64c3814
==== TABLE OF CONTENTS ====
An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//consequences.tex
An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//finite-chain.tex
An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//group-example.tex
An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//introduction.tex
An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//nuclear-finiteness.tex
An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//paper.tex
An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//spectral-preliminaries.tex
An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//ultrapower-contradiction.tex
An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//references.bib
===========================
==== BEGIN An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//consequences.tex ====
\section{Consequences for existential closure and nuclear approximation}
\label{sec:consequences}
We spell out the model-theoretic consequences of the negative answer to
Kirchberg's embedding problem. In the unital $C^*$-algebra language,
quantifier-free formulas are continuous combinations of norms of
$*$-polynomials; variables range over prescribed norm balls.
A unital algebra $E$ is \emph{existentially closed} if, for every unital
inclusion $E\subseteq F$, every finite parameter tuple $a$ from $E$,
and every quantifier-free formula $\varphi(a,y)$ in a finite tuple $y$,
\[
\inf_{y\in E}\varphi(a,y)=\inf_{y\in F}\varphi(a,y),
\]
where $y$ ranges over the corresponding products of prescribed norm balls.
Thus an extension
cannot improve the infimum of a finite quantifier-free test over $E$
\cite[Section~2.1]{GoldbringSinclair2015}.
A parameter-free finite condition consists of finitely many strict
inequalities $\varphi_j(x)<r_j$ with quantifier-free $\varphi_j$.
It is \emph{satisfiable} when some tuple in a unital $C^*$-algebra
satisfies them. The last assertion below makes the failure of good nuclear
witnesses explicit as a positive lower bound on the approximation error
of every finite-dimensional completely positive contractive factorization
of every realizing tuple.
The KEP equivalences of
Goldbring--Sinclair~\cite[Proposition~3.2 and Theorems~3.3 and~3.7]{GoldbringSinclair2015}
give these direct consequences of \cref{thm:main}.
\begin{corollary}[Consequences of negative KEP]\label{cor:kep-consequences}
Neither $\Otwo$ nor any exact or nuclear unital $C^*$-algebra is
existentially closed among unital $C^*$-algebra extensions.
For every free ultrafilter $\omega$ on $\N$, there are a separable
unital $C^*$-algebra $D$ and $\alpha\in\operatorname{Aut}(D)$ such that
$D$ embeds unitally into $\Otwo^\omega$ but $D\rtimes_\alpha\Z$ does not
embed unitally there.
Some satisfiable finite condition
$p(x)=\{\varphi_j(x)<r_j:1\le j\le s\}$ in the unital $C^*$-algebra
language, with $x=(x_1,\ldots,x_m)$, quantifier-free $\varphi_j$ and
$r_j>0$, lacks good nuclear witnesses: for some $\varepsilon_0>0$,
every realization $a=(a_i)_{i=1}^m$ of $p$ in every unital $C^*$-algebra
$E$ satisfies $\max_{1\le i\le m}\|\psi\phi(a_i)-a_i\|\ge\varepsilon_0$
for every finite-dimensional $C^*$-algebra $F$ and all completely positive
contractive maps $\phi:E\to F$, $\psi:F\to E$, not necessarily unital.
\end{corollary}
\begin{proof}
For each fixed free $\omega$, \cref{thm:main} with $B=\Otwo$ negates
unital KEP. The existential-closure claims follow from
\cite[Theorem~3.3]{GoldbringSinclair2015}, using that nuclearity implies
exactness. Proposition~3.2 of the same
paper, with its preceding construction for separable unital algebras,
gives the crossed-product claim; Definition~3.6 and Theorem~3.7 give
the finite-condition claim. These two witnesses are existential,
not identified by our explicit group construction.
\end{proof}
==== END An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//consequences.tex ====
==== BEGIN An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//finite-chain.tex ====
\begin{figure}[htbp]
\centering
\begin{tikzpicture}[>=Stealth, every node/.style={font=\small}]
\node (start) at (0,0) {$[\tau_0]=[\pi]$};
\node (tail) at (2,0) {$\cdots$};
\node (cycle) at (3.7,0) {$[\tau_l]$};
\node (middle) at (5.4,0) {$\cdots$};
\node (repeat) at (7.4,0) {$[\tau_{l+p}]$};
\draw[->] (start) -- (tail);
\draw[->] (tail) -- (cycle);
\draw[->] (cycle) -- (middle);
\draw[->] (middle) -- (repeat);
\draw[dashed] (cycle.north) to[out=55,in=125]
node[above] {same equivalence class} (repeat.north);
\node at (1.7,-0.65) {initial segment};
\node at (5.55,-0.65) {$p$ predecessor steps};
\end{tikzpicture}
\caption{A chain inside one finite set $D$. An arrow from $[\tau_i]$ to
$[\tau_{i+1}]$ means $\tau_i\prec\tau_{i+1}\circ\theta$.
Repetition gives $\tau_l\simeq\tau_l\circ\theta^p$ by equality of
dimensions. The proof then shows that each pullback of $\tau_l$ is
irreducible, allowing this periodicity to pass back to $\pi$.
The initial segment can be empty.}
\label{fig:finite-chain}
\end{figure}
==== END An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//finite-chain.tex ====
==== BEGIN An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//group-example.tex ====
\section{The group and a finite-dimensional representation}
\label{sec:group}
We construct a finite-dimensional representation inequivalent to each of
its positive pullbacks along a fixed group endomorphism. The same group
configuration appears in the nonfiniteness arguments of
Sauers~\cite[Section~2.4]{Sauers2026NonMF} and
Eckhardt~\cite[Theorem~3.1 and Example~3.2]{Eckhardt2026}.
De Cornulier~\cite[Proposition~1.1 and its proof]{deCornulier2007}
records the property~(T) affine group and its finite-index scalar
self-embeddings. We use the dyadic case.
We use column vectors. In the group $G$ of Theorem~\ref{thm:main},
write elements as $(v,M,k)$, so that
\[
(v,M,k)(w,N,\ell)=(v+2^kMw,MN,k+\ell).
\]
Set
\[
\Gamma=\{(v,M,0):v\in\Z^3,\ M\in\SL_3(\Z)\},
\qquad t=(0,I,1).
\]
We suppress the final coordinate when writing elements of $\Gamma$, and
define
\[
\theta:\Gamma\longrightarrow\Gamma,\qquad
\theta(v,M)=(2v,M),
\qquad \Lambda=\theta(\Gamma).
\]
\begin{lemma}\label{lem:group-data}
The group $\Gamma$ is finitely generated and has property~(T). The map
$\theta$ is injective,
\[
[\Gamma:\Lambda]=8,
\qquad t\gamma t^{-1}=\theta(\gamma)\quad(\gamma\in\Gamma).
\]
There is a nonzero finite-dimensional irreducible unitary representation
$\pi_*$ of $\Gamma$ that factors through reduction modulo~$2$ and is
nontrivial on the translation subgroup $\Z^3$. In particular,
\begin{equation}\label{eq:group-aperiodic}
\pi_*\not\simeq\pi_*\circ\theta^j\qquad(j\geq1).
\end{equation}
\end{lemma}
\begin{proof}
Property~(T) is the case $n=3$, $m=1$ of
\cite[Proposition~1.1]{deCornulier2007}, with $\SL_3(\Z)$ acting on one
copy of its standard rank-three integral module. For finite generation,
take the three standard translations and the elementary matrices
$I+E_{ij}$, $i\ne j$, where $E_{ij}$ is the matrix unit. These matrices
generate $\SL_3(\Z)$ by integer row reduction.
The group law shows that $\theta$ is an injective homomorphism and that
conjugation by $t$ implements it. Its image is
$\Lambda=2\Z^3\rtimes\SL_3(\Z)$. The elements
\[
r_a=(a,I),\qquad a\in\{0,1\}^3,
\]
are representatives of its eight left cosets: writing $v=a+2w$ gives
$(v,M)=r_a(2w,M)$, and two such representatives give the same coset
only when they have the same residue modulo~$2$.
Let $F$ be the finite image of the reduction homomorphism
\[
\Gamma\longrightarrow (\Z/2\Z)^3\rtimes\SL_3(\Z/2\Z),
\qquad (v,M)\longmapsto(v\bmod2,M\bmod2).
\]
The translation subgroup $(\Z/2\Z)^3$ acts nontrivially in the left
regular representation of $F$. Decomposing this representation into
irreducibles therefore gives an irreducible constituent that is
nontrivial on translations. Its pullback to $\Gamma$ is the required
$\pi_*$. Indeed, $\pi_*$ factors through reduction modulo~$2$, whereas
$\theta^j(v,I)=(2^jv,I)$. Thus $\pi_*\circ\theta^j$ is trivial on all
translations for every $j\geq1$, and $\pi_*$ is not. This proves
\eqref{eq:group-aperiodic}.
\end{proof}
The group $G$ is countable, so its full group $C^*$-algebra $C^*(G)$ is
separable and unital. Every unitary representation of $G$ extends to
$C^*(G)$. We will use this universal property after inducing $\pi_*$
from $\Gamma$ to $G$.
==== END An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//group-example.tex ====
==== BEGIN An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//introduction.tex ====
\section{The embedding problem}
\label{sec:introduction}
Throughout, $C^*$-algebras are complex, and an \emph{embedding} is an
injective $*$-homomorphism. We state unit preservation explicitly.
For a nonzero unital $C^*$-algebra $B$ and a free ultrafilter $\omega$ on
$\N$, its \emph{norm ultrapower} is
\[
B^\omega=\ell^\infty(\N,B)/I_\omega(B),\qquad
I_\omega(B)=\{(b_n):\lim_\omega\|b_n\|=0\}.
\]
Here $\ell^\infty(\N,B)$ is the algebra of bounded sequences, and the
quotient uses the operator norm. The Cuntz algebra $\Otwo$ is the universal
unital $C^*$-algebra generated by two isometries $s_1,s_2$ satisfying
$s_1s_1^*+s_2s_2^*=1$; it was introduced by Cuntz~\cite{Cuntz1977}.
The embedding theorem of Kirchberg--Phillips~\cite{KirchbergPhillips2000}
states that every separable unital exact $C^*$-algebra embeds unitally
into $\Otwo$. Here exactness means that minimal tensoring preserves
short exact sequences. Passing from $\Otwo$ to its norm ultrapower
asks how far this universality extends beyond exact algebras. The
techniques behind the embedding theorem led Kirchberg to consider this
possibility, as recounted in \cite[Section~1]{GoldbringSinclair2015}.
Kirchberg's embedding problem asks whether every separable unital
$C^*$-algebra admits a unital embedding into $\Otwo^\omega$; see
Goldbring--Sinclair~\cite[Section~3.1]{GoldbringSinclair2015}. We give an explicit
counterexample. The unital formulation is equivalent to the formulation
without unit preservation~\cite[Section~3.1]{GoldbringSinclair2015}.
This question concerns norm ultrapowers of a nuclear
$C^*$-algebra. It is distinct from the tensor-norm formulation of
Kirchberg's conjecture and from Connes' embedding problem, as emphasized
in~\cite[Section~1]{GoldbringSinclair2015}.
\Needspace{14\baselineskip}
\begin{theorem}\label{thm:main}
Let
\[
G=\Z[1/2]^3\rtimes\bigl(\SL_3(\Z)\times\Z\bigr),
\qquad (M,k)\cdot v=2^kMv,
\]
and let $A=C^*(G)$ be its full group $C^*$-algebra. Then $A$ is separable
and unital. For every nonzero unital nuclear $C^*$-algebra $B$ and every
free ultrafilter $\omega$ on $\N$, there is no unital embedding
\[
A\longrightarrow B^\omega.
\]
In particular, $A$ does not embed unitally into $\Otwo^\omega$.
\end{theorem}
The group is countable and independent of both $B$ and $\omega$;
no separability assumption on $B$ is needed.
The proof treats an arbitrary fixed nuclear $B$ throughout. Specializing
to the nuclear algebra $\Otwo$ gives the negative answer to the embedding
problem; for its nuclearity, see~\cite[Section~4, pp.~80--82]{PaulsenZheng2016}.
Section~\ref{sec:consequences} gives the consequences for existential
closure, permanence of $\Otwo^\omega$-embeddability under crossed products,
and good nuclear witnesses, with the relevant notions defined there.
\subsection{Previous work and the new obstruction}
\begingroup
\setlength{\emergencystretch}{2em}
Recent approaches study computability and tensor norms.
Fox--Goldbring--Hart~\cite{FoxGoldbringHart2024} derive
computability-theoretic consequences of a positive answer, using the
computable presentation of $\Otwo$. Goldbring--Sinclair give a conditional
route to a negative answer through a \emph{weakly explicit minimal tensor
norm}: an effective choice of universal formulas approximating the
minimal tensor-norm predicates to a fixed accuracy.
They leave this property of $\Otwo$ open
\cite[Definition~4.16, Theorem~4.18 and Question~4.20]{GoldbringSinclair2025Tensor}.
Our obstruction instead comes from the representation theory of an
explicit discrete group.\par
\endgroup
The group configuration already appears in the nonfiniteness arguments of
Sauers~\cite[Section~2.4]{Sauers2026NonMF} and
Eckhardt~\cite[Theorem~3.1 and Example~3.2]{Eckhardt2026}.
They show that the full algebra of the ascending HNN extension $G$ is
nonfinite: it contains a proper isometry $v$, so $v^*v=1$ but $vv^*\ne1$.
Its property~(T) base group
$\Gamma=\Z^3\rtimes\SL_3(\Z)$ and its finite-index scalar
self-embeddings are recorded in
de Cornulier~\cite[Proposition~1.1 and its proof]{deCornulier2007}.
Sauers and Eckhardt also construct non-MF groups: groups admitting no injective
homomorphism into the unitary group of any norm matrix corona
$\prod_j M_{d_j}/\bigoplus_j M_{d_j}$, with positive integer dimensions
$d_j$. Here the product consists of bounded sequences and the direct sum
is the ideal of sequences tending to zero in norm
\cite[Theorem~4.3]{Eckhardt2026}\cite[Sections~3--5]{Sauers2026NonMF}.
Eckhardt's example is an associated wreath product. The target class here is different:
nonfiniteness of $C^*(G)$ alone cannot exclude a unital embedding into
$\Otwo^\omega$, which itself contains proper isometries.
The additional ingredient is a finiteness statement for approximate
finite-dimensional representation types inside a nuclear algebra
(Lemma~\ref{lem:nuclear-finiteness}). Its tensor-product argument follows
Wassermann's method~\cite{Wassermann1991}, as explained
by Manuilov--Thomsen~\cite[Section~2.1]{ManuilovThomsen2007}.
The statement applies to an arbitrary tuple of unitaries, even when the
tuple satisfies no group relations. This flexibility is essential because
coordinate lifts of an ultrapower representation are only approximately
multiplicative. A dimension-independent estimate allows the representation
type to vary with the coordinate. Finite-index induction then relates
members of the resulting finite sets by composition with the self-embedding.
These two passages turn the group configuration into the norm-ultrapower
obstruction.
\subsection{Proof outline}
The subgroup $\Gamma=\Z^3\rtimes\SL_3(\Z)$ of $G$ has property~(T).
Conjugation by the generator $t$ of the last $\Z$-factor sends $\Gamma$
onto the index-eight subgroup
$\theta(\Gamma)$, where $\theta(v,M)=(2v,M)$.
There is a finite-dimensional irreducible representation $\pi_*$ of
$\Gamma$ such that $\pi_*\circ\theta^j$ is inequivalent to $\pi_*$ for
every $j\ge1$. Section~\ref{sec:group} verifies these facts.
Fix a finite generating set $S$ of $\Gamma$ for which property~(T)
gives a spectral gap. Suppose a unital embedding $A\to B^\omega$ exists,
and choose unitary lifts $x_n(g)\in B$ of its group unitaries. For each
$n$, consider the nonzero finite-dimensional irreducible unitary representations
$\pi$ for which the tuple $(x_n(s)\otimes\bar\pi(s))_{s\in S}$ has
spectral minimum below a fixed small threshold for the sum of squared
displacements from~$1$. Here $\bar\pi$ is the conjugate representation.
Sections~\ref{sec:spectral} and~\ref{sec:nuclear} specify the threshold
and show that the resulting set $D_n$ of representation classes is finite.
Write $\pi\prec\rho$ when $\pi$ occurs as a unitary subrepresentation
of $\rho$, and call $\pi'$ a predecessor of $\pi$ if
$\pi\prec\pi'\circ\theta$. Faithfulness of the embedding places
$[\pi_*]$ in $D_n$ on a set belonging to $\omega$.
In Section~\ref{sec:contradiction}, the spectral gap for
\emph{every sequence} of finite-dimensional representations supplies,
on one set belonging to $\omega$, the simultaneous implication
\[
[\pi]\in D_n\quad\Longrightarrow\quad
\text{some }[\pi']\in D_n\text{ satisfies }\pi\prec\pi'\circ\theta.
\]
Starting at $\pi_*$ and repeatedly choosing a predecessor must give a
repetition in the finite set $D_n$. Irreducibility then makes $\pi_*$
periodic under composition with $\theta$, a contradiction.
==== END An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//introduction.tex ====
==== BEGIN An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//nuclear-finiteness.tex ====
\section{Finiteness of approximate representation matches}
\label{sec:nuclear}
We prove that, for a fixed unitary tuple $\mathbf u$ in a nuclear algebra,
only finitely many finite-dimensional irreducible representation classes
satisfy $d(\mathbf u,\pi)<a$.
The tuple need not satisfy any group relations. This distinction will allow
us to apply the result to coordinate lifts from a norm ultrapower.
We retain the notation and constants of Section~\ref{sec:spectral}.
For $C^*$-algebras $B$ and $C$, the minimal tensor product
$B\otimes_{\min}C$ is realized by faithful representations on separate
Hilbert spaces; the maximal tensor product $B\otimes_{\max}C$ is
characterized by pairs of commuting representations on one Hilbert space.
We use the tensor-norm characterization of nuclearity developed by
Takesaki and Lance~\cite{Takesaki1964,Lance1973}: $B$ is nuclear if
these two tensor norms agree for every $C$.
We first obtain a lower bound when the second tensor factor has no
finite-dimensional subrepresentations. The passage from a vector to its
reduced density operator removes the arbitrary unitaries in the first
factor.
\begin{lemma}[Square-root trace inequality]\label{lem:square-root-trace}
For positive Hilbert--Schmidt operators $X,Y$ on any Hilbert space,
\[
\|X-Y\|_2^2\leq\|X^2-Y^2\|_1.
\]
Here $\|\cdot\|_1$ and $\|\cdot\|_2$ denote the trace and
Hilbert--Schmidt norms, respectively.
\end{lemma}
\begin{proof}
This is the square-root trace inequality of Powers--St\o rmer
\cite[Lemma~4.1]{PowersStormer1970};
we include the short argument needed here.
Put $F=X-Y$, and let $J=\operatorname{sgn}(F)$, with value zero on
$\ker F$. Since $X^2-Y^2=FX+YF$ and $JF=FJ=|F|$,
\begin{align*}
\|X^2-Y^2\|_1
&\geq \operatorname{Tr}\bigl(J(X^2-Y^2)\bigr)\\
&=\operatorname{Tr}\bigl(|F|(X+Y)\bigr)\\
&\geq\operatorname{Tr}(F_+F)+\operatorname{Tr}(F_-(-F))
=\operatorname{Tr}(F^2).
\end{align*}
For the last inequality, use $X+Y-F=2Y\geq0$ and
$X+Y+F=2X\geq0$, and pair these inequalities with $F_+$ and $F_-$,
respectively.
All these traces are well defined in infinite dimension. Products of two
Hilbert--Schmidt operators are trace class; their traces are cyclic, by
finite-rank approximation and
$\|AB\|_1\leq\|A\|_2\|B\|_2$.
The trace of a product of two positive Hilbert--Schmidt operators is
nonnegative, as is seen by diagonalizing one factor. These observations
also justify the cyclic rearrangement with the bounded operator $J$.
The operators involved vanish on the orthogonal complement of a separable
subspace, so no separability hypothesis on the ambient Hilbert space is
needed.
\end{proof}
Bekka~\cite{Bekka1990} proved that if a unitary representation of a
property~(T) group on a Hilbert space $K$ admits a state on
$\mathcal B(K)$ invariant under conjugation by the representation,
then it has a nonzero finite-dimensional subrepresentation; see
also~\cite[proof of Proposition~2]{Valette1997}. The next lemma gives
the explicit spectral bound needed here for an arbitrary unitary tuple
in the first tensor factor.
\Needspace{7\baselineskip}
\begin{lemma}[Uniform spectral bound]
\label{lem:weak-mixing-bound}
Let $\Gamma,S,\kappa$ satisfy \eqref{eq:kazhdan-gap}. Let
$\sigma:\Gamma\to\mathcal U(K)$ be a unitary representation with no
nonzero finite-dimensional subrepresentation. For any Hilbert space $E$
and any unitaries $(U_s)_{s\in S}$ on $E$,
\[
L\bigl((U_s\otimes\sigma(s))_{s\in S}\bigr)\geq b1,
\qquad b=\frac{\kappa^2}{4|S|}.
\]
No relations among the $U_s$ are required.
\end{lemma}
\begin{proof}
Conjugation by $\sigma$ is a unitary representation on the Hilbert space
$\mathcal S_2(K)$ of Hilbert--Schmidt operators. It has no nonzero invariant
vector: if $A\neq0$ commutes with every $\sigma(g)$, then so does the
nonzero positive compact operator $|A|$. The eigenspace of any nonzero
eigenvalue of $|A|$ is a nonzero finite-dimensional $\sigma$-invariant
subspace.
Use inner products linear in the first variable. Fix a unit vector
$\xi\in E\otimes K$. Its reduced density operator
$T$ on $K$ is characterized by
\[
\operatorname{Tr}(TC)=\langle(1\otimes C)\xi,\xi\rangle
\qquad(C\in\mathcal B(K)).
\]
Concretely, write $\xi=\sum_i e_i\otimes v_i$ with $(e_i)$ orthonormal
and $\sum_i\|v_i\|^2=1$. Then
$T=\sum_i |v_i\rangle\langle v_i|$, with convergence in trace norm;
in particular, $T\geq0$ and $\operatorname{Tr}(T)=1$. Such an expansion
uses only countably many vectors, even if the Hilbert spaces are
nonseparable.
For $s\in S$, put $\xi_s=(U_s\otimes\sigma(s))\xi$ and
$\delta_s=\|\xi_s-\xi\|$. The reduced density operator of $\xi_s$ is
$T_s=\sigma(s)T\sigma(s)^*$. Testing the two vector states against
$1\otimes C$, for $\|C\|\leq1$, gives
\[
\|T_s-T\|_1\leq2\delta_s.
\]
Lemma~\ref{lem:square-root-trace} therefore gives
\[
\|\sigma(s)T^{1/2}\sigma(s)^*-T^{1/2}\|_2^2
\leq\|T_s-T\|_1\leq2\delta_s.
\]
Since $\|T^{1/2}\|_2=1$, applying \eqref{eq:kazhdan-gap} to the
conjugation representation and then Cauchy--Schwarz yields
\[
\kappa\leq2\sum_{s\in S}\delta_s
\leq2\sqrt{|S|}\left(\sum_{s\in S}\delta_s^2\right)^{1/2}.
\]
Thus $\sum_s\delta_s^2\geq b$. Taking all unit vectors $\xi$ proves
the operator inequality; when $E\otimes K=\{0\}$ it is vacuous.
\end{proof}
The following quotient argument uses Wassermann's property~(T)
method~\cite{Wassermann1991}; see the exposition
in~\cite[Section~2.1]{ManuilovThomsen2007}. The preceding uniform bound
allows us to use an arbitrary unitary tuple in the first factor.
Nuclearity will transfer the bound to the coordinate quotient and
contradict the assumed small spectral minima.
\begin{lemma}[Nuclear finiteness]\label{lem:nuclear-finiteness}
Let $\Gamma,S,\kappa,R,b,a$ be as above. For every nonzero unital
nuclear $C^*$-algebra $B$ and every unitary $S$-tuple $\mathbf u$ in $B$,
the set
\[
D(\mathbf u)=\{[\pi]:d(\mathbf u,\pi)<a\}
\]
is finite, where $\pi$ ranges over the nonzero finite-dimensional
irreducible unitary representations of $\Gamma$ and $[\pi]$ denotes its
unitary equivalence class. No group relations on $\mathbf u$ are assumed.
\end{lemma}
\begin{proof}
Suppose that $(\pi_j)_{j\geq1}$ are pairwise inequivalent representations
with $[\pi_j]\in D(\mathbf u)$. Write $M_j=\mathcal B(\bar V_{\pi_j})$, and form
\[
Q=\frac{\prod_{j\geq1}M_j}{\bigoplus_{j\geq1}M_j},
\qquad q(g)=[(\bar\pi_j(g))_j]\quad(g\in\Gamma).
\]
The product consists of bounded sequences, and the direct sum is the ideal
of sequences whose norms tend to zero in the ordinary sense. In
particular, this auxiliary quotient does not involve an ultrafilter.
Choose a faithful unital representation $\eta:Q\to\mathcal B(K)$, and
put $\sigma=\eta\circ q$. We first show that $\sigma$ has no nonzero
finite-dimensional subrepresentations. Fix a
finite-dimensional irreducible representation $\tau$ of $\Gamma$.
Under the identification
\[
\bar V_{\pi_j}\otimes\bar V_\tau
\simeq\mathcal S_2(V_\tau,\bar V_{\pi_j}),
\]
invariant vectors for $\bar\pi_j\otimes\bar\tau$ correspond to
intertwiners from $\tau$ to $\bar\pi_j$. By irreducibility and
inequivalence, such an intertwiner can exist for at most one $j$.
Consequently,
\[
L\bigl((\bar\pi_j(s)\otimes\bar\tau(s))_{s\in S}\bigr)
\geq\kappa1
\]
outside at most one coordinate. Fixed-size matrix amplification commutes
with this quotient, entry by entry:
\[
Q\otimes\mathcal B(\bar V_\tau)
\simeq
\frac{\prod_j(M_j\otimes\mathcal B(\bar V_\tau))}
{\bigoplus_j(M_j\otimes\mathcal B(\bar V_\tau))}.
\]
The exceptional coordinate disappears, and hence
\begin{equation}\label{eq:quotient-excludes-finite-parts}
L\bigl((q(s)\otimes\bar\tau(s))_{s\in S}\bigr)\geq\kappa1.
\end{equation}
If $\sigma$ contained $\tau$, then
$\sigma\otimes\bar\tau$ would have an invariant vector, contradicting
\eqref{eq:quotient-excludes-finite-parts}. Every nonzero finite-dimensional
unitary representation contains an irreducible subrepresentation, so
$\sigma$ has none.
Represent $B$ faithfully and unitally on a Hilbert space $E$. By
Lemma~\ref{lem:weak-mixing-bound},
\begin{equation}\label{eq:spatial-lower-bound}
L\bigl((u_s\otimes q(s))_{s\in S}\bigr)\geq b1
\quad\text{in }B\otimes_{\min}Q.
\end{equation}
We now use nuclearity to pass this inequality to the coordinate quotient.
Let
\[
C=\frac{\prod_j(B\otimes M_j)}{\bigoplus_j(B\otimes M_j)}.
\]
There are commuting unital $*$-homomorphisms
\[
\alpha:B\longrightarrow C,\quad
\alpha(x)=[(x\otimes1_{M_j})_j],
\qquad
\beta:Q\longrightarrow C,\quad
\beta([(y_j)_j])=[(1_B\otimes y_j)_j].
\]
The map $\beta$ is well defined because
$\|1_B\otimes y_j\|=\|y_j\|$. The universal property of the maximal
tensor product gives a unital $*$-homomorphism
\[
\Psi:B\otimes_{\max}Q\longrightarrow C,
\qquad \Psi(x\otimes q)=\alpha(x)\beta(q).
\]
Since $B$ is nuclear,
$B\otimes_{\min}Q=B\otimes_{\max}Q$, so \eqref{eq:spatial-lower-bound}
passes through $\Psi$.
Put $L_j=L((u_s\otimes\bar\pi_j(s))_{s\in S})$. We have proved
$[(L_j)_j]\geq b1$ in $C$. Continuous functional calculus for
$f(t)=(b-t)_+$ now gives
\[
[(f(L_j))_j]=f([(L_j)_j])=0,
\qquad
\|f(L_j)\|=(b-d(\mathbf u,\pi_j))_+\longrightarrow0.
\]
Thus $\liminf_j d(\mathbf u,\pi_j)\geq b$, whereas every term is
less than $a<b$. This contradiction proves finiteness.
\end{proof}
The only use of nuclearity was to identify the minimal and maximal tensor
products before applying $\Psi$. In particular, the proof does not assume
an identification of the minimal tensor product with the coordinate
quotient, and it imposes no bound on the dimensions of the $\pi_j$.
==== END An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//nuclear-finiteness.tex ====
==== BEGIN An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//paper.tex ====
\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern,amsmath,amssymb,amsthm,mathtools}
\usepackage[margin=1in]{geometry}
\usepackage{microtype}
\usepackage{needspace,tikz}
\usetikzlibrary{arrows.meta}
\usepackage[hidelinks]{hyperref}
\usepackage[all]{hypcap}
\hypersetup{pdftitle={An explicit obstruction to nuclear norm-ultrapower embeddings},pdfauthor={OpenAI}}
\usepackage[nameinlink,capitalise,noabbrev]{cleveref}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newcommand{\Otwo}{\mathcal O_2}
\newcommand{\C}{\mathbb C}
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\title{An explicit obstruction to nuclear norm-ultrapower embeddings}
\author{OpenAI}
\date{September 23, 2026}
\begin{document}
\maketitle
\begin{abstract}
We give a negative answer to Kirchberg's norm-ultrapower embedding problem.
We construct an explicit separable unital full group $C^*$-algebra that
admits no unital embedding into $B^\omega$, for any nonzero unital nuclear
$C^*$-algebra $B$ and any free ultrafilter $\omega$ on $\mathbb N$.
In particular, it does not embed unitally into $\mathcal O_2^\omega$.
\end{abstract}
\tableofcontents
\input{introduction}
\input{group-example}
\input{spectral-preliminaries}
\input{nuclear-finiteness}
\input{ultrapower-contradiction}
\input{consequences}
\bibliographystyle{plain}
\bibliography{references}
\end{document}
==== END An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//paper.tex ====
==== BEGIN An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//spectral-preliminaries.tex ====
\section{A spectral test for representation matching}
\label{sec:spectral}
We now introduce the spectral quantity used to form the finite sets in
the proof outline. The discussion applies to any finitely generated
discrete group $\Gamma$ with property~(T).
Choose a nonempty finite generating set $S$ containing a Kazhdan set.
By property~(T), after decreasing the constant if necessary, there is
$0<\kappa\le4|S|$ such that every unitary representation $\sigma$ with
no nonzero invariant vector satisfies
\begin{equation}\label{eq:kazhdan-gap}
\sum_{s\in S}\|\sigma(s)\xi-\xi\|^2\ge\kappa\|\xi\|^2.
\end{equation}
For a tuple $\mathbf z=(z_s)_{s\in S}$ of unitaries in a nonzero unital
$C^*$-algebra $C$, set
\begin{equation}\label{eq:laplacian}
L(\mathbf z)=\sum_{s\in S}(z_s-1)^*(z_s-1),\qquad R=4|S|.
\end{equation}
Thus $0\le L(\mathbf z)\le R1$, without any group relations on
$\mathbf z$.
\begin{lemma}\label{lem:spectral-gap}
If $z_s=\sigma(s)$ for a unitary representation
$\sigma:\Gamma\to\U(C)$ in a nonzero unital $C^*$-algebra, then
\[
\spec L(\mathbf z)\subseteq\{0\}\cup[\kappa,R].
\]
\end{lemma}
\begin{proof}
Represent $C$ faithfully and unitally on a Hilbert space. The invariant
vectors for $\sigma$ and their orthogonal complement are reducing
subspaces. On the former $L$ vanishes; on the latter
\eqref{eq:kazhdan-gap} gives $L\ge\kappa1$. A faithful unital
representation preserves spectra, proving the claim.
\end{proof}
Every finite-dimensional representation below acts on a nonzero Hilbert
space. If $\pi:\Gamma\to\U(V_\pi)$ is such a representation, its
conjugate $\bar\pi$ acts on the conjugate Hilbert space $\bar V_\pi$ by
$\bar\pi(g)\bar v=\overline{\pi(g)v}$. Define
\begin{equation}\label{eq:matching-distance}
d(\mathbf z,\pi)=
\min\spec L\bigl((z_s\otimes\bar\pi(s))_{s\in S}\bigr),
\end{equation}
where the spectrum is taken in
$C\otimes\mathcal B(\bar V_\pi)$. Matrix algebras have a unique
$C^*$-tensor norm. On any faithful unital Hilbert-space realization $C\subseteq
\mathcal B(H)$, the number in \eqref{eq:matching-distance} equals
\[
\inf_{\substack{\xi\in H\otimes\bar V_\pi\\\|\xi\|=1}}
\sum_{s\in S}\|(z_s\otimes\bar\pi(s))\xi-\xi\|^2.
\]
In such a faithful realization, if $z_s=\rho(s)$ comes from a unitary
representation and $\pi$ is irreducible, then $d(\mathbf z,\pi)=0$
exactly when $\pi$ occurs in $\rho$, and otherwise
$d(\mathbf z,\pi)\ge\kappa$. Indeed, invariant vectors of
$\rho\otimes\bar\pi$ correspond to intertwiners from $V_\pi$ to $H$,
and the spectral gap separates their orthogonal complement from zero.
The number $d(\mathbf z,\pi)$ is unchanged by unitary equivalence of
$\pi$. For a finite direct sum, the same formula gives
\begin{equation}\label{eq:block-minimum}
d(\mathbf z,\pi_1\oplus\cdots\oplus\pi_r)
=\min_{1\le i\le r}d(\mathbf z,\pi_i).
\end{equation}
Fix once and for all
\begin{equation}\label{eq:thresholds}
b=\frac{\kappa^2}{4|S|},\qquad 0<a<\min\{\kappa,b\}.
\end{equation}
The next section proves that, for a tuple in a nuclear algebra, only
finitely many irreducible $\pi$ can satisfy $d(\mathbf z,\pi)<a$.
No group relations on that tuple will be needed.
==== END An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//spectral-preliminaries.tex ====
==== BEGIN An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//ultrapower-contradiction.tex ====
\section{From an ultrapower embedding to a finite contradiction}
\label{sec:ultrapower-contradiction}
\label{sec:contradiction}
To prove Theorem~\ref{thm:main}, fix a nonzero unital nuclear
$C^*$-algebra $B$ and a free
ultrafilter $\omega$ on $\mathbb N$. A set of indices is called
$\omega$-large if it belongs to $\omega$.
Suppose, towards a contradiction,
that there is a unital injective $*$-homomorphism
\[
\Phi:C^*(G)\longrightarrow B^\omega.
\]
We will associate a finite set of irreducible representations of
$\Gamma$ to each coordinate of $\Phi$. Conjugation by $t$ will force
these sets to contain predecessors under restriction by $\theta$.
The resulting finite chains contradict the choice of $\pi_*$.
\subsection{Unitary lifts and a gap for varying matrix dimensions}
Write $u_g$ for the canonical unitary of $g\in G$ in $C^*(G)$.
Each unitary in $B^\omega$ has a representative consisting of unitaries
in $B$. Indeed, if $(y_n)$ represents a unitary, then
\[
\lim_\omega\|y_n^*y_n-1\|
=\lim_\omega\|y_ny_n^*-1\|=0.
\]
On an $\omega$-large set both norms are less than $1/2$, so $y_n$ is
invertible and $y_n(y_n^*y_n)^{-1/2}$ is unitary. The difference between
this unitary and $y_n$ tends to zero along $\omega$, by continuous
functional calculus. Set the representative equal to $1$ outside this
set.
Choose such representatives $x_n(g)\in\mathcal U(B)$ of $\Phi(u_g)$
for every $g\in G$, with $x_n(e)=1$. For each fixed pair $g,h\in G$,
\begin{equation}
\lim_\omega\|x_n(g)x_n(h)-x_n(gh)\|=0.
\label{eq:lift-defects}
\end{equation}
These choices impose no group relations at any individual coordinate.
All uses of \eqref{eq:lift-defects} below involve finitely many fixed
group elements; no intersection over all pairs is needed.
Put $\mathbf x_n=(x_n(s))_{s\in S}$. Among the unitary equivalence
classes of nonzero finite-dimensional irreducible representations
of $\Gamma$, define
\begin{equation}
D_n=\{[\pi]:d(\mathbf x_n,\pi)<a\}.
\label{eq:coordinate-D}
\end{equation}
Here brackets denote unitary equivalence classes. Each $D_n$ is finite
by Lemma~\ref{lem:nuclear-finiteness}, which applies to arbitrary unitary
$S$-tuples.
The next observation provides uniform control even when the dimensions
of the representations increase with $n$.
\begin{lemma}\label{lem:varying-gap}
For every sequence $(\pi_n)$ of nonzero finite-dimensional unitary
representations of $\Gamma$, with no bound on their dimensions,
\begin{equation}
\lim_\omega d(\mathbf x_n,\pi_n)\in\{0\}\cup[\kappa,R].
\label{eq:varying-gap}
\end{equation}
In particular, if
$\{n:d(\mathbf x_n,\pi_n)<a\}\in\omega$, then this limit is zero.
\end{lemma}
\begin{proof}
Let $C_n=B\otimes\mathcal B(\overline{V_{\pi_n}})$, and form the norm
ultraproduct
\[
C_\omega=\prod_n C_n\big/
\{(c_n):\lim_\omega\|c_n\|=0\}.
\]
The product consists of bounded sequences. For every such sequence,
\begin{equation}
\|[(c_n)]\|=\lim_\omega\|c_n\|.
\label{eq:ultraproduct-norm}
\end{equation}
For the lower bound, adding an $\omega$-null sequence does not change
the ultralimit of the norms. For the upper bound, if this ultralimit is
$c$, replace $c_n$ by $0$ outside
$\{n:\|c_n\|<c+\varepsilon\}\in\omega$ and then let
$\varepsilon\downarrow0$.
Equation~\eqref{eq:lift-defects} shows that
\[
g\longmapsto[(x_n(g)\otimes\overline{\pi_n}(g))_n],
\qquad g\in\Gamma,
\]
is a genuine unitary representation in the nonzero unital algebra
$C_\omega$. By Lemma~\ref{lem:spectral-gap}, its Laplacian has
spectrum in $\{0\}\cup[\kappa,R]$. Write
\[
P_n=L\big((x_n(s)\otimes\overline{\pi_n}(s))_{s\in S}\big).
\]
Since $0\leq P_n\leq R1$, equation~\eqref{eq:ultraproduct-norm} gives
\begin{align*}
\min\operatorname{spec}[(P_n)]
&=R-\|R1-[(P_n)]\|\\
&=\lim_\omega\bigl(R-\|R1-P_n\|\bigr)
=\lim_\omega d(\mathbf x_n,\pi_n).
\end{align*}
This proves \eqref{eq:varying-gap}. In the final assertion the limit
is at most $a<\kappa$, and hence must be zero.
\end{proof}
\subsection{The initial representation occurs}
We next use injectivity of $\Phi$ to ensure that $[\pi_*]\in D_n$
for an $\omega$-large set of indices.
\begin{lemma}\label{lem:initial-match}
We have $\lim_\omega d(\mathbf x_n,\pi_*)=0$, and consequently
$\{n:[\pi_*]\in D_n\}\in\omega$.
\end{lemma}
\begin{proof}
Induce $\pi_*$ from $\Gamma$ to $G$ on the Hilbert direct sum of
copies of $V_{\pi_*}$ indexed by the left cosets $G/\Gamma$.
Explicitly, choose representatives $q_i$, including the identity:
when $gq_i=q_j\gamma$ with $\gamma\in\Gamma$, the action sends
coordinate $i$ to coordinate $j$ by $\pi_*(\gamma)$. This defines
a unitary representation. The identity coordinate is invariant
under $\Gamma$ and carries $\pi_*$. Thus the restriction of this representation
of $G$ contains $\pi_*$. Tensoring with $\overline{\pi_*}$ gives a
nonzero $\Gamma$-invariant vector: on this coordinate it is
$\sum_j e_j\otimes\overline{e_j}$ for an orthonormal basis $(e_j)$ of
$V_{\pi_*}$.
Every unitary representation of $G$ gives a representation of the
full group algebra $C^*(G)$. The preceding invariant vector shows that
\[
\min\operatorname{spec}
L\big((u_s\otimes\overline{\pi_*}(s))_{s\in S}\big)=0.
\]
Indeed, an invertible positive element could not acquire a kernel
under a unital representation. The matrix amplification of $\Phi$
is injective and preserves spectra, so the corresponding spectral
minimum in $B^\omega\otimes\mathcal B(\overline{V_{\pi_*}})$ is also
zero.
For each fixed positive integer $d$ there is the canonical
identification
\[
M_d(B^\omega)\cong (M_d(B))^\omega.
\]
To check its kernel, a sequence of $d$-by-$d$ matrices tends to zero in
norm along $\omega$ precisely when each of its finitely many entries
does. Entrywise lifts give surjectivity. Applying this identification
with $d=\dim\pi_*$ and the calculation of spectral minima in
Lemma~\ref{lem:varying-gap} proves the assertion. This fixed matrix
amplification is distinct from the varying algebras used in that lemma.
\end{proof}
\subsection{A predecessor for every representation at one coordinate}
For representations $\alpha$ and $\beta$ of the same group, write
$\alpha\prec\beta$ when $\alpha$ is unitarily equivalent to a
subrepresentation of $\beta$. The required quantifier order is the
following: on one $\omega$-large set, every member of $D_n$ has a
predecessor in that same finite set.
\begin{lemma}\label{lem:uniform-predecessor}
The set
\[
\{n:(\forall[\pi]\in D_n)(\exists[\pi']\in D_n)
\quad\pi\prec\pi'\circ\theta\}
\]
belongs to $\omega$.
\end{lemma}
\begin{proof}
Suppose otherwise. On an $\omega$-large set $F$, choose an
irreducible $\pi_n$ with $[\pi_n]\in D_n$ for which no such predecessor
exists. Outside $F$, choose the trivial representation. By
Lemma~\ref{lem:varying-gap},
\begin{equation}
\lim_\omega d(\mathbf x_n,\pi_n)=0.
\label{eq:chosen-zero}
\end{equation}
Set $\alpha_n=\pi_n\circ\theta^{-1}$, a representation of $\Lambda$,
and let $\rho_n=\operatorname{Ind}_{\Lambda}^{\Gamma}\alpha_n$.
We will prove that $\lim_\omega d(\mathbf x_n,\rho_n)=0$ and that
every irreducible constituent $\pi'$ of $\rho_n$ satisfies
$\pi_n\prec\pi'\circ\theta$.
The block-minimum identity~\eqref{eq:block-minimum} will then select a
constituent belonging to $D_n$ on an $\omega$-large set, and the second
assertion will make that constituent a predecessor of $\pi_n$.
\emph{Transport to $\Lambda$.}
Represent $B$ faithfully and unitally on a Hilbert space $\mathcal H$,
and use the same notation for its unitaries on $\mathcal H$. Choose
unit vectors $\xi_n\in\mathcal H\otimes\overline{V_{\pi_n}}$ such that
\[
\sum_{s\in S}
\|(x_n(s)\otimes\overline{\pi_n}(s))\xi_n-\xi_n\|^2
<d(\mathbf x_n,\pi_n)+1/n.
\]
Because $\omega$ is free, $1/n$ tends to zero along $\omega$. Thus
\begin{equation}
\lim_\omega
\|(x_n(g)\otimes\overline{\pi_n}(g))\xi_n-\xi_n\|=0
\qquad(g\in\Gamma).
\label{eq:fixed-word-invariance}
\end{equation}
Here is the fixed-word justification. Put
$U_n(g)=x_n(g)\otimes\overline{\pi_n}(g)$. For each $s\in S$,
$x_n(s^{-1})-x_n(s)^*$ tends to zero by
\eqref{eq:lift-defects}; a unitary and its inverse have the same
displacement norm on a vector. For a fixed word
$g=s_1^{\varepsilon_1}\cdots s_k^{\varepsilon_k}$, repeated use of
\eqref{eq:lift-defects} replaces $U_n(g)$ by the corresponding product
of $U_n(s_j)^{\varepsilon_j}$ with an error tending to zero.
The displacement of that product is
at most the sum of the $k$ displacements of its factors. Both the
word and $k$ are fixed while $n$ varies.
Put $\eta_n=(x_n(t)\otimes1)\xi_n$. These are unit vectors. For
$h\in\Lambda$ put $g=\theta^{-1}(h)$. The identity $ht=tg$ and
\eqref{eq:lift-defects} give
\[
\lim_\omega\|x_n(h)x_n(t)-x_n(t)x_n(g)\|=0.
\]
Together with \eqref{eq:fixed-word-invariance}, this proves
\begin{equation}
\lim_\omega
\|(x_n(h)\otimes\overline{\alpha_n}(h))\eta_n-\eta_n\|=0
\qquad(h\in\Lambda).
\label{eq:Lambda-invariance}
\end{equation}
\emph{Induction to $\Gamma$.}
We transfer these vectors from $\Lambda$ to $\Gamma$ by a fixed
eight-coordinate induction. Use the left coset representatives
$r_1=e,r_2,\ldots,r_8$ of $\Gamma/\Lambda$ given by the translations
in $\{0,1\}^3$. For $g\in\Gamma$ and $1\leq i\leq8$, define
$j(g,i)$ and $h(g,i)\in\Lambda$ uniquely by
\[
gr_i=r_{j(g,i)}h(g,i).
\]
The induced representation $\rho_n$ acts on
$\bigoplus_{i=1}^8 V_{\pi_n}$: $\rho_n(g)$ sends coordinate
$i$ to coordinate $j(g,i)$ by $\alpha_n(h(g,i))$. The relation
\[
h(g_1g_2,i)=h(g_1,j(g_2,i))h(g_2,i)
\]
verifies the representation law. Permutation of the coordinates and
unitarity of $\alpha_n$ verify unitarity. Its conjugate has the same
coordinate rule with $\overline{\alpha_n}$ in place of $\alpha_n$.
Define $\zeta_n\in\mathcal H\otimes\overline{V_{\rho_n}}$ by
\[
(\zeta_n)_i=8^{-1/2}(x_n(r_i)\otimes1)\eta_n.
\]
The eight summands are orthogonal and each has norm $8^{-1/2}$, so
$\|\zeta_n\|=1$. Fix $s\in S$ and $i$, and write
$j=j(s,i)$ and $h=h(s,i)$. The component sent from $i$ to $j$ by
$x_n(s)\otimes\overline{\rho_n}(s)$ differs from $(\zeta_n)_j$
by a vector whose norm is at most
\begin{align*}
8^{-1/2}\big(&\|x_n(s)x_n(r_i)-x_n(r_j)x_n(h)\|\\
&+\|(x_n(h)\otimes\overline{\alpha_n}(h))\eta_n-\eta_n\|\big).
\end{align*}
The first term tends to zero by \eqref{eq:lift-defects} and
$sr_i=r_jh$; the second does so by \eqref{eq:Lambda-invariance}.
There are only $8|S|$ fixed choices involved. Hence
\[
\lim_\omega\sum_{s\in S}
\|(x_n(s)\otimes\overline{\rho_n}(s))\zeta_n-\zeta_n\|^2=0.
\]
Thus $\lim_\omega d(\mathbf x_n,\rho_n)=0$, and in particular
$d(\mathbf x_n,\rho_n)<a$ on an $\omega$-large set.
All error bounds are independent of $\dim\pi_n$.
\emph{Selecting a predecessor.}
Decompose the finite-dimensional unitary representation $\rho_n$ into
irreducible summands. The corresponding Laplacian is a direct sum,
so its spectral minimum is the minimum of the spectral minima of
these summands. Consequently, on an $\omega$-large set, some
irreducible constituent $\pi'_n$ of $\rho_n$ has $[\pi'_n]\in D_n$.
Every irreducible constituent $\pi'$ of $\rho_n$, including the one
just obtained, satisfies $\pi_n\prec\pi'\circ\theta$. This is the
finite-index form of Frobenius reciprocity. To verify it for every
constituent, let $W$ be its invariant subspace
and let $P_1:W\to V_{\pi_n}$ be projection onto the identity
coordinate. This map is $\Lambda$-equivariant for the target
representation $\alpha_n$. It is nonzero: if $w\in W$ has a nonzero
$i$th coordinate, then
\[
P_1\rho_n(r_i^{-1})w=w_i\ne0.
\]
Since $\alpha_n$ is irreducible, $P_1|_W$ is surjective. Its adjoint
is an injective $\Lambda$-intertwiner; after normalization by a
positive scalar it is an isometric intertwiner, because
$(P_1|_W)(P_1|_W)^*$ commutes with the irreducible $\alpha_n$.
It follows that $\alpha_n\prec\pi'|_\Lambda$; composing with
$\theta:\Gamma\to\Lambda$ gives the asserted containment.
Intersect the $\omega$-large set on which such a $[\pi'_n]$ belongs to
$D_n$ with $F$. This contradicts the choice of $\pi_n$ and proves
the lemma.
\end{proof}
\subsection{Finite chains force periodicity}
The analytic argument is complete. We finish with the elementary
consequence of finiteness that makes the predecessor property
incompatible with $\pi_*$. Consider a chain of finite-dimensional
irreducible unitary representations $\tau_0=\pi,\tau_1,\ldots$ of
$\Gamma$ whose classes lie in a
finite set $D$ and satisfy $\tau_i\prec\tau_{i+1}\circ\theta$.
Finiteness gives indices $l\geq0$ and $p\geq1$ with
$\tau_l\simeq\tau_{l+p}$.
Figure~\ref{fig:finite-chain} separates the initial segment of this
chain from the repeated segment.
The key point is that periodicity at the repeated class forces every
pullback of that representation to remain irreducible, so that
periodicity reaches the starting class.
\input{figures/finite-chain}
\begin{lemma}\label{lem:finite-predecessor-cycle}
Let $\theta:\Gamma\to\Gamma$ be a homomorphism, and let $D$ be a
finite set of equivalence classes of nonzero finite-dimensional
irreducible unitary representations of $\Gamma$. Suppose that every
$[\pi]\in D$ has $[\pi']\in D$ with $\pi\prec\pi'\circ\theta$.
Then for every $[\pi]\in D$ there is an integer $p\geq1$ such that
$\pi\simeq\pi\circ\theta^p$.
\end{lemma}
\begin{proof}
Starting from $\tau_0=\pi$, choose a chain in $D$ with
$\tau_i\prec\tau_{i+1}\circ\theta$. Finiteness gives
$\tau_l\simeq\tau_{l+p}$ for some $l\geq0$ and $p\geq1$.
Composing the inclusions along this segment yields
\[
\tau_l\prec\tau_{l+p}\circ\theta^p
\simeq\tau_l\circ\theta^p.
\]
The first and last representations have the same dimension, so
\begin{equation}
\tau_l\simeq\tau_l\circ\theta^p.
\label{eq:cycle-periodicity}
\end{equation}
It follows by iteration that $\tau_l\circ\theta^{kp}\simeq\tau_l$
for every integer $k\geq0$.
For every $j\geq0$, the representation $\tau_l\circ\theta^j$ is
irreducible. Indeed, if it had a nonzero proper invariant subspace,
that subspace would remain invariant for
$\tau_l\circ\theta^{kp}$ whenever $kp\geq j$, because
$\theta^{kp}(\Gamma)\subseteq\theta^j(\Gamma)$. This contradicts
the irreducibility of $\tau_l\circ\theta^{kp}\simeq\tau_l$.
The initial segment of the chain gives
$\pi\prec\tau_l\circ\theta^l$. Its right-hand side is irreducible,
so $\pi\simeq\tau_l\circ\theta^l$. Composing
\eqref{eq:cycle-periodicity} with $\theta^l$ therefore gives
\[
\pi\circ\theta^p
\simeq\tau_l\circ\theta^{l+p}
\simeq\tau_l\circ\theta^l
\simeq\pi.
\]
\end{proof}
Choose one index $n$ in the intersection of the $\omega$-large sets
from Lemmas~\ref{lem:initial-match} and~\ref{lem:uniform-predecessor}.
Apply Lemma~\ref{lem:finite-predecessor-cycle} to $D_n$ and $\pi_*$.
It gives $\pi_*\simeq\pi_*\circ\theta^p$ for some $p\geq1$.
But $\pi_*$ acts nontrivially on the translations $\mathbb Z^3$,
whereas $\pi_*\circ\theta^p$ acts trivially on them: $\theta^p$
multiplies these translations by $2^p$, and $\pi_*$ factors through
reduction modulo $2$. This is the required contradiction.
The argument used only that $B$ is nonzero, unital and nuclear, and
that $\omega$ is free. The group $G$ and the algebra $C^*(G)$ are
independent of these choices. Thus the same algebra has no unital
embedding into $B^\omega$ for any such $B$ and any free $\omega$.
Taking $B=\mathcal O_2$ proves the stated negative answer to
Kirchberg's embedding problem.
==== END An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//ultrapower-contradiction.tex ====
==== BEGIN An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//references.bib ====
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}
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url = {https://arxiv.org/abs/math/0502140v4},
note = {Proposition~1.1 is cited from \href{https://arxiv.org/abs/math/0502140v4}{arXiv:math/0502140v4}}
}
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@misc{Sauers2026NonMF,
author = {{Sauers}},
title = {Non-{MF} Groups},
year = {2026},
howpublished = {\href{https://github.com/SauersML/group-approximation/blob/ae55c9f357d476922fd61881517c892140c93be6/non_mf_groups_exist.tex}{Public draft}},
url = {https://github.com/SauersML/group-approximation/blob/ae55c9f357d476922fd61881517c892140c93be6/non_mf_groups_exist.tex},
note = {Repository revision of 20 September 2026}
}
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@misc{GoldbringSinclair2025Tensor,
author = {Goldbring, Isaac and Sinclair, Thomas},
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archivePrefix = {arXiv},
primaryClass = {math.OA},
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note = {\href{https://arxiv.org/abs/2509.15086v1}{arXiv:2509.15086v1}, 18 September 2025}
}
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==== END An-explicit-obstruction-to-nuclear-norm-ultrapower-embeddings//references.bib ====