44.3 Affiliated Operators

Proof. (1) $\Rightarrow$ (2): Let $v \in U(M')$, then for any $\xi, \eta \in D(x^{*}x)$,

so $x^{*}x$ is affiliated with $M$. As such, $(x^{*}x-i)(x^{*}x+i)^{-1}\in M'' = M$ by the Bicommutant Theorem. By construction of the spectral measure, $E(B) \in M$ for all $B \in \cb_{\real}$, and $f(x^{*}x)$ is affiliated with $M$ for all $f \in B^{\infty}([0, \infty); \complex)$. In particular, $|x|$ is affiliated with $M$ as well.

As such, for any $v \in U(M')$, $uv|x| = u|x|v = xv = vx = vu|x|$, so $vu = uv$ on $\ol{|x|(H)}$. On the other hand, for any $\xi \in \ker(|x|)$, $|x|v\xi = v|x|\xi = 0$, so $v(\ker(|x|)) \subset \ker(|x|)$ and $\ker(uv) \supset \ker(|x|)$. As $H = \ker(|x|) \oplus \ol{|x|(H)}$, $uv = vu$ on $H$.$\square$

Proof, [Proposition 7.2.4, AP]. Let $\Gamma(x) = \bracsn{(\xi, x\xi)|\xi \in D(x)}$ and $\Gamma(y) = \bracsn{(\xi, y\xi)|\xi \in D(y)}$ be the graphs of $x$ and $y$, respectively. Since $x$ and $y$ are closed, $\Gamma(x)$ and $\Gamma(y)$ are closed subspaces of $H \oplus H$. Let $P_{x}, P_{y} \in B(H \oplus H)$ be the orthogonal projections of $H \oplus H$ onto $\Gamma(x)$ and $\Gamma(y)$, respectively.

Let $M_{2}(M) = M_{2}(\complex) \otimes M \subset B(H \oplus H)$, then

For any $u \in U(M')$, $ux = xu$, so for any $\xi \in D(x)$, $(u\xi, ux\xi) = (u\xi, xu\xi) \in \Gamma(x)$. On the other hand, if $(\xi, \eta) \in \Gamma(x)^{\perp}$, then for any $\theta \in D(x)$,

As $(u^{*}\theta, xu^{*}\theta) \in \Gamma(x)$, $(u\xi, u\eta) \in \Gamma(x)^{\perp}$. In other words, $\Gamma(x)$ and $\Gamma(x)^{\perp}$ are both invariant under $I_{M_2(\complex)}\otimes u$. As such, $\Gamma(x)$ is reducing for $I_{M_2(\complex)}\otimes u$, and $u$ commutes with $P_{x}$. Thus $P_{x} \in M_{2}(M)$. Similarly, $P_{y} \in M_{2}(M)$ as well.

Let $p = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} \in M_{2}(M)$, then

and

By Lemma 43.9.11, $P_{x} \sim p \sim P_{y}$. Since $M$ is tracial, $M_{2}(M)$ admits a faithful tracial state $\tau_{2} := \frac{1}{2}\text{Tr}_{2} \otimes \tau$. Thus $P_{x} \le P_{y}$ and $\tau_{2}(P_{x}) = \tau_{2}(P_{y})$ imply that $P_{x} = P_{y}$. Therefore $\Gamma(x) = \Gamma(y)$, and $x = y$.$\square$

Proof, [Lemma 7.2.5, AP]. Let $x = u|x|$ be the polar decomposition of $x$ and $E: \cb_{[0, \infty)}\to B(H)$ be the resolution of the identity for $|x|$, then $E(B)(H) \subset D(x)$ for all bounded $B \in \cb_{[0, \infty)}$. By Proposition 44.3.2, $E(B) \in\text{Proj}(M)$ for all $B \in \cb_{[0, \infty)}$. Since $\tau$ is normal, $\limv{n}\tau(E([0, n])) = \tau(1_{M}) = 1$, so there exists $n \in \natp$ with $\tau(E([0, n])) > 1 - \eps$. In which case, $E([0, n])H \subset D(x)$, and $\norm{xE([0, n])}_{M}\le n$.$\square$

Proof. For any $p, q \in \text{Proj}(M)$ with $pH, qH \subset V$, $(p \vee q)(H) = \ol{pH + qH}\subset \ol V$ and $\tau(p \vee q) \ge \tau(p) \vee \tau(q)$. As such, there exists an increasing sequence $\seq{p_n}\subset \text{Proj}(M)$ such that $p_{n}H \subset \ol V$ for all $n \in \natp$ and $\tau(p_{n}) \upto 1$ as $n \to \infty$.

Since $\tau$ is faithful and normal, $\tau(p_{n}) \upto 1$ implies that $\bigvee_{n \in \natp}p_{n} = 1_{M}$. By Theorem 43.3.1, $1_{M} = \sotlim_{n \to \infty}p_{n}$. As $p_{n}H \subset \ol V$ for all $n \in \natp$, $H \subset \ol V$.$\square$

Proof, [Extracted from Theorem 7.2.8, AP]. Let $r \in M$ be the projection onto $\ker((1_{M} - p)yq)$ and $s = r \wedge q$.

(1): For any $\xi \in \ker((1_{M} - p)yq)$, $yq\xi \in \ker(1_{M} - p) = pH \subset D(x)$, so $rH \subset D(xyq)$, and $(q \wedge r)H \subset D(xy)$. By definition, $yrH \subset \ker(1_{M} - p) = pH$.

(2): Let $(1 - p)yq = v |(1_{M} - p)yq|$ be the polar decomposition of $(1_{M} - p)yq$, then $v \in M$ is a partial isometry with initial space $\ol{|(1_M - p)yq|H}$ and final space $\ol{(1_M - p)yqH}= \ker((1_{M} - p)yq)^{\perp}$. Thus $v^{*}v = (1 - r)$, $vv^{*} \le 1_{M} - p$, and $1_{M} - r \prec 1_{M} - p$. Therefore

$\square$

Proof. Let $\Gamma(x) = \bracsn{(\xi, x\xi)|\xi \in D(x)}$ be the graph of $x$, $\ol{x}$ be the closure of $x$, and $\Gamma(\ol{x})$ be its graph. For each $u \in U(M')$,

$\square$

Proof, [Theorem 7.2.8, AP]. (1): By Proposition 43.8.10, $x^{*}$ is closed and densely defined.

Let $u \in U(M')$, $\xi \in D(x)$, and $\eta \in D(x^{*})$, then

so $u^{*}\eta \in D(x^{*})$ and $\eta \in D(x^{*}u^{*})$ with $x^{*}u^{*}\eta = u^{*}x^{*}\eta$. Conversely, if $\eta \in D(x^{*}u^{*})$, then

so $\eta \in D(x^{*})$ with $x^{*}u^{*}\eta = u^{*}x^{*}\eta$. Therefore $u^{*}x^{*} = x^{*}u^{*}$ for all $u \in U(M')$, and $x^{*}$ is affiliated with $M$.

(2): For each $\eps > 0$, by Lemma 44.3.4, there exists $p, q \in \text{Proj}(M)$ with $pH \subset D(x)$, $qH \subset D(y)$, and $\tau(p), \tau(q) \ge 1 - \eps$. As such, $(p \wedge q)H \subset D(x) \cap D(y) = D(x + y)$ with $\tau(p \wedge q) \ge 1 - 2\eps$. By Lemma 44.3.5, $D(x + y)$ is dense in $H$, so $x + y$ is densely defined.

By (1) and the above case, $x^{*} + y^{*}$ is also densely defined, so $(x^{*} + y^{*})^{*}$ is a closed extension of $x + y$, and $x + y$ is closable. By Lemma 44.3.7, the closure of $x + y$ lies in $\td M$.

(3): For each $\eps > 0$, by Lemma 44.3.4, there exists $p, q \in \text{Proj}(M)$ with $pH \subset D(x)$, $qH \subset D(y)$, $yq \in M$, and $\tau(p), \tau(q) \ge 1 - \eps$.

By Lemma 44.3.6, there exists $r \in \text{Proj}(M)$ with $rH \subset D(xy)$ and

As such an $r$ exists for all $\eps > 0$, Lemma 44.3.5 implies that $D(xy)$ is dense in $H$, and $xy$ is densely defined.

By (1) and the above case, $y^{*}x^{*}$ is also densely defined, so $(y^{*}x^{*})^{*}$ is a closed extension of $xy$, and $xy$ is closable. By Lemma 44.3.7, the closure of $xy$ lies in $\td M$.

(Algebra): For each $x, y, z \in \td M$,

and

Let $\ol{xy + xz}$ denote the closure of $xy + xz$, then

By (2) and (3), $\ol{xy + xz}\in \td M$, so Proposition 44.3.3 implies that

Similarly, $(x \ol{\cdot}y)^{*} = (xy)^{*} \supset y^{*}x^{*}$ and $y^{*} \ol{\cdot}x^{*} \supset y^{*}x^{*}$. If $\ol{y^*x^*}$ is the closure of $y^{*}x^{*}$, then by (1), (3), and Proposition 44.3.3, $(x \ol{\cdot}y)^{*} = \ol{y^*x^*}= y^{*} \ol{\cdot}x^{*}$.$\square$