==== TABLE OF CONTENTS ====
Kadisons-similarity-theorem-through-uniform-derivation-estimates//finite-factors.tex
Kadisons-similarity-theorem-through-uniform-derivation-estimates//global.tex
Kadisons-similarity-theorem-through-uniform-derivation-estimates//hankel.tex
Kadisons-similarity-theorem-through-uniform-derivation-estimates//introduction.tex
Kadisons-similarity-theorem-through-uniform-derivation-estimates//main.tex
Kadisons-similarity-theorem-through-uniform-derivation-estimates//preamble.tex
Kadisons-similarity-theorem-through-uniform-derivation-estimates//preliminaries.tex
Kadisons-similarity-theorem-through-uniform-derivation-estimates//propagation.tex
Kadisons-similarity-theorem-through-uniform-derivation-estimates//tensors.tex
Kadisons-similarity-theorem-through-uniform-derivation-estimates//unicode-map.tex
Kadisons-similarity-theorem-through-uniform-derivation-estimates//references.bib
===========================

==== BEGIN Kadisons-similarity-theorem-through-uniform-derivation-estimates//finite-factors.tex ====
\section{Transfer to finite factors}
\label{sec:finite-factors}

We now transfer Theorem~\ref{thm:free-corner} to matrix commutators in
finite sums of the standard representation of a separable-predual
\(\mathrm{II}_1\) factor.  All traces on matrix amplifications are normalized.

We first average the operator over an irreducible subfactor generated by
an increasing sequence of full matrix algebras. After matrix amplification,
the ultrapower of the amplified subfactor contains another subfactor
generated by increasing full matrix algebras and free from the constant
copy of the amplified ambient factor. Extending and compressing the
averaged operator then puts it in the setting of
Theorem~\ref{thm:free-corner}. Compression to constant vectors transfers
the estimate back to the matrix test, with the fixed error from the
initial averaging.

The existence of an irreducible hyperfinite subfactor in a
separable-predual \(\mathrm{II}_1\) factor is due to Popa
\citep{Popa1981}; we include a direct proof of the increasing-matrix
form used below.

\begin{lemma}[An irreducible increasing-matrix subfactor]
\label{lem:irreducible-hyperfinite}
Every separable-predual \(\mathrm{II}_1\) factor \(M\) contains a unital
subfactor \(R\) with \(R'\cap M=\C1\), generated by an increasing sequence
of unital full matrix algebras whose sizes tend to infinity.
\end{lemma}

\begin{proof}
We first give an approximation step in an arbitrary \(\mathrm{II}_1\)
factor \(N\).  If \(y=y^*\in N\) and \(\eta>0\), there is a unital full
matrix algebra \(F\subset N\), of size at least two, such that
\[
 \norm{E_{F'\cap N}(y)-\tau(y)1}_2<\eta.
\]
Indeed, approximate \(y\) in \(L^2\) by a spectral step
\(y_0=\sum_{i=1}^k\lambda_i p_i\), where the \(p_i\) partition the
identity.  For a large integer \(b\), split off
\(\lfloor b\tau(p_i)\rfloor\) subprojections of trace \(1/b\) in each
\(p_i\).  The remaining projection has trace
\[
 \frac{b-\sum_{i=1}^k\lfloor b\tau(p_i)\rfloor}{b}<\frac{k}{b}.
\]
Its numerator is an integer, so it too splits into projections of trace
\(1/b\).  These pieces form a partition of the identity by \(b\)
equivalent projections and hence extend to matrix units for a unital
\(F\cong M_b(\C)\).  Define \(y_1\in F\) to have value \(\lambda_i\)
on the pieces taken from \(p_i\), and value zero on the remaining pieces.
Then
\[
 \norm{y-y_1}_2
 \leq \norm{y-y_0}_2+\norm{y_0}\sqrt{k/b}.
\]
The expectation of \(y_1\) onto \(F'\cap N\) is \(\tau(y_1)1\).
Contractivity of this expectation on \(L^2\), and
\(\abs{\tau(y-y_1)}\leq\norm{y-y_1}_2\), give the required estimate
by choosing the spectral approximation and then \(b\).

Choose a countable \(L^2\)-dense subset \(\{x_j:j\geq1\}\) of the
self-adjoint unit ball of \(M\), and enumerate all requirements
\((j,k)\in\N^2\).  Begin with \(F_0=\C1\).  If the current stage is
a unital full matrix algebra \(F_n\), then
\(N_n=F_n'\cap M\) is a \(\mathrm{II}_1\) factor: matrix units identify
\(M\) with a matrix algebra over a corner and \(N_n\) with that corner.
For the next requirement \((j,k)\), apply the approximation step in
\(N_n\) to \(y=E_{N_n}(x_j)\), with error less than \(1/k\), obtaining
\(F\subset N_n\).  Set \(F_{n+1}=F_n\vee F\).  This is again a full
matrix algebra, its size is at least twice that of \(F_n\), and
\(N_{n+1}=F'\cap N_n\).  The identity
\(E_{N_{n+1}}=E_{N_{n+1}}E_{N_n}\) gives
\[
 \norm{E_{N_{n+1}}(x_j)-\tau(x_j)1}_2<1/k.
\]
All previously obtained bounds persist because the relative commutants
decrease and their trace expectations compose.

Let \(R=(\bigcup_nF_n)''\) and \(N=R'\cap M=\bigcap_nN_n\).
Since \(E_N=E_NE_{N_n}\), every requirement remains valid with \(N\)
in place of \(N_{n+1}\).  Thus \(E_N(x_j)=\tau(x_j)1\) for every
\(j\).  Density gives \(E_N(x)=\tau(x)1\) for all \(x\in M\), so
\(N=\C1\).  In particular \(R\) is a factor.  It is finite and contains
full matrix algebras of unbounded size, hence is of type
\(\mathrm{II}_1\), with the asserted increasing-matrix presentation.
\end{proof}

For a finite tracial von Neumann algebra $P$ and a nonprincipal
ultrafilter $\omega$ on $\N$, write
$P^\omega=\ell^\infty(\N,P)/\mathcal I_\omega$, where
$\mathcal I_\omega$ consists of the uniformly bounded sequences
$(a_i)$ with $\lim_\omega\|a_i\|_2=0$. Its trace is
$\tau_\omega([a_i])=\lim_\omega\tau(a_i)$, and $P$ embeds by
constant sequences. The lemma below places a free subfactor in such
an ultrapower while retaining its inclusion in the prescribed
subalgebra $Q^\omega$.

\begin{lemma}[Free position in the full ultrapower]
\label{lem:free-position}
Let \(Q\subset P\) be a unital irreducible inclusion of
\(\mathrm{II}_1\) factors, with \(P\) of separable predual.  For any
nonprincipal ultrafilter \(\omega\) on \(\N\), there is a subfactor
\(D\subset Q^\omega\), generated by increasing unital full matrix
algebras of unbounded sizes, which is free from the constant copy of
\(P\) in the tracial ultrapower \(P^\omega\).
\end{lemma}

\begin{proof}
We use the ultraproduct-of-subalgebras case of Popa's independence
theorem \citep[Theorem~0.1(a); see also Theorem~4.3]{Popa2014}.
For von Neumann subalgebras $A,B\subset M$, the notation
$A\prec_M B$ means Popa's intertwining of a nonzero corner of $A$
into a finite matrix amplification of $B$.
Its hypotheses are finite factors \(M_i\) with
\(\dim M_i\longrightarrow\infty\), and unital von Neumann subalgebras
\(Q_i\subset M_i\) satisfying
\(Q_i\nprec_{M_i}(Q_i'\cap M_i)\).  Write
\(\mathbf M=\prod_\omega M_i\), \(\mathbf Q=\prod_\omega Q_i\), and
\(C=\mathbf Q'\cap\mathbf M\).  Given an \(L^2\)-separable subspace
\(X\subset\mathbf M\ominus C\), the theorem supplies a diffuse abelian
von Neumann algebra \(A\subset\mathbf Q\) such that
\[
 E_C(x_0a_1x_1\cdots a_nx_n)=0
\]
whenever \(n\geq1\), \(x_0\in X\cup\{1\}\), \(x_i\in X\) for
\(1\leq i\leq n\), and \(a_i\in A\ominus\C1\).
In particular, if \(C=\C1\) and \(T\subset\mathbf M\) is a
separable-predual von Neumann subalgebra, take
\(X=T\ominus\C1\).  The moment identities, traciality, and decomposition
into scalar and centered parts show that \(A\) is free from \(T\).
Here separability is for \(L^2\), as stipulated in
\citet[Section~1.1]{Popa2014}.

We check these hypotheses for the constant sequences \(M_i=P\) and
\(Q_i=Q\).  The dimension condition holds because these factors are
infinite-dimensional. If \(Q\prec_P\C1\), the intertwining data
would include a nonzero normal \(*\)-homomorphism from a nonzero
corner of \(Q\) into a finite-dimensional matrix algebra.  Such a
homomorphism from a factor is faithful, whereas a
\(\mathrm{II}_1\) factor has arbitrarily large families of nonzero
orthogonal projections.  This is impossible.  Since \(Q'\cap P=\C1\),
the non-intertwining hypothesis follows.

We also verify the full ultrapower relative commutant:
\[
                 (Q^\omega)'\cap P^\omega=\C1.
\]
For a uniformly bounded representative \((x_i)\) of an element of this
commutant, commuting with all sequences of unitaries from \(Q\) implies
\[
 d_i:=\sup_{v\in\mathcal U(Q)}
             \norm{vx_iv^*-x_i}_2\longrightarrow_\omega0.
\]
Otherwise one could choose a witnessing unitary at each coordinate
on an \(\omega\)-large set.  The \(L^2\)-closed convex hull of the
\(Q\)-unitary orbit of \(x_i\) has a unique vector of minimum norm.
It is \(Q\)-central, has trace \(\tau(x_i)\), and belongs to \(P\):
the hull stays in the operator-norm ball of radius \(\norm{x_i}\),
which is \(L^2\)-closed by weak-* compactness and trace pairings.
Irreducibility makes this vector \(\tau(x_i)1\).  Every point in the
hull is within \(d_i\) of \(x_i\), so
\(\norm{x_i-\tau(x_i)1}_2\leq d_i\).  Bounded scalar sequences equal
their scalar ultralimit in the tracial ultrapower.  This proves the
claim, which concerns all of \(Q^\omega\), rather than merely the
constant copy of \(Q\).

Choose a Haar unitary \(u_1\) in the diffuse abelian algebra obtained
by applying Popa's theorem to the constant \(P\).  Then
\(u_1\in Q^\omega\) is free from \(P\).
The algebra \(W^*(P,u_1)\) has separable \(L^2\) space: countably
many words in generators of \(P\), in \(u_1\), and in their adjoints
have linear span dense in \(L^2\).
A second application of the same theorem therefore gives a Haar
unitary \(u_2\in Q^\omega\) free from \(W^*(P,u_1)\).
Expanding centered reduced words shows that
\(P,W^*(u_1),W^*(u_2)\) are jointly free.  Thus
\(W^*(u_1,u_2)\) is free from \(P\) and, by its free Haar moments,
is trace-isomorphic to \(L(\mathbb F_2)\).  This is a separable-predual
\(\mathrm{II}_1\) factor: the nontrivial conjugacy classes of
\(\mathbb F_2\) are infinite, so a central square-summable Fourier
coefficient function is supported at the identity.  Apply
Lemma~\ref{lem:irreducible-hyperfinite} inside \(W^*(u_1,u_2)\) to
obtain \(D\).  Freeness passes to its subalgebra \(D\), proving the
lemma.  The ambient tracial ultrapower throughout is \(P^\omega\).
\end{proof}

\begin{proposition}[Finite standard multiplicities]
\label{prop:finite-copies}
Let \(M\) be a separable-predual \(\mathrm{II}_1\) factor, and let
\(M\) act on \(H_r=L^2(M)^r\) by common left multiplication, where
\(r\geq1\) is finite.  For every \(Y\in\mathcal B(H_r)\), every
finite \(h\geq1\), and every \(X\in M_h(M)\),
\[
 \norm{[Y^{(h)},X]}
       \leq (3c_1+2)\,g_M(Y)\,\norm{X}.
\]
Here \(X\) denotes its entrywise action on \(H_r^h\); the constant
is independent of \(M,r,h\).
\end{proposition}

\begin{proof}
Put \(\delta=g_M(Y)\).  By homogeneity it suffices to consider
\(\norm{X}\leq1\).  We first arrange a symmetry test on a square
array of standard copies.  Choose \(h_0\geq h\) so that
\(m=2h_0\geq r\), extend \(Y\) by zero to \(H_m\), and retain the
notation \(Y\).  The copy projection reduces the common action of
\(M\), so this extension still has scalar commutator bound \(\delta\).
For \(X'=X\oplus0\in M_{h_0}(M)\), define in \(P=M_m(M)\)
\[
 s=
 \begin{pmatrix}
  (1-X'X'^*)^{1/2}&X'\\
  X'^*&-(1-X'^*X')^{1/2}
 \end{pmatrix}.
\]
Functional calculus gives
\((1-X'X'^*)^{1/2}X'=X'(1-X'^*X')^{1/2}\), and hence
\(s=s^*=s^{-1}\).  Traciality gives equal moments for
\(X'X'^*\) and \(X'^*X'\), including the zeroth moment; polynomial
approximation on \([0,1]\) gives equal traces for their displayed
square roots.  Consequently \(\tau_P(s)=0\).

Write vectors of \(H_m^m\) as arrays \((\xi_{ij})\), with \(i\)
the outer amplification coordinate and \(j\) the copy coordinate.
The map \((\xi_{ij})\mapsto\sqrt m\,(\xi_{ij})\) is a unitary from
\(H_m^m\) to \(L^2(P)\), since
\(\norm{(\xi_{ij})}_2^2=m^{-1}\sum_{i,j}\norm{\xi_{ij}}_2^2\)
on the latter space.  Under this identification, the matrix action
of \(P\) is left multiplication.  The upper-right block of
\([Y^{(m)},L_s]\) is the commutator with \(X'\).  Restricting its
row coordinates to the original \(h\) and its copy coordinates to
the original \(r\) recovers \([Y^{(h)},X]\) for the original operator.
It is therefore enough to bound \([Y^{(m)},L_s]\).

Choose \(R\subset M\) from Lemma~\ref{lem:irreducible-hyperfinite},
with matrix stages \(F_n\).  On \(H_m\), average \(Y\) over the compact
unitary group of each \(F_n\), using common left multiplication,
and take a weak operator cluster point \(Y_0\) as \(n\) tends to
infinity.  For every such unitary \(v\),
\(\norm{L_vYL_v^*-Y}\leq\delta\).  The averages, and hence their
cluster point, are within \(\delta\) of \(Y\).  An average over a
later stage commutes with every earlier stage, so
\[
 [Y_0,L_R]=0,\qquad
 \norm{Y_0-Y}\leq\delta,\qquad g_M(Y_0)\leq3\delta.
\]
The first identity extends from the matrix stages to \(R\) by strong
density; the norm bounds pass to the cluster point by weak operator
closedness of norm balls.

On \(L^2(P)=H_m^m\), set \(S=Y_0^{(m)}\) and \(Q=M_m(R)\).
Matrix units and \(R'\cap M=\C1\) imply \(Q'\cap P=\C1\).
Moreover \(S\) commutes with \(L_Q\), since it has the same operator
\(Y_0\) on every row and \(Y_0\) commutes with every entry from \(R\).
For \(e_0=e_{11}\otimes1\), the space \(e_0L^2(P)\) is one row of
\(m\) standard copies, with \(e_0Pe_0\cong M\) acting by common
left multiplication.  Since \(S\) commutes with \(L_{e_0}\), it follows
on the full Hilbert space that
\[
 \norm{[S,L_z]}\leq3\delta\norm{z}\qquad(z\in e_0Pe_0).
\]
If \(f\in Q\) is a projection with \(\tau_P(f)\leq1/m\), projection
comparison in the factor \(Q\) gives a unitary \(v\in Q\) with
\(f\leq ve_0v^*\).  Conjugating by \(L_v\), which commutes with
\(S\), yields
\[
 \norm{[S,L_z]}\leq3\delta\norm{z}
       \qquad(z\in fPf,\ f\in Q,\ \tau_P(f)\leq1/m).
\]

We have obtained the same small-corner estimate for every projection
of $Q$ of trace at most $1/m$. The next step transports that estimate
to the ultrapower, where Lemma~\ref{lem:free-position} supplies the
free subfactor. Fix a nonprincipal ultrafilter $\omega$ on $\N$.
Recall the tracial quotient and distinguish it from the Hilbert
ultrapower:
\[
 \begin{aligned}
 P^\omega&=\ell^\infty(\N,P)/\mathcal I_\omega,
 &\mathcal I_\omega&=
 \{(a_i):\lim_\omega\norm{a_i}_2=0\},\\
 \mathcal H&=(L^2(P))^\omega_{\mathrm{Hilb}}.
 \end{aligned}
\]
The Hilbert ultrapower uses bounded Hilbert-space sequences modulo
those whose Hilbert norms tend to zero along \(\omega\).  Sending
the class of a uniformly operator-bounded sequence \((a_i)\) to its
Hilbert-ultrapower class is a trace-preserving isometry on the
dense subspace \(P^\omega\subset L^2(P^\omega)\).  Its extension
identifies \(\mathcal C=L^2(P^\omega)\) with a closed subspace of
\(\mathcal H\).

For a bounded sequence \(a=(a_i)\) in \(P\), its coordinatewise left
action on \(\mathcal H\), and the adjoint action corresponding to
\((a_i^*)\), preserve the bounded-algebra-sequence subspace of
\(\mathcal C\).  Thus \(\mathcal C\) reduces both actions.  If
\(a\in\mathcal I_\omega\), then for every operator-bounded sequence
\((b_i)\),
\[
 \norm{a_ib_i}_2\leq\norm{b_i}\norm{a_i}_2
                       \longrightarrow_\omega0.
\]
It follows by density that the action on \(\mathcal C\) factors
through the tracial quotient and is precisely left multiplication
by \([a_i]\).  This factoring assertion is needed only on
\(\mathcal C\), not on the whole Hilbert ultrapower.

Let \(S^{\omega}_{\mathrm{Hilb}}\) be the coordinatewise action of
the fixed bounded operator \(S\) on \(\mathcal H\), and set
\[
 \widehat S=P_{\mathcal C}S^{\omega}_{\mathrm{Hilb}}
                              \big|_{\mathcal C}.
\]
Since \(S\) commutes with every left multiplier from \(Q\), its
Hilbert ultrapower commutes with every bounded left \(Q\)-sequence.
Compression to the reducing subspace \(\mathcal C\) proves
\([\widehat S,L_{Q^\omega}]=0\).

Set \(t=1/(2m)\).  We claim that
\begin{equation}
 \label{eq:ultrapower-corner}
 \begin{gathered}
 \norm{[\widehat S,L_z]}\leq3\delta\norm{z},\\
 e\in Q^\omega\text{ a projection},\qquad
       \tau(e)=t,\qquad z\in eP^\omega e.
 \end{gathered}
\end{equation}
To lift \(e\), take a bounded self-adjoint representative \((a_i)\)
in \(Q\) and set \(f_i=1_{[1/2,\infty)}(a_i)\).  The scalar inequality
\(\abs{x-1_{[1/2,\infty)}(x)}\leq2\abs{x^2-x}\) gives
\(\norm{f_i-a_i}_2\longrightarrow_\omega0\), so \([f_i]=e\) and
\(\tau(f_i)\longrightarrow_\omega t\).  The strict gap
\(t<1/m\) allows us to replace \(f_i\) by zero off an
\(\omega\)-large set and assume \(\tau(f_i)\leq1/m\) for every \(i\).

Given \(z\in eP^\omega e\) and \(\eta>0\), the \(C^*\)-quotient
norm gives a representative \((b_i)\) with
\(\sup_i\norm{b_i}\leq\norm{z}+\eta\).  The sequence
\(z_i=f_ib_if_i\) represents \(z\), lies in the required corners,
and has the same norm bound.  Consequently
\[
 \norm{[S,L_{z_i}]}\leq3\delta(\norm{z}+\eta)
 \qquad\text{for every }i.
\]
This estimate passes to the Hilbert ultrapower.  Its compression to
\(\mathcal C\) is \([\widehat S,L_z]\), since \(\mathcal C\) reduces
the bounded left sequence action.  Letting \(\eta\) decrease to zero
proves \eqref{eq:ultrapower-corner}.

By Lemma~\ref{lem:free-position}, choose \(D\subset Q^\omega\)
free from the constant \(P\), and put
\(\mathcal L=W^*(D,P)\subset P^\omega\).  Faithfulness of the trace
and the reduced-word inner products identify this algebra with the
tracial free product \(D*P\).  Let
\(G=P_{L^2(\mathcal L)}\widehat S|_{L^2(\mathcal L)}\).
The subspace \(L^2(\mathcal L)\) reduces left multiplication by
\(\mathcal L\).  Thus \(G\) commutes with \(L_D\), and
\eqref{eq:ultrapower-corner} gives the corner hypotheses of
Theorem~\ref{thm:free-corner} with
\(A_0=P\), \(\epsilon=3\delta\), and \(t=1/(2m)\).

Finally, every constant \(L^2(P)\)-vector belongs to \(\mathcal C\),
by approximation by bounded vectors from \(P\), and
\(S^{\omega}_{\mathrm{Hilb}}\) sends the constant vector \(\xi\) to
the constant vector \(S\xi\).  Both of these vectors also belong to
\(L^2(\mathcal L)\).  Therefore both compressions preserve the action
of \(S\) on the constant subspace.  As \(L_s\) preserves that subspace,
for \(\xi,\eta\in L^2(P)\) we have exactly
\[
 \langle[G,L_s]\xi,\eta\rangle
        =\langle[S,L_s]\xi,\eta\rangle.
\]
Theorem~\ref{thm:free-corner}, applied to the trace-zero symmetry
\(s\in P\), consequently yields on the original \(L^2(P)\)
\[
                   \norm{[S,L_s]}\leq3c_1\delta.
\]
Replacing \(Y_0\) by \(Y\) costs at most
\(2\norm{Y_0-Y}\norm{s}\leq2\delta\), since Hilbert amplification
preserves the difference norm.  The initial block compression now
gives the claimed bound \((3c_1+2)\delta\) for the original finite
\(r,h\).
\end{proof}
==== END Kadisons-similarity-theorem-through-uniform-derivation-estimates//finite-factors.tex ====

==== BEGIN Kadisons-similarity-theorem-through-uniform-derivation-estimates//global.tex ====
\section{From finite factors to arbitrary homomorphisms}\label{sec:global}

We now remove the restrictions on the algebra and its representation, and
then pass from inner commutators to arbitrary derivations.  Throughout this
section set
\begin{equation}\label{eq:global-constants}
 C_{\mathrm{fac}}
   =\max\{3c_1+2,\,1+2c_{\mathrm{row}},\,2\},
 \qquad C=3C_{\mathrm{fac}}+2.
\end{equation}
These constants depend only on the numerical constants of the preceding
sections.

\subsection{Representations of factors}

\begin{proposition}\label{prop:factor-representations}
Let $P\subset\B(K)$ be a unital factor acting on a separable Hilbert space.
For every $Y\in\B(K)$, every $h\ge1$, and every $X\in M_h(P)$,
\[
 \norm{[Y^{(h)},X]}
   \le C_{\mathrm{fac}}\,g_P(Y)\,\norm{X}.
\]
\end{proposition}

\begin{proof}
We first treat a type $\mathrm{II}_1$ factor $M$ in an arbitrary normal
representation on a separable Hilbert space.  Its predual is separable.
Every cyclic normal representation embeds into the tracial standard
representation.  Indeed, if $\xi$ is a cyclic vector, its normal positive
functional has a trace density $a\in L^1(M)_+$
\citep[Corollary~5.14 and Remark~5.15]{Hiai2020}.
The map on cyclic orbits
\[
 x\xi\longmapsto xa^{1/2}\in L^2(M)
\]
is isometric, because both squared norms equal the functional evaluated at
$x^*x$.  Its extension intertwines the representations and has a reducing
range.  Decomposing into orthogonal cyclic reducing subspaces, of which
there are at most countably many, embeds the given representation into
$\widetilde K=L^2(M)^{\oplus\N}$.  These are standard consequences of the normal
representation theory of finite von Neumann algebras; see
\citet{Takesaki1979}.

Extend $Y$ by zero off the embedded reducing subspace.  The scalar
commutator bound is unchanged.  On the countable amplification, let $p_n$
be the projection onto the first $n$ copies.  Since $p_n$ commutes with the
common left action,
\[
 [p_nYp_n,L_x]=p_n[Y,L_x]p_n\qquad(x\in M).
\]
Proposition~\ref{prop:finite-copies} applies to the compression of $Y$ to
$p_n\widetilde K$, with constant $3c_1+2$.  For any fixed finite matrix test, the
corresponding compressed commutators converge strongly to the full
commutator as $n\to\infty$.  A uniform operator-norm bound passes to this
strong limit.  Restricting back to the original reducing subspace proves
the assertion for type $\mathrm{II}_1$ factors.

If $P$ is properly infinite, choose isometries $v_1,\ldots,v_h\in P$ with
orthogonal ranges.  The row $V=[v_1\ \cdots\ v_h]$ satisfies $V^*V=I_{K^h}$.
Apply the row and column derivation estimates of
Section~\ref{sec:preliminaries} to $x\mapsto[Y,x]$.  With
$\delta=g_P(Y)$, they give
\[
 \norm{YV-VY^{(h)}}\le c_{\mathrm{row}}\delta,
 \qquad
 \norm{Y^{(h)}V^*-V^*Y}\le c_{\mathrm{row}}\delta.
\]
For $X\in M_h(P)$, the identity
\begin{align*}
 [Y,VXV^*]
  &=(YV-VY^{(h)})XV^*
    +V[Y^{(h)},X]V^*\\
  &\hspace{2em}+VX(Y^{(h)}V^*-V^*Y)
\end{align*}
and $\norm{VXV^*}=\norm{X}$ imply
\[
 \norm{[Y^{(h)},X]}
 \le(1+2c_{\mathrm{row}})\delta\norm{X}.
\]
Here conjugation by the isometry $V:K^h\to K$ preserves the norm of the
middle term, regardless of whether its range is all of $K$.

Finally, if $P$ is a finite type I factor, average $Y$ over its compact
unitary group to obtain $Y_0\in P'$ with $\norm{Y-Y_0}\le\delta$.
Then
\[
 [Y^{(h)},X]=[(Y-Y_0)^{(h)},X],
\]
which gives the constant $2$.  These three cases exhaust the factors and
prove the proposition.
\end{proof}

\subsection{Central decomposition}

\begin{proposition}\label{prop:separable-estimate}
Let $P\subset\B(K)$ be any unital von Neumann algebra on a separable
Hilbert space.  Then
\[
 \norm{[Y^{(h)},X]}\le C\,g_P(Y)\,\norm{X}
 \qquad(Y\in\B(K),\ X\in M_h(P),\ h\ge1).
\]
\end{proposition}

\begin{proof}
Write $\delta=g_P(Y)$.  Average $Y$ over the compact unitary groups of the
finite-dimensional algebras generated by finite families of central
projections.  Direct these families by inclusion and take a cofinal weak
operator cluster limit.  The resulting operator $Y_1$ commutes with every
central projection, hence with $Z(P)$, and satisfies
\begin{equation}\label{eq:center-average}
 \norm{Y_1-Y}\le\delta,
 \qquad g_P(Y_1)\le3\delta.
\end{equation}
For the first inequality, each unitary being averaged belongs to $P$, so
its conjugation changes $Y$ by at most $\delta$.  The inequality is
preserved by averages and weak operator limits.  The second follows from
$g_P(Y_1)\le g_P(Y)+2\norm{Y_1-Y}$.

Use the central direct-integral decomposition on a separable Hilbert
space \citep{Takesaki1979}; precise section locators for the
factor fibers, decomposable operators, and norm formula are
\citet[Sections~III.1.6.1--4 and III.5.1.10--15]{Blackadar2006}:

\[
 K=\int^{\oplus}K_\alpha\,d\mu(\alpha),
 \qquad
 P=\int^{\oplus}P_\alpha\,d\mu(\alpha),
\]
where the $P_\alpha$ are factors almost everywhere and the fibers are
separable.  The center acts as the scalar diagonal algebra, so its
commutant consists of the decomposable operators.  In particular, $Y_1$
has fibers $Y_1(\alpha)$.  We verify carefully that
\begin{equation}\label{eq:fiber-commutator}
 g_{P_\alpha}\bigl(Y_1(\alpha)\bigr)\le3\delta
 \quad\text{for almost every }\alpha.
\end{equation}

The separable direct-integral decomposition admits countably many bounded
measurable sections whose values generate $P_\alpha$ almost everywhere;
fix such a family. Take all rational complex $*$-polynomials in
these sections, including the rational scalar constants.  For each such
section $b\in P$, form the contraction
\[
 c_b=b(1+b^*b)^{-1/2}\in P.
\]
On every fiber where the generators have the stated property, these
contractions are norm dense in the contraction ball of the generated
$C^*$-algebra.  To see this, the polynomials are norm dense, the displayed
transformation is norm continuous, and its range contains the open unit
ball: its inverse there is
\[
 c\longmapsto c(1-c^*c)^{-1/2}.
\]
The closed unit ball is the norm closure of the open one.  Kaplansky
density then makes these countably many fiber contractions strongly
dense in the unit ball of $P_\alpha$.

For each $c_b$, the bound $\norm{[Y_1,c_b]}\le3\delta$ implies the same
fiber bound outside a null set, by the essential-supremum norm formula.
Removing the union of these countably many null sets leaves a single
conull set on which every test holds.  Strong approximation by the
contraction tests passes the bound to every element of the fiber unit
ball, proving \eqref{eq:fiber-commutator}.

Apply Proposition~\ref{prop:factor-representations} on that conull set.
For any fixed $X\in M_h(P)$, decomposability of its entries and the same
norm formula give
\[
 \norm{[Y_1^{(h)},X]}
 \le3C_{\mathrm{fac}}\delta\norm{X}.
\]
Restoring $Y$ costs at most $2\delta\norm{X}$ by
\eqref{eq:center-average}, giving $C=3C_{\mathrm{fac}}+2$.
\end{proof}

\subsection{Removing separability}

\begin{proof}[Proof of Theorem~\ref{thm:universal}]
Let $P\subset\B(K)$ be arbitrary, and fix $h\ge1$, $X\in M_h(P)$, and a
unit vector $\xi=(\xi_1,\ldots,\xi_h)\in K^h$.  Let $A_X\subset P$ be
the unital $C^*$-algebra generated by the entries of $X$.  The
$C^*$-algebra in $\B(K)$ generated by these entries and $Y$ is norm
separable.  The closed span of its orbits of the finitely many $\xi_i$
is therefore a separable common reducing subspace $K_0$ for $A_X$ and
$Y$.  Denote restriction of $A_X$ to $K_0$ by $\sigma$, put
$Y_0=Y|_{K_0}$, and set $P_0=\sigma(A_X)''$.

For $a\in A_X$,
\[
 \norm{[Y_0,\sigma(a)]}\le g_P(Y)\norm{a}.
\]
The $C^*$-quotient norm of the image gives, for every $b\in\sigma(A_X)$,
\[
 \norm{[Y_0,b]}
 \le g_P(Y)\inf_{\sigma(a)=b}\norm{a}
 =g_P(Y)\norm{b}.
\]
By Kaplansky density the same scalar commutator bound holds on $P_0$.
Proposition~\ref{prop:separable-estimate}, applied to $P_0$ and
$\sigma_h(X)$, now yields
\[
 \norm{[Y_0^{(h)},\sigma_h(X)]}
 \le C g_P(Y)\norm{\sigma_h(X)}
 \le C g_P(Y)\norm{X}.
\]
Since $K_0$ reduces all the operators in this test and contains every
component of $\xi$, the restricted commutator on $\xi$ equals the original
one.  Taking the supremum over unit test vectors proves
\eqref{eq:universal}.  If $K=\{0\}$, the estimate is vacuous.
\end{proof}

\subsection{Derivations and similarity}

Theorem~\ref{thm:universal} now controls inner commutators in every
representation. To reach Kirchberg's criterion we must implement an
arbitrary bounded derivation. Cyclic-domain implementation supplies
finitely many columns at a time; the uniform commutator estimate lets
us pass from these finite collections to the full derivation.

\begin{proposition}\label{prop:all-derivations}
Let $A$ be a unital $C^*$-algebra and $\sigma:A\to\B(K)$ a unital
$*$-representation.  Every bounded $\sigma$-derivation
$\Delta:A\to\B(K)$ satisfies
\[
 \norm{\Delta}_{\mathrm{cb}}\le C\norm{\Delta}
\]
and is implemented by a bounded operator on $K$.
\end{proposition}

\begin{proof}
Decompose $K$ into cyclic reducing subspaces for $\sigma$:
\[
 K=\bigoplus_{i\in I}K_i.
\]
Such a decomposition is obtained by taking a maximal orthogonal family
of cyclic reducing subspaces; a nonzero vector in its orthogonal
complement would generate another one.  The index set $I$ can have any
cardinality.  Let $q_i$ denote the projection onto $K_i$.

The map $a\mapsto\Delta(a)|_{K_i}$ is a rectangular derivation with cyclic
domain representation $\sigma|_{K_i}$ and full output representation
$\sigma$.  Lemma~\ref{lem:cyclic-implementation} gives a bounded map
$V_i:K_i\to K$ such that
\[
 \Delta(a)|_{K_i}=V_i\sigma(a)|_{K_i}-\sigma(a)V_i.
\]
For a finite subset $F\subset I$, put
\[
 q_F=\sum_{i\in F}q_i,
 \qquad V_F=\sum_{i\in F}V_iq_i\in\B(K).
\]
Every $V_F$ is bounded, and the reducing property gives the exact identity
\begin{equation}\label{eq:finite-column-identity}
 [V_F,\sigma(a)]=\Delta(a)q_F\qquad(a\in A).
\end{equation}
Only the scalar commutator norm, rather than the norm of $V_F$, needs to
be uniform in $F$.  Indeed, taking the infimum over lifts of an element
$b\in\sigma(A)$ in \eqref{eq:finite-column-identity} gives
\[
 \norm{[V_F,b]}\le\norm{\Delta}\norm{b}.
\]
This uses the $C^*$-quotient norm of the $*$-image and does not require
$\sigma$ to be faithful.  Kaplansky density then implies
\[
 g_{\sigma(A)''}(V_F)\le\norm{\Delta}.
\]

Apply Theorem~\ref{thm:universal} to $V_F$ and $\sigma(A)''$.  For every
$X\in M_h(A)$,
\begin{align*}
 \norm{\Delta_h(X)q_F^{(h)}}
   &=\norm{[V_F^{(h)},\sigma_h(X)]}\\
   &\le C\norm{\Delta}\norm{\sigma_h(X)}
    \le C\norm{\Delta}\norm{X}.
\end{align*}
The projections $q_F$ converge strongly to the identity as a net indexed
by finite subsets of $I$.  Hence the displayed bound passes to
$\Delta_h(X)$ for every fixed $h$ and $X$.  This proves the claimed
complete boundedness.  Lemma~\ref{lem:cb-implementation} then supplies a
bounded implementing operator.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:similarity}]
For every faithful unital $*$-representation of $A$, each bounded
derivation into the ambient bounded operators is inner by
Proposition~\ref{prop:all-derivations}.  Thus condition~(2) of
Theorem~\ref{thm:kirchberg} holds for $A$.  Its equivalent similarity
property gives, for every bounded unital homomorphism
$\pi:A\to\B(H)$, a bounded invertible $S$ with bounded inverse such
that $S\pi(\cdot)S^{-1}$ is a $*$-homomorphism.
The identity isomorphism treats the zero Hilbert space as well.
\end{proof}

\subsection{Hyperreflexivity}\label{sec:hyperreflexivity}

The uniform commutator estimate also controls distance to any von Neumann
algebra. We apply it to the commutant, whose projections describe the
algebra's invariant subspaces.

\begin{proof}[Proof of Corollary~\ref{cor:hyperreflexivity}]
Put \(P=M'\). Since \(M\) is self-adjoint, its invariant subspaces are
reducing and their orthogonal projections are exactly the projections in
\(P\). Hence \(\alpha_M\) is the usual off-diagonal distance seminorm.
For every projection \(e\in P\), the two off-diagonal blocks give
\[
 \|[T,e]\|=\max\{\|(1-e)Te\|,\|eT(1-e)\|\}\le\alpha_M(T),
\]
where the last inequality also uses the projection \(1-e\).
Spectral step approximations of a self-adjoint contraction \(a\in P\)
express it, modulo a scalar, as a norm limit of linear combinations of
projections in \(P\) whose coefficients have total absolute value at most two.
Thus \(\|[T,a]\|\le2\alpha_M(T)\). The real and imaginary parts of
any contraction in \(P\) are self-adjoint contractions, so
\(g_P(T)\le4\alpha_M(T)\).

Arveson's commutant-distance formula, in the form recalled in
\citet[Equation~(2.1)]{ChristensenSinclairSmithWhite2010}, gives
\[
 2\dist(T,M)=\|\operatorname{ad}_T|_P\|_{\mathrm{cb}},
 \qquad \operatorname{ad}_T(a)=[T,a],
\]
since \(P'=M\). Taking the supremum over finite matrix sizes in
Theorem~\ref{thm:universal} therefore yields
\[
 2\dist(T,M)\le Cg_P(T)\le4C\alpha_M(T),
\]
which proves Equation~\eqref{eq:universal-hyperreflexivity}.
\end{proof}
==== END Kadisons-similarity-theorem-through-uniform-derivation-estimates//global.tex ====

==== BEGIN Kadisons-similarity-theorem-through-uniform-derivation-estimates//hankel.tex ====
\section{Free compression and the Hankel test}
\label{sec:hankel}

Retain $e,p,d,w,u$, the isometries $U_k$, and the subspace $\mathcal K$
from the preceding section. We construct $h\in e\cL e$ whose norm is
of order $u$ and whose compression to $\mathcal K$ has a Hankel component
of the same order, with a uniform lower bound after division by $u$.

\subsection{The dihedral corner}

Let $\mathcal C=W^*(d,s)$. Freeness of the centered symmetries $d,s$
shows that their alternating reduced words, together with $1$, form an
orthonormal word basis of $L^2(\mathcal C)$. These are precisely the
group words of the infinite dihedral group. In particular,
\[
  w=ds\text{ is a Haar unitary},\qquad dwd=w^{-1},\qquad p=(1+d)/2.
\]
Write $m$ for Haar probability measure on $\T$.

\begin{lemma}[The abelian corner and corner freeness]
\label{lem:dihedral-corner}
The algebra $p\mathcal C p$ is identified with the even functions on the
circle by
\begin{equation}
  \begin{gathered}
    p\mathcal C p
      =\{f(w)p:f\in L^\infty(\T,m),\ f(z)=f(z^{-1})\},\\
    \tau_p(f(w)p)=\int_{\T}f\,dm.
  \end{gathered}
  \label{eq:dihedral-corner}
\end{equation}
The map $f\mapsto f(w)p$ in this formula is a faithful normal unital
$*$-isomorphism onto the corner. Moreover, $pDp$ and $p\mathcal C p$
are free with respect to the normalized trace $\tau_p$.
\end{lemma}

\begin{proof}
Every group word has the form $w^l$ or $w^ld$, $l\in\mathbb Z$.
Since $dp=p$ and $dw^ld=w^{-l}$,
\[
  pw^ldp=pw^lp=\frac{w^l+w^{-l}}2p.
\]
An even function of $w$ commutes with $d$ and hence with $p$. Thus
$f\mapsto f(w)p$ is a normal $*$-homomorphism on the even-function
algebra, with unit $p$. The group trace gives
\[
  \tau_p(f(w)p)
      =2\tau(f(w)p)=\tau(f(w))+\tau(f(w)d)=\int_{\T}f\,dm,
\]
initially for trigonometric polynomials and then for bounded functions
by $L^2$ approximation. In particular the map is faithful, and its
range is a von Neumann algebra. The compressed group-word span is
strongly dense in $p\mathcal C p$ and consists of even Laurent
polynomials times $p$. This proves \eqref{eq:dihedral-corner} and the
isomorphism assertion. It also gives
\[
  \|f(w)p\|=\|f\|_\infty,\qquad
  \|f(w)p\|_{2,\tau_p}=\|f\|_{L^2(m)}.
\]

For freeness, let $a_1,\ldots,a_j\in pDp$ be centered for $\tau_p$.
Then $\tau(a_i)=0$ and $a_id=a_i$. The centered trigonometric span in
$p\mathcal C p$ is generated by the elements $pw^lp$, $l\ge1$.
For $l_1,\ldots,l_j\ge1$, absorbing the corner projections and using
traciality gives
\begin{align*}
 &\tau_p(a_1pw^{l_1}p\cdots a_jpw^{l_j}p)\\
 &\qquad=2\tau(a_1w^{l_1}\cdots a_jw^{l_j})\\
 &\qquad=2\tau\big(a_1s(ds)^{l_1-1}\cdots
                              a_js(ds)^{l_j-1}\big)=0.
\end{align*}
All the displayed $D$ letters are centered, and $s$ is centered and
free from the whole algebra $D$, by $\cL=D*A_0$. Thus the last word is
a reduced alternating word of trace zero. Fej\'er means of an even
bounded zero-mean function are even zero-mean trigonometric polynomials,
are uniformly bounded, and converge in $L^2(m)$. Approximating each
corner letter in this manner extends the vanishing to all cyclic
alternating products of even length. Products of odd alternating
length have their endpoints in the same algebra. Move one endpoint
cyclically next to the other, split their product into its centered and
scalar parts, and apply the even-length case and induction on the
length. The length-one case is centeredness. This proves freeness.
\end{proof}

\subsection{The expectation and the norm of a free compression}

\begin{lemma}[Free compression]
\label{lem:free-compression}
For $b=b^*\in p\mathcal C p$ with $\tau_p(b)=0$,
\begin{equation}
  E_{\mathcal C}(ebe)=u^2b,\qquad
  \|ebe\|\le2\sqrt u\,\|b\|_{2,\tau_p}+u\|b\|,
  \qquad u=\tau_p(e).
  \label{eq:free-compression}
\end{equation}
Here $E_{\mathcal C}$ is the trace-preserving conditional expectation
from $\cL$ onto $\mathcal C$.
\end{lemma}

\begin{proof}
Put $\mathcal A=pDp$, $\mathcal Z=p\mathcal C p$, and
$a=e-up\in\mathcal A$, so $\tau_p(a)=0$. Since $p\in\mathcal C$,
bimodularity shows that $E_{\mathcal C}(ebe)$ belongs to $\mathcal Z$.
On $p\cL p$, its restriction is the trace-preserving expectation onto
$\mathcal Z$ for $\tau_p$. Expand
\[
  ebe=u^2b+u(ab+ba)+aba.
\]
Freeness gives zero expectation for $ab$ and $ba$. For the last term,
test against any $c\in\mathcal Z$ and write
$c=c^\circ+\tau_p(c)p$. Freeness and traciality give
\[
  \tau_p(c^\circ aba)=0,
  \qquad \tau_p(aba)=\tau_p(ba^2)
                  =\tau_p(b)\tau_p(a^2)=0.
\]
Thus $E_{\mathcal Z}(aba)=0$ also. This proves the expectation formula.

For the norm estimate work in
\[
  H_c=L^2(W^*(\mathcal A,\mathcal Z),\tau_p).
\]
This representation is faithful, so its operator norm is the algebra
norm. The free-word decomposition, grouping a possible initial
$\mathcal A$ letter, takes the form
\[
  H_c=L^2(\mathcal A,\tau_p)\otimes F_0,
  \qquad F_0=\C\Omega\oplus F_0^\circ,
\]
where $F_0^\circ$ is the closed span of reduced words starting in the
centered $L^2$ subspace $\mathcal Z^\circ$. Let $Q_0$
be the projection of $F_0$ onto the complement of the vacuum, and let
$\Pi$ project in $H_c$ onto words starting in $\mathcal Z^\circ$.
Then
\[
  \Pi=|1_{\mathcal A}\rangle\langle1_{\mathcal A}|\otimes Q_0,
  \qquad L_e=\lambda_{\mathcal A}(e)\otimes I_{F_0},
\]
where $1_{\mathcal A}=p$ has $L^2(\tau_p)$ norm one. Consequently
\begin{equation}
  \|L_e\Pi\|=\|\Pi L_e\|\le\|e\|_{2,\tau_p}=\sqrt u.
  \label{eq:corner-overlap}
\end{equation}

The following three blocks are the standard creation, annihilation,
and diagonal decomposition of a centered free-product letter; see
\citet[Section~1, Lemmas~2.1--2.3 and Corollary~2.4]{RicardXu2006}.
We give the bounds needed here directly.
Let $T=L_b$ on $H_c$. Because $b$ is centered, on the vacuum and on
words starting in $\mathcal A^\circ$ its left action simply prepends
the letter $b$. Therefore
\[
  (1-\Pi)T(1-\Pi)=0,\qquad
  \|\Pi T(1-\Pi)\|\le\|b\|_{2,\tau_p}.
\]
Self-adjointness gives the same bound for $(1-\Pi)T\Pi$, and
$\|\Pi T\Pi\|\le\|b\|$. Decomposing $T$ into these three blocks,
using $\|L_e\|\le1$, and applying \eqref{eq:corner-overlap}, we obtain
\begin{align*}
  \|ebe\|=\|L_eTL_e\|
   &\le\|L_e\Pi T(1-\Pi)L_e\|
      +\|L_e(1-\Pi)T\Pi L_e\|
      +\|L_e\Pi T\Pi L_e\|\\
   &\le\sqrt u\,\|b\|_{2,\tau_p}
      +\sqrt u\,\|b\|_{2,\tau_p}+u\|b\|.
\end{align*}
This proves \eqref{eq:free-compression}.
\end{proof}

\subsection{Fourier coefficients of the compressed operator}

The preceding lemma gives both the expectation and the upper norm
bound for a corner compression. We now compute its matrix entries on
$\mathcal K$; their dependence on the sum of the indices will isolate
the commutator with $L_s$.

Let $f\in L^\infty(\T,m)$ be real, even, and of mean zero, and set
\[
  b_f=f(w)p,\qquad h=eb_fe,\qquad
  \gamma_l=\int_{\T}z^lf(z)\,dm(z)\quad(l\in\mathbb Z).
\]
Then $b_f$ meets the hypotheses of Lemma~\ref{lem:free-compression}.
We claim that, for $j,k\ge0$,
\begin{equation}
  \begin{aligned}
    U_j^*L_hU_k
      &=L_{t^{-1}E_{A_0}(w^jhw^{-k})}\big|_B\\
      &=u\big(\gamma_{j-k}I_B+\gamma_{j+k+1}L_s|_B\big).
  \end{aligned}
  \label{eq:hankel-compression}
\end{equation}
For the first equality, test on $v,z\in A_0$. Since $eh=he=h$,
\begin{align*}
  \langle L_hU_kv,U_jz\rangle
    &=t^{-1}\tau(z^*w^jhw^{-k}v)\\
    &=\left\langle
        t^{-1}E_{A_0}(w^jhw^{-k})v,z\right\rangle.
\end{align*}
Density then gives the operator identity on $B$.

The reduced-word decomposition shows that on $W^*(D,s)$ the expectation
$E_{A_0}$ equals the expectation onto $W^*(s)=\C1+\C s$: every reduced
word containing a centered $D$ letter is orthogonal to $L^2(A_0)$.
Because $W^*(s)\subset\mathcal C$, this latter expectation factors
through $E_{\mathcal C}$. By \eqref{eq:free-compression} and
bimodularity,
\[
  E_{\mathcal C}(w^jhw^{-k})=u^2w^jb_fw^{-k}.
\]
Using $dw^{-k}=w^kd$ and $d=ws$, compute
\[
  w^jpw^{-k}
     =\frac{w^{j-k}+w^{j+k}d}{2}
     =\frac{w^{j-k}+w^{j+k+1}s}{2}.
\]
The group trace implies
\[
  E_{W^*(s)}(f(w)w^l)=\gamma_l1,\qquad
  E_{W^*(s)}(f(w)w^ls)=\gamma_ls.
\]
For instance this follows on Laurent polynomials because only the
identity power contributes, and then by $L^2$ approximation. Hence
\[
  t^{-1}E_{A_0}(w^jhw^{-k})
     =\frac{u^2}{2t}
          (\gamma_{j-k}1+\gamma_{j+k+1}s)
     =u(\gamma_{j-k}1+\gamma_{j+k+1}s),
\]
which proves \eqref{eq:hankel-compression}, including its normalization
and its Hankel index.

The matrices
\[
  T_f=(\gamma_{j-k})_{j,k\ge0},\qquad
  H_f=(\gamma_{j+k+1})_{j,k\ge0}
\]
define bounded operators on $\ell^2(\N_0)$ of norm at most
$\|f\|_\infty$. To see this directly, decompose
\[
  L^2(\T,m)=H_-\oplus H_+,\qquad
  H_-=\overline{\operatorname{span}}\{z^{-j}:j\ge0\},\quad
  H_+=\overline{\operatorname{span}}\{z^{k+1}:k\ge0\}.
\]
In the indicated orthonormal bases, $T_f$ and $H_f$ are respectively
the $H_-,H_-$ and $H_-,H_+$ blocks of multiplication by $f$.
Equation~\eqref{eq:hankel-compression}, first on finite coordinate
vectors and then by continuity, therefore gives
\begin{equation}
  \iota^*L_h\iota
      =u(T_f\otimes I_B+H_f\otimes L_s|_B).
  \label{eq:toeplitz-hankel-form}
\end{equation}
As $J=I\otimes B_0$, it follows that
\begin{equation}
  \big\|[J,\iota^*L_h\iota]\big\|
       =u\|H_f\|\,\|[B_0,L_s|_B]\|.
  \label{eq:hankel-commutator}
\end{equation}
Here we used the operator norm identity
$\|A\otimes B\|=\|A\|\|B\|$ for Hilbert-space tensor products.

\begin{proof}[Completion of the proof of Theorem~\ref{thm:free-corner}]
Put $\theta=u/4=1/(2N)$, so $0<\theta\le1/4$, and use angular coordinates
$z=e^{i\alpha}$, $-\pi<\alpha\le\pi$, to define
\[
  f(e^{i\alpha})=\mathbf1_{[-\theta,\theta]}(\alpha)-\frac\theta\pi.
\]
This function is real, even, and of mean zero. The trace formula in
\eqref{eq:dihedral-corner} gives
\[
  \|b_f\|\le1,\qquad
  \|b_f\|_{2,\tau_p}^2
       =\frac\theta\pi\left(1-\frac\theta\pi\right)\le u.
\]
Consequently \eqref{eq:free-compression} gives
\begin{equation}
  \|h\|=\|eb_fe\|\le3u.
  \label{eq:narrow-arc-norm}
\end{equation}
For every $l\ge1$ the Fourier coefficient is
\[
  \gamma_l=\frac{\sin(l\theta)}{\pi l}.
\]
If $0\le j,k<N$, then $1\le j+k+1\le2N-1$ and
$(j+k+1)\theta<1<\pi/2$. The inequality
$\sin x\ge2x/\pi$ on $[0,\pi/2]$ therefore gives
\[
  (H_f)_{jk}\ge\frac{2\theta}{\pi^2}.
\]
Testing the leading $N$ by $N$ block on its constant unit vector,
and using $\theta N=1/2$, proves the uniform bound
\begin{equation}
  \|H_f\|\ge\frac{2\theta N}{\pi^2}\ge\frac1{2\pi^2}.
  \label{eq:narrow-arc-hankel}
\end{equation}
Combine \eqref{eq:hankel-commutator}, \eqref{eq:compression-bound},
\eqref{eq:narrow-arc-norm}, and \eqref{eq:narrow-arc-hankel} to obtain
\[
  \frac{u}{2\pi^2}\|[B_0,L_s|_B]\|
       \le3u(1+4c_0)\epsilon.
\]
Cancelling $u>0$ yields
\[
  \|[B_0,L_s|_B]\|
      \le6\pi^2(1+4c_0)\epsilon=c_1\epsilon.
\]
The level-zero case of \eqref{eq:repeated-letter} identifies $G|_B$
with $B_0$, so this is exactly the required compression estimate for
the symmetrized operator. Lemma~\ref{lem:free-symmetrization} preserves
the compression of $[G,L_s]$ between vectors of $B$. The same estimate
therefore holds for the original operator in
Theorem~\ref{thm:free-corner}, completing its proof.
\end{proof}
==== END Kadisons-similarity-theorem-through-uniform-derivation-estimates//hankel.tex ====

==== BEGIN Kadisons-similarity-theorem-through-uniform-derivation-estimates//introduction.tex ====
\section{Introduction}
\label{sec:introduction}

Kadison's similarity problem asks whether the adjoint operation of a
\(C^*\)-algebra is respected, after a bounded change of Hilbert space
coordinates, by every bounded representation of its underlying algebra.
Kadison formulated the problem in 1955 in his study of the
orthogonalization of operator representations \citep{Kadison1955}.
Its force lies in the absence of any assumed relation between the
algebraic representation and the involution: multiplicativity and
boundedness alone must produce one equivalent Hilbert space inner
product for which all represented adjoints are correct.  This paper
resolves the problem positively in its unrestricted bounded unital form.

\begin{theorem}[Similarity]\label{thm:similarity}
Let \(A\) be any unital complex \(C^*\)-algebra, let \(H\) be any complex
Hilbert space, and let \(\pi:A\to\mathcal B(H)\) be a bounded,
complex-linear, unital algebra homomorphism.  There exists an operator
\(S\in\mathcal B(H)\), invertible with bounded inverse, such that
\[
  \rho(a)=S\pi(a)S^{-1}\qquad(a\in A)
\]
is a \(*\)-homomorphism.  In particular,
\(\rho(a^*)=\rho(a)^*\) for every \(a\in A\).
\end{theorem}

The theorem imposes no separability, nuclearity, finite generation,
faithfulness, or normality condition.  The similarity may depend on the
whole homomorphism.  The result concerns the existence of this single
bounded invertible operator, and does not require an a priori bound for
its condition number.

\subsection{Historical context}

The nuclear case was established by Bunce and Christensen
\citep{Bunce1981,Christensen1981}; see also the precise quantitative
comparison in \citet[Theorem~1]{Pisier2006Nuclear}.
Haagerup's theorem that every nuclear $C^*$-algebra is amenable
\citep[Theorem~3.1]{Haagerup1983Amenable} connects this case with the
cohomology of operator algebras. Complete boundedness provided the
principal analytic framework for the general similarity problem.
Haagerup solved the cyclic representation case and identified
complete boundedness as the criterion for similarity to a
\(*\)-representation \citep{Haagerup1983}. He also proved the similarity
property for $C^*$-algebras without tracial states
\citep[Proposition~1.8]{Haagerup1983}, and Christensen established it
for type~\(\mathrm{II}_1\) factors with property~\(\Gamma\)
\citep{Christensen1986}; see also
\citet[Section~1.2, Theorem~1.3]{Ozawa2006}. These cases explain the role
of the finite-factor argument below, where the estimate must be uniform
in the factor and its representation multiplicity.

Christensen established the
corresponding criterion for derivations: a derivation into the ambient
bounded operators is inner precisely when it is completely bounded
\citep{Christensen1977}.  Paulsen's similarity theorem extends the
homomorphism criterion to unital operator algebras and gives the optimal
condition number \citep[p.~226]{Paulsen1984}.  Kirchberg then proved that,
for each unital \(C^*\)-algebra, the similarity property is equivalent to
innerness of derivations in every faithful \(*\)-representation
\citep{Kirchberg1996}; we use the explicit per-algebra formulation in
\citet[Theorem~1.1]{Ozawa2006}.  Ringrose's automatic continuity theorem
allows this derivation property to be stated using bounded derivations
\citep{Ringrose1972}.

A von Neumann algebra is \emph{hyperreflexive} if the distance from
any operator to the algebra is bounded by a fixed multiple of the
supremum of its off-diagonal compression norms over the algebra's
invariant subspaces. More recently,
\citet[Theorem~2.4]{EleftherakisPaulsen2026} characterized a positive
answer to the similarity problem by preservation of hyperreflexivity
when adjoining a projection: whenever $M\subset\mathcal B(H)$ is
a hyperreflexive von Neumann algebra and $q\in\mathcal B(H)$ is a
projection, $W^*(M,q)$ is hyperreflexive. Their result is an
equivalence with this extension property and with the assertion that
every von Neumann algebra is hyperreflexive. Corollary~\ref{cor:hyperreflexivity}
below makes the latter consequence quantitative with one universal constant.

Pisier organized quantitative similarity estimates through the
similarity degree and its interpretation as the length of matrix
factorizations \citep{Pisier1999Degree,Pisier2001Remarks}.
His characterization of nuclearity by a similarity exponent below
three distinguishes the strength of the nuclear case
\citep[Theorem~1]{Pisier2006Nuclear}. These results concern quantitative
bounds for homomorphisms. The proof below is organized around a
uniform bound for derivations, and concludes with existence of a
similarity as stated in Theorem~\ref{thm:similarity}.

The companion result on discrete groups identifies unitarizability with
amenability~\cite[Theorem~1.1]{OpenAIUnitarizability2026}.
Unlike the homomorphisms in Theorem~\ref{thm:similarity}, a uniformly bounded
group representation need not have a bounded extension to the full or
reduced group $C^*$-algebra; the automatic extension is to $\ell^1(G)$.

\subsection{The uniform estimate and proof strategy}

An inner derivation is a map $a\mapsto[Y,a]$. Complete boundedness
means uniform boundedness of all entrywise finite matrix amplifications.
Our main estimate controls those amplifications of an inner derivation
by its norm at the scalar level.  For a represented unital
von Neumann algebra \(P\subset\mathcal B(K)\) and
\(Y\in\mathcal B(K)\), put
\[
  g_P(Y)=\sup_{a\in P,\,\|a\|\le1}\|Ya-aY\|,
\]
and write \(Y^{(h)}\) for the diagonal amplification of \(Y\) on
\(K^h\).

\begin{theorem}[Uniform commutator estimate]\label{thm:universal}
There is an absolute constant \(C<\infty\) such that, for every complex
Hilbert space \(K\), every unital von Neumann algebra
\(P\subset\mathcal B(K)\), every \(Y\in\mathcal B(K)\), every integer
\(h\ge1\), and every \(X\in M_h(P)\),
\begin{equation}\label{eq:universal}
  \|[Y^{(h)},X]\|\le C\,g_P(Y)\,\|X\|.
\end{equation}
Here \(M_h(P)\) acts naturally on \(K^h\), and \([T,U]=TU-UT\).
\end{theorem}

\begin{corollary}[Universal hyperreflexivity]\label{cor:hyperreflexivity}
Let \(H\) be any complex Hilbert space and let
\(M\subset\mathcal B(H)\) be a unital von Neumann algebra. For
\(T\in\mathcal B(H)\), set
\[
 \alpha_M(T)=\sup\{\|(1-e)Te\|:e\in M',\ e=e^*=e^2\}.
\]
Then, with the absolute constant \(C\) from Theorem~\ref{thm:universal},
\begin{equation}\label{eq:universal-hyperreflexivity}
 \dist(T,M)\le 2C\,\alpha_M(T)\qquad(T\in\mathcal B(H)).
\end{equation}
Thus every such \(M\) is hyperreflexive with constant at most \(2C\),
independently of the algebra, its representation, and the Hilbert space.
No optimality of this constant is asserted.
\end{corollary}

Applying the commutator estimate to \(M'\) gives this distance bound
through Arveson's commutant-distance formula. The proof is given in
Section~\ref{sec:hyperreflexivity}.


The difficult part is obtaining a constant independent of matrix size
for finite factors. Section~\ref{sec:preliminaries} first combines
Dickson's row estimate \citep[Lemma~2.9]{Dickson2014} with its column
counterpart to implement rectangular derivations with cyclic domain.
Here rectangular derivations take values in operators between two
representation spaces. The implementation norm is controlled by the derivation norm with an
absolute constant.

Sections~\ref{sec:tensors}--\ref{sec:hankel} use this cyclic-domain
estimate in a tracial free product \(\mathcal L=D*A_0\), where \(D\)
is a \({\rm II}_1\) factor generated by increasing full matrix
algebras and \(A_0\) is a finite von Neumann algebra with a faithful
normal tracial state. The input is an operator \(G\) on
\(L^2(\mathcal L)\) commuting with left multiplication by \(D\),
whose commutators with elements of a corner \(e\mathcal L e\)
have norm at most \(\epsilon\) times the element norm. Here
\(e\in D\) is a projection of trace \(1/N\), for an integer
\(N\ge2\). The output bounds the compression to \(L^2(A_0)\) of its
commutator with any trace-zero self-adjoint unitary in \(A_0\), by
an absolute multiple of \(\epsilon\) independent of the corner
trace. This is Theorem~\ref{thm:free-corner}.

The reduced-word model belongs to the free-product framework developed
by Ching and Voiculescu \citep{Ching1973,Voiculescu1985}.
Its role here is to turn the corner hypothesis into a statement about
operators on Hilbert tensor products.

In Section~\ref{sec:tensors}, averaging over the finite matrix unitary
groups reveals the structure of the operator on reduced words.
Finite-dimensional Schur--Weyl
duality supplies invariant cycle-trace coefficients; the endpoint
\(L^1\) bound and boundedness on Hilbert tensor products eliminate all
contractions except permutations between equally many source and test
letters. At each word level, repeated-letter tensors consequently carry an
action on their remaining factors that is independent of the repeated
$D$-letter. Section~\ref{sec:propagation} approximates
\(G\), on each sufficiently small right support, by a bounded right
multiplier. This propagates the repeated-letter action through
arbitrarily long words with a uniform error.
Section~\ref{sec:hankel} compresses a multiplication operator built
from two free trace-zero self-adjoint unitaries. The resulting Hankel
matrix has norm bounded below as the corner trace tends to zero.
The trace factors in the upper and lower bounds cancel, completing
the free-product corner estimate.

Section~\ref{sec:finite-factors} uses Popa's relative independence
theorem \citep[Theorem~0.1(a)]{Popa2014} to place this free product
inside the tracial ultrapower of an arbitrary separable-predual
\({\rm II}_1\) factor after matrix amplification. The
irreducible-inclusion case already appears in Popa's free-independent
sequence construction
\citep[Theorem~2.1 and proof of Corollary~2.2]{Popa1995}.
The
operator under study is passed through a Hilbert space ultrapower and
compressed to the tracial \(L^2\)-space, preserving the necessary
commutator coefficients. Section~\ref{sec:global} extends the estimate
to arbitrary normal factor representations, then uses central
decomposition and separable reducing subspaces to obtain
Theorem~\ref{thm:universal}. The resulting constant is numerical and
independent of matrix size, representation, and Hilbert space.
Finally, using the cyclic-domain estimate again, finite collections
of cyclic columns of a \(*\)-representation implement a given
derivation locally.  Their scalar commutator norms
are uniformly bounded, so \eqref{eq:universal} gives complete
boundedness of the full derivation.  Paulsen's theorem implements it,
and Kirchberg's equivalence proves Theorem~\ref{thm:similarity}.
==== END Kadisons-similarity-theorem-through-uniform-derivation-estimates//introduction.tex ====

==== BEGIN Kadisons-similarity-theorem-through-uniform-derivation-estimates//main.tex ====
\documentclass[11pt]{article}
\input{preamble}
\title{Kadison's similarity theorem through\\uniform derivation estimates}
\author{OpenAI}
\date{September 23, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove that every bounded complex-linear unital algebra homomorphism from a unital
complex $C^*$-algebra into the bounded operators on an arbitrary Hilbert
space is similar to a $*$-homomorphism. This resolves Kadison's similarity
conjecture positively. We also obtain one universal hyperreflexivity constant
for all unital von Neumann algebras on arbitrary complex Hilbert spaces.
\end{abstract}
\input{sections/introduction}
\input{sections/preliminaries}
\input{sections/tensors}
\input{sections/propagation}
\input{sections/hankel}
\input{sections/finite-factors}
\input{sections/global}
\begingroup
\raggedright
\setlength{\bibsep}{5pt}
\bibliographystyle{abbrvnat}
\bibliography{references}
\endgroup
\end{document}
==== END Kadisons-similarity-theorem-through-uniform-derivation-estimates//main.tex ====

==== BEGIN Kadisons-similarity-theorem-through-uniform-derivation-estimates//preamble.tex ====
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==== END Kadisons-similarity-theorem-through-uniform-derivation-estimates//preamble.tex ====

==== BEGIN Kadisons-similarity-theorem-through-uniform-derivation-estimates//preliminaries.tex ====
\section{Rows, cyclic modules, and derivations}
\label{sec:preliminaries}

This section supplies two inputs to the proof: a uniform implementer
for a rectangular derivation with cyclic domain, and the per-algebra
derivation criterion that will conclude similarity. The cyclic estimate
will first be used to extend control from a small free-product corner
to an entire right-supported subspace.

All Hilbert spaces and algebras are complex, and inner products are linear
in the first variable. All algebra homomorphisms are complex-linear, and
all \(*\)-representations are unital unless stated otherwise. No
separability or faithfulness is implicit. We write \(E_1\) for the closed
unit ball of a normed space \(E\). For a unital self-adjoint operator
algebra \(P\subset\mathcal B(K)\) and \(Y\in\mathcal B(K)\), set
\begin{equation}
 g_P(Y)=\sup_{a\in P_1}\|Ya-aY\|.
 \label{eq:commutator-seminorm}
\end{equation}
The same notation will be used when \(P\) is a specified represented
copy of an abstract algebra.

For a linear map \(u:A\to\mathcal B(K,F)\), its amplification
\(u_{p,q}\) acts entrywise on \(M_{p,q}(A)\), with values in
\(\mathcal B(K^q,F^p)\). Rectangular matrix norms on \(A\) are the usual
\(C^*\)-matrix norms. We abbreviate \(u_{h,h}\) to \(u_h\), and set
\[
 \|u\|_{\mathrm{cb}}=\sup_{h\ge1}\|u_h\|,\qquad
 \|u\|_{\mathrm{row}}=\sup_{h\ge1}\|u_{1,h}\|,\qquad
 \|u\|_{\mathrm{col}}=\sup_{h\ge1}\|u_{h,1}\|.
\]
A map is completely bounded when its cb norm is finite. For an operator
\(Y\), the notation \(Y^{(h)}\) means the diagonal operator on \(h\)
copies of its Hilbert space.

If \((P,\tau)\) is a finite von Neumann algebra with a faithful normal
tracial state, \(L^2(P)=L^2(P,\tau)\) is the completion of \(P\) for
\(\langle x,y\rangle=\tau(y^*x)\). We write \(\|\cdot\|_2\) for this
norm, and \(\|x\|_1=\tau(|x|)\) for \(x\in P\). Left and right
multiplication are denoted by \(L_a\) and \(R_a\), with
\(R_a\xi=\xi a\); thus \(R_aR_b=R_{ba}\). The tracial commutation
theorem identifies \(L(P)'\) with \(\{R_a:a\in P\}\). The support
notation for an \(L^2\)-vector is understood through its affiliated
operator: if \(\xi=v|\xi|\) is its polar decomposition, its left and
right support projections are \(vv^*\) and \(v^*v\), respectively.
They are equivalent in \(P\), and in particular have the same trace.
For a projection \(r\in P\), the condition that \(R_b\) vanish on
\(L^2(P)(1-r)\) is exactly \(b=rb\).
We use these standard facts, and standard-form theory for von Neumann
algebras, without separability restrictions
\citep{Takesaki1979,Haagerup1975}.

\subsection{Uniform row and column estimates}

Dickson's argument derives the row estimate from Haagerup's little
Grothendieck inequality and credits Christensen with the estimate in
this form \citep[Lemmas~2.8--2.9]{Dickson2014}; this is part of the
factorization approach to similarity developed by Pisier
\citep{Pisier1999Degree,Pisier2001Remarks}.
The estimate states that every bounded unital homomorphism
\(\varphi:A\to\mathcal B(K)\), for any unital \(C^*\)-algebra \(A\),
satisfies
\begin{equation}
 \|\varphi\|_{\mathrm{row}}
       \le\sqrt2\,\|\varphi\|^2
 \qquad\text{\citep[Lemma~2.9, Equation~(19)]{Dickson2014}}.
 \label{eq:dickson-row}
\end{equation}
The map \(\varphi^\sharp(a)=\varphi(a^*)^*\) is complex-linear and has
the same norm as \(\varphi\). It is again a homomorphism, since
\[
 \varphi^\sharp(ab)
 =\bigl(\varphi(b^*)\varphi(a^*)\bigr)^*
 =\varphi^\sharp(a)\varphi^\sharp(b).
\]
Apply \eqref{eq:dickson-row} to \(\varphi^\sharp\) on the row
\([a_1^*,\ldots,a_h^*]\), and take the rectangular adjoint. This gives
the same bound for \(\|\varphi\|_{\mathrm{col}}\). Consequently we fix
the absolute constant
\begin{equation}
 c_{\rm row}=4\sqrt2,
 \label{eq:row-constant}
\end{equation}
which bounds both norms whenever \(\|\varphi\|\le2\).

Let \(A\) be a unital \(C^*\)-algebra. A \emph{rectangular derivation}
for \(*\)-representations
\(\sigma:A\to\mathcal B(K)\) and
\(\lambda:A\to\mathcal B(F)\) is a linear map
\(\Delta:A\to\mathcal B(K,F)\) satisfying
\begin{equation}
 \Delta(ab)=\Delta(a)\sigma(b)+\lambda(a)\Delta(b)
 \qquad(a,b\in A).
 \label{eq:rectangular-derivation}
\end{equation}
The derivation identity and unitality imply \(\Delta(1)=0\).

\begin{lemma}[Row and column bounds]
\label{lem:row-column}
Every bounded rectangular derivation satisfies
\begin{equation}
 \max\{\|\Delta\|_{\mathrm{row}},\|\Delta\|_{\mathrm{col}}\}
       \le c_{\rm row}\|\Delta\|.
 \label{eq:rectangular-row-column}
\end{equation}
\end{lemma}

\begin{proof}
The assertion is immediate when \(\Delta=0\). Otherwise put
\(d=\|\Delta\|>0\), and consider the upper triangular map
\begin{equation}
 \Phi(a)=
 \begin{pmatrix}
  \lambda(a)&d^{-1}\Delta(a)\\
  0&\sigma(a)
 \end{pmatrix}
 \in\mathcal B(F\oplus K).
 \label{eq:triangular-homomorphism}
\end{equation}
It is unital and multiplicative by \eqref{eq:rectangular-derivation}.
Its diagonal map is contractive, and its off-diagonal map has norm one;
hence \(\|\Phi\|\le2\). The off-diagonal rectangular compression of
\(\Phi_{p,q}\), from \(K^q\) to \(F^p\), is
\(d^{-1}\Delta_{p,q}\). The row and column estimates for \(\Phi\)
give \eqref{eq:rectangular-row-column}.
\end{proof}

\subsection{Completely bounded implementation}

We use the quantitative completely bounded similarity theorem of
Paulsen: if a unital homomorphism \(\varphi:A\to\mathcal B(H)\) is
completely bounded, there is an invertible \(S\in\mathcal B(H)\) such
that \(S\varphi(\cdot)S^{-1}\) is completely contractive and
\(\|S\|\|S^{-1}\|\le\|\varphi\|_{\mathrm{cb}}\)
\citep[p.~226]{Paulsen1984}. For a unital \(C^*\)-algebra the
resulting homomorphism is a \(*\)-homomorphism: a unitary and its inverse
are sent to contractions, so the image is a unitary; unitaries linearly
span the algebra. For the later cyclic-domain estimate, set
\begin{equation}
 c_0=(1+4c_{\rm row})^2.
 \label{eq:cyclic-constant}
\end{equation}

\begin{lemma}[Completely bounded rectangular derivations]
\label{lem:cb-implementation}
Let \(\sigma,\lambda,\Delta\) satisfy
\eqref{eq:rectangular-derivation}. If \(\Delta\) is completely bounded,
then some \(V\in\mathcal B(K,F)\) satisfies
\begin{equation}
 \Delta(a)=V\sigma(a)-\lambda(a)V\qquad(a\in A).
 \label{eq:rectangular-implementation}
\end{equation}
If \(d=\|\Delta\|>0\), one may choose
\[
 \|V\|\le d\bigl(1+\|\Delta\|_{\mathrm{cb}}/d\bigr)^2.
\]
\end{lemma}

\begin{proof}
We use the triangular-metric argument recalled in
\citet[Lemma~1.2]{Ozawa2006}, allowing different domain and codomain
representations and keeping track of the implementer norm.
For \(\Delta=0\), take \(V=0\). Otherwise the map \(\Phi\) in
\eqref{eq:triangular-homomorphism} is completely bounded, with
\[
 \|\Phi\|_{\mathrm{cb}}
 \le 1+\|\Delta\|_{\mathrm{cb}}/d=:G.
\]
Choose the similarity \(S\) supplied by Paulsen's theorem, and put
\(D=S^*S\). The fact that \(S\Phi(\cdot)S^{-1}\) preserves adjoints
is equivalent to
\begin{equation}
 D\Phi(a)=\Phi(a^*)^*D\qquad(a\in A).
 \label{eq:intertwining-metric}
\end{equation}
Write \(D=(D_{ij})_{i,j=1}^2\) relative to \(F\oplus K\). Its first
row in \eqref{eq:intertwining-metric} gives
\begin{align}
 D_{11}\lambda(a)&=\lambda(a)D_{11},\notag\\
 d^{-1}D_{11}\Delta(a)+D_{12}\sigma(a)&=\lambda(a)D_{12}.
 \label{eq:metric-blocks}
\end{align}
Since \(D\ge\|D^{-1}\|^{-1}I\), its compression \(D_{11}\) is
positive invertible on \(F\), with
\(\|D_{11}^{-1}\|\le\|D^{-1}\|\). Its inverse also commutes with
\(\lambda(A)\). Hence
\[
 V=-dD_{11}^{-1}D_{12}
\]
satisfies \eqref{eq:rectangular-implementation}, by
\eqref{eq:metric-blocks}, and
\[
 \|V\|
 \le d\,\|D^{-1}\|\|D\|
 =d\,\bigl(\|S^{-1}\|\|S\|\bigr)^2
 \le dG^2.
\]
If either Hilbert space is zero, \(\Delta=0\), so it has already been
covered.
\end{proof}

\subsection{Cyclic domains}

The cyclic hypothesis needed below concerns only the domain
\(*\)-representation. The output representation may have arbitrary
multiplicity.

\begin{lemma}[Column factorization in a cyclic representation]
\label{lem:cyclic-factorization}
Let \(P\subset\mathcal B(K)\) be a unital von Neumann algebra with a
cyclic vector. For every integer \(h\ge1\) and every unit vector
\(\xi=(\xi_1,\ldots,\xi_h)\in K^h\), there are a unit vector
\(\zeta\in K\) and a contraction \(v=(v_i)\in M_{h,1}(P)\) such
that \(\xi_i=v_i\zeta\) for every \(i\).
\end{lemma}

\begin{proof}
Put \(N=P'\), and choose a cyclic vector \(\Omega\) for \(P\).
It separates \(N\): if \(x\in N\) and \(x\Omega=0\), then
\(xa\Omega=ax\Omega=0\) for every \(a\in P\), whence \(x=0\).
Let \(L=\overline{N\Omega}\subset K\). This subspace reduces \(N\),
and restriction \(r:N\to\mathcal B(L)\) is faithful and normal.
Indeed, an element in its kernel annihilates \(\Omega\); normality
follows from compression to a reducing subspace. Thus it is a normal
\(*\)-isomorphism onto the von Neumann algebra \(r(N)\), and its
inverse is normal. The vector \(\Omega\) is cyclic and separating for
\(r(N)\).

The standard-form theorem for a representation with a cyclic separating
vector realizes every normal positive functional on \(r(N)\) as a
vector functional, with its vector in the natural cone
\citep[Lemma~2.10(1), pp.~278--279]{Haagerup1975}.
Apply it, through the normal isomorphism \(r\),
to
\[
 \omega(x)=\sum_{i=1}^h\langle x\xi_i,\xi_i\rangle,
 \qquad x\in N.
\]
We obtain \(\zeta\in L\subset K\) such that
\(\langle x\zeta,\zeta\rangle=\omega(x)\) for every \(x\in N\).
In particular \(\|\zeta\|^2=\omega(1)=1\). The assignment
\[
 x\zeta\longmapsto(x\xi_1,\ldots,x\xi_h),\qquad x\in N,
\]
is isometric, since both squared norms equal \(\omega(x^*x)\).
It extends to an isometry from \(\overline{N\zeta}\) into \(K^h\).
Extend it by zero on \(\overline{N\zeta}^{\perp}\), obtaining a
contraction \(v:K\to K^h\). The subspace \(\overline{N\zeta}\)
reduces \(N\), and the isometry intertwines the \(N\)-actions.
Therefore \(vx=x^{(h)}v\) for every \(x\in N\). Each coordinate
\(v_i\) lies in \(N'=P\), and \(v\zeta=\xi\), as required.
All these constructions, including the standard-form assertion, are
valid on nonseparable Hilbert spaces.
\end{proof}

\begin{lemma}[Implementation on a cyclic domain]
\label{lem:cyclic-implementation}
Let \(A\) be a unital \(C^*\)-algebra, let
\(\sigma:A\to\mathcal B(K)\) and
\(\lambda:A\to\mathcal B(F)\) be \(*\)-representations, and suppose
that \(\sigma\) has a cyclic vector. Every bounded rectangular
derivation \(\Delta:A\to\mathcal B(K,F)\) is completely bounded,
with
\begin{equation}
 \|\Delta\|_{\mathrm{cb}}\le4c_{\rm row}\|\Delta\|.
 \label{eq:cyclic-cb}
\end{equation}
Moreover there is \(V\in\mathcal B(K,F)\) such that
\begin{equation}
 \Delta(a)=V\sigma(a)-\lambda(a)V\quad(a\in A),\qquad
 \|V\|\le c_0\|\Delta\|.
 \label{eq:cyclic-implementation}
\end{equation}
\end{lemma}

\begin{proof}
Zero spaces and the zero derivation are immediate, so assume they do
not occur. Put \(P=\sigma(A)''\). Fix \(h\ge1\),
\(X\in M_h(A)_1\), and a unit vector \(\xi\in K^h\).
Lemma~\ref{lem:cyclic-factorization} gives a unit vector \(\zeta\in K\)
and a contraction \(v\in M_{h,1}(P)\) with \(v\zeta=\xi\).

Matrix Kaplansky density gives a net of column contractions
\(z_\alpha\in M_{h,1}(\sigma(A))\) converging strongly to \(v\).
To see the rectangular assertion directly, regard \(v\) as the first
column of a matrix in \(M_h(P)\), with all other columns zero.
Approximate that matrix strongly by contractions in
\(M_h(\sigma(A))\), and compress to its first input coordinate.
The \(*\)-image \(\sigma(A)\) is the \(C^*\)-quotient
\(A/\ker\sigma\), and the same quotient identification holds at each
square matrix level. Consequently each \(z_\alpha\) lifts to
\(w_\alpha\in M_{h,1}(A)\) with
\begin{equation}
 \|w_\alpha\|\le2,\qquad
 \sigma_{h,1}(w_\alpha)=z_\alpha,\qquad
 \sigma_{h,1}(w_\alpha)\zeta\longrightarrow\xi.
 \label{eq:cyclic-column-lifts}
\end{equation}
For example, first lift the square matrix whose first column is
\(z_\alpha\) and whose other columns vanish to a matrix of norm less
than \(2\), then compress its first column. Only fixed slack in the
quotient norm is being used.

Entrywise application of \eqref{eq:rectangular-derivation} gives
\[
 \Delta_h(X)\sigma_{h,1}(w_\alpha)
   =\Delta_{h,1}(Xw_\alpha)
       -\lambda_h(X)\Delta_{h,1}(w_\alpha).
\]
Since \(\lambda\) is completely contractive, Lemma~\ref{lem:row-column}
and \eqref{eq:cyclic-column-lifts} imply
\[
 \bigl\|\Delta_h(X)\sigma_{h,1}(w_\alpha)\zeta\bigr\|
 \le c_{\rm row}\|\Delta\|
       \bigl(\|Xw_\alpha\|+\|\lambda_h(X)\|\|w_\alpha\|\bigr)
 \le4c_{\rm row}\|\Delta\|.
\]
At the fixed finite level \(h\), \(\Delta_h(X)\) is already a bounded
operator. We may therefore take the limit in
\eqref{eq:cyclic-column-lifts}, obtaining the same bound on
\(\|\Delta_h(X)\xi\|\). Taking the supremum over \(\xi,X,h\)
proves \eqref{eq:cyclic-cb}. The quotient argument did not require
\(\Delta\) to vanish on \(\ker\sigma\); its identity was applied to
the chosen lifts in \(A\). Lemma~\ref{lem:cb-implementation} and
\eqref{eq:cyclic-constant} now give
\eqref{eq:cyclic-implementation}.
\end{proof}

\subsection{The derivation criterion for similarity}

For the last step of the proof we use Kirchberg's equivalence in its
per-algebra form. We state the quantifiers because the final
\(*\)-representation used to test derivations is independent of the
bounded homomorphism in the similarity problem.

\begin{theorem}[Kirchberg's similarity--derivation equivalence]
\label{thm:kirchberg}
For a unital \(C^*\)-algebra \(A\), the following properties are
equivalent.
\begin{enumerate}
 \item Every bounded unital homomorphism
 \(\pi:A\to\mathcal B(H)\), on any Hilbert space \(H\), is similar
 by a bounded invertible operator to a \(*\)-homomorphism.
 \item For every faithful unital \(*\)-representation
 \(\sigma:A\to\mathcal B(K)\), on any Hilbert space \(K\), every
 bounded linear map \(\Delta:A\to\mathcal B(K)\) satisfying
 \[
  \Delta(ab)=\Delta(a)\sigma(b)+\sigma(a)\Delta(b)
 \]
 has the form \(\Delta(a)=V\sigma(a)-\sigma(a)V\) for some
 \(V\in\mathcal B(K)\).
\end{enumerate}
\end{theorem}

This is the theorem of \citet{Kirchberg1996}; the exact formulation
with every faithful representation is stated in
\citet[Theorem~1.1]{Ozawa2006}. The usual derivation property is stated
without a continuity hypothesis. Ringrose's automatic continuity
theorem for derivations from \(C^*\)-algebras into Banach bimodules
\citep[Theorem~2, p.~434]{Ringrose1972} makes it equivalent to the
bounded version above.
The triangular metric computation in
Lemma~\ref{lem:cb-implementation} uses only Paulsen's established
completely bounded theorem; the unrestricted similarity conclusion
will follow from Theorem~\ref{thm:kirchberg} after all bounded
derivations have been treated.
==== END Kadisons-similarity-theorem-through-uniform-derivation-estimates//preliminaries.tex ====

==== BEGIN Kadisons-similarity-theorem-through-uniform-derivation-estimates//propagation.tex ====
\section{Propagation from small corners}
\label{sec:propagation}

Continue with the notation and hypotheses of Theorem~\ref{thm:free-corner},
and set $H=L^2(\cL)$. After Lemma~\ref{lem:free-symmetrization}, we may use
the symmetrized operator $G$ and the operators $B_k\in\mathcal B(J_k)$
from Lemma~\ref{lem:word-structure}. Our goal is to compare the action of
$B_k$ with that of $B_0$ along repeated letters, with an error independent
of $k$. We first extract a consequence of the small-corner hypothesis
that needs only the exact commutation with $L_D$.

\begin{lemma}[Full columns from a small corner]
\label{lem:full-column}
For every projection $r\in\cL$ with $\tau(r)\le t$, there is $b\in\cL$
with $b=rb$ such that
\begin{equation}
  \big\|(G-R_b)|_{Hr}\big\|\le c_0\epsilon.
  \label{eq:full-column}
\end{equation}
The same assertion holds for $G^*$.
\end{lemma}

\begin{proof}
The case $r=0$ is immediate. Fix $r\ne0$ and a projection $e\in D$ with
$\tau(e)=t$. The factoriality established in Lemma~\ref{lem:free-factor}
and comparison of projections give $r'\le e$ and a partial isometry
$v\in\cL$ such that $v^*v=r$ and $vv^*=r'$. Right multiplication by $v^*$
is a unitary left $e\cL e$-module map
\[
  eHr\longrightarrow eHr',\qquad \xi\longmapsto\xi v^*.
\]
Indeed its inverse is right multiplication by $v$, and
\[
  \|\xi v^*\|_2^2
    =\tau(v\xi^*\xi v^*)
    =\tau(\xi^*\xi r)=\|\xi\|_2^2.
\]
These equalities also hold for $L^2$ vectors by continuity. The vector $r'$
is cyclic for the left action of $e\cL e$ on $eHr'$, since
$(e\cL e)r'=e\cL r'$ is dense in that Hilbert space. Thus $eHr$ is a
cyclic domain representation of the unital algebra $e\cL e$.

Since $G$ commutes with $L_e$, its restriction defines a bounded map
$G_e:eHr\to eH$. On these domain and codomain representations the map
\[
  \Delta(a)=G_eL_a|_{eHr}-L_aG_e,\qquad a\in e\cL e,
\]
is a rectangular derivation of norm at most $\epsilon$, by
\eqref{eq:free-corner-hypothesis}. Lemma~\ref{lem:cyclic-implementation}
therefore supplies an exact intertwiner $T:eHr\to eH$ with
\[
  TL_a|_{eHr}=L_aT\quad(a\in e\cL e),\qquad
  \|G_e-T\|\le c_0\epsilon.
\]

The value $t=1/N$ allows an extension that preserves this norm. Choose
$v_1,\ldots,v_N\in D$ with
\[
  v_i^*v_i=e,\qquad v_iv_i^*\perp v_jv_j^*\ (i\ne j),\qquad
  \sum_{i=1}^N v_iv_i^*=1.
\]
The maps $(\xi_i)\mapsto\sum_i v_i\xi_i$ identify $(eHr)^N$ with $Hr$
and $(eH)^N$ with $H$ unitarily. Define
\[
  \widetilde T\left(\sum_{i=1}^N v_i\xi_i\right)
      =\sum_{i=1}^N v_iT\xi_i,\qquad \xi_i\in eHr.
\]
In these coordinates $\widetilde T$ is $T^{(N)}$, and $G|_{Hr}$ is
$G_e^{(N)}$, because $G$ commutes with each $L_{v_i}$. Consequently
\[
  \|G|_{Hr}-\widetilde T\|=\|G_e-T\|\le c_0\epsilon.
\]
For $a\in\cL$, its left action has matrix entries $L_{v_i^*av_j}$ in
both coordinate systems. Every $v_i^*av_j$ lies in $e\cL e$, so the
intertwining identity for $T$ implies
$\widetilde T L_a|_{Hr}=L_a\widetilde T$.

Extend $\widetilde T$ by zero on $H(1-r)$. Since $R_r$ commutes with all
left multiplications, this extension belongs to $L(\cL)'$ and hence is
$R_b$ for a bounded $b\in\cL$, by the tracial commutation theorem. Its
vanishing on $H(1-r)$ says $R_b=R_bR_r$, which, with the convention
$R_b\xi=\xi b$, gives $b=rb$. This proves \eqref{eq:full-column}.
Finally $G^*$ commutes with $L_D$ and
$[G^*,L_a]=-[G,L_{a^*}]^*$, so all the same hypotheses and estimates
hold for $G^*$.
\end{proof}

\begin{lemma}[Propagation along repeated letters]
\label{lem:propagation}
For every $k\ge1$ and $v\in B=L^2(A_0)$,
\begin{equation}
  \big\|B_k(s^{\otimes k}\otimes v)
       -s^{\otimes k}\otimes B_0v\big\|
       \le 2c_0\epsilon\|v\|_2.
  \label{eq:propagation}
\end{equation}
This also holds with $B_k^*$ and $B_0^*$ in place of $B_k$ and $B_0$.
\end{lemma}

\begin{proof}
Fix $k\ge1$ and first take $v\in A_0$; the zero vector is harmless.
Choose a nonzero projection $q\in D$ with $a=\tau(q)\le t$. The left
support of $qv$ is dominated by $q$. Its right support $r$ is equivalent
to its left support by polar decomposition, so $\tau(r)\le a\le t$.
Choose $b=rb$ as in Lemma~\ref{lem:full-column} for this $r$, and put
\[
  y=qvb-qB_0v.
\]
The level-zero identity in \eqref{eq:repeated-letter} gives
$G(qv)=qB_0v$, and the tensor decomposition gives
$\|qv\|_2=\sqrt a\,\|v\|_2$. It follows that
\begin{equation}
  qy=y,\qquad \|y\|_2\le c_0\epsilon\sqrt a\,\|v\|_2.
  \label{eq:propagation-error-vector}
\end{equation}
The vector $B_0v$ need only belong to $L^2(A_0)$; every expression here
uses bounded left multiplication on this vector.

Let $z$ be a Haar unitary of $qDq$ for its normalized trace, and set
$x_n=a^{-1/2}z^n$ for $n\ge1$. Then
\[
  \tau(x_n)=0,\quad \|x_n\|_2=1,\quad
  x_n^*x_n=x_nx_n^*=a^{-1}q,
  \qquad x_n,x_n^*\rightharpoonup0\ \text{in }L^2(D).
\]
The weak convergence follows, for example, from orthonormality of the
positive and negative powers. The reduced words
$\xi_n=(sx_n)^kv$ have norm $\|v\|_2$ and right support at most $r$,
because their last $x_n$ satisfies $x_nq=x_n$. With
$A_n=(sx_n)^{k-1}s$, we have
\[
  \xi_nb-(sx_n)^kB_0v=A_nx_ny.
\]
Thus \eqref{eq:full-column} yields
\begin{equation}
  \|G\xi_n-(sx_n)^kB_0v\|_2
    \le c_0\epsilon\|v\|_2+\|A_nx_ny\|_2.
  \label{eq:propagation-prefix}
\end{equation}

To estimate the second term, write \eqref{eq:word-decomposition} as
$H=E\otimes F$, where
\[
  F=\bigoplus_{l\ge0}(E^\circ)^{\otimes l}\otimes J_l,
\]
and let $P_{\mathrm{sc}}$ be projection onto $\C1\otimes F$. This projection keeps every word level
and selects the scalar component of its first $D$ factor. On this decomposition
left multiplication by $x_n$ acts on $E$. With inner products linear in
the first variable, define $\ell_n(\eta)=\langle\eta,x_n^*\rangle$.
Then
\[
  P_{\mathrm{sc}}(x_ny)=1\otimes (\ell_n\otimes I_F)y,
  \qquad \|\ell_n\|=1.
\]
For each elementary tensor this expression tends to zero in norm, since
$x_n^*\rightharpoonup0$. Approximating the fixed vector $y$ by finite
tensor sums, and using the common bound $\|\ell_n\|=1$, proves
\begin{equation}
  \|P_{\mathrm{sc}}(x_ny)\|_2\longrightarrow0.
  \label{eq:vanishing-tensor-coefficient}
\end{equation}

On $(1-P_{\mathrm{sc}})H$, multiplication by $A_n$ prepends a reduced centered word
ending in $s$ to a leading centered $D$ letter. Each prefix letter has
$L^2$ norm one, so this restriction is an isometry. On $P_{\mathrm{sc}}H$ we use
the bound $\|L_{A_n}\|\le a^{-(k-1)/2}$. Also $qy=y$ gives
\[
  \|x_ny\|_2^2
     =\langle L_{x_n^*x_n}y,y\rangle=a^{-1}\|y\|_2^2.
\]
Consequently
\[
  \|A_nx_ny\|_2
   \le \|(1-P_{\mathrm{sc}})x_ny\|_2
        +a^{-(k-1)/2}\|P_{\mathrm{sc}}x_ny\|_2,
\]
and \eqref{eq:vanishing-tensor-coefficient} implies
\[
  \limsup_n\|A_nx_ny\|_2
       \le a^{-1/2}\|y\|_2\le c_0\epsilon\|v\|_2.
\]
Here $q,k,v,b,y$ are all fixed before $n$ tends to infinity. The
possibly large prefix bound therefore disappears with the vanishing
coefficient; a convergence rate uniform in $a$ or $k$ is unnecessary.

By the full-output identity \eqref{eq:repeated-letter}, the left side
of \eqref{eq:propagation-prefix} equals
\[
  \big\|B_k(s^{\otimes k}\otimes v)
            -s^{\otimes k}\otimes B_0v\big\|
\]
for every $n$: its endpoint and repeated $D$ factors all have norm one.
Taking the limit superior proves \eqref{eq:propagation} for bounded $v$.
Density of $A_0$ in $B$ and boundedness of the operators extend it to
all $v\in B$. Applying the same argument to $G^*$, using the adjoint
identities of Lemma~\ref{lem:word-structure}, gives the final assertion.
\end{proof}

The propagation estimate now controls all word levels with the same
error. We assemble orthogonal copies of $B$ from these levels; this
will allow a single compressed commutator to test $B_0$.

Choose $e\in D$ with $\tau(e)=t$ and $p\in D$ with $e\le p$ and
$\tau(p)=1/2$; this is possible since $t\le1/2$. Set
\begin{equation}
  d=2p-1,\qquad w=ds,\qquad u=\tau_p(e)=2t,\qquad
  \tau_p=2\tau\big|_{p\cL p}.
  \label{eq:dihedral-parameters}
\end{equation}
The elements $d$ and $s$ are free centered symmetries. For $k\ge0$ define
\[
  U_k:B\longrightarrow eH,\qquad
  U_kv=t^{-1/2}ew^{-k}v=t^{-1/2}e(sd)^kv.
\]
In \eqref{eq:word-decomposition}, this is the tensor with endpoint
$t^{-1/2}e$, repeated $D$ letter $d$, and last factor
$s^{\otimes k}\otimes v$. Hence each $U_k$ is an isometry, its range
belongs to level $k$, and these ranges are mutually orthogonal. Put
\[
  \mathcal K=\overline{\bigoplus_{k\ge0}U_kB}\subset eH.
\]
Via the $U_k$, identify $\mathcal K$ with
$\ell^2(\N_0)\otimes B$, where $\N_0=\{0,1,\ldots\}$, and set
$J=I\otimes B_0$. Denote the inclusion $\mathcal K\hookrightarrow H$
by $\iota$.

\begin{lemma}[Compression of the commutator]
\label{lem:compressed-commutator}
For every $h\in e\cL e$,
\begin{equation}
  \big\|[J,\iota^*L_h\iota]\big\|
       \le (1+4c_0)\epsilon\|h\|.
  \label{eq:compression-bound}
\end{equation}
\end{lemma}

\begin{proof}
Lemma~\ref{lem:propagation} and \eqref{eq:repeated-letter} give
\[
  \|GU_kv-U_kB_0v\|_2\le2c_0\epsilon\|v\|_2,
\]
with zero error for $k=0$. Each error lies in level $k$ by
Lemma~\ref{lem:word-structure}. Their squared norms therefore add on
finite orthogonal sums. Passing to the closure, and repeating the
argument for $G^*$, gives the full-codomain estimates
\begin{equation}
  \|G\iota-\iota J\|\le2c_0\epsilon,
  \qquad \|G^*\iota-\iota J^*\|\le2c_0\epsilon.
  \label{eq:two-inclusion-errors}
\end{equation}
Write $R=G\iota-\iota J$, $R'=G^*\iota-\iota J^*$, and
$T_h=\iota^*L_h\iota$. Taking the adjoint of the second error identity
gives $\iota^*G=J\iota^*+(R')^*$. A direct expansion now yields
\[
  \iota^*[G,L_h]\iota
     =[J,T_h]+(R')^*L_h\iota-\iota^*L_hR.
\]
Apply \eqref{eq:free-corner-hypothesis} to the first term and
\eqref{eq:two-inclusion-errors} to the other two. Their bounds are
$\epsilon\|h\|$, $2c_0\epsilon\|h\|$, and
$2c_0\epsilon\|h\|$, proving \eqref{eq:compression-bound}.
\end{proof}
==== END Kadisons-similarity-theorem-through-uniform-derivation-estimates//propagation.tex ====

==== BEGIN Kadisons-similarity-theorem-through-uniform-derivation-estimates//tensors.tex ====
\section{Free-product geometry and invariant tensors}
\label{sec:tensors}

Let $D$ be a ${\rm II}_1$ factor generated as a von Neumann algebra
by an increasing sequence
$(D_j)$ of unital full matrix subalgebras whose sizes tend to infinity.
Let $A_0$ be a finite von Neumann algebra with a faithful normal tracial
state, and form the tracial von Neumann free product
\[
  \mathcal L=D*A_0.
\]
We use $\tau$ for the free-product trace and its restrictions.  Freeness
here is scalar tracial freeness: the trace of an alternating product of
centered letters from the two algebras is zero.  Its Hilbert-space
consequence is the orthogonal decomposition into reduced words, with
each reduced-word space isometric to the corresponding Hilbert tensor
product. This model comes from the free-product framework of
\citet{Ching1973} and \citet{Voiculescu1985}. The corner estimate below
is proved here from that word decomposition.

This section and the next two prove the following estimate.  Its constant
does not depend on the trace of the corner.

\begin{theorem}[Free-product corner estimate]\label{thm:free-corner}
Let $\mathcal L=D*A_0$ be as above, let $t=1/N$ for an integer $N\geq2$,
and suppose that $G\in\mathcal B(L^2(\mathcal L))$ commutes with $L_D$.
Suppose that $\epsilon\geq0$ and
\begin{equation}\label{eq:free-corner-hypothesis}
  \|[G,L_x]\|\leq\epsilon\|x\|
  \qquad
  (x\in e\mathcal L e,
   \ e\in D\text{ a projection with }\tau(e)=t).
\end{equation}
If $s=s^*=s^{-1}\in A_0$ and $\tau(s)=0$, then
\begin{equation}\label{eq:free-corner-conclusion}
 \left\|P_{L^2(A_0)}[G,L_s]\big|_{L^2(A_0)}\right\|
       \leq c_1\epsilon,
 \qquad c_1=6\pi^2(1+4c_0).
\end{equation}
\end{theorem}

Since equal-trace projections in $D$ are unitarily conjugate,
commutation with $L_D$ makes it equivalent to assume
\eqref{eq:free-corner-hypothesis} for one such projection.  Moreover,
for $x\in e\mathcal L e$ the commutator in that equation is supported on
$eL^2(\mathcal L)$.  The proof of the theorem is completed after the
Fourier compression calculation in Section~\ref{sec:hankel}.

\subsection{Factoriality and symmetrization}

\begin{lemma}\label{lem:free-factor}
One has $D'\cap\mathcal L=\mathbb C1$.  In particular, $\mathcal L$ is a
${\rm II}_1$ factor.
\end{lemma}

\begin{proof}
Put $E=L^2(D)$, $E^\circ=E\ominus\mathbb C1$, and
$B^\circ=L^2(A_0)\ominus\mathbb C1$.  Regrouping reduced words by their
first and last $A_0$-letters gives the $D$--$D$ bimodule decomposition
\[
 L^2(\mathcal L)\ominus L^2(D)
 \cong E\otimes\mathcal V\otimes E,
 \qquad
 \mathcal V=
 \bigoplus_{n\geq1}
 B^\circ\otimes(E^\circ\otimes B^\circ)^{\otimes(n-1)}.
\]
The endpoints in $E$ are unrestricted; expanding each as
$\mathbb C1\oplus E^\circ$ recovers every reduced word containing an
$A_0$-letter exactly once.  Left and right $D$ act on the first and last
factors, respectively.

Choose a Haar unitary $v$ in a diffuse abelian subalgebra of $D$ and put
$u_n=v^n$.  These unitaries converge ultraweakly to zero in $D$.
Indeed, the restriction of any normal functional to
$W^*(v)\cong L^\infty(\mathbb T)$ has an $L^1(\mathbb T)$ density,
whose Fourier coefficients tend to zero.  Consequently
$L_{u_n}$ and $R_{u_n^*}$ converge weakly to zero on $E$: first test
against bounded vectors in $D$, and then use their $L^2$ density.
On elementary tensors the matrix coefficients of
\[
 L_{u_n}R_{u_n^*}
  =L_{u_n}\otimes I_{\mathcal V}\otimes R_{u_n^*}
\]
therefore tend to zero.  The same holds on the whole complement by
density and uniform boundedness.  A $D$-central vector in this
complement is fixed by all these operators, and hence is zero.

If $z\in D'\cap\mathcal L$, its component perpendicular to $L^2(D)$
is $D$-central, because the displayed decomposition is a bimodule
decomposition.  Thus $z=E_D(z)\in D$, where $E_D$ is the
trace-preserving conditional expectation.  It follows that
$z\in Z(D)=\mathbb C1$.  The center of $\mathcal L$ is contained in
this relative commutant.  Its faithful finite trace and its diffuse
subfactor $D$ then make $\mathcal L$ a ${\rm II}_1$ factor.
\end{proof}

For $u\in\mathcal U(D)$, let $\theta_u$ be the trace-preserving
free-product automorphism which equals $\operatorname{Ad}u$ on $D$
and is the identity on $A_0$.  Its Hilbert-space implementation $W_u$
acts by $\theta_u$ on bounded words.  In particular,
\[
 W_uL_xW_u^*=L_{\theta_u(x)},\qquad
 W_u\xi=\xi\quad(\xi\in L^2(A_0)).
\]

\begin{lemma}[Symmetrization]\label{lem:free-symmetrization}
In proving Theorem~\ref{thm:free-corner}, one may suppose in addition
that $G$ commutes with $W_u$ for every
$u\in\bigcup_j\mathcal U(D_j)$.  This replacement preserves the
compressed commutator in \eqref{eq:free-corner-conclusion} exactly,
and preserves \eqref{eq:free-corner-hypothesis} for both $G$ and $G^*$.
\end{lemma}

\begin{proof}
Average the original operator over each compact matrix unitary group:
\[
 G_j=\int_{\mathcal U(D_j)}W_uGW_u^*\,d\mu_j(u).
\]
This weak operator integral is well defined.  The unitary
representation $u\mapsto W_u$ is strongly continuous on each group,
as is seen first on bounded reduced words and then by density.
Every conjugate in the integral commutes with $L_D$.  It also obeys
\eqref{eq:free-corner-hypothesis}, since $\theta_u$ permutes the
$D$-corners of trace $t$.  Convexity and weak operator closedness of
norm balls preserve these properties under integration and under a
weak operator cluster limit $\widetilde G$ as $j\to\infty$.
Such a limit exists because $\|G_j\|\leq\|G\|$.
For fixed $u\in\mathcal U(D_i)$, all $G_j$ with $j\geq i$ commute
with $W_u$, by Haar invariance.  Thus $\widetilde G$ has every stated
symmetry.

The operators $W_u$ fix $L^2(A_0)$ pointwise and commute with $L_s$.
For each conjugate, and therefore for the averages and their limit,
\[
 P_{L^2(A_0)}[W_uGW_u^*,L_s]\big|_{L^2(A_0)}
   =P_{L^2(A_0)}[G,L_s]\big|_{L^2(A_0)}.
\]
Finally,
$[G^*,L_x]=-[G,L_{x^*}]^*$, so the hypothesis for a self-adjoint
corner algebra also gives its adjoint version.  The same applies to
$\widetilde G$.
\end{proof}

For the remainder of the proof of Theorem~\ref{thm:free-corner},
$G$ denotes the symmetrized operator.  At the end we recover the
original compressed commutator using
Lemma~\ref{lem:free-symmetrization}.

\subsection{The word decomposition}

Set
\[
 E=L^2(D),\quad E^\circ=E\ominus\mathbb C1,
 \qquad B=L^2(A_0),\quad B^\circ=B\ominus\mathbb C1,
 \qquad J_k=(B^\circ)^{\otimes k}\otimes B.
\]
We use the following ordering of the reduced-word factors:
\begin{equation}\label{eq:word-decomposition}
 L^2(\mathcal L)
   =\bigoplus_{k\geq0}\mathcal H_k,
 \qquad
 \mathcal H_k=E\otimes(E^\circ)^{\otimes k}\otimes J_k.
\end{equation}
The tensor
\[
 d_0\otimes d_1\otimes\cdots\otimes d_k
       \otimes(a_1\otimes\cdots\otimes a_k\otimes z)
\]
represents the word $d_0a_1d_1\cdots a_kd_kz$.  Initially the
letters may be taken bounded, with $d_i\in D\ominus\mathbb C1$
and $a_i\in A_0\ominus\mathbb C1$ for $i\geq1$, and unrestricted
$d_0\in D$, $z\in A_0$.  Expanding the endpoints into scalar and
centered parts produces all the usual reduced-word spaces exactly
once.  This proves both orthogonality and completeness in
\eqref{eq:word-decomposition}; the tensor notation then extends to
arbitrary $L^2$-letters by the corresponding isometries.
On each level, $L_D$ acts on the first factor, while $W_u$ conjugates
all $k+1$ factors coming from $D$ and fixes $J_k$.

We will show that $G$ preserves each word level $\mathcal H_k$ and, on
tensors with a repeated centered $D$ letter, changes only the $J_k$ factor.
Scalar slices against fixed vectors in the $J_k$ and $J_l$ factors reduce
these assertions to the following intertwiner calculation.
Write $C_u\xi=u\xi u^*$ on $E$ and on its invariant
subspace $E^\circ$.

\begin{lemma}[Invariant intertwiners]\label{lem:invariant-intertwiners}
Let $k,l\geq0$, and let
\[
 Y:E\otimes(E^\circ)^{\otimes k}
       \longrightarrow E\otimes(E^\circ)^{\otimes l}
\]
be bounded.  Suppose that $Y$ intertwines the left $D$ actions and
that
\[
 Y C_u^{\otimes(k+1)}=C_u^{\otimes(l+1)}Y
 \qquad(u\in\bigcup_j\mathcal U(D_j)).
\]
Then $Y=0$ if $k\ne l$.  If $k=l$, there are scalars
$(a_\sigma)_{\sigma\in S_k}$ such that
\[
 Y=I_E\otimes\sum_{\sigma\in S_k}a_\sigma U_\sigma,
 \qquad
 U_\sigma(x_1\otimes\cdots\otimes x_k)
   =x_{\sigma^{-1}(1)}\otimes\cdots\otimes x_{\sigma^{-1}(k)}.
\]
For $k=0$ the sum consists of the scalar multiple of the identity
corresponding to the empty permutation.
\end{lemma}

\begin{proof}
For a vector or algebra element $x\in E$, put
$x^\circ=x-\tau(x)1$.  On unrestricted variables
$a,x_1,\ldots,x_k,z_1,\ldots,z_l\in D$, define the complex
multilinear form
\[
 F(a,x,z)=
 \left\langle
 Y(1\otimes x_1^\circ\otimes\cdots\otimes x_k^\circ),
 a^*\otimes(z_1^*)^\circ\otimes\cdots\otimes(z_l^*)^\circ
 \right\rangle.
\]
It is multilinear with our convention that inner products are linear
in the first argument.  Fixing the tail vectors and taking their
Hilbert-space slice of $Y$ gives an operator $T:E\to E$ commuting
with $L_D$.  The tracial commutation theorem gives $T=R_b$ for a
bounded $b\in D$, with
\[
 \|b\|=\|T\|
  \leq\|Y\|\prod_i\|x_i^\circ\|_2
                 \prod_j\|(z_j^*)^\circ\|_2.
\]
Since $F(a,x,z)=\tau(ab)$, this yields the endpoint estimate
\begin{equation}\label{eq:coefficient-bound}
 |F(a,x,z)|\leq
 \|Y\|\,\|a\|_1
       \prod_{i=1}^k\|x_i\|_2
       \prod_{j=1}^l\|z_j\|_2.
\end{equation}
The same slice also identifies coefficients with arbitrary bounded
first endpoints: replacing $1,a^*$ by $d,c$ replaces $a$ by $c^*d$,
because $\langle db,c\rangle=\tau(c^*db)$.

\smallskip\noindent
\emph{Finite matrix invariants and stable coefficients.}
Put $p=1+k+l$ and restrict all variables to a stage
$D_j\cong M_m(\mathbb C)$.  The intertwining assumption says that
$F$ is invariant under simultaneous unitary conjugation.  To see
precisely the finite-dimensional consequence, label its variables
$v_0,\ldots,v_{p-1}$.  Every multilinear form on $M_m^p$ has a unique
expression
\[
 \operatorname{Tr}\bigl(A(v_0\otimes\cdots\otimes v_{p-1})\bigr)
 \qquad
 (A\in\operatorname{End}((\mathbb C^m)^{\otimes p})),
\]
by the nondegenerate trace pairing.  Conjugation invariance is
equivalent to $A$ commuting with every $u^{\otimes p}$.
Finite-dimensional Schur--Weyl duality identifies this commutant
with the span of the tensor permutation operators; they are linearly
independent when $m\geq p$
\citep[Theorem~3.1]{GrinkoOzols2024}.
Taking their trace pairings gives the cycle-trace forms
\[
 \Phi_\pi(v_0,\ldots,v_{p-1})
   =\prod_{(i_1\,\cdots\,i_r)\text{ a cycle of }\pi}
       \tau(v_{i_1}\cdots v_{i_r}),
 \qquad \pi\in S_p.
\]
Here cycles follow the successor convention of the displayed
product; reversing the convention only reindexes the permutations.
The normalization from $\operatorname{Tr}$ to $\tau$ multiplies
each form by a nonzero scalar, so these forms are also independent
for $m\geq p$.  Thus
\[
 F\big|_{D_j^p}=\sum_{\pi\in S_p}\lambda_{\pi,j}\Phi_\pi.
\]
This identity is on \emph{unrestricted} matrix variables: the
centering occurs inside the definition of $F$.
Under a unital inclusion of full matrix stages, normalized traces
agree, and hence every $\Phi_\pi$ restricts to the identically
defined form on the smaller stage.  Uniqueness on any smaller stage
of size at least $p$ proves that $\lambda_{\pi,j}$ is independent
of $j$ from that point onward.  Denote it by $\lambda_\pi$.

\smallskip\noindent
\emph{Isolating coefficients before restricting to centered tails.}
The invariant-theory step has expressed every coefficient through
finitely many cycle-trace forms with coefficients fixed across matrix
stages. We now use the endpoint bound to remove long cycles, then
Hilbert-space boundedness to remove pairs within one side.
Fix $\pi\in S_p$.  Choose distinct matrix indices for the slots and
put $v_i=e_{i,\pi(i)}$.  In a nonzero product of these matrix units,
the variable in slot $i$ must be followed by the variable whose
initial index is $\pi(i)$.  Closing each product under its trace
therefore forces exactly the cycles of $\pi$.  Consequently
\[
 \Phi_\rho(v_0,\ldots,v_{p-1})
  =\begin{cases}
      m^{-c(\pi)},&\rho=\pi,\\
      0,&\rho\ne\pi,
    \end{cases}
\]
where $c(\pi)$ is the number of cycles.  Every chosen matrix unit
has $L^1$ norm $m^{-1}$ and $L^2$ norm $m^{-1/2}$, including
diagonal units.  Applying \eqref{eq:coefficient-bound} gives
\[
 |\lambda_\pi|m^{-c(\pi)}
       \leq\|Y\|m^{-1-(k+l)/2}.
\]
The coefficient is fixed while the stage size tends to infinity.
It follows that
\[
 \lambda_\pi=0\qquad\text{if}\qquad
 c(\pi)<1+\frac{k+l}{2}.
\]

Now restrict the $k+l$ tail variables to centered matrices.  Any
form with a singleton tail cycle vanishes.  For a permutation with
no such singleton, let $r$ be the length of the cycle containing the
distinguished variable $a=v_0$.  Its number of cycles is at most
\[
 1+\frac{k+l-(r-1)}2.
\]
If $r>1$, this is below the required threshold.  If $r=1$, the
threshold is attained only when every tail cycle has length two.
Thus the only remaining terms have $a$ as a singleton and pair all
tail slots.  On centered tails,
\[
 F(a,x,z)=\tau(a)
       \sum_{\mathcal P}\lambda_{\mathcal P}
          \prod_{\{i,j\}\in\mathcal P}\tau(v_i v_j),
\]
where $\mathcal P$ ranges over pairings of the tail slots.  There
are no terms when the number of tail slots is odd.

Using $a=c^*d$ gives the corresponding simple-tensor coefficients
with first endpoints $d,c$ in the union of the matrix stages.
These formulas extend to arbitrary endpoints in $E$ and arbitrary
centered $L^2$ tail letters.  Indeed, that union is $L^2$ dense,
the left side is continuous in its individual $L^2$ letters, and
each factor on the right is continuous by
$|\tau(uv)|\leq\|u\|_2\|v\|_2$.  At this point the assertion is
continuity of coefficients on simple tensors, not boundedness of
each pairing on the completed tensor product.

\smallskip\noindent
\emph{Excluding pairings within one side.}
Choose an infinite orthonormal family $(h_n)$ of self-adjoint
centered vectors in $E$.  For example, normalized real
trigonometric functions of a Haar unitary give such a family.
Then $\tau(h_nh_m)=\delta_{nm}$.
Fix a pairing $\mathcal P$, with $r_s$ pairs entirely in the source
slots and $r_t$ entirely in the test slots.  Assign a disjoint
infinite pool of the $h_n$ to each pair.  For each source--test pair,
use one fixed member of its pool in both slots.  For each
source--source pair, sum the source tensor over $M$ members of its
pool, placing the same member in its two slots, independently for
the different pairs.  Form the test tensor analogously for the
test--test pairs.  Take both first endpoints to be $1$.

In any term of the expanded coefficient every label occurs exactly
twice, in the positions assigned by $\mathcal P$, and different
pairs have different labels.  Hence all other pairings contribute
zero.  The resulting coefficient is exactly
$\lambda_{\mathcal P}M^{r_s+r_t}$.  The source summands are
orthonormal, as are the test summands, so the product of the two
tensor norms is $M^{(r_s+r_t)/2}$.  Boundedness of $Y$ gives
\[
 |\lambda_{\mathcal P}|M^{r_s+r_t}
       \leq\|Y\|M^{(r_s+r_t)/2}.
\]
Letting $M\to\infty$ proves $\lambda_{\mathcal P}=0$ whenever
$r_s+r_t>0$.  The argument also applies when one side has no tail
slots.

All surviving pairs therefore join a source slot to a test slot,
which requires $k=l$.  If these numbers differ, all simple-tensor
coefficients vanish and $Y=0$.  Otherwise these pairings are
indexed by bijections $\sigma\in S_k$, and the coefficients are
\[
 \tau(c^*d)\sum_{\sigma\in S_k}a_\sigma
       \prod_{i=1}^k\tau(x_i z_{\sigma(i)}).
\]
Since $\tau(x_i z_j)=\langle x_i,z_j^*\rangle$, these are precisely
the coefficients of
$I_E\otimes\sum_\sigma a_\sigma U_\sigma$.  Equality against all
elementary test tensors, and their density, proves the asserted
operator identity.
\end{proof}

\begin{lemma}[Word structure]\label{lem:word-structure}
The symmetrized operator $G$ preserves every level $\mathcal H_k$
in \eqref{eq:word-decomposition}.  There are operators
$B_k\in\mathcal B(J_k)$, with $\|B_k\|\leq\|G\|$, such that, for
all $d_0\in E$, $x\in E^\circ$, and $\alpha\in J_k$,
\begin{equation}\label{eq:repeated-letter}
 \begin{aligned}
 G(d_0\otimes x^{\otimes k}\otimes\alpha)
     &=d_0\otimes x^{\otimes k}\otimes B_k\alpha,\\
 G^*(d_0\otimes x^{\otimes k}\otimes\alpha)
     &=d_0\otimes x^{\otimes k}\otimes B_k^*\alpha.
 \end{aligned}
\end{equation}
These are identities in the full space $L^2(\mathcal L)$.
The operators $B_k$ are independent of the repeated letter $x$;
in particular, $x$ may be any complex centered $L^2$ vector.
\end{lemma}

\begin{proof}
Let $P_k$ be the level projection.  It commutes with $L_D$ and all
$W_u$.  For $\alpha\in J_k$ and $\beta\in J_l$, take the scalar
slice in these factors of $P_lGP_k$ to obtain an operator
\[
 Y_{\alpha,\beta}:
 E\otimes(E^\circ)^{\otimes k}
   \longrightarrow E\otimes(E^\circ)^{\otimes l},
 \qquad
 \|Y_{\alpha,\beta}\|\leq\|G\|\|\alpha\|\|\beta\|.
\]
It satisfies the hypotheses of
Lemma~\ref{lem:invariant-intertwiners}.  For $k\ne l$ every such
slice vanishes, so $P_lGP_k=0$.  Thus the levels reduce $G$.

Fix a unit vector $h\in E^\circ$ and let
$V_k:J_k\to\mathcal H_k$ be the isometry
$V_k\alpha=1\otimes h^{\otimes k}\otimes\alpha$.
Define $B_k=V_k^*GV_k$.  On an equal-level scalar slice, the lemma
gives
\[
 Y_{\alpha,\beta}
    =I_E\otimes\sum_{\sigma\in S_k}
                        a_\sigma(\alpha,\beta)U_\sigma,
 \qquad
 \langle B_k\alpha,\beta\rangle
       =\sum_{\sigma\in S_k}a_\sigma(\alpha,\beta).
\]
Every $U_\sigma$ fixes $x^{\otimes k}$ for every $x\in E^\circ$.
Consequently the last scalar is independent of the unit letter
used to define $V_k$.  The same slice identity, now tested against
arbitrary endpoint and tail vectors and arbitrary $\beta$, gives
the first equality in \eqref{eq:repeated-letter}.  These tests
determine the whole output on level $k$, and the other levels have
already been excluded.  This also proves the assertion for every
endpoint $d_0\in E$, with no boundedness requirement on its
representative.  No self-adjointness of $x$ is used: the relevant
cross contraction is $\tau(xz)=\langle x,z^*\rangle$.

The same argument applies to $G^*$.  Its induced operator on $J_k$
is $V_k^*G^*V_k=(V_k^*GV_k)^*=B_k^*$, proving the second equality.
\end{proof}

At level zero the lemma says $G|_{\mathcal H_0}=I_E\otimes B_0$.
Since $B=L^2(A_0)$ is the subspace $\mathbb C1\otimes B$ there,
the quantity to be estimated in
Theorem~\ref{thm:free-corner}, after symmetrization, is exactly
\[
 \left\|P_B[G,L_s]\big|_B\right\|
      =\|[B_0,L_s|_B]\|.
\]
==== END Kadisons-similarity-theorem-through-uniform-derivation-estimates//tensors.tex ====

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