44.1 Traces
Definition 44.1.1 (Trace).label Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\tau: M_{+} \to [0, \infty]$, then $\tau$ is a trace on $M$ if:
- (1)
For each $x, y \in M_{+}$ and $\lambda \ge 0$, $\tau(\lambda x + y) = \lambda \tau(x) + \tau(y)$.
- (2)
For each $x \in M_{+}$ and $u \in U(M)$, $\tau(uxu^{*}) = \tau(x)$.
Definition 44.1.2 (Tracial von Neumann Algebra).label Let $M$ be a von Neumann algebra and $\tau: M \to \complex$ be a faithful tracial state, then $(M, \tau)$ is a tracial von Neumann algebra if $\tau$ is normal.
Proposition 44.1.3.label Let $M$ be a unital $C^{*}$-algebra, $\tau: M \to \complex$ be a faithful tracial state, and $(L^{2}(M, \tau), \pi, 1_{M})$ be its GNS triple, then:
- (1)
The uniform structure of pointwise convergence[1] on $\ol{B_M(0, 1)}\subset B(L^{2}(M, \tau))$ coincides with the uniform structure of $\ol{B_M(0, 1)}\subset L^{2}(M, \tau)$ induced by the $L^{2}(M, \tau)$ norm.
- (2)
$\pi(M) \subset B(L^{2}(M, \tau))$ is a von Neumann algebra if and only if $\ol{B_M(0, 1)}$ is complete in $L^{2}(M, \tau)$. In which case, $\tau$ is normal.
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 $B(L^{2}(M, \tau))$.$\square$
Proposition 44.1.4.label Let $(M, \tau)$ be a tracial von Neumann algebra, then the following are equivalent:
- (1)
$M$ is separable.
- (2)
$\ol{B_M(0, 1)}$ is strong-operator separable.
- (3)
$\ol{B_M(0, 1)}$ is separable in $L^{2}(M, \tau)$.
- (4)
$L^{2}(M, \tau)$ is separable.
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 44.1.3.
(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.$\square$
- Strong operator topology.keyboard_return
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