \qquad\#(w,Z)=q-r,
 \qquad S=\{Z=0\}.
\]
Choose any linear coordinates on $\slm_n$. The Jacobian of the
$r$ functions $G-N$ with respect to $\ell$ is invertible at $p$.
The formal implicit function theorem therefore gives a unique
formal solution
\[
 \ell=\phi(w,Z),\qquad \phi(0,0)=0,
 \qquad G(\phi(w,Z),w,Z)=N.
\]
In particular, for the exact partial fiber $\mathcal Y=G^{-1}(N)$,
\begin{equation}\label{indep2:partial-completion}
 \widehat{\mathcal O}_{\mathcal Y,p}
       \simeq K[[w,Z]].
\end{equation}
The correction $\phi$ may have a linear term. That causes no
problem: its entries have zero constant term, and no $Z$
coordinate is replaced or eliminated.

We verify the initial trace term after this correction explicitly.
Let $\mathfrak m$ be the maximal ideal in the completed chart
at $p$. Equation~\eqref{indep2:normal-expansion} gives
\begin{equation}\label{indep2:trace-initial}
 \widetilde h
    \equiv H_v(D^*,X_0,Y_0,Z)
                    \pmod{\mathfrak m^{v+1}},
 \qquad \widetilde h\in\mathfrak m^v.
\end{equation}
To prove the congruence, every monomial in $H_v$ already has
$Z$-degree $v$. Replacing a coefficient variable among $D,X,Y$
by its value at $p$ changes that coefficient by an element of
$\mathfrak m$, raising the total order by at least one. The terms
$H_j$ with $j>v$ have total order at least $v+1$ before any such
replacement. The polynomial is independent of $A$, so that
coordinate introduces no further term. Upon substituting
$\ell=\phi(w,Z)$, every centered coefficient variable remains in
$(w,Z)$ and the normal coordinates stay exactly $Z$. Hence
\eqref{indep2:trace-initial} becomes
\begin{equation}\label{indep2:trace-on-fiber}
 \widetilde h|_{\mathcal Y}
   \equiv H_v(D^*,X_0,Y_0,Z)
                   \pmod{(w,Z)^{v+1}}
       \quad\hbox{in }K[[w,Z]].
\end{equation}
Its right-hand side is nonzero. Thus the first surviving normal
trace term is also the full lowest total Taylor term on the
\emph{exact} partial fiber; an implicit correction cannot cancel it.

The surjective differential of $G$ makes
$R=\mathcal O_{\mathcal Y,p}$ a regular local ring of dimension
$q-r=mn^2-n^2+1$. In particular, $R$ is a domain. The completion
map is injective (equivalently here, completion is faithfully flat
and the maximal-ideal filtration is separated).
Equation~\eqref{indep2:trace-on-fiber} shows that
$\widetilde h|_{\mathcal Y}$ is a nonzero element of $R$.
It lies in the maximal ideal because $F(p)=N$ has trace zero,
and, since $R$ is a domain, it is a non-zero-divisor. The full
fiber over $N$ is obtained by adding this one trace equation to
$G=N$, so
\[
 \dim\mathcal O_{F^{-1}(N),p}
      =(q-r)-1=mn^2-n^2.
\]
Moreover, the $r$ traceless equations form a regular sequence in
the regular local ring of the chart: their independent linear
parts can be completed to a regular system of parameters.
Appending the non-zero-divisor $\widetilde h$ gives a regular
sequence of length $n^2$. The full fiber is therefore a local
complete intersection at $p$.

The target $M_n$ and the chart source are regular, and the latter
is Cohen--Macaulay. The displayed fiber dimension is the difference
of their dimensions, so miracle flatness proves that $F$ is flat
at $p$. Since $\Psi$ is \'etale at $p$, the original fiber at
$a=\Psi(p)$ has the same local dimension and is a local complete
intersection. Flatness descends through this local faithfully flat
change of source; equivalently, miracle flatness applies directly
to the original regular local source of dimension $mn^2$ and its
fiber of dimension $mn^2-n^2$. Thus the asserted point belongs to
the original map $f$, rather than just to a restriction or a
quotient representative.

Flatness for morphisms of finite presentation is open on the
source, and a flat morphism locally of finite presentation is open.
It follows that the actual image contains an open neighborhood of
$N$ in $M_n$. Indeed, each $K$-point in this open set has a nonempty
finite-type fiber, which has a $K$-point because $K$ is algebraically
closed. The degeneration in Theorem~\ref{local} now yields every nonscalar target.
The scalar part of that theorem applies to the \emph{original
ordinary multilinear} polynomial $f$; we spell it out to keep its
scope explicit.

Choose Weyl matrices $U,V\in M_n$ with $UV=\zeta VU$ and
$U^n=V^n=I_n$, where $\zeta$ is a primitive $n$th root of unity.
The matrices $U^aV^b$, $0\le a,b<n$, form a basis, and their
traces are zero except for $U^0V^0=I_n$. Since $\tr f\ne0$, some
original input tuple has nonzero output trace. Expanding that
tuple multilinearly in the Weyl basis shows that a tuple of basis
matrices has nonzero output trace. Every word of $f$ uses exactly
the same original inputs once, so at this basis tuple every word
is a scalar multiple of the same Weyl basis matrix, of degree
$\sum_i(a_i,b_i)$ in $(\mathbb Z/n\mathbb Z)^2$.
Its nonzero trace forces that degree to be zero. The output is
therefore a nonzero scalar matrix. Scaling one original input
gives every scalar matrix. Together with the open-image argument
this proves $\Val_n(f)=M_n$, as a set of single evaluations.

Finally, if $v=1$, \eqref{indep2:trace-on-fiber} is a nonzero
linear form on the tangent space of $\mathcal Y$. Thus
$d\widetilde h_p$ is nonzero on $\ker dG_p$, and the combined
map $(G,\widetilde h)$ has rank $r+1=n^2$. This is equivalent to
$dF_p$ having full rank. If $v\ge2$,
\eqref{indep2:trace-initial} gives $d\widetilde h_p=0$, whereas
$dG_p$ still has rank $r$. Hence $dF_p$ has rank exactly $n^2-1$.
The \'etale chart preserves these ranks for $df_a$. The
first case is smooth, and the second is flat but not smooth.
\end{proof}

\begin{proof}[Completion of the proof of Theorem~\ref{indep2:main}]
If some ordered pair of original positions has a nonzero complete
common-$D$ restriction, Theorem~\ref{complete2:main} gives both the
actual-image classification and the required original local fiber.
For the remaining case every such complete restriction is zero.
Proposition~\ref{indep2:normal} supplies a nonzero independent
pair, and \eqref{indep2:endpoints} gives its zero endpoints. Apply Theorem~\ref{indep2:seed} and
Proposition~\ref{indep2:trace-transfer} at the specified size $n$.
These two cases exhaust proper ordinary $J^2\setminus J^3$.
Throughout, the test deciding the target is whether the \emph{actual}
polynomial $\tr f$ on this $M_n^m$ is zero. No conclusion about that
test is inferred merely from its value on the commuting-passive
slice, and its answer is allowed to depend on $n$.
\end{proof}

\subsection{Higher-ideal corrections and the exact second-layer boundary}
\label{indep2:scope}

\begin{lemma}[Vanishing on the whole commuting-passive slice]
\label{indep2:commuting-vanishing}
Let $R$ be ordinary multilinear and belong to $J^r$. If fewer than
$r$ original positions are active and all other original inputs
commute with one another, then $R$ vanishes for arbitrary matrices
in the active positions. In particular, every ordinary multilinear
$R\in J^3$ vanishes on the entire two-active,
commuting-passive source in every matrix size.
\end{lemma}
\begin{proof}
Write $R$ as a sum of products of $r$ commutator-ideal factors,
allowing arbitrary word contexts, and take its component of degree
one in each original variable. In every surviving summand the
variable sets of the factors and contexts are disjoint. With fewer
than $r$ active positions, at least one of the $r$ ideal factors
uses only passive inputs. Those matrices lie in a commutative
subalgebra, on which the commutator ideal evaluates to zero.
That factor, and therefore the whole summand including its
surrounding words, vanishes.

One can also see that no cancellation or choice of presentation is
being assumed by multiplying the active inputs by independent
central parameters $t_1,\ldots,t_k$, where $k<r$. Modulo the ideal
$(t_1,\ldots,t_k)$ the evaluation takes values in a commutative
algebra, so it sends $J^r$ into $(t_1,\ldots,t_k)^r$.
Ordinary multilinearity makes the evaluated $R$ equal to
$t_1\cdots t_k$ times its unscaled value, of total parameter degree
$k<r$. Its membership in that ideal forces the unscaled value to
be zero. The identity holds on the entire slice and hence also
along its simultaneous conjugates.
\end{proof}

\begin{corollary}[Same-degree corrections with no lost trace terms]
\label{indep2:higher-corrections}
Let $F$ be proper ordinary multilinear of degree $m$ with
$F\in J^2\setminus J^3$. For every same-degree proper ordinary
multilinear $R\in J^3$, the corrected polynomial $f=F+R$ has the
conclusions of Theorem~\ref{indep2:main} in each size $n\ge3$.
Its complete independent symbol pair is unchanged for every
selected active pair. Its target and its compatible nilpotent
input are determined using the actual corrected polynomial.
\end{corollary}
\begin{proof}
The class of $F+R$ in $J^2/J^3$ is the nonzero class of $F$.
Lemma~\ref{indep2:commuting-vanishing} shows that the entire
selected commuting-passive restriction, not just one triangular
value, is unchanged. Thus its complete independent symbols are
unchanged. Apply Theorem~\ref{indep2:main}.

The restriction identity is the only use made of the vanishing
of $R$ on this slice. Its transverse derivatives need not vanish.
Every one of them and every higher transverse trace coefficient
is retained in \eqref{indep2:trace-pullback} and
\eqref{indep2:normal-expansion} for $f=F+R$. The first surviving
order and the bilinear form \eqref{indep2:trace-detector} may
therefore change, and the seed is reselected for that form.
The assertion does not say that a previously chosen triangular
witness stays flat for every correction. Cross-partition terms
outside $J^3$ are likewise not removed: their contributions are
already in the full coefficient arrays of
\eqref{indep2:symbols}.
\end{proof}

\begin{remark}[The alternating quartic lies in the independent branch]
\label{indep2:s4}
Use the standard convention
\[
 s_4(x_1,x_2,x_3,x_4)
   =\sum_{\sigma\in S_4}\operatorname{sgn}(\sigma)
      x_{\sigma(1)}x_{\sigma(2)}x_{\sigma(3)}x_{\sigma(4)}.
\]
Its commutator expansion is
\[
 \begin{split}
 s_4={}&[x_1,x_2][x_3,x_4]+[x_3,x_4][x_1,x_2]\\
       &-[x_1,x_3][x_2,x_4]-[x_2,x_4][x_1,x_3]\\
       &+[x_1,x_4][x_2,x_3]+[x_2,x_3][x_1,x_4].
 \end{split}
\]
Set $x_1=D_1$, $x_2=D_2$, $x_3=X$, $x_4=Y$, with $D_1,D_2$
diagonal. The first two products vanish, and the complete symbols
in the convention of \eqref{indep2:path} are exactly
\begin{equation}\label{indep2:s4-symbols}
 P(a,b)=a_2b_1-a_1b_2,\qquad
 Q(a,b)=a_1b_2-a_2b_1=-P(a,b).
\end{equation}
Indeed, the $XY$-ordered terms contribute
$-a_1b_2+a_2b_1$, and the $YX$-ordered terms contribute its negative.
Both common-$D$ specializations are zero, both endpoints
$P(a,-a),Q(a,-a)$ are zero, and $P$ is not zero.
Writing $c=-a-b$, its triangle norm is
\[
 P(a,b)P(b,c)P(c,a)+Q(a,b)Q(b,c)Q(c,a)=0.
\]
Thus this example uses the size-three zero-row and size-four
cycle-kernel branch of Theorem~\ref{indep2:seed}; it is not covered
by a nonzero common-$D$ restriction. In fact every common-$D$
chosen-pair restriction of $s_4$ vanishes, because two of its
original inputs coincide.

The polynomial is proper: inserting $I$ in one position makes
all its commutator products vanish. The expansion places it in
$J^2$. Its nonzero independent restriction, together with
Lemma~\ref{indep2:commuting-vanishing}, proves $s_4\notin J^3$.
Cyclically moving the first letter of a four-letter word to the
last position preserves its trace and changes the alternating
sign by $(-1)^3=-1$. Summing in $s_4$ gives
$\tr s_4=-\tr s_4$, hence $\tr s_4=0$ in characteristic zero.
Theorem~\ref{indep2:main} consequently gives actual image $\slm_n$
and an original smooth nonzero rank-one square-zero input for
every $n\ge3$; its local fiber dimension is $3n^2+1$.

For a small exact instance, take
\[
 D_1=\diag(0,1,2),\quad D_2=\diag(0,1,4),\quad
 X=E_{23}+E_{32},\quad Y=E_{21}+E_{31}.
\]
Formula~\eqref{indep2:s4-symbols} gives the actual value
$N=-2E_{21}+2E_{31}$, which has rank one and square zero.
The all-size smoothness assertion above follows from the proved
seed and trace arguments, not from extrapolation of this one
value or from a numerical rank calculation.
\end{remark}

\section{Further questions}\label{paper:further}

\subsection{The size-two boundary}

The image classification on $M_2$ is due to Kanel-Belov, Malev and
Rowen \cite[Theorem~2]{BMR2012}. The local statement of
Theorem~\ref{indep2:main} has a separate size-two boundary.
By the Amitsur--Levitzki theorem \cite{AL}, $s_4$ is an identity
on $M_2$, although $s_4\in J^2\setminus J^3$. Moreover, with zero
endpoints in \eqref{indep2:path}, every surviving term requires
three distinct vertices. Thus the independent-diagonal
restriction used in that branch vanishes on $M_2$. For noncentral
polynomials on $M_2$, it remains to determine the scope of the
smooth and flat nilpotent-fiber conclusions.

\subsection{The third commutator layer}

The classification in degrees four through six is given in
\cite[Theorem~A]{Pan46}. The higher-arity study in \cite{PanHigher}
includes local-fiber results for the entire six-slot third layer,
for classes arising from $s_6$ with same-degree $J^4$ corrections
in sizes $n\ge4$, and for specified families of higher arity.
The general proper third layer $J^3/J^4$ in higher arity remains
open. The fixed-size identity and central cases must be
distinguished from the noncentral case. On $M_3$, higher-arity
third-layer classes can have nonzero scalar image
\cite[section \emph{Central weighted third classes in size three}]{PanHigher}.

For an ordinary multilinear $f\in J^3\setminus J^4$ and $n\ge4$,
some restriction with three active original variables and independently
chosen diagonal passive inputs is nonzero. Indeed,
$\operatorname{Id}(UT_4)=J^4$ by \cite[Theorem~2.1]{UT}.
In a nonzero evaluation on $UT_4$, expand each input into its diagonal
part and its strictly upper triangular part.
Terms with fewer than three strictly upper triangular inputs vanish
by Lemma~\ref{indep2:commuting-vanishing}. Terms with at least four
vanish because the fourth power of the strictly upper triangular
ideal of $UT_4$ is zero. A nonzero term therefore has exactly three
active inputs, and it embeds into $M_n$. By the same lemma, every
same-degree multilinear correction in $J^4$ vanishes on each
three-active restriction.

\Needspace{7\baselineskip}
In size three, the three-active restrictions can all vanish on a
nonidentity in $J^3\setminus J^4$. An example in seven variables is
\[
 \sum_{i=1}^{5}(-1)^{i-1}
 s_4(x_1,\ldots,\widehat{x_i},\ldots,x_5)[[x_6,x_7],x_i],
\]
where the hat indicates omission
\cite[section \emph{Higher third symbols and original correction trace
fibers}]{PanHigher}. In this case one needs a construction using a
larger set of original input directions.

For a noncentral polynomial $f$ in the third layer, write
$h=\tr f$ and $g=f-n^{-1}hI_n$ on the full original source $M_n^m$.
The following question specifies a sufficient extension of the
method of this paper.

\begin{problem}[Original fibers in the third layer]
Let $n\ge3$ and let $f$ be a proper ordinary multilinear polynomial in
$J^3\setminus J^4$ with at least one nonscalar value on $M_n$. Is there
an original input $a$ with $f(a)=N$, $\rank N=1$, and $N^2=0$ such that
\[
 \rank(dg_a)=n^2-1,\qquad
 h|_{g^{-1}(N)}\ne0\ \text{in }\mathcal O_{g^{-1}(N),a}
 \quad\text{if } h\not\equiv0?
\]
\end{problem}

For $h\equiv0$, the differential condition gives the smooth
traceless fiber. For $h\not\equiv0$, Proposition~\ref{higher-trace}
gives the flat full-matrix fiber. Thus the rank condition and the
trace condition must hold at the same original input. Nonvanishing
on $UT_4$ supplies a restriction, and the conditions in the problem
concern the local ring of the original fiber.

Suppose $h\not\equiv0$ and $h$ vanishes on a three-active slice.
In an original-input chart, the first nonzero transverse trace
coefficient is trilinear in the three active matrices. A sufficient
statement for extending Proposition~\ref{indep2:trace-transfer} is
that, for every passive tuple in a nonempty regular-diagonal open
set, the active tensors $X\otimes Y\otimes Z$ span $M_n^{\otimes 3}$
as $(X,Y,Z)$ ranges over inputs with nonzero rank-one square-zero
value at which the slice differential, with simultaneous
conjugation, surjects onto $\slm_n$. This would permit the choice of an input on which the
first nonzero trace coefficient is nonzero. The exact traceless
equations could then be solved along the slice with the transverse
parameters fixed. Degree eight is the first degree
in which a nonzero $J^4$ correction can occur, so it is the first
case in which this trace condition must allow a nonzero correction
with the same original variables. Polynomials lying wholly in $J^4$
require a separate treatment.

\subsection{Beyond the proper second layer}

For nonproper polynomials in higher degree, a unit restriction can
supply a surjective traceless differential along the retained
variables at a rank-one nilpotent value. After restoring the
deleted inputs, one must control the trace on the exact traceless
fiber of the original map, as in the bounded-degree argument of
\cite{Pan46}. The constructions for traceless
containment in \cite{PanMesyanHigher} address the strengthened
Mesyan conjecture in higher degrees. The local-fiber question here
also requires full-matrix coverage for the traceful noncentral
branch. The arbitrary-degree L'vov--Kaplansky conjecture remains
open.

\Needspace{7\baselineskip}
\section*{Acknowledgments}
The author used ChatGPT to assist with manuscript drafting, revision,
and mathematical computations. The author is responsible for the
mathematical content of the paper. This work was supported by the
Beijing Natural Science Foundation (Grant Nos.~1262017, 1252011).

\Needspace{32\baselineskip}
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\end{document}