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\topmatter
\title On the Statistical Independence of Algebras of Observables \endtitle
\author Martin Florig and Stephen J. Summers \endauthor
\affil Department of Mathematics \\
University of Florida \\
Gainesville, FL 32611 \\
USA \endaffil
\date{May 20, 1996}
\enddate
\abstract{We re-examine various notions of statistical independence presently
in use in algebraic quantum theory, establishing alternative characterizations
for such independence, some of which are also valid without assuming that
the observable algebras mutually commute. In addition, in the
context which holds in concrete applications to quantum theory, the equivalence
of three major notions of statistical independence is proven.}
\endabstract
\endtopmatter
\document
\heading Introduction \endheading
The notions of ``independence of two systems'' are legion in quantum
theory. This is quite understandable, since the physical concept is central in
many aspects of quantum theory, and the various formalizations of independence
capture different qualitative and quantitative aspects for possibly different
ends.\footnote{See \cite{8}\cite{12}\cite{20} for recent
applications
of these notions.} Those notions which have appeared in the literature and
have formulations in algebraic quantum theory have been extensively reviewed
in \cite{24}, where their logical interrelationships have been discussed.
However, some issues of logical relation were left open in that review. One
of the goals of this paper is to settle the few remaining conjectures in
\cite{24}. \par
Representing the algebras of observables associated to the two subsystems
by $\As$ and $\Bs$, respectively, one of the most commonly used expressions
of independence is the requirement that the algebras mutually commute
elementwise. Another is expressed heuristically in the condition that each
system can be prepared independently of the other. It is known
\cite{10}\cite{18}\cite{5}\cite{16}\cite{6} that this latter notion
of statistical independence is logically independent of the requirement
that the algebras $\As$ and $\Bs$ commute. In \cite{24} the various versions
of statistical independence utilized in the literature were discussed
almost exclusively in the context of commuting pairs of algebras. Another
purpose of this paper is to provide further information about statistical
independence in the more general circumstance that the observable algebras do
not necessarily commute. \par
We shall provide alternative characterizations of $C^*$- and
$W^*$-independence (see below for definitions), which are also valid if the
algebras do not necessarily commute. In addition, we shall prove that in the
category of commuting pairs of von Neumann algebras on separable Hilbert
spaces, where all three notions are applicable, $C^*$-independence, strict
locality and $W^*$-independence are equivalent. This is precisely the setting
most often met in applications to quantum theory. Moreover, we shall
furnish some new results about the notion of $C^*$-independence in the
product sense, shall show that $W^*$-independence is strictly weaker than
$W^*$-independence in the product sense, and shall close with some comments
concerning the question of additional conditions sufficient to conclude the
mutual commutativity of a pair of $C^*$-independent algebras. \par
Throughout this paper a few assumptions will be made tacitly, since
they obtain in the applications to theoretical physics known to us. All
algebras are assumed to have identities (represented by 1), and subalgebras
considered here will always contain the identity of the original algebra.
The symbol $\Bs(\Hs)$ will be used to denote the algebra of all bounded
operators on the Hilbert space $\Hs$. If $\As$ is a subset of $\Bs(\Hs)$, then
$\As'$ represents the von Neumann algebra of all elements of $\Bs(\Hs)$ which
commute with every element of $\As$. If $\phi$ is a state on $\Cs$ and
$\As \subset \Cs$ is a subalgebra, then the restriction of $\phi$ to $\As$
will be represented by $\phi \mid \As$. In addition, if $\As$ and $\Bs$ are
two $C^*$- , resp. $W^*$-, subalgebras, then $\As\vee\Bs$ will denote the
$C^*$- , resp. $W^*$-, algebra they generate. \par
\heading Statistical Independence \endheading
In quantum theory, observables (or effects) are represented by
selfadjoint operators and preparations by states on the *-algebra
generated by the observables. Basic observables are represented in
algebraic quantum theory by positive, selfadjoint elements of a $C^*$-algebra
with norm bounded by 1, of which projection operators (or decision effects)
are only a special case. More general observables are assembled from such
basic observables, usually in the form of measures taking such operators
as values, of which the projection-valued measures found in the spectral
theorem for normal operators are a special case (see, {\it e.g.}
\cite{4}\cite{2}). The probability of measuring ``yes'' for the
basic observable $F$ after the preparation represented by the state $\phi$
is $\phi(F)$, while the probability for measuring ``no'' under the same
conditions is $\phi(1-F)$. \par
In \cite{11} Haag and Kastler introduced a notion they called
statistical independence. If $\As$ and $\Bs$ represent the algebras generated
by the observables associated with two quantum subsystems, the statistical
independence of $\As$ and $\Bs$ can be loosely construed as follows:
{\it any} two partial states on the two subsystems can be realized by the
{\it same} preparation procedure (assuming that arbitrary states on the
algebra of observables for the entire quantum system can
actually be prepared); or alternatively, no choice of a state prepared on one
subsystem can prevent the preparation of any state on the other subsystem.
There are different mathematical formulations of this notion. We begin with
the most natural version in the category of $C^*$-algebras. At many points in
the sequel we shall utilize without further comment a basic fact about
$C^*$-algebras: If $a \in \CC$ is any spectral value of a selfadjoint element
$A$ of a $C^*$-algebra $\As$, there exists a state $\varphi$ on $\As$ such
that $\varphi(A) = a$; hence for any normal element $A \in \As$ there exists
a state $\varphi$ on $\As$ such that $\vert\varphi(A)\vert = \Vert A \Vert$.
\proclaim{Definition} Let $\Cal A$ and $\Cal B$ be subalgebras
of a $C^*$-algebra $\Cal C$. The pair $(\As,\Bs)$ is (or $\Cal A$ and $\Cal B$
are) said to be $C^*$-independent if for every state $\phi_1$ on $\Cal A$ and
every state $\phi_2$ on $\Cal B$ there exists a state $\phi$
on $\Cal C$ such that $\phi \mid \Cal A = \phi_1$ and
$\phi \mid \Cal B = \phi_2$.
\endproclaim
The following characterization of $C^*$-independent pairs of
{\it commuting} algebras was proven by Roos \cite{21}. The algebraic tensor
product of two $C^*$-algebras $\As$ and $\Bs$ is represented by $\As\odot\Bs$,
while the minimal $C^*$-tensor product is denoted by $\As\otimes\Bs$.
\proclaim{Theorem 1}{\rm\cite{21}} Let $\Cal A$ and $\Cal B$ be
commuting subalgebras of the $C^*$-algebra $\Cal C$. Then the
following are equivalent. \par
(i) $\Cal A$ and $\Cal B$ are $C^*$-independent. \par
(ii) $0\neq A \in \Cal A$ and $0\neq B \in \Cal B$ imply that $AB\neq 0$.
\par
(iii) The map $\eta : \As\Bs \rightarrow \As \odot \Bs$ defined by
$\eta(AB) = A \otimes B$, $A \in \As$, $B \in \Bs$ is an isomorphism continuous
in the minimal $C^*$-cross norm on $\As \odot \Bs$ and can therefore be
continuously extended to a surjective homomorphism $\overline{\eta} :
\Cal A \vee \Cal B \rightarrow \Cal A \otimes \Cal B$. \par
(iv) For every state $\phi_1$ on $\Cal A$ and every state $\phi_2$ on
$\Cal B$ there exists a state $\phi$ on $\Cal C$ such that
$\phi (AB) = \phi _1 (A)\,\phi _2 (B) = \phi(A)\phi(B)$ for all $A\in \Cal A$
and all $B\in \Cal B$.
\endproclaim
Condition (ii) in Theorem 1 is called the Schlieder property by algebraic
quantum field theorists. We recall that Murray and von Neumann \cite{19} showed
that if $\As$ is a von Neumann factor on a Hilbert space $\Hs$, then the pair
$(\As,\As')$ satisfies the Schlieder condition and is therefore
$C^*$-independent. States $\phi$ on $\Cal C$ with the property that
$\phi (AB) = \phi(A)\,\phi(B)$ for all $A\in \Cal A$ and all
$B\in \Cal B$ are called product states. Roos' theorem therefore establishes
that for $C^*$-independent pairs $(\As,\Bs)$, arbitrary states on $\As$ and
$\Bs$ can be simultaneously extended to a {\it product} state. This suggests
an alternative characterization of $C^*$-independent pairs of commuting
algebras. \par
\proclaim{Proposition 2} If $\As$ and $\Bs$ are commuting subalgebras of a
$C^*$-algebra $\Cs$, then the following are equivalent. \par
(i) $(\As,\Bs)$ is $C^*$-independent. \par
(ii) $\Vert AB \Vert = \Vert A \Vert\Vert B \Vert$, for all $A \in \As$
and $B \in \Bs$.
\endproclaim
\demo{Proof} $(i) \Rightarrow (ii)$: By Roos \cite{21}, if
$(\As,\Bs)$ is $C^*$-independent, then $\As$ and $\Bs$ are algebraically
independent, so by Turumaru \cite{27}, the *-algebra generated by $\As$
and $\Bs$ is isomorphic to $\As \odot \Bs$. Hence the norm on $\Cs$
induces a $C^*$-norm $\alpha$ on $\As \odot \Bs$ by
$\alpha(\sum_{i=1}^n A_i \otimes B_i) = \Vert \sum_{i=1}^n A_iB_i \Vert$.
By Corollary 11.3.10 in \cite{14}, every $C^*$-norm on
$\As \odot \Bs$ is a cross norm, i.e.
$\alpha(A \otimes B) = \Vert A \Vert \Vert B \Vert$ for all
$A \in \As$ and $B \in \Bs$. Hence, assertion (ii) follows. \par
$(ii) \Rightarrow (i)$: This condition trivially implies the
Schlieder property and thus the $C^*$-independence of $(\As,\Bs)$.
\hfill\qed\enddemo
However, the notion of $C^*$-independence as defined above is not
restricted to commuting pairs of $C^*$-algebras, and it is of interest to
master this notion also in the noncommuting case. Note that if $\As$ and $\Bs$
do not commute, we can appeal to none of the above-cited results. Nonetheless,
one can still prove that conditions (i) and (ii) in Proposition 2 are
equivalent. With $A \in \As$, $\sigma(A)$ will denote the spectrum of $A$ in
$\As$.
\proclaim{Proposition 3} If $\As$ and $\Bs$ are (not necessarily commuting)
subalgebras of a $C^*$-algebra $\Cs$, then the following are equivalent. \par
(i) $(\As,\Bs)$ is $C^*$-independent. \par
(ii) $\Vert AB \Vert = \Vert A \Vert\Vert B \Vert$, for all $A \in \As$
and $B \in \Bs$.
\endproclaim
\demo{Proof} $(i) \Rightarrow (ii)$: Assume that $(\As,\Bs)$ is
$C^*$-independent, and let $A \in \As$ and $B \in \Bs$. The hypothesis of
$C^*$-independence implies the existence of a state $\varphi$ on
$\As \vee \Bs$ such that $\varphi(\vert A \vert ) = \Vert A \Vert$ and
$\varphi(\vert B \vert ) = \Vert B \Vert$. Then one observes that the
Cauchy-Schwarz inequality implies
$$\Vert A \Vert^2 = \varphi(\vert A \vert )^2 \leq
\varphi(1)\varphi(\vert A \vert^2) \leq \Vert \vert A \vert^2 \Vert =
\Vert A \Vert^2 \, . $$
\nind In other words, one has the equality
$\varphi(\vert A \vert^2) = \Vert A \Vert^2$. This entails that for any
$C \in \As \vee \Bs$,
$$\align
\vert \varphi((\vert A \vert - \Vert A \Vert \cdot 1)C)\vert^2
&\leq \varphi((\vert A \vert - \Vert A \Vert \cdot 1)^2 )\varphi(C^*C) \\
&= \{ \varphi(\vert A \vert^2) - 2\Vert A \Vert\varphi(\vert A \vert) +
\Vert A \Vert^2 \} \varphi(C^*C) = 0 \, .
\endalign $$
\nind Hence one has $\varphi(\vert A \vert C) = \Vert A \Vert \varphi(C)$
for all $C \in \As \vee \Bs$. Thus, it follows that
$\varphi(\vert A \vert\vert B \vert) = \Vert A \Vert \varphi(\vert B \vert)
= \Vert A \Vert \Vert B \Vert$. One therefore can conclude that
$$\align
\Vert AB \Vert &= \Vert B^* A^* A B \Vert^{\frac{1}{2}} =
\Vert B^* \vert A \vert^2 B \Vert^{\frac{1}{2}} =
\Vert \, \vert A \vert B \Vert \\
&= \Vert \, \vert A \vert B (\vert A \vert B)^* \Vert^{\frac{1}{2}} =
\Vert \,\vert A \vert\vert B^* \vert^2 \vert A \vert\,\Vert^{\frac{1}{2}} \\
&= \Vert \,\vert A \vert\vert B^* \vert\,\Vert =
\Vert A \Vert\Vert B^* \Vert \\
&= \Vert A \Vert\Vert B \Vert \, .
\endalign $$
$(ii) \Rightarrow (i)$: Assume that
$\Vert AB \Vert = \Vert A \Vert\Vert B \Vert$, for all $A \in \As$
and $B \in \Bs$. Let $A = A^* \in \As$ and $B = B^* \in \Bs$ be arbitrary.
The first step is to show that for any
$\lambda \in [\min\sigma(A),\max\sigma(A)]$ and any
$\mu \in [\min\sigma(B),\max\sigma(B)]$, there exists a state $\varphi$
on $\Cs$ such that $\varphi(A) = \lambda$ and $\varphi(B) = \mu$. Assume
without loss of generality (by adding suitable multiples of the identity to
$A$ and $B$) that $A \geq 0$, $0 \in \sigma(A)$, $B \geq 0$ and
$0 \in \sigma(B)$. By taking suitable convex combinations of states, it
will also suffice to take $\lambda = \Vert A \Vert$ and $\mu = \Vert B \Vert$.
\par
There exists a state $\varphi$ on $\Cs$ with
$$\varphi((A^{\frac{1}{2}}B^{\frac{1}{2}})^* A^{\frac{1}{2}}B^{\frac{1}{2}})
= \Vert A^{\frac{1}{2}}B^{\frac{1}{2}} \Vert^2 =
(\Vert A^{\frac{1}{2}} \Vert\Vert B^{\frac{1}{2}} \Vert)^2 =
\Vert A \Vert \Vert B \Vert \, , $$
\nind where the second equality uses the hypothesis and the third standard
properties of $C^*$-norms. Squaring both sides of this equality, the
Cauchy-Schwarz inequality then entails that one has
$$\align
\Vert A \Vert^2 \Vert B \Vert^2 =
\varphi((A^{\frac{1}{2}}B^{\frac{1}{2}})^*A^{\frac{1}{2}}B^{\frac{1}{2}})^2 &=
\varphi((B^{\frac{1}{2}}AB^{\frac{1}{4}})B^{\frac{1}{4}})^2 \\
&\leq \varphi(B^{\frac{1}{2}}AB^{\frac{1}{2}}AB^{\frac{1}{2}})\varphi(B^{\frac{1}{2}}) \\
&\leq \Vert A \Vert^2 \Vert B \Vert^{\frac{3}{2}}\varphi(B^{\frac{1}{2}}) \\
&\leq \Vert A \Vert^2 \Vert B \Vert^2 \, .
\endalign $$
\nind But this entails
$\varphi(B^{\frac{1}{2}}) = \Vert B^{\frac{1}{2}} \Vert$,
and squaring the above inequalities also yields $\varphi(B) = \Vert B \Vert$,
since $\varphi(1 \cdot B^{\frac{1}{2}})^2 \leq \varphi(B)$. Moreover,
$$\align
\vert \varphi((\Vert B \Vert^{\frac{1}{2}} \cdot 1 -
B^{\frac{1}{2}})C)\vert^2
&\leq \varphi((\Vert B \Vert^{\frac{1}{2}}\cdot 1 -
B^{\frac{1}{2}})^2)\varphi(C^* C)\\
&= \varphi(\Vert B \Vert \cdot 1 - 2\Vert B \Vert^{\frac{1}{2}} B^{\frac{1}{2}} + B)
\varphi(C^*C) \\
&= 0 \, ,
\endalign $$
\nind for arbitrary $C \in \Cs$, which entails
$\varphi(B^{\frac{1}{2}}C) = \Vert B \Vert^{\frac{1}{2}} \varphi(C)$. One
therefore observes that
$$\varphi(B^{\frac{1}{2}}AB^{\frac{1}{2}}) = \Vert B \Vert \varphi(A) \leq
\Vert A \Vert \Vert B \Vert \, , $$
\nind so that $\varphi(A) = \Vert A \Vert$ and $\varphi(B) = \Vert B \Vert$.
\par
As the next step, consider, for fixed $B = B^* \in \Bs$ and
$\mu \in [\min\sigma(B),\max\sigma(B)]$, the set of states ($\Ss(\As)$
represents the set of all states on the algebra $\As$)
$$\Vs \equiv \{ \psi \in \Ss(\As) \mid \psi = \varphi \mid \As \} \, , $$
\nind for any $\varphi \in \Ss(\Cs)$ satisfying $\varphi(B) = \mu$.
Due to the weak*-compactness of $\Ss(\Cs)$, every net of states
$\varphi_{\nu}$ on $\Cs$ whose restrictions to $\As$ are Cauchy in the
weak*-topology and satisfy $\varphi_{\nu}(B) = \mu$ has a convergent subnet,
the limit of which when restricted to $\As$ is
an element of $\Vs$. Hence, $\Vs$ is weak*-closed and convex.
If there exists an element $\xi \in \Ss(\As)$ which is not contained in
$\Vs$, then, by the Hahn-Banach Theorem, there exists a selfadjoint element
$A \in \As$ such that $\xi(A) \neq \psi(A)$ for all $\psi \in \Vs$. But this
contradicts the first step of this proof, since, necessarily,
$\xi(A) \in [\min\sigma(A),\max\sigma(A)]$. Hence, one must have
$\Vs = \Ss(\As)$. Repeating similar arguments for $B$, the stated
conclusion follows easily.
\hfill\qed\enddemo
If we have subalgebras of a $W^*$-algebra, which are replete with
projections, indeed are generated by their projections, then we can establish
the following equivalences. We denote the set of all projections of an algebra
$\As$ by $\proj(\As)$.
\proclaim{Proposition 4} Let $\As$ and $\Bs$ be (not necessarily commuting)
subalgebras of a $W^*$-algebra $\Cs$. The following are equivalent. \par
(i) $(\As,\Bs)$ is $C^*$-independent. \par
(ii) For any nontrivial projections $P \in \As$, $Q \in \Bs$, and
$\lambda,\mu \in [0,1]$, there exists a state $\varphi$ on $\Cs$ such that
$\varphi(P) = \lambda$ and $\varphi(Q) = \mu$. \par
(iii) For any nonzero projection $P \in \As$ and every state
$\varphi$ on $\Bs$, there exists a state $\phi$ on $\Cs$ such that
$\phi(P) = 1$ and $\phi \mid \Bs = \varphi$.
\endproclaim
\demo{Proof} The implication $(i) \Rightarrow (iii)$ is trivial. To see the
validity of the implication $(iii) \Rightarrow (ii)$, simply choose states
$\phi_0, \phi_1$ on $\Cs$ such that $\phi_0(1-P) = 1 = \phi_1(P)$ and
$\phi_0(Q) = \mu = \phi_1(Q)$. Then set
$\varphi = (1-\lambda)\phi_0 + \lambda\phi_1$. \par
Hence, proof of the implication $(ii) \Rightarrow (i)$ will suffice
to verify Prop. 4. Note that choosing $\lambda = \mu = 1$, the resulting
equalities $\varphi(P) = 1 = \varphi(Q)$ entail
$$\align
\Vert PQ \Vert \geq \vert\varphi(PQ)\vert &= \vert\varphi(Q) -
\varphi((1-P)Q)\vert \\
&\geq 1 - \varphi(1-P)^{\frac{1}{2}}\varphi(Q)^{\frac{1}{2}} = 1 \, ,
\endalign $$
\nind and therefore one has $\Vert PQ \Vert = 1$ for any choice of nontrivial
projections $P \in \proj(\As)$ and $Q \in \proj(\Bs)$. \par
Let $A \neq 0 \neq B$ with $A \in \As$ and $B \in \Bs$. By considering the
spectral projections of $\vert A \vert$ and $\vert B^* \vert$, one can conclude
that for any sufficiently small $\epsilon > 0$ there exist nonzero
projections $R \in \proj(\As)$ and $S \in \proj(\Bs)$ such that
$$R \leq (\Vert A \Vert^2 - \epsilon)^{-1}A^* A \quad \text{and} \quad
S \leq (\Vert B \Vert^2 - \epsilon)^{-1}B B^* \, . $$
\nind Then one has
$$\align
1 &= \Vert RS \Vert = \Vert S^* R^* RS \Vert^{\frac{1}{2}} =
\Vert SRS \Vert^{\frac{1}{2}} \leq (\Vert A \Vert^2 - \epsilon)^{-\frac{1}{2}}
\Vert SA^*AS\Vert^{\frac{1}{2}} \\
&= (\Vert A \Vert^2 - \epsilon)^{-\frac{1}{2}} \Vert AS\Vert =
(\Vert A \Vert^2 - \epsilon)^{-\frac{1}{2}} \Vert ASA^*\Vert^{\frac{1}{2}} \\
&\leq (\Vert A \Vert^2 - \epsilon)^{-\frac{1}{2}}
(\Vert B \Vert^2 - \epsilon)^{-\frac{1}{2}} \Vert ABB^*A^*\Vert^{\frac{1}{2}}\\
&= (\Vert A \Vert^2 - \epsilon)^{-\frac{1}{2}}
(\Vert B \Vert^2 - \epsilon)^{-\frac{1}{2}} \Vert AB\Vert \, .
\endalign $$
\nind But this entails $\Vert AB \Vert = \Vert A \Vert \Vert B \Vert$, and
appeal to Prop. 3 completes the proof.
\hfill\qed\enddemo
Of course, this argument also is valid for those $C^*$-algebras $\Cs$
which are not $W^*$-algebras but which are generated by their projections -
each selfadjoint element of $\Cs$ has a spectral resolution in $\Cs$
(and there do exist such algebras - see \cite{7}\cite{13}). A natural
formulation of statistical independence in the category of $W^*$-algebras is
discussed next.
\proclaim{Definition} Let $\Cal A$ and $\Cal B$ be subalgebras
of a $W^*$-algebra $\Cal C$. The pair $(\As,\Bs)$ is (or $\Cal A$ and $\Cal B$
are) said to be
$W^*$-independent if for every normal state $\phi_1$ on $\Cal A$ and
every normal state $\phi_2$ on $\Cal B$ there exists a normal state
$\phi$ on $\Cal C$ such that $\phi \mid \Cal A = \phi_1$ and
$\phi \mid \Cal B = \phi_2$. \endproclaim
There are some related and strictly stronger conditions in use in the
literature - see \cite{24} for an overview. In particular, in the light of
Roos' product state characterization of $C^*$-independence of commuting
subalgebras, it would be natural to ask if the same is true of
$W^*$-independence. This is, however, not the case. In fact, if the normal
extension $\phi$ in the above definition is a product state over the commuting
pair $(\As,\Bs)$, then\footnote{if $\phi$ has central support 1 in
$\As\vee\Bs$} $\As\vee\Bs$ is isomorphic to the (unique) $W^*$-product
$\As\overline{\otimes}\Bs$ \cite{25}\cite{3}. This property was called
$W^*$-independence in the product sense in \cite{24}. We shall see below that
$W^*$-independence of commuting subalgebras is strictly weaker than
$W^*$-independence in the product sense. \par
And of course the notion of $C^*$-independence is also applicable to this
setting of a pair of subalgebras of a $W^*$-algebra. It was established in
Corollary 3.5 of \cite{24} that the $W^*$-independence of commuting
subalgebras of a $W^*$-algebra $\Cs$ implies that they are also
$C^*$-independent. (It shall be
shown below that this is also true if the subalgebras do not mutually commute.)
Here we wish to show that these properties are actually equivalent if $\Cs$
can be realized as a von Neumann algebra on a separable Hilbert space, in other
words, if $\Cs$ is $\sigma$-finite.\footnote{The term countably decomposable is
used synonymously in the literature.} We prepare the way with the following
technical remark, which we shall use more than once in the sequel.
\proclaim{Lemma 5} Let $\As$ be a von Neumann algebra on a Hilbert space
$\Hs$, and let $\Omega \in \Hs$ be an arbitrary nonzero vector. Let
$\omega$ be the functional defined on $\As$ by
$\omega(A) = (\Omega,A\Omega)$ for all $A \in \As$. Then for any linear
functional $\varphi$ on $\As$ such that $0 \leq \varphi \leq c\omega$,
for some positive real $c$, there exists a positive element $A' \in \As'$ such
that
$\varphi(A) = \omega(A'A) = (A'{}^{\frac{1}{2}}\Omega,AA'{}^{\frac{1}{2}}\Omega)$ for all $A \in \As$. Moreover, for any
positive element $A' \in \As'$, the linear functional defined by
$\varphi(A) \equiv \omega(A'A) = (A'{}^{\frac{1}{2}}\Omega,AA'{}^{\frac{1}{2}}\Omega)$ for all $A \in \As$ is
positive and satisfies $\varphi \leq \Vert A' \Vert\omega$.
\endproclaim
\demo{Proof} Let $\varphi$ be a linear functional on $\As$ such that
$0 \leq \varphi \leq c\omega$, for some positive real $c$, and let
$P' \in \As'$ be the projection onto the closure of $\As\Omega$. Moreover, let
$(\Hs_{\omega},\pi_{\omega},\Omega)$ be the GNS-representation
of $(\As,\omega)$. According to Theorem 29.2 \cite{9}, there then exists
a positive $T \in \pi_{\omega}(\As)' = (\As_{P'})' = (\As')_{P'}$ such that
$\varphi(A) = \omega(TA)$ for all $A \in \As$. This
entails that $T = P'A'_1P' \mid_{P'\Hs}$ for some positive $A_1' \in \As'$, so,
in fact, one has $\varphi(A) = \omega(P'A_1{}'P'A)$ for all $A \in \As$. The
desired element is therefore $A' = P'A'_1P'$. The second assertion is
demonstrated in the proof of Theorem 29.2 in \cite{9}.
\hfill\qed\enddemo
The next step is an alternative characterization of $W^*$-independent
subalgebras of a $W^*$-algebra.
\proclaim{Lemma 6} Let $\As$ and $\Bs$ be (not necessarily commuting)
subalgebras of a $W^*$-algebra $\Cs$ acting on a Hilbert space $\Hs$. The
following are equivalent. \par
(i) $(\As,\Bs)$ is $W^*$-independent. \par
(ii) For any nonzero vectors $\Phi,\Psi \in \Hs$, there exist $A' \in \As'$
and $B' \in \Bs'$ such that $A'\Phi = B'\Psi \neq 0$.
\endproclaim
\demo{Proof} $(i) \Rightarrow (ii)$: Without loss of generality one may assume
$\Vert\Phi\Vert = \Vert\Psi\Vert = 1$, and let $\phi$ and $\psi$ be the
corresponding normal states on $\As$ and $\Bs$, respectively, {\it i.e.}
$\phi(A) = (\Phi,A\Phi)$ and $\psi(B) = (\Psi,B\Psi)$ for all $A \in \As$ and
$B \in \Bs$. By assumption, there exists a normal state $\omega$ on $\Cs$ such
that $\omega \mid \As = \phi$ and $\omega \mid \Bs = \psi$. Choose a
suitable nonzero $\Xi \in \Hs$ such that the corresponding positive linear
functional $\xi$ on $\Cs$, defined by $\xi(C) = (\Xi,C\Xi)$ for all
$C \in \Cs$, satisfies the inequality $\xi \leq \omega$. By Lemma 5 there
exist positive elements $A_1' \in \As'$ and $B_1' \in \Bs'$ such that
$\xi(A) = \phi(A_1'A) = (A_1'{}^{\frac{1}{2}}\Phi,AA_1'{}^{\frac{1}{2}}\Phi)$
and
$\xi(B) = \psi(B_1'B) = (B_1'{}^{\frac{1}{2}}\Psi,BB_1'{}^{\frac{1}{2}}\Psi)$
for all $A \in \As$ and $B \in \Bs$. Let $P,Q \in \As'$ be the projections
onto the closures of $\As A_1'{}^{\frac{1}{2}}\Phi$ and $\As\Xi$, respectively.
Since the GNS representation induced by a state is unique up to unitary
equivalence, there exists a partial isometry $U = QUP \in \Bs(\Hs)$ such that
$UA_1'{}^{\frac{1}{2}}\Phi = \Xi$, $U^*U = P$, $UU^* = Q$ and
$$A\tilde{A}\Xi = UA\tilde{A}A_1^{\frac{1}{2}}\Phi =
UAP\tilde{A}A_1^{\frac{1}{2}}\Phi
= UAU^*U\tilde{A}A_1^{\frac{1}{2}}\Phi = (UAU^*)\tilde{A}\Xi \, ,
$$
\nind for all $A,\tilde{A} \in \As$. One concludes $U^*A = AU^*$, in other
words, one has $U \in \As'$. Similarly, one obtains the existence of a partial
isometry $V \in \Bs'$ such that
$VB_1'{}^{\frac{1}{2}}\Psi = \Xi = UA_1'{}^{\frac{1}{2}} \Phi$. So
$A' \equiv UA_1'{}^{\frac{1}{2}}$ and $B' \equiv VB_1'{}^{\frac{1}{2}}$ are
the desired elements. \par
$(ii) \Rightarrow (i)$: Let $\phi$ and $\psi$ be normal states on $\As$
and $\Bs$, respectively. Consider the set
$$\Us \equiv \{ \{\varphi_{\alpha}\} \mid \sum_{\alpha} \varphi_{\alpha} \in
\Cs_*, \varphi_{\alpha} \geq 0 , (\sum_{\alpha} \varphi_{\alpha} \mid \As) \leq
\phi , (\sum_{\alpha} \varphi_{\alpha}\mid \Bs) \leq \psi \} \, , $$
\nind which can be partially ordered by set inclusion. It will be shown that
an arbitrary linearly ordered subset
$\{ \Gs_{\beta} = \{\varphi_{\alpha} \mid \alpha \in \Fs_{\beta} \}\}$ of
$\Us$ has an upper bound in $\Us$. (Note that for arbitrary indices
$\beta_1$ and $\beta_2$ entering into this linearly ordered subset, one must
have either $\Fs_{\beta_1} \subset \Fs_{\beta_2}$ or
$\Fs_{\beta_2} \subset \Fs_{\beta_1}$.) For any $n \in \IN$ there exists only
a finite number of states $\varphi_{\alpha} \in \cup_{\beta}\Gs_{\beta}$
satisfying $\varphi_{\alpha}(1) > \frac{1}{n}$, since otherwise there would
exist a $\beta_0$ with
$\sum_{\alpha \in \Fs_{\beta_0}} \varphi_{\alpha}(1) > 1 = \phi(1)$. Hence,
the set $\{ \varphi_{\alpha} \mid \alpha \in \cup_{\beta}\Fs_{\beta} \}$ is
countable. But
$\Vert \sum_{\alpha = n}^m \varphi_{\alpha}\Vert =
\sum_{\alpha = n}^m \varphi_{\alpha}(1)$ and
$\sum_{\alpha \leq n}\varphi_{\alpha}(1) \leq 1$ then imply the norm
convergence of $\sum_{\alpha}\varphi_{\alpha}$, and also the normality
of the limit, since $\Cs_*$ is norm-closed. Therefore,
$\{\varphi_{\alpha} \mid \alpha \in \cup_{\beta}\Fs_{\beta} \}$ is an element
of $\Us$ and an upper bound for
$\{ \Gs_{\beta} = \{\varphi_{\alpha} \mid \alpha \in \Fs_{\beta} \}\}$ in
$\Us$. One may now employ Zorn's Lemma to conclude the existence of a
maximal element $\{\chi_{\alpha}\}$ in $\Us$. Let
$\chi \equiv \sum_{\alpha}\chi_{\alpha}$. By construction, $\chi$ is a normal
positive linear functional on $\Cs$ whose restriction to $\As$, resp. $\Bs$,
is bounded by $\phi$, resp. $\psi$. \par
Assume that $\chi \mid \As < \phi$ and $\chi \mid \Bs < \psi$. The
normality of $\phi - \chi > 0$, resp. $\psi - \chi > 0$, entails the existence
of a nonzero vector $\Xi_1 \in \Hs$, resp. $\Xi_2 \in \Hs$, such that the
corresponding positive linear functional $\omega_{\As}$, resp. $\omega_{\Bs}$,
defined on $\As$, resp. $\Bs$, by $\omega_{\As}(A) = (\Xi_1,A\Xi_1)$ for all
$A \in \As$, resp. $\omega_{\Bs}(B) = (\Xi_2,B\Xi_2)$ for all $B \in \Bs$,
satisfies the inequality $\omega_{\As} + \chi\mid\As \leq \phi$, resp.
$\omega_{\Bs} + \chi\mid\Bs \leq \psi$. By assumption, there exist
$A' \in \As'$ and $B' \in \Bs'$ such that
$0 \neq A'\Xi_1 = B'\Xi_2$. Define a positive linear functional $\omega$ on
$\Cs$ by $\omega(C) = (A'\Xi_1,CA'\Xi_1) = (B'\Xi_2,CB'\Xi_2)$ for all
$C \in \Cs$. Then $\omega(A) = (A'{}^*A'\Xi_1,A\Xi_1)$,
for all $A \in \As$. From Lemma 5 one sees that this entails the
inequality $\omega\mid\As \leq \Vert A'{}^*A'\Vert\omega_{\As}$. Similarly,
one proves that $\omega\mid\Bs \leq \Vert B'{}^*B'\Vert\omega_{\Bs}$. Let
$\mu$ be the larger of the numbers $\Vert A'{}^*A'\Vert$,
$\Vert B'{}^*B'\Vert$. Then $\{\chi_{\alpha}\} \cup \{c\omega\}$
is an element of $\Us$ for any $0 < c \leq \frac{1}{\mu}$. Since
$\{\chi_{\alpha}\}$ is countable, the maximality of $\{\chi_{\alpha}\}$ is
contradicted for some choice of $c$ as indicated (note $\omega$ is nonzero).
\par
Thus one can conclude that either $\chi\mid\As = \phi$ or
$\chi\mid\Bs = \psi$. Assume without loss of generality that the former
obtains. Then $\psi \geq \chi\mid\Bs$ and $\psi(1) = 1 = \phi(1)$ imply
the equalities
$$\Vert\psi - (\chi\mid\Bs)\Vert = \psi(1) - (\chi\mid\Bs)(1) =
\phi(1) - (\chi\mid\As)(1) = 0 \, . $$
\hfill\qed\enddemo
We need one more technical lemma.
\proclaim{Lemma 7} Let $\As$ be a von Neumann algebra in standard form
on the Hilbert space $\Hs$ (in other words, $\As$ has a cyclic and
separating vector $\Omega \in \Hs$). Then for every nonzero vector
$\Phi \in \Hs$ there exist elements $A \in \As$ and $A' \in \As'$ such that
$A'\Phi = A\Omega \neq 0$.
\endproclaim
\demo{Proof} Consider the linear form $l$ on $\As'$ defined by
$l(A') = (\Omega,A'\Phi)$ for $A' \in \As'$. Since $\Omega$ is cyclic
with respect to $\As'$, the linear functional $l$ is nonzero. From the
polar decomposition of normal functionals (see, {\it e.g.} Theorem III.4.2 of
\cite{26}) one may conclude the existence of a partial isometry $U' \in \As'$
such that the functional on $\As'$ given by
$(\Omega,A'U'\Phi) = (A'{}^*\Omega,U'\Phi)$ is positive on $\As'$ and
nontrivial. By Prop. 2.5.27 (1) of \cite{1} there exists a positive,
selfadjoint operator $Q$ affiliated with $\As$ such that $U'\Phi = Q\Omega$.
\par
So consider the positive linear functional on $\As$ defined by this
vector: $(U'\Phi,AU'\Phi) = (Q\Omega,AQ\Omega)$ for all $A \in \As$. By
Theorem 2.5.31 in \cite{1}, this normal positive linear functional is
implemented on $\As$ by a (unique) vector $\Xi$ in the natural positive cone
associated to $(\As,\Omega)$, {\it i.e.}
$(U'\Phi,AU'\Phi) = (Q\Omega,AQ\Omega) = (\Xi,A\Xi)$ for all $A \in \As$.
As in the proof of Lemma 6, since the corresponding GNS-representation is
unique up to unitary equivalence, there exists a partial isometry
$V' \in \As'$ such that $V'U'\Phi = V'Q\Omega = \Xi$ is contained in the
natural positive cone associated to $(\As,\Omega)$ and such
that $V'{}^*V'Q\Omega = Q\Omega$. Then since the modular conjugation $J$
associated with $(\As,\Omega)$ leaves the natural positive cone pointwise
invariant, one has $V'U'\Phi = JV'U'\Phi = JV'Q\Omega = (JV'J)(JQJ)\Omega$.
\par
By Lemma 2.5.8 in \cite{1} one can choose a suitable spectral
projection $P' \in \As'$ of the selfadjoint $JQJ$ (which is affiliated with
$\As'$) such that $P'(JQJ) \in \As'$ and $P'(JQJ)\Omega \neq 0$. Hence one
sees that
$P'V'U'\Phi = P'(JV'J)(JQJ)\Omega = (JV'J)(P'(JQJ))\Omega = JV'(JP'J)Q\Omega
\neq 0$, utilizing the facts that $JV'J \in \As$ and
$V' : \overline{\As Q\Omega} \mapsto \overline{\As\Xi}$ is bijective. Using
this vector to induce a normal positive linear functional on $\As$, the same
argument again implies the existence of a partial isometry $W' \in \As'$ such
that $0 \neq W'P'V'U'\Phi$ is contained in the natural positive cone. Hence
one finds
$$0 \neq W'P'V'U'\Phi = JW'(P'V'U'\Phi) = JW'(JV'J)P'(JQJ)\Omega =
V'(JW'J)(JP'J)Q\Omega \, , $$
\nind since $JW'J \in \As$. Noting that $(JW'J)(JP'J)Q\Omega \in \As Q\Omega$,
one therefore has
$$\align
0 &\neq V'{}^*W'P'V'U'\Phi = V'{}^*V'(JW'J)(JP'J)Q\Omega \\
&= (JW'J)(JP'J)V'{}^*V'Q\Omega = (JW'J)(JP'J)Q\Omega \, .
\endalign $$
\nind The claim then follows after noting that
$A' \equiv V'{}^*W'P'V'U' \in\As'$ and
$A \equiv (JW'J)(JP'J)Q = (JW'J)J(P'JQJ)J \in \As$.
\hfill\qed\enddemo
We may now prove that $C^*$- and $W^*$-independence are equivalent in
a context which arises quite commonly in quantum physics. In particular, the
hypothesis of the next proposition obtains when $\As$ and $\Bs$ are commuting
von Neumann algebras acting on a separable Hilbert space $\Hs$.
\par
\proclaim{Proposition 8} Let $\As$ and $\Bs$ be commuting subalgebras of
a $\sigma$-finite $W^*$-algebra $\Cs$. The following are equivalent. \par
(i) $(\As,\Bs)$ is $C^*$-independent. \par
(ii) $(\As,\Bs)$ is $W^*$-independent.
\endproclaim
\demo{Proof} The implication $(ii) \Rightarrow (i)$ was proven in
Corollary 3.5 in \cite{24}. So let $(\As,\Bs)$ be $C^*$-independent.
$\Cs$ can be realized as a von Neumann algebra in standard form on a Hilbert
space $\Hs$ (see, {\it e.g.} Prop. 2.5.6 in \cite{1}). Let
$\Omega \in \Hs$ be the corresponding cyclic and separating vector. In
addition, let $P' \in \As'$ be the projection onto the closure of $\As\Omega$.
Since $\Omega$ is separating for $\As$, the von Neumann algebra
$\As_{P'}$ is isomorphic to $\As$. Hence, one may consider normal states on
$\As_{P'}$ without loss of generality. Similarly, one may consider normal
states on $\Bs_{Q'}$, where $Q' \in \Bs'$ is the projection onto the closure of
$\Bs\Omega$. Note that $\As_{P'}$, resp. $\Bs_{Q'}$, has $\Omega$ as a
cyclic and separating vector in the Hilbert space $P'\Hs$, resp.
$Q'\Hs$. Since every normal positive linear functional on a von
Neumann algebra in standard form can be realized as the vector functional
induced by a vector in the natural positive cone of the algebra (see
{\it e.g.} Theorem 2.5.31 in \cite{1}), it is no loss of generality to
take the nonzero vector $\Phi$, resp. $\Psi$, in Lemma 6 to lie in $P'\Hs$,
resp. $Q'\Hs$. \par
In the one case, Lemma 7 entails the existence of $A \in \As$
and $A' = P'A'P' \in \As'$ with
$0 \neq A\Omega = P'AP'\Omega = P'A'P'\Phi = A'\Phi$, with similar results
in the other case. So consider the vectors $0 \neq A\Omega = A'\Phi$ in the
one case and $0 \neq B\Omega = B'\Psi$, for some $B \in \Bs$ and $B' \in \Bs'$,
in the other. Since $A \neq 0 \neq B$, the hypothesis and the equivalence
of $C^*$-independence and the Schlieder property entail that $AB \neq 0$.
But then, since $\overline{\Cs'\Omega} = \Hs$, there exists an element
$C' \in \Cs'$ such that $ABC'\Omega \neq 0$. Then
$BC'A'\Phi = BC'A\Omega = ABC'\Omega = AC'B\Omega = AC'B'\Psi \neq 0$ and
$BC'A' \in \As'$ with $AC'B' \in \Bs'$ complete the proof, by Lemma 6.
\hfill\qed\enddemo
A further notion of statistical independence suitable for the
category of $W^*$-algebras is called strict locality and was introduced in
\cite{15} on physical grounds (see also \cite{17} for an earlier
version).
\proclaim{Definition} Let $(\As,\Bs)$ be an ordered pair of
subalgebras of the $W^*$-algebra $\Cal C$. $(\As,\Bs)$ is said to be strictly
local if for any nonzero projection $P \in \Cal A$
and any state $\varphi \in \Cal B_*$ there exists a state $\phi \in
(\Cal A \vee \Cal B)_* $ such that $\phi (P) = 1$ and
$\phi \mid \Cal B = \varphi$. \endproclaim
Interpreting $P \in \text{Proj}(\As)$ as a decision effect, we see that
if the pair $(\As,\Bs)$ is strictly local, no preparation on subsystem $\Bs$
can exclude the occurrence of any probability that $P$ is true in subsystem
$\As$. Indeed, for $1-P \in \text{Proj}(\As)$ there also exists a state
$\tilde{\phi} \in (\As\vee\Bs)_*$ such that $\tilde{\phi} \mid \Bs =
\varphi$ and $\tilde{\phi}(1-P) = 1$. Hence for any $\lambda \in [0,1]$ the
state $\phi_{\lambda} \equiv \lambda\phi + (1-\lambda)\tilde{\phi}$ satisfies
$\phi_{\lambda} \mid \Bs = \varphi$ and $\phi_{\lambda}(P) = \lambda$.
See \cite{15}\cite{24} for a further discussion of this property. \par
It is obvious that if $(\As,\Bs)$ is $W^*$-independent, then it is also
strictly local. It was shown in \cite{24} that the strict locality of a pair
of {\it commuting} subalgebras of a $W^*$-algebra implies the
$C^*$-independence of the pair. Here we show that, in fact, the converse is
also true. Hence, at least for commuting pairs of $W^*$-subalgebras, if
$(\As,\Bs)$ is strictly local, then $(\Bs,\As)$ is also strictly local.
\proclaim{Proposition 9} Let $\As$ and $\Bs$ be commuting subalgebras of a
$W^*$-algebra $\Cs$. The pair $(\As,\Bs)$ is strictly local if and only if
it is $C^*$-independent.
\endproclaim
\demo{Proof} The implication $\Rightarrow$ was proven in Prop. 4.2 in
\cite{24}. So let $(\As,\Bs)$ be $C^*$-independent, and let
$P \in \As$ be a nonzero projection and $\varphi \in \Cal B_*$ be a state.
Moreover, let $\Cs$ be realized as a von Neumann algebra acting upon a
Hilbert space $\Hs$. Since by assumption one has $P \in \Bs'$, the canonical
*-homomorphism $\tau : \Bs \mapsto \Bs_P = P\Bs P \mid P\Hs$ given by
$\tau(B) = PBP = BP$ is well-defined, and by the
Schlieder property is a *-isomorphism onto a von Neumann algebra. $\tau$
is therefore $\sigma$-weakly bicontinuous, so that $\varphi \circ \tau^{-1}$
is a normal state on $\tau(B)$. There must therefore exist a density matrix
$\rho$ on $\Hs$ such that $\rho = P \rho P$ implements the state
$\varphi \circ \tau^{-1}$. Hence, $\psi(C) \equiv \tr(\rho C)$ defines a
normal state on $\Cs$ satisfying $\psi(P) = 1$ and
$$\psi(B) = \psi(BP) = \tr(\rho BP) = (\varphi \circ \tau^{-1})(\tau(B)) =
\varphi(B) \, , $$
\nind for any $B \in \Bs$.
\hfill\qed\enddemo
We note that strict locality implies $C^*$-independence even if the
pair of subalgebras of the $W^*$-algebra $\Cs$ does not necessarily commute.
\proclaim{Corollary 10} Let $\As$ and $\Bs$ be (not necessarily commuting)
subalgebras of a $W^*$-algebra $\Cs$. If $(\As,\Bs)$ is strictly local, then
it is also $C^*$-independent.
\endproclaim
\demo{Proof} Observe that the argument $(iii) \Rightarrow (ii) \Rightarrow (i)$
in the proof of Prop. 4 is also valid if the states in (iii) and (ii) are
required to be normal.
\hfill\qed\enddemo
We have therefore proven the following theorem, establishing the
equivalence of three standard notions of statistical independence in a
context where they can be compared and which commonly occurs in physical
applications, and providing alternative characterizations of these properties.
However, as shown in \cite{24}, there are other versions of statistical
independence which are not equivalent to these. \par
\proclaim{Theorem 11} Let $\As$ and $\Bs$ be commuting subalgebras of
a $\sigma$-finite $W^*$-algebra $\Cs$. The following are equivalent. \par
(i) $(\As,\Bs)$ is $C^*$-independent. \par
(ii) $(\As,\Bs)$ is strictly local. \par
(iii) $(\As,\Bs)$ is $W^*$-independent. \par
(iv) For every state $\phi_1$ on $\Cal A$ and every state $\phi_2$ on
$\Cal B$ there exists a state $\phi$ on $\Cal C$ such that
$\phi (AB) = \phi _1 (A)\,\phi _2 (B)$ for all $A\in \Cal A$ and all
$B\in \Cal B$. \par
(v) $\Vert AB \Vert = \Vert A \Vert\Vert B \Vert$, for all $A \in \As$
and $B \in \Bs$. \par
(vi) $0\neq A \in \Cal A$ and $0\neq B \in \Cal B$ imply that $AB\neq 0$.
\par
(vii) The map $\eta : \As\Bs \rightarrow \As \odot \Bs$ defined by
$\eta(AB) = A \otimes B$, $A \in \As$, $B \in \Bs$ is an isomorphism
continuous in the minimal $C^*$-cross norm on $\As \odot \Bs$ and can
therefore be continuously extended to a surjective homomorphism
$\overline{\eta} : \Cal A \vee \Cal B \rightarrow \Cal A \otimes \Cal B$.
\par
(viii) For any nonzero vectors $\Phi,\Psi \in \Hs$, there exist
$A' \in \As'$ and $B' \in \Bs'$ such that $A'\Phi = B'\Psi \neq 0$. \par
(ix) For any nontrivial projections $P \in \As$, $Q \in \Bs$, and
$\lambda,\mu \in [0,1]$, there exists a state $\varphi$ on $\Cs$ such that
$\varphi(P) = \lambda$ and $\varphi(Q) = \mu$. \par
(x) For any nonzero projection $P \in \As$ and every state
$\varphi$ on $\Bs$, there exists a state $\phi$ on $\Cs$ such that
$\phi(P) = 1$ and $\phi \mid \Bs = \varphi$.
\endproclaim
We note, in particular, that it follows that the algebras of observables
associated to complementary wedges in irreducible vacuum representations in
quantum field theory are $W^*$-independent. Hence, such pairs of algebras are
$W^*$-independent, but not $W^*$-independent in the product sense. It was
previously known that they were $C^*$-independent and strictly local, but this
is the first proof that they are also $W^*$-independent. Such tangent wedge
algebras provide examples of algebras of observables satisfying
$W^*$-independence but still maximally violating Bell's inequalities in every
normal state (see \cite{22}\cite{23}\cite{24} for further explanation). \par
These examples therefore demonstrate that $W^*$-independence is
strictly weaker than $W^*$-independence in the product sense. Without using the
rather involved arguments concerning Bell's inequalities, we point out also
that if $\As$ is a type III factor acting on a separable Hilbert space, then
$(\As,\As')$ is a $C^*$-independent, hence $W^*$-independent pair of
$W^*$-algebras which is not $W^*$-independent in the product sense. It is
remarkable that for such pairs all normal partial states have normal
extensions, {\it none} of which is allowed to be a product state, and also
all partial states have extensions to product states, {\it none} of which can
be normal. \par
It was pointed out in \cite{24} that a property called there
$C^*$-independence in the product sense, namely the property that
$\Cal A \vee \Cal B$ is isomorphic to the minimal $C^*$-product
$\Cal A \otimes \Cal B$, is strictly stronger than $C^*$-independence. We
present some further results concerning this notion. We first show that
the existence of a faithful product state over the pair $(\As,\Bs)$ implies
that $(\As,\Bs)$ is $C^*$-independent in the product sense. \par
\proclaim{Proposition 12} Let $\As$ and $\Bs$ be commuting subalgebras
of a $C^*$-algebra $\Cs$. If for some state $\phi_1$ on $\As$ and some
state $\phi_2$ on $\Bs$ there exists a faithful state $\phi$ on
$\As \vee \Bs$ such that $\phi(AB) = \phi_1(A)\phi_2(B)$ for all $A \in \As$
and $B \in \Bs$, then $(\As,\Bs)$ is $C^*$-independent in the product sense.
\endproclaim
\demo{Proof} For any states $\phi_1$ on $\As$ and $\phi_2$ on $\Bs$ there
exists a state $\psi$ on $\As \otimes \Bs$ such that
$\psi(A \otimes B) = \phi_1(A)\phi_2(B) = \phi(AB)$ for all $A \in \As$
and $B \in \Bs$. It was shown in Corollary 3.3 in \cite{24} that the
existence of a faithful product state implies that the pair $(\As,\Bs)$ is
$C^*$-independent. Hence, using the map
$\overline{\eta} : \Cal A \vee \Cal B \rightarrow \Cal A \otimes \Cal B$ from
Theorem 1 (or Theorem 11) above, one has
$\psi \circ \overline{\eta} (AB) = \psi(A \otimes B) = \phi(AB)$. The
faithfulness of $\phi$ entails that $\overline{\eta}$ is an isomorphism.
\hfill\qed\enddemo
It is next shown that in the context most commonly met in quantum
theory, the converse of Prop. 12 is also true.
\proclaim{Proposition 13} Let $\As$ and $\Bs$ be commuting $C^*$-algebras
acting on a separable Hilbert space $\Hs$. If $(\As,\Bs)$ is $C^*$-independent
in the product sense, then there exists a faithful product state $\phi$ on
$\As \vee \Bs$.
\endproclaim
\demo{Proof} Assume that $\As \vee \Bs$ is isomorphic to $\As \otimes \Bs$.
By Prop. II.3.19 in \cite{26} there are faithful normal states on
$\As''$ and $\Bs''$. The corresponding product state on
$\As'' \overline{\otimes} \Bs''$ is also faithful (Corollary IV.5.12 in
\cite{26}). Since $\As \otimes \Bs$ is isomorphic to a subalgebra of
$\As'' \overline{\otimes} \Bs''$ and, by assumption, $\As\vee\Bs$ is itself
isomorphic to $\As\otimes\Bs$, the assertion follows.
\hfill\qed\enddemo
We comment that if $\As$ is a hyperfinite type III factor, then the
pair $(\As,\As')$ is $C^*$-independent in the product sense (since $\As$ is
semidiscrete), but it is not $W^*$-independent in the product sense. On
the other hand, $W^*$-independence in the product sense implies
$C^*$-independence in the product sense \cite{24}.
\heading Remarks on Statistical and Kinematical Independence \endheading
As mentioned earlier, the various notions of the statistical independence
of the pair $(\As,\Bs)$ are logically independent of the requirement that
the algebras $\As$ and $\Bs$ commute elementwise, which was called
kinematical independence in \cite{24}. In \cite{24} a conjecture due to
Werner was mentioned: if there exists a faithful product state
across the $C^*$-independent pair $(\As,\Bs)$, then $\As$ and $\Bs$ commute.
The following example shows that this conjecture is false. Consider the
projections $E$ and $F$ on $\Hs = \CC^6$:
$$
E=\pmatrix
1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 0\endpmatrix \quad , \quad
F=\pmatrix
1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2}\\
0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \endpmatrix \, . $$
\nind The operators $E$ and $F$ clearly do not commute, so the algebras,
$\Cal A$ and $\Cal B$, they respectively generate do not commute. Moreover,
any state on $\Cal A$, resp. $\Cal B$, is uniquely determined by its value on
$E$, resp. $F$. Noting that any such value must lie between 0
and 1, if we take $a \equiv \phi_1 (E)$ and $b \equiv \phi_2 (F)$
then the following matrix
$$
\rho=\left( \matrix
\frac{a(b + b^2)}{2} & 0 & 0 & 0 & 0 & 0\\
0 & \frac{(1-a)(b + b^2)}{2}) & 0 & 0 & 0 & 0\\
0 & 0 & \frac{a(2-3b +b^2)}{2}) & 0 & 0 & 0\\
0 & 0 & 0 & \frac{(1-a)(2-3b +b^2)}{2}) & 0 & 0\\
0 & 0 & 0 & 0 & \ssize{a(b - b^2)} & 0\\
0 & 0 & 0 & 0 & 0 & \ssize{(1-a)(b-b^2)} \endmatrix \right)$$
\nind is a density matrix determining a state on $\Cal C$ which is a
common extension of the states $\phi _1$ and $\phi _2$. Hence $\As$ and $\Bs$
are $C^*$-independent. Moreover, this state is a faithful product state as long
as $\phi_1$ and $\phi_2$ are chosen so that
$\phi_1(E),\phi_2(F) \not\in \{ 0,1 \}$. \par
Some conditions sufficient for $C^*$-independence of $(\As,\Bs)$ to
imply that $\As$ and $\Bs$ commute were proven in \cite{18}\cite{16}.
These conditions are fairly unmanageable, and we would like to emphasize
that it would be of interest to find more useful sufficient conditions. We
point out, however, that in the light of Prop. 2, this is not obvious,
since if $P$ and $Q$ are projections on a Hilbert space $\Hs$ of at least six
dimensions such that the restriction of $P$ to a four-dimensional subspace
$\Ks$ is diagonal with entries $1,1,0,0$, say, and the restriction of $Q$ to
the same subspace is diagonal with entries $1,0,1,0$, then the algebras
respectively generated by these projections are $C^*$-independent no matter
what the projections $P$ and $Q$ are doing on the orthogonal complement of
$\Ks$ in $\Hs$. \par
\bigpagebreak
\heading References \endheading
\roster
\item{}O. Bratteli and D.W. Robinson, {\it Operator Algebras and
Quantum Statistical Mechanics I}, Berlin, Heidelberg, New York:
Springer-Verlag, 1979.
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