text:
\begin{document}
\title[Maximal Tracial Algebras]{Maximal Tracial Algebras}
\author[D. Hadwin \MakeLowercase{and} H. Yousefi]{Don Hadwin$^{1}$ \MakeLowercase{and} Hassan Yousefi$^{2\ast}$}
\address{$^{1}$Department of Mathematics \& Statistics, University of New Hampshire, 33
Academic Way, Durham, NH 03824\\
$^{2}$Department of Mathematics, California State University Fullerton, CA
92831, USA.}
\email{\textcolor[rgb]{0.00,0.00,0.84}{hyousefi@fullerton.edu}}
\subjclass[2010]{Primary 46L05; Secondary 47L10, 46M07, 46L10.}
\keywords{Tracial Algebras, Dual Pairs, Multiplier Pairs, Tracial Ultraproducts.}
\date{Received: xxxxxx; Revised: yyyyyy; Accepted: zzzzzz. }
\date{\indent$^{\ast}$Corresponding author}
\date{Received: xxxxxx; Revised: yyyyyy; Accepted: zzzzzz. }
\date{\indent$^{\ast}$Corresponding author}

\begin{abstract}
We introduce and study the notion of \emph{maximal tracial algebras.} We prove
several results in a general setting based on dual pairs and multiplier pairs.
In a special case that $X$ is a Banach space we determine the abelian
subalgebras of $\mathcal{B}\left(  X\right)  $ that are maximal tracial for
rank-one tensors. In another special case that $\mathcal{H\ }$is a Hilbert
space we show that a unital weak-operator closed subalgebra $\mathcal{A}$ of
$\mathcal{B}\left(  \mathcal{H}\right)  $ is abelian and transitive if and
only if it is maximal $e\otimes e$-tracial for every unit vector $e$ in
$\mathcal{H}$. We also make slight connections between our ideas and the
Kadison Similarity Problem and also the Connes' Embedding Problem.

\end{abstract}
\maketitle

\setcounter{page}{1}

\let\thefootnote\relax\footnote{Copyright 2016 by the Tusi Mathematical
Research Group.}

\section{Introduction and Definitions}

Algebras with traces have played an important role in linear algebra and the
theory of operator algebras \cite{Russo}, \cite{Cuntz}, \cite{Kad2}. We
consider a situation where $\mathcal{A}$ is a unital subalgebra of an algebra
$\mathcal{B}$ and $\varphi$ is a functional on $\mathcal{B}$ with
$\varphi\left(  1\right)  =1$ that is tracial on $\mathcal{A}$. We study the
case when $\mathcal{A}$ is a maximal such algebra for $\varphi.$ For the case
when $\mathcal{B}$ is the algebra of $2\times2$ matrices over an arbitrary
field, we completely characterize the pairs $\left(  \mathcal{A}
,\varphi\right)  $ so that $\mathcal{A}$ is maximal $\varphi$-tracial. In
particular we show some connections between our ideas in this paper and two
very famous problems in Functional Analysis, namely the \emph{Kadison
Similarity Problem} \cite{Kad} and the \emph{Connes' Embedding Problem}
\cite{Connes}.

Throughout this paper $\mathcal{H}$ is a separable Hilbert space over the
field of complex numbers $\mathbb{C}$ and $\mathcal{B}\left(  \mathcal{H}
\right)  $ denotes the set of all bounded operators on $\mathcal{H}.$ We show
the set of all $k\times k$ matrices over a field $\mathbb{F}$ by $M_{k}\left(
\mathbb{F}\right)  $ and for $A\in M_{k}\left(  \mathbb{F}\right)  $ we use
$Tr\left(  A\right)  $ to denote the usual trace of $A$. The identity matrix
in $M_{k}\left(  \mathbb{F}\right)  $ is shown by $I_{k}$ and by $e_{ij}$
($1\leq i,j\leq k$) we mean the standard matrix units in $M_{k}\left(
\mathbb{F}\right)  $, i.e. the $k\times k$ matrix whose all entries are $0$'s
except the $\left(  i,j\right)  $-entry which is $1$. For $x,y\in\mathcal{H}$,
the rank-one operator $x\otimes y$ is defined by $\left(  x\otimes y\right)
h=\left\langle h,y\right\rangle x$. Note that if $A,B\in\mathcal{B}\left(
\mathcal{H}\right)  $, then $A\left(  x\otimes y\right)  B=Ax\otimes B^{\ast
}y.$ If $\mathcal{B}$ is an algebra over a field $\mathbb{F}$ and
$\mathcal{S}\subseteq\mathcal{B}$, then by the commutant $\mathcal{S}^{\prime
}$ of $\mathcal{S}$ in $\mathcal{B}$ we mean $\mathcal{S}^{\prime}=\left\{
B\in\mathcal{B}:BS=SB,\forall S\in\mathcal{S}\right\}  .$ If the algebra
$\mathcal{B}$ is unital, then we show its unit by $1.$

Section one of this paper contains some definitions, notations, terminologies,
and a motivating example.

Most of our main results are in section two where we prove several theorems in
very general cases for \emph{dual pairs}. Theorem \ref{10} is one of the main
results of this section in which we characterize when an abelian algebra is
maximal $\varphi$-tracial provided that $\varphi$ is a rank-one functional.
This result, for instance, is used in a special case in Corollary \ref{25} to
determine which abelian subalgebras of $\mathcal{B}\left(  X\right)  $ are
maximal tracial for rank-one tensors $x\otimes\alpha,$ where $X$ is a Banach
space, $x\in X,$ and $\alpha$ is a unital linear functional in the normed dual
of $X.$ In this section, we also show that a unital weak-operator closed
subalgebra $\mathcal{A}$ of $\mathcal{B}\left(  \mathcal{H}\right)  $ is
maximal $e\otimes e$-tracial for every unit vector $e$ in $\mathcal{H}$ if and
only if $\mathcal{A}$ is abelian and transitive. We also provide several
interesting examples in this section.

Section three deals with Multiplier Pairs. We prove a general statement in
Theorem \ref{35} about multiplier pairs that gives a lot of examples of
maximal tracial algebras. For instance we use Theorem \ref{35} in section four
where maximal tracial algebras of von Neumann algebras are studied.

In section four, after proving a few theorems, we raise the following
question: If a von Neumann algebra $\mathcal{M}$ on a Hilbert space
$\mathcal{H}$ is $\left(  e\otimes f\right)  $-tracial, where $e,f\in
\mathcal{H}$ are cyclic vectors for $\mathcal{M}$ and $\left\langle
e,f\right\rangle =1$, then should $\mathcal{M}$ be maximal $\left(  e\otimes
f\right)  $-tracial? For the case where $\mathcal{H}$ is separable, we will
reduce the problem to the case where $\mathcal{M}$ is a finite factor von
Neumann algebra, and such a factor has a unique norm continuous unital tracial
functional. We will also show a slight relationship between some of our ideas
in this section and the Kadison Similarity Problem \cite{Kad}.

In the last section of our paper, by introducing a new type of ultraproduct,
we make a connection between some of our results in this paper and the Connes'
Embedding Problem \cite{Connes}. We will show in Theorem \ref{60} that the
analogue of Connes' embedding problem has an affirmative answer in the setting
of maximal tracial ultraproducts.

\begin{definition}
Suppose $\mathcal{A}$ and $\mathcal{B}$ are unital algebras over a field
$\mathbb{F}$ with $1\in\mathcal{A}\subset\mathcal{B}$ and suppose
$\varphi:\mathcal{B}\rightarrow\mathbb{F}$ is a linear map such that
$\varphi\left(  1\right)  =1$. We say that $\mathcal{A}$ is a \emph{tracial
algebra} for $\varphi$ or $\mathcal{A}$ is $\varphi$\emph{-tracial }(or
$\varphi$ is a \emph{tracial functional for }$\mathcal{A}$) if, for every
$x,y\in\mathcal{A}$,
\[
\varphi\left(  xy\right)  =\varphi\left(  yx\right)  \text{ .}
\]
We say that $\mathcal{A}$ is a \emph{maximal tracial algebra} for $\varphi$ in
$\mathcal{B}$ if $\mathcal{A}$ is tracial for $\varphi$ and no larger algebra
of $\mathcal{A}$ is tracial for $\varphi$.
\end{definition}

The following is our first simple, yet useful, result. If $\mathcal{B}$ is
unital algebra over a Hausdorff field $\mathbb{F}$, then by an \emph{algebra
topology} on $\mathcal{B}$ we mean a topology for which addition, scalar
multiplication and the multiplication in each variable are all continuous.

\begin{proposition}
Suppose $1\in\mathcal{B}$ is an algebra over a field $\mathbb{F}$,
$\varphi:\mathcal{B}\rightarrow\mathbb{F}$ is linear. Then

\begin{enumerate}
\item If $\mathcal{A}\subset\mathcal{B}$ is a $\varphi$-tracial algebra, then
$\mathcal{A+}\mathbb{F}1$ is also $\varphi$-tracial.

\item If $\left\{  \mathcal{A}_{i}:i\in I\right\}  $ is an increasingly
directed family of $\varphi$-tracial algebras, then so is $\bigcup_{i\in
I}\mathcal{A}_{i}.$

\item Every $\varphi$-tracial algebra is contained in a maximal $\varphi
$-tracial algebra.

\item Every maximal $\varphi$-tracial algebra is unital.

\item If $\mathbb{F}$ is a Hausdorff field, $\mathcal{A}$ is $\varphi
$-tracial, and $\varphi$ is continuous with respect to some Hausdorff algebra
topology $\mathcal{T}$ on $\mathcal{B}$ , then $\mathcal{A}^{-\mathcal{T}}$ is
also $\varphi$-tracial.

\item If $\mathbb{F}$ is a Hausdorff field, $\mathcal{A}$ is $\varphi
$-tracial, and $\varphi$ is continuous with respect to some Hausdorff algebra
topology $\mathcal{T}$ on $\mathcal{B}$, then every maximal $\varphi$-tracial
algebra in $\mathcal{B\ }$is $\mathcal{T}$-closed.
\end{enumerate}
\end{proposition}

\begin{proof}
$\left(  1\right)  .$ This follows from the fact that for every $a,b\in
\mathcal{A}$ and every $\alpha,\beta\in\mathbb{F}$ we have
\[
\left(  a+\alpha1\right)  \left(  b+\beta1\right)  -\left(  b+\beta1\right)
\left(  a+\alpha1\right)  =ab-ba.
\]

$\left(  2\right)  $-$\left(  4\right)  $. These are trivial.

$\left(  5\right)  .$ Suppose $a,b\in\mathcal{A}^{-\mathcal{T}}$. Then there
are nets $\left\{  a_{n}\right\}  ,$ $\left\{  b_{m}\right\}  $ in
$\mathcal{A}$ such that $a_{n}\rightarrow a$ and $b_{m}\rightarrow b$ with
respect to $\mathcal{T}$. Thus, for each $m$,
\[
a_{n}b_{m}\rightarrow ab_{m}\text{ and }b_{m}a_{n}\rightarrow b_{m}a.
\]
Thus
\[
\varphi\left(  b_{m}a\right)  =\lim_{n}\varphi\left(  b_{m}a_{n}\right)
=\lim_{n}\varphi\left(  a_{n}b_{m}\right)  =\varphi\left(  ab_{m}\right)  .
\]
Similarly,
\[
\varphi\left(  ba\right)  =\lim_{m}\varphi\left(  b_{m}a\right)  =\lim
_{m}\varphi\left(  ab_{m}\right)  =\varphi\left(  ab\right)
\]

$\left(  6\right)  .$ This easily follows from $\left(  5\right)  $.
\end{proof}

We now look at a simple matrix example which was the motivating example for
this paper.

\begin{example}
\label{matrix}Suppose $\mathbb{F}$ is any field. A linear functional
$\varphi:M_{2}\left(  \mathbb{F}\right)  \rightarrow\mathbb{F}$ is given by a
matrix $K\in M_{2}\left(  \mathbb{F}\right)  $ and is defined by
$\varphi\left(  T\right)  =Tr\left(  TK\right)  ,$ where $Tr$ is the usual
trace. In this case we write $\varphi=\varphi_{K}$. Here
\[
K=\left(
\begin{array}
[c]{cc}
\varphi\left(  e_{11}\right)  & \varphi\left(  e_{21}\right) \\
\varphi\left(  e_{12}\right)  & \varphi\left(  e_{22}\right)
\end{array}
\right)  ,
\]
where $e_{ij}$ ($1\leq i,j\leq2$) are the standard matrix units for
$M_{2}\left(  \mathbb{F}\right)  $. The condition $\varphi_{K}\left(
1\right)  =1$ implies $Tr\left(  K\right)  =1.$ Also, if $\mathcal{A}$ is a
unital algebra in $M_{2}\left(  \mathbb{F}\right)  $ and $S\in M_{2}\left(
\mathbb{F}\right)  $ is invertible, then $\varphi_{K}$ is tracial for
$S\mathcal{A}S^{-1}$ if and only if $\varphi_{S^{-1}KS}$ is tracial for
$\mathcal{A}$. Hence the problem of finding the tracial unital functionals for
$\mathcal{A}$ is the same as for $S\mathcal{A}S^{-1}$.

For a given unital subalgebra $\mathcal{A}$ of $M_{2}\left(  \mathbb{F}
\right)  $, we want to find all functionals $\varphi$ for which $\mathcal{A}$
is maximal $\varphi\,$-tracial.

If $\dim\left(  \mathcal{A}\right)  =1$, then $\mathcal{A}$ is abelian, but
not maximal abelian, so $\mathcal{A}$ is not maximal tracial for any functional.

If $\dim\left(  \mathcal{A}\right)  =4$, then the only tracial functional on
$\mathcal{A}$ is $\lambda Tr\left(  {}\right)  $ for some $\lambda$ in
$\mathbb{F}$. If the characteristic of $\mathbb{F}$ is $2$, then $\lambda
Tr\left(  1\right)  =0\neq1$, and no unital $\varphi$ exists, and if the
characteristic of $\mathbb{F}$ is not $2,$ then $\varphi=\frac{1}{2}Tr\left(
{}\right)  $ is the only tracial unital functional on $\mathcal{A}$.

Next, suppose $\dim\left(  \mathcal{A}\right)  =2$. Then there is a matrix
$T\notin\mathbb{F}I_{2}$ such that $\mathcal{A}=\mathbb{F}I_{2}+\mathbb{F}T,$
that is, the algebra of polynomials in $T.$ We want to find the maximal
$\varphi$-tracial algebras in $M_{2}\left(  \mathbb{F}\right)  $. Let
$p\left(  \lambda\right)  =\lambda^{2}-Tr\left(  T\right)  \lambda+\det\left(
T\right)  $ be the characteristic polynomial of $T$.

If $p\left(  T\right)  $ splits, then either $\mathcal{A}$ is similar to
\[
\mathcal{D}_{2}=\left\{  \left(
\begin{array}
[c]{cc}
a & 0\\
0 & b
\end{array}
\right)  :a,b\in\mathbb{F}\right\}  \text{ or }\mathcal{T}_{2}=\left\{
\left(
\begin{array}
[c]{cc}
a & b\\
0 & a
\end{array}
\right)  :a,b\in\mathbb{F}\right\}  .
\]
If $\varphi\neq\frac{1}{2}Tr\left(  {}\right)  $, then $\mathcal{D}_{2}$ or
$\mathcal{T}_{2}$ is maximal $\varphi$-tracial only if there is a
$3$-dimensional algebra $\mathcal{D}$ that contains $\mathcal{D}_{2}$ or
$\mathcal{T}_{2}$ for which $\varphi$ is tracial. The only $3$-dimensional
algebras containing $\mathcal{D}_{2}$ are
\[
\mathcal{U}_{2}=\left\{  \left(
\begin{array}
[c]{cc}
a & b\\
0 & c
\end{array}
\right)  :a,b,c\in\mathbb{F}\right\}  \text{ or }\mathcal{L}_{2}=\left\{
\left(
\begin{array}
[c]{cc}
a & 0\\
b & c
\end{array}
\right)  :a,b\in\mathbb{F}\right\}  .
\]
Since $e_{11}e_{12}=e_{12}$ and $e_{12}e_{11}=0,$ then a functional $\varphi$
is tracial on $\mathcal{U}_{2}$ if and only if $\varphi\left(  e_{12}\right)
=0.$ Similarly, a functional $\varphi$ is tracial on $\mathcal{L}_{2}$ if and
only if $\varphi\left(  e_{21}\right)  =0.$ Thus $\mathcal{D}_{2}$ is maximal
tracial for $\varphi$ if and only if $\varphi\left(  e_{12}\right)  \neq0$ and
$\varphi\left(  e_{21}\right)  \neq0$. Also the only $3$-dimensional algebra
containing $\mathcal{T}_{2}$ is $\mathcal{U}_{2}$, so $\mathcal{T}_{2}$ is
maximal $\varphi$-tracial if and only if $\varphi\left(  e_{12}\right)
\neq0.$

If $p\left(  \lambda\right)  $ doesn't split, then $\mathcal{A}$ is isomorphic
to $\mathbb{F}\left[  x\right]  /\left\langle p\left(  x\right)  \right\rangle
$, which is a field. If $\mathcal{D}$ is an algebra and $\mathcal{A}
\varsubsetneq\mathcal{D}$, then
\[
\dim_{\mathbb{F}}\left(  \mathcal{D}\right)  =2\dim_{\mathcal{A}}\left(
\mathcal{D}\right)  \geq4.
\]
Hence $\mathcal{A}$ is a maximal subalgebra of $M_{2}\left(  \mathbb{F}
\right)  $. Thus if $\varphi\neq\frac{1}{2}Tr\left(  {}\right)  $, then
$\mathcal{A}$ is maximal $\varphi$-tracial.

If $\dim\left(  \mathcal{A}\right)  =3,$ then the characteristic polynomial of
every $T\in\mathcal{A}$ splits, and $\mathcal{A}$ must be similar to
$\mathcal{U}_{2}$, which is maximal $\varphi$-tracial if and only if
$\varphi\neq\frac{1}{2}Tr\left(  {}\right)  $ and $\varphi\left(
e_{12}\right)  \neq0.$
\end{example}

\section{General Dual Pairs}

Dual pairs over a Hausdorff field were studied in \cite{GenRef}. Suppose that
$\mathbb{F}$ is a topological field with a Hausdorff topology, e.g., a
subfield of the complex numbers with the usual topology, or an arbitrary field
with the discrete topology. Next suppose that $X$ is a vector space over
$\mathbb{F}$ and $Y$ is a vector space of linear maps from $X$ to $\mathbb{F}$
such that $\cap\left\{  \ker f:f\in Y\right\}  =\left\{  0\right\}  $ (that
is, $Y$ separates the points of $X$). Such a pair $(X,Y)$ is called a
\emph{dual pair} over $\mathbb{F}$. In \cite{GenRef} dual pairs were used to
unify some of the results on different versions of reflexivity for algebras
and linear spaces of $\mathcal{B}\left(  \mathcal{H}\right)  $. In this
section, for a dual pair $(X,Y)$, we will prove some theorems for continuous
linear maps on $X$ and $Y$ that will provide several important results in
special cases of dual pairs.

\begin{definition}
Let $X$ be a vector space over a field $\mathbb{F}$, $X^{\prime}$ be the dual
space of $X,$ and let $L\left(  X\right)  $ be the set of linear maps on $X.$
For a linear map $\alpha:X\rightarrow\mathbb{F}$ and $x_{0}\in X$ we define
the linear maps $x_{0}\otimes\alpha:L\left(  X\right)  \rightarrow\mathbb{F}$
and $\alpha\otimes x_{0}:L\left(  X^{\prime}\right)  \rightarrow\mathbb{F}$ by
\[
\left(  x_{0}\otimes\alpha\right)  \left(  T\right)  =\alpha\left(  T\left(
x_{0}\right)  \right)  \text{ and }\left(  \alpha\otimes x_{0}\right)  \left(
S\right)  =S\left(  \alpha\right)  \left(  x_{0}\right)  \text{.}
\]
For a dual pair $(X,Y)$ over a Hausdorff field $\mathbb{F}$ we let
$\sigma\left(  X,Y\right)  $-topology on $X$ be the smallest topology on $X$
that makes all of the maps in $Y$ continuous on $X$, and we let the
$\sigma\left(  Y,X\right)  $-topology on $Y$ be the smallest topology for
which the map $f\rightarrow f\left(  x\right)  $ is continuous on $Y$ for each
$x$ in $X$. For $A\subseteq X$ and $B\subseteq Y$, we define
\[
A^{\bot}=\left\{  f\in Y:f|_{A}=0\right\}  \text{ and }B_{\bot}=\cap\left\{
\ker\left(  f\right)  :f\in B\right\}  .
\]
For a subset $D$ of either $X$ or $Y$, we denote the linear span of $D$ by
$sp(D).$
\end{definition}

It was shown in \cite[Proposition 1.1]{GenRef} that
\[
\left(  A^{\perp}\right)  _{\perp}=sp\left(  A\right)  ^{-\sigma\left(
X,Y\right)  }\text{ and }\left(  B_{\perp}\right)  ^{\perp}=sp\left(
B\right)  ^{-\sigma\left(  Y,X\right)  }
\]
always hold. In particular, if $0\neq x\in X$, there is an $f\in Y$ such that
$f\left(  x\right)  =1$. It was also shown in \cite[Proposition 1.1]{GenRef}
that if $f:X\rightarrow\mathbb{F}$ is linear and $\sigma\left(  X,Y\right)  $
continuous, then $f\in Y$.

Suppose $\left(  X,Y\right)  $ is a dual pair over a Hausdorff field
$\mathbb{F}$. Let $L_{\sigma\left(  X,Y\right)  }\left(  X\right)  $ denote
the linear transformations $T:X\rightarrow X$ that are $\sigma\left(
X,Y\right)  $-$\sigma\left(  X,Y\right)  $ continuous. We give $L\left(
X\right)  $ the weak operator topology (WOT) defined as the topology of
pointwise $\sigma\left(  X,Y\right)  $ convergence. This means that the weak
operator topology is the weak topology on $L_{\sigma\left(  X,Y\right)
}\left(  X\right)  $ induced by the set of linear functionals $\mathcal{E}
=\left\{  x\otimes y:x\in X,y\in Y\right\}  $. Consequently, if $\mathcal{W}
=sp\left(  \mathcal{E}\right)  $, then $\left(  L_{\sigma\left(  X,Y\right)
}\left(  X\right)  ,\mathcal{W}\right)  $ is a dual pair and the weak operator
topology is the $\sigma\left(  L_{\sigma\left(  X,Y\right)  }\left(  X\right)
,\mathcal{W}\right)  $ topology. Thus, by \cite[Proposition 1.1]{GenRef}, a
linear map $\varphi:L_{\sigma\left(  X,Y\right)  }\left(  X\right)
\rightarrow\mathbb{F}$ is WOT continuous if and only if it is in $\mathcal{W}
$, i.e., of the form $\varphi=\sum_{k=1}^{n}x_{k}\otimes\alpha_{k}$ for
$x_{1},\ldots,x_{n}\in X$ and $\alpha_{1},\ldots,\alpha_{n}\in Y$.

If we denote the duality of $X$ and $Y$ with the notation
\[
\alpha\left(  x\right)  =\left\langle x,\alpha\right\rangle
\]
for $x\in X$ and $\alpha\in Y$, then, for every $T\in L_{\sigma\left(
X,Y\right)  }\left(  X\right)  $, there is a $T^{\#}\in L_{\sigma\left(
Y,X\right)  }\left(  Y\right)  $ such that, for every $x\in X$ and every
$\alpha\in Y$,
\[
\left\langle Tx,\alpha\right\rangle =\left\langle x,T^{\#}\alpha\right\rangle
\text{.}
\]
Clearly, $T^{\#\#}=T,$ $1^{\#}=1,$ $\left(  ST\right)  ^{\#}=T^{\#}S^{\#},$
$\left(  \lambda S+T\right)  ^{\#}=\lambda S^{\#}+T^{\#}$. If $\mathcal{S}
\subset L_{\sigma\left(  X,Y\right)  }\left(  X\right)  $, we let
$\mathcal{S}^{\#}$ denote $\left\{  S^{\#}:S\in\mathcal{S}\right\}  $. If
$\varphi$ is a WOT continuous linear functional on $L_{\sigma\left(
X,Y\right)  }\left(  X\right)  $, we define the adjoint functional
$\varphi^{\#}$ on $L_{\sigma\left(  Y,X\right)  }\left(  Y\right)  $ by
\[
\varphi^{\#}\left(  T^{\#}\right)  =\varphi\left(  T\right)  .
\]
Clearly $\left(  x\otimes\alpha\right)  ^{\#}=\alpha\otimes x$.

The following lemma can be proved easily.

\begin{lemma}
\label{5}Suppose that $(X,Y)$ is a dual pair, $\mathcal{A}$ is a unital
subalgebra of $L_{\sigma\left(  X,Y\right)  }\left(  X\right)  ,$ and that
$\varphi$ is a WOT continuous linear functional such that $\varphi\left(
1\right)  =1$. Then $\mathcal{A}$ is $\varphi$-tracial if and only if
$\mathcal{A}^{\#}$ is $\varphi^{\#}$-tracial and $\mathcal{A}$ is maximal
$\varphi$-tracial if and only if $\mathcal{A}^{\#}$ is maximal $\varphi^{\#}$-tracial.
\end{lemma}

Note that every abelian algebra is $\varphi$-tracial for every $\varphi$. The
next theorem characterizes when an abelian algebra is maximal $\varphi
$-tracial provided $\varphi$ is a rank-one functional.

\begin{theorem}
\label{10}Suppose that $(X,Y)$ is a dual pair, $\mathcal{A}$ is an abelian
unital subalgebra of $L_{\sigma\left(  X,Y\right)  }\left(  X\right)  $, and
that $x\in X,$ $\alpha\in Y$ with $\left(  x\otimes\alpha\right)  \left(
1\right)  =\alpha\left(  x\right)  =1.$ Then $\mathcal{A}$ is maximal tracial
with respect to $x\otimes\alpha$ if and only if the following conditions hold:

\begin{enumerate}
\item $\mathcal{A}$ is maximal abelian in $L_{\sigma\left(  X,Y\right)
}\left(  X\right)  ,$

\item $\left[  \mathcal{A}x\right]  ^{-\sigma\left(  X,Y\right)  }=X,$ and

\item $\left[  \mathcal{A}^{\#}\alpha\right]  ^{-\sigma\left(  Y,X\right)
}=Y$.
\end{enumerate}
\end{theorem}

\begin{proof}
Suppose $\mathcal{A}$ is maximal tracial with respect to $x\otimes\alpha$.
Clearly, $\mathcal{A}$ is maximal abelian. Assume, via contradiction, that
$M=\left[  \mathcal{A}x\right]  ^{-\sigma\left(  X,Y\right)  }\neq X$. Let
\[
\mathcal{A}_{1}=\left\{  T\in L_{\sigma\left(  X,Y\right)  }\left(  X\right)
:T\left(  M\right)  \subset M\text{, }T|_{M}\in\mathcal{A}|_{M}\right\}  .
\]
Then $\mathcal{A}_{1}$ is tracial with respect to $x_{0}\otimes\alpha$ and,
since $M\neq X$, $\mathcal{A}_{1}$ is not abelian. This contradiction implies
$\left[  \mathcal{A}x\right]  ^{-\sigma\left(  X,Y\right)  }=X$.

From Lemma \ref{5} we know that if $\mathcal{A}$ is maximal $\left(
x\otimes\alpha\right)  $-tracial, then $\mathcal{A}^{\#}$ is maximal $\left(
x\otimes\alpha\right)  ^{\#}$-tracial. Since $\left(  x\otimes\alpha\right)
^{\#}=\alpha\otimes x$, we conclude $\left[  \mathcal{A}^{\#}\alpha\right]
^{-\sigma\left(  Y,X\right)  }=Y$.

Now suppose $\left(  1\right)  $-$\left(  3\right)  $ hold. Suppose $T\in
L_{\sigma\left(  X,Y\right)  }\left(  X\right)  $ and the algebra generated by
$\mathcal{A}\cup\left\{  T\right\}  $ is tracial for $x\otimes\alpha$. Then,
for every $A,B,C\in\mathcal{A}$, we have
\begin{align*}
\left(  x\otimes\alpha\right)  \left(  B\left(  AT-TA\right)  C\right)   &
=\left(  x\otimes\alpha\right)  \left(  \left(  AT-TA\right)  CB\right) \\
&  =\left(  x\otimes\alpha\right)  \left(  AT\left(  CB\right)  -T\left(
CB\right)  A\right) \\
&  =\left(  x\otimes\alpha\right)  \left(  \left(  CB\right)  AT-T\left(
CB\right)  A\right) \\
&  =0.
\end{align*}
This means
\[
\left[  AT-TA\right]  \left(  Cx\right)  \in\ker\left(  B^{\prime}
\alpha\right)
\]
for every $B,C\in\mathcal{A}$. It follows from $\left(  2\right)  $ and
$\left(  3\right)  $ that $AT=TA$. Hence, by $\left(  1\right)  $,
$T\in\mathcal{A}$.
\end{proof}

A subalgebra $\mathcal{A}$ of $L_{\sigma\left(  X,Y\right)  }\left(  X\right)
$ is called \emph{transitive} if the only $\sigma\left(  X,Y\right)  $-closed
$\mathcal{A}$-invariant linear subspaces are $\left\{  0\right\}  $ and $X$.
It is easily shown that $M$ is $\mathcal{A}$-invariant if and only if
$M^{\perp}$ is $\mathcal{A}^{\#}$-invariant. Thus $\mathcal{A}$ is transitive
in $L_{\sigma\left(  X,Y\right)  }\left(  X\right)  $ if and only if
$\mathcal{A}^{\#}$ is transitive in $L_{\sigma\left(  Y,X\right)  }\left(
Y\right)  $.

\begin{theorem}
\label{15}A subalgebra $\mathcal{A}$ of $L_{\sigma\left(  X,Y\right)  }\left(
X\right)  $ is maximal abelian and transitive if and only if $\mathcal{A}$ is
maximal $x\otimes\alpha$-tracial for every $x\in X$ and $\alpha\in Y$ with
$\alpha\left(  x\right)  =\left(  x\otimes\alpha\right)  \left(  1\right)  =1$.
\end{theorem}

\begin{proof}
Suppose $\mathcal{A}$ is maximal $x\otimes\alpha$-tracial for every $x\in X$
and $\alpha\in Y$ with $\alpha\left(  x\right)  =\left(  x\otimes
\alpha\right)  \left(  1\right)  =1$. Let $S,T\in\mathcal{A}$. Assume, via
contradiction, that $ST\neq TS.$ Then there is an $x\in X$ such that $\left(
ST-TS\right)  x\neq0$. We know that there are $\alpha_{1},\alpha_{2}\in Y$
such that $\alpha_{1}\left(  \left(  ST-TS\right)  x\right)  \neq0$ and
$\alpha_{2}\left(  x\right)  \neq0$. Thus there is an $\alpha\in\left\{
\alpha_{1},\alpha_{2},\alpha_{1}+\alpha_{2}\right\}  $ such that
$\alpha\left(  x\right)  \neq0$ and $\alpha\left(  \left(  ST-TS\right)
x\right)  \neq0$. By dividing by $\alpha\left(  x\right)  $ we can assume that
$\alpha\left(  x\right)  =1$ and thus
\[
\left(  x\otimes\alpha\right)  \left(  ST-TS\right)  =\alpha\left(  \left(
ST-TS\right)  x\right)  \neq0,
\]
a contradiction. Thus $\mathcal{A}$ is abelian. Since $\mathcal{A}$ is maximal
tracial for every $x\otimes\alpha$ with $\alpha\left(  x\right)  =1,$ it
follows that $\mathcal{A}$ is maximal abelian.

Next suppose $x\in X$ and $x\neq0.$ Then there is an $\alpha\in Y$ such that
$\alpha\left(  x\right)  =1.$ Since $\mathcal{A}$ is abelian and maximal
$\left(  x\otimes\alpha\right)  $-tracial, we know from Theorem \ref{10} that
$\left[  \mathcal{A}x\right]  ^{-\sigma\left(  X,Y\right)  }=X$. Thus
$\mathcal{A}$ is transitive.

The proof of the other direction is straightforward by using Theorem \ref{10}.
\end{proof}

\begin{corollary}
\label{20}Suppose $\mathcal{H}$ is a Hilbert space and $\mathcal{A}\subset
B\left(  \mathcal{H}\right)  $ is a unital weak-operator closed algebra. Then
$\mathcal{A}$ is maximal $e\otimes e$-tracial for every unit vector $e$ in
$\mathcal{H}$ if and only if $\mathcal{A}$ is abelian and transitive.
\end{corollary}

\begin{proof}
Following the proof of Theorem \ref{15} , we just need to show $\mathcal{A}$
is abelian. If $S,T\in\mathcal{A}$ and $ST-TS\neq0,$ then the numerical range
of $ST-TS$ is not $\left\{  0\right\}  $, which means that there is a unit
vector $e\in\mathcal{H}$ such that
\[
\left(  e\otimes e\right)  \left(  ST-TS\right)  =\left\langle \left(
ST-TS\right)  e,e\right\rangle \neq0,
\]
which contradicts the assumption that $\mathcal{A}$ is $\left(  e\otimes
e\right)  $-tracial. Hence $\mathcal{A}$ is abelian and the rest follows as in
the proof of Theorem \ref{15} .
\end{proof}

We need to interpret the preceding results for Banach spaces. Suppose $X$ is a
Banach space and $Y$ is the dual space of $X.$ Then the $\sigma\left(
X,Y\right)  $-topology on $X$ is the weak topology and the $\sigma\left(
Y,X\right)  $-topology on $Y$ is the weak*-topology. Hence $L_{\sigma\left(
X,Y\right)  }\left(  X\right)  $ is the set of all linear transformations $T$
on $X$ that are weak-weak continuous, which, by the closed graph theorem,
simply means that $L_{\sigma\left(  X,Y\right)  }\left(  X\right)  $ is the
set of all linear transformations $T$ on $X$ that are bounded on $X$. However,
$L_{\sigma\left(  Y,X\right)  }\left(  Y\right)  $ is the set of all linear
transformations on $Y$ that are weak*-weak* continuous, which are precisely
the adjoints of transformations in $B\left(  X\right)  $. Thus $L_{\sigma
\left(  Y,X\right)  }\left(  Y\right)  \neq B\left(  Y\right)  ,$ except when
$X$ is a reflexive Banach space. If $T\in L_{\sigma\left(  X,Y\right)
}\left(  X\right)  $, then $T^{\#}$ is the usual Banach-space adjoint of $T$.
However, if $S\in L_{\sigma\left(  Y,X\right)  }\left(  Y\right)  $, then
$S=T^{\#}$ for some $T\in B\left(  X\right)  $, and our definition of $S^{\#}$
is $S^{\#}=T$. Nonetheless, the Banach-space adjoint of $S$ acts on normed
dual of $Y,$ which is the second dual of $X$. For example, C. Read \cite{Read}
has constructed a transitive operator $T$ on $\ell^{1}$. But no operator $A$
is transitive on the dual $\ell^{\infty}$ of $\ell^{1}$, since, for any
nonzero $x\in\ell^{\infty}$ the closed linear span $M$ of $\left\{
x,Ax,\ldots\right\}  $ is nonzero, and not $\ell^{\infty}$ ($M$ is separable)
and $A$-invariant. However, $T^{\#}$ has no nontrivial weak*-closed invariant
subspaces of $\ell^{\infty}$.

\begin{corollary}
\label{25}Suppose $X$ is a Banach space with normed dual $Y$, $\mathcal{A}
\subset B\left(  X\right)  $ is an abelian algebra, and $x\in X,$ $\alpha\in
Y$ with $\alpha\left(  x\right)  =1$. Then a maximal abelian algebra
$\mathcal{A}\subset B\left(  X\right)  $ is a maximal $x\otimes\alpha$-tracial
algebra if and only if $\left[  \mathcal{A}x\right]  ^{-\left\Vert
.\right\Vert }=X$ and $\left[  \mathcal{A}^{\#}\alpha\right]  ^{-\text{weak*}
}=Y$.
\end{corollary}

\begin{example}
Suppose $\left(  \Omega,\Sigma,\mu\right)  $ is a probability space and
$\alpha$ is a normalized gauge norm on $L^{\infty}\left(  \mu\right)  ,$ i.e.,
$\alpha$ is a norm such that $\alpha\left(  1\right)  =1,$ and for every $f\in
L^{\infty}\left(  \mu\right)  ,$ $\alpha\left(  f\right)  =\alpha\left(
\left\vert f\right\vert \right)  $. We can extend the domain of $\alpha$ to
every measurable function $f$ by
\[
\alpha\left(  f\right)  =\sup\left\{  \alpha\left(  h\right)  :h\in L^{\infty
}\left(  \mu\right)  ,\left\vert h\right\vert \leq\left\vert f\right\vert
\text{ a.e. }\left(  \mu\right)  \right\}  .
\]
Then $\mathcal{L}^{\alpha}\left(  \mu\right)  =\left\{  f:\alpha\left(
f\right)  <\infty\right\}  $ is a Banach space and $L^{\alpha}\left(
\mu\right)  $ is defined as the $\alpha$-closure of $L^{\infty}\left(
\mu\right)  $. We say that the gauge norm $\alpha$ is \emph{continuous} if
\[
\lim_{\mu\left(  E\right)  \rightarrow0}\alpha\left(  \chi_{E}\right)  =0.
\]
It is well-known that if $f\in L^{\alpha}\left(  \mu\right)  $ and $h\in
L^{\infty}\left(  \mu\right)  ,$ then
\[
\alpha\left(  hf\right)  \leq\alpha\left(  f\right)  \left\Vert h\right\Vert
_{\infty}.
\]
Thus $L^{\infty}\left(  \mu\right)  $ acts, via multiplication, as operators
on $L^{\alpha}\left(  \mu\right)  $. It is shown that $L^{\infty}\left(
\mu\right)  $ is actually a maximal abelian algebra of operators on
$L^{\alpha}\left(  \mu\right)  $. The norm $\alpha$ has a \emph{dual norm}
$\alpha^{\prime}$ on $L^{\infty}\left(  \mu\right)  $ defined by
\[
\alpha^{\prime}\left(  f\right)  =\sup\left\{  \left\Vert fh\right\Vert
_{1}:h\in L^{\infty}\left(  \mu\right)  ,\alpha\left(  h\right)
\leq1\right\}  \text{ .}
\]
If $\left\Vert .\right\Vert _{1}\leq\alpha$, then $\alpha^{\prime}$ is a
normalized gauge norm, and, for $f\in L^{\alpha}\left(  \mu\right)  $ and
$h\in\mathcal{L}^{\alpha^{\prime}}\left(  \mu\right)  ,$
\[
\left\Vert fh\right\Vert _{1}\leq\alpha\left(  f\right)  \alpha^{\prime
}\left(  h\right)  \text{.}
\]
If, in addition, $\alpha$ is continuous, then the normed dual of $L^{\alpha
}\left(  \mu\right)  $ is $\mathcal{L}^{\alpha^{\prime}}\left(  \mu\right)  $,
given for $h\in\mathcal{L}^{\alpha^{\prime}}\left(  \mu\right)  ,$ the
functional $\psi_{h}$ on $L^{\alpha}\left(  \mu\right)  $ defined by
\[
\psi_{h}\left(  f\right)  =\int_{\Omega}fhd\mu\text{ .}
\]
Suppose $f\in L^{\alpha}\left(  \mu\right)  $ and $h\in\mathcal{L}
^{\alpha^{\prime}}\left(  \mu\right)  $, then $L^{\infty}\left(  \mu\right)  $
is maximal $f\otimes h$-tracial if and only if both $L^{\infty}\left(
\mu\right)  f$ is dense in $L^{\alpha}\left(  \mu\right)  $ and $L^{\infty
}\left(  \mu\right)  h$ is weak*-dense in $\mathcal{L}^{\alpha^{\prime}
}\left(  \mu\right)  $. This is equivalent to $f\left(  \omega\right)  \neq0$
a.e. $\left(  \mu\right)  $ and $h\left(  \omega\right)  \neq0$ a.e. $\left(
\mu\right)  $. To see this, suppose $f\neq0$ a.e. $\left(  \mu\right)  $.
Suppose $\psi\in L^{\alpha}\left(  \mu\right)  ^{\#}$ and $\psi|_{L^{\infty
}\left(  \mu\right)  f}=0.$ Then there is an $h\in\mathcal{L}^{\alpha^{\prime
}}\left(  \mu\right)  $ such that, $\psi=\psi_{h}.$ Hence, for every $u\in
L^{\infty}\left(  \mu\right)  $,
\[
\int_{\Omega}ufhd\mu=0.
\]
Since $fh\in L^{1}\left(  \mu\right)  $, this means $fh=0$ a.e. $\left(
\mu\right)  ,$ which implies $h=0$ a.e. $\left(  \mu\right)  $. Thus $\psi=0.$
A similar proof shows that if $h\in\mathcal{L}^{\alpha^{\prime}}\left(
\mu\right)  $, then $L^{\infty}\left(  \mu\right)  h\ $is weak*-dense in
$\mathcal{L}^{\alpha}\left(  \mu\right)  $.
\end{example}

\begin{example}
Let $H^{2}$ be the classical Hardy space on the unit disk. It is known that
$H^{\infty}$ (acting as multiplications) is a maximal abelian algebra in
$B\left(  H^{2}\right)  $. The cyclic vectors for $H^{\infty}$ are the outer
functions (see \cite{Helson}). There are many cyclic vectors in $H^{2}$ for
$\left(  H^{\infty}\right)  ^{\#}$. We see that $H^{\infty}$ is $f\otimes
h$-maximal tracial if and only if $f$ is outer and $h$ is a cyclic vector for
$\left(  H^{\infty}\right)  ^{\#}$. Since $1$ is not cyclic for $\left(
H^{\infty}\right)  ^{\#}$, we see that $H^{\infty}$ is \textbf{not}
$1\otimes1$-maximal tracial.
\end{example}

\begin{example}
In the preceding example, $H^{\infty}$ is the unital weakly closed algebra
generated by the unilateral shift operator on $\ell^{2}$. Suppose
$X\in\left\{  c_{0}\right\}  \cup\left\{  \ell^{p}:1\leq p<\infty\right\}  $,
and let
\[
e_{0}=\left(  1,0,\ldots\right)  ,e_{1}=\left(  0,1,0,\ldots\right)
,\ldots\text{ .}
\]
Suppose $\left\{  \beta_{n}\right\}  _{n\geq0}$ is a bounded sequence of
positive numbers, and define $Te_{n}=\beta_{n}e_{n+1},$ i.e.
\[
T\left(  a_{0},a_{1},a_{2},\ldots\right)  =\left(  0,\beta_{0}a_{0},\beta
_{1}a_{1},\ldots\right)  .
\]
Then $T$ is a bounded linear operator and $\mathcal{A}=\left\{  T\right\}
^{\prime}=\left\{  p\left(  T\right)  :p\in\mathbb{C}\left[  z\right]
\right\}  ^{-WOT}$ is a maximal abelian algebra in $B\left(  X\right)  $.
Clearly, $e_{0}$ is a cyclic vector for $T$. There are also many weak*-cyclic
vectors for $T^{\#}$ of the form $h=\left(  1,d_{1},d_{2},\ldots\right)  $.
Thus $\mathcal{A}$ is maximal $e_{0}\otimes h$-tracial for many choices of $h$.
\end{example}

\begin{example}
Suppose $I$ is an uncountable set. Then $\ell^{\infty}\left(  I\right)  $ is a
maximal abelian algebra of operators on $\ell^{2}\left(  I\right)  $, but,
since any vector in $\ell^{2}\left(  I\right)  $ has countable support,
$\ell^{\infty}\left(  I\right)  $ has no cyclic vector. Hence $\ell^{\infty
}\left(  I\right)  $ is not maximal tracial for $x\otimes y$ for any
$x,y\in\ell^{2}\left(  I\right)  $, or for any weak*-continuous linear
$\varphi$ functional on $B\left(  \ell^{2}\left(  I\right)  \right)  $ since
such a $\varphi$ involves only countably many vectors.
\end{example}

\begin{definition}
Suppose $\left(  X,Y\right)  $ is a dual pair over a Hausdorff field
$\mathbb{F}$, and $\mathcal{A}$ is a WOT-closed unital subalgebra of
$L_{\sigma\left(  X,Y\right)  }\left(  X\right)  $. We say that $\mathcal{A}$
has a \emph{complemented invariant subspace lattice} if and only if, for every
$\mathcal{A}$-invariant $\sigma\left(  X,Y\right)  $-closed linear subspace
$M$ of $X$, there is an idempotent $P\in L_{\sigma\left(  X,Y\right)  }\left(
X\right)  $ such that $P\left(  X\right)  =M$ and $P\in\mathcal{A}^{\prime}$,
where $\mathcal{A}^{\prime}$ denotes the commutant of $\mathcal{A}.$
\end{definition}

\begin{theorem}
\label{30}Suppose $\left(  X,Y\right)  $ is a dual pair over a Hausdorff field
$\mathbb{F}$, and $\mathcal{A}$ is a WOT-closed unital subalgebra of
$L_{\sigma\left(  X,Y\right)  }\left(  X\right)  $ with a complemented
invariant subspace lattice. If $\mathcal{A}$ is maximal tracial for $e\otimes
f,$ with $e\in X,$ $f\in Y$ and $f\left(  e\right)  =1$, then $\left[
\mathcal{A}e\right]  ^{-\sigma\left(  X,Y\right)  }=X$ and $\left[
\mathcal{A}^{\#}f\right]  ^{-\sigma\left(  Y,X\right)  }=Y$.
\end{theorem}

\begin{proof}
Let $M=\left[  \mathcal{A}e\right]  ^{-\sigma\left(  X,Y\right)  }$. Then $M$
is $\mathcal{A}$-invariant. Thus there is an idempotent $P\in L_{\sigma\left(
X,Y\right)  }\left(  X\right)  $ such that $P\left(  X\right)  =M$ and
$P\in\mathcal{A}^{\prime}$. With respect to the decomposition $X=M\oplus\ker
P$, we can write every $T\in B\left(  X\right)  $ as an operator matrix
\[
T=\left(
\begin{array}
[c]{cc}
B & C\\
D & E
\end{array}
\right)  .
\]
If $A\in\mathcal{A}$ we have
\[
A=\left(
\begin{array}
[c]{cc}
A_{11} & 0\\
0 & A_{22}
\end{array}
\right)  .
\]
Let $\mathcal{S}$ be the set of all the operators of the form $\left(
\begin{array}
[c]{cc}
0 & C\\
0 & 0
\end{array}
\right)  $ and let $\mathcal{D}$ $=\mathcal{A+S}$. Since $e\in M,$ then it
follows that $\mathcal{S}$ is a two-sided ideal in $\mathcal{D}$ and that
$\mathcal{S}\subset\ker\left(  e\otimes f\right)  $. Thus for every
$A,B\in\mathcal{A}$ and every $S_{1},S_{2}\in\mathcal{S}$ we have
\[
\left(  A+S_{1}\right)  \left(  B+S_{2}\right)  -\left(  B+S_{2}\right)
\left(  A+S_{1}\right)  -\left(  AB-BA\right)  \in\mathcal{S}\text{.}
\]
Hence $\mathcal{D}$ is $e\otimes f$-tracial. Since $\mathcal{A}$ is maximal
$e\otimes f$-tracial, then $P=1$ and $M=X$.

Similarly, suppose $N=\left[  \mathcal{A}^{\#}f\right]  ^{-\sigma\left(
Y,X\right)  }.$ Then $N_{\perp}$ is $\mathcal{A}$-invariant. Thus there is an
idempotent $P\in L_{\sigma\left(  X,Y\right)  }\left(  X\right)
\cap\mathcal{A}^{\prime}$ such that $P\left(  X\right)  =N_{\perp}\subset\ker
f$. Consequently $\left(  1-P^{\#}\right)  \left(  Y\right)  =N$. Writing the
matrix representations of the elements of $L_{\sigma\left(  X,Y\right)
}\left(  X\right)  $ and $\mathcal{A}$ as above, we can let $\mathcal{T}$ be
the set of all operators of the form $\left(
\begin{array}
[c]{cc}
0 & 0\\
D & 0
\end{array}
\right)  .$ Since the range of any operator in $\mathcal{T}$ is contained in
$N_{\perp}\subset\ker f$, we see that $\mathcal{T}\subset\ker\left(  e\otimes
f\right)  $. Therefore $\mathcal{E=A}+T$ is $e\otimes f$-tracial, and, since
$\mathcal{A}$ is maximal $\left(  e\otimes f\right)  $-tracial, we see that
$N_{\perp}=\left\{  0\right\}  $, and $N=\left(  N_{\perp}\right)  ^{\perp}=Y$.
\end{proof}

\section{Multiplier Pairs}

Multiplier pairs were studied in \cite{HN}. A \emph{multiplier pair} is a pair
$\left(  X,Y\right)  $ where $X$ is a Banach space, $Y$ is a Hausdorff
topological vector space, $X\subset Y$, and there is a multiplication on $X$
with values in $Y$, i.e., a bilinear map
\[
\cdot:X\times X\rightarrow Y
\]
such that

\begin{enumerate}
\item $\cdot$ is continuous in each variable

\item There is an \emph{identity element} $e\in X$ such that, for every $a\in
X,$
\[
a\cdot e=e\cdot a=a
\]

\item The sets $\mathcal{L}_{0}=\left\{  a\in X:a\cdot X\subset X\right\}  $
and $\mathcal{R}_{0}=\left\{  b\in X:X\cdot b\subset X\right\}  $ are dense in
$X$,

\item There are dense subsets $E\subset\mathcal{L}_{0}$, $F\subset X$,
$G\subset\mathcal{R}_{0}$, such that, for all $a\in E,b\in F,c\in G$, we have
\[
\left(  a\cdot b\right)  \cdot c=a\cdot\left(  b\cdot c\right)  .
\]

\end{enumerate}

If $a\in\mathcal{L}_{0}$ and $b\in\mathcal{R}_{0}$, the operators $L_{a}$ and
$R_{b}$ defined on $X$ by
\[
L_{a}x=a\cdot x\text{ and }R_{b}x=x\cdot b
\]
are bounded on $X$ (see \cite[Theorem 1]{HN}). Moreover, it was shown in
\cite[Theorem 1]{HN} that if $\mathcal{L}=\left\{  L_{a}:a\in\mathcal{L}
_{0}\right\}  $ and $\mathcal{R=}\left\{  R_{b}:b\in\mathcal{R}_{0}\right\}
$, then $\mathcal{L},\mathcal{R}\subset B\left(  X\right)  $ and
\[
\mathcal{L}^{\prime}=\mathcal{R}\text{ and }\mathcal{R}^{\prime}
=\mathcal{L}\text{ .}
\]
In particular, if the multiplication is commutative, then $\mathcal{L}
=\mathcal{R}$ is maximal abelian in $B\left(  X\right)  $. In this case the
next theorem reduces to Theorem \ref{10}.

We now prove a result about multiplier pairs that gives a lot of examples of
maximal tracial algebras. In many of the examples in \cite{HN}, condition
$\left(  1\right)  $ in the following theorem holds.

\begin{theorem}
\label{35}Suppose $\left(  X,Y\right)  $ is a multiplier pair with identity
element $e$, and suppose $\alpha\in X^{\#}$ and $\alpha\left(  e\right)  =1$.
Also suppose

\begin{enumerate}
\item $\left\{  R_{B}:B\in\mathcal{L}_{0}\cap\mathcal{R}_{0}\right\}  $ is
WOT-dense in $\mathcal{R}$

\item $\left[  \mathcal{L}^{\#}\alpha\right]  ^{-\text{\textrm{weak*}}}
=X^{\#}$, and

\item $\mathcal{L}$ is $\left(  e\otimes\alpha\right)  $-tracial
\end{enumerate}

Then $\mathcal{L}$ is maximal $\left(  e\otimes\alpha\right)  $-tracial.
\end{theorem}

\begin{proof}
Suppose $B\in\mathcal{L}_{0}\cap\mathcal{R}_{0}$.

\textbf{Claim:} $L_{B}^{\#}\alpha=R_{B}^{\#}\alpha.$

Proof of Claim: Suppose $A\in\mathcal{L}_{0}$. Then
\begin{align*}
\left(  L_{B}^{\#}\alpha\right)  \left(  A\right)   &  =\alpha\left(
BA\right)  =\left(  e\otimes\alpha\right)  \left(  L_{B}L_{A}\right) \\
&  =\left(  e\otimes\alpha\right)  \left(  L_{A}L_{B}\right)  =\alpha\left(
AB\right) \\
&  =\left(  R_{B}^{\#}\alpha\right)  \left(  A\right)  .
\end{align*}
Since $\mathcal{L}_{0}$ is dense in $X$, the claim is proved.

Suppose $\mathcal{D}\subset B\left(  X\right)  $ is a $\left(  e\otimes
\alpha\right)  $-tracial algebra containing $\mathcal{L}$, and suppose
$T\in\mathcal{D}$. Choose $B\in\mathcal{L}_{0}\cap\mathcal{R}_{0}$ and
$C,D\in\mathcal{L}_{0}$. We then have
\begin{align*}
\left\langle \left(  TR_{B}\right)  \left(  C\right)  ,L_{D}^{\#}
\alpha\right\rangle  &  =\left\langle T\left(  CB\right)  ,L_{D}^{\#}
\alpha\right\rangle =\left\langle TL_{C}L_{B}e,L_{D}^{\#}\alpha\right\rangle
\\
&  =\left\langle L_{D}TL_{C}L_{B}e,\alpha\right\rangle =\left(  e\otimes
\alpha\right)  \left(  L_{D}TL_{C}L_{B}\right) \\
&  =\left(  e\otimes\alpha\right)  \left(  L_{B}L_{D}TL_{C}\right)
=\left\langle L_{B}L_{D}TL_{C}e,\alpha\right\rangle \\
&  =\left\langle L_{D}TL_{C}e,L_{B}^{\#}\alpha\right\rangle =\left\langle
L_{D}TL_{C}e,R_{B}^{\#}\alpha\right\rangle \\
&  =\left\langle R_{B}L_{D}TL_{C}e,\alpha\right\rangle =\left\langle
L_{D}R_{B}TL_{C}e,\alpha\right\rangle \\
&  =\left\langle \left(  R_{B}T\right)  \left(  C\right)  ,L_{D}^{\#}
\alpha\right\rangle .
\end{align*}
Since $\mathcal{L}^{\#}\alpha$ is weak* dense in $X^{\#}$ and $\mathcal{L}
_{0}$ is dense in $X$, we see that
\[
TR_{B}=R_{B}T
\]
for every $B\in\mathcal{L}_{0}\cap\mathcal{R}_{0}$. It follows from $\left(
1\right)  $ that $T\in\mathcal{R}^{\prime}=\mathcal{L}$. Thus $\mathcal{L}$ is
maximal $\left(  e\otimes\alpha\right)  $-tracial.
\end{proof}

\section{von Neumann Algebras}

Suppose $\mathcal{M}$ is a von Neumann algebra acting on a Hilbert space $H$
with a faithful normal tracial state $\tau$. There is a notion of
\emph{convergence in measure }on $\mathcal{M}$ (see \cite{Helson}), defined by
saying a net $\left\{  T_{\lambda}\right\}  $ in $\mathcal{M}$ converges to
$0$ in measure, denoted by $T_{\lambda}\rightarrow0$ $\left(  \tau\right)  $,
if and only if there is a net $\left\{  P_{\lambda}\right\}  $ of projections
in $\mathcal{M}$ such that $\tau\left(  P_{\lambda}\right)  \rightarrow0$ and
$\left\Vert T_{\lambda}\left(  1-P_{\lambda}\right)  \right\Vert
\rightarrow0.$ The completion of $\mathcal{M}$, denoted by $\mathcal{\hat{M}}
$, with respect to this topology, is a $\ast$-algebra, in which all of the
algebraic operations on $\mathcal{M}$ have continuous extensions. Moreover,
every $A\in\mathcal{\hat{M}}$ has a polar decomposition $A=U\left\vert
A\right\vert $ where $U$ is a unitary in $\mathcal{M}$ and $\left\vert
A\right\vert =\left(  A^{\ast}A\right)  ^{1/2}$. A norm $\beta$ on
$\mathcal{M}$ is called a \emph{normalized unitarily invariant norm} on
$\mathcal{M}$ if, for every $T\in\mathcal{M}$ and all unitary operators
$U,V\in\mathcal{M}$,
\[
\beta\left(  UTV\right)  =\beta\left(  T\right)  \text{.}
\]
Examples of such norms are the $p$-norms, $1\leq p<\infty$, defined on
$\mathcal{M}$ by
\[
\left\Vert T\right\Vert _{p}=\tau\left(  \left\vert T\right\vert ^{p}\right)
^{1/p},
\]
where $\left\vert T\right\vert =\left(  T^{\ast}T\right)  ^{1/2}$. We define
$L^{\beta}\left(  \mathcal{M},\tau\right)  $ to be the completion of
$\mathcal{M}$ with respect to $\beta.$ When $\beta\geq\left\Vert .\right\Vert
_{1}$, we get $L^{\beta}\left(  \mathcal{M},\tau\right)  \subset L^{1}\left(
\mathcal{M},\tau\right)  \subset\mathcal{\hat{M}}$. It was shown in \cite{HN}
that $\left(  L^{\beta}\left(  \mathcal{M},\tau\right)  ,\mathcal{\hat{M}
}\right)  $ is a multiplier pair and $\mathcal{L}_{0}=\mathcal{R}
_{0}=\mathcal{M}$ and $e=1$. Suppose $T\in\mathcal{M}$. We define $\varphi
_{T}\in L^{\beta}\left(  \mathcal{M},\tau\right)  ^{\#}$, by
\[
\varphi_{T}\left(  A\right)  =\tau\left(  TA\right)  \text{.}
\]
If we let $\alpha=\varphi_{1}$, we see that $1\otimes\alpha=\tau$ is tracial
on $\mathcal{M}$. Also, if $A\in L^{p}\left(  \mathcal{M},\tau\right)  \subset
L^{1}\left(  \mathcal{M},\tau\right)  $ with polar decomposition $U\left\vert
A\right\vert $, and if
\[
\left(  L_{D}^{\#}\alpha\right)  \left(  A\right)  =0
\]
for every $D\in\mathcal{M}$, then
\[
0=L_{U^{\ast}}^{\#}\left(  A\right)  =\tau\left(  U^{\ast}A\right)
=\tau\left(  \left\vert A\right\vert \right)  .
\]
This implies $A=0.$ Thus $\left\{  L_{D}^{\#}\alpha:D\in\mathcal{L}
_{0}\right\}  $ is weak* dense in $L^{\beta}\left(  \mathcal{M},\tau\right)
$. If we apply Theorem \ref{35}, we obtain the following result.

\begin{theorem}
\label{40}Let $\mathcal{M}$ be a von Neumann algebra with a faithful normal
tracial state. If $\beta\geq\left\Vert .\right\Vert _{1}$ is a normalized
unitarily invariant norm, then $\mathcal{M=}\left\{  L_{A}:A\in\mathcal{M}
\right\}  $ is a maximal $1\otimes\varphi_{1}$-tracial algebra in $B\left(
L^{\beta}\left(  \mathcal{M},\tau\right)  \right)  .$
\end{theorem}

An example of a von Neumann algebra with a faithful normal tracial state is
$L^{\infty}\left(  \mu\right)  $ where $\left(  \Omega,\mu\right)  $ is a
probability space and $\tau:L^{\infty}\left(  \mu\right)  \rightarrow
\mathbb{C}$ is defined by
\[
\tau\left(  f\right)  =\int_{\Omega}fd\mu\text{ .}
\]
A normalized unitarily invariant norm $\beta$ on $L^{\infty}\left(
\mu\right)  $ is a \emph{normalized gauge norm}, i.e., a norm such that
$\beta\left(  1\right)  =1$ and $\beta\left(  f\right)  =\beta\left(
\left\vert f\right\vert \right)  $ for every $f\in L^{\infty}\left(
\mu\right)  $. If $\beta\geq\left\Vert .\right\Vert _{1}$, we have that
$L^{\beta}\left(  \mu\right)  $ is a multiplier pair and $\mathcal{L}
_{0}=\mathcal{R}_{0}=L^{\infty}\left(  \mu\right)  $ is maximal abelian in
$B\left(  L^{\beta}\left(  \mu\right)  \right)  $. Thus Theorem \ref{10}
applies. A vector $f\in L^{\beta}\left(  \mu\right)  $ is cyclic for
$L^{\infty}\left(  \mu\right)  $ if and only if $f\neq0$ a.e. $\left(
\mu\right)  $.

\begin{corollary}
\label{45}Suppose $\left(  \Omega,\mu\right)  $ is a probability space,
$\beta\geq\left\Vert .\right\Vert _{1}$ is a normalized gauge norm on
$L^{\infty}\left(  \mu\right)  ,$ $f\in L^{\beta}\left(  \mu\right)  $ and
$\alpha\in L^{\beta}\left(  \mu\right)  $ satisfy
\[
f\neq0\text{ a.e. }\left(  \mu\right)  \text{ and }L^{\infty}\left(
\mu\right)  ^{\#}\alpha\text{ is weak* dense in }L^{\beta}\left(  \mu\right)
^{\#}.
\]
Then $L^{\infty}\left(  \mu\right)  $ is maximal $\left(  f\otimes
\alpha\right)  $-tracial.
\end{corollary}

We now consider an important problem. Suppose $\mathcal{M}$ is a von Neumann
algebra on a Hilbert space $\mathcal{H}$ and $e,f\in\mathcal{H}$ and
$\left\langle e,f\right\rangle =1.$ If $e\otimes f$ is tracial on
$\mathcal{M}$, when is $\mathcal{M}$ maximal $\left(  e\otimes f\right)  $-tracial?

We first note that if $E$ is an invariant subspace for $\mathcal{M}$, then
$E^{\perp}$ is also invariant, so Theorem \ref{30} implies that a necessary
condition for $\mathcal{M}$ to be maximal $\left(  e\otimes f\right)
$-tracial is that $e$ and $f$ be cyclic vectors for $\mathcal{M}$. The real
question is whether the converse is true.

\begin{problem}
\label{50}Suppose that $\mathcal{M}$ is a von Neumann algebra on a Hilbert
space $\mathcal{H}$ and $e,f$ are cyclic vectors for $\mathcal{M}$ with
$\left\langle e,f\right\rangle =1.$ If $\mathcal{M}$ is $\left(  e\otimes
f\right)  $-tracial, then is $\mathcal{M}$ maximal $\left(  e\otimes f\right)
$-tracial?
\end{problem}

In the case where $\mathcal{H}$ is separable, we will reduce the problem to
the case where $\mathcal{M}$ is a finite factor von Neumann algebra, and such
a factor has a unique norm continuous unital tracial functional.

Suppose $\mathcal{M}$ is a von Neumann algebra on a separable Hilbert space
$\mathcal{H}$ and $e,f$ are cyclic vectors for $\mathcal{M}$ with
$\left\langle e,f\right\rangle =1.$ Let
\begin{align*}
\mathcal{J}  &  =\left\{  A\in\mathcal{M}:\left(  e\otimes f\right)  \left(
AB\right)  =0\text{ for all }B\in\mathcal{M}\right\} \\
&  =\left\{  A\in\mathcal{M}:Ae\perp\mathcal{M}f\right\} \\
&  =\left\{  A\in\mathcal{M}:Ae=0\right\}  .
\end{align*}
Clearly, $\mathcal{J}$ is a WOT-closed linear space. Also, if $e\otimes f$ is
tracial on $\mathcal{M}$, then $\mathcal{J}$ is a two-sided ideal in
$\mathcal{M}$. Hence there is a projection $P$ in the center $\mathcal{Z}
\left(  \mathcal{M}\right)  $ such that $\mathcal{J=}P\mathcal{M}$. Thus
$P\mathcal{M}e=\mathcal{M}Pe=$ $0$. Since $\mathcal{M}e$ is dense in
$\mathcal{H}$, it follows that $P=0$ and $\mathcal{J}=0.$ Similarly, if
$A\in\mathcal{M}$ and $Af=0$, then $A=0.$ Thus $e$ and $f$ are cyclic
separating vectors for $\mathcal{M}$.

If we write the central decomposition for $\mathcal{M}$, we get
\[
\mathcal{H}=\int_{\Omega}^{\oplus}\mathcal{H}_{\omega}d\mu\left(
\omega\right)  ,\text{ }\mathcal{M}=\int_{\Omega}^{\oplus}\mathcal{M}_{\omega
}d\mu\left(  \omega\right)  ,\text{ }
\]
\[
e=\int_{\Omega}^{\oplus}e_{\omega}d\mu\left(  \omega\right)  ,\text{ }
f=\int_{\Omega}^{\oplus}f_{\omega}d\mu\left(  \omega\right)  \text{, and}
\]
and
\[
\mathcal{Z}\left(  \mathcal{M}\right)  =\int_{\Omega}^{\oplus}
\mathcal{\mathbb{C}}I_{\omega}d\mu\left(  \omega\right)  \text{ ,}
\]
where each $\mathcal{M}_{\omega}$ is a factor von Neumann algebra. Since $e$
and $f$ are separating vectors for $\mathcal{M}$, we have $e_{\omega}\neq0$
and $f_{\omega}\neq0$ almost everywhere. We also get $e_{\omega}$ and
$f_{\omega}$ are cyclic and separating for $\mathcal{M}_{\omega}$ a.e.
$\left(  \mu\right)  $. We also have $e_{\omega}\otimes f_{\omega}$ is tracial
for almost every $\omega$. Since a factor von Neumann algebra $\mathcal{M}
_{\omega}$ can have at most one nonzero continuous unital tracial state
$\tau_{\omega}$, we know that almost every $\mathcal{M}_{\omega}$ is a finite
factor and has a unique tracial functional $\tau_{\omega}$ of norm $1.$ We see
that $\left\langle e_{\omega},f_{\omega}\right\rangle \neq0$ a.e. $\left(
\mu\right)  $. Thus $\tau_{\omega}=\frac{1}{\left\langle e_{\omega},f_{\omega
}\right\rangle }e_{\omega}\otimes f_{\omega}$.

Suppose now that $\mathcal{M}\subset\mathcal{D}$ and $\mathcal{D}$ is $\left(
e\otimes f\right)  $-tracial. Suppose $P$ is a central projection in
$\mathcal{M}$. Then, for all $A,B\in\mathcal{M}$, we have
\begin{align*}
\left(  e\otimes f\right)  \left(  A\left(  PT-TP\right)  B\right)   &
=\left(  e\otimes f\right)  \left(  APTB\right)  -\left(  e\otimes f\right)
\left(  ATPB\right) \\
&  =\left(  e\otimes f\right)  \left(  \left(  BAPT\right)  \right)  -\left(
e\otimes f\right)  \left(  \left(  BPAT\right)  \right) \\
&  =\left(  e\otimes f\right)  \left(  \left(  PBAT\right)  \right)  -\left(
e\otimes f\right)  \left(  \left(  PBAT\right)  \right) \\
&  =0.
\end{align*}
Thus
\[
\left\langle \left(  PT-TP\right)  \left(  Be\right)  ,\left(  A^{\ast
}f\right)  \right\rangle =0
\]
for every $A,B\in\mathcal{M}$. Since $e$ and $f$ are cyclic, we see that
$PT-TP=0$ for every projection $P\in\mathcal{Z}\left(  \mathcal{M}\right)  $.
Thus $T\in\mathcal{Z}\left(  \mathcal{M}\right)  ^{\prime}$. Hence we can
write $T=\int_{\Omega}^{\oplus}T_{\omega}d\mu\left(  \omega\right)  $. Since
the algebra generated by $T$ and $\mathcal{M}$ is $\left(  e\otimes f\right)
$-tracial, it follows that, for almost all $\omega\in\Omega$, the algebra
generated by $T_{\omega}$ and $\mathcal{M}_{\omega}$ is $\left(  e_{\omega
}\otimes f_{\omega}\right)  $-tracial. It now easily follows that
$\mathcal{M}$ is maximal $\left(  e\otimes f\right)  $-tracial if and only if
$\mathcal{M}_{\omega}$ is maximal $\left(  \frac{1}{\left\langle e_{\omega
},f_{\omega}\right\rangle }e_{\omega}\otimes f_{\omega}\right)  $-tracial a.e.
$\left(  \mu\right)  $. Hence when $\mathcal{H}$ is separable, this reduces
the problem to the case where $\mathcal{M}$ is a finite factor von Neumann algebra.

Suppose now that $\mathcal{M}$ is a finite factor and $\tau$ is the unique
continuous unital tracial functional on $\mathcal{M}$. Since $e$ and $f$ are
separating cyclic vectors, we can assume, via unitary equivalence \cite{Kad0},
that $\mathcal{H}=L^{2}\left(  \mathcal{M},\tau\right)  $ and $e,f\in
L^{2}\left(  \mathcal{M},\tau\right)  \subset L^{1}\left(  \mathcal{M}
,\tau\right)  \subset\mathcal{\hat{M}}$ as above. We now have $e\otimes
f=\tau$ on $\mathcal{M}$. This means that, for every $A\in\mathcal{M}$,
\[
\tau\left(  f^{\ast}Ae\right)  =\tau\left(  Aef^{\ast}\right)  =\tau\left(
A\right)  \text{.}
\]
Thus
\[
\tau\left(  A\left(  1-ef^{\ast}\right)  \right)  =0.
\]
It follows $ef^{\ast}=1$ (where the multiplication is in $\mathcal{\hat{M}}$).
Since $\mathcal{M}$ is finite, $f^{\ast}e=1$. Thus $e,e^{-1}\in L^{2}\left(
\mathcal{M},\tau\right)  $ and $f^{\ast}=e^{-1},$ so $f=\left(  e^{-1}\right)
^{\ast}$. Thus $e\otimes f=e\otimes\left(  e^{-1}\right)  ^{\ast}$. Now the
question asks whether $\mathcal{M}$ is maximal $\left(  e\otimes\left(
e^{-1}\right)  ^{\ast}\right)  $-tracial.

If our question has an affirmative answer, it would imply that if
$\mathcal{M}\subset\mathcal{B}\left(  \mathcal{H}\right)  $ is a von Neumann
algebra and $S\in\mathcal{B}\left(  \mathcal{H}\right)  $ is invertible and
$\mathcal{A}=S^{-1}\mathcal{M}S$ and $\left(  e\otimes f\right)  $ is a unital
tracial functional on $\mathcal{A}$, then $\mathcal{A}$ is maximal $\left(
e\otimes f\right)  $-tracial.

There is a slight relationship between these ideas and a famous problem in
C*-algebra theory called the \emph{Kadison Similarity Problem} \cite{Kad},
which asks if every bounded unital homomorphism $\rho:\mathcal{A}\rightarrow
B\left(  \mathcal{H}\right)  $ from a C*-algebra $\mathcal{M}$ is similar to a
$\ast$-homomorphism. It is enough to prove the case when $\mathcal{A}$ is
separable. This is equivalent to the question when $\mathcal{A}$ is a von
Neumann algebra acting on a separable Hilbert space, and $\rho$ is weak*-weak*
continuous \cite{Kad}. It follows from \cite{Kad} that the problem reduces to
the case when $\mathcal{A}$ is a $II_{1}$ factor von Neumann algebra on a
separable Hilbert space. It follows from a theorem of R. Kadison, that when
$\mathcal{A}$ is a $II_{1}$ factor and $\rho:\mathcal{A}\rightarrow B\left(
\mathcal{H}\right)  $ is weak*-weak* continuous, that $\rho\left(
\mathcal{A}\right)  $ is similar to a von Neumann algebra.

\textbf{Question:} Suppose $\mathcal{A}$ is a $II_{1}$ factor von Neumann
algebra, $\rho:\mathcal{A}\rightarrow B\left(  \mathcal{H}\right)  $ is a
weak*-weak* continuous bounded $\ast$-homomorphism, and $e,f\in\mathcal{H}$
satisfy $\left\langle e,f\right\rangle =1,$ $\rho\left(  \mathcal{A}\right)  $
is $\left(  e\otimes f\right)  $-tracial, and $e$ is cyclic for $\rho\left(
\mathcal{A}\right)  $ and $f$ is cyclic for $\rho\left(  \mathcal{A}\right)
^{\ast}$, must $\mathcal{\rho}\left(  \mathcal{A}\right)  $ be maximal
$\left(  e\otimes f\right)  $-tracial? Is the lattice of invariant subspaces
of $\rho\left(  \mathcal{A}\right)  $ complemented?

\section{Maximal Tracial Ultraproducts}

The theory of ultraproducts has played a fundamental role in many areas of
mathematics, logic, Banach spaces, von Neumann algebras and C*-algebras. For
von Neumann algebras, the tracial ultraproducts have been extremely important.
We define a new type of ultraproduct. Suppose $\left\{  \mathcal{A}_{i}:i\in
I\right\}  $ is an infinite collection of unital C*-algebras. Also suppose
$\alpha$ is a nontrivial ultrafilter on $I$. We let $\mathcal{A}=

{\displaystyle\prod_{i\in I}}

\mathcal{A}_{i}$ be the C*-direct product, and we define
\[
\mathcal{J}=\left\{  \left\{  a_{i}\right\}  \in\mathcal{A}:\lim
_{i\rightarrow\alpha}\left\Vert a_{i}\right\Vert =0\right\}  \text{.}
\]
Then $\mathcal{J}$ is a closed two-sided ideal in $\mathcal{A}$, and
$\mathcal{A}/\mathcal{J}$ is called the\emph{ ultraproduct }of the C*-algebras
$\mathcal{A}_{i}$, and is denoted by
\[

{\displaystyle\prod_{i\in I}^{\alpha}}

\mathcal{A}_{i}\text{ .}
\]
Next suppose for each $i\in I$ we have $\tau_{i}$ is a tracial state on
$\mathcal{A}_{i}$. If we define
\[
\mathcal{J}_{\alpha}=\left\{  \left\{  a_{i}\right\}  \in\mathcal{A}
:\lim_{i\rightarrow\alpha}\tau_{i}\left(  a_{i}^{\ast}a_{i}\right)
=0\right\}  \text{,}
\]
then $\mathcal{J}_{\alpha}$ is a closed two-sided ideal in $\mathcal{A}$, and
$\mathcal{A}/\mathcal{J}_{\alpha}$ is called the\emph{ tracial ultraproduct
}of the C*-algebras $\mathcal{A}_{i}$, and is denoted by
\[

{\displaystyle\prod_{i\in I}^{\alpha}}

\left(  \mathcal{A}_{i},\tau_{i}\right)  \text{.}
\]
S. Sakai \cite{Sakai} proved that if each $\mathcal{A}_{i}$ is a factor von
Neumann algebra, then $

{\displaystyle\prod_{i\in I}^{\alpha}}

\left(  \mathcal{A}_{i},\tau_{i}\right)  $ is a factor von Neumann algebra.
More recently, the first author and W. H. Li \cite{HadLi} proved that every
tracial ultraproduct of C*-algebras is a von Neumann algebra.

Another way to view a tracial ultraproduct, is by defining a tracial state
$\tau_{\alpha}$ on $

{\displaystyle\prod_{i\in I}}

\mathcal{A}_{i}$ by
\[
\tau_{\alpha}\left(  \left\{  a_{i}\right\}  \right)  =\lim_{i\rightarrow
\alpha}\tau_{i}\left(  a_{i}\right)  .
\]
Then $\mathcal{J}_{\alpha}=\left\{  a\in

{\displaystyle\prod_{i\in I}}

\mathcal{A}_{i}:\tau_{\alpha}\left(  a^{\ast}a\right)  =0\right\}  $, which is
the first part of the GNS construction for $\tau_{\alpha}$.

We now consider the case where, for each $i\in I$, $\varphi_{i}$ is a state on
$\mathcal{A}_{i}$. We define $\varphi_{\alpha}$ on $

{\displaystyle\prod_{i\in I}}

\mathcal{A}_{i}$ by
\[
\varphi_{\alpha}\left(  \left\{  a_{i}\right\}  \right)  =\lim_{i\rightarrow
\alpha}\varphi_{i}\left(  a_{i}\right)  .
\]
Let $\mathcal{M}\subset\mathcal{A}$ be a maximal tracial C*-algebra with
respect to $\varphi$. We define the maximal tracial ultraproduct with respect
to $\mathcal{M}$, denoted by
\[

{\displaystyle\prod_{\mathcal{M}}^{\alpha}}

\left(  \mathcal{A}_{i},\varphi_{i}\right)  ,
\]
to be $\mathcal{M}/\mathcal{J}_{\alpha}\left(  \mathcal{M}\right)  $, where
$\mathcal{J}\left(  \alpha,\mathcal{M}\right)  =\left\{  a\in\mathcal{M}
:\varphi_{\alpha}\left(  a^{\ast}a\right)  =0\right\}  $. Since $\varphi
_{\alpha}|_{\mathcal{M}}$ is tracial, $\varphi_{\alpha}$ induces a tracial
state $\hat{\varphi}_{\alpha}.$

If each $\varphi_{i}$ is tracial, then $\mathcal{A}$ is $\varphi$-tracial and
$

{\displaystyle\prod_{\mathcal{A}}^{\alpha}}

\left(  \mathcal{A}_{i},\varphi_{i}\right)  $ is the usual tracial
ultraproduct. If $\varphi$ is tracial on $\mathcal{A}$, then $

{\displaystyle\prod_{\mathcal{A}}^{\alpha}}

\left(  \mathcal{A}_{i},\varphi_{i}\right)  $ is the tracial ultraproduct
defined by H. Ando and E. Kirchberg \cite{AndoKir}.

One of the biggest open problems in the theory of von Neumann algebras is
\emph{Connes' Embedding Problem} \cite{Connes}, which asks if every von
Neumann algebra $\mathcal{R}$ acting on a separable Hilbert space with a
faithful normal tracial state $\tau$ can be tracially embedded into a tracial
ultraproduct $

{\displaystyle\prod_{i\in I}^{\alpha}}

\left(  \mathcal{A}_{i},\tau_{i}\right)  $ with each $\mathcal{A}_{i}$
finite-demensional. This means there is a unital $\ast$-homomorphism
$\pi:\mathcal{R}\rightarrow

{\displaystyle\prod_{i\in I}^{\alpha}}

\left(  \mathcal{A}_{i},\tau_{i}\right)  $ such that $\tau=\hat{\tau}_{\alpha
}\circ\pi$.

The analogue of Connes' embedding problem has an affirmative answer in the
setting of maximal tracial ultraproducts.

\begin{theorem}
\label{60}Every von Neumann algebra $\mathcal{R}$ acting on a separable
Hilbert space with a faithful normal tracial state $\tau$ can be tracially
embedded into a tracial ultraproduct of finite-dimensional algebras.
\end{theorem}

\begin{proof}
We can assume that $\mathcal{R}\subset B\left(  \mathcal{H}\right)  $ for some
separable infinite-dimensional Hilbert space $\mathcal{H}$ and that there is a
unit vector $f\in\mathcal{H}$ such that, for every $T\in\mathcal{R}$,
\[
\tau\left(  T\right)  =\left\langle Tf,f\right\rangle .
\]
We can also assume that $f$ is a cyclic vector for $\mathcal{R}$. Let
$E=\left\{  e_{1},e_{2},\ldots\right\}  $ be an orthonormal basis for
$\mathcal{H}$ such that $e_{1}=f$. For each positive integer $n$, let
$P_{n}\ $be the orthogonal projection onto the linear span of $\left\{
e_{1},\ldots,e_{n}\right\}  $. Let $\mathcal{A}_{n}=P_{n}B\left(  H\right)
P_{n}$ and define a state $\varphi_{n}:\mathcal{A}_{n}\rightarrow\mathbb{C}$
by
\[
\varphi_{n}\left(  A\right)  =\left\langle Af,f\right\rangle =\left\langle
Ae_{1},e_{1}\right\rangle .
\]
Let $\alpha$ be any free ultrafilter on $\mathbb{N}$. Let
\[
\mathcal{M}_{0}=\left\{  \left\{  A_{n}\right\}  \in

{\displaystyle\prod_{n\in\mathbb{N}}}

\mathcal{A}_{n}:\left\{  A_{n}\right\}  \text{ is *SOT convergent, }
\lim_{n\rightarrow\infty}A_{n}\in\mathcal{R}\right\}  \text{.}
\]
Since $\tau$ is a trace on $\mathcal{R}$, for $S=\left\{  A_{n}\right\}
,T=\left\{  B_{n}\right\}  \in\mathcal{M}_{0}$ with $A_{n}\rightarrow
A\in\mathcal{R}$ and $B_{n}\rightarrow B\in\mathcal{R}$ $\left(
\text{*SOT}\right)  $, we have
\[
A_{n}B_{n}\rightarrow AB\text{ and }B_{n}A_{n}\rightarrow BA\text{ }\left(
\text{*SOT}\right)  .
\]
Thus
\[
\varphi_{\alpha}\left(  ST\right)  =\tau\left(  AB\right)  =\tau\left(
BA\right)  =\varphi_{\alpha}\left(  TS\right)  .
\]
Hence $\varphi_{\alpha}$ is tracial for the C*-algebra $\mathcal{M}_{0}$. Then
there is a maximal $\varphi_{\alpha}$-tracial C*-algebra $\mathcal{M}\subset

{\displaystyle\prod_{n\in\mathbb{N}}}

\mathcal{A}_{n}.$

Let $\eta:\mathcal{M\rightarrow M}/\left(  \mathcal{M}\cap\mathcal{J}_{\alpha
}\right)  =

{\displaystyle\prod_{\mathcal{M}}^{\alpha}}

\left(  \mathcal{A}_{n},\varphi_{n}\right)  $ be the quotient map and define
$\pi:\mathcal{R}\rightarrow

{\displaystyle\prod_{\mathcal{M}}^{\alpha}}

\left(  \mathcal{A}_{n},\varphi_{n}\right)  $ by
\[
\gamma\left(  A\right)  =\eta\left(  \left\{  P_{n}AP_{n}\right\}  \right)  .
\]
We easily see that, for every $A\in\mathcal{R}$,
\[
\hat{\varphi}_{\alpha}\left(  \pi\left(  A\right)  \right)  =\lim_{\alpha
}\left\langle P_{n}AP_{n}f,f\right\rangle =\left\langle Af,f\right\rangle
=\tau\left(  A\right)  .
\]
Thus $\hat{\varphi}_{\alpha}\circ\pi=\tau$. Since $\tau$ is faithful on
$\mathcal{R}$, $\pi$ is an embedding.
\end{proof}


text:
\section{Projection Robust Wasserstein Distance}\label{sec:background}
In this section, we present the basic setup and optimality conditions for the computation of the projection robust 2-Wasserstein (PRW) distance between two discrete probability measures with at most $n$ components. We also review basic ideas in Riemannian optimization.

\subsection{Structured max-min optimization model}
In this section we define the PRW distance~\citep{Paty-2019-Subspace} and show that computing the PRW distance between two discrete probability measures supported on at most $n$ points reduces to solving a structured max-min optimization model over the Stiefel manifold and the transportation polytope.

Let $\PScr(\br^d)$ be the set of Borel probability measures in $\br^d$ and let $\PScr_2(\br^d)$ be the subset of $\PScr(\br^d)$ consisting of probability measures that have finite second moments. Let $\mu, \nu \in \PScr_2(\br^d)$ and $\Pi(\mu, \nu)$ be the set of couplings between $\mu$ and $\nu$. The 2-Wasserstein distance~\citep{Villani-2008-Optimal} is defined by
\begin{equation*}
\WCal_2(\mu, \nu) \ \mydefn \ \left(\inf_{\pi \in \Pi(\mu, \nu)} \int \|x-y\|^2 \ d\pi(x, y)\right)^{1/2}.
\end{equation*}
To define the PRW distance, we require the notion of the push-forward of a measure by an operator. Letting $\XCal, \YCal \subseteq \br^d$ and $T: \XCal \rightarrow \YCal$, the push-forward of $\mu \in \PScr(\XCal)$ by $T$ is defined by $T_{\#} \mu \in \mathscr{P}(\YCal)$. In other words, $T_{\#} \mu$ is the measure satisfying $T_{\#} \mu(A) = \mu(T^{-1}(A))$ for any Borel set in $\YCal$.
\begin{definition}
For $\mu, \nu \in \PScr_2(\br^d)$, let $\GCal_k = \{E \subseteq \br^d \mid \dim(E) = k\}$ be the Grassmannian of $k$-dimensional subspace of $\br^d$ and let $P_E$ be the orthogonal projector onto $E$ for all $E \in \GCal_k$.  The $k$-dimensional PRW distance is defined as $\PCal_k(\mu, \nu) \mydefn \sup_{E \in \GCal_k} \WCal_2(P_{E\#}\mu, P_{E\#}\nu)$.
\end{definition}
\citet[Proposition~5]{Paty-2019-Subspace} have shown that there exists a subspace $E^* \in \GCal_k$ such that $\PCal_k(\mu, \nu) = \WCal_2(P_{E^*\#}\mu, P_{E^*\#}\nu)$ for any $k \in [d]$ and $\mu, \nu \in \PScr_2(\br^d)$. For any $E \in \GCal_k$, the mapping $\pi \mapsto \int \|P_E(x-y)\|^2 \ d\pi(x, y)$ is lower semi-continuous. This together with the compactness of $\Pi(\mu, \nu)$ implies that the infimum is a minimum. Therefore, we obtain a structured max-min optimization problem:
\begin{equation*}
\PCal_k(\mu, \nu) = \max_{E \in \GCal_k} \min_{\pi \in \Pi(\mu, \nu)} \left(\int \|P_E(x-y)\|^2 \ d\pi(x, y)\right)^{1/2}.
\end{equation*}
Let us now consider this general problem in the case of discrete probability measures, which is the focus of the current paper. Let $\{x_1, x_2, \ldots, x_n\} \subseteq \br^d$ and $\{y_1, y_2, \ldots, y_n\} \subseteq \br^d$ denote sets of $n$ atoms, and let $(r_1, r_2, \ldots, r_n) \in \Delta^n$ and $(c_1, c_2, \ldots, c_n) \in \Delta^n$ denote weight vectors.  We define discrete probability measures $\mu \mydefn \sum_{i=1}^n r_i\delta_{x_i}$ and $\nu \mydefn \sum_{j=1}^n c_j\delta_{y_j}$. In this setting, the computation of the $k$-dimensional PRW distance between $\mu$ and $\nu$ reduces to solving a structured max-min optimization model where the maximization and minimization are performed over the Stiefel manifold $\St(d, k) : = \{U \in \br^{d \times k} \mid U^\top U = I_k\}$ and the transportation polytope $\Pi(\mu, \nu) : = \{\pi \in \br_+^{n \times n} \mid r(\pi) = r, \ c(\pi) = c\}$ respectively. Formally, we have
\begin{equation}\label{prob:main}
\max\limits_{U \in \br^{d \times k}} \min\limits_{\pi \in \br_+^{n \times n}} \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j}\|U^\top x_i - U^\top y_j\|^2 \quad \st \ U^\top U = I_k, \ r(\pi) = r, \ c(\pi) = c.
\end{equation}
The computation of this PRW distance raises numerous challenges. Indeed, there is no guarantee for finding a global Nash equilibrium as the special case of nonconvex optimization is already NP-hard~\citep{Murty-1987-Some}; moreover, Sion's minimax theorem~\citep{Sion-1958-General} is not applicable here due to the lack of quasi-convex-concave structure. More practically, solving Eq.~\eqref{prob:main} is expensive since (i) preserving the orthogonality constraint requires the singular value decompositions (SVDs) of a $d \times d$ matrix, and (ii) projecting onto the transportation polytope results in a costly quadratic network flow problem. To avoid this, ~\citep{Paty-2019-Subspace} proposed a convex surrogate for Eq.~\eqref{prob:main}:
\begin{equation}\label{prob:main-CCP}
\max\limits_{0 \preceq \Omega \preceq I_d} \min\limits_{\pi \in \br_+^{n \times n}} \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j}(x_i - y_j)^\top\Omega(x_i - y_j), \quad \st \ \trac(\Omega)=k, \ r(\pi) = r, \ c(\pi) = c.
\end{equation}
Eq.~\eqref{prob:main-CCP} is intrinsically a bilinear minimax optimization model which makes the computation tractable. Indeed, the constraint set $\RCal = \{\Omega \in \br^{d \times d} \mid 0 \preceq \Omega \preceq I_d, \trac(\Omega)=k\}$ is convex and the objective function is bilinear since it can be rewritten as $\langle \Omega, \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j} (x_i - y_j)(x_i - y_j)^\top\rangle$.  Eq.~\eqref{prob:main-CCP} is, however, only a convex relaxation of Eq.~\eqref{prob:main} and its solutions are not necessarily good approximate solutions for the original problem. Moreover, the existing algorithms for solving Eq.~\eqref{prob:main-CCP} are also unsatisfactory---in each loop, we need to solve a OT or entropic regularized OT exactly and project a $d \times d$ matrix onto the set $\RCal$ using the SVD decomposition, both of which are computationally expensive as $d$ increases (see Algorithm~1 and~2 in \cite{Paty-2019-Subspace}).
\begin{algorithm}[!t]
\caption{Riemannian Gradient Ascent with Sinkhorn Iteration (RGAS)}\label{alg:grad-sinkhorn}
\begin{algorithmic}[1]
\STATE \textbf{Input:} $\{(x_i, r_i)\}_{i \in [n]}$ and $\{(y_j, c_j)\}_{j \in [n]}$, $k = \bigOtil(1)$, $U_0 \in \St(d, k)$ and $\epsilon$.
\STATE \textbf{Initialize:} $\widehat{\epsilon} \leftarrow \frac{\epsilon}{10\|C\|_\infty}$, $\eta \leftarrow \frac{\epsilon\min\{1, 1/\bar{\theta}\}}{40\log(n)}$ and $\gamma \leftarrow \frac{1}{(8 L_1^2 + 16L_2)\|C\|_\infty + 16 \eta^{-1}L_1^2\|C\|_\infty^2}$.
\FOR{$t = 0, 1, 2, \ldots$}
\STATE Compute $\pi_{t+1} \leftarrow \textsc{regOT}(\{(x_i, r_i)\}_{i \in [n]}, \{(y_j, c_j)\}_{j \in [n]}, U_t, \eta, \widehat{\epsilon})$.
\STATE Compute $\xi_{t+1} \leftarrow P_{\Tg_{U_t}\St}(2V_{\pi_{t+1}}U_t)$.
\STATE Compute $U_{t+1} \leftarrow \retr_{U_t}(\gamma\xi_{t+1})$.
\ENDFOR
\end{algorithmic}
\end{algorithm}

\subsection{Entropic regularized projection robust Wasserstein}
Eq.~\eqref{prob:main} has structure that can be exploited. Indeed, fixing a $U \in \St(d, k)$, the problem reduces to minimizing a linear function over the transportation polytope, i.e., the OT problem. Therefore, we can reformulate Eq.~\eqref{prob:main} as the maximization of the function $f(U) \mydefn \min_{\pi \in \Pi(\mu, \nu)} \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j} \|U^\top x_i - U^\top y_j\|^2$ over the Stiefel manifold $\St(d, k)$.

Since the OT problem admits multiple optimal solutions, $f$ is not differentiable which makes the optimization over the Stiefel manifold hard~\citep{Absil-2019-Collection}. Computations are greatly facilitated by adding smoothness, which allows the use of gradient-type and adaptive gradient-type algorithms. This inspires us to consider an entropic regularized version of Eq.~\eqref{prob:main}, where an entropy penalty is added to the PRW distance. The resulting optimization model is as follows:
\begin{equation}\label{prob:main-regularized}
\max\limits_{U \in \br^{d \times k}} \min\limits_{\pi \in \br_+^{n \times n}} \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j}\|U^\top x_i - U^\top y_j\|^2 - \eta H(\pi) \quad \st \ U^\top U = I_k, \ r(\pi) = r, \ c(\pi) = c,
\end{equation}
where $\eta > 0$ is the regularization parameter and $H(\pi) \mydefn - \langle \pi, \log(\pi) - \one_n\one_n^\top\rangle$ denotes the entropic regularization term. We refer to Eq.~\eqref{prob:main-regularized} as the computation of \emph{entropic regularized PRW distance}. Accordingly, we define the function $f_\eta = \min_{\pi \in \Pi(\mu, \nu)} \{\sum_{i=1}^n \sum_{j=1}^n \pi_{i, j} \|U^\top x_i - U^\top y_j\|^2 - \eta H(\pi)\}$ and reformulate Eq.~\eqref{prob:main-regularized} as the maximization of the differentiable function $f_\eta$ over the Stiefel manifold $\St(d, k)$. Indeed, for any $U \in \St(d, k)$ and a fixed $\eta > 0$, there exists a unique solution $\pi^* \in \Pi(\mu, \nu)$ such that $\pi \mapsto \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j}\|U^\top x_i - U^\top y_j\|^2 - \eta H(\pi)$ is minimized at $\pi^*$. When $\eta$ is large, the optimal value of Eq.~\eqref{prob:main-regularized} may yield a poor approximation of Eq.~\eqref{prob:main}. To guarantee a good approximation, we scale the regularization parameter $\eta$ as a function of the desired accuracy of the approximation. Formally, we consider the following relaxed optimality condition for $\widehat{\pi} \in \Pi(\mu, \nu)$ given $U \in \St(d, k)$.
\begin{definition}\label{def:approx_transportation_plan}
The transportation plan $\widehat{\pi} \in \Pi(\mu, \nu)$ is called an \emph{$\epsilon$-approximate optimal transportation plan for a given $U \in \St(d, k)$} if the following inequality holds:
\begin{equation*}
\sum_{i=1}^n \sum_{j=1}^n \widehat{\pi}_{i, j} \|U^\top x_i - U^\top y_j\|^2 \ \leq \ \min\limits_{\pi \in \Pi(\mu, \nu)} \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j} \|U^\top x_i - U^\top y_j\|^2 + \epsilon.
\end{equation*}
\end{definition}

\subsection{Optimality condition}
Recall that the computation of the PRW distance in Eq.~\eqref{prob:main} and the entropic regularized PRW distance in Eq.~\eqref{prob:main-regularized} are equivalent to
\begin{equation}\label{prob:Stiefel-nonsmooth}
\max\limits_{U \in \St(d, k)} \ \left\{f(U) \mydefn \min\limits_{\pi \in \Pi(\mu, \nu)} \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j} \|U^\top x_i - U^\top y_j\|^2\right\},
\end{equation}
and
\begin{equation}\label{prob:Stiefel-smooth}
\max\limits_{U \in \St(d, k)} \ \left\{f_\eta(U) \mydefn \min\limits_{\pi \in \Pi(\mu, \nu)} \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j} \|U^\top x_i - U^\top y_j\|^2 - \eta H(\pi)\right\}.
\end{equation}
Since $\St(d, k)$ is a compact matrix submanifold of $\br^{d \times k}$~\citep{Boothby-1986-Introduction}, Eq.~\eqref{prob:Stiefel-nonsmooth} and Eq.~\eqref{prob:Stiefel-smooth} are both special instances of the Stiefel manifold optimization problem. The dimension of $\St(d, k)$ is equal to $dk - k(k+1)/2$ and the tangent space at the point $Z \in \St(d, k)$ is defined by $\Tg_Z\St \mydefn \{\xi \in \br^{d \times k}: \xi^\top Z + Z^\top\xi = 0\}$. We endow $\St(d, k)$ with Riemannian metric inherited from the Euclidean inner product $\langle X, Y\rangle$ for any $X, Y \in \Tg_Z\St$ and $Z \in \St(d, k)$. Then the projection of $G \in \br^{d \times k}$ onto $\Tg_Z\St$ is given by~\citet[Example~3.6.2]{Absil-2009-Optimization}: $P_{\Tg_Z\St}(G) = G - Z(G^\top Z + Z^\top G)/2$. We make use of the notion of a \textit{retraction}, which is the first-order approximation of an exponential mapping on the manifold and which is amenable to computation~\citep[Definition~4.1.1]{Absil-2009-Optimization}. For the Stiefel manifold, we have the following definition:
\begin{definition}\label{def:retraction}
A retraction on $\St \equiv \St(d, k)$ is a smooth mapping $\retr: \Tg\St \rightarrow \St$ from the tangent bundle $\Tg\St$ onto $\St$ such that the restriction of $\retr$ onto $\Tg_Z\St$, denoted by $\retr_Z$, satisfies that (i) $\retr_Z(0) = Z$ for all $Z \in \St$ where $0$ denotes the zero element of $\textnormal{T}\St$, and (ii) for any $Z \in \St$, it holds that $\lim_{\xi \in \Tg_Z\St, \xi \rightarrow 0} \|\retr_Z(\xi) - (Z + \xi)\|_F/\|\xi\|_F = 0$.
\end{definition}
The retraction on the Stiefel manifold has the following well-known properties~\citep{Boumal-2019-Global, Liu-2019-Quadratic} which are important to subsequent analysis in this paper.
\begin{proposition}\label{prop:retraction}
For all $Z \in \St \equiv \St(d, k)$ and $\xi \in \Tg_Z\St$, there exist constants $L_1 > 0$ and $L_2 > 0$ such that the following two inequalities hold:
\begin{eqnarray*}
\|\retr_Z(\xi) - Z\|_F & \leq & L_1\|\xi\|_F, \\
\|\retr_Z(\xi) - (Z + \xi)\|_F & \leq & L_2\|\xi\|_F^2.
\end{eqnarray*}
\end{proposition}
For the sake of completeness, we provide four popular restrictions~\citep{Edelman-1998-Geometry, Wen-2013-Feasible, Liu-2019-Quadratic, Chen-2020-Proximal} on the Stiefel manifold in practice. Determining which one is the most efficient in the algorithm is still an open question; see the discussion after~\citet[Theorem~3]{Liu-2019-Quadratic} or before~\citet[Fact~3.6]{Chen-2020-Proximal}.
\begin{itemize}
\item \textbf{Exponential mapping.} It takes $8dk^2 + \bigO(k^3)$ flops and has the closed-form expression:
\begin{equation*}
\retr_Z^{\exp}(\xi) \ = \ \begin{bmatrix} Z & Q\end{bmatrix}\exp\left(\begin{bmatrix} -Z^\top\xi & -R^\top \\ R & 0 \end{bmatrix}\right)\begin{bmatrix} I_k \\ 0 \end{bmatrix}.
\end{equation*}
where $QR = -(I_k - ZZ^\top)\xi$ is the unique QR factorization.
\item \textbf{Polar decomposition.} It takes $3dk^2 + \bigO(k^3)$ flops and has the closed-form expression:
\begin{equation*}
\retr_Z^{\textnormal{polar}}(\xi) \ = \ (Z + \xi)(I_k + \xi^\top\xi)^{-1/2}.
\end{equation*}
\item \textbf{QR decomposition.} It takes $2dk^2 + \bigO(k^3)$ flops and has the closed-form expression:
\begin{equation*}
\retr_Z^{\textnormal{qr}}(\xi) \ = \ \textnormal{qr}(Z + \xi),
\end{equation*}
where $\textnormal{qr}(A)$ is the Q factor of the QR factorization of $A$.
\item \textbf{Cayley transformation.} It takes $7dk^2 + \bigO(k^3)$ flops and has the closed-form expression:
\begin{equation*}
\retr_Z^{\textnormal{cayley}}(\xi) \ = \ \left(I_n - \frac{1}{2}W(\xi) \right)^{-1}\left(I_n + \frac{1}{2}W(\xi) \right)Z,
\end{equation*}
where $W(\xi) = (I_n - ZZ^\top/2)\xi Z^\top - Z\xi^\top(I_n - ZZ^\top/2)$.
\end{itemize}
We now present a novel approach to exploiting the structure of $f$. We begin with several definitions.
\begin{algorithm}[!t]
\caption{Riemannian Adaptive Gradient Ascent with Sinkhorn Iteration (RAGAS)}\label{alg:adagrad-sinkhorn}
\begin{algorithmic}[1]
\STATE \textbf{Input:} $\{(x_i, r_i)\}_{i \in [n]}$ and $\{(y_j, c_j)\}_{j \in [n]}$, $k = \bigOtil(1)$, $U_0 \in \St(d, k)$, $\epsilon$ and $\alpha \in (0, 1)$.
\STATE \textbf{Initialize:} $p_0 = \zero_d$, $q_0 = \zero_k$, $\widehat{p_0} = \alpha\|C\|_\infty^2\one_d$, $\widehat{q_0} = \alpha\|C\|_\infty^2\one_k$, $\widehat{\epsilon} \leftarrow \frac{\epsilon\sqrt{\alpha}}{20\|C\|_\infty}$, $\eta \leftarrow \frac{\epsilon\min\{1, 1/\bar{\theta}\}}{40\log(n)}$ and $\gamma \leftarrow \frac{\alpha}{16L_1^2 + 32L_2 + 32\eta^{-1}L_1^2\|C\|_\infty}$.
\FOR{$t = 0, 1, 2, \ldots$}
\STATE Compute $\pi_{t+1} \leftarrow \textsc{regOT}(\{(x_i, r_i)\}_{i \in [n]}, \{(y_j, c_j)\}_{j \in [n]}, U_t, \eta, \widehat{\epsilon})$.
\STATE Compute $G_{t+1} \leftarrow P_{\Tg_{U_t}\St}(2V_{\pi_{t+1}}U_t)$.
\STATE Update $p_{t+1} \leftarrow \beta p_t + (1-\beta)\diag(G_{t+1}G_{t+1}^\top)/k$ and $\widehat{p}_{t+1} \leftarrow \max\{\widehat{p}_t, p_{t+1}\}$.
\STATE Update $q_{t+1} \leftarrow \beta q_t + (1-\beta)\diag(G_{t+1}^\top G_{t+1})/d$ and $\widehat{q}_{t+1} \leftarrow \max\{\widehat{q}_t, q_{t+1}\}$.
\STATE Compute $\xi_{t+1} \leftarrow P_{\Tg_{U_t}\St}(\Diag(\widehat{p}_{t+1})^{-1/4}G_{t+1}\Diag(\widehat{q}_{t+1})^{-1/4})$.
\STATE Compute $U_{t+1} \leftarrow \retr_{U_t}(\gamma\xi_{t+1})$.
\ENDFOR
\end{algorithmic}
\end{algorithm}
\begin{definition}\label{def:C}
The \emph{coefficient matrix} between $\mu = \sum_{i=1}^n r_i\delta_{x_i}$ and $\nu = \sum_{j=1}^n c_j\delta_{y_j}$ is defined by $C = (C_{ij})_{1 \leq i, j \leq n} \in \br^{n \times n}$ with each entry $C_{ij} = \|x_i - y_j\|^2$.
\end{definition}
\begin{definition}\label{def:V}
The \emph{correlation matrix} between $\mu = \sum_{i=1}^n r_i\delta_{x_i}$ and $\nu = \sum_{j=1}^n c_j\delta_{y_j}$ is defined by $V_\pi = \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j} (x_i - y_j)(x_i - y_j)^\top \in \br^{d \times d}$.
\end{definition}
The first lemma shows that the structure of the function $f$ is not very bad regardless of nonconvexity and the lack of smoothness.
\begin{lemma}\label{lemma:obj-weak-concave}
The function $f$ is $2\|C\|_\infty$-weakly concave.
\end{lemma}
\begin{proof}
By~\citet[Proposition~4.3]{Vial-1983-Strong}, it suffices to show that the function $f(U) - \|C\|_\infty\|U\|_F^2$ is concave for any $U \in \br^{d \times k}$. By the definition of $f$, we have
\begin{equation*}
f(U) \ = \ \min\limits_{\pi \in \Pi(\mu, \nu)} \trac\left(U^\top V_\pi U\right).
\end{equation*}
Since $\{x_1, x_2, \ldots, x_n\} \subseteq \br^d$ and $\{y_1, y_2, \ldots, y_n\} \subseteq \br^d$ are two given groups of $n$ atoms in $\br^d$, the coefficient matrix $C$ is independent of $U$ and $\pi$. Furthermore, $\sum_{i=1}^n \sum_{j=1}^n \pi_{i, j} = 1$ and $\pi_{i, j} \geq 0$ for all $i, j \in [n]$ since $\pi \in \Pi(\mu, \nu)$. Putting these pieces together with Jensen's inequality, we have
\begin{equation*}
\|V_\pi\|_F \ \leq \ \sum_{i=1}^n \sum_{j=1}^n \pi_{i, j} \|(x_i - y_j)(x_i - y_j)^\top\|_F \ \leq \ \max_{1 \leq i, j \leq n} \|x_i - y_j\|^2 \ = \ \|C\|_\infty.
\end{equation*}
This implies that $U \mapsto \trac(U^\top V_\pi U) - \|C\|_\infty\|U\|_F^2$ is concave for any $\pi \in \Pi(\mu, \nu)$. Since $\Pi(\mu, \nu)$ is compact, Danskin's theorem~\citep{Rockafellar-2015-Convex} implies the desired result.
\end{proof}
The second lemma shows that the subdifferential of the function $f$ is independent of $U$ and bounded by a constant $2\|C\|_\infty$.
\begin{lemma}\label{lemma:bound-subdiff}
Each element of the subdifferential $\partial f(U)$ is bounded by $2\|C\|_\infty$ for all $U \in \St(d, k)$.
\end{lemma}
\begin{proof}
By the definition of the subdifferential $\partial f$, it suffices to show that $\|V_\pi U \|_F \leq \|C\|_\infty$ for all $\pi \in \Pi(\mu, \nu)$ and $U \in \St(d, k)$. Indeed, by the definition, $V_\pi$ is symmetric and positive semi-definite. Therefore, we have
\begin{equation*}
\max_{U \in \St(d, k)} \|V_\pi U \|_F \ \leq \ \|V_\pi\|_F \ \leq \ \|C\|_\infty.
\end{equation*}
Putting these pieces together yields the desired result.
\end{proof}
\begin{remark}
Lemma~\ref{lemma:obj-weak-concave} implies there exists a concave function $g: \br^{d \times k} \rightarrow \br$ such that $f(U) = g(U) + \|C\|_\infty\|U\|_F^2$ for any $U \in \br^{d \times k}$. Since $g$ is concave, $\partial g$ is well defined and~\citet[Proposition~4.6]{Vial-1983-Strong} implies that $\partial f(U) = \partial g(U) + 2\|C\|_\infty U$ for all $U \in \br^{d \times k}$.
\end{remark}
This result together with~\citet[Proposition~4.5]{Vial-1983-Strong} and~\citet[Theorem~5.1]{Yang-2014-Optimality} lead to the Riemannian subdifferential defined by $\subdiff f(U) = P_{\Tg_U\St}(\partial f(U))$ for all $U \in \St(d, k)$.
\begin{definition}\label{def:stationarity}
The subspace projection $\widehat{U} \in \St(d, k)$ is called an \emph{$\epsilon$-approximate optimal subspace projection} of $f$ over $\St(d, k)$ in Eq.~\eqref{prob:Stiefel-nonsmooth} if it satisfies  $\dist(0, \subdiff f(\widehat{U})) \leq \epsilon$.
\end{definition}
\begin{definition}\label{def:optimal-pair}
The pair of subspace projection and transportation plan $(\widehat{U}, \widehat{\pi}) \in \St(d, k) \times \Pi(\mu, \nu)$ is an \emph{$\epsilon$-approximate pair of optimal subspace projection and optimal transportation plan} for the computation of the PRW distance in Eq.~\eqref{prob:main} if the following statements hold true: (i) $\widehat{U}$ is an $\epsilon$-approximate optimal subspace projection of $f$ over $\St(d, k)$ in Eq.~\eqref{prob:Stiefel-nonsmooth}. (ii) $\widehat{\pi}$ is an $\epsilon$-approximate optimal transportation plan for the subspace projection $\widehat{U}$.
\end{definition}
The goal of this paper is to develop a set of algorithms which are guaranteed to converge to a pair of approximate optimal subspace projection and optimal transportation plan, which stand for a stationary point of the max-min optimization model in Eq.~\eqref{prob:main}. In the next section, we provide the detailed scheme of our algorithm as well as the finite-time theoretical guarantee.

config: arxiv


TITLE: Why can you transition from a Riemann Sum to an Intergal?
QUESTION [0 upvotes]: I understand that the integral is defined through a Riemann Sum, as in
$$\int_a^bf(x)\,dx=\lim_{n\to\infty}\sum_{i=0}^{n-1}f\left(a+i\cdot\frac{b-a}{n}\right)\frac{b-a}{n}$$
But I don't understand how this definition is made. I understand the structure of the Riemann Sum which is summing up infinetely many infinietely thin rectangles to approximate the area under a curve, but I don't understand why this turns into an integral and then can be evaluated the way an integral is. Any help is appreciated.

REPLY [1 votes]: Perhaps a good way to think about that formula is the following:
There exist some functions f defined over the interval [a,b] which have the property that the limit on the right hand side exists and the sequence of sums converges to that limit. In the case where that happens we will name that limit, or call that limit by the symbology on the left hand side.
So then in the Analysis class or Advanced Calculus class, (depending on how the classes are named, in the old days and still in some places it goes by the name Calculus), you will prove that these limits exist for various families of functions, and you will also prove what is known as the fundamental theorem of calculus.
When you say evaluation you are applying the fundamental theorem of calculus and some of the other theorems which you will prove in the Analysis or the Advanced Calculus class.
So when you say:
"but I don't understand why this turns into an integral and then can be evaluated the way an integral is."
I interpret that to mean that you are not confused as to what the evaluation rules are and how to evaluate per se, but rather why you are allowed to use those rules to evaluate.
The answer to that query is the subject matter of what could be a first Analysis course and exceeds the length and detail allowed in this format. I encourage you to look into it further as time allows.
If time is an issue and you are trying to get through your current course, you can be satisfied applying the evaluation rules to the best of your ability to all of the functions you will encounter in your course and later, should you perchance progress further in your study, the mystery of why you are allowed to use those evaluation rules will be cleared up.


TITLE: Finding kernel for a matrix
QUESTION [2 upvotes]: I'm having trouble understanding what it means to find the kernel of a matrix.
I have a matrix as follows and I need to determine the kernel.
$\begin{bmatrix}2 & 4 & 5 \\ 1 & 2 & 3 \\ -1 & -1 & -3 \end{bmatrix}$
I'm able to get the augmented matrix down to echelon form as follows.
$\begin{bmatrix}1 & 2 & 3 & 0 \\ 0 & 5 & 4 & 0\end{bmatrix}$
Where do I take it from here? What would the form of the solution be?
Would I just write a couple of equations like the following two?
$x_1+2x_2+3x_3=0$
$0x_1+5x_2+4x_3=0$

REPLY [1 votes]: I found that this Khan Academy video was the best resource for helping me understand this problem more.
In the end I wrote the following for my answer.
$x_1=-\frac{7}{5}x_3+\frac{1}{5}x_4$
$x_2=-\frac{4}{5}x_3-\frac{3}{5}x_4$
I wasn't certain how to express this with $t$, as Dovah-king and the video mentioned, because we have two free variables, unlike the example in the video.


TITLE: Sequences and subsequences convergence
QUESTION [0 upvotes]: Prove or disprove:

Let $a_n$ sequence, such that for any prime number p, the subsequence $a_{kp}$ is converging, so $a_n$ is necessarily converging.

I tried to find a counter example, such that there are two different sub sequences which converge to different limits, but didn't succeed. I think that the claim is wrong, but can't show it.
Please help, thank you!

REPLY [0 votes]: Hint: Let $n_j = (\prod_{i=1}^j p_i) + 1$ for the first $j$ primes $p_i$ and define this special subsequence $a_{n_j}$ to be $1$, and let $a_n = 0$ for all other $n$ that are not of the form. Show that this sequence has all subsequences $\{a_{kp}\}_{k=1}^\infty$ converging to $0$ but that the special subsequence $a_{n_j}$ converges to 1 to so that the overall sequence $a_n$ is not convergent.