44.3 The Hyperfinite $\vnII_{1}$ Factor

Proof. Let $(M_{0}, \bracsn{\phi^i|i \in I})$ be the inductive limit of $(\bracsn{M_i}_{i \in I}, \bracsn{\phi^i_j|i \le j})$ as vector spaces. Since $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are trace-preserving and $\bracsn{\tau_i}_{i \in I}$ are faithful, $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are also unital isometric embeddings. As such, $\bracsn{M_i}_{i \in I}$ may be identified as subspaces of $M_{0}$, the norms on $\bracsn{M_i}_{i \in I}$ extend to $M_{0}$, and $\phi^{i}: M_{i} \to M_{0}$ are isometric embeddings for all $i \in I$.

For each $i, j \in I$ with $i \le j$, $\phi^{i}_{j}: M_{i} \to M_{j}$ is trace-preserving. By universal property of the inductive limit, there exists a linear functional $\tau: M_{0} \to \complex$ such that $\tau \circ \phi^{i} = \tau_{i}$ for all $i \in I$. Similarly, $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are *-homomorphisms, the products and involutions from $\bracsn{M_i}_{i \in I}$ extend to $M_{0}$, and $\phi^{i}: M_{i} \to M_{0}$ is a unital trace-preserving *-homomorphism for all $i \in I$. In particular, $1_{M_0}= \phi^{i}(1_{M_i})$ for all $i \in I$.

For each $i \in I$, $\tau_{i}: M_{i} \to \complex$ is a faithful tracial state. As such, for any $x \in M_{0}$, $x = 0$ if and only if $\tau(x^{*}x) = 0$, so the pairing $\dpn{x, y}{\tau}= \tau(y^{*}x)$ is an inner product on $M_{0}$. Let $H$ be the completion of $M_{0}$ with respect to $\dpn{\cdot, \cdot}{\tau}$. For each $x \in M_{0}$, let $L_{x}: M_{0} \to M_{0}$ be defined by $L_{x}(y) = xy$ for all $y \in M_{0}$. Then, for any $x, y, z \in M_{0}$,

and

with

Thus the mapping $M_{0} \to B(H)$ with $x \mapsto L_{x}$ allows identifying $M_{0}$ as an involutive subalgebra of $B(H)$.

Finally, let $M = M_{0}'' \subset B(H)$ and extend $\tau$ to $M$ by $\tau(x) = \dpn{x1_{M_0}, 1_{M_0}}{\tau}$, then $\tau$ is a vector state on $M$.

(1): By von Neumann’s Bicommutant Theorem, $M$ is a von Neumann algebra. As $\tau(z) = \dpn{z1_{M_0}, 1_{M_0}}{\tau}$ for all $z \in M$, $\tau$ is strong-operator continuous, and the mapping $(x, y) \mapsto \tau(xy)$ is separately strong-operator continuous. Since $\tau$ is the extension of $\bracsn{\tau_i}_{i \in I}$, $\tau|_{M_i}$ is tracial for all $i \in I$, and $\tau|_{M_0}$ is also tracial. By strong-operator density of $M_{0}$ in $M$, $\tau$ is a tracial state.

(2), (3): By definition of $\tau$ and universal property of inductive limits for vector spaces.

(4): Let $(K, \pi, \xi)$ be the GNS triple associated with $\sigma$, then since $\sigma$ is a faithful state, the representation $\pi: N \to B(K)$ is faithful. Using Lemma 43.4.10, identify $N = \pi(N) \subset B(K)$ as a von Neumann subalgebra of $B(K)$.

By universal property of inductive limits of vector spaces, there exists a trace-preserving *-homomorphism $\phi_{0}: M_{0} \to N$ such that the following diagram commutes

for all $i \in I$. Let

then for any $x \in M_{0}$, $\norm{UL_x1_{M_0}}_{K}^{2} = \sigma(\phi_{0}(x^{*}x)) = \tau(x^{*}x) = \norm{L_x1_{M_0}}_{H}^{2}$. Thus $U$ extends into an isometry from $H$ to $K$. Moreover, for each $x \in M_{0}$ and $y, z \in N$,

so $x \mapsto \dpn{\phi_0(x)\pi(y)\xi, \pi(z)\xi}{K}$ is a weak-operator continuous functional on $M_{0}$. Since $\pi(N)\xi$ is dense in $K$, by the Arzelà-Ascoli Theorem, the weak operator topology on $\ol{B_{\pi(N)}(0, 1)}$ is induced by maps of the form $x \mapsto \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{K}$ with $y, z \in N$. As such, $\phi_{0}: M_{0} \to N$ is weak-operator continuous when restricted to bounded sets.

By the Kaplansky Density Theorem and Theorem 6.7.6, $\phi_{0}$ extends into a *-homomorphism $\phi: M \to N$. Since $(N, \sigma)$ is a tracial von Neumann algebra, $\phi$ is also trace-preserving. Finally, if $\Phi: M \to N$ is another such map, then it is also weak-operator continuous on bounded sets by Lemma 43.4.10. Therefore $\Phi = \phi$ by uniqueness of continuous extensions.$\square$

Proof. For each $n \in \natp$, identify $M_{2^n}(\complex)$ as a subalgebra of $M_{2}(\complex)^{\ol \otimes \infty}$. Let $M_{0} = \bigcup_{n \in \natp}M_{2^n}(\complex)$, then by (4) of Definition 44.3.1, $M_{2}(\complex)^{\ol \otimes \infty}= M_{0}''$.

Let $(H, \pi, \xi)$ be the GNS triple associated with $\tau$. Using Lemma 43.4.10, identify $M_{2}(\complex)^{\ol \otimes \infty}$ as a von Neumann subalgebra of $B(H)$. Let $x \in Z(M_{2}(\complex)^{\ol \otimes \infty})$, then by the Kaplansky Density Theorem, there exists a net $\angles{x_\alpha}_{\alpha \in A}\subset M_{0}$ such that $\sup_{\alpha \in A}\normn{x_\alpha}_{M_2(\complex)^{\ol \otimes \infty}}< \infty$ and $x_{\alpha} \to x$ in the strong operator topology.

Fix $\alpha \in A$, then there exists $n \in \natp$ such that $x_{\alpha} \in M_{2^n}(\complex)$. As $M_{2^n}(\complex)$ is finite-dimensional^[1], Haar’s Theorem implies that there exists a left-translation invariant Borel probability measure $\mu: \cb_{U(M_{2^n}(\complex))}\to [0, 1]$. Let $y_{\alpha} = \int_{U(M_{2^n}(\complex))}ux_{\alpha} u^{*} \mu(du)$, then for each $v \in U(M_{2^n}(\complex))$,

so $vy_{\alpha} = y_{\alpha} v$. By Corollary 39.7.5, $y_{\alpha} \in Z(M_{2^n}(\complex)) = \complex I_{M_{2^n}(\complex)}$, so there exists $\lambda_{\alpha} \in \complex$ with $y_{\alpha} = \lambda_{\alpha} I_{M_{2^n}(\complex)}$. Since $x \in Z(M_{2}(\complex)^{\ol \otimes \infty})$ and $\mu$ is a probability measure^[2],

where

By the Cauchy-Schwarz inequality,

Thus $\limsup_{\alpha \in A}\normn{(x - \lambda_\alpha)\xi}_{H} \le \lim_{\alpha \in A}\norm{(x - x_\alpha)\xi}_{H} = 0$. As $x \in Z(M_{2}(\complex)^{\ol \otimes \infty})$, the above shows that $\lambda_{\alpha} \to x$ pointwise on $M_{2}(\complex)^{\ol \otimes \infty}H$. Since $\sup_{\alpha \in A}|\lambda_{\alpha}| \le \sup_{\alpha \in A}\normn{x_\alpha}_{M_2(\complex)^{\ol \otimes \infty}}< \infty$, the Arzelà-Ascoli Theorem implies that $\lambda_{\alpha} \to x$ in the strong operator topology. Therefore $x \in \complex I_{M_2(\complex)^{\ol \otimes \infty}}$, and $M_{2}(\complex)^{\ol \otimes \infty}$ is a factor.$\square$