7
$\begingroup$

In his "Noncommutative Geometry" book Connes asserts (on p. 539) that for two standard probability spaces $(X,\mu_X)$, $(Y,\nu_Y)$ an $N$-$M$-bimodule for $M=L^\infty(X,\mu_X)$ and $N=L^\infty(Y,\mu_Y)$ is given by a measure class $\mu$ on $X\times Y$ with marginal projections $\mathrm{Pr_X}(\mu)$, $\mathrm{Pr_Y}(\mu)$ absolutely continuous w.r.t. $\mu_X$, $\mu_Y$ and by a $\mu$-measurable function $n: X\times Y \rightarrow \mathbb{Z}$. There is no proof for this fact in the book, and it is not clear for me, why does $\mu$ appear to be a countably-additive measure, not just a finitely-additive one.

It follows from the definition that $N$-$M$-bimodule is a representation $\pi$ of maximal $C^\star$-tensor product $N\otimes_{max} M^{o}$ such that restrictions of $\pi$ on $N$ and $M^o$ are both normal. Functionals of the form $$ \mu_\pi: z\in N\otimes_{max} M^{o} \rightarrow \langle \pi(z)\xi,\xi\rangle $$ where $\xi$ is a cyclic vector, are exactly binormal states, i.e. such normed positive functionals on $N\otimes_{max} M^{o}$ that the maps $(f,g)\rightarrow \mu_\pi(f\otimes g)$ are normal separately in both arguments.

In the commutative ("measure-theoretical") case binormal states can be associated (via the appropriate form of Riesz representation theorem) with finitely-additive probability measures on $X\times Y$ having countably-additive marginal projections, absolutely continuous w.r.t. reference measures $\mu_X$, $\mu_Y$ respectively. It is not a priori clear, why they are countably-additive themselves.

$\endgroup$
4
  • 3
    $\begingroup$ I'm not so sure about your last paragraph. Are you sure this is the association given by the Riesz representation theorem? $\endgroup$ Feb 12, 2015 at 18:43
  • $\begingroup$ I'm apologize for an inaccurate formulation. Of course, I mean that with any binormal state we can associate some set function of the specified type, but, as follows from the Connes' assertion, not every such set function corresponds to a binormal state. $\endgroup$ Feb 13, 2015 at 15:31
  • 2
    $\begingroup$ Yes, but that is what is unclear to me. How are you associating a finite-additive probability measure on $X \times Y$ to a bi-normal functional on $N \otimes_{max} M^o$? I think this is where the confusion lies. $\endgroup$ Feb 13, 2015 at 20:54
  • $\begingroup$ You are right, thank you. I got confused exactly at this point. $\endgroup$ Feb 14, 2015 at 16:36

1 Answer 1

1
$\begingroup$

The answer to my own question. Many thanks to Jesse Peterson for pointing out the confusing place.

Let $\mathcal A$, $\mathcal B$ be sigma-algebras of subsets of $X$ and $Y$ respectively. Define the following two algebras of subsets on $X\times Y$

  1. $\mathcal A \times \mathcal B$ be an algebra of all step-sets: finite unions of disjoint subsets of the form $A\times B$, $A\in \mathcal A$, $B\in \mathcal B$ ("measurable rectangles").
  2. $\mathcal A \otimes \mathcal B$ be a $\sigma$-algebra generated by the algebra of step-sets.

When we restrict a normalized state on $L^\infty(\mu_X) \otimes_{max} L^\infty(\mu_Y)$ to indicators of step-sets, we obtain a probability finitely-additive measure on the algebra $\mathcal A \times \mathcal B$ (due to positivity and the normalizing condition of a state). Its restrictions on $\mathcal A$, $\mathcal B$ appear to be countably additive absolutely continuous measures due to binormality: countable additivity corresponds normality, and absolute continuity follows from the Riesz representation of $(L^\infty(\mu_X))^*$ as the space of bounded absolutely continuous finitely-additive measures.

Then we should extend our finitely-additive measure from $\mathcal A \times \mathcal B$ to $\mathcal A \otimes \mathcal B$. It is not hard to prove, that countable additivity of marginals implies countable additivity of our measure on $\mathcal A \times \mathcal B$ (see "Lemma" below). By Hahn-Kolmogorov theorem, it has the unique countably-additive extension on $\mathcal A \otimes \mathcal B$.

It is straightforward to check that any measure of the specified type defines a binormal state on $L^\infty(\mu_X) \otimes_{max} L^\infty(\mu_Y)$ via integration.

Lemma. Any finitely additive measure $\gamma$ on $\mathcal A \times \mathcal B$ with countably additive marginals is countably additive on $\mathcal A \times \mathcal B$.

Proof. Since $\gamma$ is finitely additive, it follows that for a countable family $\{C_i\}$ of disjoint elements of $\mathcal A\times \mathcal B$ with $\bigcup_{i=1}^\infty C_i\in \mathcal A\times \mathcal B$ $$ \gamma\left(\bigcup_{i=1}^\infty C_i\right)\geq \sum_{i=1}^\infty\gamma(C_i) $$ Note that for a countable family $\{A_k\}$ of disjoint elements of $\mathcal A$ and any $B\in\mathcal B$ we have $$ \gamma\left(\bigcup_{k=1}^\infty A_k \times B\right)=\gamma\left(\bigcup_{k=1}^\infty A_k \times Y\right)-\gamma\left(\bigcup_{k=1}^\infty A_k \times (Y\setminus B)\right) $$ Then we can use countable additivity of the marginal $$ \gamma\left(\bigcup_{k=1}^\infty A_k \times B\right)+\gamma\left(\bigcup_{k=1}^\infty A_k \times (Y\setminus B)\right)=\gamma\left(\bigcup_{k=1}^\infty A_k \times Y\right)=\sum_{k=1}^\infty\gamma(A_k\times Y) $$ Since it is forbidden for $\gamma$ to have a strict inequality of the form $$ \gamma\left(\bigcup_{k=1}^\infty A_k \times (Y\setminus B)\right)< \sum_{i=1}^\infty\gamma(A_k\times (Y\setminus B)) $$ we conclude that $$ \gamma\left(\bigcup_{k=1}^\infty A_k \times B\right)= \sum_{k=1}^\infty\gamma(A_k\times B) $$ and hence all measures of the form $\mu_B(A):=\gamma(A\times B)$, $B\in \mathcal B$ on $\mathcal A$ are countably additive. By analogous argument all measures $\nu_A(B):=\gamma(A\times B)$, $A\in \mathcal A$ on $\mathcal B$ are also countably additive.

Let $\{A_n\}$ be a countable family of disjoint elements of $\mathcal A$, $\{B_k\}$ be a countable family of disjoint elements of $\mathcal B$, $A=\bigcup_{n=1}^\infty A_n$, $B=\bigcup_{k=1}^\infty B_k$, $\bigcup_{k=1}^\infty \bigcup_{n=1}^\infty (A_n\times B_k)\in \mathcal A \times \mathcal B$, then $$ \gamma\left(\bigcup_{k=1}^\infty \bigcup_{n=1}^\infty A_n\times B_k\right)=\gamma\left(\bigcup_{k=1}^\infty A\times B_k\right)=\sum_{n=1}^\infty\gamma(A\times B_k)=\sum_{k=1}^\infty\sum_{n=1}^\infty\gamma(A_n\times B_k) $$ Since $$ \sum_{k=1}^\infty\sum_{n=1}^\infty\gamma(A_n\times B_k)=\gamma\left(\bigcup_{k=1}^\infty \bigcup_{n=1}^\infty A_n\times B_k\right)=\gamma\left(\bigcup_{k=1}^\infty \bigcup_{n\neq k} A_n\times B_k\right)+\gamma\left(\bigcup_{n=1}^\infty A_n\times B_n\right) $$ and $$ \gamma\left(\bigcup_{k=1}^\infty \bigcup_{n\neq k} A_n\times B_k\right)\geq \sum_{k=1}^\infty\sum_{n\neq k}\gamma(A_n\times B_k) $$ we conclude that $$ \gamma\left(\bigcup_{n=1}^\infty A_n\times B_n\right)=\sum_{n=1}^\infty\gamma(A_n\times B_n) $$

Let $C=\bigcup_{n=1}^\infty D_n$, $C=\bigcup_{j=1}^N C_j$, $D_n=\bigcup_{i=1}^{M_n} D_{n,i}$, where the families $\{C_j\}$ and $\{D_{n,i}\}$ are families of disjoint measurable rectangles. It follows that $D_{n,i,j}=D_{n,i}\cap C_j$ is also a family of disjoint measurable rectangles. Since $C_j=\bigcup_{n=1}^\infty \bigcup_{i=1}^{M_n} D_{n,i,j}$, we can use the obtained result to show that $$ \gamma(C_j)=\sum_{n=1}^\infty \sum_{i=1}^{M_n} \gamma(D_{n,i,j}) $$ Since $D_{n,j}=\bigcup_{j=1}^{N} D_{n,i,j}$, it is obvious that $\gamma(D_{n,j})=\sum_{j=1}^N \gamma(D_{n,i,j})$. It is also true that $\gamma(C)=\sum_{j=1}^N \gamma(C_j)$, and it follows that $$ \gamma(C)=\sum_{i=1}^\infty \gamma(D_n) $$ which completes the proof of the lemma.

$\endgroup$
5
  • $\begingroup$ I don't follow the last part of the argument --- where does the decomposition $D_n = \bigcup_{i=1}^{M_n} D_{n,i}$ come from? But it seems to me that the usual proof of this lemma for product measures, where you integrate characteristic functions, should work. $\endgroup$
    – Nik Weaver
    Feb 14, 2015 at 19:29
  • $\begingroup$ $D_n$ is an element of the algebra of step-sets, hence it can be represented as a finite union of disjoint rectangles. I agree that a shorter proof for the lemma is definitely possible, and probably it already exists. $\endgroup$ Feb 15, 2015 at 9:06
  • $\begingroup$ I really don't think this proof is right. The result about $\gamma(\bigcup A_n \times B_n)$ is only proven when the sequences $\{A_n\}$ and $\{B_k\}$ are pairwise disjoint. $\endgroup$
    – Nik Weaver
    Feb 15, 2015 at 16:44
  • $\begingroup$ It is very likely that I am wrong, but I can't see the mistake. After we obtain the result for pairwise disjoint rectangles, we check that $\gamma(\bigcup D_n)=\sum \gamma(D_n)$ for arbitrary sequence of step-sets $\{D_n\}$. To show that, we split $\bigcup D_n$ into a union of disjoint rectangles $\{D_{n,i,j}\}$ and apply the result about pairwise disjoint sequences to it. $\endgroup$ Feb 16, 2015 at 15:07
  • $\begingroup$ The result is not obtained for pairwise disjoint rectangles, it is obtained for sequences $\{A_n\}$ and $\{B_k\}$ each of which is pairwise disjoint. $\endgroup$
    – Nik Weaver
    Feb 16, 2015 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.