45.1 Traces

Proof, [Proposition 2.6.4, AP]. (1): For any $x, y \in \ol{B_M(0, 1)}$,

so the uniform structure of pointwise convergence is stronger than the uniformity induced by the $L^{2}(M, \tau)$ norm.

On the other hand, by the Arzelà-Ascoli Theorem, the uniform structure of pointwise convergence on $\ol{B_M(0, 1)}$ coincides with the uniform structure of pointwise convergence on any dense subset of $L^{2}(M, \tau)$. In particular, this includes $M \subset L^{2}(M, \tau)$. For each $x, y \in \ol{B_M(0, 1)}$ and $t \in M$, since $\tau$ is tracial,

so the uniformity induced by the $L^{2}(M, \tau)$ norm is stronger than the uniform structure of pointwise convergence.

(2): Using (1), it is sufficient to show that $\ol{B_M(0, 1)}\subset B(L^{2}(M, \tau))$ is complete if and only if it is strong-operator closed. By Proposition 6.7.3, $\ol{B_M(0, 1)}$ is complete if and only if it is closed in $L^{2}(M, \tau)^{L^2(M, \tau)}$ with respect to the product topology. Finally, the Arzelà-Ascoli Theorem implies that $\ol{B_M(0, 1)}$ is closed in $L^{2}(M, \tau)^{L^2(M, \tau)}$ if and only if it is strong-operator closed in $B(L^{2}(M, \tau))$.$\square$

Proof. (1) $\Rightarrow$ (2): Let $H$ be a separable Hilbert space with $M \subset B(H)$ and $D \subset H$ be countable and dense. By the Arzelà-Ascoli Theorem, the strong operator topology on $\ol{B_M(0, 1)}$ coincides with the topology of pointwise convergence on $D$. By Theorem 6.3.10 and Proposition 5.5.7, $H^{D}$ is metrisable and separable. Therefore $\ol{B_M(0, 1)}\subset H^{D}$ is also metrisable and separable.

(2) $\Rightarrow$ (3): By Proposition 45.1.4.

(3) $\Rightarrow$ (4): By Lemma 43.4.10, the standard representation $(L^{2}(M, \tau), \pi, 1_{M})$ is faithful and normal. As $M$ is dense in $L^{2}(M, \tau)$, $\ol{B_M(0, 1)}$ being separable in $L^{2}(M, \tau)$ implies that $L^{2}(M, \tau)$ is separable.$\square$

Proof. (1): Suppose that $M$ has a minimal projection. By Theorem 43.9.15, there exists a complex Hilbert space $H$ such that $M \iso B(H)$. Let $\seqi{e}\subset H$ be an orthonormal basis, then $e_{i} \otimes e_{i} \in \text{Proj}(M)$ is minimal for all $i \in I$, and all such rank one projections are equivalent. As such, for any $j \in I$,

so $I$ must be finite, and $M \iso M_{|I|}(\complex)$.

(2): Suppose that $M$ has no minimal projections. By Lemma 43.9.13, $M$ has no abelian projections. As $\tau$ is a faithful tracial state, $1_{M}$ is finite. Therefore $M$ is a $\vnII_{1}$ factor.$\square$

Proof, [Lemma 4.1.1, AP]. Since $p$ is not minimal, there exists a projection $q \le p$ with $0 \ne q \ne p$. As $M$ is a factor, $Z(q)Z(p - q) = 1_{M}$. By Proposition 43.8.5, there exist non-zero projections $p_{1}, p_{2} \in \text{Proj}(M)$ with $p_{1} \sim p_{2}$, $p_{1} \le q$, and $p_{2} \le p - q$.$\square$

Proof, [Lemma 4.1.1, AP]. Let

and $\mathcal{M}\subset \mathcal{F}$ be a maximal family such that $\bracsn{q, r|(q,r) \in \mathcal{M}}$ are all mutually orthogonal. Let $q = \sum_{(q, r) \in \mathcal{M}}q$ and $r = \sum_{(q, r) \in \mathcal{M}}r$, then by Lemma 43.8.10, $q \sim r$, $q \perp r$, and $q + r \le p$.

If $q + r \ne p$, then by Lemma 45.1.7 applied to $p - (q + r)$, there exist non-zero projections $q', r' \in \text{Proj}(M)$ with $q' \sim r'$, $q' \perp r'$, and $q' + r' \le p - (q + r)$, which contradicts the maximality of $\mathcal{M}$.$\square$

Proof. By Lemma 45.1.8.$\square$

Proof, [Proposition 4.1.3, AP]. Using Proposition 45.1.6, assume without loss of generality that $M$ is diffuse.

Let $p \in \text{Proj}(M)$ with $0 \le p < 1_{M}$. Let $n \in \natp$, then by Corollary 45.1.9, there exist mutually orthogonal equivalent projections $\bracsn{p_j}_{1}^{2^n}\subset \text{Proj}(M)$ such that $\sum_{j = 1}^{2^n}p_{j} = 1$. By the corollary:vna-factor-comparability, there exists a unique $0 \le k \le 2^{n} - 1$ such that

As such, for any tracial state $\tau: M \to \complex$,

so $\tau(p)$ is independent of the choice of $\tau$. By Theorem 43.7.2, $\tau$ is unique.$\square$

Proof, [Proposition 4.1.4, AP]. If $M$ is a factor, then $\tau$ is unique by Proposition 45.1.10. Suppose that $M$ is not a factor, then $M$ admits a non-trivial central projection $z \in Z(M)$. In which case, for any $\alpha \in (0, 1)$,

is also a normal faithful tracial state. In particular,

so if $\alpha \ne \tau(z)$, then $\tilde \tau \ne \tau$.$\square$

Proof, [Proposition 4.1.5, AP]. Let $I \subset M$ be a non-zero two-sided ideal. By Proposition 43.7.4, $I$ admits a non-zero projection $e \in I$. Let $\mathcal{P}\subset \text{Proj}(M)$ be a maximal family of mutually orthogonal projections equivalent to $e$. Since $\tau$ is faithful, $\tau(p) > 0$ for all $p \in \mathcal{P}$, so $\mathcal{P}$ is finite.

Let $q = 1_{M} - \sum_{p \in \mathcal{P}}p$, then by the comparability theorem, either $q \preceq e$ or $e \prec q$. If $e \prec q$, then there exists a subprojection $p \le q$ with $e \sim p$, which contradicts the maximality of $\mathcal{P}$. Thus $q \preceq e$.

Let $\mathcal{Q}= \mathcal{P}\sqcup \bracsn{q}$, then $p \preceq e$ for all $p \in \mathcal{Q}$ and $1_{M} = \sum_{p \in \mathcal{Q}}p$. For each $p \in \mathcal{Q}$, let $v_{p} \in M$ be a partial isometry such that $v_{p}v_{p}^{*} = p$ and $v_{p}^{*}v_{p} \le e$. As such, $p = v_{p}ev_{p}^{*} \in I$ for all $p \in \mathcal{Q}$. Therefore $1_{M} = \sum_{p \in \mathcal{Q}}p \in I$, and $I = M$.$\square$

Proof, [Proposition 4.1.6, AP]. (Injective): Let $p, q \in \text{Proj}(M)$ with $\tau(p) = \tau(q)$. By the comparability theorem, either $p \prec q$, $p \sim q$, or $q \prec p$. Since $\tau$ is faithful, $p \prec q$ implies that $\tau(p) < \tau(q)$, and $q \prec p$ implies that $\tau(q) < \tau(p)$. Therefore the only possibility is $p \sim q$.

(Surjective): By Lemma 45.1.8, there exists a family $\bracsn{p_r|r \in \mathbb{D} \cap [0, 1]}\subset \text{Proj}(M)$ such that:

1. (i)

   For each dyadic rational number $r \in \mathbb{D} \cap [0, 1]$, $\tau(p_{r}) = r$.

2. (ii)

   For each $r, s \in \mathbb{D} \cap [0, 1]$ with $r \le s$, $p_{r} \le p_{s}$.

For each $t \in [0, 1]$, let $p_{t} = \sup\bracsn{p_r|r \in \mathbb{D} \cap [0, 1], r \le t}$, then since $\tau$ is normal,

$\square$