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Bokuan Li
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\section{The Hyperfinite $\vnII_1$ Factor}
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\label{section:hyperfinite-ii1-factor}
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\begin{definition}[Inductive Limit of Tracial von Neumann Algebras]
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\label{definition:inductive-limit-tracial-von-neumann-algebra}
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Let $(\bracsn{(M_i, \tau_i)}_{i \in I}, \bracsn{\phi^i_j|i, j \in I, i \le j})$ be an upward-directed system of tracial von Neumann algebras, where $\phi^i_j: M_i \to M_j$ is a trace-preserving *-homomorphism for each $i, j \in I$ with $i \le j$, then there exists a pair $((M, \tau), \bracsn{\phi^i}_{i \in I})$ such that:
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\begin{enumerate}
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\item $(M, \tau)$ is a tracial von Neumann algebra.
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\item For each $i \in I$, $\phi^i: M_i \to M$ is a trace-preserving *-homomorphism.
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\item For each $i, j \in I$ with $i \le j$, the following diagram commutes:
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\[
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\xymatrix{
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M_i \ar@{->}[d]_{\phi^i_j} \ar@{->}[rd]^{\phi^i} & \\
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M_j \ar@{->}[r]_{\phi^j} & M
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}
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\]
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\item[(U)] For any pair $((N, \sigma), \bracsn{\psi^i}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique trace-preserving *-homomorphism $\psi: M \to N$ such that the following diagram commutes:
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\[
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\xymatrix{
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M_i \ar@{->}[d]_{\phi^i} \ar@{->}[rd]^{\psi^i} & \\
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M \ar@{->}[r]_{\psi} & N
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}
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\]
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for all $i \in I$.
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\item $\braks{\bigcup_{i \in I}\phi^i(M_i)}'' = M$, where the bicommutant is taken in $M$.
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\end{enumerate}
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The pair $((M, \tau), \bracsn{\phi^i}_{i \in I}) = \lim_{\longleftarrow}(M_i, \tau_i)$ is the \textbf{inductive limit} of $(\bracsn{(M_i, \tau_i)}_{i \in I}, \bracsn{\phi^i_j|i, j \in I, i \le j})$.
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\end{definition}
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\begin{proof}
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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 $i \in I$.
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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 $i \in I$.
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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 $x, y, z \in M_0$,
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\[
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\dpn{L_xy, L_xy}{\tau} = \tau(y^*x^*xy) \le \norm{x}_{M_0}^2 \tau(y^*y) = \norm{x}_{M_0}^2 \dpn{y, y}{\tau}
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\]
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and
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\[
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\dpn{L_xy, z}{\tau} = \tau(z^*xy) = \tau((x^*z)^*y) = \dpn{y, L_{x^*}z}{\tau}
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\]
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with
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\[
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\dpn{L_x 1_{M_0}, L_x 1_{M_0}}{\tau} = \tau(x^*x) \le \tau(\normn{x}_{M_0}^2) = \norm{x}_{M_0}^2
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\]
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Thus the mapping $M_0 \to B(H)$ with $x \mapsto L_x$ allows identifying $M_0$ as an involutive subalgebra of $B(H)$.
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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 $M$.
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(1): By \hyperref[von Neumann's Bicommutant Theorem]{theorem:bicommutant}, $M$ 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.
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(2), (3): By definition of $\tau$ and universal property of inductive limits for vector spaces.
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(4): Let $(K, \pi, \xi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\sigma$, then since $\sigma$ is a faithful state, the representation $\pi: N \to B(K)$ is faithful. Using \autoref{lemma:tracial-von-neumann-gns}, identify $N = \pi(N) \subset B(K)$ as a von Neumann subalgebra of $B(K)$.
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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
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\[
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\xymatrix{
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M_i \ar@{->}[d]_{\phi^i} \ar@{->}[rd]^{\psi^i} & \\
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M_0 \ar@{->}[r]_{\psi_0} & N
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}
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\]
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for all $i \in I$. Let
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\[
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U: H \to K \quad \dpn{UL_x1_{M_0}, \pi(y)\xi}{K} = \sigma(y^*\phi_0(x))
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\]
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then for any $x \in M$, $\norm{UL_x1_{M_0}}_K^2 = \sigma(\phi_0(x^*x)) = \tau(x^*x) = \norm{L_x1_{M_0}}_H^2$. Thus $U$ extends into an isometry from $H$ to $K$. Moreover, for each $x \in M_0$ and $y, z \in N$,
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\begin{align*}
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\dpn{(\pi \circ \psi_0)(x)\pi(y)\xi, \pi(z)\xi}{K} &= \sigma(z^*\phi_0(x)y) = \sigma(yz^*\phi_0(x)) = \sigma((zy^*)^*\phi_0(x)) \\
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&= \dpn{UL_x1_{M_0}, \pi(zy^*)\xi}{K} = \dpn{L_x1_{M_0}, U^*\pi(zy^*)\xi}{H}
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\end{align*}
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so $x \mapsto \dpn{\psi_0(x)\pi(y)\xi, \pi(z)\xi}{K}$ is a weak-operator continuous functional on $M_0$. Since $\pi(N)\xi$ is dense in $K$, by the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, 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, $\psi_0: M_0 \to N$ is weak-operator continuous when restricted to bounded sets.
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By the \hyperref[Kaplanski Density Theorem]{theorem:kaplansky-density} and \autoref{theorem:uniform-continuous-extension}, $\psi_0$ extends into a *-homomorphism $\phi: M \to N$. Since $(N, \sigma)$ is a tracial von Neumann algebra, $\psi$ 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 \autoref{lemma:tracial-von-neumann-gns}. Therefore $\Phi = \phi$ by uniqueness of continuous extensions.
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\end{proof}
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\begin{definition}[Hyperfinite $\vnII_1$ Factor]
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\label{definition:hyperfinite-ii1}
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For each $n \in \natp$, equip $M_{2^n}(\complex)$ with the normalised trace $\tau_n: M_{2^n}(\complex) \to \complex$, and
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\[
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\phi_n: M_{2^n}(\complex) \to M_{2^{n+1}}(\complex) \quad x \mapsto \begin{bmatrix}
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x & 0 \\
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0 & x
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\end{bmatrix}
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\]
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Then, $\phi_n$ is a trace-preserving *-homomorphism, and the resulting inductive limit $\lim_{\longrightarrow} (M_{2^n}(\complex), \tau_n)$ is a factor.
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The tracial von Neumann algebra $(M_2(\complex)^{\ol \otimes \infty}, \tau) := \lim_{\longrightarrow} (M_{2^n}(\complex), \tau_n)$ is the \textbf{hyperfinite $\vnII_1$ factor}.
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\end{definition}
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\begin{proof}
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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 \autoref{definition:inductive-limit-tracial-von-neumann-algebra}, $M_2(\complex)^{\ol \otimes \infty} = M_0''$.
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Let $(H, \pi, \xi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\tau$. Using \autoref{lemma:tracial-von-neumann-gns}, 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 \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, 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.
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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\footnote{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.}, \hyperref[Haar's Theorem]{theorem:haar} 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 $v \in U(M_{2^n}(\complex))$,
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\begin{align*}
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vy_\alpha v^* &= \int_{U(M_{2^n}(\complex))} vux_\alpha u^*v^* \mu(du) = \int_{U(M_{2^n}(\complex))} (vu)x_\alpha (vu)^* \mu(du) \\
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&= \int_{U(M_{2^n}(\complex))} ux_\alpha u^* \mu(du) = y_\alpha
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\end{align*}
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so $vy_\alpha = y_\alpha v$. By \autoref{corollary:sum-of-unitaries}, $y_\alpha \in Z(M_{2^n}(\complex)) = \complex I_{M_{2^n}(\complex)}$, 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\footnote{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. },
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\[
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(x - \lambda_\alpha)\xi = \int_{U(M_{2^n}(\complex))} (x - ux_\alpha u^*)\xi \mu(du) = \int_{U(M_{2^n}(\complex))} u(x - x_\alpha) u^*\xi \mu(du)
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\]
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where
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\begin{align*}
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\norm{u(x - x_\alpha) u^*}_H^2 &= \tau(u(x - x_\alpha)^*u^*u(x - x_\alpha)u^*) \\
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&= \tau((x - x_\alpha)^*(x - x_\alpha)uu^*uu^*) \\
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&= \tau((x - x_\alpha)^*(x - x_\alpha)) = \norm{(x - x_\alpha)\xi}_H^2
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\end{align*}
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By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz},
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\begin{align*}
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\normn{(x - \lambda_\alpha)\xi}_H^2 &= \int_{U(M_{2^n}(\complex))} \dpn{u(x - \lambda_\alpha)u^*\xi, (x - \lambda_\alpha)\xi}{H} \mu(du) \\
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&\le \norm{(x - \lambda_\alpha)\xi}_H\braks{\int_{U(M_{2^n}(\complex))}\normn{u(x - x_\alpha) u^*\xi}_H^2 \mu(du)} \\
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&= \normn{(x - \lambda_\alpha)\xi}_H \norm{(x - x_\alpha)\xi}_H
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\end{align*}
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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 \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} 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.
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\end{proof}
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