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\title[Universal modules and natural pairings]{Universal Modules, Natural Pairings, and Weighted Isometries for Noncommutative \texorpdfstring{$L^p$}{Lp}-Spaces}
\author{Zhenyu Sun}
\author{Jilin Wang}
\date{}
\subjclass[2020]{46L52, 46L07, 46M15, 18A40}
\keywords{noncommutative $L^p$-space, universal property, natural transformation,
conditional expectation, modular automorphism, operator space}
\hypersetup{
 pdftitle={Universal Modules, Natural Pairings, and Weighted Isometries for Noncommutative Lp-Spaces},
 pdfauthor={Zhenyu Sun and Jilin Wang},
 pdfsubject={Universal modules, natural transformations, and quantitative weighted isometries},
 pdfkeywords={noncommutative Lp, universal property, natural bilinear form, modular compatibility}}

\begin{document}
\begin{abstract}
For $1\le p<\infty$, we consider the functor $L^p$ on finite tracial von
Neumann algebras with trace-preserving embeddings, including nonunital ones.
For a functor of unital contractive left modules with compatible distinguished
vectors $\xi_M$, finite direct sums of scalars determine the best constant
in $\|x\xi_M\|\le K\|x\|_p$. The pointed $L^p$ functor is initial among
the families satisfying this estimate with $K=1$.
Retaining only the bound on scalar points gives an atomic norm. On
$\C^n$ with unit weights the norm of the identity from $\ell^p_n$ to this
atomic space is asymptotic to $p^{-1}(\log n)^{1-1/p}$ for $p>1$.
For $1\le p,q<\infty$, every objectwise bounded natural bilinear family
on $L^p\times L^q$ has the form $c\,\tau(xy)+d\,\tau(x)\tau(y)$ for
$x,y\in M$, with $c,d$ independent of the object and $c=0$ when
$1/p+1/q>1$. A uniformly bounded natural inner product
on every $L^p(M,\tau)$ exists exactly at $p=2$.
For finite-dimensional unital embeddings preserving a faithful positive
functional, we determine the nearest modularly compatible target density
both in Hilbert--Schmidt norm and in the Hilbert--Schmidt distance between
positive square roots. The minimum squared distance between square roots lies
between a weighted sum of the losses in squared $L^2$ norm on matrix
units and twice that sum. Both constants are sharp.
\end{abstract}

\maketitle
\section{Introduction}

For a finite von Neumann algebra $M$ with faithful normal finite trace
$\tau$, each nonzero projection $e$ defines a trace-preserving embedding
of the scalars into $M$ by $z\mapsto ze$, provided the scalar trace has
mass $\tau(e)$. Such embeddings relate the value of a functor on a scalar
algebra to its behavior on individual projections in $M$. We use them
to characterize the pointed $L^p$-module functor and to classify the
scalar pairings that are natural for trace-preserving embeddings. In
finite dimensions we also ask how much an embedding can distort the
$L^2$ norm when it preserves a positive functional with a noncentral
density.

Throughout the tracial part of the paper, morphisms are normal injective
$*$-homomorphisms preserving the given traces. We allow nonunital
morphisms and all positive finite trace masses, and take
$1\le p,q<\infty$. For background on tracial integration, general
noncommutative $L^p$-spaces, and their interpolation theory, see
\cite{Segal1953,Nelson1974,Haagerup1979,Kosaki1984,Terp1982,PisierXu2003}.

\subsection*{Scalar tests and universal modules}

Leinster characterized the classical $L^p$ functor on finite measure
spaces by a universal property and obtained integration from the case
$p=1$ \cite{Leinster2023}. Liu, Liu, Huang, and Zhou studied categorical
integration for normed modules over finite-dimensional algebras
\cite{LiuLiuHuangZhou2026}. For the theory of noncommutative $L^p$ modules,
see Junge and Sherman \cite{JungeSherman2005}.

Let $F$ be a functor to Banach spaces and linear contractions. Give
each $F(M)$ a unital contractive left $M$-action and a distinguished
vector $\xi_M$. Assume that the maps $F(\pi)$ intertwine the actions
and satisfy
\[
 F(\pi)\xi_M=\pi(1_M)\xi_N
 \qquad\bigl(\pi:(M,\tau)\to(N,\sigma)\bigr).
\]
Write $\C_s$ for the scalars with trace $z\mapsto sz$. On a finite
direct sum $D_{\mathbf s}=\bigoplus_j\C_{s_j}$ with coordinate
projections $e_j$, consider the inequalities
\begin{equation}\label{eq:intro-scalar-tests}
 \left\|\sum_j z_je_j\xi_{D_{\mathbf s}}\right\|
 \le K\left(\sum_j s_j|z_j|^p\right)^{1/p}.
\end{equation}
The optimal constant equals the supremum of
$\|x\xi_M\|/\|x\|_{p,\tau}$ over all $(M,\tau)$ and $0\ne x\in M$.
When finite, it is the uniform norm of the unique natural family of
bounded module maps $T_M:L^p(M,\tau)\to F(M)$ sending $1_M$ to $\xi_M$.
The proof uses spectral approximation for $|x|$ and the module action
of the partial isometry in the polar decomposition of $x$. With $K=1$,
this gives the initial property of the pointed $L^p$ functor
(\cref{thm:mod-scalar-testing,cor:mod-initial-reconstruction}).

Suppose we retain only $\|\xi_{\C_s}\|\le s^{1/p}$ from
\eqref{eq:intro-scalar-tests}. The initial module is then the completion
of $M$ in the norm
\[
 \alpha_{p,\tau}(x)=
 \inf_{x=\sum_j a_je_j}\sum_j\|a_j\|_\infty\,\tau(e_j)^{1/p},
\]
where $a_j\in M$ and the $e_j$ are arbitrary projections. For $p=1$
this is the trace norm. On finite scalar algebras the formula gives a
weighted Lorentz norm. For $p>1$ and unit weights, the norm of the
identity $\ell^p_n\to(\C^n,\alpha_p)$ is asymptotic to
$p^{-1}(\log n)^{1/p'}$, where $p'=p/(p-1)$
(\cref{thm:mod-atomic-initial,prop:mod-atomic-lorentz}).
The role of the module action is already visible on $M_2(\C)$.
\Cref{thm:mod-action-obstruction} constructs a pointed functor
satisfying a direct-sum estimate for which the prescribed values on
rank-one projections admit no linear extension.

\subsection*{Natural linear maps and scalar pairings}

A family of maps is objectwise bounded if each component is bounded
and uniformly bounded if one bound holds for all components.
The objectwise bounded natural linear functionals on $L^p$ are the
multiples of the trace. The objectwise bounded natural linear maps $L^p\to L^q$
are the multiples of the finite-trace inclusion for $p\ge q$, and zero
for $p<q$ (\cref{lem:objectwise-scalar-naturality,thm:linear-naturality}).
For $p>1$, a uniformly bounded family of linear functionals $\ell_M$
satisfies the sharp defect estimate of \cref{prop:approximate-scalar-naturality},
\[
 \sup_M\|\ell_M\|\le
 \sup_{\pi:M\to N}\|\ell_N\circ L^p(\pi)-\ell_M\|.
\]

Izumi studied bilinear and sesquilinear forms associated with
interpolation realizations of noncommutative $L^p$-spaces
\cite{Izumi1998}. Here naturality asks that pullback along every tracial
embedding preserve the pairing on its source.
If $B_M:L^p(M,\tau)\times L^q(M,\tau)\to\C$ is such a family of
bounded bilinear forms, then on $M\times M$,
\begin{equation}\label{eq:intro-bilinear}
 B_M(x,y)=c\,\tau(xy)+d\,\tau(x)\tau(y)
\end{equation}
for scalars $c,d$ independent of the object. The product of traces
extends continuously for every pair $(p,q)$. The single-trace term
extends for $1/p+1/q\le1$, and its coefficient vanishes otherwise.
To obtain the two terms, we use inner automorphisms of $M_2(M)$ to
isolate a balanced bilinear form. The remaining form is natural in two
independently varying algebras
(\cref{thm:objectwise-bilinear-classification}).

Varying the scalar trace mass in \eqref{eq:intro-bilinear} gives the
uniform classification. Only $c\,\tau(xy)$ survives when $1/p+1/q=1$,
only $d\,\tau(x)\tau(y)$ when $p=q=1$, and all other uniformly bounded
families vanish. Thus a uniformly bounded natural family of inner
products exists exactly when $p=2$. Normalizing the form to take value
one at $(1,1)$ on $\C_1$ gives $(x,y)\mapsto\tau(x^*y)$
(\cref{cor:uniform-bilinear-classification,cor:natural-hilbert-exponent}).

\subsection*{Weighted embeddings}

Tracial embeddings induce complete isometries with completely contractive
left inverses (\cref{sec:operator-spaces}). We use the predual of the
opposite algebra at the $L^1$ endpoint
\cite[Section~2]{JungeRuanSherman2005}, \cite[Section~9]{Pisier2014}.
Complete isometries of noncommutative $L^p$-spaces and their relation to
conditional expectations are studied in
\cite{JungeRuanSherman2005,Sherman2006}. For extensions of maps in
the nontracial setting, see \cite{HaagerupJungeXu2010}.

Consider now finite-dimensional algebras with faithful densities $h,k$
and an injective $*$-homomorphism $\pi$ satisfying
$\Tr_N(k\pi(x))=\Tr_M(hx)$. The symmetric weighted norm is
$\|x\|_{p,h}=\|h^{1/(2p)}xh^{1/(2p)}\|_p$, with unnormalized matrix
traces. The embedding preserves the weighted $L^2$ norm precisely when
its support commutes with $k$ and it intertwines the modular groups.
This criterion is related to the self-polar forms of Haagerup and
St\o rmer \cite{HaagerupStormer1995} and to Takesaki's theorem on
conditional expectations \cite{Takesaki1972}.
\Cref{thm:weighted-modular-equivalence} gives a finite-dimensional proof.
For a unital embedding, in multiplicity coordinates, the criterion reads
\[
 \pi(x)_j=\bigoplus_i x_i\otimes I_{r_{ji}},
 \qquad k_j=\bigoplus_i h_i\otimes\rho_{ji},
 \qquad \rho_{ji}>0,\quad \sum_j\Tr\rho_{ji}=1 \quad\text{for each }i.
\]
Such an embedding preserves the weighted norms for all $1\le p<\infty$.

Fix a unital $\pi$ and $h$. Among compatible densities $\ell$ with this
source density, \cref{thm:weighted-nearest-densities} gives the unique
minimizer of $\|k-\ell\|_2$ and the unique minimizer of
$\|\sqrt{k}-\sqrt{\ell}\|_2$. Let $V_h$ be the subspace of block
matrices $\bigoplus_{j,i}\sqrt{h_i}\otimes z_{ji}$, where
$z_{ji}\in M_{r_{ji}}$, and put
$\delta(k)=\dist_2(\sqrt{k},V_h)^2$. The minimum of
$\|\sqrt{k}-\sqrt{\ell}\|_2^2$ lies between $\delta(k)$ and
$2\delta(k)$, with both constants sharp over finite dimensions.
In an eigenbasis of $h_i$ with eigenvalues $\lambda_{ia}$ and matrix
units $e_{ab}^{(i)}$, we have
\begin{equation}\label{eq:intro-finite-defect}
 \delta(k)=\sum_i\frac1{\Tr h_i}\sum_{a,b}
 \sqrt{\lambda_{ia}\lambda_{ib}}\,
 \bigl(\|e_{ab}^{(i)}\|_{2,h}^2
       -\|\pi(e_{ab}^{(i)})\|_{2,k}^2\bigr).
\end{equation}
Each summand is nonnegative. Thus equality of the weighted norms on these
matrix units characterizes modular compatibility. The same identity bounds
the distance to the compatible densities in terms of the largest
relative loss of squared norm. For a nonunital
embedding, the squared norm loss at $1_M$ is exactly
$\tfrac12\|[\sqrt{k},\pi(1_M)]\|_2^2$
(\cref{prop:weighted-finite-tests,thm:weighted-density-stability,prop:weighted-support-stability}).

\section{Tracial embeddings and dense cores}\label{sec:preliminaries}

Let $\Nrmone$ and $\Banone$ denote the categories of complex normed
spaces and complex Banach spaces, respectively, with linear contractions
as morphisms.  Write $U:\Banone\to\Nrmone$ for the forgetful functor.
We fix a universe of objects and form functor categories in a larger universe.
An object of $\TrEmb$ is a nonzero von Neumann algebra $M$ with a
faithful normal finite trace $\tau$.  Write $\mathcal P(M)$ for its
projections.  An arrow
$\pi:(M,\tau)\to(N,\sigma)$ is a normal injective $*$-homomorphism satisfying
$\sigma\circ\pi=\tau$.  We allow nonunital embeddings.  One may also
adjoin the zero algebra, with $L^p(0)=0$.  In that convention formulas
involving the total trace mass refer to nonzero objects.

For $1\le p<\infty$ put
\[
 \|x\|_{p,\tau}=\tau(|x|^p)^{1/p}\quad(x\in M),
 \qquad
 L^p(M,\tau)=\overline{(M,\|\cdot\|_{p,\tau})}.
\]
We use the standard tracial integration, H\"older, and duality theorems
\cite{Segal1953,Nelson1974,PisierXu2003}.  In particular,
\begin{equation}\label{eq:basic-module-bound}
 \|axb\|_{p,\tau}\le \|a\|_\infty\|x\|_{p,\tau}\|b\|_\infty,
 \qquad
 \|a\|_{p,\tau}\le \tau(1)^{1/p}\|a\|_\infty .
\end{equation}
The first inequality extends left and right multiplication by continuity
to operators $L_M(a)$ and $R_M(b)$ on $L^p(M,\tau)$, with norms at most
$\|a\|_\infty$ and $\|b\|_\infty$, respectively.  We identify $M$ with
its dense image in $L^p(M,\tau)$ and use $\tau$ also for the continuous
extension of the trace.  Its norm is
\begin{equation}\label{eq:trace-functional-norm}
 \|\tau:L^p(M,\tau)\to\C\|=\tau(1)^{1-1/p}.
\end{equation}
H\"older's inequality gives the upper bound, and evaluation at
$1_M/\|1_M\|_{p,\tau}$ gives equality.

\begin{lemma}\label{lem:tracial-functor}
Every arrow $\pi:(M,\tau)\to(N,\sigma)$ induces an isometry
$L^p(\pi):L^p(M,\tau)\to L^p(N,\sigma)$ extending $\pi$.
These extensions define a functor $L^p:\TrEmb\to\Banone$.
\end{lemma}
\begin{proof}
With $e=\pi(1)$, functional calculus in the corner $eNe$ gives
$|\pi(x)|^p=\pi(|x|^p)$.  Hence
$\|\pi(x)\|_{p,\sigma}^p=\sigma(\pi(|x|^p))=\tau(|x|^p)$.
Thus $\pi$ extends uniquely to an isometry.  The identity and composition
laws pass from the dense algebras to their completions.
\end{proof}

For $s>0$ we write $\C_s=(\C,\tau_s)$, where $\tau_s(z)=sz$.
Both summand embeddings into
$(M,\tau)\oplus(N,\sigma)=(M\oplus N,\tau\oplus\sigma)$ preserve traces.
If $e\ne0$ is a projection of $M$, the map
\begin{equation}\label{eq:scalar-projection-embedding}
 \C_{\tau(e)}\longrightarrow(M,\tau),\qquad z\longmapsto ze
\end{equation}
is an arrow.  We give $M_n(M)$ the trace $\Tr_n\otimes\tau$, with
$\Tr_n$ the unnormalized matrix trace.  Each matrix-corner map
$x\mapsto e_{jj}\otimes x$ is then an arrow of $\TrEmb$.

\begin{proposition}\label{prop:projection-cores}
The set of finite sums of pairwise orthogonal projections with complex
coefficients is precisely the set of finite-spectrum normal elements.
It is a complex vector space if and only if $M$ is abelian.
The linear span of all projections,
\[
 \mathcal P_{\rm lin}(M)=\operatorname{span}_{\C}\{e\in M:e=e^*=e^2\},
\]
is preserved by embeddings in $\TrEmb$.  It is dense in $M$ for the
operator norm and in $L^p(M,\tau)$ for the $L^p$ norm.
\end{proposition}
\begin{proof}
The first assertion is the finite-spectrum spectral theorem.
If projections $e,f$ do not commute, then
$(e+if)^*(e+if)-(e+if)(e+if)^*=2i(ef-fe)\ne0$.
Thus closure under addition forces all projections to commute and
hence, by spectral approximation, forces $M$ to be abelian.
Conversely, in an abelian algebra a common refinement of two finite
spectral partitions gives a finite spectral partition for the sum.
For arbitrary $M$, approximate the real and imaginary parts of any
$x\in M$ in operator norm by finite-spectrum selfadjoint elements.
Their linear combinations belong to $\mathcal P_{\rm lin}(M)$.
Equation~\eqref{eq:basic-module-bound} then gives $L^p$-density.
Functoriality follows because $*$-homomorphisms preserve projections.
\end{proof}

\begin{lemma}[Functorial completion]\label{lem:completion}
Let $\mathcal C$ be a small category and let $A:\mathcal C\to\Nrmone$.
Objectwise completion defines a functor $\widehat A:\mathcal C\to\Banone$.
For every $F:\mathcal C\to\Banone$, restriction along the dense inclusions
gives a bijection
\[
 \operatorname{Nat}_{\Banone}(\widehat A,F)
 \simeq
 \operatorname{Nat}_{\Nrmone}(A,UF).
\]
The same assertion holds for matrix-normed spaces and complete contractions.
\end{lemma}
\begin{proof}
Each component of a transformation on the right extends uniquely to a
contraction on the completion.  The two sides of every naturality square
agree on a dense subspace, so the extensions are natural.
Conversely, restriction recovers the original transformation.
For operator spaces the same extension is contractive at every matrix
level, since finite matrices over a dense subspace are dense.
\end{proof}

Leinster uses this completion adjunction in the proof of the classical
$L^p$ universal property \cite[Theorem~3.7]{Leinster2023}.

\section{Scalar tests and universal pointed modules}\label{sec:mod-scalar-tests}

Fix $1\leq p<\infty$.  Given a distinguished vector $\xi_M$ in a
Banach left $M$-module, consider the map $x\mapsto x\xi_M$ on $M$.
Under the compatibility conditions below, its optimal uniform bound
with respect to the $L^p$ norm can be computed on finite direct sums
of scalar algebras.

\subsection{Supported points and scalar testing}

We abbreviate $F(M,\tau)$ to $F(M)$ when the trace is understood.
A \emph{supported pointed module family} consists of a functor
$F:\TrEmb\to\Banone$, a unital contractive Banach left $M$-module
structure on every $F(M)$, and a vector $\xi_M\in F(M)$ for every object.
The compatibility conditions are
\begin{align}
 F(\pi)(a\cdot u)&=\pi(a)\cdot F(\pi)(u),
       &&a\in N,\ u\in F(N),\label{eq:mod-equivariance}\\
 F(\pi)(\xi_N)&=\pi(1_N)\cdot\xi_M,
       &&\pi:(N,\sigma)\longrightarrow(M,\tau).
       \label{eq:mod-supported-point}
\end{align}
Here contractivity of the action means
$\|a\cdot u\|\leq\|a\|_\infty\|u\|$.
A morphism of such families is a natural transformation whose components
are contractive module maps preserving the distinguished vectors.

For a finite tuple
$\mathbf s=(s_1,\ldots,s_n)$ of positive numbers, put
\[
 D_{\mathbf s}=\bigoplus_{j=1}^n\C_{s_j},
 \qquad
 \iota_j:\C_{s_j}\longrightarrow D_{\mathbf s}.
\]
Define the scalar test map
\begin{equation}\label{eq:mod-scalar-test-map}
 V_{\mathbf s}(z)
 =\sum_{j=1}^n z_jF(\iota_j)\xi_{\C_{s_j}},
 \qquad z=(z_1,\ldots,z_n)\in\C^n.
\end{equation}
If $e_j$ is the $j$th coordinate projection of $D_{\mathbf s}$, then
\eqref{eq:mod-supported-point} gives
$F(\iota_j)\xi_{\C_{s_j}}=e_j\xi_{D_{\mathbf s}}$.  Thus
$V_{\mathbf s}(z)=z\cdot\xi_{D_{\mathbf s}}$.

\begin{lemma}\label{lem:mod-polar-and-scalar}
For every supported pointed module family, the following statements hold.
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.5em]
\item For $x\in M$,
\[
 \|x\cdot\xi_M\|=\||x|\cdot\xi_M\|.
\]
\item For any $K\geq0$, the scalar bounds
\[
 \|\xi_{\C_s}\|\leq K s^{1/p}\qquad(s>0)
\]
are equivalent to
\[
 \|\xi_M\|\leq K\tau(1_M)^{1/p}
 \qquad((M,\tau)\in\TrEmb).
\]
Under these bounds,
$\|e\cdot\xi_M\|\leq K\tau(e)^{1/p}$ for every projection $e\in M$.
\end{enumerate}
\end{lemma}

\begin{proof}
Write $x=v|x|$ for the polar decomposition in $M$.  The identities
$x=v|x|$ and $|x|=v^*x$, together with contractivity of the action, prove
\textup{(i)}.  For \textup{(ii)}, use the unital trace-preserving embedding
$\C_{\tau(1_M)}\to M$ in \eqref{eq:mod-supported-point} and take
norms.  The converse is the case $M=\C_s$.  The estimate for
$e\ne0$ follows from the scalar corner embedding
\eqref{eq:scalar-projection-embedding}, and the case $e=0$ is immediate.
\end{proof}

\begin{theorem}[Scalar testing with the optimal constant]
\label{thm:mod-scalar-testing}
For a supported pointed module family, define
\begin{align*}
 C_{\mathrm{sc},p}
 &=\sup_{n,\,\mathbf s,\,z\ne0}
 \frac{\|V_{\mathbf s}(z)\|}
      {(\sum_{j=1}^n s_j|z_j|^p)^{1/p}},\\*
 C_{\mathrm{alg},p}
 &=\sup_{(M,\tau),\,0\ne x\in M}
 \frac{\|x\cdot\xi_M\|}{\tau(|x|^p)^{1/p}}.
\end{align*}
These quantities agree, with $+\infty$ allowed.  In the first supremum
it suffices to take $z_j\geq0$ for every $j$.

If their common value is finite, there is a unique natural family of
bounded module maps
\[
 T_M:L^p(M,\tau)\longrightarrow F(M),
 \qquad T_M(1_M)=\xi_M,
\]
and its uniform norm is
\begin{equation}\label{eq:mod-optimal-extension-bound}
 \sup_{(M,\tau)}\|T_M\|=C_{\mathrm{sc},p}.
\end{equation}
Conversely, a uniformly bounded natural family with these properties
can exist only if $C_{\mathrm{sc},p}<\infty$.
\end{theorem}

\begin{proof}
Since $V_{\mathbf s}(z)=z\xi_{D_{\mathbf s}}$, each scalar test occurs
in the supremum defining $C_{\mathrm{alg},p}$.  Moreover, replacing
$z$ by $|z|$ preserves the test ratio by
\cref{lem:mod-polar-and-scalar}\textup{(i)}.  It remains to prove
$C_{\mathrm{alg},p}\leq C_{\mathrm{sc},p}$ when $C_{\mathrm{sc},p}<\infty$.

Let $a=\sum_{j=1}^n\lambda_j e_j$ be a positive finite-spectrum element
of $M$, with nonzero pairwise orthogonal projections $e_j$ and
$\lambda_j\geq0$.  Set $s_j=\tau(e_j)$.  The map
\[
 \pi:D_{\mathbf s}\longrightarrow M,
 \qquad(z_j)\longmapsto\sum_j z_je_j
\]
is a morphism of $\TrEmb$, including when $\sum_j e_j<1_M$.
Equivariance and the supported-point condition give
\[
 F(\pi)V_{\mathbf s}(\lambda)
 =\sum_j\lambda_je_j\xi_M=a\xi_M.
\]
Since $F(\pi)$ is contractive,
\[
 \|a\xi_M\|
 \leq C_{\mathrm{sc},p}
       \left(\sum_j\lambda_j^p\tau(e_j)\right)^{1/p}
 =C_{\mathrm{sc},p}\|a\|_{p,\tau}.
\]
Every positive element of $M$ is an operator-norm limit of positive
finite-spectrum elements.  The action is operator-norm continuous, and
$\|b\|_{p,\tau}\leq\tau(1_M)^{1/p}\|b\|_\infty$.
The preceding inequality therefore holds for all positive $a\in M$.
For arbitrary $x$, apply it to $|x|$ and use
\cref{lem:mod-polar-and-scalar}\textup{(i)}.  This proves
$C_{\mathrm{alg},p}\leq C_{\mathrm{sc},p}$.

The map $x\mapsto x\xi_M$ now has a unique bounded extension $T_M$
to $L^p(M,\tau)$.  On $N\subseteq L^p(N,\sigma)$,
\[
 F(\pi)T_N(x)
 =\pi(x)\pi(1_N)\xi_M
 =T_M(\pi(x)),
\]
so the extensions are natural.  The identity $T_M(ax)=aT_M(x)$ also
extends by density.  A module map sending $1_M$ to $\xi_M$ must agree
with $T_M$ on $M$, which proves uniqueness and bounds each scalar test
by the uniform norm of any such family.  Since extension preserves
operator norms, \eqref{eq:mod-optimal-extension-bound} follows.
\end{proof}

Let $\mathsf{PM}_p^0$ be the category of supported pointed module
families satisfying $\|\xi_{\C_s}\|\leq s^{1/p}$.
Let $\mathsf{PM}_p$ be its full subcategory defined by the scalar
partition inequalities
\begin{equation}\label{eq:mod-scalar-partition}
 \left\|\sum_j z_jF(\iota_j)\xi_{\C_{s_j}}\right\|
 \leq\left(\sum_j s_j|z_j|^p\right)^{1/p}.
\end{equation}

\begin{corollary}[Reconstruction of the pointed $L^p$-module functor]
\label{cor:mod-initial-reconstruction}
The left regular module functor $(L^p,L,\mathbf1)$ is initial in
$\mathsf{PM}_p$.  If $p=1$, the scalar bounds alone imply
\eqref{eq:mod-scalar-partition}, so
$\mathsf{PM}_1^0=\mathsf{PM}_1$.
More generally,
\[
 C_{\mathrm{sc},1}
 =\sup_{s>0}\frac{\|\xi_{\C_s}\|}{s}.
\]
\end{corollary}

\begin{proof}
For the left regular family, the norm in
\eqref{eq:mod-scalar-partition} is exactly the displayed weighted
$\ell^p$ norm.  The initial morphism to any other object is supplied
by \cref{thm:mod-scalar-testing}.  For $p=1$, the triangle inequality
and contractivity of the $F(\iota_j)$ give
\[
 \|V_{\mathbf s}(z)\|
 \leq\sum_j|z_j|\|\xi_{\C_{s_j}}\|.
\]
Taking $n=1$ gives the reverse inequality between the two suprema.
\end{proof}

\begin{proposition}[Exact cyclic representation]
\label{prop:mod-exact-cyclic}
Let $(M,\tau)$ be fixed, let $X$ be a unital contractive Banach left
$M$-module, and let $\xi\in X$.  The following are equivalent.
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.5em]
\item For every finite orthogonal family of projections $e_j$ and
every $\lambda_j\geq0$,
\[
 \left\|\sum_j\lambda_je_j\xi\right\|^p
 =\sum_j\lambda_j^p\tau(e_j).
\]
\item The map $x\mapsto x\xi$ extends to an isometric $M$-module
isomorphism from $L^p(M,\tau)$ onto $\overline{M\xi}$.
\end{enumerate}
In particular, if $\overline{M\xi}=X$ and \textup{(i)} holds, the
isomorphism onto $X$ is uniquely determined by $1_M\mapsto\xi$.
\end{proposition}

\begin{proof}
Positive spectral approximation and the polar-decomposition argument
of \cref{lem:mod-polar-and-scalar} turn \textup{(i)} into
$\|x\xi\|=\|x\|_{p,\tau}$ for every $x\in M$.
The isometric extension has range $\overline{M\xi}$ and is
module-linear by density.  The converse follows by applying the
isometry to positive finite-spectrum elements.  A module map with the
prescribed distinguished value is determined on $M$.
\end{proof}

\begin{remark}\label{rem:mod-full-direct-sum}
For a family in $\mathsf{PM}_p^0$, a sufficient condition for
\eqref{eq:mod-scalar-partition} is the direct-sum estimate
\begin{equation}\label{eq:mod-full-direct-sum}
 \|F(i)u+F(j)v\|
 \leq(\|u\|^p+\|v\|^p)^{1/p},
\end{equation}
for arbitrary vectors in $F(M)$ and $F(N)$, where
$i:M\to M\oplus N$ and $j:N\to M\oplus N$ are the summand maps.
Iterating this estimate on scalar summands and applying
$\|\xi_{\C_s}\|\leq s^{1/p}$ gives \eqref{eq:mod-scalar-partition}.  For $p>1$, \eqref{eq:mod-full-direct-sum} is strictly
stronger than the defining conditions of $\mathsf{PM}_p$.
Indeed,
\[
 F(M)=L^p(M,\tau)\oplus_\infty L^1(M,\tau),
 \qquad \xi_M=(1_M,0),
\]
with the diagonal action and induced functorial maps, belongs to
$\mathsf{PM}_p$.  Taking $M=N=\C_1$ and $u=v=(0,1)$ makes the
left side of \eqref{eq:mod-full-direct-sum} equal to $2$, while its
right side is $2^{1/p}$.  The scalar tests constrain precisely the
cyclic submodules generated by the distinguished vectors.
\end{remark}

\subsection{The initial object without the partition estimate}

For $x\in M$, define
\begin{equation}\label{eq:mod-atomic-norm}
 \alpha_{p,\tau}(x)
 =\inf\left\{
 \sum_{j=1}^n\|a_j\|_\infty\tau(e_j)^{1/p}:
 x=\sum_{j=1}^n a_je_j,\quad
 a_j\in M,\quad e_j\in\mathcal P(M)
 \right\}.
\end{equation}
The projections $e_j$ need not be orthogonal.  The one-term
decomposition $x=x1_M$ is always admissible.
Write $\mathcal A_p(M,\tau)$ for the completion in this norm.

\begin{theorem}[The scalar-bounded initial module]
\label{thm:mod-atomic-initial}
The function $\alpha_{p,\tau}$ is a norm, and
\begin{equation}\label{eq:mod-atomic-comparison}
 \|x\|_{p,\tau}\leq\alpha_{p,\tau}(x)
 \leq\tau(1_M)^{1/p}\|x\|_\infty.
\end{equation}
Left multiplication extends to a unital contractive action of $M$ on
$\mathcal A_p(M,\tau)$.  Trace-preserving embeddings extend to
contractions and make $\mathcal A_p$ a functor on $\TrEmb$.
With distinguished vector $1_M$, this functor is initial in
$\mathsf{PM}_p^0$.  For every projection $e$,
\begin{equation}\label{eq:mod-atomic-projections}
 \alpha_{p,\tau}(e)=\tau(e)^{1/p}.
\end{equation}
For $p=1$, the equality $\alpha_{1,\tau}(x)=\|x\|_{1,\tau}$ holds for
all $x\in M$.
\end{theorem}

\begin{proof}
Homogeneity and the triangle inequality follow by scaling and
concatenating decompositions.  For every summand in
\eqref{eq:mod-atomic-norm},
\[
 \|a_je_j\|_{p,\tau}
 \leq\|a_j\|_\infty\tau(e_j)^{1/p}.
\]
The triangle inequality in $L^p$ gives the lower bound in
\eqref{eq:mod-atomic-comparison}, including definiteness of the norm.
The decomposition $x=x1_M$ gives the upper bound.  Replacing $a_j$ by
$ba_j$ in a decomposition proves
\[
 \alpha_{p,\tau}(bx)\leq\|b\|_\infty\alpha_{p,\tau}(x).
\]
The module action therefore extends to the completion.  If
$\pi:(N,\sigma)\to(M,\tau)$ is an embedding, then $\|\pi(a_j)\|_\infty=\|a_j\|_\infty$
and $\tau(\pi(e_j))=\sigma(e_j)$.  Applying $\pi$ to a decomposition
of $x$ therefore gives a decomposition of $\pi(x)$ with the same sum
in \eqref{eq:mod-atomic-norm}.  Taking the infimum gives
$\alpha_{p,\tau}(\pi(x))\leq\alpha_{p,\sigma}(x)$.
Uniqueness of continuous extensions proves functoriality and
equivariance.  The identity
$\mathcal A_p(\pi)(1_N)=\pi(1_N)\cdot1_M$ proves the supported-point
condition.  The lower bound in \eqref{eq:mod-atomic-comparison} and
the single-atom decomposition $e=1_Me$ give
\eqref{eq:mod-atomic-projections}, including the required scalar bounds.

Let $(F,\xi)$ belong to $\mathsf{PM}_p^0$.  By
\cref{lem:mod-polar-and-scalar}, $\|e\xi_M\|\leq\tau(e)^{1/p}$.
Hence every atomic decomposition gives
\[
 \|x\xi_M\|
 \leq\sum_j\|a_j\|_\infty\|e_j\xi_M\|
 \leq\sum_j\|a_j\|_\infty\tau(e_j)^{1/p}.
\]
Taking the infimum shows that $x\mapsto x\xi_M$ extends to a
contraction $\mathcal A_p(M,\tau)\to F(M)$.  As in
\cref{thm:mod-scalar-testing}, equivariance and the supported-point
identity give naturality on the dense algebras, hence on their
completions.  Each component is a module map and sends $1_M$ to
$\xi_M$.  These two properties determine it on $M$, proving uniqueness.

For $p=1$, a positive finite-spectrum element
$a=\sum_j\lambda_je_j$ satisfies
$\alpha_{1,\tau}(a)\leq\sum_j\lambda_j\tau(e_j)=\|a\|_{1,\tau}$.
Positive spectral approximation is valid in the atomic norm by the
upper bound in \eqref{eq:mod-atomic-comparison}.  Finally, the
contractive module action and polar decomposition give
$\alpha_{1,\tau}(x)=\alpha_{1,\tau}(|x|)$.
This proves equality with the $L^1$ norm for all $x$.
\end{proof}

The initial morphism $\mathcal A_p\to L^p$ extends the identity on each
algebra by continuity.  On a finite scalar algebra its inverse is
bounded, and its norm can be computed explicitly for unit weights.

\begin{proposition}[Finite scalar formula and the sharp distortion]
\label{prop:mod-atomic-lorentz}
Let $D_{\mathbf s}=\bigoplus_{j=1}^n\C_{s_j}$.
Relabel coordinates so that
$a_1=|z_1|\geq\cdots\geq a_n=|z_n|$, and set
$a_{n+1}=0$, $S_0=0$, and $S_k=\sum_{j=1}^k s_j$.
Then
\begin{align}
 \alpha_{p,\mathbf s}(z)
 &=\sum_{k=1}^n(a_k-a_{k+1})S_k^{1/p}
   \label{eq:mod-finite-lorentz}\\
 &=\int_0^\infty
       \left(\sum_{\{j:|z_j|>t\}}s_j\right)^{1/p}\,dt.
       \label{eq:mod-lorentz-distribution}
\end{align}
Thus $\alpha_{p,\mathbf s}$ is the weighted $L^{p,1}$ norm, normalized
so that a characteristic function of mass $s$ has norm $s^{1/p}$.
In Lorentz's notation this is the $\Lambda(\varphi,1)$ norm with
$\varphi(u)=p^{-1}u^{1/p-1}$ \cite{Lorentz1951}.

For $p>1$, put $p'=p/(p-1)$.  On $\C^n$ with unit coordinate weights,
the norm of the identity from $\ell^p_n$ to the atomic norm is exactly
\begin{equation}\label{eq:mod-sharp-distortion}
 D_{p,n}
 =\left(\sum_{k=1}^n
       \bigl(k^{1/p}-(k-1)^{1/p}\bigr)^{p'}\right)^{1/p'}.
\end{equation}
In particular,
\begin{equation}\label{eq:mod-distortion-asymptotic}
 D_{p,n}\sim \frac1p(\log n)^{1/p'}\qquad(n\to\infty).
\end{equation}
\end{proposition}

\begin{proof}
Multiplication by a diagonal unitary is an atomic-norm isometry, so
we may assume $z_j=a_j\geq0$ in the stated order.  Let $E_k$ be the
projection onto the first $k$ coordinates.  The decomposition
$z=\sum_k(a_k-a_{k+1})E_k$ gives the upper bound in
\eqref{eq:mod-finite-lorentz}.

For the reverse bound, put
$d_k=S_k^{1/p}-S_{k-1}^{1/p}$ and define
$\ell(w)=\sum_k d_kw_k$.
For every subset $J\subseteq\{1,\ldots,n\}$, concavity of
$g(t)=t^{1/p}$ gives
\begin{equation}\label{eq:mod-subset-increments}
 \sum_{k\in J}d_k
 \leq\left(\sum_{k\in J}s_k\right)^{1/p}.
\end{equation}
Indeed, for $k\in J$ put $t_k=\sum_{j\in J,\,j<k}s_j$.  Since
$t_k\leq S_{k-1}$, concavity gives
$d_k\leq g(t_k+s_k)-g(t_k)$.  Summing over $J$ in increasing order
gives \eqref{eq:mod-subset-increments} by telescoping.
If $e_J$ is the projection corresponding to $J$, then
\[
 |\ell(be_J)|
 \leq\|b\|_\infty\sum_{k\in J}d_k
 \leq\|b\|_\infty\tau(e_J)^{1/p}.
\]
It follows from the definition of $\alpha$ that
$|\ell(w)|\leq\alpha_{p,\mathbf s}(w)$.
Summation by parts gives
$\ell(z)=\sum_k(a_k-a_{k+1})S_k^{1/p}$, proving equality.
On each nonempty interval $(a_{k+1},a_k)$ the distribution function
has value $S_k$, which gives \eqref{eq:mod-lorentz-distribution}.

For unit weights, write
$\delta_k=k^{1/p}-(k-1)^{1/p}$.
Formula \eqref{eq:mod-finite-lorentz} becomes
$\alpha_p(z)=\sum_k\delta_k a_k$.
H\"older's inequality gives the upper bound $D_{p,n}$ in
\eqref{eq:mod-sharp-distortion}.  The sequence $\delta_k$ is
decreasing, and equality is attained by the decreasing sequence
$a_k=\delta_k^{p'-1}$, up to an overall scalar.
Finally,
$\delta_k^{p'}\sim p^{-p'}k^{-1}$, so summation gives
\eqref{eq:mod-distortion-asymptotic}.
\end{proof}

\begin{corollary}\label{cor:mod-partition-necessary}
For $p>1$, the pointed $L^p$-module functor is not initial in
$\mathsf{PM}_p^0$.  There is no uniformly bounded natural family of
module maps $L^p\to\mathcal A_p$ preserving the distinguished vectors.
\end{corollary}

\begin{proof}
Every such component would be the identity on the dense algebra.
Already on $\C_1\oplus\C_1$,
\[
 \alpha_p(2,1)=1+2^{1/p}>(2^p+1)^{1/p}=\|(2,1)\|_p,
\]
where strictness follows from the strict triangle inequality in
$\ell^p_2$ for the nonproportional vectors $(1,0)$ and $(1,1)$.
Thus the required initial morphism cannot be contractive.
The constants in \eqref{eq:mod-sharp-distortion} tend to infinity,
which excludes every uniform bound.
\end{proof}

\subsection{A functorial obstruction without the module action}

We now drop the module action and retain the conditions on embeddings,
direct sums, and distinguished vectors specified below.  On $M_2(\C)$
these conditions allow maps that are linear on each commutative
subalgebra but have no common linear extension.

Let $\mathsf{PF}_p$ denote the category of pairs $(F,\xi)$, where
$F:\TrEmb\to\Banone$ is a functor and the distinguished vectors
$\xi_M\in F(M)$ satisfy
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.5em]
\item $F(\pi)\xi_N=\xi_M$ for every \emph{unital} embedding
$\pi:N\to M$;
\item $\xi_{M\oplus N}=F(i)\xi_M+F(j)\xi_N$ for the summand maps;
\item $\|\xi_M\|\leq\tau(1_M)^{1/p}$;
\item the direct-sum estimate \eqref{eq:mod-full-direct-sum} holds.
\end{enumerate}
Morphisms are point-preserving natural contractions.  The pointed
functor $(L^p,\mathbf1)$ belongs to $\mathsf{PF}_p$.

\begin{theorem}\label{thm:mod-action-obstruction}
For every $1\leq p<\infty$, there is an object $(F,\xi)$ of
$\mathsf{PF}_p$ for which no point-preserving natural transformation
from $L^p$ to $F$ has linear components.  In particular,
$(L^p,\mathbf1)$ is not initial in $\mathsf{PF}_p$.
\end{theorem}

\begin{proof}
Let $\mathcal D$ be the full subcategory consisting of the zero algebra,
if present, and the algebras isomorphic to $\C$, $\C^2$, or $M_2(\C)$,
with arbitrary faithful finite traces.  If an algebra embeds into an object of $\mathcal D$,
then it also belongs to $\mathcal D$.  Indeed, a proper nonzero
$*$-subalgebra of $M_2(\C)$ is either one-dimensional or unitarily
conjugate to the diagonal algebra.  Outside $\mathcal D$ set
$F(M)=0$ and $\xi_M=0$.  Send every arrow with target outside
$\mathcal D$ to zero.  It remains to define $F$ on $\mathcal D$.

On the commutative objects of $\mathcal D$, set $F(M)=L^p(M,\tau)$
and $\xi_M=1_M$.  To define $F$ on a two-by-two matrix algebra,
write $\tau=c\operatorname{Tr}_2$, $c>0$, and let
\[
 \mathbb S_M
 =\{a\in M:a=a^*,\ \operatorname{Tr}_2(a)=0,\ a^2=1_M\}.
\]
This is the unit sphere in the three-dimensional real space of
traceless selfadjoint matrices, for the inner product
$\frac12\operatorname{Tr}_2(ab)$.
For $n\in\mathbb S_M$, define
\[
 f_n(a)=\left(\tfrac12\operatorname{Tr}_2(na)\right)^3,
 \qquad a\in\mathbb S_M,
\]
and let $H_M$ be the complex span of the functions $f_n$ in
$C(\mathbb S_M)$.  The space $H_M$ is finite dimensional because its
elements are restrictions of homogeneous cubics in three variables.
Each $f_n$ has supremum norm one, and $f_{-n}=-f_n$.
Set
\[
 F(M)=\C\oplus H_M,
 \qquad
 \|(b,h)\|=(2c)^{1/p}\max\{|b|,\|h\|_\infty\},
 \qquad \xi_M=(1,0).
\]

Every normal $x\in M$ can be written as
$x=\lambda e+\mu(1_M-e)$ with a rank-one projection $e$.
Put $n=2e-1_M$ and define
\begin{equation}\label{eq:mod-cubic-spectral-map}
 Q_M(x)=\left(\frac{\lambda+\mu}{2},
                   \frac{\lambda-\mu}{2}f_n\right).
\end{equation}
Swapping the two spectral projections leaves this expression
unchanged, and it is independent of $e$ when $x$ is scalar.
Thus $Q_M$ is well defined on normal elements and is complex-linear
on every commutative $*$-subalgebra.  H\"older's inequality gives
\begin{align*}
 \|Q_M(x)\|
 &=\frac{(2c)^{1/p}}{2}
       \max\{|\lambda+\mu|,|\lambda-\mu|\}\\
 &\leq c^{1/p}(|\lambda|^p+|\mu|^p)^{1/p}
 =\|x\|_{p,\tau}.
\end{align*}
For $p=1$ this is simply the triangle inequality.

Between commutative objects of $\mathcal D$, use the usual $L^p$
maps.  For an embedding $\pi:N\to M$ with $N$ commutative and
$M\cong M_2(\C)$, set $F(\pi)=Q_M\circ\pi$.  The range of $\pi$ is commutative, so this
composition is linear, and the preceding estimate makes it contractive.
An embedding $\pi:N\to M$ between two-by-two matrix algebras is a
unital $*$-isomorphism and preserves the trace coefficient $c$.  It
induces an isometry $\mathbb S_N\to\mathbb S_M$.  Set
$F(\pi)(b,h)=(b,h\circ\pi^{-1})$.  Since
$f_n\circ\pi^{-1}=f_{\pi(n)}$, this action intertwines $Q_N$ and $Q_M$.
The resulting maps respect composition in $\mathcal D$.  No arrow
enters $\mathcal D$ from outside, so extension by zero gives a functor
on $\TrEmb$.

The distinguished points are preserved by unital embeddings, since
$Q_M(1_M)=(1,0)$.  They have norm at most $\tau(1_M)^{1/p}$.
If a direct sum of two nonzero algebras lies in $\mathcal D$, both
summands are scalar.  In this case additivity of the points is the
identity $(1,1)=(1,0)+(0,1)$, and the weighted $\ell^p$ norm gives
the direct-sum estimate.  For a direct sum outside $\mathcal D$ its
target space is zero, so both conditions hold there as well.
The zero algebra, if included, contributes only the zero space.
Thus $(F,\xi)$ is an object of $\mathsf{PF}_p$.

Suppose that a point-preserving natural transformation $T:L^p\to F$
had linear components.  On $\C_1$, point preservation forces
$T_{\C_1}=\mathrm{id}$.  For a rank-one projection
$e\in M_2(\C)$ with the usual trace, naturality along
$\C_1\to M_2(\C)$ gives $T_{M_2}(e)=Q_{M_2}(e)$.
Let
\[
 z=\begin{pmatrix}1&0\\0&-1\end{pmatrix},
 \qquad
 x=\begin{pmatrix}0&1\\1&0\end{pmatrix},
 \qquad n=\frac{x+z}{\sqrt2}.
\]
For $a\in\{x,z,n\}$, the projection $e=(1+a)/2$ has rank one.
Thus $T_{M_2}(a)=2Q_{M_2}(e)-\xi_{M_2}=(0,f_a)$.
Linearity would give $f_x+f_z=\sqrt2f_n$.
Evaluating this identity at $z\in\mathbb S_{M_2}$ gives $1=1/2$.
This contradiction proves the assertion.
\end{proof}

\section{Scalar and linear natural transformations}
\label{sec:linear-naturality}

A natural family is \emph{objectwise bounded} if each component is
bounded and \emph{uniformly bounded} if the component norms have a common
finite bound.  In $\TrEmb$ the total trace mass can be any $s>0$.

\subsection{Scalar naturality}

\begin{lemma}\label{lem:objectwise-scalar-naturality}
Fix $1\leq u<\infty$.  Let
\[
  \ell_{M,\tau}:L^u(M,\tau)\longrightarrow\C
\]
be a bounded linear functional at each object, and suppose that
\[
  \ell_{M,\tau}\circ L^u(\pi)=\ell_{N,\sigma}
\]
for every $\pi:(N,\sigma)\to(M,\tau)$ in $\TrEmb$.  There is a unique
$c\in\C$ such that
\[
  \ell_{M,\tau}(x)=c\,\tau(x)
  \qquad(x\in L^u(M,\tau)),
\]
where $\tau$ denotes its continuous extension to $L^u(M,\tau)$.  The
component norms are
\[
  \|\ell_{M,\tau}\|=|c|\,\tau(1_M)^{1-1/u}.
\]
\end{lemma}

\begin{proof}
For $s>0$, put $a(s)=\ell_{\C_s}(1)$.  Naturality for the diagonal map
$\C_{s+t}\to\C_s\oplus\C_t$ and the two summand embeddings gives
\[
  a(s+t)=a(s)+a(t).
\]
Let $A=L^\infty([0,1])$ with its integral trace, and for $0<s\leq1$ let
$e_s$ be the characteristic function of $[0,s]$.  The map
$\C_s\to A$, $\lambda\mapsto\lambda e_s$, is a morphism.  Hence
\[
  |a(s)|=|\ell_A(e_s)|\leq\|\ell_A\|s^{1/u}.
\]
Thus $a$ extends to a continuous additive function on $\mathbb R$, so
$a(s)=cs$ with $c=a(1)$.

For each nonzero projection $e\in M$, naturality for
$\C_{\tau(e)}\to M$, $\lambda\mapsto\lambda e$, gives
$\ell_{M,\tau}(e)=c\tau(e)$.  Since the linear span of the projections is
dense in $L^u(M,\tau)$ by \cref{prop:projection-cores}, the formula holds
on all of $L^u(M,\tau)$.  The norm now follows from
\eqref{eq:trace-functional-norm}.
\end{proof}

\begin{corollary}\label{cor:uniform-scalar-naturality}
For $p=1$, the uniformly bounded scalar natural families are precisely the
multiples $c\tau$, with uniform norm $|c|$.  They are contractive exactly
when $|c|\le1$.  For $1<p<\infty$, the zero family is the only uniformly
bounded scalar natural family.
\end{corollary}
\begin{proof}
By \cref{lem:objectwise-scalar-naturality}, the component norm at total
mass $s$ is $|c|s^{1-1/p}$.  The assertion follows by allowing $s$ to
range over all positive real numbers.
\end{proof}

\subsection{Natural maps between different exponents}

\begin{theorem}\label{thm:linear-naturality}
Fix $1\le p,q<\infty$.  Let
\[
 T_{M,\tau}:L^p(M,\tau)\longrightarrow L^q(M,\tau)
\]
be an objectwise bounded natural family of complex linear maps.
If $p\ge q$, there is a unique $c\in\C$ such that every component is
$c$ times the canonical inclusion $L^p(M,\tau)\subseteq L^q(M,\tau)$,
and
\begin{equation}\label{eq:linear-natural-component-norm}
 \|T_{M,\tau}\|=|c|\tau(1_M)^{1/q-1/p}.
\end{equation}
If $p<q$, the family is zero.

Consequently, the uniformly bounded natural families $L^p\to L^q$
are $c\,\id$ for $p=q$, with uniform norm $|c|$, and zero for
$p\ne q$.
\end{theorem}

\begin{proof}
Write $T_{\C_s}(z)=a(s)z$.  In the object $\C_s\oplus\C_t$, naturality
for the two summand embeddings gives
\[
 T_{\C_s\oplus\C_t}(1,0)=(a(s),0),\qquad
 T_{\C_s\oplus\C_t}(0,1)=(0,a(t)).
\]
Naturality for the unital diagonal embedding
$\C_{s+t}\to\C_s\oplus\C_t$ now gives
\[
 (a(s),a(t))=a(s+t)(1,1).
\]
Thus $a(s)=c$ is independent of $s>0$.  The projection embeddings
\eqref{eq:scalar-projection-embedding} imply $T_{M,\tau}(e)=ce$ for
every projection $e\in M$.  For $x\in M$, take a sequence in
$\mathcal P_{\rm lin}(M)$ converging to $x$ in operator norm
(\cref{prop:projection-cores}).  This convergence holds in both $L^p$
and $L^q$, so linearity and continuity give $T_{M,\tau}(x)=cx$.

If $p\ge q$, H\"older's inequality gives
\[
 \|x\|_{q,\tau}\le\tau(1_M)^{1/q-1/p}\|x\|_{p,\tau}.
\]
Equality holds at $1_M$.  The inclusion therefore extends from $M$ to
$L^p$, and density gives \eqref{eq:linear-natural-component-norm}.
If $p<q$, work in $A=L^\infty([0,1])$ with the integral trace and
projections $e_s$ of mass $s$.  Boundedness of $T_A$ yields
\[
 |c|s^{1/q}\le\|T_A\|s^{1/p}\qquad(0<s\le1).
\]
Letting $s\downarrow0$ forces $c=0$.  Finally, when $p>q$, the
component norms in \eqref{eq:linear-natural-component-norm} are
unbounded over arbitrary trace masses unless $c=0$.
\end{proof}

On the subcategory of unital embeddings, $x\mapsto\tau(x)1_M/\tau(1_M)$
is also a natural endomorphism.  For a proper corner embedding $\pi$,
applying naturality to the unit would equate a nonzero multiple of the
target unit with $\pi(1)$.

\subsection{A sharp estimate for approximate scalar naturality}

\begin{proposition}\label{prop:approximate-scalar-naturality}
Fix $1<p<\infty$, and let
$\ell_{M,\tau}:L^p(M,\tau)\to\C$ be an arbitrary family of bounded linear
functionals with
\[
 K:=\sup_{(M,\tau)}\|\ell_{M,\tau}\|<\infty.
\]
Define its naturality defect by
\[
 \delta:=\sup_{\pi:(N,\sigma)\to(M,\tau)}
 \|\ell_{M,\tau}\circ L^p(\pi)-\ell_{N,\sigma}\|.
\]
Then
\begin{equation}\label{eq:sharp-scalar-naturality-defect}
 K\leq\delta.
\end{equation}
The coefficient $1$ in this inequality is optimal.
\end{proposition}

\begin{proof}
Fix $(N,\sigma)$, $x\in L^p(N,\sigma)$, and a positive integer $n$.
Equip $N^{\oplus n}$ with the sum trace, and let $i_j:N\to N^{\oplus n}$
be its $j$th summand embedding.  By the definition of $\delta$,
\[
 \left|\sum_{j=1}^n\ell_{N^{\oplus n}}(L^p(i_j)x)
           -n\ell_N(x)\right|
 \leq n\delta\|x\|_p.
\]
The sum of the embedded vectors is $(x,\ldots,x)$ and has norm
$n^{1/p}\|x\|_p$.  Consequently,
\[
 |\ell_N(x)|
 \leq\bigl(Kn^{1/p-1}+\delta\bigr)\|x\|_p.
\]
Letting $n\to\infty$ gives $|\ell_N(x)|\le\delta\|x\|_p$ at every
object, which proves \eqref{eq:sharp-scalar-naturality-defect}.

To see that equality can occur, fix $\varepsilon>0$, put $p'=p/(p-1)$,
and set, at each nonzero object,
\[
 \ell_{M,\tau}
 =\varepsilon\,\tau(1_M)^{-1/p'}\tau.
\]
Each such component has norm $\varepsilon$.  For an embedding
$\pi:(N,\sigma)\to(M,\tau)$ with nonzero source, let $S=\sigma(1_N)$ and
$T=\tau(1_M)$.  Then $0<S\leq T$, and
\[
 \|\ell_{M,\tau}\circ L^p(\pi)-\ell_{N,\sigma}\|
 =\varepsilon\left(1-(S/T)^{1/p'}\right)
 \leq\varepsilon.
\]
Taking summand embeddings with $T/S\to\infty$ makes this norm tend to
$\varepsilon$.  Thus $K=\delta=\varepsilon$ for this family.
\end{proof}

\section{Natural bilinear forms and the Hilbert exponent}
\label{sec:natural-bilinear}

\subsection{Classification of bilinear forms}

\begin{definition}\label{def:natural-bilinear-family}
Let $1\leq p,q<\infty$.  A \emph{natural bilinear family} of exponents
$(p,q)$ consists of bounded complex bilinear forms
\[
 B_{M,\tau}:L^p(M,\tau)\times L^q(M,\tau)\longrightarrow\C
\]
such that
\begin{equation}\label{eq:bilinear-naturality}
 B_{M,\tau}\bigl(L^p(\pi)x,L^q(\pi)y\bigr)
 =B_{N,\sigma}(x,y)
\end{equation}
for every morphism $\pi:(N,\sigma)\to(M,\tau)$ and every
$x\in L^p(N,\sigma)$, $y\in L^q(N,\sigma)$.  Its uniform norm, possibly
infinite, is
\[
 \|B\|_{\mathrm{unif}}:=\sup_{(M,\tau)}\|B_{M,\tau}\|.
\]
\end{definition}

For $\alpha=p^{-1}+q^{-1}\leq1$, the form $(x,y)\mapsto\tau(xy)$ on
$M\times M$ has a unique bounded extension to
$L^p(M,\tau)\times L^q(M,\tau)$.  Indeed, writing $T=\tau(1_M)$,
\begin{equation}\label{eq:bilinear-holder-finite}
 |\tau(xy)|\leq T^{1-\alpha}\|x\|_{p,\tau}\|y\|_{q,\tau}.
\end{equation}
We use the same notation for this extension.  The separate trace product
$(x,y)\mapsto\tau(x)\tau(y)$ is bounded
for every $1\le p,q<\infty$, by \eqref{eq:trace-functional-norm}.

\begin{theorem}\label{thm:objectwise-bilinear-classification}
Let $B$ be a natural bilinear family of exponents $(p,q)$, and set
$\alpha=p^{-1}+q^{-1}$.
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.5em,beginpenalty=10000]
\item If $\alpha\leq1$, there are unique $c,d\in\C$ such that
\begin{equation}\label{eq:bilinear-classification-formula}
 B_{M,\tau}(x,y)
 =c\,\tau(xy)
  +d\,\tau(x)\tau(y)
\end{equation}
at every object and for all $x\in L^p(M,\tau)$,
$y\in L^q(M,\tau)$.
\item If $\alpha>1$, there is a unique $d\in\C$ such that
\[
 B_{M,\tau}(x,y)
 =d\,\tau(x)\tau(y).
\]
\end{enumerate}
Conversely, each of these formulas defines an objectwise bounded natural
bilinear family.
\end{theorem}

\begin{proof}
Work first with $x,y\in M$ and suppress $\tau$ from the notation for $B$.
Equip $M_2(M)$ with the trace $\operatorname{Tr}_2\otimes\tau$, where
$\operatorname{Tr}_2$ is unnormalized.  Write $E_{ij}(x)=e_{ij}\otimes x$ and set
\begin{align}
 C_M(x,y)&:=B_{M_2(M)}\bigl(E_{12}(x),E_{21}(y)\bigr),
       \label{eq:bilinear-C-definition}\\
 D_M(x,y)&:=B_{M_2(M)}\bigl(E_{11}(x),E_{22}(y)\bigr).
       \label{eq:bilinear-D-definition}
\end{align}
Since $|E_{ij}(x)|^u=E_{jj}(|x|^u)$, the map $x\mapsto E_{ij}(x)$
preserves the $L^u$-norm.  Hence $C_M$ and $D_M$ extend to bounded
bilinear forms on $L^p(M,\tau)\times L^q(M,\tau)$.

\emph{The matrix identities.}
For selfadjoint $H\in M_2(M)$, conjugation by $\exp(itH)$ is a
trace-preserving automorphism.  Differentiating
\eqref{eq:bilinear-naturality} at $t=0$ gives
\begin{equation}\label{eq:bilinear-infinitesimal-invariance}
 B_{M_2(M)}([H,X],Y)+B_{M_2(M)}(X,[H,Y])=0.
\end{equation}
The derivatives of the two arguments exist in operator norm, hence also
in the respective $L^p$- and $L^q$-norms.  Linearity in $H$ extends the
identity to every $H\in M_2(M)$.

The flip matrix interchanges the two coordinates.  Naturality under
conjugation by this unitary gives
\begin{align}
 B_{M_2(M)}\bigl(E_{22}(x),E_{11}(y)\bigr)&=D_M(x,y),
    \label{eq:bilinear-flip-D}\\
 B_{M_2(M)}\bigl(E_{21}(x),E_{12}(y)\bigr)&=C_M(x,y).
    \label{eq:bilinear-flip-C}
\end{align}
In \eqref{eq:bilinear-infinitesimal-invariance}, choose
\[
 H=E_{12}(a),\qquad X=E_{21}(x),\qquad Y=E_{11}(y).
\]
The commutators are
\[
 [H,X]=E_{11}(ax)-E_{22}(xa),
 \qquad [H,Y]=-E_{12}(ya).
\]
Using naturality for the diagonal corner embeddings and
\eqref{eq:bilinear-flip-D}--\eqref{eq:bilinear-flip-C}, we obtain
\begin{equation}\label{eq:bilinear-matrix-identity-one}
 B_M(ax,y)-D_M(xa,y)-C_M(x,ya)=0.
\end{equation}
Taking $a=1_M$ gives
\begin{equation}\label{eq:bilinear-C-D-decomposition}
 B_M(x,y)=C_M(x,y)+D_M(x,y).
\end{equation}
The second choice
\[
 H=E_{12}(a),\qquad X=E_{11}(x),\qquad Y=E_{21}(y)
\]
gives
\[
 [H,X]=-E_{12}(xa),
 \qquad [H,Y]=E_{11}(ay)-E_{22}(ya),
\]
and hence
\begin{equation}\label{eq:bilinear-matrix-identity-two}
 -C_M(xa,y)+B_M(x,ay)-D_M(x,ya)=0.
\end{equation}

Conjugation by $\operatorname{diag}(u,1_M)$, with $u$ unitary in $M$,
shows that $D_M(uxu^*,y)=D_M(x,y)$.  Conjugation by
$\operatorname{diag}(1_M,u)$ gives the corresponding invariance in the
second variable.  Differentiating and complexifying as above yields
\begin{equation}\label{eq:bilinear-D-separate-traces}
 D_M(ax,y)=D_M(xa,y),\qquad
 D_M(x,ay)=D_M(x,ya).
\end{equation}
Substituting \eqref{eq:bilinear-C-D-decomposition} in
\eqref{eq:bilinear-matrix-identity-one} and
\eqref{eq:bilinear-matrix-identity-two}, the $D_M$ terms cancel by
\eqref{eq:bilinear-D-separate-traces}, leaving
\begin{equation}\label{eq:bilinear-C-balancing}
 C_M(ax,y)=C_M(x,ya),\qquad
 C_M(xa,y)=C_M(x,ay).
\end{equation}

\emph{The single-trace part.}
Set $f_M(z)=C_M(1_M,z)$.  Setting $x=1_M$ in the two identities of
\eqref{eq:bilinear-C-balancing} gives
\begin{equation}\label{eq:bilinear-C-tracial-functional}
 C_M(x,y)=f_M(xy),\qquad f_M(xy)=f_M(yx).
\end{equation}
Moreover,
\begin{equation}\label{eq:bilinear-f-bound}
 |f_M(z)|\leq\|B_{M_2(M)}\|\tau(1_M)^{1/p}\|z\|_{q,\tau}.
\end{equation}
Thus $f_M$ extends to a bounded linear functional on $L^q(M,\tau)$.

If $\pi:(N,\sigma)\to(M,\tau)$ is a morphism, so is its entrywise
amplification $\pi_2:M_2(N)\to M_2(M)$.  Equations
\eqref{eq:bilinear-C-definition} and \eqref{eq:bilinear-naturality} give
\[
 C_M(\pi(x),\pi(y))=C_N(x,y).
\]
Writing $e=\pi(1_N)$ and using
\eqref{eq:bilinear-C-tracial-functional}, we obtain
\[
 f_M(\pi(z))
 =f_M(e\pi(z))
 =C_M(e,\pi(z))
 =C_N(1_N,z)
 =f_N(z).
\]
The extensions of $f_M$ therefore form an objectwise bounded scalar
natural family.  By \cref{lem:objectwise-scalar-naturality}, there is
$c\in\C$, independent of $M$, such that
\begin{equation}\label{eq:bilinear-C-final}
 C_M(x,y)=c\tau(xy)\qquad(x,y\in M).
\end{equation}

\emph{The separate-trace part.}
For $(A,\rho)$ and $(B,\sigma)$, equip $A\oplus B$ with the sum trace and set
\[
 \Delta_{A,B}(x,y)
 :=B_{A\oplus B}(x\oplus0,0\oplus y).
\]
This is a bounded bilinear form on $L^p(A,\rho)\times L^q(B,\sigma)$.
An embedding $\pi:A\to A'$ induces
$\pi\oplus\id_B:A\oplus B\to A'\oplus B$, so $\Delta$ is natural in
the first variable.  The same argument applies to the second variable.
Fix $B$ and $y\in L^q(B,\sigma)$.  By \cref{lem:objectwise-scalar-naturality},
\[
 \Delta_{A,B}(x,y)=\rho(x)\,g_B(y),
 \qquad g_B(y)=\Delta_{\C_1,B}(1,y).
\]
Since $g_B(y)=\Delta_{\C_1,B}(1,y)$, the functionals $g_B$ are bounded
and natural in $B$.  Applying \cref{lem:objectwise-scalar-naturality}
once more gives $g_B(y)=d\sigma(y)$ for a common $d\in\C$.  Thus
\begin{equation}\label{eq:bilinear-Delta-final}
 \Delta_{A,B}(x,y)=d\rho(x)\sigma(y).
\end{equation}
The diagonal block embedding
$M\oplus M\to M_2(M)$ is trace-preserving.  Naturality for this embedding
identifies $D_M$ with $\Delta_{M,M}$, so
\begin{equation}\label{eq:bilinear-D-final}
 D_M(x,y)=d\tau(x)\tau(y).
\end{equation}
Combining \eqref{eq:bilinear-C-D-decomposition},
\eqref{eq:bilinear-C-final}, and \eqref{eq:bilinear-D-final} establishes
\eqref{eq:bilinear-classification-formula} on $M\times M$.

\emph{The exponent restriction and extension.}
Let $A=L^\infty([0,1])$ carry the integral trace and let
$e_s=\mathbf 1_{[0,s]}$ for $0<s\le1$.  If $\alpha>1$, then
\[
 |cs+ds^2|=|B_A(e_s,e_s)|\leq\|B_A\|s^\alpha.
\]
Dividing by $s$ and letting $s\downarrow0$ gives $c=0$.  When $\alpha\leq1$,
\eqref{eq:bilinear-holder-finite} bounds the single-trace term.
For all $p,q$, the separate-trace term is bounded in modulus by
\[
 |d|\,\tau(1_M)^{2-\alpha}\|x\|_{p,\tau}\|y\|_{q,\tau}.
\]
Density now extends the formula to the stated product of $L^p$-spaces.

For uniqueness, evaluate on the coordinate projections of
$\C_1\oplus\C_1$.  The ordered pair of distinct projections gives $d$,
and taking both arguments equal to the first projection gives $c+d$.
Conversely, trace preservation and multiplicativity give naturality of
the indicated forms on $M\times M$, hence on the completions.
\end{proof}

\begin{corollary}\label{cor:uniform-bilinear-classification}
A uniformly bounded natural bilinear family of exponents $(p,q)$ has one
of the following forms.
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.5em]
\item If $p^{-1}+q^{-1}=1$, then
\[
 B_{M,\tau}(x,y)=c\tau(xy)
\]
for some $c\in\C$.
\item If $p=q=1$, then
\[
 B_{M,\tau}(x,y)=d\,\tau(x)\tau(y)
\]
for some $d\in\C$.
\item For all other pairs of exponents, $B=0$.
\end{enumerate}
The uniform norms in \textup{(i)} and \textup{(ii)} are $|c|$ and $|d|$,
respectively.  In every case $B_{M,\tau}$ is symmetric on $M\times M$.
\end{corollary}

\begin{proof}
Write $\alpha=p^{-1}+q^{-1}$ and $K=\|B\|_{\mathrm{unif}}$.  Evaluating
\cref{thm:objectwise-bilinear-classification} on $(1,1)$ in $\C_s$ gives
\begin{equation}\label{eq:uniform-bilinear-scaling}
 |cs+ds^2|\leq Ks^\alpha\qquad(s>0).
\end{equation}
If $\alpha<2$, divide by $s^2$ and let $s\to\infty$ to obtain $d=0$.
For $\alpha<1$, the same limit after division by $s$ gives $c=0$.
For $\alpha>1$, the objectwise classification already gives $c=0$.
Thus a single-trace term can survive only for $\alpha=1$, and a
separate-trace term only for $\alpha=2$, that is, $p=q=1$.  H\"older's
inequality gives the stated norm bounds, with equality at $(1,1)$ in $\C_1$.
\end{proof}

\begin{remark}\label{rem:bilinear-local-versus-uniform}
Conjugation in $M_2(M)$ separates the two trace terms for every
finite tracial algebra, including those that are not hyperfinite.
Uniform boundedness then restricts the exponents by varying the trace
mass on $\C$.
\end{remark}

\subsection{Natural sesquilinear forms}

\begin{corollary}\label{cor:natural-hilbert-exponent}
Fix $1\leq p<\infty$.  Suppose that each $L^p(M,\tau)$ is equipped with a
bounded sesquilinear form $H_{M,\tau}$, conjugate-linear in the first
variable, and that
\[
 H_{M,\tau}\bigl(L^p(\pi)x,L^p(\pi)y\bigr)
 =H_{N,\sigma}(x,y)
\]
for every morphism $\pi:(N,\sigma)\to(M,\tau)$.  Assume
$\sup_{(M,\tau)}\|H_{M,\tau}\|<\infty$.  Then
\[
 H_{M,\tau}(x,y)=
 \begin{cases}
 c\,\tau(x^*y),&p=2,\\[2pt]
 d\,\overline{\tau(x)}
       \tau(y),&p=1,\\[2pt]
 0,&p\notin\{1,2\},
 \end{cases}
\]
with a single coefficient independent of the object.  If every form is
positive and nondegenerate, then $p=2$ and $c>0$.  If, in addition,
$H_{\C_1}(1,1)=1$, then $c=1$.
\end{corollary}

\begin{proof}
The adjoint map on $M$ extends to a conjugate-linear isometry on every
$L^p(M,\tau)$ and commutes with the maps induced by $*$-monomorphisms.
Hence
\[
 B_{M,\tau}(x,y):=H_{M,\tau}(x^*,y)
\]
is a uniformly bounded natural complex bilinear family.  Apply
\cref{cor:uniform-bilinear-classification} with $q=p$.
Under positivity, evaluation at $(1,1)$ in $\C_1$ makes the coefficient
nonnegative.  When $p=1$, the nonzero vector $(1,-1)\in\C_1\oplus\C_1$
has trace zero and pairs to zero with every vector.  Nondegeneracy
therefore forces $p=2$ and $c>0$.  In that case $H_{\C_1}(1,1)=c$.
\end{proof}

\section{Complete isometries and conditional expectations}\label{sec:operator-spaces}

For an operator space $E$, the opposite space $E^{\rm op}$ has matrix norms
$\|[x_{ij}]\|_{M_n(E^{\rm op})}=\|[x_{ji}]\|_{M_n(E)}$.
Under the trace pairing $x\mapsto(a\mapsto\tau(ax))$, the standard
operator-space structures are given by
\begin{equation}\label{eq:standard-os}
 L^1_{\rm os}(M,\tau)=(M^{\rm op})_*,
 \qquad
 L^p_{\rm os}(M,\tau)=[M,L^1_{\rm os}(M,\tau)]_{1/p}
 \quad(1<p<\infty).
\end{equation}
At the Banach space level the pairing identifies $L^1(M,\tau)$
isometrically with $M_*$.  The $L^1$ endpoint carries the matrix norms
of the opposite predual, and the interpolation formula holds at every
matrix level \cite[Section~2, (2.7)]{JungeRuanSherman2005},
\cite[Section~9]{Pisier2014}.

\begin{remark}\label{rem:opposite}
For $M=M_n$ with the usual trace, consider the subspace of first-row
matrices.  Its endpoint spaces in \eqref{eq:standard-os} are the row
space $R_n$ and the column space $C_n$.  At $p=2$ it is the operator
Hilbert space $OH_n$.  Transporting the matrix norms of $M_*$ through
the same pairing gives $R_n$ at both endpoints
\cite[Section~9]{Pisier2014}.
\end{remark}

For the embeddings in $\TrEmb$, the conditional expectation on the image
corner gives an explicit left inverse.  For the general theory of
noncommutative $L^p$-isometries, see \cite{JungeRuanSherman2005,Sherman2006}.

\begin{proposition}\label{prop:complete-isometry}
For every $\pi:(M,\tau)\to(N,\sigma)$ in $\TrEmb$ and every
$1\le p<\infty$, $L^p(\pi)$ is a complete isometry for
\eqref{eq:standard-os}.  It has a canonical completely contractive left
inverse $R_{\pi,p}$.  If $\rho:(N,\sigma)\to(P,\omega)$ is another arrow,
then
\[
 R_{\rho\pi,p}=R_{\pi,p}R_{\rho,p}.
\]
\end{proposition}
\begin{proof}
Put $e=\pi(1)$.  The finite tracial inclusion $\pi(M)\subseteq eNe$
admits a unique normal conditional expectation $E_\pi$ preserving
$\sigma|_{eNe}$ \cite{Takesaki1972}.  Set
\[
 R_\pi(a)=\pi^{-1}E_\pi(eae),\qquad a\in N.
\]
Compression and conditional expectation are normal completely positive
maps, so $R_\pi$ is normal, unital and completely positive.  It satisfies
$R_\pi\pi=\id$.  Trace preservation and the bimodule property of $E_\pi$
\cite{Tomiyama1957} give
\begin{equation}\label{eq:trace-adjoint}
 \tau(xR_\pi(a))=\sigma(\pi(x)a)
 \qquad(x\in M,\ a\in N).
\end{equation}
At $p=\infty$ the maps are $\pi$ and $R_\pi$.
At $p=1$, under \eqref{eq:standard-os}, the embedding is
$(R_\pi^{\rm op})_*$ and its left inverse is $(\pi^{\rm op})_*$.
Both preadjoints are completely contractive.  By \eqref{eq:trace-adjoint}
they agree with $\pi$ and $R_\pi$, respectively, on $M$ and $N$.
Interpolation gives complete contractions $L^p(\pi)$ and $R_{\pi,p}$
with $R_{\pi,p}L^p(\pi)=\id$ by density.  At each matrix level this left
inverse gives the reverse norm inequality, so $L^p(\pi)$ is a complete isometry.

Applying \eqref{eq:trace-adjoint} twice gives
\[
 \tau(xR_\pi R_\rho(a))=\omega(\rho\pi(x)a).
\]
The nondegeneracy of the trace pairing identifies $R_\pi R_\rho$
with $R_{\rho\pi}$ on $P$.  Density proves the asserted identity on $L^p$.
\end{proof}

\begin{corollary}\label{cor:complete-expectation}
If $M\subseteq N$ is a unital finite tracial inclusion, its
trace-preserving conditional expectation extends to a completely
contractive $M$-bimodule projection
\[
 E_p:L^p(N,\sigma)\longrightarrow L^p(M,\sigma|_M).
\]
For a commutative square of unital inclusions $M\subseteq N$ and
$M'\subseteq N'$ with vertical arrows $\alpha:M\to M'$ and $\beta:N\to N'$,
if $E'\beta=\alpha E$, then $E'_pL^p(\beta)=L^p(\alpha)E_p$.
\end{corollary}
\begin{proof}
Apply \cref{prop:complete-isometry} to the inclusion.  The bimodule
identity extends by \eqref{eq:basic-module-bound}, and the identity for
the square extends by continuity.
\end{proof}

A commutative square of inclusions need not satisfy $E'\beta=\alpha E$.
For example, take the inclusions $\C I\subseteq M_2$ and
$D_2\subseteq M_2$ with normalized traces, where $D_2$ is the diagonal
algebra.  If $\alpha$ is the scalar inclusion and $\beta=\id_{M_2}$,
then $\alpha E(e_{11})=\tfrac12 I$, whereas $E'\beta(e_{11})=e_{11}$.

Let $\mathcal M_p^{\rm os}$ send $(M,\tau)$ to the dense subspace
$M\subseteq L^p_{\rm os}(M,\tau)$ with its induced matrix norms.
By \cref{lem:completion,prop:complete-isometry}, its objectwise
operator-space completion is $L^p_{\rm os}$.  Equivalently, every
natural complete contraction from these dense cores to a functor of
complete operator spaces extends uniquely to $L^p_{\rm os}$.

\section{Symmetric weighted embeddings and modular compatibility}
\label{sec:weighted}

For a unital inclusion $A\subseteq B$ and a faithful normal state $\varphi$
on $B$, the self-polar forms of $\varphi$ and $\varphi|_A$ agree on $A$
precisely when there is a $\varphi$-preserving conditional expectation onto $A$
\cite[Theorem~4.2]{HaagerupStormer1995}.  Junge, Ruan, and Sherman use this
criterion in their classification of $2$-isometries
\cite[proof of Theorem~4.9]{JungeRuanSherman2005}.  Takesaki characterizes
the existence of the expectation by invariance under the modular group
\cite{Takesaki1972}.  For finite-dimensional embeddings we give a direct
proof that also identifies the obstruction on the support of a nonunital
embedding.  The resulting block decomposition lets us compute the distance
to the nearest compatible density.

\subsection{The qualitative criterion}

Let
\[
 M=\bigoplus_{i=1}^r M_{n_i}(\mathbb C),
 \qquad N=\bigoplus_{j=1}^s M_{m_j}(\mathbb C)
\]
carry the sums of the unnormalized matrix traces on their summands.
Fix positive invertible densities $h\in M$ and $k\in N$.  For
$1\le p<\infty$, write
\[
 \varphi_h(x)=\Tr_M(hx),\qquad
 \varphi_k(a)=\Tr_N(ka),\qquad
 \|x\|_{p,h}=\|h^{1/(2p)}xh^{1/(2p)}\|_p.
\]
Define $\|\cdot\|_{p,k}$ on $N$ in the same way.  All unweighted
$L^p$ norms in this section use the unnormalized traces.  Let
$\pi:M\to N$ be an injective $*$-homomorphism satisfying
\begin{equation}\label{eq:weighted-preservation}
 \varphi_k\pi=\varphi_h.
\end{equation}
We allow $\Tr_N\pi\ne\Tr_M$.  Put $e=\pi(1_M)$.
The trace adjoint $\pi^\dagger:N\to M$ is characterized by
\[
 \Tr_M(x\pi^\dagger(a))=\Tr_N(\pi(x)a).
\]
It is completely positive, and \eqref{eq:weighted-preservation} is equivalent
to $\pi^\dagger(k)=h$.

\begin{lemma}\label{lem:weighted-l2-contraction}
The map $\pi:(M,\|\cdot\|_{2,h})\to(N,\|\cdot\|_{2,k})$ is a contraction.
Define
\begin{equation}\label{eq:weighted-adjoint}
 R_{h,k}(a)=h^{-1/2}\pi^\dagger(k^{1/2}ak^{1/2})h^{-1/2},
 \qquad S_{h,k}=R_{h,k}\pi.
\end{equation}
Then $R_{h,k}$ is unital and completely positive, and $S_{h,k}$ is a
positive self-adjoint contraction for the inner product
\[
 \langle x,y\rangle_h=\Tr_M(x^*h^{1/2}yh^{1/2}).
\]
In particular,
\begin{equation}\label{eq:weighted-defect-form}
 \mathcal D_\pi(x):=\|x\|_{2,h}^2-\|\pi(x)\|_{2,k}^2
 =\langle x,(I-S_{h,k})x\rangle_h\ge0,
\end{equation}
and $\pi$ preserves the weighted $L^2$ norm if and only if $S_{h,k}=I$.
\end{lemma}

\begin{proof}
The formula for $R_{h,k}$ gives complete positivity, and
$R_{h,k}(1_N)=h^{-1/2}\pi^\dagger(k)h^{-1/2}=1_M$.
The trace adjoint identity gives, for $x,y\in M$,
\[
 \langle x,S_{h,k}y\rangle_h
 =\Tr_N(\pi(x)^*k^{1/2}\pi(y)k^{1/2})
 =\langle\pi(x),\pi(y)\rangle_k.
\]
Thus $S_{h,k}$ is positive and self-adjoint on the weighted Hilbert space.
As a map on the $C^*$-algebra $M$, it is completely positive and subunital since
$S_{h,k}(1_M)=R_{h,k}(e)\le1_M$.  Its operator norm on the
$C^*$-algebra $M$ is therefore at most one, so its eigenvalues lie in
$[0,1]$.  Hence $0\le S_{h,k}\le I$ on the weighted Hilbert space.
Taking $y=x$ gives \eqref{eq:weighted-defect-form} and the criterion for
equality.
\end{proof}

After unitary changes of coordinates in the summands of $N$, every such
embedding has the form
\begin{equation}\label{eq:weighted-multiplicity}
 \mathbb C^{m_j}=
 \left(\bigoplus_{i:r_{ji}>0}\mathbb C^{n_i}\otimes\mathbb C^{r_{ji}}\right)
 \oplus H_{j,0},
 \qquad
 \pi(x)_j=\left(\bigoplus_i x_i\otimes I_{r_{ji}}\right)\oplus0.
\end{equation}
Terms with $r_{ji}=0$ are omitted throughout.  For $a\in N$, let
$a_{j,ii}$ denote its compression to
$\mathbb C^{n_i}\otimes\mathbb C^{r_{ji}}$.  The ordinary partial trace
over the second tensor factor gives
\[
 \pi^\dagger(a)_i=
 \sum_j\Tr_{r_{ji}}(a_{j,ii}).
\]

\begin{theorem}\label{thm:weighted-modular-equivalence}
Under \eqref{eq:weighted-preservation}, the following conditions are equivalent.
\begin{enumerate}
\item $\|\pi(x)\|_{2,k}=\|x\|_{2,h}$ for all $x\in M$.
\item $[k,e]=0$ and there is a $\varphi_k|_{eNe}$-preserving conditional
expectation from $eNe$ onto $\pi(M)$.
\item $[k,e]=0$ and
\[
 k^{it}\pi(x)k^{-it}=\pi(h^{it}xh^{-it})
 \qquad(x\in M,\ t\in\mathbb R).
\]
\item In \eqref{eq:weighted-multiplicity}, the density has the form
\begin{equation}\label{eq:weighted-factorization}
 k_j=\left(\bigoplus_i h_i\otimes\rho_{ji}\right)\oplus k_{j,0},
 \qquad \rho_{ji}>0,\quad k_{j,0}>0,\quad
 \sum_j\Tr(\rho_{ji})=1\quad(1\le i\le r).
\end{equation}
The complement term is omitted when $H_{j,0}=0$.
\end{enumerate}
When these conditions hold, $\|\pi(x)\|_{p,k}=\|x\|_{p,h}$ for every
$x\in M$ and $1\le p<\infty$.  The conditional expectation in
\textup{(2)} is unique and equals $\pi R$, where
$\omega_\rho(b)=\Tr(\rho b)$ and
\begin{equation}\label{eq:weighted-expectation-formula}
 R(a)_i=\sum_j
   (\mathrm{id}_{M_{n_i}}\otimes\omega_{\rho_{ji}})(a_{j,ii}),
 \qquad a\in eNe.
\end{equation}
\end{theorem}

\begin{proof}
Assume \textup{(1)}.  Since
$\Tr_M(h)=\Tr_N(ke)$, evaluating \eqref{eq:weighted-defect-form} at $1_M$
gives
\begin{equation}\label{eq:weighted-support-identity}
 0=\Tr_N(ke)-\Tr_N(ek^{1/2}ek^{1/2})
   =\|(1-e)k^{1/2}e\|_2^2.
\end{equation}
Hence $[k,e]=0$.  By \cref{lem:weighted-l2-contraction}, the restriction
$R_{h,k}:eNe\to M$ is a unital completely positive left inverse of $\pi$.
Equality in the completely positive Schwarz inequality on $\pi(M)$ implies
\[
 R_{h,k}(\pi(x)a\pi(y))=xR_{h,k}(a)y
 \qquad(x,y\in M,\ a\in eNe).
\]
Consequently $\pi R_{h,k}$ is a conditional expectation.  For $a\in eNe$,
\[
 \varphi_h(R_{h,k}(a))
 =\Tr_N(ek^{1/2}ak^{1/2})=\varphi_k(a),
\]
which proves \textup{(2)}.

Assume \textup{(2)}, let $E$ be the stated expectation, and put
$R=\pi^{-1}E$.  Bimodularity over the central projections of $M$ implies
that $R$ vanishes on blocks joining distinct source summands.
On the $(j,i)$ diagonal block, bimodularity over $M_{n_i}$ gives
\[
 R_i(x\otimes b)=x\,\omega_{ji}(b)
\]
for a positive functional $\omega_{ji}$ on $M_{r_{ji}}$.
Indeed, $R_i(I\otimes b)$ commutes with every matrix in $M_{n_i}$ and is
therefore scalar.  Unitality gives $\sum_j\omega_{ji}(I)=1$.
Write $\omega_{ji}(b)=\Tr(\rho_{ji}b)$.  Evaluating
$\varphi_hR=\varphi_k|_{eNe}$ on each block gives
$k_{j,ii}=h_i\otimes\rho_{ji}$ and zero off-diagonal blocks in $eNe$.
Since $k$ is invertible, each $\rho_{ji}$ is positive definite.  Together
with $[k,e]=0$, this proves \textup{(4)}.  The blocks of $k$ determine the
functionals $\omega_{ji}$ uniquely, so \eqref{eq:weighted-expectation-formula}
also proves uniqueness of $E$.

Taking imaginary powers in \eqref{eq:weighted-factorization} gives
\textup{(3)}.  Conversely, apply \textup{(3)} to the minimal central
projections of $M$.  Their images commute with every $k^{it}$, hence
with $k$.  On a $(j,i)$ block, differentiation at $t=0$ now gives
\[
 [\log k_{j,ii}-\log h_i\otimes I,\ x\otimes I]=0
 \qquad(x\in M_{n_i}).
\]
The commutant is $I\otimes M_{r_{ji}}$.  Exponentiation therefore yields
$k_{j,ii}=h_i\otimes\rho_{ji}$ with $\rho_{ji}>0$.
Weight preservation gives $\sum_j\Tr(\rho_{ji})=1$, proving
\textup{(4)}.

Finally, \textup{(4)} implies
\[
 k_j^{1/(2p)}\pi(x)_jk_j^{1/(2p)}
 =\left(\bigoplus_i
 h_i^{1/(2p)}x_ih_i^{1/(2p)}\otimes\rho_{ji}^{1/p}\right)\oplus0.
\]
Taking $p$th powers of Schatten norms and summing over $j,i$ gives
$\|\pi(x)\|_{p,k}^p=\|x\|_{p,h}^p$.
This includes \textup{(1)} and completes the proof.
\end{proof}

\begin{corollary}\label{cor:weighted-complete-isometry}
Under the equivalent conditions of
\cref{thm:weighted-modular-equivalence}, equip $M$ and $N$ with the
operator-space structures transported from their standard tracial
$L^p$-spaces through
\[
 J_{p,h}(x)=h^{1/(2p)}xh^{1/(2p)},
 \qquad J_{p,k}(a)=k^{1/(2p)}ak^{1/(2p)}.
\]
Then $\pi$ is a complete isometry for every $1\le p<\infty$.
\end{corollary}

\begin{proof}
In the coordinates of \eqref{eq:weighted-factorization}, the map
\[
 T_p=J_{p,k}\pi J_{p,h}^{-1}:
 L^p(M,\Tr_M)\longrightarrow L^p(N,\Tr_N)
\]
has the form
\[
 T_p(x)_j=\left(\bigoplus_i x_i\otimes\rho_{ji}^{1/p}\right)\oplus0.
\]
Consequently, for every positive integer $n$ and
$X\in L^p(M_n\otimes M)$,
\[
 \|(\id_{M_n}\otimes T_p)(X)\|_p^p
 =\sum_i\|X_i\|_p^p\sum_j\Tr(\rho_{ji})=\|X\|_p^p.
\]
By \cite[(2.8)--(2.9)]{JungeRuanSherman2005}, the corresponding
operator-space matrix norm is
\[
 \|X\|_{M_n(L^p(M))}
 =\sup_{\substack{\|a\|_{S_{2p}^n}\le1\\
                   \|b\|_{S_{2p}^n}\le1}}
 \|(a\otimes1_M)X(b\otimes1_M)\|_{L^p(M_n\otimes M)},
\]
where $S_r^n=L^r(M_n,\Tr_n)$.
Entrywise application of $T_p$ commutes with multiplication by the
scalar matrices $a$ and $b$.  The preceding norm equality therefore
gives equality of the operator-space matrix norms at every level.
\end{proof}

\begin{example}\label{ex:weighted-amplification}
For any invertible density matrix $\rho\in M_d$ with $\Tr(\rho)=1$,
the embedding $\pi:M_n\to M_n\otimes M_d$, $\pi(x)=x\otimes I_d$,
with target density $k=h\otimes\rho$, satisfies
\eqref{eq:weighted-factorization}.  Hence
\[
 \|h^{1/(2p)}xh^{1/(2p)}\otimes\rho^{1/p}\|_p
 =\|x\|_{p,h}.
\]
These embeddings scale the ordinary trace by $d$.
\end{example}

\subsection{Nearest compatible densities}

In this subsection and the next, assume $\pi$ is unital,
so that the complement spaces in \eqref{eq:weighted-multiplicity} vanish.
Fix $h$, and let $\mathcal K_{\pi,h}$ be the set of all invertible positive
$\ell\in N$ with $\pi^\dagger(\ell)=h$ that satisfy the equivalent conditions
of \cref{thm:weighted-modular-equivalence}.  In these coordinates,
\begin{equation}\label{eq:weighted-compatible-set}
 \mathcal K_{\pi,h}
 =\left\{\bigoplus_{j,i}h_i\otimes\rho_{ji}:
 \rho_{ji}>0,\quad\sum_j\Tr(\rho_{ji})=1\ \text{for every }i\right\}.
\end{equation}
Define
\[
 d_{\mathrm{mat}}(k)=\inf_{\ell\in\mathcal K_{\pi,h}}\|k-\ell\|_2,
 \qquad
 d_{\mathrm{sq}}(k)=\inf_{\ell\in\mathcal K_{\pi,h}}
                    \|k^{1/2}-\ell^{1/2}\|_2.
\]
The second distance compares positive square roots in the given matrix algebra.

Set $q=k^{1/2}$, $t_i=\Tr(h_i)$, $u_i=\Tr(h_i^2)$, and define
\begin{equation}\label{eq:weighted-projection-coefficients}
 \begin{split}
 c_{ji}&=u_i^{-1}\Tr_{n_i}[(h_i\otimes I)k_{j,ii}],\\
 b_{ji}&=t_i^{-1}\Tr_{n_i}[(h_i^{1/2}\otimes I)q_{j,ii}],
 \qquad r_i=\sum_j\Tr(b_{ji}^2).
 \end{split}
\end{equation}
Here $\Tr_{n_i}$ traces over the source factor.  The partial trace in
$\pi^\dagger$ traces over the multiplicity factor.

\begin{theorem}\label{thm:weighted-nearest-densities}
The matrices $c_{ji}$ and $b_{ji}$ are positive definite and satisfy
\[
 \sum_j\Tr(c_{ji})=1,\qquad 0<r_i\le1.
\]
The unique minimizers of $d_{\mathrm{mat}}(k)$ and $d_{\mathrm{sq}}(k)$ are,
respectively,
\begin{equation}\label{eq:weighted-two-minimizers}
 \bar k=\bigoplus_{j,i}h_i\otimes c_{ji},
 \qquad
 \widehat k=\bigoplus_{j,i}h_i\otimes\frac{b_{ji}^2}{r_i}.
\end{equation}
The minimum squared distances are
\begin{align}
 d_{\mathrm{mat}}(k)^2
 &=\|k\|_2^2-\sum_i u_i\sum_j\Tr(c_{ji}^2),
 \label{eq:weighted-density-distance}\\
 d_{\mathrm{sq}}(k)^2
 &=2\sum_i t_i(1-\sqrt{r_i}).
 \label{eq:weighted-square-root-distance}
\end{align}
Let
\[
 V_h=\left\{\bigoplus_{j,i}h_i^{1/2}\otimes z_{ji}:
 z_{ji}\in M_{r_{ji}}\right\}\subseteq N
\]
and put
\begin{equation}\label{eq:weighted-delta}
 \delta(k)=\operatorname{dist}_2(q,V_h)^2
          =\sum_i t_i(1-r_i).
\end{equation}
Then
\begin{equation}\label{eq:weighted-delta-comparison}
 \delta(k)\le d_{\mathrm{sq}}(k)^2\le2\delta(k),
\end{equation}
and $\delta(k)=0$ if and only if $k\in\mathcal K_{\pi,h}$.
\end{theorem}

\begin{proof}
The Hilbert--Schmidt orthogonal projection of $k$ onto
\[
 W_h=\left\{\bigoplus_{j,i}h_i\otimes z_{ji}:
                 z_{ji}\in M_{r_{ji}}\right\}
\]
is $\bar k$, by the defining property of the partial trace and
$\|h_i\|_2^2=u_i$.  Each coefficient is positive definite because
\[
 \Tr_{n_i}[(h_i\otimes I)k_{j,ii}]
 =\Tr_{n_i}[(h_i^{1/2}\otimes I)k_{j,ii}(h_i^{1/2}\otimes I)]>0.
\]
Moreover, $\pi^\dagger(k)=h$ gives
\[
 \sum_j\Tr(c_{ji})
 =u_i^{-1}\Tr\left(h_i\sum_j\Tr_{r_{ji}}k_{j,ii}\right)=1.
\]
Thus $\bar k\in\mathcal K_{\pi,h}$.  Since this set is contained in $W_h$,
$\bar k$ is its unique point nearest to $k$.  The Pythagorean identity
gives \eqref{eq:weighted-density-distance}.

Similarly, the orthogonal projection of $q$ onto $V_h$ is
$q_0=\bigoplus_{j,i}h_i^{1/2}\otimes b_{ji}$.
Applying the same partial trace argument to $q_{j,ii}$ and $h_i^{1/4}$
gives $b_{ji}>0$.
Let $P_{ji}$ be the projection onto the $(j,i)$ block.
Orthogonal projection of each diagonal block gives
\[
 t_i r_i\le\sum_j\|q_{j,ii}\|_2^2
 \le\sum_j\|q_jP_{ji}\|_2^2
 =\Tr_N(k\pi(1_{M_{n_i}}))=t_i.
\]
Hence $0<r_i\le1$.  Since $\Tr_N(k)=\Tr_M(h)=\sum_i t_i$,
subtracting $\|q_0\|_2^2=\sum_i t_i r_i$ from $\|q\|_2^2$ gives
\eqref{eq:weighted-delta}.

Every $\ell\in\mathcal K_{\pi,h}$ has
$\ell^{1/2}=\bigoplus_{j,i}h_i^{1/2}\otimes a_{ji}$ with $a_{ji}>0$ and
$\sum_j\Tr(a_{ji}^2)=1$.  For each $i$, the Hilbert--Schmidt
Cauchy--Schwarz inequality gives
\[
 \sum_j\Tr(b_{ji}a_{ji})\le\sqrt{r_i},
\]
with equality precisely when $a_{ji}=b_{ji}/\sqrt{r_i}$ for every $j$.
Since $\Tr(\ell)=\Tr(k)=\sum_i t_i$, minimizing
\[
 \|q-\ell^{1/2}\|_2^2
 =2\sum_i t_i-2\sum_i t_i\sum_j\Tr(b_{ji}a_{ji})
\]
gives the stated $\widehat k$ and \eqref{eq:weighted-square-root-distance}.
Finally, for $0<r\le1$,
\[
 1-r\le2(1-\sqrt r)\le2(1-r).
\]
Summing with weights $t_i$ gives \eqref{eq:weighted-delta-comparison}.
Since the minimum defining $d_{\mathrm{sq}}$ is attained, $\delta(k)=0$
exactly when $k\in\mathcal K_{\pi,h}$.
\end{proof}

\begin{example}\label{ex:weighted-sharp-square-root}
Both constants in \eqref{eq:weighted-delta-comparison} are optimal uniformly
over finite dimensions.  For $n\ge2$, take $M=M_n$, $h=I_n/n$,
$N=M_n\otimes M_n$, and $\pi(x)=x\otimes I_n$.
For $\Omega=n^{-1/2}\sum_{a=1}^n e_a\otimes e_a$ and $0<\eta<1$, set
\[
 k_\eta=(1-\eta)|\Omega\rangle\langle\Omega|+\eta I_{n^2}/n^2.
\]
These densities are faithful and have source marginal $h$.  Let
$P=|\Omega\rangle\langle\Omega|$ and $q=k_\eta^{1/2}$.  Since $q$ is a
linear combination of $P$ and $I$, and $\Tr_n(P)=I_n/n$, its partial
trace over the source is $(\Tr(q)/n)I_n$.  Thus the single coefficient
$r$ in \eqref{eq:weighted-projection-coefficients} is
$(\Tr(q)/n)^2$.  It is less than one for $0<\eta<1$ by strict
Cauchy--Schwarz, since $\Tr(q^2)=1$ and $q$ is nonscalar.
As $\eta\downarrow0$, $q$ tends to $P$ and $r$ tends to $1/n^2$.  Hence
\[
 \frac{d_{\mathrm{sq}}(k_\eta)^2}{\delta(k_\eta)}
 =\frac{2}{1+\sqrt r}\longrightarrow\frac{2n}{n+1}.
\]
Letting $n\to\infty$ proves optimality of the upper constant.  For
fixed $n$, letting $\eta\uparrow1$ gives $q\to I_{n^2}/n$ and
$r\to1$.  The ratio then tends to one.
\end{example}

\subsection{Finite tests and quantitative stability}

In each summand choose matrix units $e_{ab}^{(i)}$ with respect to an
eigenbasis of $h_i$, with eigenvalues $\lambda_{i1},\ldots,\lambda_{in_i}$.
Define the maximal relative loss of squared $L^2$ norm by
\begin{equation}\label{eq:weighted-relative-loss}
 \varepsilon(k)=\sup_{x\ne0}
 \frac{\mathcal D_\pi(x)}{\|x\|_{2,h}^2},
 \qquad
 t=\Tr(h),\qquad
 \mu_h=\min_i\frac{\lambda_{\min}(h_i)^2}{t_i}.
\end{equation}
Here $0\le\varepsilon(k)<1$ and $\mu_h>0$.  Indeed, injectivity of
$\pi$ makes the Gram operator $S_{h,k}$ positive definite.

\begin{proposition}\label{prop:weighted-finite-tests}
The losses of squared norm on the matrix units determine $\delta(k)$ by
\begin{equation}\label{eq:weighted-finite-test-identity}
 \delta(k)=\sum_i\frac1{t_i}\sum_{a,b=1}^{n_i}
 \sqrt{\lambda_{ia}\lambda_{ib}}\,
 \mathcal D_\pi(e_{ab}^{(i)}).
\end{equation}
Consequently,
\begin{equation}\label{eq:weighted-distortion-bounds}
 \frac{\delta(k)}t\le\varepsilon(k)\le\frac{\delta(k)}{\mu_h},
 \qquad
 \frac{d_{\mathrm{sq}}(k)^2}{2t}
 \le\varepsilon(k)\le\frac{d_{\mathrm{sq}}(k)^2}{\mu_h}.
\end{equation}
In particular, if $\pi$ preserves the weighted $L^2$ norm of every
$e_{ab}^{(i)}$, it preserves that norm on all of $M$.
\end{proposition}

\begin{proof}
Let $Q_{ji,a}$ denote the $(a,a)$ multiplicity block of $q_{j,ii}$ in the
chosen basis.  Matrix multiplication gives
\[
 \|\pi(e_{ab}^{(i)})\|_{2,k}^2
 =\sum_j\Tr(Q_{ji,a}Q_{ji,b}),
 \qquad
 \|e_{ab}^{(i)}\|_{2,h}^2=\sqrt{\lambda_{ia}\lambda_{ib}}.
\]
Weight the first formula by $\sqrt{\lambda_{ia}\lambda_{ib}}$ and sum
over $a,b$.  By \eqref{eq:weighted-projection-coefficients} this gives
\[
 \sum_j\Tr\!\left[\left(\sum_a\sqrt{\lambda_{ia}}Q_{ji,a}\right)^2\right]
 =t_i^2r_i.
\]
The same weighted sum for the second formula is $t_i^2$.  Subtract,
divide by $t_i$, and sum over $i$ to obtain
\eqref{eq:weighted-finite-test-identity}.

The vectors
$f_{ab}^{(i)}=(\lambda_{ia}\lambda_{ib})^{-1/4}e_{ab}^{(i)}$
form an orthonormal basis of $(M,\langle\cdot,\cdot\rangle_h)$.
By \cref{lem:weighted-l2-contraction},
$A=I-S_{h,k}$ is positive, with $\|A\|=\varepsilon(k)$, and
\[
 \delta(k)=\sum_{i,a,b}\frac{\lambda_{ia}\lambda_{ib}}{t_i}
           \langle f_{ab}^{(i)},Af_{ab}^{(i)}\rangle_h.
\]
The coefficients have sum $t$ and are at least $\mu_h$.
It follows that
$\mu_h\|A\|\le\mu_h\operatorname{tr}(A)\le\delta(k)\le t\|A\|$,
where $\operatorname{tr}$ is the trace of a linear operator on this
finite-dimensional Hilbert space.  The remaining inequalities follow from
\eqref{eq:weighted-delta-comparison}.
\end{proof}

\begin{theorem}\label{thm:weighted-density-stability}
Let $m=\lambda_{\min}(k)$, $M_0=\|k\|_\infty$, and
$H_0=\max_i\|h_i\|_\infty$.  Then
\begin{equation}\label{eq:weighted-two-distance-bounds}
 \sqrt m\,d_{\mathrm{sq}}(k)
 \le d_{\mathrm{mat}}(k)
 \le(\sqrt{M_0}+\sqrt{H_0})d_{\mathrm{sq}}(k).
\end{equation}
In particular,
\begin{equation}\label{eq:weighted-density-delta-bounds}
 \sqrt{m\delta(k)}\le d_{\mathrm{mat}}(k)
 \le\sqrt2(\sqrt{M_0}+\sqrt{H_0})\sqrt{\delta(k)},
\end{equation}
and
\begin{equation}\label{eq:weighted-density-distortion-bounds}
 \frac{d_{\mathrm{mat}}(k)^2}
 {2t(\sqrt{M_0}+\sqrt{H_0})^2}
 \le\varepsilon(k)
 \le\frac{d_{\mathrm{mat}}(k)^2}{m\mu_h}.
\end{equation}
\end{theorem}

\begin{proof}
For any positive $\ell$, write $X=k^{1/2}-\ell^{1/2}$.  The identity
\[
 k-\ell=k^{1/2}X+X\ell^{1/2}
\]
and the Hilbert--Schmidt inner product give
\[
 \sqrt m\|X\|_2^2
 \le\Tr(X^*k^{1/2}X)+\Tr(X^*X\ell^{1/2})
 \le\|X\|_2\|k-\ell\|_2.
\]
Taking the infimum over $\mathcal K_{\pi,h}$ proves the lower bound.
For the upper bound, take $\ell=\widehat k$ from
\cref{thm:weighted-nearest-densities}.  Its multiplicity densities have
total trace one for each $i$, so $\|\widehat k\|_\infty\le H_0$.
The same algebraic identity yields
\[
 d_{\mathrm{mat}}(k)\le\|k-\widehat k\|_2
 \le(\sqrt{M_0}+\sqrt{H_0})\|k^{1/2}-\widehat k^{1/2}\|_2.
\]
Equations \eqref{eq:weighted-delta-comparison} and
\eqref{eq:weighted-distortion-bounds} give the remaining estimates.
\end{proof}

For $M=M_n$ and $h=I_n/n$, the bounds in
\eqref{eq:weighted-distortion-bounds} reduce to
$\delta(k)\le\varepsilon(k)\le n^2\delta(k)$.
Here the factor $n^2$ is the reciprocal of the least coefficient in
the weighted sum for $\delta(k)$ over the orthonormal matrix units.

\subsection{The support defect for nonunital embeddings}

For a possibly nonunital $\pi$, testing the unit measures the part of
$k^{1/2}$ that couples $e=\pi(1_M)$ to its complement.

\begin{proposition}\label{prop:weighted-support-stability}
Put $\Delta_e=\mathcal D_\pi(1_M)$.  Then
\begin{equation}\label{eq:weighted-support-defect}
 \Delta_e=\|(1-e)k^{1/2}e\|_2^2
         =\tfrac12\|[k^{1/2},e]\|_2^2.
\end{equation}
The pinched density
\[
 k_e=eke+(1-e)k(1-e)
\]
is positive definite, has the same trace as $k$, and satisfies
$\pi^\dagger(k_e)=h$.  It is the unique Hilbert--Schmidt nearest density
commuting with $e$, and
\begin{equation}\label{eq:weighted-pinching-distance}
 \|k-k_e\|_2^2=\|[k,e]\|_2^2.
\end{equation}
If $mI\le k\le M_0I$, then
\begin{equation}\label{eq:weighted-support-spectral-bounds}
 8m\Delta_e\le\|k-k_e\|_2^2\le8M_0\Delta_e.
\end{equation}
\end{proposition}

\begin{proof}
Subtracting $\Tr_N(ek^{1/2}ek^{1/2})$ from $\Tr_N(ke)$, as in
\eqref{eq:weighted-support-identity}, gives the first equality in
\eqref{eq:weighted-support-defect}.
The two off-diagonal blocks of $[k^{1/2},e]$ are adjoints up to sign, giving
the second equality.  Pinching is the Hilbert--Schmidt orthogonal projection
onto the algebra commuting with $e$.  It preserves lower spectral bounds
and the trace.  Since $\pi(x)=e\pi(x)e$, replacing $k$ by $k_e$ also
preserves the functional on $\pi(M)$.  The two off-diagonal blocks of
$k-k_e$ agree up to sign with those of $[k,e]$, proving
\eqref{eq:weighted-pinching-distance}.

In an eigenbasis of $k$, with eigenvalues $\kappa_a$, the entries of the
two commutators satisfy
\[
 [k,e]_{ab}=(\sqrt{\kappa_a}+\sqrt{\kappa_b})
            [k^{1/2},e]_{ab}.
\]
Squaring, summing, and using
$2\sqrt m\le\sqrt{\kappa_a}+\sqrt{\kappa_b}\le2\sqrt{M_0}$
proves \eqref{eq:weighted-support-spectral-bounds}.
\end{proof}

The embedding $M\to eNe$ is unital, with target density $eke$.
Let $\ell_e$ be the nearest compatible density to $eke$ given by
\eqref{eq:weighted-two-minimizers} for the matrix distance in $eNe$.
By \eqref{eq:weighted-factorization}, a compatible density on $N$ may
have any positive definite complementary block.  The off-diagonal
part $k-k_e$ is Hilbert--Schmidt orthogonal to both diagonal corners.
Thus the unique nearest compatible density on $N$ is
$\ell_e+(1-e)k(1-e)$, and its squared distance from $k$ is
$\|k-k_e\|_2^2+\|eke-\ell_e\|_2^2$.

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\end{thebibliography}

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