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\begin{document}

\title[Non-MF group notes]{Non-MF group notes}
\subjclass[2020]{Primary 46L05; Secondary 20F05, 22D25, 20E26}
\keywords{MF groups, approximate unitary representations, property~(T),
group $C^*$-algebras, sofic groups}

\maketitle
\draftnotice
\noindent\textbf{Scope.}  These notes contain an earlier affine/Clifford
construction.  They are not a companion to the current manuscript
\emph{Non-MF Groups}, whose proof uses a different group and a different
argument.  References to the main paper below refer to the earlier
manuscript version and are retained only as part of this research record.
\medskip

\setcounter{tocdepth}{1}
\tableofcontents
\par\clearpage

\section{A literal presentation and affine certificate}
\label{sec:literal-presentation}

The earlier manuscript version defined $E$ conceptually from the affine
group $\mathbb Z^3\rtimes\SL_3(\mathbb Z)$ and the doubling endomorphism.
This section gives the literal presentation used by the Lean development
and an explicit matrix certificate for its affine base.

First define the \emph{linear presentation}
\begin{equation}\label{eq:sl3presentation}
 \mathcal R=\left\langle x,y,z\ \middle|\
 \begin{gathered}
 x^3=y^3=z^2=(xz)^3=(yz)^3=1,\\
 (x^{-1}zxy)^2=(y^{-1}zyx)^2=(xy)^6=1
 \end{gathered}\right\rangle,
\end{equation}
and the \emph{base}
\begin{equation}\label{eq:gamma-presentation}
\mathcal B=\left\langle v_1,v_2,v_3,x,y,z\ \middle|\
\begin{gathered}
\text{the eight relations in \eqref{eq:sl3presentation}},\\
[v_1,v_2]=[v_1,v_3]=[v_2,v_3]=1,\\
xv_1x^{-1}=v_3,\ xv_2x^{-1}=v_1,\
xv_3x^{-1}=v_2,\\
yv_1y^{-1}=v_1,\ yv_2y^{-1}=v_2^{-1}v_3,\
yv_3y^{-1}=v_1v_2^{-1},\\
zv_1z^{-1}=v_2v_3^{-1},\ zv_2z^{-1}=v_1v_3^{-1},\
zv_3z^{-1}=v_3^{-1}
\end{gathered}\right\rangle.
\end{equation}
Write $\iota\colon\mathcal B\to E$ for the homomorphism induced by the first
six generators of the presentation below.

\begin{definition}\label{def:E}
Use the six base generators and the relations displayed in
\eqref{eq:gamma-presentation}.  Define
\begin{equation}\label{eq:E}
  E=\left\langle v_1,v_2,v_3,x,y,z,t,c\ \middle|\
  \begin{gathered}
   \textup{(B)}\quad\text{all relations in \eqref{eq:gamma-presentation}},\\
   \textup{(T)}\quad tv_i t^{-1}=v_i^2\ (1\le i\le3),\\
   \phantom{\textup{(T)}\quad}
     txt^{-1}=x,\quad tyt^{-1}=y,\quad tzt^{-1}=z,\\
   \textup{(C)}\quad c^2=1,\\
   \phantom{\textup{(C)}\quad}
     [c,g]=1\ \ (g\in\{v_1,v_2,v_3,x,y,z\}),\\
   \textup{(W)}\quad [w,g]=1\quad(g\in\{v_1,v_2,v_3,x,y,z,t,c\})
  \end{gathered}\right\rangle,
\end{equation}
where $w$ abbreviates the word
$\bigl[\,tct^{-1},\,v_1(tct^{-1})v_1^{-1}\bigr]$.
This is an explicit finite presentation with eight generators and
forty-one relators: twenty in family~\textup{(B)}, six in
family~\textup{(T)}, seven in family~\textup{(C)}, and eight in
family~\textup{(W)}.
If $a$ and $b$ are involutions and $[a,b]$ is central, then
$[a,b]^{-1}=[b,a]=a[a,b]a^{-1}=[a,b]$.  Applying this to the two
involutions in the definition of $w$ gives $w^2=1$.
Families~\textup{(T)} and~\textup{(C)} give, respectively,
\leanverified{Sofic/LiteralNonMFPresentation}{GroupApproximation.LiteralNonMFPresentation.manuscriptLiteralPresentation}
\[
 t\iota(\mathcal B)t^{-1}\subseteq\iota(\mathcal B),
 \qquad [c,\iota(\mathcal B)]=1.
\]
\end{definition}

\subsection{Property~\textup{(T)} of the base}

\begin{proposition}\label{prop:literal-base-T}
The twenty-relator group $\mathcal B$ has property~\textup{(T)}, in both
the real-orthogonal and complex-unitary formulations.
\leanverified{Sofic/LiteralBaseP13PropertyTBridge}{GroupApproximation.LiteralBaseP13PropertyTBridge.manuscriptBaseHasKazhdanPropertyT}
\end{proposition}

\begin{proof}
By Remark~\ref{rem:classical-base} the presentation
\eqref{eq:gamma-presentation} defines $\mathbb Z^3\rtimes\SL_3(\mathbb
Z)$, which has property~\textup{(T)}
\cite[Corollary~1.4.16 and Theorem~1.7.1]{BHV}.  Let $\rho$ be an
orthogonal representation on a real Hilbert space $H_{\mathbb R}$.  Its
complexification $H_{\mathbb R}\otimes_{\mathbb R}\mathbb C$ has the unitary
representation $\rho\otimes1$.  The invariant vectors and displacement
norms are those of $\rho$ in each real coordinate, so every Kazhdan pair for
the unitary formulation is also one for the orthogonal formulation.
\end{proof}

\begin{remark}[the classical identification]\label{rem:classical-base}
The eight relations \eqref{eq:sl3presentation} present
$\SL_3(\mathbb Z)$ \cite[Theorem~2]{CRW}; the corresponding matrices
are also recorded in \cite{CLV}.  So
$\mathcal B\cong\mathbb Z^3\rtimes\SL_3(\mathbb Z)$, and the
surjection $\mathcal B\to\bar\Gamma$ of Lemma~\ref{lem:linear} onto
the matrix realization is an isomorphism.
\leanverified{Monsters/LiteralBaseCompleteness}{GroupApproximation.LiteralBaseCompleteness.baseAffineEquiv}
\end{remark}

\begin{lemma}[linear model]\label{lem:linear}
For $M\in\GL_3(\mathbb Q)$ and $u\in\mathbb Q^3$, write
\[
 A(M,u)=\begin{pmatrix}M&u\\0&1\end{pmatrix}.
\]
Assign $v_i\mapsto A(I_3,e_i)$ and
\[
\begin{aligned}
 x&\mapsto A\!\left(\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix},0\right),
&y&\mapsto A\!\left(\begin{pmatrix}1&0&1\\0&-1&-1\\0&1&0\end{pmatrix},0\right),\\[3pt]
 z&\mapsto A\!\left(\begin{pmatrix}0&1&0\\1&0&0\\-1&-1&-1\end{pmatrix},0\right).
\end{aligned}
\]
Let $\bar\Gamma\le\GL_4(\mathbb Q)$ be the subgroup generated by these six
matrices, and put
$D=\operatorname{diag}(2,2,2,1)$.  The six matrices satisfy all twenty
relators of $\mathcal B$, and
\[
 \bar\alpha(u)=DuD^{-1}
\]
is an injective endomorphism of $\bar\Gamma$ which squares the three
translation generators and fixes $x,y,z$.  Moreover the matrix assigned to
$v_1$ does not lie in $\operatorname{range}(\bar\alpha)$.
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptLinearModel}
\end{lemma}

\begin{proof}
Each of the twenty relators reduces to an equality between explicit
$4\times4$ rational matrices.  Conjugation by $D$ gives the six stable-letter equations
and is injective because it is conjugation by an invertible matrix.  Every
displayed generator and its explicitly computed inverse has integral entries;
equivalently, all six generators lie in $\GL_4(\mathbb Z)$.  So
$\bar\Gamma\le\GL_4(\mathbb Z)$.  However $D^{-1}\bar v_1D$ has entry $1/2$
in position $(1,4)$, so it is not in $\bar\Gamma$.  This proves the last
assertion.
\end{proof}

Form the direct limit $T=\varinjlim(\bar\Gamma,\bar\alpha)$ and let
$\tau$ denote its shift automorphism.  Set $\Sigma=T\rtimes\mathbb Z$.
Let $j\colon\bar\Gamma\hookrightarrow\Sigma$ be the level-zero map.  Put
$X=\Sigma/j(\bar\Gamma)$, with base point $o=j(\bar\Gamma)$.  Since
$D^{-1}\bar v_1D\notin\bar\Gamma$ (Lemma~\ref{lem:linear}), we have
$\tau o\ne j(\bar v_1)\tau o$, and the Clifford generators at these two
points anticommute.

\begin{proposition}\label{prop:witness}
$w\neq1$ in $E$.
\leanverified{Sofic/LiteralNonMFLinearWitness}{GroupApproximation.LiteralNonMFLinearWitness.literal_mark_ne_one}
\end{proposition}

\begin{proof}
Let $\Cl(X)$ be generated by a central involution $\zeta$ and involutions
$c_\xi$, $\xi\in X$, with
$[c_\xi,c_\eta]=\zeta$ for $\xi\ne\eta$, and put
$W=\Cl(X)\rtimes\Sigma$.
Define a map on the generators of \eqref{eq:E} by
\[
\begin{gathered}
  g\mapsto j(\bar g)\in\Sigma\quad
  (g\in\{v_1,v_2,v_3,x,y,z\}),\\
  t\mapsto\tau,\qquad c\mapsto c_o .
\end{gathered}
\]
Lemma~\ref{lem:linear} gives the twenty base and six stable-letter
relations.  The level-zero subgroup fixes $o$, so $c_o$
commutes with every base generator, and $c_o^2=1$.
Under this assignment,
\[
  tct^{-1}\mapsto c_{\tau o},\qquad
  v_1(tct^{-1})v_1^{-1}\mapsto c_{j(\bar v_1)\tau o},
\]
and the two points are distinct by Lemma~\ref{lem:linear}, so
\[
  w\mapsto[c_{\tau o},c_{j(\bar v_1)\tau o}]=-1=\zeta .
\]
Since $\zeta$ is central in $W$, the remaining relations $[w,g]=1$ for
$g\in\{v_1,v_2,v_3,x,y,z,t,c\}$ are also satisfied.  So the
assignment extends to a group homomorphism $E\to W$ sending $w$ to
$\zeta\neq1$, so $w\neq1$ in $E$.
\end{proof}

\subsection{The presentation-level obstruction}

Korchagin's local microstate criterion for finitely presented groups
\cite[Proposition~8]{Korchagin} is the corresponding positive result.
For the non-MF presentation here, the same viewpoint gives the following
finite obstruction.

Since $E$ is finitely presented, the forty-one relators of
\eqref{eq:E} serve as the test set, and the hypothesis becomes a
condition on an eight-tuple of unitaries.

\begin{corollary}[a uniform obstruction on the presentation]
\label{cor:relator-uniform}
There is a $\delta>0$ with the following property.  For every $d\ge1$
and every eight-tuple $V_1,\dots,V_8\in\U(d)$, if
\[
 \opnorm{r(V_1,\dots,V_8)-1}\le\delta
 \qquad\text{for each of the forty-one relators $r$ of \eqref{eq:E},}
\]
then $\opnorm{w(V_1,\dots,V_8)-1}<1$.
\leanverified{Sofic/LiteralRelatorObstruction}{GroupApproximation.LiteralRelatorObstruction.literal_relator_uniform_obstruction}
\end{corollary}

\begin{proof}
If no such $\delta$ exists, then for every $n$ there are a dimension
$d_n$ and a tuple $V^{(n)}$ satisfying every relator to within $1/n$
and with $\opnorm{w(V^{(n)})-1}\ge1$.  Each tuple defines a
homomorphism $\varphi_n\colon F_8\to\U(d_n)$, since $F_8$ is free.  Their
unitary classes define a homomorphism
$\Phi\colon F_8\to\U(\mathcal Q((d_n)))$.  Each of the forty-one
relators lies in $\ker\Phi$, its images converging to $1$ in operator
norm, so $\Phi$ factors as $\Theta\circ\pi$ through the quotient map
$\pi\colon F_8\to E$.  Corollary~\ref{cor:marked-obstruction} gives $\Theta(w)=1$, that
is $\opnorm{w(V^{(n)})-1}\to0$, contradicting the lower bound $1$.
\end{proof}

The compactness proof is ineffective.  An explicit value of $\delta$ and an
explicit test set $F_0$ would require an effective Kazhdan constant for the
base and effective compression estimates.


\section{Consequences of the example}\label{sec:position}

Throughout, $\hat c=tct^{-1}$ and $u=[\hat c,\iota(v_1)]$, so that $w=u^2$.
For a countable group $G$, write $\operatorname{Res}_{\mathrm{MF}}(G)$
for its \emph{MF residual}: the intersection of the kernels of all its
homomorphisms to $\U(\mathcal Q)$ (Definition~\ref{def:radical}).
Proposition~\ref{prop:univquot} shows that $G$ is MF exactly when this
subgroup is trivial.  By Corollary~\ref{cor:marked-obstruction},
$1\ne w\in\operatorname{Res}_{\mathrm{MF}}(E)$.

\subsection*{The quotient by the central involution}

\begin{mainthm}[the quotient of $E$ by $\langle w\rangle$]\label{thm:signfree}
The quotient $E/\langle w\rangle$ is not MF\@.  The image of
$[\iota(v_1),\hat c]=u^{-1}$ is nontrivial in $E/\langle w\rangle$ and lies
in $\operatorname{Res}_{\mathrm{MF}}(E/\langle w\rangle)$; consequently
$\{1,w\}$ is a proper subgroup of
$\operatorname{Res}_{\mathrm{MF}}(E)$.
\leanverified{Sofic/LiteralSignFreeQuotient}{GroupApproximation.LiteralSignFreeQuotient.signFree_collapse}
\leanverified{Sofic/LiteralSignFreeQuotient}{GroupApproximation.LiteralSignFreeQuotient.signFreeQuotient_not_isCDEOperatorMF}
\leanverified{Sofic/LiteralSignFreeQuotient}{GroupApproximation.LiteralSignFreeQuotient.commutator_not_zpow_mark}
\end{mainthm}

The MF residuals of $E$ and $E/\langle w\rangle$ are computed exactly in
Section~\ref{sec:exact-residual}.

\noindent
Theorem~\ref{thm:collapse} states that if an involution commutes with the
conjugate copy of a Kazhdan subgroup and its conjugates under the
original subgroup commute pairwise, then every homomorphism to
$\U(\mathcal Q)$ maps all of those conjugates to the same element.
In $E$ the pairwise commutators of those conjugates of $\hat c$ lie in
$\langle w\rangle$, so in $E/\langle w\rangle$ the element $\hat c$
satisfies these hypotheses, and every homomorphism
$E/\langle w\rangle\to\U(\mathcal Q)$ maps $u$ to $1$.  Thus
$E/\langle w\rangle$ is a second finitely presented non-MF group.

\subsection*{A stably finite non-MF group algebra}
Every MF algebra is stably finite \cite{BK}; the converse fails.  The
canonical trace on $\Cred(E)$ is faithful, so $\Cred(E)$ is stably
finite (Lemma~\ref{lem:faithfultrace}).
Abstract examples follow from the failure of the Connes embedding
problem \cite{FGH}; the reduced algebra of the finitely presented group
$E$ is an explicit one (Theorem~\ref{thm:reduced}).  For
$\Cred(\SL_3(\mathbb Z))$ and $\Cred(\SL_4(\mathbb Z))$ the MF
property is undecided; only the stronger PMF property is known to
fail, for $\SL_4(\mathbb Z)$ \cite{MdlS}.

\begin{mainthm}[the reduced group algebra of $E$]\label{thm:reduced}
The algebra $\Cred(E)$ is unital and separable, has a faithful
tracial state, so it is stably finite, but it is not an MF
algebra.
\leanverified{Sofic/LiteralNonMFEndpoint}{GroupApproximation.LiteralNonMFEndpoint.manuscriptTheoremD}
\end{mainthm}

\begin{proof}
$\Cred(E)$ is unital and separable because $E$ is countable, and by
Lemma~\ref{lem:faithfultrace} the canonical trace is faithful, so the
algebra is stably finite.  It is not an MF algebra by
Corollary~\ref{cor:marked-obstruction}.
\end{proof}

The block normal form of Section~\ref{sec:algebras-sofic} shows in
addition that $E$, and so $\Cred(E)$, is exact.
The canonical map $\mathcal B\to E$ is injective
(Section~\ref{sec:literal-presentation}), so $E$ contains an infinite Kazhdan group
and is nonamenable.
\leanverified{Analysis/NuclearityAmenability}{GroupApproximation.NuclearityAmenability.manuscriptNotAmenableOfBaseEmbeds}
By Lance's theorem, nonamenability of a discrete group is equivalent
to nonnuclearity of its reduced group $C^*$-algebra, so
$\Cred(E)$ is not nuclear \cite{Lance}.  In the direction used here,
nuclearity is read as the completely positive approximation property:
a discrete group whose reduced algebra has that property admits an
invariant mean, and $E$ does not.
\leanverified{Analysis/LanceNuclearity}{GroupApproximation.CStarExactness.reducedGroupCStar_not_nuclear}
Since $\Cred(E)$ is nonnuclear, this example does not address the
nuclear form of the Blackadar--Kirchberg problem, whether every stably
finite separable nuclear $C^*$-algebra is quasidiagonal.

\subsection*{A sofic example}

\begin{mainthm}[the explicit group is sofic]\label{thm:Esofic}
The group $E$ of Definition~\ref{def:E} is sofic.
\leanverified{Sofic/LiteralSoficAssembly}{GroupApproximation.LiteralSoficAssembly.markedGroup_isSofic}
\end{mainthm}

\noindent
So $E$ is hyperlinear.  LEF groups are MF \cite{CDE} and $E$ is not
MF (Corollary~\ref{cor:marked-obstruction}), so $E$ is not LEF; a residually finite group
is LEF, so $E$ is not residually finite either.
\leanverified{Sofic/LiteralSoficConsequences}{GroupApproximation.LiteralSoficAssembly.markedGroup_finitelyPresented_hyperlinear_nonMF}
\leanverified{Sofic/LiteralSoficConsequences}{GroupApproximation.LiteralSoficAssembly.markedGroup_sofic_not_isLEF_unconditional}
\leanverified{Sofic/LiteralSoficConsequences}{GroupApproximation.LiteralSoficAssembly.markedGroup_finitelyPresented_sofic_nonMF}

Let $E_0$ be the kernel of the exponent sum in $t$.  For every finite
subset of $E_0$, the $T$-coordinates lie in a single level of the
ascending chain and the lamp coordinates involve finitely many
conjugates of $c$; the subgroup they generate embeds in a residually
finite group.  So $E_0$ is LEF (Section~\ref{sec:Esofic}).  Since
$E$ is a split extension of $E_0$ by $\Z$, it is sofic by the
Elek--Szab\'o amenable-extension theorem \cite{ElekSzabo}.  This
extension does not preserve LEF: LEF groups are MF \cite{CDE}, so
$E_0$ is LEF whereas $E$ is not.

MF groups are not closed under extensions or quotients: a
Clifford quotient of $E$ is a non-MF extension of an MF group by an
MF group, and the MF group $F_8$ surjects onto $E$
(Section~\ref{sec:algebras-sofic}, Corollary~\ref{cor:quotclosure}).

\subsection*{A hyperlinear non-MF trace}

We use Shulman's terminology \cite{ShulmanTraces}.  Hyperlinear and MF
traces are defined by the same pointwise bounded, trace-correct matrix
models, with the algebraic defects measured in normalized
Hilbert--Schmidt norm and operator norm, respectively.  The
approximating maps themselves need not be linear, positive, unital, or
completely positive.  Every MF trace is hyperlinear; Shulman asks
whether the converse holds.

\begin{lemma}
\label{lem:mftrace-group}
Let $G$ be a countable group and let $\tau_G$ be the canonical trace on
$\Cmax(G)$, determined by $\tau_G(u_g)=1$ for $g=1$ and $\tau_G(u_g)=0$
otherwise.  If $\tau_G$ is an MF trace, then $G$ is MF\@.
\leanverified{Sofic/TraceSeparationEndpoint}{GroupApproximation.manuscriptMFTraceGroupBridge}
\end{lemma}

\begin{proof}
Let $\varphi_n\colon\Cmax(G)\to M_{k_n}(\C)$ witness that $\tau_G$ is MF
and put $\mathcal Q=\prod_n M_{k_n}(\C)\big/\bigoplus_n M_{k_n}(\C)$,
the quotient of the bounded product by the operator-norm $c_0$ ideal.
Pointwise boundedness puts $(\varphi_n(a))_n$ in the bounded product for
every $a$.  The operator-norm defect conditions make
$\Phi(a)=[(\varphi_n(a))_n]$ a $*$-homomorphism
$\Phi\colon\Cmax(G)\to\mathcal Q$.  Set $p=\Phi(1)$, a projection with
$\Phi(\Cmax(G))\subseteq p\mathcal Qp$.  Then
$\Theta(g)=\Phi(u_g)+(1-p)$ defines a homomorphism
$\Theta\colon G\to\U(\mathcal Q)$.  We claim that $\Theta$ is
injective.  If $g\ne1$ and $\Theta(g)=1$, then $\Phi(u_g)=\Phi(1)$, so
$\lVert\varphi_n(u_g)-\varphi_n(1)\rVert\to0$; the normalized trace is
contractive for the operator norm, whereas trace-correctness gives
$\operatorname{tr}_{k_n}(\varphi_n(u_g))\to0$ and
$\operatorname{tr}_{k_n}(\varphi_n(1))\to1$, a contradiction.  Thus $G$
embeds in the unitary group of a norm matrix corona and is MF.
\end{proof}

\noindent
Schafhauser \cite[Definition~1.1 and Proposition~2.2]{Schafhauser}
uses a stronger notion of MF group and relates it to the canonical
trace on $\Cred(G)$.  Here we use the Carri\'on--Dadarlat--Eckhardt
notion and the canonical trace on $\Cmax(G)$, as in
Lemma~\ref{lem:mftrace-group}.

\begin{mainthm}[a hyperlinear non-MF trace]\label{thm:trace}
The canonical trace $\tau_E$ on $\Cmax(E)$ is hyperlinear but not
MF\@.  Consequently, there is a separable unital $C^*$-algebra carrying
a hyperlinear non-MF trace.  In particular, this gives a negative
answer to Shulman's question whether every hyperlinear trace is MF
\cite{ShulmanTraces}.
\leanverified{Sofic/TraceSeparationEndpoint}{GroupApproximation.manuscriptTraceSeparation}
\end{mainthm}

\begin{proof}
By Theorem~\ref{thm:Esofic}, $E$ is sofic.  Let
$\sigma_n\colon E\to\operatorname{Sym}(d_n)$ be a sofic approximation
and $U_n(g)$ the associated permutation unitaries.  For a free
ultrafilter $\omega$, let
\[
 \oplus_{2,\omega}M_{d_n}(\C)
 =\bigl\{(x_n):\lim_\omega\lVert x_n\rVert_{2,\tr_{d_n}}=0\bigr\}
\]
inside the bounded product, and write
$\operatorname{tr}_\omega([(x_n)])=\lim_\omega\tr_{d_n}(x_n)$ for the
induced trace on the tracial ultraproduct.  The classes
$[(U_n(g))_n]$ define a homomorphism
\[
E\longrightarrow
\U\!\left(\prod_n M_{d_n}(\C)\big/\oplus_{2,\omega}M_{d_n}(\C)\right).
\]
Moreover, asymptotic freeness of the sofic approximation gives
\[
\operatorname{tr}_\omega([(U_n(g))_n])=\delta_{g,1},
\]
because the normalized trace of a permutation matrix is the proportion
of its fixed points.  By the universal property of $\Cmax(E)$ this
extends to a $*$-homomorphism $\pi$ with
$\tau_E=\operatorname{tr}_\omega\circ\pi$, so $\tau_E$ is hyperlinear
by Shulman's tracial-ultraproduct characterization
\cite{ShulmanTraces}.

If $\tau_E$ were an MF trace, Lemma~\ref{lem:mftrace-group} would imply
that $E$ is MF, contradicting Corollary~\ref{cor:marked-obstruction}\@.  Separability of
$\Cmax(E)$ follows from countability of $E$.
\end{proof}

\section{Weighted asymptotic-commutant invariance}\label{supp:transport-variants}

\begin{theorem}[one-sided conjugation at an arbitrary weight]
\label{thm:transport-variants}
Let $H$ be countable, let $L\leq H$ have property~\textup{(T)}, and let
$t\in H$ satisfy $tLt^{-1}\leq L$.  Let $U_n\colon H\to\U(d_n)$ be an
operator-norm asymptotic representation, let $(\nu_n)$ be nonnegative
real weights, and let $x_n\in M_{d_n}(\C)$ be matrices for which there is
$C$ with $\operatorname{Tr}(x_n^*x_n)\le C\,\nu_n$ for all $n$ and,
for every $\ell\in L$ and every $\varepsilon>0$, eventually
\[
 \operatorname{Tr}\,\bigl|x_n-U_n(\ell)\,x_n\,
   U_n(\ell)^*\bigr|^2\le\varepsilon\,\nu_n.
\]
Then $U_n(t)\,x_nU_n(t)^*$ and $U_n(t)^*\,x_nU_n(t)$ satisfy both
conditions as well.
\leanverified{Sofic/ManuscriptSubgroupSpecializations}{GroupApproximation.ManuscriptExactWrappers.manuscriptWeightedTransportSubgroup}
\end{theorem}

\begin{proof}
Equip the $n$th coordinate space with the inner product
\[
 \langle x,y\rangle_n=
 \begin{cases}
  \operatorname{Tr}(y^*x)/\nu_n,&\nu_n>0,\\
  \operatorname{Tr}(y^*x),&\nu_n=0.
 \end{cases}
\]
If $\nu_n=0$, the bound $\operatorname{Tr}(x_n^*x_n)\le C\nu_n$
forces $x_n=0$, so choosing the ordinary Hilbert--Schmidt
inner product on that coordinate does not affect the distinguished
vector.  Conjugation by a unitary is unitary for every one of these
inner products.  The bound makes $(x_n)$ a bounded ultraproduct vector,
and the adjoint actions remain operator-norm almost multiplicative
because the scalar normalization of the Hilbert norm does not enter
the operator norm.  Let $\pi$ be the induced representation on the Hilbert
space ultraproduct, let $P$ project onto $\operatorname{Fix}\pi(L)$, and put
$V=\pi(t)$.  The inclusion $tLt^{-1}\leq L$ gives $P\leq VPV^*$.
Both projections belong to the norm ultraproduct of the matrix algebras and
are unitarily equivalent there.  That algebra is finite, so $P=VPV^*$.
So both $V[x_n]$ and $V^*[x_n]$ are fixed by $\pi(L)$, which gives the
asserted estimates.
\end{proof}

\section{Obstructions from a one-sided conjugation datum}\label{sec:conj}

\begin{definition}[Kazhdan conjugation datum]\label{def:pattern}
Let $H$ be a countable group.  A \emph{Kazhdan conjugation datum} in
$H$ is a tuple $(\Lambda,\iota,t,c)$, where $\Lambda$ is a countable
group with property~\textup{(T)}, $\iota\colon\Lambda\to H$ is a group
homomorphism, possibly with nontrivial kernel, and $t,c\in H$,
subject to
\begin{enumerate}
  \item[(M1)] $t\,\iota(\lambda)\,t^{-1}\in\iota(\Lambda)$ for every
    $\lambda\in\Lambda$;
  \item[(M2)] $c\,\iota(\lambda)=\iota(\lambda)\,c$ for every
    $\lambda\in\Lambda$.
\end{enumerate}
Write $\hat c:=tct^{-1}$.  The datum determines the normal closure
\leanverified{Sofic/ConjugationDatumAnyUniverse}{GroupApproximation.ManuscriptExactWrappers.manuscriptMarkedKazhdanPattern_anyUniverse}
\[
  N_{\mathrm{conj}}
  \;=\;\bigl\langle\!\bigl\langle\,
  [\,\hat c,\iota(\lambda)\,]\;:\;\lambda\in\Lambda
  \,\bigr\rangle\!\bigr\rangle_H\;\trianglelefteq\;H .
\]
\end{definition}

For a finitely generated $\Lambda$, conditions (M1)--(M2) are imposed by
finitely many relations once generating sets are fixed.  Conditions
(M1)--(M2) imply that $c$ commutes with $\iota(\Lambda)$ and that $\hat c$
commutes with the subgroup $t\iota(\Lambda)t^{-1}$.

\begin{definition}[universal $2$-norm triviality]\label{def:invisible}
Let $U=(U_n)$ be an operator-norm asymptotic representation of a
countable group $H$ and let $\omega$ be an ultrafilter refining the
cofinite filter.  Write
\[
 K_2^{\omega}(U)\;=\;\bigl\{g\in H:\
 \textstyle\lim_\omega\hsnorm{U_n(g)-1}=0\bigr\},
\]
the kernel of the homomorphism induced by $U$ into the tracial
ultraproduct of the matrix blocks along $\omega$, and so a normal
subgroup of $H$.  An element $g$ is \emph{universally $2$-norm trivial} if
$g\in K_2^{\omega}(U)$ for every such $U$ and every such $\omega$.
If $\hsnorm{U_n(g)-1}\to0$ along the full sequence, then
$g\in K_2^{\omega}(U)$ for every such $\omega$.
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptHSInvisibleCharacterization}
\end{definition}

\begin{theorem}[finite-normal obstruction criterion]\label{thm:criterion}
Let $H$ be a countable group with a Kazhdan conjugation datum
$(\Lambda,\iota,t,c)$, and let $F\trianglelefteq H$ be a finite
normal subgroup with $F\subseteq N_{\mathrm{conj}}$.  Then every
homomorphism $\Theta\colon H\to\U(\mathcal Q)$ satisfies
$\Theta(f)=1$ for all $f\in F$.
\leanverified{Sofic/FiniteNormalAnyUniverse}{GroupApproximation.ManuscriptExactWrappers.manuscriptFiniteNormalObstructionCriterion_anyUniverse}
\end{theorem}

\begin{proof}
\emph{Universal $2$-norm triviality of $N_{\mathrm{conj}}$.}  Let
$U_n\colon H\to\U(d_n)$ be an operator-norm asymptotic representation
of $H$ (for instance, coordinate unitary lifts of a homomorphism to
$\U(\mathcal Q)$), write $U_{g,n}=U_n(g)$, and put
\[
  K_2=\{g\in H:\hsnorm{U_{g,n}-1}\longrightarrow0\}.
\]
Unitary invariance and asymptotic multiplicativity make
$K_2\trianglelefteq H$.  Put $L=\iota(\Lambda)$, which has
property~\textup{(T)} because it is a quotient of $\Lambda$.
Since $c$ centralizes $L$, the sequence $(U_{c,n})$ belongs to the
asymptotic commutant of $L$ in normalized Hilbert--Schmidt norm.
Theorem~\ref{thm:transport-variants}, with $\nu_n=d_n$, applied to
$x_n=U_{c,n}$, shows
that $(U_{\hat c,n})$ for $\hat c=tct^{-1}$ lies in the same asymptotic
commutant.  So
$[\hat c,\iota(\lambda)]\in K_2$ for every $\lambda\in\Lambda$; since
$K_2$ is normal, it contains the normal closure of these commutators:
$N_{\mathrm{conj}}\subseteq K_2$, and so $F\subseteq K_2$.
Convergence of the full sequence gives triviality along every
ultrafilter refining the cofinite filter, so every element of
$N_{\mathrm{conj}}$ is universally $2$-norm trivial
(Definition~\ref{def:invisible}).

\emph{The coordinate averaging corner.}  Suppose a homomorphism $\Theta$ to $\U(\mathcal Q)$ is nontrivial on $F$; choose coordinate unitary
lifts $(V_{g,n})$ of $\Theta$ and form the coordinate averaging operators
\[
 p_n=\frac1{|F|}\sum_{f\in F}V_{f,n}.
\]
Because the lifts are asymptotically multiplicative and $F$ is finite,
$(p_n)$ is asymptotically self-adjoint and idempotent.  Since $F$ is
normal, $(p_n)$ asymptotically commutes with every lift of
$\Theta(H)$.  Continuous functional calculus applied to the Hermitian
parts yields genuine projections at operator-norm distance $o(1)$ from
$p_n$, and these projections are asymptotically central as well.
Let $q_n$ be their complements.  Nontriviality of
$\Theta$ on $F$ forces $q_n\ne0$ along infinitely many coordinates ---
otherwise the averages converge to $1$ and $\Theta|_F$ is trivial.
Retain exactly those coordinates and relabel them by $\N$.  Compress
$V_{g,n}$ to $q_n\mathbb C^{d_n}$ and take its polar part.  Asymptotic
centrality of $q_n$ makes the polar correction tend to zero in operator
norm.  The corrected operators form an operator-norm asymptotic
representation $(W_{g,n})$ of $H$ on the corners
$q_nM_{d_n}(\C)q_n\cong M_{r_n}(\C)$, $r_n=\operatorname{rank}q_n>0$.
Since $q_n$ is the complement of a projection within $o(1)$ of $p_n$,
the finite average vanishes on the corner:
\[
 \Bigl\lVert\sum_{f\in F}W_{f,n}\Bigr\rVert\longrightarrow0
 \quad\text{in operator norm.}
 \tag{\ensuremath{\ast}}
\]

\emph{The contradiction.}  Since $F\subseteq N_{\mathrm{conj}}$ and
$(W_{g,n})$ is itself an operator-norm asymptotic representation of
$H$, every $f\in F$ satisfies $\hsnorm{W_{f,n}-1}\to0$.  Because $F$
is finite, these convergences are uniform over $f\in F$, so
\[
 \sum_{f\in F}W_{f,n}\longrightarrow |F|1
 \quad\text{in normalized Hilbert--Schmidt norm},
\]
each block with its own normalized trace.  This contradicts
\((\ast)\), because operator-norm convergence dominates normalized
Hilbert--Schmidt convergence.  So every homomorphism
$H\to\U(\mathcal Q)$ is trivial on $F$.
\end{proof}

\begin{corollary}[the marked obstruction]\label{cor:marked-obstruction}
\[
 1\ne w\in\operatorname{Res}_{\mathrm{MF}}(E).
\]
\leanverified{Sofic/LiteralNonMFEndpoint}{GroupApproximation.LiteralNonMFEndpoint.literal_mark_mem_manuscriptCoronaMFResidual}
\leanverified{Sofic/LiteralNonMFLinearWitness}{GroupApproximation.LiteralNonMFLinearWitness.literal_mark_ne_one}
\end{corollary}

\begin{proof}
Proposition~\ref{prop:witness} gives $w\ne1$.  Put
$L=\iota(\mathcal B)$ and $\hat c=tct^{-1}$.  The element
$u=[\hat c,\iota(v_1)]$ belongs to $N_{\mathrm{conj}}$, and $w=u^2$.
The subgroup $\langle w\rangle$ is finite and normal because $w$ is a central
involution.  Theorem~\ref{thm:criterion}, applied with
$F=\langle w\rangle$, implies that every homomorphism from $E$ to a norm
matrix corona maps $w$ to the identity.
\end{proof}

The same conclusion holds for every finite normal subgroup $F$ contained
in a subgroup $D\le H$ whose elements lie in $K_2^{\omega}(U)$ for every
operator-norm asymptotic representation $U$ of $H$ and every $\omega$.
Finiteness of $F$ is essential: universal $2$-norm triviality alone does
not force triviality in the corona (Section~\ref{sec:limits}).

For $F=\{1,w\}$, the projection is
$p_n=\tfrac12(1+V_{w,n})$; in the corona corner the image of $w$ is
$-1$ exactly, while the finite-stage compressions satisfy
$\opnorm{W_{w,n}+1}\to0$.

\subsection{An intrinsic form}

For a subgroup $L\le H$, let
\[
 P_L=\{s\in H:sLs^{-1}\subseteq L\},
\]
the \emph{compression semigroup} of $L$ in the terminology of
Kun--Thom \cite{KT}, and put $G^{+}(L)=\langle P_L\rangle$.  Then set
\begin{equation}\label{eq:intrinsicdefect}
 \mathfrak D(H,L)=
 \left\langle\!\left\langle
 [gzg^{-1},\ell]:g\in G^{+}(L),\ z\in C_H(L),\ \ell\in L
 \right\rangle\!\right\rangle_H,
\end{equation}
which depends only on the pair $(H,L)$, not on a choice of datum.

\begin{corollary}[transport by the compression semigroup]\label{cor:generaltransport}
Let $(r_n)$ and $V_{g,n}\in\U(r_n)$ be an operator-norm asymptotic
representation of $H$, let $\Lambda$ have property~\textup{(T)}, let
$\iota\colon\Lambda\to H$ be a homomorphism, and put $L=\iota(\Lambda)$.
Suppose that $(x_n\in M_{r_n}(\C))$ is uniformly bounded in operator norm
and belongs to the asymptotic commutant in normalized Hilbert--Schmidt norm of $L$:
\[
 \hsnorm{x_n-V_{\iota(\lambda),n}x_nV_{\iota(\lambda),n}^*}
 \longrightarrow0
 \qquad(\lambda\in\Lambda).
\]
Then for every $g\in G^{+}(L)$, conjugation and inverse conjugation
both preserve that asymptotic commutant.  Explicitly, both
\[
 V_{g,n}x_nV_{g,n}^*
 \quad\text{and}\quad
 V_{g,n}^*x_nV_{g,n}
\]
satisfy the preceding convergence for every $\lambda\in\Lambda$.
\leanverified{Sofic/DefectActionAnyUniverse}{GroupApproximation.KazhdanAsymptoticCommutant.compressionGroup_transport_both_anyUniverse}
\end{corollary}

\begin{proof}
For a one-sided conjugator $s$, Theorem~\ref{thm:transport-variants}, with
$\nu_n=d_n$, applied to $L=\iota(\Lambda)$ gives the conclusion for
$V_{s,n}x_nV_{s,n}^*$.  The subgroup $L$ has property~\textup{(T)}
because it is a quotient of $\Lambda$.  The equality of the
fixed-space projection with its conjugate established in that proof
also gives invariance under the inverse conjugation, so the
conclusion for $V_{s,n}^*x_nV_{s,n}$.  The set of $g\in H$ for which
conjugation by $V_{g,n}$ and by $V_{g,n}^*$ both preserve the
asymptotic commutant is a subgroup, and it contains every one-sided
conjugator; so it contains $G^{+}(L)$.
\end{proof}

\begin{theorem}[the subgroup $\mathfrak D$ and the MF residual]
\label{thm:compression-radical}
Let $H$ and $\Gamma$ be countable groups, let $\Gamma$ have
property~\textup{(T)}, let
$\iota\colon\Gamma\to H$ be a homomorphism, and put
$L=\iota(\Gamma)$.  If $F\trianglelefteq H$ is finite and
$F\le\mathfrak D(H,L)$, then every homomorphism $H\to\U(\mathcal Q)$ maps every element of $F$ to the identity.
\leanverified{Sofic/DefectRadicalAnyUniverse}{GroupApproximation.KazhdanAsymptoticCommutant.manuscriptCompressionRadical_anyUniverse}
\end{theorem}

\begin{proof}
Let $(U_{h,n})$ be coordinate lifts of an operator-norm asymptotic
representation of $H$ and put
$K_2=\{h:\hsnorm{U_{h,n}-1}\to0\}$, a normal subgroup of $H$.
If $z\in C_H(L)$, then $(U_{z,n})$ lies in the asymptotic commutant of
$L$, and by Corollary~\ref{cor:generaltransport} so does
$(U_{gzg^{-1},n})$ for every $g\in G^{+}(L)$.  So each
generator $[gzg^{-1},\ell]$, $\ell\in L$, displayed in
\eqref{eq:intrinsicdefect} belongs to $K_2$; since
$K_2\trianglelefteq H$, it contains their normal closure:
$\mathfrak D(H,L)\le K_2$.  So every element of $\mathfrak D(H,L)$ is
universally $2$-norm trivial in every operator-norm asymptotic
representation of $H$.  If a homomorphism $H\to\U(\mathcal Q)$ were
nontrivial on a finite normal $F\le\mathfrak D(H,L)$, the
averaging-corner argument of Theorem~\ref{thm:criterion} would
contradict this universal $2$-norm triviality.
\end{proof}

\subsection{The normal-Kazhdan form of the obstruction}

A finite group has property~\textup{(T)}, and its averaging operator is
the projection onto its fixed space.  The spectral projection of a
Kazhdan average replaces that operator for an arbitrary normal
property-\textup{(T)} subgroup, which may be infinite, noncentral and
torsion-free.

\begin{theorem}[obstruction from universal $2$-norm triviality]\label{thm:abstract-nk}
Let $H$ be a countable group and let $D\le H$ be a subgroup every
element of which is universally $2$-norm trivial
(Definition~\ref{def:invisible}).  Let $K\trianglelefteq H$ be a normal subgroup with
property~\textup{(T)} such that $K\le D$.  Then every homomorphism $\Theta\colon H\to\U(\mathcal Q)$ satisfies
$\Theta(k)=1$ for all $k\in K$.
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptAbstractNormalKazhdanObstruction}
\end{theorem}

\begin{proof}
Suppose a homomorphism $\Theta$ to $\U(\mathcal Q)$ is nontrivial on
$K$.  Its image $\bar H=\Theta(H)$ is countable and, as a subgroup of
$\U(\mathcal Q)$, is MF.  Choose an operator-norm
asymptotic representation
$(V_{g,n})$ in which every fixed nontrivial element eventually stays
at operator-norm distance at least $1$ from the identity.  These
models provide operator-norm separation, but at growing dimension this
yields no lower bound on normalized Hilbert--Schmidt distances.  Every
operator-norm asymptotic representation of $\bar H$ composes with
$\Theta$ to one of $H$, so every element of $\Theta(D)$ is again
universally $2$-norm trivial, while $\bar K=\Theta(K)\le\Theta(D)$ is normal,
Kazhdan and nontrivial.  We now work in $\bar H$.

\emph{The Kazhdan corner.}  Fix a free ultrafilter $\omega$.  Let
$H_\omega$ be the Hilbert-space ultraproduct of the coordinate spaces
$\C^{d_n}$, and let $B_\omega=\prod_\omega B(\C^{d_n})$ act on
$H_\omega$.  The classes
$\pi(g)=[V_{g,n}]_\omega$ define a unitary representation of $\bar H$
on $H_\omega$.  Fix a finite symmetric generating Kazhdan set
$S\subseteq\bar K$ containing $1$, with constant $\varepsilon_0$, put
$\theta=1-\varepsilon_0^{2}/(4|S|)$, and let
\[
 h=\frac1{|S|}\sum_{a\in S}\pi(a)\in B_\omega .
\]
The Kazhdan inequality implies that the spectrum of $h$ lies in
$[-1,1-\varepsilon_0^{2}/(2|S|)]\cup\{1\}$ and so in
$[-1,\theta]\cup\{1\}$, so the spectral projection $P=\chi_{\{1\}}(h)$
belongs to $B_\omega$; its range is $\operatorname{Fix}\pi(\bar K)$,
where $h=1$.  Put $q=1-P$.

If $q=0$ then $\pi(k)=1$ for every $k\in\bar K$, against the
operator-norm separation of a fixed nontrivial element of $\bar K$.
For $g\in\bar H$ the projection $\pi(g)P\pi(g)^{*}$ has range
$\operatorname{Fix}\pi(g\bar Kg^{-1})=\operatorname{Fix}\pi(\bar K)$ by
normality, so $P$ and $q$ commute with $\pi(\bar H)$ exactly.  Lift $q$
to projections $q_n\in M_{d_n}(\C)$, nonzero along $\omega$ and with
$\opnorm{[q_n,V_{g,n}]_{-}}\to_\omega0$ for every $g\in\bar H$, and lift
$h$ to the self-adjoint coordinates
\[
 h_n=\frac1{2|S|}\sum_{a\in S}\bigl(V_{a,n}+V_{a,n}^{*}\bigr)
 \in M_{d_n}(\C),
\]
which represent $h$ because $S$ is symmetric.  The compression of $h$
to the corner satisfies $qhq\le\theta q$ in $B_\omega$, and, writing
$x_{+}$ for the positive part of a self-adjoint $x$, the positive part
is a continuous function of $x$ and the norm ultraproduct computes it
coordinatewise, so that inequality is exactly the coordinate statement
\begin{equation}\label{eq:corner-spectral}
 \opnorm{\bigl(q_nh_nq_n-\theta q_n\bigr)_{+}}\to_\omega0 .
\end{equation}

To obtain ordinary convergence before compressing, pass to a
subsequence.  Enumerate
$\bar H=\{g_1,g_2,\dots\}$.  For each $j$ the set
\[
 A_j=\Bigl\{n\ :\ q_n\ne0,\ \ \max_{i\le j}\opnorm{[q_n,V_{g_i,n}]_{-}}<1/j,\ \
 \opnorm{\bigl(q_nh_nq_n-\theta q_n\bigr)_{+}}<1/j\Bigr\}
\]
is again a finite intersection of sets in $\omega$ --- the last
constraint by \eqref{eq:corner-spectral} --- so lies in $\omega$ and
in particular is infinite.  Choose $n_1<n_2<\cdots$ with $n_j\in A_j$
and relabel along $(n_j)_j$.  Then $q_n\ne0$ for every $n$.  For every
fixed $g\in\bar H$, the element $g=g_i$ occurs in the constraints
defining $A_j$ for all $j\ge i$, so
$\opnorm{[q_n,V_{g,n}]_{-}}\to0$.  The last constraint defining $A_j$
also implies that the left-hand side of \eqref{eq:corner-spectral}
tends to $0$.  Compress $V_{g,n}$ to $q_n\mathbb C^{d_n}$ and take its
polar part.  The polar correction tends to zero in operator norm because
$q_n$ is asymptotically central.  The corrected operators form an
operator-norm asymptotic representation $(W_{g,n})$ of $\bar H$ on the
corners $q_nM_{d_n}(\C)q_n$.

\emph{A $2$-norm separation on the corner.}  Write
$r_n=\operatorname{rank}q_n$ and equip the corner
$q_nM_{d_n}(\C)q_n\cong M_{r_n}(\C)$ with its renormalized trace
$\tr_{r_n}$.  Applying $\tr_{r_n}$ to the ordinary form of
\eqref{eq:corner-spectral} gives
\[
 \frac1{|S|}\sum_{a\in S}\operatorname{Re}\tr_{r_n}\bigl(q_nV_{a,n}q_n\bigr)
 \;=\;\tr_{r_n}\bigl(q_nh_nq_n\bigr)\;\le\;\theta+o(1),
\]
and polar correction moves each $q_nV_{a,n}q_n$ by $o(1)$ in operator
norm, so by $o(1)$ in $\tr_{r_n}$, so the same bound holds with
$W_{a,n}$ in its place: the average over $S$
of the real parts of the renormalized traces of the compressed models
is at most $\theta+o(1)$.  So for each $n$, at least one generator in
$S$ has compressed model with real renormalized trace at
most $\theta+o(1)$, and so normalized Hilbert--Schmidt distance at
least $\sqrt{2(1-\theta)}-o(1)$ from the corner identity.  Since $S$
is finite, a subsequence realizes this at a single generator $s_0$.

\emph{The contradiction.}  The corner models form an operator-norm
asymptotic representation of $\bar H$, and $s_0$ lies in
$\bar K\le\Theta(D)$.  So $s_0$ is universally $2$-norm trivial in the corner
models: $\hsnorm{W_{s_0,n}-1}\to0$, contradicting the positive lower
bound above.
\end{proof}

\begin{theorem}[normal-Kazhdan obstruction]\label{thm:normal-kazhdan}
Let $H$ be a countable group with a Kazhdan conjugation datum
$(\Lambda,\iota,t,c)$, and let $K\trianglelefteq H$ be a normal
subgroup with property~\textup{(T)} such that
$K\subseteq N_{\mathrm{conj}}$.  Then every homomorphism $\Theta\colon H\to\U(\mathcal Q)$ satisfies $\Theta(k)=1$ for all
$k\in K$.
\leanverified{Sofic/NormalKazhdanAnyUniverse}{GroupApproximation.ManuscriptExactWrappers.manuscriptNormalKazhdanObstruction_anyUniverse}
\end{theorem}

Since finite groups have property~\textup{(T)},
Theorem~\ref{thm:criterion} is the finite case of
Theorem~\ref{thm:normal-kazhdan}.

\begin{proof}
The first step of the proof of Theorem~\ref{thm:criterion} shows that
every element of $N_{\mathrm{conj}}$ is universally $2$-norm trivial
(Definition~\ref{def:invisible}).  Theorem~\ref{thm:abstract-nk} with
$D=N_{\mathrm{conj}}$ gives the conclusion.
\end{proof}

Every element of $\mathfrak D(H,L)$ is universally $2$-norm trivial, so
Theorem~\ref{thm:abstract-nk} gives the following normal-Kazhdan
analogue of Theorem~\ref{thm:compression-radical}:

\begin{corollary}[normal-Kazhdan form of the intrinsic residual]\label{cor:intrinsic-nk}
Let $H$ and $\Gamma$ be countable groups, let $\Gamma$ have
property~\textup{(T)}, let $\iota\colon\Gamma\to H$ be a homomorphism,
and put $L=\iota(\Gamma)$.  If $K\trianglelefteq H$ is a normal
subgroup with property~\textup{(T)} and $K\le\mathfrak D(H,L)$, then
every homomorphism $H\to\U(\mathcal Q)$ maps every
element of $K$ to the identity.
\leanverified{Sofic/NormalKazhdanAnyUniverse}{GroupApproximation.ManuscriptExactWrappers.manuscriptIntrinsicNormalKazhdanRadical_anyUniverse}
\end{corollary}

\begin{proof}
The proof of Theorem~\ref{thm:compression-radical} shows that every
element of $\mathfrak D(H,L)$ is universally $2$-norm trivial in every
operator-norm asymptotic representation of $H$; apply
Theorem~\ref{thm:abstract-nk} with $D=\mathfrak D(H,L)$.
\end{proof}

So the subgroup generated by \emph{all} normal property-\textup{(T)}
subgroups of $\mathfrak D(H,L)$ also lies in the kernel of
every homomorphism to $\U(\mathcal Q)$.
\leanverified{Sofic/NormalKazhdanAnyUniverse}{GroupApproximation.ManuscriptExactWrappers.manuscriptIntrinsicNormalKazhdanPart_anyUniverse}

\section{The finite-dimensional obstruction}\label{sec:findim}

Every finite-dimensional representation of $E$ over any field maps
$w$ to the identity by Theorem~\ref{thm:exactfd-intrinsic}.
Consequently finite-dimensional linear representations do not separate
the points of $E$, and every finite quotient maps $w$ to the
identity.  Since the
Clifford quotient of Section~\ref{sec:literal-presentation} sends $w$ to a
nontrivial element, that quotient is infinite.  The central involution
$w$ lies in the kernel of every homomorphism to a finite group, so $E$
is a group of \emph{Deligne type} in the terminology of
\cite[Definition~1.2]{BDL}.  Such groups were proposed there as the
natural candidate counterexamples for all three remaining approximation
problems, and $E$ is such a counterexample for the operator norm.

\begin{theorem}[the exact intrinsic obstruction]\label{thm:exactfd-intrinsic}
Let $k$ be a field, $V$ a finite-dimensional $k$-vector space, $H$ a
group, and $\Gamma\le H$ a subgroup.  Then every homomorphism
$\pi\colon H\to\GL(V)$ is trivial on the subgroup
$\mathfrak D(H,\Gamma)$ of \eqref{eq:intrinsicdefect}.
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptTheoremB}
\end{theorem}

\begin{proof}
Let $C$ be the commutant of $\pi(\Gamma)$.  The elements $g\in H$ for
which conjugation by $\pi(g)$ preserves $C$ in both directions form a
subgroup.  If $s\in P_\Gamma$, then
\[
 C\subseteq \pi(s)C\pi(s)^{-1},
\]
because $s\Gamma s^{-1}\leq\Gamma$.  Both spaces have the same finite
dimension, so equality holds.  Thus every element of $P_\Gamma$ preserves
$C$ in both directions, and so does all of
$G^{+}(\Gamma)=\langle P_\Gamma\rangle$.  Thus, for $g\in
G^{+}(\Gamma)$ and $z\in C_H(\Gamma)$, the operator
$\pi(gzg^{-1})$ still belongs to $C$ and commutes with every
$\pi(\ell)$, $\ell\in\Gamma$.  So every displayed generator in
\eqref{eq:intrinsicdefect} lies in $\ker\pi$; since
$\ker\pi\trianglelefteq H$, the whole normal closure
$\mathfrak D(H,\Gamma)$ does as well.
\end{proof}

\begin{corollary}\label{cor:notRFD}
Every finite-dimensional linear representation of the group $E$ of
Definition~\ref{def:E}, over any field, maps $w$ to the identity, and
every homomorphism from $E$ to a finite group does as well.  In
particular $E$ admits no injective homomorphism into $\GL(V)$ for any
finite-dimensional vector space $V$ over any field.
\leanverified{Sofic/LiteralFiniteDimensionalObstruction}{GroupApproximation.LiteralFiniteDimensionalObstruction.manuscriptFiniteDimensionalConsequences}
\end{corollary}

\begin{proof}
Apply Theorem~\ref{thm:exactfd-intrinsic} with $H=E$ and
$\Gamma=\iota(\mathcal B)$.  The defining relations put
$t$ in $P_\Gamma$, $c$ in $C_E(\Gamma)$, and
$u=[tct^{-1},\iota(v_1)]$ in $\mathfrak D(E,\Gamma)$.  So every
finite-dimensional representation maps $u$, and so $w=u^2$, to the
identity.  Moreover,
$w\neq1$ by Proposition~\ref{prop:witness}.  For a finite target,
compose with its faithful left regular representation over
$\mathbb Q$.  The composite is a finite-dimensional linear
representation and so maps $w$ to the identity; faithfulness of the
left regular representation then implies that the original
homomorphism also maps $w$ to the identity.
\end{proof}

\section{The cyclic comparison}\label{sec:necessity}

Fritz's marked $\mathrm{BS}(2,3)$ example is the context for this
comparison \cite{Fritz,SlofstraVidick}.

% This is an ordinary numbered theorem, not a headline result; cite it by its
% content label \ref{thm:cyclic}.
\begin{theorem}[the cyclic comparison]\label{thm:cyclic}
Define the presentation
\[
 E_{\mathrm{BS}}=\left\langle\gamma_0,t,c\ \middle|\
 \begin{aligned}
  &t\gamma_0t^{-1}=\gamma_0^2,\quad c^2=1,\quad[c,\gamma_0]=1,\\
  &w_{\mathrm{BS}}^2=1,\quad
    [w_{\mathrm{BS}},\gamma_0]=[w_{\mathrm{BS}},t]
      =[w_{\mathrm{BS}},c]=1
 \end{aligned}\right\rangle,
\]
where
$w_{\mathrm{BS}}=[tct^{-1},\gamma_0(tct^{-1})\gamma_0^{-1}]$.
Then $E_{\mathrm{BS}}$ is finitely presented and $w_{\mathrm{BS}}\ne1$.
There is an explicit surjective homomorphism from $E_{\mathrm{BS}}$ onto
the subgroup of the Clifford semidirect product generated by the
displayed images, and it maps $w_{\mathrm{BS}}$ to $\zeta\ne1$.
Nevertheless every finite-dimensional representation of
$E_{\mathrm{BS}}$ over every field maps $w_{\mathrm{BS}}$ to the identity.
\leanverified{Monsters/LiteralCyclicCalibration}{GroupApproximation.LiteralCyclicCalibration.manuscriptCyclicCalibration}
\end{theorem}

\begin{proof}[Proof of Theorem~\ref{thm:cyclic}]
The relator $w_{\mathrm{BS}}^2=1$ is redundant: $tct^{-1}$
and $\gamma_0(tct^{-1})\gamma_0^{-1}$ are involutions, so centrality
forces their commutator to have order two by the calculation following
Definition~\ref{def:E}.  Deleting it gives a
six-relator presentation of the same group.
The stable relation gives
\(t\langle\gamma_0\rangle t^{-1}\subseteq
\langle\gamma_0\rangle\), and \(c\) centralizes this cyclic base.
By Theorem~\ref{thm:exactfd-intrinsic}, every
finite-dimensional representation over every field maps
\(w_{\mathrm{BS}}\) to the identity.

For nontriviality, use the affine realization of
\(\mathrm{BS}(1,2)\) and its coset action.  The base point is fixed by the
cyclic base, while the two points represented by \(t\) and \(\gamma_0t\)
are distinct.  Attach Clifford generators to these points and map \(c\) to
the generator at the base point.  All seven relators hold, and
\[
 w_{\mathrm{BS}}\longmapsto
 [c_{t\langle\gamma_0\rangle},
  c_{\gamma_0t\langle\gamma_0\rangle}]
 =-1.
\]
Thus \(w_{\mathrm{BS}}\ne1\).  Restricting the target to the subgroup
generated by the three displayed images gives the asserted surjection
onto a Clifford quotient.
\end{proof}

\paragraph{Sharpness of the Kazhdan hypothesis.}
Running the Clifford representation of the preceding proof over
finite site sets gives explicit unitary models of the seven relators
in which $w_{\mathrm{BS}}$ is the scalar $-1$.  Fix $m\ge0$, put
$M=2m+3$ and $\zeta=e^{2\pi i/M}$, and take the coordinates
$\mathbb Z/2\times\mathbb Z/M$, the first for the Clifford sign and
the second for the spectrum of the base.  Let $X$ and $Z$ be the flip and
the sign on the first coordinate, put $D=2^{-1/2}(X+Z)$, a
self-adjoint unitary, and set
\[
 \gamma_0(s,k)=(-1)^{s}\zeta^{k},\qquad c=t^{-1}Dt,
\]
where $\gamma_0$ is the displayed diagonal unitary, $b_0=0$,
$b_1=m+1$, and $t$ is the pullback unitary
\[
 (t\xi)(s,k)=\xi(s,2k+b_s),
 \qquad t e_{(s,\,2k+b_s)}=e_{(s,k)}.
\]
This is well defined because $(s,k)\mapsto(s,2k+b_s)$ is a permutation
when $M$ is odd.  Here $\gamma_0$ and $t$
are monomial and $c$ is an equal-weight sum of two monomial matrices.
Since $tct^{-1}=D$ and $\gamma_0^2(s,k)=\zeta^{2k}$ is independent of
the spin coordinate, the distinguished word is
\[
 w_{\mathrm{BS}}=\bigl[\,D,\;ZDZ\,\bigr]=(D\cdot ZDZ)^2=-1 ,
\]
because $ZDZ=2^{-1/2}(Z-X)$ and $D\cdot ZDZ=\tfrac12(XZ-ZX)$ squares
to $-1$.  Being a scalar, $w_{\mathrm{BS}}$ satisfies exactly the four
relators $w_{\mathrm{BS}}^2=1$ and
$[w_{\mathrm{BS}},\gamma_0]=[w_{\mathrm{BS}},t]=[w_{\mathrm{BS}},c]=1$;
with $c^2=1$ this is five of the seven.  The remaining two hold
approximately.  From $2(m+1)=M-1$ we get
$\zeta^{m+1}=-e^{-i\pi/M}$, and
\[
 \opnorm{t\gamma_0t^{-1}-\gamma_0^{2}}=\bigl|1+\zeta^{m+1}\bigr|
 =2\sin\frac{\pi}{2M},
 \qquad
 \opnorm{[c,\gamma_0]-1}\le2\bigl|1+\zeta^{m+1}\bigr| ;
\]
$D$ commutes with $\gamma_0^2$ exactly, so the lamp relator inherits
the doubling defect and nothing else.  Both are $O(1/m)$.  Assembling
the models over $m$ gives
$\Theta\colon E_{\mathrm{BS}}\to\U(\mathcal Q)$ with
$\Theta(w_{\mathrm{BS}})\ne1$, and the image of $\Theta$ is MF
because it embeds in $\U(\mathcal Q)$.  So the
property~\textup{(T)} hypothesis in the central involution
obstruction is essential: with the cyclic base
$\langle\gamma_0\rangle$ in place of a Kazhdan base, some
homomorphism to $\U(\mathcal Q)$ is nontrivial on the distinguished
word.  Equivalently, in
the residual language of Section~\ref{sec:consequences},
$w_{\mathrm{BS}}\notin\operatorname{Res}_{\mathrm{MF}}(E_{\mathrm{BS}})$.
\leanverified{Sofic/CliffordBSAmenableMF}{GroupApproximation.CliffordBSAmenableMF.manuscriptSharpnessOfKazhdanHypothesis}

The realized Clifford quotient is amenable: it is a subgroup of
the semidirect product of the locally finite Clifford group on the
coset space by the solvable affine group
$\mathrm{BS}(1,2)\cong\mathbb Z[1/2]\rtimes\mathbb Z$.  So it is MF: the reduced $C^*$-algebra of every countable
amenable group embeds into the universal UHF algebra
\cite[Theorem~B]{SchafhauserAF}.  The models above give
$\Theta(w_{\mathrm{BS}})\ne1$ for an MF image
$\Theta(E_{\mathrm{BS}})$.

Exact finite-dimensional representations behave differently.  In a finite image, the
doubling relation gives
$\pi(\langle\gamma_0^2\rangle)=\pi(\langle\gamma_0\rangle)$; since
$\pi(tct^{-1})$ centralizes the first subgroup, it centralizes
$\pi(\gamma_0)$, and the image of $w_{\mathrm{BS}}$ is the identity.
To transfer this obstruction to exact local embeddings, take a finite
set containing the generators together with every intermediate prefix
of every defining word and of $w_{\mathrm{BS}}$.  Any exact local
embedding of this set evaluates those words correctly and so
satisfies the same equations.  Then neither
$E_{\mathrm{BS}}$ nor its Clifford quotient is LEF, and the latter is
not residually finite.
\leanverified{Monsters/CyclicBaseLEFObstruction}{GroupApproximation.CyclicBaseLEFObstruction.cyclicBase_exactModel_package}

\paragraph{The scaling family.}
For $m\ge2$, define $E_m$ by replacing the three relations
$tv_i t^{-1}=v_i^2$ in the eight-generator presentation of $E$ with
$tv_i t^{-1}=v_i^m$ and leaving every other relator unchanged.  Then
$E_2=E$.  Write $w_m$ for the image of the same
distinguished word $w$.

\begin{corollary}[the scaling family]\label{cor:scaling-family}
For every natural number $m\ge2$, the group $E_m$ is finitely presented and
$w_m$ is a nontrivial central involution mapped to the identity by every
homomorphism $E_m\to\U(\mathcal Q)$.  Consequently $E_m$ is not MF,
and neither $\Cmax(E_m)$ nor $\Cred(E_m)$ is an MF $C^*$-algebra.
\leanverified{Sofic/ScalingFamilyEndpoint}{GroupApproximation.ScalingFamilyEndpoint.manuscriptTheoremFamily}
\end{corollary}

\begin{proof}
Replace the affine dilation $\operatorname{diag}(2,2,2,1)$ by
$\operatorname{diag}(m,m,m,1)$ (both acting as in
Lemma~\ref{lem:linear}).  It conjugates each integral translation
$v_i$ to $v_i^m$ and fixes the three linear generators.  Conjugating
$v_1$ by the inverse dilation gives translation by $e_1/m$, which is
not integral.  So the two relevant cosets remain distinct, and the
Clifford representation sends $w_m$ to $-1$.  The Kazhdan and
central-involution arguments are unchanged for every $m\ge2$.
\end{proof}

\section{The MF residual}\label{sec:consequences}

\begin{definition}\label{def:radical}
For a countable group $G_1$, the \emph{MF residual} is
\[
  \operatorname{Res}_{\mathrm{MF}}(G_1)
  \;=\;\bigcap_{(d_n)}\ \,\bigcap_{\Theta\colon G_1\to\U(\mathcal Q((d_n)))}\ker\Theta,
\]
a normal subgroup of $G_1$; the inner intersection is over all
homomorphisms and the outer one over all sequences $(d_n)$ of positive
integers.
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptMFRadical}
\end{definition}

If $G_1/N$ is MF, compose $G_1\to G_1/N$ with an injective homomorphism
$G_1/N\to\U(\mathcal Q)$.  The resulting homomorphism has kernel $N$,
so $\operatorname{Res}_{\mathrm{MF}}(G_1)\le N$.  Thus the MF residual
is contained in every normal subgroup whose quotient is MF\@.
Proposition~\ref{prop:univquot} shows that its own quotient is MF\@.
It is the operator-norm analogue of the invisible subgroups
of \cite[Definitions~2.5--2.7]{SlofstraCorrelations} and of the
metric-ultraproduct approximation framework of \cite{NST}.

\begin{lemma}[functoriality]\label{lem:portable}
For every homomorphism $\varphi\colon G_1\to G_2$ of countable
groups,
\[
  \varphi\bigl(\operatorname{Res}_{\mathrm{MF}}(G_1)\bigr)
  \;\subseteq\;\operatorname{Res}_{\mathrm{MF}}(G_2).
\]
In particular, if
$\varphi\colon E\to G_2$ satisfies $\varphi(w)\neq1$, then $G_2$ is
not MF\@.
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptRadicalPortability}
\end{lemma}

\begin{proof}
If $x\in\operatorname{Res}_{\mathrm{MF}}(G_1)$ and $\Theta$ is a
homomorphism $G_2\to\U(\mathcal Q)$, then $\Theta\circ\varphi$ is a homomorphism $G_1\to\U(\mathcal Q)$, so $\Theta(\varphi(x))=1$.  For
the second statement, Corollary~\ref{cor:marked-obstruction} gives
$w\in\operatorname{Res}_{\mathrm{MF}}(E)$.  So
$1\neq\varphi(w)\in\operatorname{Res}_{\mathrm{MF}}(G_2)$.  An
injective homomorphism $G_2\to\U(\mathcal Q)$ would map $\varphi(w)$
to $1$, a contradiction.
\end{proof}

\begin{proposition}[the universal MF quotient]\label{prop:univquot}
Let $G_1$ be a countable group and
$R=\operatorname{Res}_{\mathrm{MF}}(G_1)$.
\begin{enumerate}
\item $R$ is the kernel of a single homomorphism $G_1\to\U(\mathcal Q)$.
\item $G_1/R$ is MF\@.
\item Every homomorphism from $G_1$ to an MF group factors through
$G_1/R$.
\end{enumerate}
In particular $G_1$ is MF if and only if $R=\{1\}$.
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptUniversalMFQuotient}
\end{proposition}

\begin{proof}
(3) If $\varphi\colon G_1\to M$ with $M$ MF and
$j\colon M\to\U(\mathcal Q((d_n)))$ is an injective homomorphism to $\U(\mathcal Q)$,
then $j\circ\varphi$ is a homomorphism $G_1\to\U(\mathcal Q)$, so
$j(\varphi(R))=1$ and, by injectivity of $j$, $\varphi(R)=1$.

(1) and (2).  Set $Q=G_1/R$.  For every $\bar g\ne1$ in $Q$, the
definition of $R$ gives a homomorphism
$\Theta_{\bar g}\colon Q\to\U(\mathcal Q)$ with
$\Theta_{\bar g}(\bar g)\ne1$.  Enumerate the nonidentity elements
$\bar g_1,\bar g_2,\dots$ and choose coordinate unitary lifts of the
corresponding homomorphisms.  At stage $k$, choose in the $i$th lift,
for each $i\le k$, a coordinate at which multiplicativity holds within
$1/k$ on the first $k$ group elements.  Choose the same coordinate so
that the designated element $\bar g_i$ remains separated by its fixed
positive constant.  Such coordinates exist because
$\Theta_{\bar g_i}$ separates $\bar g_i$ along infinitely many
coordinates.  Take the block sum of
the $k$ chosen coordinates.  Block operator norms are maxima, so the
single block coming from $\Theta_{\bar g_i}$ already separates
$\bar g_i$, so the sum separates all of
$\bar g_1,\dots,\bar g_k$ at once.  The resulting sequence is an
operator-norm asymptotic representation separating every nonidentity
element from the identity.  The corresponding unitary classes give a
faithful homomorphism $Q\to\U(\mathcal Q)$.  This proves (2), and composing with
$G_1\twoheadrightarrow Q$ proves (1).  Finally, if $G_1$ is MF then an
injective homomorphism to $\U(\mathcal Q)$ gives $R=\{1\}$; if $R=\{1\}$ then
(2) says $G_1$ is MF\@.
\end{proof}

\begin{corollary}[exact residual from a candidate quotient]\label{cor:exactradical}
Let $N\trianglelefteq G_1$ satisfy
$N\le\operatorname{Res}_{\mathrm{MF}}(G_1)$.  Every homomorphism $G_1\to\U(\mathcal Q)$ factors uniquely through $G_1/N$.  If in
addition $G_1/N$ is MF, then
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptExactRadicalFromCandidateQuotient}
\[
 \operatorname{Res}_{\mathrm{MF}}(G_1)=N.
\]
\end{corollary}

\begin{proof}
The containment $N\le\ker\Theta$ gives the unique quotient
factorization for every such $\Theta$.  For the
reverse containment in the displayed equality, compose the quotient
map $G_1\to G_1/N$ with an injective homomorphism $G_1/N\to\U(\mathcal Q)$:
its kernel is exactly $N$, while the MF residual lies in the kernel of
every homomorphism to $\U(\mathcal Q)$.
\end{proof}

\begin{corollary}[residual reduction to the quotient]\label{cor:pullback}
Let $N\trianglelefteq G_1$ satisfy
$N\le\operatorname{Res}_{\mathrm{MF}}(G_1)$ and let
$q\colon G_1\to G_1/N$ be the quotient map.  Then
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptRadicalReductionToQuotient}
\[
 \operatorname{Res}_{\mathrm{MF}}(G_1)
 \;=\;
 q^{-1}\bigl(\operatorname{Res}_{\mathrm{MF}}(G_1/N)\bigr).
\]
\end{corollary}

\begin{proof}
For the forward containment, Lemma~\ref{lem:portable} gives
$q(\operatorname{Res}_{\mathrm{MF}}(G_1))
\subseteq\operatorname{Res}_{\mathrm{MF}}(G_1/N)$.  Conversely, suppose
$q(x)\in\operatorname{Res}_{\mathrm{MF}}(G_1/N)$ and let $\Theta$ be
a homomorphism $G_1\to\U(\mathcal Q)$.  Since
$N\le\operatorname{Res}_{\mathrm{MF}}(G_1)$, $\Theta$ factors as
$\bar\Theta\circ q$ with $\bar\Theta$ a homomorphism
$G_1/N\to\U(\mathcal Q)$; so $\Theta(x)=\bar\Theta(q(x))=1$.
\end{proof}

\begin{theorem}[complexity of MF-radical membership]
\label{thm:mf-radical-arithmetic}
Uniformly in a finite presentation $P$ and a word $v$,
\[
 v\notin\operatorname{Res}_{\mathrm{MF}}(G_P)\quad\hbox{is }\Pi^0_2,
 \qquad
 v\in\operatorname{Res}_{\mathrm{MF}}(G_P)\quad\hbox{is }\Sigma^0_2.
\]
The same $\Sigma^0_2$ upper bound holds for the set of radical words after
fixing $P$.
\end{theorem}

\begin{proof}
Let $D(P,v,n,d)$ assert the existence of $d$-dimensional generator unitaries
with relator defect at most $2^{-n-10}$ and $\|v(U)-1\|\geq1/2$.  This is
decidable over the real closed field.  Marked corona extraction and
amplification, in one direction, and the diagonal norm-corona construction,
in the other, give
\[
 v\notin\operatorname{Res}_{\mathrm{MF}}(G_P)
 \quad\Longleftrightarrow\quad
 \forall n\,\exists d\;D(P,v,n,d),
\]
and complementation proves the claim.
\end{proof}

\begin{theorem}[undecidable radical membership in one fixed group]
\label{thm:fixed-radical-membership}
There is one fixed finitely presented non-MF group $U$ for which membership in
$\operatorname{Res}_{\mathrm{MF}}(U)$ is undecidable.
\end{theorem}

\begin{proof}
Let $A$ be Kharlampovich's finitely presented solvable group of derived length
three with undecidable word problem \cite{Kharlampovich}.  It is amenable, so
$\Cred(A)$ is quasidiagonal~\cite{TWW} and embeds in a norm matrix
corona, and $A$ is MF\@.  Set $U=E*A$.  This group is finitely presented and non-MF, since
the free factor $E$ embeds in it.

Let $\pi_A\colon E*A\to A$ be the canonical retraction.  Since $A$ is MF,
Proposition~\ref{prop:univquot} gives
$\operatorname{Res}_{\mathrm{MF}}(A)=\{1\}$.  Functoriality then gives
\[
 \operatorname{Res}_{\mathrm{MF}}(U)\leq\ker\pi_A.
\]
For every $a\in A$ it follows that
\[
 i_A(a)\in\operatorname{Res}_{\mathrm{MF}}(U)
 \quad\Longleftrightarrow\quad a=1\text{ in }A.
\]
Indeed the forward implication follows by applying $\pi_A$, and the reverse
implication is trivial.  The inclusion of words in the $A$ factor is
computable, so a radical-membership algorithm for $U$ would decide the word
problem of $A$.
\end{proof}

Here actual residual membership varies.  A different phenomenon occurs even
when membership in the residual is promised in advance.

\begin{theorem}[a fixed computer inside the MF residual]
\label{thm:fixed-radical-computer}
There are a fixed finitely presented non-MF group $V$ and a computable family
of words $(r_p)_{p\in\N^2}$ such that
\[
 r_p\in
 [\operatorname{Res}_{\mathrm{MF}}(V),V]
 \subseteq\operatorname{Res}_{\mathrm{MF}}(V)
 \qquad(p\in\N^2),
\]
but the set $\{p:r_p=1\text{ in }V\}$ is undecidable.
\leanverified{Computability/MFRadicalComputer}{GroupApproximation.MFRadicalComputer.closed_package}
\leanverified{Computability/MFRadicalComputer}{GroupApproximation.MFRadicalComputer.word_mem_commutator_residual}
\end{theorem}

\begin{proof}
Choose a fixed finitely presented group $A$ with a computable family
$(a_p)_{p\in\N^2}$ for which the predicate $a_p=1$ is undecidable
\cite{Boone}.  Let $w\ne1$ be the marked element of $E$ lying in
$\operatorname{Res}_{\mathrm{MF}}(E)$, put $V=E*A$, and define
\[
 r_p=[w,a_p],
\]
where the two entries are viewed in the corresponding free factors.
Functoriality puts the image of $w$ in
$\operatorname{Res}_{\mathrm{MF}}(V)$, so
$r_p\in[\operatorname{Res}_{\mathrm{MF}}(V),V]$ for every $p$.
Free-product normal form and $w\ne1$ give
$[w,a_p]=1$ if and only if $a_p=1$.  Thus identity testing on the promised
residual-valued family $(r_p)$ is undecidable.  Finally, $V$ is finitely
presented and contains the non-MF free factor $E$, so it is not MF.
\end{proof}

Thus Theorem~\ref{thm:fixed-radical-membership} makes residual membership
undecidable, whereas Theorem~\ref{thm:fixed-radical-computer} makes equality
with the identity undecidable inside a computable family already known to
lie in the residual.

\begin{corollary}\label{cor:nofaithful}
Let $A$ be a unital $C^*$-algebra admitting an injective, possibly
nonunital, $*$-homomorphism into a matrix quotient.  Then the group $E$
admits no injective homomorphism into $\U(A)$.
\leanverified{Sofic/LiteralNonMFConsequences}{GroupApproximation.LiteralNonMFConsequences.literal_no_faithful_corona_subalgebra_target}
\end{corollary}

\begin{proof}
Let $j\colon A\to\mathcal Q$ be the injective, possibly nonunital,
$*$-homomorphism and let $\phi\colon E\to\U(A)$.  Define
\[
   \widehat j(u)=j(u)+1-j(1_A)\in\U(\mathcal Q).
\]
Then \(\widehat j\) is an injective homomorphism on \(\U(A)\), so
\(\widehat j\circ\phi\) is a homomorphism \(E\to\U(\mathcal Q)\).
By Corollary~\ref{cor:marked-obstruction}, \((\widehat j\circ\phi)(w)=1\); injectivity
of \(\widehat j\) gives \(\phi(w)=1\).  Proposition~\ref{prop:witness} has
\(w\neq1\), so \(\phi\) cannot be injective.
\end{proof}

\subsection{Permanence and its failure}

\begin{lemma}[permanence]\label{lem:permanence}
Let all groups be countable.
\begin{enumerate}
\item Every subgroup of an MF group is MF\@.
\item Every residually finite group is MF\@.
\item Every locally finite group is MF\@.
\leanverified{Sofic/OperatorMFPositiveControls}{GroupApproximation.IsOperatorMF.subgroup}
\leanverified{Sofic/OperatorMFPositiveControls}{GroupApproximation.isOperatorMF_of_residuallyFinite}
\leanverified{Sofic/LocallyFiniteMF}{GroupApproximation.isOperatorMF_of_locallyFinite}
\end{enumerate}
\end{lemma}

\begin{proof}
(1)  Restrict an MF embedding of the ambient group to the subgroup.
(2)  For a finite test set $F$, choose a finite quotient that is injective on
$F$ and compose it with the left regular permutation action.  This gives an
exact finite-dimensional representation.  Distinct permutations differ in a
column of their permutation matrices and so have operator-norm distance
at least $1$, so the group is MF.
(3)  A finite test set $F$ lies in a finite subgroup $K$.  The left
regular representation $\lambda_K\colon K\to\U(\C^{K})$ is a
finite-dimensional unitary representation of $K$ by permutation
matrices.  Because a local model is defined on all of $G$, set
\[
 V(g)=\lambda_K(g)\ \ (g\in K),\qquad V(g)=1\ \ (g\notin K).
\]
On the test set $F\subseteq K$, this map is exactly multiplicative,
since products of test elements lie in $K$ and $\lambda_K$ is a
homomorphism; the values outside $K$ do not enter the two conditions.
It separates as in (2).
\end{proof}

\begin{corollary}[quotients]\label{cor:quotclosure}
MF groups are not closed under quotients: the free group
\(F_8\) is MF, and the explicit group \(E\) is its quotient
but is not MF\@.
\leanverified{Sofic/LiteralMFQuotientControls}{GroupApproximation.LiteralMFQuotientControls.manuscriptQuotientNonclosure}
\end{corollary}

\begin{proof}
Map the eight free generators onto the eight displayed generators of \(E\).
The resulting homomorphism is surjective because those generators define the
displayed presentation.  The group \(F_8\) is residually finite, so MF by
Lemma~\ref{lem:permanence}(2), while \(E\) is not MF by Corollary~\ref{cor:marked-obstruction}.
\end{proof}

\section{Commuting orbits and the MF residual}\label{subsec:collapse}

\begin{theorem}[a commuting orbit forces commutation]\label{thm:projection-collapse}
Let $H$ be countable, let $L\le H$ have property~\textup{(T)}, let
$s\in H$ satisfy $s\gamma s^{-1}\in L$ for every $\gamma\in L$, and let
$\Theta\colon H\to\U(\mathcal Q)$ be a homomorphism.  Let
$p\in\mathcal Q$ be a projection such that $p$ commutes with
$\Theta(s\gamma s^{-1})$ for every $\gamma\in L$ and the projections
$\Theta(\gamma)\,p\,\Theta(\gamma)^*$, $\gamma\in L$, commute pairwise.
Then $p$ commutes with $\Theta(L)$.
\leanverified{Sofic/ProjectionCompressionCollapse}{GroupApproximation.ProjectionCompressionCollapse.corona_projection_collapse}
\end{theorem}

\begin{proof}
Property~\textup{(T)} makes $L$ finitely generated; fix a finite
symmetric generating set $S=\{a_1,\dots,a_m\}$.  Write
$p_g=\Theta(g)\,p\,\Theta(g)^{*}$ and $\Delta_g=p_g-p$ for $g\in L$, and
suppose $\Delta_{g_0}\ne0$ for some $g_0\in L$.

Since $p_g$ and $p$ are commuting projections, $\Delta_g$ is self-adjoint
with $\Delta_g^{3}=\Delta_g$, so $\Delta_g^{2}$ is a projection and $\Delta_g\Delta_g^{2}=\Delta_g$;
write $q_g=\Delta_g^{2}$.  All the $p_g$ commute, and then so do all the $\Delta_g$
and all the $q_g$.  Put
\[
 q=1-\prod_{a\in S}(1-q_a),
\]
the join of the projections $q_a$, $a\in S$.  The elements $g$ with
$p_g=p$ form a subgroup of $L$, so if $q=0$, that is, $\Delta_a=0$ for
every $a\in S$, then $\Delta_g=0$ for every $g\in L$, contradicting
$\Delta_{g_0}\ne0$.  Thus $q\ne0$.

\emph{The $q$-normalized Hilbert space.}  Take a self-adjoint bounded
lift $(x_n)$ of $q$.  Since $\opnorm{x_n^{2}-x_n}\to0$, for all
sufficiently large $n$ the spectrum of $x_n$ avoids $[1/4,3/4]$.
Choose a continuous function $f\colon\mathbb R\to[0,1]$ with
$f=0$ on $(-\infty,1/4]$ and $f=1$ on $[3/4,\infty)$.  Continuous
functional calculus then gives, for all sufficiently large $n$, a
projection $Q_n=f(x_n)$ with $\opnorm{Q_n-x_n}\to0$; define $Q_n=0$
on the finitely many remaining coordinates.  Thus $(Q_n)$ is a
projection lift of $q$.

Let $J=\{n:Q_n\ne0\}$.  Because $q\ne0$, the set $J$ is infinite.
Restriction of coordinates induces a $*$-homomorphism
\[
 \rho_J\colon
 \prod_nM_{d_n}(\mathbb C)/\bigoplus_nM_{d_n}(\mathbb C)
 \longrightarrow
 \prod_{n\in J}M_{d_n}(\mathbb C)/\bigoplus_{n\in J}M_{d_n}(\mathbb C).
\]
Replace $\Theta$ by $\rho_J\circ\Theta$ and every element introduced
above by its image under $\rho_J$.  All projection, commutation and
cocycle identities are preserved, while the new image of $q$ is still
nonzero because its lift consists of nonzero projections and so
has norm $1$ at every coordinate of $J$.  Relabel $J$ by $\mathbb N$;
then, without changing notation, we have
$k_n=\operatorname{rank}Q_n\ge1$ for every $n$.

Give $M_{d_n}(\C)$ the inner product
$\langle x,y\rangle_n=\operatorname{Tr}(y^{*}x)/k_n$, write
$\lVert x\rVert_{\mathrm F}=\operatorname{Tr}(x^{*}x)^{1/2}$ for the
unnormalized Hilbert--Schmidt norm, and let $K_\omega$ be the
Hilbert-space ultraproduct of these spaces along a free ultrafilter
$\omega$.

\emph{The corner-supported ideal.}  Let $\mathcal I_q$ be the
algebraic two-sided ideal generated by $q$, namely the set of finite
sums $\sum_{j\le r}x_jqy_j$ with $x_j,y_j\in\mathcal Q$.  It is
invariant under inner automorphisms.  It is not closed in norm; the
finite-sum description gives the rank bound \eqref{eq:rank-frobenius}
below.

For $z=\sum_{j\le r}x_jqy_j\in\mathcal I_q$, choose bounded coordinate
lifts $X_{j,n}$ of $x_j$ and $Y_{j,n}$ of $y_j$ and put
$Z_n=\sum_{j\le r}X_{j,n}Q_nY_{j,n}$; then
$\operatorname{rank}(Z_n)\le rk_n$.  The square of the Frobenius norm
is the sum of the squares of the singular values; at most
$\operatorname{rank}(Z_n)$ of them are nonzero and each is at most the
operator norm, so
\begin{equation}\label{eq:rank-frobenius}
 \lVert Z_n\rVert_{\mathrm F}^{2}
 \;\le\;\operatorname{rank}(Z_n)\opnorm{Z_n}^{2}
 \;\le\;r\,k_n\opnorm{Z_n}^{2}
\end{equation}
and $(Z_n)$ is a bounded sequence for the inner products
$\langle\cdot,\cdot\rangle_n$.  Its class in $K_\omega$ depends only on
$z=\sum_jx_jqy_j$: a sequence $Z_n'$ obtained from a second
decomposition, with $r'$ summands, satisfies $\opnorm{Z_n-Z_n'}\to0$
and $\operatorname{rank}(Z_n-Z_n')\le(r+r')k_n$, so
$\lVert Z_n-Z_n'\rVert_{\mathrm F}^{2}/k_n\to0$ by
\eqref{eq:rank-frobenius}.  Write $\Lambda(z)=[Z_n]_\omega$.  The map
$\Lambda\colon\mathcal I_q\to K_\omega$ is linear,
\[
 \lVert\Lambda(q)\rVert
 =\lim_\omega\Bigl(\frac{\operatorname{Tr}Q_n}{k_n}\Bigr)^{1/2}=1,
\]
and $\Lambda(z)=0$ implies $\Lambda(xzy)=0$ for all
$x,y\in\mathcal Q$, since bounded lifts of $x$ and $y$ multiply
$\lVert Z_n\rVert_{\mathrm F}$ by a bounded factor.

\emph{An exact cocycle.}  For $a\in S$ the projection $q_a$ annihilates
$\prod_{b\in S}(1-q_b)$, so $q_aq=q_a$ and
$\Delta_a=\Delta_aq_a=\Delta_aq\in\mathcal I_q$.  From
$p_{gh}=\Theta(g)\,p_h\,\Theta(g)^{*}$,
\begin{equation}\label{eq:collapse-cocycle}
 \Delta_{gh}=\Delta_g+\Theta(g)\,\Delta_h\,\Theta(g)^{*}\qquad(g,h\in L),
\end{equation}
and $\mathcal I_q$ is inner-invariant, so $\Delta_g\in\mathcal I_q$ for
every $g\in L$.

Choose unitary coordinate lifts $U_n(h)$ of $\Theta(h)$.  The formula
$\pi(h)=[\Ad U_n(h)]_\omega$ defines a unitary representation of $H$ on
$K_\omega$: conjugation by a unitary preserves each
$\langle\cdot,\cdot\rangle_n$, and
$\opnorm{\Ad u-\Ad v}\le2\opnorm{u-v}$.  Moreover,
$\pi(h)\Lambda(z)=\Lambda\bigl(\Theta(h)\,z\,\Theta(h)^{*}\bigr)$ on
the closed invariant subspace
$K_q=\overline{\Lambda(\mathcal I_q)}$.  By
\eqref{eq:collapse-cocycle} the map $\beta(g)=\Lambda(\Delta_g)$ satisfies
\[
 \beta(gh)=\beta(g)+\pi(g)\beta(h)\qquad(g,h\in L),
\]
a $1$-cocycle for $\pi|_L$ with values in $K_q$.

\emph{The cocycle is nonzero.}  Suppose $\beta(a)=0$ for every $a\in S$.  Put $r_1=1$ and
$r_i=\prod_{j<i}(1-q_{a_j})$ for $i\ge2$.  The products
$e_i=r_iq_{a_i}$ are projections with $e_i=r_i-r_{i+1}$, so they are
pairwise orthogonal and $\sum_ie_i=1-r_{m+1}=q$.  Each $\Delta_{a_i}$
commutes with $r_i$, so $e_i=(\Delta_{a_i}r_i)\,\Delta_{a_i}$ and
$\Lambda(e_i)=0$, whence $\Lambda(q)=0$ against
$\lVert\Lambda(q)\rVert=1$.  So $\beta$ is not identically zero.

\emph{Property~\textup{(T)}.}  By the Delorme--Guichardet theorem
\cite{Delorme,Guichardet}, $\beta$ is a coboundary: there is $y\in K_q$
with $\beta(g)=y-\pi(g)y$ for every $g\in L$.  The hypothesis on $p$
gives $\Delta_{s\gamma s^{-1}}=0$, so
$\pi(s\gamma s^{-1})y=y$ for every $\gamma\in L$.

Normalize the Hilbert--Schmidt norms by $k_n=\operatorname{rank}Q_n$
and repeat, with this weight, the fixed-space
comparison in the proof of Theorem~\ref{thm:transport-variants}.  The
projection $P$ onto $\operatorname{Fix}\pi(L)$ and its conjugate
$Q=\pi(s)P\pi(s)^{*}$ onto
$\operatorname{Fix}\pi(sLs^{-1})$ lie in the same finite norm
ultraproduct, with $P\le Q$ and $P\sim Q$.  Finiteness of the norm
ultraproduct gives $P=Q$.  So $\pi(g)y=y$ and
$\beta(g)=0$ for every $g\in L$, contradicting the previous paragraph.
So $\Delta_g=0$ for every $g\in L$ in the original corona; equivalently,
$p$ commutes with $\Theta(g)$ for every $g\in L$.
\end{proof}

Equivalently, every projection of $\Theta(sLs^{-1})'\cap\mathcal Q$
whose $\Theta(L)$-conjugates commute pairwise already lies in
$\Theta(L)'\cap\mathcal Q$.

\begin{definition}[involutive orbit witness]\label{def:invwitness}
Let $H$ be a group, $L\le H$ a subgroup, and $s\in H$.  An element
$k\in H$ is an \emph{involutive orbit witness} for $(L,s)$ if
\begin{enumerate}
\item[(W1)] $k^2=1$;
\item[(W2)] $s\gamma s^{-1}$ commutes with $k$ for every $\gamma\in L$;
\item[(W3)] the conjugates $\gamma_1k\gamma_1^{-1}$ and
  $\gamma_2k\gamma_2^{-1}$ commute for all $\gamma_1,\gamma_2\in L$.
\end{enumerate}
The pair $(L,s)$ determines the normal subgroup
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptInvolutiveCollapsePattern}
\[
  D_{\mathrm{coll}}(L,s)
  \;=\;
  \bigl\langle\!\bigl\langle\,
  [\gamma,k]\;:\;\gamma\in L,\
  k\text{ a witness for }(L,s)
  \,\bigr\rangle\!\bigr\rangle_H .
\]
\end{definition}

In a permutational wreath product $K^{(X)}\rtimes G$ with $L\le G$
conjugated into itself by $s$ and $x_0\in X$ a point fixed by
$sLs^{-1}$, every involution in the copy of $K$ supported at $x_0$ is
a witness: distinct coordinate copies commute, and $sLs^{-1}$ fixes
$x_0$.

\begin{theorem}[the orbit of an involutive witness]\label{thm:collapse}
Let $H$ be countable, let $L\le H$ be a subgroup with
property~\textup{(T)}, and let $s\in H$ satisfy
$s\gamma s^{-1}\in L$ for every $\gamma\in L$.  Then
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptInvolutiveCollapse}
\[
  D_{\mathrm{coll}}(L,s)\;\le\;\operatorname{Res}_{\mathrm{MF}}(H).
\]
\end{theorem}

\begin{proof}
Let $\Theta\colon H\to\U(\mathcal Q)$ be a homomorphism and $k$ a
witness for $(L,s)$.  By (W1) the unitary $u=\Theta(k)$ is
self-adjoint, so $p=\tfrac12(1-u)$ is a projection, and
\[
 \Theta(\gamma)\,p\,\Theta(\gamma)^{*}
 =\tfrac12\bigl(1-\Theta(\gamma k\gamma^{-1})\bigr)
 \qquad(\gamma\in L).
\]
By (W2) the projection $p$ commutes with $\Theta(s\gamma s^{-1})$ for
$\gamma\in L$, and by (W3) the displayed projections commute pairwise.
By Theorem~\ref{thm:projection-collapse}, $p$ commutes with
$\Theta(L)$; then so does $u=1-2p$, and $\Theta([\gamma,k])=1$ for every
$\gamma\in L$.  Kernels are normal, so $\Theta$ is trivial on
$D_{\mathrm{coll}}(L,s)$.
\end{proof}

\begin{remark}[finite-order witnesses]\label{rem:torsion-collapse}
Condition (W1) can be relaxed to finite order.  Let $H$ be countable,
let $L\le H$ have property~\textup{(T)}, let $s\in H$ satisfy
$s\gamma s^{-1}\in L$ for every $\gamma\in L$, and let $k\in H$ have
finite order $m$.  If $sLs^{-1}$ centralizes $k$ and the conjugates
$\gamma k\gamma^{-1}$, $\gamma\in L$, commute pairwise, then every
homomorphism $\Theta\colon H\to\U(\mathcal Q)$ satisfies
$\Theta([\gamma,k])=1$ for every $\gamma\in L$.  For
$j\in\mathbf{Z}/m$, let
\[
 p_j=\frac1m\sum_{t\in\mathbf{Z}/m}
 e^{-2\pi ijt/m}\Theta(k)^t.
\]
These are the spectral projections of $\Theta(k)$.  The hypotheses
imply that $\Theta(sLs^{-1})$ centralizes every $p_j$ and that their
$\Theta(L)$-conjugates commute pairwise.  So every $p_j$ commutes with $\Theta(L)$ by
Theorem~\ref{thm:projection-collapse}.  Fourier
reconstruction,
$\Theta(k)=\sum_j e^{2\pi ij/m}p_j$, then shows that $\Theta(k)$ also
commutes with $\Theta(L)$.  The normal closure of all such commutators,
over all finite-order witnesses for $(L,s)$, lies in
$\operatorname{Res}_{\mathrm{MF}}(H)$, with equality whenever the
corresponding quotient of $H$ is MF\@.
\leanverified{Sofic/TorsionSpectralCollapse}{GroupApproximation.TorsionSpectralCollapse.actualCoronaMFInvisible_of_torsionWitness}
\leanverified{Sofic/TorsionSpectralCollapse}{GroupApproximation.TorsionSpectralCollapse.torsionCollapseDefect_le_actualCoronaMFResidual_of_kazhdan}
\leanverified{Sofic/TorsionSpectralCollapse}{GroupApproximation.TorsionSpectralCollapse.actualCoronaMFResidual_eq_torsionCollapseDefect_of_kazhdan}
\end{remark}

\begin{proof}[Proof of Theorem~\ref{thm:signfree}]
Write $q\colon E\to E/\langle w\rangle$, $\bar c=q(\hat c)$, and
$\bar L=q(\iota(\mathcal B))$.  The element $\bar c$ is an involution
because $c^2=1$, and $q(t)\bar Lq(t)^{-1}$ centralizes $\bar c$
because conjugating the relation $[c,\iota(\mathcal B)]=1$
of \eqref{eq:E} by $t$ gives
$[\,t\iota(\mathcal B)t^{-1},\,\hat c\,]=1$ in $E$, where $\hat c=tct^{-1}$.  The
subgroup that centralizes $\hat c$ is the conjugate one, which is why the
relation is conjugated by $t$.  Property~\textup{(T)} passes to $\bar L$,
and $q(t)$ conjugates it into itself.  Thus (W1) and (W2) of Definition~\ref{def:invwitness} hold, and only
(W3) remains.

\emph{Pairwise commutation of the orbit.}  The linear generators $x,y,z$
commute with $\hat c$, and conjugation by
$t$ sends each $v_i$ to $v_i^2$, so $\iota(v_i^2)$ commutes with $\hat c$
as well.  So the conjugate $\iota(g)\,\hat c\,\iota(g)^{-1}$
depends only on the exponent vector of the translation part of $g$
modulo $2$, so there are exactly eight such conjugates, indexed by
$(\Z/2)^3$.  By Definition~\ref{def:E}, the distinguished word is
$w=[\hat c,\iota(v_1)\,\hat c\,\iota(v_1)^{-1}]$, the commutator attached to
the class of $v_1$.  The linear part acts through
$\SL_3(\mathbb Z/2)$, which is transitive on the nonzero parity
classes; the seven words
$1,x,x^2,z,yx,y^2x,xyx$ in the linear generators send the class of $v_1$
to all seven nonzero classes, and each commutes with $\hat c$.
Conjugating the displayed identity by the corresponding $\iota(r)$
shows that every nonzero parity commutator is a conjugate of $w$; the
zero class gives $[\hat c,\hat c]=1$.  Since $w$ is central, all eight lie in
$\langle w\rangle$.  Conjugating the commutator of two arbitrary
conjugates by the inverse of one of them reduces it to this case.  So all conjugates of
$\bar c$ under $\bar L$ commute pairwise.  Thus $\bar c$ is an
involutive orbit witness for $(\bar L,q(t))$, and by
Theorem~\ref{thm:collapse}
$[q(\iota(v_1)),\bar c]\in
\operatorname{Res}_{\mathrm{MF}}(E/\langle w\rangle)$.

The homomorphism $E\to W$ of Proposition~\ref{prop:witness} sends
$[\iota(v_1),\hat c]$ to the
product of the Clifford generators at the two distinct points $\tau o$
and $j(\bar v_1)\tau o$, and sends $\langle w\rangle$ into
$\langle\zeta\rangle$.  The product of the Clifford generators at two
distinct points does not lie in $\langle\zeta\rangle$.  So
$[\iota(v_1),\hat c]\notin\langle w\rangle$, and its image in
$E/\langle w\rangle$ is nontrivial.  This image lies in the MF
residual, so Proposition~\ref{prop:univquot} implies that the quotient
is non-MF\@.
Corollary~\ref{cor:pullback} identifies
$\operatorname{Res}_{\mathrm{MF}}(E)$ with the $q$-preimage of
$\operatorname{Res}_{\mathrm{MF}}(E/\langle w\rangle)$, so $u$ belongs
to it.
\end{proof}

\begin{corollary}[collapse computes the residual]\label{cor:collapsequot}
In the setting of Theorem~\ref{thm:collapse}, write
$q\colon H\to H/D_{\mathrm{coll}}(L,s)$ for the quotient map.  Then
\[
  \operatorname{Res}_{\mathrm{MF}}(H)
  \;=\;
  q^{-1}\bigl(\operatorname{Res}_{\mathrm{MF}}
  (H/D_{\mathrm{coll}}(L,s))\bigr),
\]
and if $H/D_{\mathrm{coll}}(L,s)$ is MF, then
$\operatorname{Res}_{\mathrm{MF}}(H)=D_{\mathrm{coll}}(L,s)$.
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptCollapseRadicalReduction}
\end{corollary}

\begin{proof}
Combine Theorem~\ref{thm:collapse} with
Corollaries~\ref{cor:pullback} and~\ref{cor:exactradical}.
\end{proof}

If $H/D_{\mathrm{coll}}(L,s)$ is MF, then it is the universal MF
quotient of $H$ (Proposition~\ref{prop:univquot}): every homomorphism
from $H$ to an MF group factors through it.

\subsection{The exact residual of $E$}\label{sec:exact-residual}

\begin{theorem}[the exact MF residual]\label{thm:exact-mf-residual}
For the group $E$ of Definition~\ref{def:E}, put $d=\hat c=tct^{-1}$,
$u=[d,v_1]$, and $D=\langle\!\langle u\rangle\!\rangle_E$.  Then
\[
 \operatorname{Res}_{\mathrm{MF}}(E)=D=N_{\mathrm{conj}},
 \qquad
 E/D\cong \Sigma*_{\mathcal B}(\mathcal B\times C_2),
\]
and $E/D$ is MF\@.  Moreover, if $q\colon E\to E/\langle w\rangle$ and
$D_{\mathrm{coll}}$ is the collapse defect used in the proof of
Theorem~\ref{thm:signfree}, then
\[
 \operatorname{Res}_{\mathrm{MF}}(E/\langle w\rangle)
   =D_{\mathrm{coll}}=D/\langle w\rangle,
 \qquad
 \operatorname{Res}_{\mathrm{MF}}(E)=q^{-1}(D_{\mathrm{coll}}).
\]
\end{theorem}

\begin{proof}
$d$ commutes with the linear part of $\mathcal B$ and with every
even translation.  So $[d,b]$ depends only on the translation part of
$b$ modulo $2$.  The linear part is transitive on the seven nonzero classes
of $(\mathbb Z/2)^3$, by the explicit seven-word calculation used in the
proof of Theorem~\ref{thm:signfree}; so every nontrivial $[d,b]$ is a
conjugate of $u$.  This proves $D=N_{\mathrm{conj}}$.  Moreover,
$u\in\operatorname{Res}_{\mathrm{MF}}(E)$ by
Theorem~\ref{thm:signfree}; since the residual is normal,
$D=\langle\!\langle u\rangle\!\rangle_E
\le\operatorname{Res}_{\mathrm{MF}}(E)$.  Universal $2$-norm triviality
alone does not imply this corona statement (Section~\ref{sec:limits}).

In the quotient by $D$, the involution $d$ commutes with all of
$\mathcal B$.  Replacing $c$ by $d$, so that $c=t^{-1}dt$, makes the old
relation $[c,\mathcal B]=1$ redundant after conjugation by $t$; the
central-mark relations are redundant as well because $w=u^2$.  The remaining
presentation is precisely
\[
 \langle\mathcal B,t,d:\ tbt^{-1}=\alpha(b),\ d^2=1,
 [d,\mathcal B]=1\rangle
 \cong \Sigma*_{\mathcal B}(\mathcal B\times C_2).
\]
The group $\Sigma$ is finitely generated linear in characteristic zero and
so residually finite \cite{Malcev}.  Put $G_0=\Sigma\times C_2$.  Choose a
separating chain of finite quotients of $G_0$, let $A$ be the $C^*$-algebra
generated by their regular representations in the norm matrix corona, and let
$\tau$ be an ultralimit of the normalized traces.  The algebra $A$ is
separable because it is generated by the countable family of group unitaries;
it is MF and
$\tau(g)=0$ for every $g\ne1$.  Put $C=C^*(\mathcal B)\subseteq A$.
Shulman's symmetric amalgam theorem \cite[Theorem~10]{Shulman} makes
the full unital amalgam $A*_{C}A$ MF\@.  The canonical homomorphism
$G_0*_{\mathcal B}G_0\to\U(A*_{C}A)$ is injective.  Indeed, in the GNS von
Neumann algebra of $(A,\tau)$, the trace-preserving conditional expectation
onto $C''$ sends every group unitary outside $\mathcal B$ to zero.  So the
reduced amalgam trace vanishes on every nontrivial Bass--Serre reduced
word, and $G_0*_{\mathcal B}G_0$ is MF\@.  Normal forms for amalgamated free products give
an embedding
\[
 \Sigma*_{\mathcal B}(\mathcal B\times C_2)
 \hookrightarrow G_0*_{\mathcal B}G_0,
\]
using the first vertex group for $\Sigma$ and the second for
$\mathcal B\times C_2$.  Subgroups of MF groups are MF, so $E/D$ is MF\@.
Corollary~\ref{cor:exactradical} now gives
$\operatorname{Res}_{\mathrm{MF}}(E)=D$.  Applying
Corollary~\ref{cor:pullback} first to $\langle w\rangle\le D$ and then to
the collapse defect gives the two identities downstairs.  Explicitly,
$q(u)^{-1}=[q(\iota(v_1)),\bar c]$ is one of the generators of
$D_{\mathrm{coll}}$, so $D/\langle w\rangle\le D_{\mathrm{coll}}$; the
reverse inclusion follows from the residual calculation just proved.
\end{proof}

\section{Spectral components and exact finite-lamp quotients}
\label{sec:spectral-visible-notes}

\subsection{Spectral-component rigidity}

\begin{theorem}[spectral-component motion]\label{thm:notes-spectral-motion}
Let $L\le H$ have property~\textup{(T)}, let $sLs^{-1}\le L$, and let
$\pi\colon H\to\U(\mathcal Q)$ be a homomorphism to a norm matrix corona.
Suppose that $v\in\mathcal Q$ is normal, commutes with
$\pi(sLs^{-1})$, and has pairwise commuting $L$-conjugates.  Then every
clopen spectral projection of $v$ commutes with $\pi(L)$ and, for
$\gamma\in L$,
\[
 \opnorm{\pi(\gamma)v\pi(\gamma)^*-v}
 \le
 \sup\{\operatorname{diam}(C):
 C\text{ is a connected component of }\sigma(v)\}.
\]
In particular, if every connected component of $\sigma(v)$ is a singleton,
then $v$ commutes with $\pi(L)$.
\leanverified{Manuscript/SpectralPaper/MainTheorems}{GroupApproximation.SpectralPaper.spectralCompressionTheorem}
\leanverified{Manuscript/SpectralPaper/MainTheorems}{GroupApproximation.SpectralPaper.zeroDimensionalSpectralCollapseTheorem}
\end{theorem}

\begin{proof}
For a clopen set $U\subseteq\sigma(v)$, continuous functional calculus gives
the projection $p_U=1_U(v)$.  Functional calculus is equivariant under
unitary conjugation, so the hypotheses on $v$ imply the hypotheses of
Theorem~\ref{thm:projection-collapse} for $p_U$.  So every $p_U$ commutes
with $\pi(L)$.

Fix $\gamma\in L$ and put $w=\pi(\gamma)v\pi(\gamma)^*$.  The algebra
$C^*(1,v,w)$ is commutative.  If $\chi$ is a character of this algebra, then
$\chi(v)$ and $\chi(w)$ belong to exactly the same clopen subsets of
$\sigma(v)$.  In a compact Hausdorff space, points which cannot be separated
by a clopen set lie in the same connected component.  Thus
\[
 |\chi(w)-\chi(v)|
 \le \sup_C\operatorname{diam}(C).
\]
Taking the supremum over all characters proves the norm estimate.  If all
components are singletons, the right-hand side is zero.
\end{proof}

\subsection{The heightwise abelian quotient}

Let
\[
 \Gamma_* = \mathbb Z^3\rtimes\SL_3(\mathbb Z),
 \qquad \alpha(v,A)=(2v,A),
\]
let $G_*$ be the ascending HNN extension of $\Gamma_*$ along $\alpha$, and
put $X=G_*/\Gamma_*$.  The height homomorphism $h\colon G_*\to\mathbb Z$
induces a map $\eta\colon X\to\mathbb Z$.  For a finite group $K$, put
\[
 W_K=K^{(X)}\rtimes G_*.
\]
Writing $K_{\mathrm{ab}}=K/[K,K]$, define
\[
 \Phi_K\colon W_K\longrightarrow
 P_K:=K_{\mathrm{ab}}^{(\mathbb Z)}\rtimes G_*
\]
by summing the abelianized lamp values over each height fibre.

\begin{theorem}[exact finite-lamp quotient]
\label{thm:notes-visible-quotient}
For every finite group $K$, every homomorphism from $W_K$ to a norm matrix
corona factors uniquely through $\Phi_K$.  Moreover
\[
 \begin{aligned}
 \ker\Phi_K
 &=\operatorname{Rad}_{\mathrm{MF}}(W_K)
  =\operatorname{Rad}_{\mathrm{fd}}(W_K)
  =\operatorname{Res}_{\mathrm{fin}}(W_K)\\
 &=\operatorname{Rad}_{\mathrm{lin}}(W_K)
  =\operatorname{Rad}_{\mathrm{Bohr}}(W_K).
 \end{aligned}
\]
Equivalently, precomposition with $\Phi_K$ is a bijection on homomorphisms
into norm matrix coronas, residually finite groups, finite-dimensional linear
groups over arbitrary fields, and compact Hausdorff groups.
\end{theorem}

\begin{proof}
Put $\Gamma_n=t^{-n}\Gamma_*t^n$ and $X_n=G_*/\Gamma_n$.  The subgroups
$\Gamma_n$ increase, and
\[
 N:=\bigcup_{n\geq0}\Gamma_n=\ker h,
 \qquad G_*/N\cong\mathbb Z.
\]
Let $\Phi_1$ abelianize the lamp values and push them from $X$ to $X_1$.
For $n\geq1$, let $\psi_n$ push
$K_{\mathrm{ab}}^{(X_n)}$ to $K_{\mathrm{ab}}^{(X_{n+1})}$, and write
$\Phi_n=\psi_{n-1}\cdots\psi_1\Phi_1$.

For $\Phi_1$, apply Theorem~\ref{thm:notes-spectral-motion} with
$L=\Gamma_*$ and $s=t$ to $k_{t\Gamma_*}$, where $k\in K$.  Every corona
image of this lamp has finite spectrum, commutes with the image of
$t\Gamma_*t^{-1}$, and has pairwise commuting $\Gamma_*$-conjugates.  Every corona
homomorphism $\pi$ then satisfies
\[
 \pi(\gamma k_{t\Gamma_*}\gamma^{-1})=\pi(k_{t\Gamma_*})
 \qquad(\gamma\in\Gamma_*).
\]
Conjugating these equalities by elements of $G_*$ identifies the lamps in
each fibre of $X\to X_1$.  Each fibre
contains distinct sites, so commutation between different site groups makes
the identified copy of $K$ abelian.  These equalities generate
$\ker\Phi_1$.  So
$\ker\Phi_1\leq\operatorname{Rad}_{\mathrm{MF}}(W_K)$.

For $n\geq1$, apply the same theorem in
$K_{\mathrm{ab}}^{(X_n)}\rtimes G_*$ with $L=\Gamma_n$ and $s=t$.  A lamp at
$t\Gamma_n$ commutes with $t\Gamma_nt^{-1}=\Gamma_{n-1}$, and its
$\Gamma_n$-conjugates commute pairwise.  Every corona homomorphism identifies
the lamps in each fibre of $X_n\to X_{n+1}$.  These equalities generate
$\ker\psi_n$, so
$\ker\psi_n\leq\operatorname{Rad}_{\mathrm{MF}}
(K_{\mathrm{ab}}^{(X_n)}\rtimes G_*)$.  Corollary~\ref{cor:pullback}, applied
inductively, gives
\[
 \ker\Phi_n\leq\operatorname{Rad}_{\mathrm{MF}}(W_K)
 \qquad(n\geq1).
\]

The map $\Phi_K$ is the pushforward to $G_*/N$.  If a finitely supported lamp
configuration belongs to $\ker\Phi_K$, choose $n$ such that any two points of
its support with the same image in $G_*/N$ already have the same image in
$X_n$.  The configuration then belongs to $\ker\Phi_n$.  Thus
\[
\ker\Phi_K=\bigcup_{n\geq1}\ker\Phi_n
 \leq\operatorname{Rad}_{\mathrm{MF}}(W_K),
\]
so every corona homomorphism annihilates $\ker\Phi_K$.

The homomorphism
\[
 P_K\longrightarrow G_*\times
 \bigl(K_{\mathrm{ab}}^{(\mathbb Z)}\rtimes\mathbb Z\bigr),
 \qquad (f,g)\longmapsto\bigl(g,(f,h(g))\bigr),
\]
is injective.  The group $G_*$ and the wreath product
$K_{\mathrm{ab}}^{(\mathbb Z)}\rtimes\mathbb Z$
are residually finite, so $P_K$ is residually finite.  Its finite quotients
separate every element outside $\ker\Phi_K$.  So the finite,
finite-dimensional unitary, and MF residuals equal
$\ker\Phi_K$.  A
finite-dimensional linear image is residually finite by Mal'cev's theorem,
and finite-dimensional unitary representations separate points of compact
groups by Peter--Weyl.  So the linear and Bohr residuals have the same
kernel.  Surjectivity of $\Phi_K$ gives uniqueness.
\end{proof}

\begin{corollary}[one relation changes the MF property]
\label{cor:notes-a5-relation}
Take $K=A_5$ and let $r$ be the one-site lamp with value $(12)(34)$.  Then
\[
 \operatorname{Rad}_{\mathrm{MF}}(W_{A_5})
 =A_5^{(X)}
 =\langle\!\langle r\rangle\!\rangle_{W_{A_5}},
 \qquad
 W_{A_5}/\langle\!\langle r\rangle\!\rangle\cong G_*.
\]
Thus $W_{A_5}$ is sofic and non-MF, while the quotient is residually finite
and MF, although every homomorphism from $W_{A_5}$ to a norm matrix corona
already maps $r$ to $1$.
\end{corollary}

\begin{proof}
The group $A_5$ is perfect, so $P_{A_5}=G_*$.  It is also simple, so the
normal closure of any nonidentity element in one site is the entire site
copy.  Transitivity of the $G_*$-action on $X$ then gives the displayed
normal-closure identity.  Theorem~\ref{thm:notes-visible-quotient} computes
the MF radical.  Telescope-window approximations make $W_{A_5}$ sofic, while
congruence quotients make $G_*$ residually finite.
\end{proof}

\subsection{The abelian boundary}

\begin{theorem}\label{thm:notes-abelian-boundary}
For every countable abelian group $A$,
\[
 A^{(X)}\rtimes G_*\text{ is MF}
 \quad\Longleftrightarrow\quad
 A\text{ is torsion-free}.
\]
\end{theorem}

\begin{proof}
If $A$ has nontrivial torsion, a finite-order one-site lamp has
zero-dimensional spectrum in every corona representation.  Theorem
\ref{thm:notes-spectral-motion} identifies its image with the image of each
$\Gamma_*$-translate in the same fibre of $X\to X_1$.  For a nonidentity
torsion element $a\in A$, choose $\gamma\in\Gamma_*$ such that
$\gamma t\Gamma_*\ne t\Gamma_*$; such an element exists because
$[\Gamma_*:t\Gamma_*t^{-1}]=8$.  Every corona homomorphism annihilates
$a_{\gamma t\Gamma_*}a_{t\Gamma_*}^{-1}$, which is nonidentity in
$A^{(X)}$.

Suppose that $A$ is torsion-free.  The embedding
$A\to A\otimes_{\mathbb Z}\mathbb Q$ is injective, and the countable
$\mathbb Q$-vector space $A\otimes_{\mathbb Z}\mathbb Q$ embeds in a
$\mathbb Q$-linear complement of $2\pi\mathbb Q$ in $\mathbb R$.  Composing
the two embeddings gives an injective additive map
$\theta\colon A\to\mathbb R$ whose image meets $2\pi\mathbb Z$ only at zero.
Put
$\Gamma_1=t\Gamma_*t^{-1}$ and identify $G_*/\Gamma_1$ with $X$ by
$g\Gamma_1\mapsto gt\Gamma_*$.  Choose representatives
$x_j=r_j\Gamma_1$ for the points of $G_*/\Gamma_1$, and let $T$ be a finite
generating set of $\Gamma_1$.

Residual finiteness of $G_*$ gives finite quotients
$q_n\colon G_*\to F_n$ which separate the first $n$ nonidentity elements of
$G_*$ and satisfy
\[
 q_n(r_j^{-1}r_i)\notin
 \{\text{$q_n(T)$-words of length at most $3n$}\}
 \quad(1\leq i\ne j\leq n).
\]
Set $K_n=q_n(\Gamma_1)$, let $\ell_n$ be word length in $K_n$ with respect
to $q_n(T)$, and define
\[
 f_n(y)=
 \begin{cases}
  \max\{1-\ell_n(y)/n,0\},&y\in K_n,\\
  0,&y\notin K_n.
 \end{cases}
\]
On $\ell^2(F_n)$ define
\[
 D_n(a)\delta_y=e^{i\theta(a)f_n(y)}\delta_y,
 \qquad
 \Lambda_n(g)\delta_y=\delta_{q_n(g)y}.
\]
For fixed $a\in A$ and $u\in\Gamma_1$,
\[
 \|\Lambda_n(u)D_n(a)\Lambda_n(u)^*-D_n(a)\|
 \leq \frac{|\theta(a)|\,|u|_T}{n}.
\]
Consequently $D(a)=[D_n(a)]$ commutes with
$\Lambda(u)=[\Lambda_n(u)]$ in the norm matrix corona.  For
$x=r\Gamma_1$, set $U_x(a)=\Lambda(r)D(a)\Lambda(r)^*$.  This is independent
of the representative, the unitaries $U_x(a)$ commute, and
\[
 \Lambda(g)U_x(a)\Lambda(g)^*=U_{gx}(a).
\]
So they define a homomorphism
\[
 A^{(G_*/\Gamma_1)}\rtimes G_*\longrightarrow\mathcal U(\mathcal Q).
\]

Let $b$ be a nonzero lamp configuration and choose
$a_j=b(x_j)\ne0$.  For all sufficiently large $n$, evaluation at
$\delta_{q_n(r_j)}$ gives the eigenvalue $e^{i\theta(a_j)}$: the separation
condition makes the other lamps in the support contribute $1$.  So the
image of $b$ differs from the identity.  If an element has nontrivial
$G_*$-component $g$, then $q_n(g)\ne1$ for all sufficiently large $n$.
The corresponding matrix is a diagonal unitary times a fixed-point-free
permutation, so its distance from the identity is at least $\sqrt2$.
The homomorphism is injective, and the group is MF.
\end{proof}

\section{Group algebras, soficity, and permanence}
\label{sec:algebras-sofic}

\subsection{The reduced group algebra of \texorpdfstring{$E$}{E}}

\begin{lemma}\label{lem:faithfultrace}
Let $G_1$ be a countable discrete group.
\begin{enumerate}
\item The canonical tracial state $\tau$ on $\Cred(G_1)$ is
faithful.
\item Every unital $C^*$-algebra $A$ with a faithful tracial state
is stably finite: for every $k\ge1$, every isometry in $M_k(A)$ is a
unitary.
\leanverified{Sofic/ManuscriptExactWrappers}{GroupApproximation.ManuscriptExactWrappers.manuscriptFaithfulTraceAndStableFiniteness}
\end{enumerate}
\end{lemma}

\begin{proof}
(1) Realize $\Cred(G_1)\subseteq B(\ell^2G_1)$ as the norm closure
of the span of the left translation unitaries $\lambda_g$.  Let
$\rho_g$ be the right translation unitaries,
$(\rho_g\xi)(h)=\xi(hg)$.  Each $\rho_g$ commutes with each
$\lambda_{g'}$, so with every element of $\Cred(G_1)$.  The
canonical trace is $\tau(x)=\langle x\delta_e,\delta_e\rangle$.  If
$\tau(x^*x)=0$ then $\lVert x\delta_e\rVert=0$, so $x\delta_e=0$,
and then
$x\delta_g=x\rho_{g^{-1}}\delta_e=\rho_{g^{-1}}x\delta_e=0$ for
every $g\in G_1$.  The vectors $\delta_g$ are total in
$\ell^2G_1$, so $x=0$.

(2) On $M_k(A)$ the functional
$\tau_k\bigl([x_{ij}]\bigr)=\tfrac1k\sum_i\tau(x_{ii})$ is a state
with the trace property
$\tau_k(xy)=\tfrac1k\sum_{i,j}\tau(x_{ij}y_{ji})=\tau_k(yx)$.
It is faithful.  Indeed, the $i$-th diagonal entry of $x^*x$ is
$\sum_jx_{ji}^*x_{ji}$.  If $\tau_k(x^*x)=0$, then
$\tau(x_{ji}^*x_{ji})=0$ and so $x_{ji}=0$ for all $i,j$.  If
$v\in M_k(A)$ satisfies $v^*v=1$, then $y:=1-vv^*$ is a projection
with
$\tau_k(y)=\tau_k(1)-\tau_k(vv^*)=\tau_k(1)-\tau_k(v^*v)=0$; since
$y=y^*y$, faithfulness gives $y=0$, i.e.\ $vv^*=1$.
\end{proof}

\subsection{The Clifford quotient and permanence}

The map $E\to W$ of Proposition~\ref{prop:witness} is surjective: its image
contains $\Sigma$ and the generator $c_o$ at the base point, so
every Clifford generator by transitivity of
the $\Sigma$-action on $X$.  Since $\Sigma$ is finitely generated,
a finite generating set for $\Sigma$ together with $c_o$ generates $W$.
The map sends
$w\in\operatorname{Res}_{\mathrm{MF}}(E)$ to the nontrivial sign
$\zeta$, so $\zeta\in\operatorname{Res}_{\mathrm{MF}}(W)$ by
Lemma~\ref{lem:portable}.  So the finitely generated group $W$ is
not MF\@.  Neither is $\Cred(W)$: if $\Cred(W)$ embedded in a matrix
quotient, the map $u\mapsto e(u)+1-e(1)$ would give an injective
homomorphism $W\to\U(\mathcal Q)$, as in the proof of
Corollary~\ref{cor:marked-obstruction}.

\paragraph{Failure of extension permanence.}
MF groups are not closed under extensions.  More precisely, $W$ fits into
an exact sequence
\[
 1\longrightarrow\Cl(X)\longrightarrow W\longrightarrow\Sigma
 \longrightarrow1
\]
whose kernel is locally finite --- so LEF, sofic and MF --- and
whose quotient is MF, while $W$ itself is finitely generated and not
MF\@.
\leanverified{Sofic/LiteralWitnessConsequences}{GroupApproximation.LiteralWitnessConsequences.literalWitness_locallyFiniteKernel_nonMF}

\begin{proof}
The Clifford group $\Cl(X)$ is locally finite: every finitely
generated subgroup involves only finitely many sites, so is contained
in the corresponding finite Clifford group.  Then $\Cl(X)$ is MF
by Lemma~\ref{lem:permanence}(3).  The group $\Sigma$ identifies with
the subgroup of
$\GL_4(\mathbb Q)$ generated by $\bar\Gamma$ and $D$: the levels
are $D^{-n}\bar\Gamma D^{n}$, and the determinant gives the
$\mathbb Z$-coordinate.  It is residually finite.  Modulo every odd
integer, the dilation matrix is invertible because its determinant is
$8$; so the congruence reduction of the integral affine model is
compatible with the dilation and extends over the direct limit and
its shift.  These finite quotients separate every nontrivial element
of $\Sigma$, consistent with Mal'cev's theorem for finitely generated
linear groups \cite{Malcev}.  Thus $\Sigma$ is MF by Lemma~\ref{lem:permanence}(2),
whereas $W$ is not MF by the preceding construction.
\end{proof}

\paragraph{Extensions by $\Z$.}
The two canonical projections define a surjection $W\to\Z$ with kernel
$\Cl(X)\rtimes T$.  Every finite subset of the kernel lies in
$L\rtimes\bar\Gamma_n$ for some finite invariant subgroup
$L\le\Cl(X)$ and some level $\bar\Gamma_n$.  This semidirect product is
residually finite because the action on $L$ factors through a finite
quotient.  So the kernel is LEF and MF\@.  Thus $W$ is a non-MF
extension of an LEF group by $\Z$.  It is nevertheless sofic by the
Elek--Szab\'o amenable-extension theorem \cite{ElekSzabo}.
\leanverified{Sofic/LEFMarkedCompression}{GroupApproximation.LiteralLEFExtension.literalWitness_LEFKernel_integerExtension_nonMF}
\leanverified{Sofic/SoficByAmenablePermanence}{GroupApproximation.SoficByAmenablePermanence.manuscriptSoficOfSoficKerOfAmenableQuotient}

\paragraph{A simple sofic envelope.}
There is a countable simple sofic group $S_{\mathrm{simp}}$ with
\[
 \operatorname{Res}_{\mathrm{MF}}(S_{\mathrm{simp}})=S_{\mathrm{simp}}.
\]
Equivalently, every homomorphism from $S_{\mathrm{simp}}$ to an MF group is
trivial.

\begin{proof}
By Elek--Szab\'o, every countable sofic group embeds in a countable simple
sofic group \cite[Theorem~1]{ElekSzaboHyper}.  Embed $E$, which is sofic
by Theorem~\ref{thm:Esofic}, in such a group.  By functoriality of
the MF residual, the image of $w$ is a nontrivial element of the MF
residual of the simple group.  This residual is a nontrivial normal
subgroup, so simplicity makes it the whole group.  Every homomorphism to an MF group maps the MF residual to $1$.
\end{proof}
\leanverified{Sofic/SimpleSoficEnvelopeUnconditional}{GroupApproximation.SimpleSoficEnvelopeUnconditional.manuscript_simpleSoficEnvelope}

\subsection{Soficity of $E$}\label{sec:Esofic}

Deleting the letter $c$ and the families \textup{(C)} and \textup{(W)}
from \eqref{eq:E} leaves the presentation
$\langle\,\mathcal B,t\mid tut^{-1}=\bar\alpha(u)\,\rangle$ of the
ascending HNN extension of the base along the doubling endomorphism.
By Remark~\ref{rem:classical-base} the base is $\bar\Gamma$, so
\[
 E/\langle\!\langle c\rangle\!\rangle\;\cong\;\Sigma
\]
for the group $\Sigma=T\rtimes\Z$ of Section~\ref{sec:literal-presentation}, with
its site set $X=\Sigma/B$, base point $o$, shift $\tau$ and level-zero
base $B=j(\bar\Gamma)$.  The levels
\[
 \bar\Gamma_n=\tau^{-n}B\tau^{n}\qquad(n\ge0)
\]
form an ascending chain of copies of $\bar\Gamma$ with union
$T$.  Put $K=\tau^{-1}B\tau$; then $B\le K$ and
$[K:B]=[\bar\Gamma:\bar\alpha(\bar\Gamma)]=8$.  The cosets
$\Sigma/K$ are the \emph{blocks}, and each block, as a set of sites, has
$[K:B]$ elements.

Let $\mathcal G$ be the graph on $X$ whose edges are the $\Sigma$-orbit
of the pair $\{\tau o,\,j(\bar v_1)\tau o\}$, whose entries are
distinct by Lemma~\ref{lem:linear}.  Two sites are adjacent in
$\mathcal G$ exactly when they are distinct and lie in a common block.
Adjacent sites share a block because the two sites are $gB$ and $hB$
for the representatives $g=\tau$ and $h=j(\bar v_1)\tau$, and
$g^{-1}h=\tau^{-1}j(\bar v_1)\tau\in K$ since $j(\bar v_1)\in B$ and
$K=\tau^{-1}B\tau$; so $gK=hK$.  (Sites are cosets, so the product
and inverse have to be taken in $\Sigma$ on representatives, not on the
sites themselves.)  Conversely, the bijection
$u\,\bar\alpha(\bar\Gamma)\mapsto\tau^{-1}j(u)\tau o$ identifies the
block of $o$ with $\bar\Gamma/\bar\alpha(\bar\Gamma)$ and the
generating edge with the pair
$\{\bar\alpha(\bar\Gamma),\,\bar v_1\bar\alpha(\bar\Gamma)\}$.  Left
multiplication by $\bar\alpha(\bar\Gamma)$ on the coset space
$\bar\Gamma/\bar\alpha(\bar\Gamma)$ is transitive on the nontrivial
cosets, so every pair of distinct sites of a block is the image of
the generating pair.  Let $L=\langle x,y,z\rangle$ be the linear part,
so $\bar\Gamma=\Z^3\rtimes L$.  Since $\bar\alpha$ squares the three
translations and fixes $x,y,z$ (Lemma~\ref{lem:linear}), its image is
$(2\Z^3)\rtimes L$.  Reduction of the translation coordinate modulo~$2$
identifies $\bar\Gamma/\bar\alpha(\bar\Gamma)$ with $(\Z/2)^3$.
Under that identification left multiplication by an
element of $\bar\alpha(\bar\Gamma)$ with linear part $N$ and
(necessarily even) translation part sends a class $\bar u$ to
$\bar N\bar u$, so the action factors through the linear action of the
reduction $\bar L\le\GL_3(\mathbb F_2)$ of $L$.  The seven words
\[
 1,\quad x,\quad x^2,\quad z,\quad yx,\quad y^2x,\quad xyx
\]
send $\bar e_1$ to
$\bar e_1$, $\bar e_3$, $\bar e_2$, $(0,1,1)$, $(1,1,0)$, $(1,1,1)$,
$(1,0,1)$, the seven nonzero classes.  This proves the required
transitivity.
\leanverified{Sofic/LiteralAffineCosetTransitivity}{GroupApproximation.LiteralAffineCosetTransitivity.conjD_cosetTransitive}

For a graph $\mathcal G$ on $X$, let $C(\mathcal G)$ be the group
presented by generators $\zeta$ and $c_\xi$ $(\xi\in X)$ and relations
\[
 \zeta^2=c_\xi^2=1,\qquad [\zeta,c_\xi]=1,\qquad
 [c_\xi,c_\eta]=\zeta\ \ \text{for each edge }\{\xi,\eta\}
 \text{ of }\mathcal G .
\]
For the complete graph on $X$ this is $\Cl(X)$, and adjoining the anticommutation
relations at the remaining pairs gives a surjection
$C(\mathcal G)\to\Cl(X)$.

\begin{proposition}[normal form for $E$]\label{prop:blocknormalform}
There is an isomorphism
\[
 E\;\cong\;C(\mathcal G)\rtimes\Sigma
\]
under which $\langle\!\langle c\rangle\!\rangle$ is the factor
$C(\mathcal G)$, the conjugate $gcg^{-1}$ is the lamp $c_{gB}$, and the
distinguished word $w$ is the sign $\zeta$.  Each level $\bar\Gamma_n$ has
finite orbits on $X$.
\leanverified{Sofic/LiteralBlockNormalForm}{GroupApproximation.LiteralBlockNormalForm.markedGroupEquivModel}
\leanverified{Sofic/LiteralBlockNormalForm}{GroupApproximation.LiteralBlockNormalForm.lampKernelEquiv}
\leanverified{Sofic/LiteralBlockNormalForm}{GroupApproximation.LiteralBlockNormalForm.manuscriptSiteLampImage}
\leanverified{Sofic/LiteralBlockNormalForm}{GroupApproximation.LiteralBlockNormalForm.toModel_mark}
\leanverified{Sofic/LiteralBlockNormalForm}{GroupApproximation.LiteralBlockNormalForm.manuscriptFiniteLevelSiteOrbit}
\leanverified{Sofic/LiteralVerticalBridge}{GroupApproximation.LiteralVerticalBridge.verticalEquiv}
\end{proposition}

\begin{proof}
Send the six base letters and $t$ to the corresponding generators of
$\Sigma$, and $c$ to $c_o$.  Families~\textup{(B)} and~\textup{(T)} of
\eqref{eq:E} hold in $\Sigma$ by construction.  For family~\textup{(C)}:
$c_o^2=1$, and $B$ fixes $o$, so each base generator fixes $c_o$.  For
family~\textup{(W)}: the pair $\{\tau o,\,j(\bar v_1)\tau o\}$ is an
edge, so the image of $w$ is
$[c_{\tau o},c_{j(\bar v_1)\tau o}]=\zeta$, which is central in
$C(\mathcal G)$ and fixed by $\Sigma$.  This defines
$E\to C(\mathcal G)\rtimes\Sigma$.

In the other direction, send $\zeta$ to $w$ and $c_{gB}$ to
$gcg^{-1}$.  The assignment is independent of the coset representative,
because the base centralizes $c$.  The images are involutions, and $w$
is central of order two.  Along an edge
$g\{\tau o,\,j(\bar v_1)\tau o\}$ the commutator of the two images is
$gwg^{-1}=w$, and $C(\mathcal G)$ imposes relations only along edges.
So the assignment respects the defining relations of
$C(\mathcal G)$.
Combined with the splitting $\Sigma\to E$ on the seven
remaining letters, this gives $C(\mathcal G)\rtimes\Sigma\to E$.  The
two composites fix the eight letters of $E$ and the generators
$\zeta,c_\xi$ of $C(\mathcal G)$, so are the identity.

For the orbits, $\bar\alpha(\bar\Gamma)$ has index eight in
$\bar\Gamma$ --- the parity classes of the translation lattice --- so
$\tau$ conjugates $B$ onto a subgroup of finite index in itself.  So every
$\bar\Gamma_n$ is commensurable with $B$, all of $\Sigma$
commensurates $B$, and orbit--stabilizer makes each $\bar\Gamma_n$-orbit in
$X=\Sigma/B$ finite.
\end{proof}

Composing the isomorphism with $C(\mathcal G)\to\Cl(X)$ gives the
surjection $E\to W$ of Proposition~\ref{prop:witness}.  The group
$\Cl(X)$ is locally finite; $C(\mathcal G)$ is not, since two lamps at
sites of different blocks generate an infinite dihedral group and three
blocks give a nonabelian free subgroup.  For a finite set $S$ of sites
write
\[
 M_S=\bigl\langle\,\zeta,\;c_\xi\ (\xi\in S)\,\bigr\rangle
 \;\le\;C(\mathcal G).
\]

This distinction prevents a misleading wreath-product identification.
The auxiliary quotient $W/\langle\zeta\rangle$ is an ordinary
permutational lamp group, but
$C(\mathcal G)/\langle\zeta\rangle$ is the free product of the elementary
abelian $2$-groups attached to the blocks of $\mathcal G$.  Thus the present
normal form does not identify $E/\langle w\rangle$ with an ordinary
permutational wreath product.  It is nevertheless closely related to
central extensions of lamplighter groups \cite{FFG23,Bodart25}, to the
generalized wreath products of Kun--Thom \cite{KT}, to the sofic-action
framework of Gao, Kunnawalkam Elayavalli, and Patchell \cite{GKEP}, and
to the halo-product framework of Alekseev--Bradford
\cite{AlekseevBradford}.
The comparison with permutational wreath products and the possible
sofic-action route were pointed out by Fournier-Facio, who also observed the
close relation between the sign-free quotient and the construction shared by
Eckhardt.

\begin{lemma}[finite-site subgroups are residually finite]\label{lem:window}
For every finite set $S$ of sites the subgroup $M_S$ is residually
finite.
\leanverified{Sofic/BlockCliffordTowerSofic}{GroupApproximation.BlockCliffordTowerSofic.manuscriptResiduallyFiniteWindow}
\leanverified{Sofic/LiteralSoficAssembly}{GroupApproximation.LiteralSoficAssembly.lampEquiv}
\end{lemma}

\begin{proof}
The quotient map by $\langle\zeta\rangle$ maps $C(\mathcal G)$ onto
the free product of the elementary abelian $2$-groups of the blocks,
and $M_S$ into the free product of the finitely many factors at blocks
meeting $S$.  A free
product of finite groups is residually finite.  The kernel of
$C(\mathcal G)\to C(\mathcal G)/\langle\!\langle\zeta\rangle\!\rangle$ is
the normal closure of $\zeta$, and $\zeta$ is central, so that normal
closure is $\langle\zeta\rangle$; so the kernel of this map on $M_S$
lies in $\langle\zeta\rangle$.  The surjection
$C(\mathcal G)\to\Cl(X)$ sends $\zeta$ to the sign of $\Cl(X)$, which
is nontrivial, and $\Cl(X)$ is locally finite, so the image of the
finitely generated subgroup $M_S$ in $\Cl(X)$ is finite.  An element of $M_S$
in the kernel of the first map lies in $\langle\zeta\rangle$, and the
second map separates the two elements of that subgroup.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:Esofic}]
Let $E_0$ be the kernel of the homomorphism $E\to\Z$ given by the
exponent sum in $t$.  By
Proposition~\ref{prop:blocknormalform}, $E_0\cong C(\mathcal G)\rtimes
T$.

Take a finite subset of $E_0$.  Its $T$-coordinates lie in one level
$\bar\Gamma_n$, the chain being ascending, and its
$C(\mathcal G)$-coordinates involve only finitely many lamp generators
and possibly the sign, so they lie in $M_{S_0}$ for some finite set
$S_0$ of sites.  Let
$S$ be the union of the $\bar\Gamma_n$-orbits of the sites in $S_0$, a
finite set by Proposition~\ref{prop:blocknormalform}.  Then $M_S$ is
$\bar\Gamma_n$-invariant and contains $M_{S_0}$, so the finite subset lies
in $M_S\rtimes\bar\Gamma_n$.

That group is residually finite.  Indeed $M_S$ is residually finite by
Lemma~\ref{lem:window}, the level $\bar\Gamma_n\cong\bar\Gamma$ is
residually finite through the congruence quotients of its matrix model
(Lemma~\ref{lem:linear}), and the action of $\bar\Gamma_n$ on $M_S$
factors through a finite permutation group $F$ because $S$ is finite.
The kernel of $\bar\Gamma_n\to F$ has finite index, and the
corresponding finite-index subgroup of the semidirect product is
$M_S\times\ker(\bar\Gamma_n\to F)$, which is residually finite;
residual finiteness passes to a group containing a residually finite
subgroup of finite index.

So $E_0$ is a directed union of subgroups that embed in residually
finite groups; that is, $E_0$ is LEF, and then sofic.  The group
$E$ is a split extension of $E_0$ by $\Z$, split by $t$, so the
Elek--Szab\'o amenable-extension theorem implies that $E$ is sofic
\cite{ElekSzabo}, and so hyperlinear.  By Corollary~\ref{cor:marked-obstruction}, $E$
is not MF\@.  LEF groups are MF \cite{CDE}, so $E$ is not LEF; a
residually finite group is LEF, so $E$ is not residually finite
either.
\end{proof}

\paragraph{Exactness.}
The same decomposition also proves that $E$ is exact.  The quotient
$C(\mathcal G)/\langle\zeta\rangle$ is the free product of the
elementary abelian $2$-groups attached to the blocks, each of order
$2^{8}$.  There are countably many blocks; enumerate them, and let
$Q_n$ be the free product of the factors attached to the first $n$.
Each factor is finite, so it is exact, and $Q_n$, a free product of
finitely many exact groups, is exact by Dykema's theorem
\cite{DykemaExact}.  The $Q_n$ increase with union the whole quotient.
In a discrete group every subgroup is open, so a group that is the
union of an increasing family of exact subgroups is exact
\cite[Lemma~2.5]{KWExact}, and the quotient is exact.  Since
$\langle\zeta\rangle$ is finite, extension permanence
\cite{KWPermanence} gives exactness of $C(\mathcal G)$.  The group $\Sigma\le\GL_4(\mathbb Q)$ is exact by
Guentner--Higson--Weinberger \cite{GHW}.  Finally
$E\cong C(\mathcal G)\rtimes\Sigma$ is an extension, so is exact by
\cite{KWPermanence}.  For a discrete group, group exactness is
equivalent to exactness of its reduced group $C^*$-algebra
\cite{KWExact}; so $\Cred(E)$ is exact.


\section{Quasi-identities and undecidability}
\label{sec:marked-horn}

\subsection{A finite universal Horn obstruction}

The presentation of $E$ yields a finite quasi-identity ---
equivalently, a universal Horn sentence --- satisfied by every MF
group.  Write $R$ for the forty-one relators of \eqref{eq:E},
regarded as words in eight variables, and $w$ for the distinguished word in those
variables.

\begin{proposition}[finite quasi-identity]\label{prop:horn}
Every MF group $G$ satisfies the finite universal Horn sentence
\[
 \forall x_1,\dots,x_8\colon\quad
 \bigwedge_{r\in R}\,r(x_1,\dots,x_8)=1
 \;\Longrightarrow\; w(x_1,\dots,x_8)=1 ,
\]
while the canonical generating tuple of $E$ satisfies every relator in $R$
and has $w\ne1$.
\leanverified{Sofic/LiteralUniversalHorn}{GroupApproximation.LiteralUniversalHorn.manuscriptLiteralUniversalHorn}
\end{proposition}

\begin{proof}
A tuple satisfying the relators induces a homomorphism $E\to G$ sending
the distinguished word of $E$ to $w(x_1,\dots,x_8)$.  If $G$ is MF,
the image is trivial, because
$w\in\operatorname{Res}_{\mathrm{MF}}(E)$ by Corollary~\ref{cor:marked-obstruction} and
the residual is functorial (Lemma~\ref{lem:portable}).
The canonical tuple of $E$ satisfies $R$ by definition, and its
distinguished word is nontrivial by Corollary~\ref{cor:marked-obstruction}.
\end{proof}

In the space of eight-generator marked groups, the groups satisfying
all relators of $R$ and $w\ne1$ form a nonempty clopen set.  Both
conditions depend on finitely many word equalities and inequalities,
and $E$ belongs to this set.  Proposition~\ref{prop:horn} excludes every
MF group from it.
\leanverified{Sofic/LiteralMarkedCylinder}{GroupApproximation.LiteralMarkedCylinder.literal_nonempty_clopen_nonMF_cylinder}

\subsection{Undecidability of MF recognition}

Lemma~\ref{lem:permanence}(1) and Corollary~\ref{cor:marked-obstruction} show that MF is
a Markov property of finitely presented groups: the trivial group is
MF, while by subgroup heredity the finitely presented group $E$
embeds in no MF group.  So the Adian--Rabin theorem implies
that MF recognition is undecidable \cite{Rabin58}.

Fix an effective coding of finite presentations by natural numbers,
under which a code determines computably its number of generators and
its finite list of relator words, and write $G_e$ for the group
presented by the code $e$.

\begin{corollary}[undecidability of MF recognition]%
\label{cor:undecidable}
Write $W$ for the predicate on pairs consisting of a presentation code
and a word in the generators of that code, holding when the word is
trivial in the presented group.  Then:
\begin{enumerate}
\item $W$ is undecidable;
\item no algorithm decides, from a presentation code, whether the
presented group is MF;
\item $W$ is recursively enumerable, whereas its complement and the
set of presentation codes of non-MF groups are not.
\end{enumerate}
\end{corollary}
\leanverified{Computability/BooneWordProblemUndecidable}{GroupApproximation.Computability.not_computablePred_wordProblemPred}
\leanverified{Computability/BooneWordProblemUndecidable}{GroupApproximation.Computability.operatorMF_recognition_not_computable}
\leanverified{Computability/WordProblemRE}{GroupApproximation.WordProblemRE.rePred_wordProblemPred}
\leanverified{Computability/WordProblemRE}{GroupApproximation.WordProblemRE.operatorMF_negative_side_not_re}
\leanverified{Computability/AdianRabinVariantTransform}{GroupApproximation.AdianRabinVariantTransform.operatorMF_recognition_undecidable_of_wordProblem}
\leanverified{Computability/AdianRabinVariantTransform}{GroupApproximation.AdianRabinVariantTransform.operatorMF_negative_side_not_re_of_wordProblem}

\begin{proof}
For the undecidability of $W$, use Boone's finitely presented group with
undecidable word problem \cite{Boone}.  A decision procedure for the uniform
predicate $W$ would decide the word problem of this one fixed presentation,
which is impossible.

For the reduction we use the effective form of the Adian--Rabin
construction, which supplies a computable map $\Delta$ from pairs (a
presentation code $e$, a word $u$ in its generators) to presentation
codes such that

\begin{itemize}
\item if $u=1$ in $G_e$, then $\Delta(e,u)$ presents an MF group: the
trivial group in the classical form, and a free group in the variant
used here, a free group being residually finite and so MF by
Lemma~\ref{lem:permanence}(2); and
\item if $u\ne1$ in $G_e$, then $G_e$ embeds in the group presented by
$\Delta(e,u)$.
\end{itemize}

\noindent
This is Rabin's theorem \cite{Rabin58} and, in the form with the
explicit computable output presentation, Adian's \cite{Adian55}; an
English translation of Adian's papers with the construction written out
is \cite{NybergBrodda}.  Apply $\Delta$ to the free product of the
input presentation with a fixed presentation of $E$ and to the input
word.  The resulting presentation code is computable from the pair and
presents an MF group exactly when the input word is trivial.  In the
trivial case the output group is free and so MF\@.  In the nontrivial
case the free product, and so $E$, embeds in the output group.
The output group is then non-MF by Lemma~\ref{lem:permanence}(1).  Composing a
decision procedure for MF recognition with this map would then
decide $W$, so no such procedure exists.  For the last assertion, the
positive instances of $W$ are exactly those carrying a finite
certificate --- a product of conjugates of relators freely equal to
the given word --- and whether a given certificate works is
decidable, so $W$ is recursively enumerable.  Were its complement
recursively enumerable as well, $W$ would be decidable, which it is
not.  Finally, the preimage of the set of non-MF presentation codes
under this computable map is exactly the set of pairs outside $W$, and
the preimage of a recursively enumerable set under a computable map is
recursively enumerable.
\end{proof}

\begin{theorem}[arithmetical complexity of MF recognition]
\label{thm:mf-arithmetic}
For finite presentation codes,
\[
 \mathrm{MF}_{\mathrm{fp}}\in\Pi^0_2\setminus\Pi^0_1,
 \qquad
 \mathrm{NONMF}_{\mathrm{fp}}\in\Sigma^0_2\setminus\Sigma^0_1.
\]
Moreover the complement of the uniform word problem many-one reduces to
$\mathrm{NONMF}_{\mathrm{fp}}$.
\end{theorem}

\begin{proof}
At scale $n$, label every word of length at most $n$ either by a finite
normal-closure proof of triviality or by a separation requirement
$\|w(U)-1\|\geq1/2$, and require relator defect at most $2^{-n-10}$.
Fixed-dimensional feasibility is decidable over the real closed field.
Separation-one normalization gives these certificates for MF groups.
Conversely, if $A$ bounds the normal-closure areas of the finitely many
multiplication words for a finite test set, choosing
$A\,2^{-n-10}<\varepsilon$ turns a certificate into a weak MF model, and so an
MF model.  Then
\[
 P\in\mathrm{MF}_{\mathrm{fp}}
 \quad\Longleftrightarrow\quad
 \forall n\,\exists c\;C_{\mathrm{MF}}(P,n,c)
\]
for a recursive $C_{\mathrm{MF}}$.  Corollary~\ref{cor:undecidable} supplies
properness over the first level and the displayed reduction.
\end{proof}

The FIN/INF switch gives exact second-level completeness for recursive
presentations, and the accompanying manuscript \emph{MF Recognition is
$\Pi^0_2$-Complete} extends it to finite presentations through HNN
extensions whose edge isomorphism is implemented by a corona unitary.

\paragraph{Multiplicity.}
There are infinitely many pairwise nonisomorphic finitely presented
non-MF groups, and continuum many pairwise nonisomorphic finitely
generated non-MF groups.
The groups $E\times\mathbb Z^{k}$ for $k\ge0$ are
finitely presented, pairwise nonisomorphic, and non-MF, each containing
$E$ (Lemma~\ref{lem:permanence}(1)).  They are pairwise nonisomorphic
because their abelianizations have distinct torsion-free ranks.
\leanverified{Sofic/LiteralProductMultiplicity}{GroupApproximation.LiteralProductMultiplicity.manuscriptInfiniteMultiplicity_literal}
For finitely generated examples, choose continuum many pairwise
nonisomorphic finitely generated groups $N$, for instance Neumann's
two-generator groups \cite{Neumann37}.  Each product $E\times N$ is
non-MF\@.  A fixed countable group $X$ can be isomorphic to
$E\times N$ for at most countably many isomorphism types of $N$,
because each such $N$ embeds in $X$ as a finitely generated subgroup
and $X$ has only countably many finitely generated subgroups.  So
the products yield $2^{\aleph_0}$ pairwise nonisomorphic finitely
generated non-MF groups.
\leanverified{Sofic/ContinuumMultiplicity}{GroupApproximation.ContinuumMultiplicity.manuscriptContinuumMultiplicity}
\leanverified{Monsters/NeumannContinuum}{GroupApproximation.NeumannContinuum.manuscriptContinuumMultiplicityUnconditional}

\section{Limitations of the operator-norm method}\label{sec:limits}

The argument requires operator-norm control.  If multiplication is
controlled only in normalized Hilbert--Schmidt norm, the induced adjoint
operators on the Hilbert--Schmidt spaces need not be close in operator norm.
The estimate $\opnorm{\Ad(U)-\Ad(V)}\le2\opnorm{U-V}$ has no converse:
unitaries can be arbitrarily close in normalized Hilbert--Schmidt norm while
their adjoint operators remain at operator-norm distance $2$.
Operator-norm control ensures that the classes
$[\Ad U_n(g)]_\omega$ are multiplicative; Hilbert--Schmidt control
alone does not provide this.
\leanverified{Sofic/HilbertSchmidtAdjointGap}{GroupApproximation.HilbertSchmidtAdjointGap.exists_hsClose_adjointFar}

Adding an identity block leaves operator-norm multiplicative defects
and separations unchanged but decreases normalized Hilbert--Schmidt
distances.  By taking the identity block sufficiently large, all
pairwise normalized Hilbert--Schmidt distances can be made smaller
than any prescribed positive bound.
\leanverified{Sofic/NormTraceGap}{GroupApproximation.NormModel.exists_hs_collapse}
Thus universal $2$-norm triviality of an element does not by itself prove a
group non-MF\@.  The central-involution argument passes to an
invariant corner and equips it with the trace normalized by its own
rank $r_n$: the $2$-norm separation on the corner is measured against
$r_n$, so $r_n/d_n$ may tend to zero without
affecting the contradiction.
The spectral projection of Theorem~\ref{thm:normal-kazhdan} is
normalized on its own range in the same way, and the proof of
Theorem~\ref{thm:projection-collapse} normalizes by the rank $k_n$.

Stable finiteness and faithful traces on the coordinate algebras do
not suffice for Theorem~\ref{thm:transport-variants}.  For a finite
factor $M$ in its standard representation on $L^2(M,\tau)$, the right
action is the commutant of the left action.  So the von Neumann
algebra generated by the two actions is
$(M\cup M')''=Z(M)'=B(L^2(M,\tau))$; in infinite dimension this is not
a finite $C^*$-algebra.  Constant coordinates
$A_n=\Cred(E)$ give an explicit counterexample: the left regular
representation embeds $E$ exactly and injectively into the norm
quotient
\[
 \textstyle\prod_nA_n\big/\bigoplus_nA_n ,
\]
the element $c$ centralizes the Kazhdan base there, and conjugation by
$t$ produces $tct^{-1}$, whose commutator with the base is the element
$u$ with $u^2=w\ne1$ of Corollary~\ref{cor:marked-obstruction}.
\leanverified{Sofic/FaithfullyTracedCoordinateNoGo}{GroupApproximation.FaithfullyTracedCoordinates.manuscriptFaithfullyTracedCoordinateNoGo}
So Theorem~\ref{thm:transport-variants} fails when its matrix
coordinates are replaced by stably finite unital $C^*$-algebras
carrying faithful tracial states and its normalized Hilbert--Schmidt
norm by the trace $2$-norm, with all its remaining hypotheses
unchanged.

Alekseev--Thom ask whether commutants arising in sofic approximations
of Kazhdan groups can be recovered from finite-dimensional coordinate
subalgebras \cite[Open Problem~6.2]{AlekseevThom}.  Their question
concerns Hamming and tracial models, whereas the adjoint spectral
estimates here require operator-norm multiplicative control, which
permutation and Hilbert--Schmidt models do not supply.

\paragraph{Unitary lifting over general $C^*$-algebras.}
If $u$ is unitary in
$\prod_nA_n/\bigoplus_nA_n$ for unital $C^*$-algebras $A_n$, and
$(x_n)$ is any bounded lift, then
$\opnorm{x_n^{*}x_n-1}\to0$ and $\opnorm{x_nx_n^{*}-1}\to0$.
Thus $x_n$ is invertible for all large $n$, and the polar correction
$u_n=x_n(x_n^{*}x_n)^{-1/2}$ is unitary with
$\opnorm{u_n-x_n}\to0$.  This argument uses only polar correction in the
coordinate algebras and requires neither real rank zero nor semiprojectivity
in the sense of Loring \cite{Loring98}.  Theorem~\ref{thm:transport-variants}
additionally uses the Hilbert-space conjugation model and finiteness of the
norm ultraproduct.

\section{A torsion-free non-MF group}
\label{sec:torsionfree}

\begin{theorem}[a torsion-free example]\label{thm:torsionfree}
There is a torsion-free group with property~\textup{(T)} that equals its
own MF radical.  Theorem~\ref{paper-thm:torsion-free} of the main paper
gives such a group that is two-generated, finitely presented, and
acylindrically hyperbolic, and shows that every nontrivial quotient of it
also equals its own MF radical.
\end{theorem}

\begin{proof}
The construction in \cite[Section~2]{FFF} starts from a torsion-free double
HNN extension and applies Hull's common-quotient theorem to obtain a
torsion-free acylindrically hyperbolic property-\textup{(T)}
quotient $G_0$.  In the notation of that construction, $G_0$ contains a
Kazhdan subgroup $\Gamma$, a stable letter $t_1$, and an embedded finitely
presented infinite simple torsion-free group $S'\le\Gamma$ such that
\[
 t_1\Gamma t_1^{-1}\subseteq\Gamma,
 \qquad t_1^{-1}S't_1\subseteq C_{G_0}(\Gamma).
\]
For $s\in S'$, put $z=t_1^{-1}st_1$.  Then
$z\in C_{G_0}(\Gamma)$ and $t_1\in P_\Gamma\subseteq G^+(\Gamma)$, so for
every $\ell\in\Gamma$,
\[
 [s,\ell]=[t_1zt_1^{-1},\ell]\in\mathfrak D(G_0,\Gamma).
\]
Thus $[S',S']\le\mathfrak D(G_0,\Gamma)$.  The infinite nonabelian simple
group $S'$ is perfect, and so
$S'\le\mathfrak D(G_0,\Gamma)$.

Put $N=\langle\!\langle S'\rangle\!\rangle_{G_0}$.  This subgroup is infinite and
normal, so s-normal.  By Osin's lemma, $N$ acts non-elementarily
\cite[Lemma~7.1]{Osin}.  The finite radical of the torsion-free group $G_0$ is
trivial, so $N$ is suitable.  Apply Theorem~7.1 of \cite{Hull} with the
elements $t_i$ equal to a finite generating set of $G_0$ and with an
injectivity radius containing a fixed nonidentity element of $S'$, and write
$q\colon G_0\twoheadrightarrow G$ for the resulting quotient.  The group $G$
is torsion-free by Theorem~7.1(e) and has property-\textup{(T)} because it
is a quotient of $G_0$.  Condition~(c) gives that the image of $N$ equals
$G$.  The image of $S'$ is nontrivial, and its kernel is a proper normal
subgroup of the simple group $S'$, so $q|_{S'}$ is injective.  It follows
that $q(S')$ normally generates $G$.

Put $L=q(\Gamma)$.  The relations above descend to
\[
 q(t_1)Lq(t_1)^{-1}\subseteq L,
 \qquad q(t_1)^{-1}q(S')q(t_1)\subseteq C_G(L).
\]
The same commutator calculation gives
$q(S')\le\mathfrak D(G,L)$.  Since $q(S')$ normally generates $G$ and
$\mathfrak D(G,L)$ is normal, $\mathfrak D(G,L)=G$.
Applying Corollary~\ref{cor:intrinsic-nk} with $K=G$ shows that every
homomorphism $G\to\U(\mathcal Q)$ is trivial.  The embedded copy of $S'$
makes $G$ nontrivial, so no such homomorphism is injective and $G$ is not
MF\@.

The normal-Kazhdan criterion is essential here because a torsion-free group
has no nontrivial finite subgroup.
\leanstep{LI.07}
\end{proof}

\begin{remark}[a small-cancellation realization]
\label{rem:ff-realization}
The construction of \cite[\S2]{FFF} gives a finitely presented torsion-free
property-\textup{(T)} group $P$ containing
$P_1\times P_2\times S$, where $P_i\cong P$ and $S$ is finitely presented,
simple, and torsion-free.  Choose an isomorphism
$\alpha\colon P\to P_1$ and $a\in P_2\setminus\{1\}$.  Define
$E(P,\alpha,a)$ from a finite presentation of $P$ by adjoining $t,c$, imposing
\[
 tpt^{-1}=\alpha(p),\qquad [c,p]=1,\qquad c^2=1,
\]
for the chosen generators $p$ of $P$, and making
$w=[tct^{-1},a(tct^{-1})a^{-1}]$ central.  The Clifford coset representation
used in Proposition~\ref{prop:witness} maps $w$ to its central involution.
Thus $w\ne1$, and $E(P,\alpha,a)$ has torsion.
\end{remark}

\section{Open questions}

The construction leaves the following questions open:
\begin{enumerate}
\item Does MF imply hyperlinearity?  An MF embedding controls
operator norm but does not require the induced tracial model to
separate nontrivial group elements, so it does not directly produce a
hyperlinear model.  This
question concerns the CDE convention fixed above; for stronger
trace-controlled MF conventions the implication is built into the definition.
\item Does the obstruction survive when the matrix blocks are replaced
by other coordinate algebras?  The proof of
Theorem~\ref{thm:transport-variants} uses finite dimensionality of
$M_{d_n}(\C)$ for the conjugation action on
$L^2(M_{d_n}(\C),\tr_{d_n})$ and for finiteness of the norm
ultraproduct.  These conclusions need not hold for arbitrary faithfully
traced coordinate algebras.  For
which separable $C^*$-algebras $A_n$ does every homomorphism
$H\to\U\bigl(\prod_nA_n/\bigoplus_nA_n\bigr)$ still map $w$ to the
identity?
\item For which groups do the criteria of
Theorems~\ref{thm:criterion},~\ref{thm:normal-kazhdan}
and~\ref{thm:projection-collapse} compute the residual exactly?
\end{enumerate}

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