Absolutely compatible pair of elements in a von Neumann algebra-II

TL;DR

This paper characterizes absolutely compatible pairs of elements within von Neumann algebras, revealing their structure and similarities to generic pairs of projections, thus advancing understanding of compatibility in operator algebras.

Contribution

It provides a complete description of absolutely compatible pairs of strict elements in von Neumann algebras, highlighting their structural form and relation to projection pairs.

Findings

Characterization of absolutely compatible pairs in von Neumann algebras

Structural resemblance to generic pairs of projections

Complete description of strict element pairs

Abstract

Let $A$ be a unital C$^*$-algebra with unity $1_A$. A pair of elements $0 \le a, b \le 1_A$ in $A$ is said to be \emph{absolutely compatible} if, $\vert a - b \vert + \vert 1_A - a - b \vert = 1_A.$ In this paper we provide a complete description of absolutely compatible pair of strict elements in a von Neumann algebra. The end form of such a pair has a striking resemblance with that of a `generic pair' of projections on a complex Hilbert space introduced by Halmos.

Equations

$$ab=0=ba=a^{\ast }b=a{b}^{\ast }.$$

$$|a-b|+|1_A-a-b|=1_A.$$

$$s(a):=\sup \{p\in \mathcal{P}(M):p\le a\}$$

$$n(a):=\sup \{p\in \mathcal{P}(M):pa=0\}.$$

$$a=p_1\oplus a_1\oplus a_2\oplus 0\oplus a_3$$

$$b=b_1\oplus p_2\oplus b_2\oplus b_3\oplus 0$$

$$a=U^{\ast }\left[\begin{array}{cc}{a}_1& (a_1{b}_1{)}^{\frac{1}{2}}\\ (a_1{b}_1{)}^{\frac{1}{2}}& p-a_1\end{array}\right]U\text{and}b=U^{\ast }\left[\begin{array}{cc}{b}_1& -(a_1{b}_1{)}^{\frac{1}{2}}\\ -(a_1{b}_1{)}^{\frac{1}{2}}& p-b_1\end{array}\right]U.$$

$$P=U^{\ast }\left[\begin{array}{cc}{C}^2& CS\\ CS& S^2\end{array}\right]U\text{and}Q=U^{\ast }\left[\begin{array}{cc}{C}^2& -CS\\ -CS& S^2\end{array}\right]U.$$

$$0\le \left[\begin{array}{cc}{p}_0& 0\\ 0& 1-p\end{array}\right]\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{12}^{\ast }& a_{22}\end{array}\right]\left[\begin{array}{cc}{p}_0& 0\\ 0& 1-p\end{array}\right]=\left[\begin{array}{cc}0& p_0{a}_{12}\\ a_{12}^{\ast }{p}_0& a_{22}\end{array}\right].$$

$$\left[\begin{array}{cc}p-p_0& 0\\ 0& 1-p\end{array}\right]a\left[\begin{array}{cc}p-p_0& 0\\ 0& 1-p\end{array}\right]=a.$$

$$a=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{12}^{\ast }& a_{22}\end{array}\right]\text{and}b=\left[\begin{array}{cc}{b}_{11}& -a_{12}\\ -a_{12}^{\ast }& b_{22}\end{array}\right]$$

$$a_{12}=u|a_{12}|$$

$$(p_1-a_{11})(p_1-b_{11})$$

$$u|a_{12}|$$

$$|a_{12}|(a_{11}u+u{a}_{22}-u{)}^{\ast }(a_{11}u+u{a}_{22}-u)=0.$$

$$p_1=u{u}^{\ast }=a_{11}u{u}^{\ast }+u{a}_{22}{u}^{\ast }=a_{11}+u{a}_{22}{u}^{\ast }$$

$$U^{\ast }\left[\begin{array}{cc}{a}_1& (a_1{b}_1{)}^{\frac{1}{2}}\\ (a_1{b}_1{)}^{\frac{1}{2}}& p-a_1\end{array}\right]U=a\text{and}{U}^{\ast }\left[\begin{array}{cc}{b}_1& -(a_1{b}_1{)}^{\frac{1}{2}}\\ -(a_1{b}_1{)}^{\frac{1}{2}}& p-b_1\end{array}\right]U=b.$$

$$a_{11}+b_{11}-p_1$$

$$u^{\ast }(a_{11}+b_{11}-p_1{)}^{k}u=(p-a_1-b_1{)}^k$$

$$r(p-a_1-b_1)=u^{\ast }r(a_1+b_{11}-p_1)u=u^{\ast }{p}_{1}u=p.$$

$$s(p-a_1-b_1)\le p-a_1-b_1\le p-a_1$$

$$s(p-a_1-b_1)\le s(p-a_1)=0$$

$$|⟨ab\eta ,\xi ⟩{|}^2$$

$$⟨\left[\begin{array}{cc}{a}^2& ab\\ ab& 1-a^2\end{array}\right]\left[\begin{array}{c}\xi \\ \eta \end{array}\right]⟩$$

$$1_{M_2(M)}-a_1=\left[\begin{array}{cc}1-a^2& -ab\\ -ab& a^2\end{array}\right]\text{and}{1}_{M_2(M)}-b_1=\left[\begin{array}{cc}1-b^2& ab\\ ab& b^2\end{array}\right]$$

$$\xi =a^2\xi +ab\eta \text{and}\eta =ab\xi +(1-a^2)\eta .$$

$$a^2\xi +ab\eta =0\text{and}ab\xi +(1-a^2)\eta =0$$

$$(a_1-b_1{)}^2={\left[\begin{array}{cc}{a}^2-b^2& 2ab\\ 2ab& b^2-a^2\end{array}\right]}^2=\left[\begin{array}{cc}(a^2+b^2{)}^2& 0\\ 0& (a^2+b^2{)}^2\end{array}\right]$$

$$|1_{M_2(M)}-a_1-b_1|$$

Full text

1. Introduction

Let $A$ be a C∗-algebra. A pair of elements $a,b\in A$ is said to be orthogonal, if

[TABLE]

Orthogonal pair of positive elements play an important role in the theory of C∗-algebras. For example, it follows from the functional calculus that every self-adjoint element $a\in A_{sa}$ has a unique decomposition: $a=a^{+}-a^{-}$ in $A^{+}$, where $a^{+}$ is algebraically orthogonal to $a^{-}$. Recently, the author proved an order theoretic characterization of algebraic orthogonality among positive elements of a C∗-algebra [7]. (Also see [5, 6].)

Orthogonal pairs of positive elements of norm $\leq 1$ exhibit an interesting property. Let $A$ be a unital C∗-algebra with unity $1_{A}$. For a pair of elements $0\leq a,b\leq 1_{A}$ in $A$, we have $ab=0$ ($a$ is algebraically orthogonal to $b$) if and only if $a+b\leq 1_{A}$ and $|a-b|+|1_{A}-a-b|=1_{A}.$ We isolate the later part and propose the following definition: A pair of elements $0\leq a,b\leq 1_{A}$ in $A$ is said to be absolutely compatible, ([7]), if

[TABLE]

It was proved in [7] that a projection $p$ in a C∗-algebra $A$ is absolutely compatible with a positive element $a$ of $A$ with $\|a\|\leq 1$ if, and only if, $ap=pa$. However, the two notions are distinct in general. The notion of absolute compatibility was introduced as an instrument to prove a spectral decomposition theorem in the context of “absolute order unit spaces” [7]. As an absolute order unit space is an order theoretic generalization of unital C∗-algebras [7], absolute compatibility appears to be a significant property.

Keeping this point of view, the author, along with Jana and Peralta, initiated a study of absolute compatibility in operator algebras [3, 4]. Let $M$ be a von Neumann algebra and let $0\leq a\leq 1_{M}$. We write

[TABLE]

and

[TABLE]

For $0\leq a\leq 1_{M}$, we say that $a$ is strict in $M$, if $s(a)=0$ and $n(a)=0$. In [3], it was proved that an absolutely compatible pair of positive elements in a von Neumann algebra has a (matricial) decomposition as a direct sum of commuting and ‘strict’ elements. Let $0\leq a,b\leq 1_{M}$ such that $a$ is absolutely compatible with $b$. Then there exist mutually orthogonal projections $p_{1},p_{2},s,n_{1},n_{2}\in\mathcal{P}(M)$ with $p_{1}+p_{2}+s+n_{1}+n_{2}=1_{M}$ such that

[TABLE]

and

[TABLE]

with respect to $\{p_{1},p_{2},s,n_{1},n_{2}\}$ with $a_{2}$ and $b_{2}$ are strict and absolutely compatible in $sMs$ [3, Theorem 2.10]. Thus the study of absolutely compatible pair of elements reduces to such strict pairs. It was further proved that

Theorem 1.1**.**

[3, Theorem 2.12]** Let $M$ be a von Neumann algebra, and let $a,b$ be elements in $[0,1]_{M}$. Then $a$ is absolutely compatible with $b$ if and only if there exists a projection $p_{1}$ in $M$ so that $a$ and $b$ have matrix representations, say a=\left(\begin{array}[]{cc}a_{11}&a_{12}\\ a_{12}^{\ast}&a_{22}\end{array}\right), and b=\left(\begin{array}[]{cc}b_{11}&b_{12}\\ b_{12}^{\ast}&b_{22}\end{array}\right) with respect to the set $\{p_{1},1-p_{1}=p_{2}\}$ (i.e., $a_{ij}=p_{i}ap_{j}$ and $b_{ij}=p_{i}bp_{j}$) satisfying:

1. $(i)$

$a_{12}+b_{12}=0$; 2. $(ii)$

$a_{12}a_{12}^{\ast}=(p_{1}-a_{11})(p_{1}-b_{11})$; 3. $(iii)$

$a_{12}^{\ast}a_{12}=a_{22}b_{22}=b_{22}a_{22}$; 4. $(iv)$

$a_{12}=a_{11}a_{12}+a_{12}a_{22}=b_{11}a_{12}+a_{12}b_{22}$.

Using these decompositions, a complete description of absolutely compatible pair of strict elements was given for the finite dimensional algebras $\mathbb{M}_{n}$ in the same paper [3, Theorem 3.9].

In the present paper we continue the investigation and provide a complete description of absolutely compatible pair of strict elements in a von Neumann algebra. However, as expected, the finite dimensional (matricial) techniques used in [3] fail to work. At this juncture, the author would like to thank Professor Kalyan B. Sinha for his suggestion to use polar decomposition. This idea works just perfectly. The main results of the paper are Theorems 1.2 and 1.3.

Theorem 1.2**.**

Let $M$ be a von Neumann algebra with the underlying Hilbert space $H$. Assume that $a,b\in[0,1]_{M}$ be strict and absolutely compatible pair. Put $p=1-r(a\circ b)$.

1. (1)

Then $pH$ is isometrically isomorphic to $(1-p)H$. In particular, $H\equiv K\oplus K$, where $K=pH$. 2. (2)

There exist strict elements $a_{1},b_{1}\in[0,p]\cap M$ with $a_{1}b_{1}=b_{1}a_{1}$, $a_{1}+b_{1}\leq p$ together with $p-(a_{1}+b_{1})$ strict in $pMp$; and a unitary $U:H\to K\oplus K$ such that

[TABLE]

Theorem 1.3**.**

Let $M$ be a von Neumann algebra with the underlying Hilbert space $H$. Assume that $a,b\in[0,1]_{M}$ be strict and commuting pair such that $a^{2}+b^{2}\leq 1$ with $1-(a^{2}+b^{2})$ strict. Put $a_{1}=\begin{bmatrix}a^{2}&ab\\ ab&1-a^{2}\end{bmatrix}$ and $b_{1}=\begin{bmatrix}b^{2}&-ab\\ -ab&1-b^{2}\end{bmatrix}$. Then $a_{1},b_{1}\in[0,1]_{M_{2}(M)}$ and $a_{1}$ is absolutely compatible with $b_{1}$.

It is interesting and surprising to note that the end form of an absolutely compatible pair of strict elements bears a striking resemblance with that of a ‘generic pair’ of projections on a complex Hilbert space (studied by Halmos [2]).

A pair of closed subspaces $M$ and $N$ of a Hilbert space $H$ is said to be in generic position, if each of the following four subspaces of $H$: $M\cap N$, $M\cap N^{\perp}$, $M^{\perp}\cap N$ and $M^{\perp}\cap N^{\perp}$ are trivial. Let $P$ and $Q$ be the projections of $H$ on $M$ and $N$ respectively. In [2], Halmos proved that if $M$ and $N$ are in generic position, then there exists a Hilbert space $K$, commuting, positive and invertible contractions $C$ and $S$ in $B(K)$ and a unitary operator $U:H\to K\oplus K$ such that

[TABLE]

This observation further signifies the importance of a pair of strict absolutely compatible elements in an operator algebra. This paper is in the sequel of [3].

2. The main results

We obtain some basic results to prove Theorems 1.2 and 1.3.

Lemma 2.1**.**

Let $M$ be a von Neumann algebra with the underlying Hilbert space $H$ and let $a\in[0,1]_{M}$ be strict. If $p\in\mathcal{P}(M)$, then, for the matrix representation $a=\begin{bmatrix}a_{11}&a_{12}\\ a_{12}^{*}&a_{22}\end{bmatrix}$ with respect to $\{p,1-p\}$, we have $a_{11}\in[0,p]$ and $a_{22}\in[0,1-p]$ are also strict in the von Neumann algebras $pMp$ and $(1-p)M(1-p)$ respectively.

Proof.

Without any loss of generality, we may assume that $p\neq 0,p\neq 1$. First, we show that $r(a_{11})=p$. Put $p-r(a_{11})=p_{0}$. Then $p_{0}a_{11}=0$ so that

[TABLE]

Thus $p_{0}a_{12}=0$ and consequently,

[TABLE]

Now, it follows that $1\leq r(a)\leq 1-p_{0}$ so that $p_{0}=0$. Thus $r(a_{11})=p$. Similarly, $r(a_{22})=1-p$. Since $n(x)=1-r(x)$ for any $x\in[0,1]_{M}$, we have $n(a_{11})=0$ and $n(a_{22})=0$.

Next, as $a$ is strict, so is $1-a=\begin{bmatrix}p-a_{11}&-a_{12}\\ -a_{12}^{*}&1-p-a_{22}\end{bmatrix}$. Thus, as above, $r(p-a_{11})=p$ and $r(1-p-a_{22})=1-p$. Therefore, $s(a_{11})=0$ and $s(a_{22})=0$. Hence $a_{11}\in[0,p]$ and $a_{22}\in[0,1-p]$ are strict in the von Neumann algebras $pMp$ and $(1-p)M(1-p)$ respectively. ∎

Lemma 2.2**.**

Let $M$ be a von Neumann algebra with the underlying Hilbert space $H$ and let $a\in[0,1]_{M}$ be strict. If $ax=0$ for some $x\in M$, then $x=0$. In particular, if $a\xi=0$ for some $\xi\in H$, then $\xi=0$.

Proof.

Let $ax=0$. Then $a|x^{*}|^{2}=axx^{*}=0$ so that $a|x^{*}|=0$ and consequently, $ar(|x^{*}|)=0$. Thus $r(|x^{*}|)\leq n(a)=0$ so that $r(|x^{*}|)=0$. Now, it follows that $x=0$.

Next, if $a\xi=0$, then $ap=0$ where $p$ is the projection of $H$ on the span of $\xi$. Now, by the first step, $p=0$ so that $\xi=0$. ∎

Lemma 2.3**.**

Let $M$ be a von Neumann algebra with the underlying Hilbert space $H$ and let $a,b\in[0,1]_{M}$ be strict.

1. (1)

Then $a^{\frac{1}{2}}$ is also strict in $M$. 2. (2)

If $ab=ba$, then $ab$ is also strict in $M$.

Proof.

1. (1)

Let $p=s(a^{\frac{1}{2}})$. Then $p=pa^{\frac{1}{2}}=a^{\frac{1}{2}}p$. Squaring, we get $p=pa=ap$ so that $p\leq s(a)=0$. Thus $p=0$ and consequently, $s(a^{\frac{1}{2}})=0$. Also, $n(a^{\frac{1}{2}})=n(a)=0$. Therefore, $a^{\frac{1}{2}}$ is strict. 2. (2)

Let $ab=ba$. Then $ab\in[0,1]_{M}$ with $ab\leq a$. Thus $s(ab)\leq ab\leq a$ so that $s(ab)\leq s(a)=0$. Therefore, $s(ab)=0$. Next, we have $abn(ab)=0$. As $a$ and $b$ are strict, a repeated use of Lemma 2.2 yields that $n(ab)=0$. Thus $ab$ is also strict.

∎

Remark 2.4**.**

Let $a\in[0,1]_{M}$. Then $a$ is strict if and only if $a^{2}$ is strict,

The following results are a compilation of operator algebra folklore which can easily be found in the literature. (See, for example, [1, Section I.5.2].)

Folklore 2.5**.**

Let $M$ be a von Neumann algebra with the underlying Hilbert space $H$ and let $x\in M$. If $x=u|x|$ is the polar decomposition of $x$, then

1. (1)

$u:(1-r(|x|))H\to r(|x|)H$* is an unitary. In particular, $u$ is a partial isometry in $M$.* 2. (2)

$|x|=u^{*}x$. 3. (3)

$r(|x|)=u^{*}u$. 4. (4)

$|x^{*}|^{k}=u|x|^{k}u^{*}$* for all $k\in\mathbb{N}$.* 5. (5)

$|x|^{k}=u^{*}|x^{*}|^{k}u$* for all $k\in\mathbb{N}$.*

Now we prove the main results of the paper.

Proof of Theorem 1.2.

1. (1)

Put $r(a\circ b)=p_{1}$ so that $p=1-p_{1}$. Now, by Theorem 1.1, $a$ and $b$ have matrix representations

[TABLE]

with respect to $\{p_{1},p\}$ such that

1. (a)

$a_{12}a_{12}^{*}=(p_{1}-a_{11})(p_{1}-b_{11})$; 2. (b)

$a_{12}^{*}a_{12}=a_{22}b_{22}$; and 3. (c)

$a_{12}=a_{11}a_{12}+a_{12}a_{22}=b_{11}a_{12}+a_{12}b_{22}$.

Since $a$ and $b$ are strict, we have $p_{1}\neq 0,p_{1}\neq 1$. Also, by Lemma 2.1, $a_{11}$ and $b_{11}$ are strict in $p_{1}Mp_{1}$ and $a_{22}$ and $b_{22}$ are strict in $pMp$. Consider the polar decomposition

[TABLE]

so that $U:u^{*}uH\to uu^{*}H$ is a unitary. By (a) and Folklore 2.5(4), we have

[TABLE]

Now, as $a_{11}$ and $b_{11}$ are strict elements in $[0,p_{1}]$ with $a_{11}b_{11}=b_{11}a_{11}$, by Lemma 2.3(2), we may conclude that $(p_{1}-a_{11})(p_{1}-b_{11})$ is also a strict element in $[0,p_{1}]$. Thus $r((p_{1}-a_{11})(p_{1}-b_{11}))=p_{1}$. Thus $uu^{*}=p_{1}=1-p$. In a similar way, by using (b) and Folklore 2.5(5), we may prove that $u^{*}u=p$. Thus $u:pH\to(1-p)H$ is a unitary. 2. (2)

By (b), we have $a_{22}b_{22}=|a_{12}|^{2}$ so that $a_{22}$, $b_{22}$ and $|a_{12}|$ commute with each other. Thus by (c), we get

[TABLE]

In other words,

[TABLE]

By Lemma 2.3, $|a_{12}|$ is strict in $pMp$ so that, by Lemma 2.2, $a_{11}u+ua_{22}=u$. Therefore,

[TABLE]

whence $ua_{22}u^{*}=p_{1}-a_{11}$. In the same way we can show that $ub_{22}u^{*}=p_{1}-b_{11}$. Now, put $a_{22}=a_{1}$, $b_{22}=b_{1}$ and $U=\begin{bmatrix}0&p\\ u^{*}&0\end{bmatrix}$. Then $a_{1}$ and $b_{1}$ are strict elements in $[0,p]$ and $U:H\to K\oplus K$ is a unitary. Also, the matrix multiplications yield that

[TABLE]

Finally, we show that $p-a_{1}-b_{1}$ is a strict element of $[0,p]$. Note that $a\circ b=\begin{bmatrix}a_{11}+b_{11}-p_{1}&0\\ 0&0\end{bmatrix}$. Thus $r(a_{11}+b_{11}-p_{1})=r(a\circ b)=p_{1}$. Next,

[TABLE]

for $upu^{*}=uu^{*}uu^{*}=p_{1}^{2}=p_{1}$. Also, then

[TABLE]

for all $k\in\mathbb{N}$. Since for any $x\in M^{+}$, $r(x)$ is limit of the increasing sequence $\{(\frac{1}{n}+x)^{-1}x\}$ in the strong operator topology, we may conclude that

[TABLE]

Thus $n(p-a_{1}-b_{1})=0$. Also

[TABLE]

so that

[TABLE]

for $p-a_{1}$ is strict. Therefore, $p-a_{1}-b_{1}$ is a strict element in $[0,p]$.

∎

Proof of Theorem 1.3.

Let $\xi,\eta\in H$. Then

[TABLE]

Thus we obtain that

[TABLE]

Therefore, $a_{1}\geq 0$. Similarly, we can also show that $b_{1}\geq 0$. Again,

[TABLE]

so that $a_{1},b_{1}\in[0,1]_{M_{2}(M)}$.

Now, we show that $a_{1}$ is strict. Let $p$ be a projection in $M_{2}(M)$ such that $p\leq a_{1}$. Then $p=a_{1}p$. For $\begin{bmatrix}\xi\\ \eta\end{bmatrix}\in p(H\oplus H)$, we have $a_{1}\begin{bmatrix}\xi\\ \eta\end{bmatrix}=\begin{bmatrix}\xi\\ \eta\end{bmatrix}$ so that

[TABLE]

By the second condition, we get $a^{2}\eta=ab\xi$ so that $a(a\eta-b\xi)=0$. Since $a$ is strict, by Lemma 2.2, we may conclude that $a\eta=b\xi$. Using this in the first relation, we get $\xi=a^{2}\xi+b^{2}\xi$ so that $(1-a^{2}-b^{2})\xi=0$. Since $1-a^{2}-b^{2}$ is also strict, invoking Lemma 2.2 again we conclude that $\xi=0$. But then, $a\eta=0$ whence $\eta=0$ as $a$ is also strict. Now, it follows that $p=0$ so that $s(a_{1})=0$.

Next, let $q$ be a projection in $M_{2}(M)$ such that $a_{1}q=0$. If $\begin{bmatrix}\xi\\ \eta\end{bmatrix}\in q(H\oplus H)$, we have $a_{1}\begin{bmatrix}\xi\\ \eta\end{bmatrix}=0$. Thus

[TABLE]

and, as above we again conclude that $q=0$. Therefore, $n(a_{1})=0$ too whence $a_{1}$ is strict. In a similar manner, we can conclude that $b_{1}$ is also strict. Finally, we show that $a_{1}$ is absolutely compatible with $b_{1}$. We have

[TABLE]

so that $|a_{1}-b_{1}|=\begin{bmatrix}a^{2}+b^{2}&0\\ 0&a^{2}+b^{2}\end{bmatrix}$. Also

[TABLE]

It follows that $|a_{1}-b_{1}|+|1_{M_{2}(M)}-a_{1}-b_{1}|=1_{M_{2}(M)}$ so that $a_{1}$ is absolutely compatible with $b_{1}$. ∎

Acknowledgements: The author is thankful to Antonio M. Peralta for introducing to him the notion of ‘strict’ elements.