\section{Traces}
\label{section:tracial-von-neumann-algebra}

\begin{definition}[Trace]
\label{definition:trace}
    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 \textbf{trace} on $M$ if:
    \begin{enumerate}
        \item For each $x, y \in M_+$ and $\lambda \ge 0$, $\tau(\lambda x + y) = \lambda \tau(x) + \tau(y)$. 
        \item For each $x \in M_+$ and $u \in U(M)$, $\tau(uxu^*) = \tau(x)$. 
    \end{enumerate}
\end{definition}

\begin{definition}[Tracial von Neumann Algebra]
\label{definition:tracial-von-neumann-algebra}
    Let $M$ be a von Neumann algebra and $\tau: M \to \complex$ be a faithful tracial state, then $(M, \tau)$ is a \textbf{tracial von Neumann algebra} if $\tau$ is normal. 
\end{definition}

\begin{definition}[Standard Representation]
\label{definition:tracial-vna-standard}
    Let $(M, \tau)$ be a tracial von Neumann algebra, then its \hyperref[GNS]{definition:gns-triple} representation on $L^2(M, \tau)$ is the \textbf{standard representation} of $(M, \tau)$.  
\end{definition}

\begin{proposition}
\label{proposition:cstar-trace-vna}
    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 \hyperref[GNS]{definition:gns-triple} triple, then:
    \begin{enumerate}
        \item The uniform structure of pointwise convergence\footnote{Strong operator topology.} 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. 
        \item $\pi(M) \subset B(L^2(M, \tau))$ 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. 
    \end{enumerate}
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition 2.6.4]{AnantharamanPopa}}}. ]
    (1): For any $x, y \in \ol{B_M(0, 1)}$, 
    \begin{align*}
    \norm{x - y}_{L^2(M, \tau)}^2 &= \tau((x - y)^*(x - y)) = \dpn{\pi(x - y)^*\pi(x - y)1_M, 1_M}{L^2(M, \tau)} \\
    &= \dpn{\pi(x - y)1_M, \pi(x - y)1_M}{L^2(M, \tau)} = \norm{\pi(x - y)1_M}_{L^2(M, \tau)}^2
    \end{align*}

    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 \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, 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, 
    \begin{align*}
        \norm{(x - y)t}_{L^2(M, \tau)}^2 &= \tau(t^*(x - y)^*(x - y)t) \le \tau(tt^*(x - y)^*(x - y)) \\
        &\le \norm{t}_M^2 \tau((x - y)^*(x - y)) = \norm{t}_M^2 \norm{x - y}_{L^2(M, \tau)}^2
    \end{align*}

    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 \autoref{proposition:complete-closed}, $\ol{B_M(0, 1)}$ 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 \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} 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 $B(L^2(M, \tau))$. 
\end{proof}

\begin{proposition}
\label{proposition:tracial-separable-condition}
    Let $(M, \tau)$ be a tracial von Neumann algebra, then the following are equivalent:
    \begin{enumerate}
        \item $M$ is separable. 
        \item $\ol{B_M(0, 1)}$ is strong-operator separable. 
        \item $\ol{B_M(0, 1)}$ is separable in $L^2(M, \tau)$. 
        \item $L^2(M, \tau)$ is separable. 
    \end{enumerate}
\end{proposition}
\begin{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 \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the strong operator topology on $\ol{B_M(0, 1)}$ coincides with the topology of pointwise convergence on $D$. By \autoref{theorem:uniform-metrisable} and \autoref{proposition:separable-product}, $H^D$ is metrisable and separable. Therefore $\ol{B_M(0, 1)} \subset H^D$ is also metrisable and separable.

    (2) $\Rightarrow$ (3): By \autoref{proposition:cstar-trace-vna}. 

    (3) $\Rightarrow$ (4): By \autoref{lemma:tracial-von-neumann-gns}, 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. 
\end{proof}

\begin{proposition}
\label{proposition:tracial-von-neumann-factor}
    Let $(M, \tau)$ be a tracial von Neumann factor, then exactly one of the following holds:
    \begin{enumerate}
        \item There exists $n \in \natp$ such that $M \iso M_n(\complex)$. 
        \item $M$ is of type $\vnII_1$. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): Suppose that $M$ has a minimal projection. By \autoref{theorem:type1-bh}, 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 $j \in I$, 
    \[
    \tau(1_M) = \sum_{i \in I}\tau(e_i \otimes e_i) = |I| \cdot \tau(e_j \otimes e_j)
    \]

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

    (2): Suppose that $M$ has no minimal projections. By \autoref{lemma:abelian-minimal-factor}, $M$ has no abelian projections. As $\tau$ is a faithful tracial state, $1_M$ is finite. Therefore $M$ is a $\vnII_1$ factor. 
\end{proof}

\begin{lemma}
\label{lemma:diffuse-factor-subsplit}
    Let $M$ be a diffuse factor, $p \in \text{Proj}(M)$ be a non-zero projection, then there exist non-zero projections $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$ and $p_1 + p_2 \le p$.  
\end{lemma}
\begin{proof}[Proof, {{\cite[Lemma 4.1.1]{AnantharamanPopa}}}. ]
    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 \autoref{proposition:central-support-mvn}, there exist non-zero projections $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$, $p_1 \le q$, and $p_2 \le p - q$. 
\end{proof}

\begin{lemma}
\label{lemma:diffuse-factor-split}
    Let $M$ be a diffuse factor, $p \in \text{Proj}(M)$ be a non-zero projection, then there exist $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$ and $p_1 + p_2 = p$. 
\end{lemma}
\begin{proof}[Proof, {{\cite[Lemma 4.1.1]{AnantharamanPopa}}}. ]
    Let
    \[
    \mathcal{F} = \bracsn{(q, r) \in \text{Proj}(M)^2|q \sim r, q \perp r, q + r \le p}
    \] 

    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 \autoref{lemma:mvn-equivalent-direct-sum}, $q \sim r$, $q \perp r$, and $q + r \le p$. 

    If $q + r \ne p$, then by \autoref{lemma:diffuse-factor-subsplit} 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 $\mathcal{M}$. 
\end{proof}

\begin{corollary}
\label{corollary:diffuse-factor-dyadic}
    Let $M$ be a diffuse factor, then for each $n \in \natp$, there exist mutually orthogonal equivalent projections $\bracsn{p_j}_1^{2^n} \subset \text{Proj}(M)$ such that $\sum_{j = 1}^{2^n}p_j = 1$. If $\tau: M \to \complex$ is a tracial state, then $\tau(p_j) = 2^{-n}$ for all $1 \le j \le 2^n$.
\end{corollary}
\begin{proof}
    By \autoref{lemma:diffuse-factor-split}. 
\end{proof}

\begin{proposition}
\label{proposition:tracial-factor-simple}
    Let $M$ be a factor, then $M$ admits at most one tracial state. 
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition 4.1.3]{AnantharamanPopa}}}. ]
    Using \autoref{proposition:tracial-von-neumann-factor}, 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 \autoref{corollary:diffuse-factor-dyadic}, 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 \hyperref[corollary:vna-factor-comparability]{theorem:vna-comparability}, there exists a unique $0 \le k \le 2^n - 1$ such that
    \[
    \sum_{j = 1}^k p_j \preceq p \prec \sum_{j = 1}^{k+1}p_j
    \]

    As such, for any tracial state $\tau: M \to \complex$, 
    \[
    \frac{k}{2^n} \le \tau(p) \le \frac{k+1}{2^n}
    \]

    so $\tau(p)$ is independent of the choice of $\tau$. By \autoref{theorem:vn-projection-norm-dense}, $\tau$ is unique.
\end{proof}

\begin{proposition}
\label{proposition:tracial-factor-unique-trace}
    Let $(M, \tau)$ be a tracial von Neumann algebra, then $M$ is a factor if and only if $\tau$ is the unique normal faithful tracial state on $M$. 
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition 4.1.4]{AnantharamanPopa}}}. ]
    If $M$ is a factor, then $\tau$ is unique by \autoref{proposition:tracial-factor-simple}. Suppose that $M$ is not a factor, then $M$ admits a non-trivial central projection $z \in Z(M)$. In which case, for any $\alpha \in (0, 1)$, 
    \[
    \tilde \tau(x) = \frac{\alpha}{\tau(z)} \tau(xz) + \frac{1 - \alpha}{1 - \tau(z)}\tau(x(1 - z))
    \]

    is also a normal faithful tracial state. In particular, 
    \[
    \tilde \tau(z) = \frac{\alpha}{\tau(z)} \tau(z) + \frac{1 - \alpha}{1 - \tau(z)}\tau(z(1 - z)) = \alpha
    \]

    so if $\alpha \ne \tau(z)$, then $\tilde \tau \ne \tau$. 
\end{proof}

\begin{proposition}
\label{proposition:tracial-factor-calkin}
    Let $(M, \tau)$ be a tracial factor, then $M$ admits no non-trivial two-sided ideal. 
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition 4.1.5]{AnantharamanPopa}}}. ]
    Let $I \subset M$ be a non-zero two-sided ideal. By \autoref{proposition:vna-two-sided-ideal}, $I$ 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 \hyperref[comparability theorem]{corollary:vna-factor-comparability}, 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 $v_pv_p^* = p$ and $v_p^*v_p \le e$. As such, $p = v_pev_p^* \in I$ for all $p \in \mathcal{Q}$. Therefore $1_M = \sum_{p \in \mathcal{Q}}p \in I$, and $I = M$.  
\end{proof}

\begin{proposition}
\label{proposition:ii1-trace-projection-equivalence-classes}
    Let $M$ be a $\vnII_1$ factor with tracial normal state $\tau: M \to \complex$, then the mapping $p \mapsto \tau(p)$ induces a bijection from the equivalence classes of $\text{Proj}(M)$ to $[0, 1]$. 
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition 4.1.6]{AnantharamanPopa}}}. ]
    (Injective): Let $p, q \in \text{Proj}(M)$ with $\tau(p) = \tau(q)$. By the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, 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 \autoref{lemma:diffuse-factor-split}, there exists a family $\bracsn{p_r|r \in \mathbb D \cap [0, 1]} \subset \text{Proj}(M)$ such that:
    \begin{enumerate}[label=(\roman*)]
        \item For each dyadic rational number $r \in \mathbb D \cap [0, 1]$, $\tau(p_r) = r$. 
        \item For each $r, s \in \mathbb D \cap [0, 1]$ with $r \le s$, $p_r \le p_s$. 
    \end{enumerate}

    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, 
    \begin{align*}
    \tau(p_t) &= \tau(\sup\bracsn{p_r|r \in \mathbb D \cap [0, 1], r \le t}) \\
    &= \sup\bracsn{\tau(p_r)|r \in \mathbb D \cap [0, 1], r \le t} = t
    \end{align*}
    
    
\end{proof}