Definition 44.3.2 (Hyperfinite $\vnII_{1}$ Factor).label For each $n \in \natp$, equip $M_{2^n}(\complex)$ with the normalised trace $\tau_{n}: M_{2^n}(\complex) \to \complex$, and
Then, $\phi_{n}$ is a trace-preserving *-homomorphism, and the resulting inductive limit $\lim_{\longrightarrow}(M_{2^n}(\complex), \tau_{n})$ is a factor.
The tracial von Neumann algebra $(M_{2}(\complex)^{\ol \otimes \infty}, \tau) := \lim_{\longrightarrow}(M_{2^n}(\complex), \tau_{n})$ is the hyperfinite $\vnII_{1}$ factor.
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.$\square$
- It seems true in general that inductive limits of $\vnII_{1}$ factors are also $\vnII_{1}$ factors. However, this compactness shortcut is not available in the general case.keyboard_return
- Even though $x$ is a priori not in $M_{2^n}(\complex)$, the integrand here surely has separable range, so the Bochner integral may be used to save some inner product computations. keyboard_return
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