45.2 The Fundamental Group
Definition 45.2.1 (Induced von Neumann Algebra).label Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $q \in \text{Proj}(M')$, then $Mq$ is a von Neumann algebra on $qH$, known as the induced von Neumann algebra of $M$ with respect to $q$.
Proposition 45.2.2.label Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $p \in \text{Proj}(M)$, then:
- (1)
For any central projection $e \in Z(pMp)$, $e = Z(e)p$.
- (2)
$Z(pMp) = Z(M)p$.
Proof, [Proposition 4.2.1, AP]. (1): Since $(p - e)Me = (p - e)pMpe = 0$, $(p - e)yey^{*} = 0$ for all $y \in U(M)$, so $(p - e)z(e) = 0$ by Proposition 43.8.4, and $e = Z(e)p$.
(2): By Theorem 43.7.2.$\square$
Proposition 45.2.3.label Let $H$ be a complex Hilbert space and $M \subset B(H)$ be a von Neumann algebra, then
- (1)
For any $p \in \text{Proj}(M)$, the commutant $(pMp)'$ in $B(pH)$ is $pM'$.
- (2)
For any $q \in \text{Proj}(M')$, the commutant $(Mq)'$ in $B(qH)$ is $qM'q$.
Proof, [Proposition 4.2.2, AP]. (1): For any $x \in M$ and $y \in M'$, $pxp \cdot yp = ypxp = yp \cdot pxp$, so $pMp \subset (M'p)'$.
For any $x \in (M'p)' \subset B(pH)$, let $\tilde x = xp = pxp \in B(H)$, then for any $y \in M'$,
By the Bicommutant Theorem, $\tilde x \in M$, $x = p\tilde x p \in pMp$, and $pMp \supset (M'p)'$.$\square$
Proposition 45.2.4.label Let $(M, \tau)$ be a $\vnII_{1}$ factor, then $M_{n}(M)$ is a $\vnII_{1}$ factor, with normalised trace $\frac{1}{n}\text{Tr}_{n} \otimes \tau$.
Definition 45.2.5 (Amplification).label Let $(M, \tau)$ be a $\vnII_{1}$ factor, $t > 0$, $n \in \natp$ with $n \ge t$, and $p \in \text{Proj}(M_{n}(M))$ with $(\text{Tr}_{n} \otimes \tau)p = t$, then the von Neumann algebra $M^{t} := pM_{n}(M)p \subset M_{n}(M)$ is an amplification of $M$.
The spatial isomorphism class of $M^{t}$ is independent of the choice of $n \ge t$ and $p \in \text{Proj}(M_{n}(M))$ with $(\text{Tr}_{n} \otimes \tau)p = t$.
Proof. By Proposition 45.1.13.$\square$
Lemma 45.2.6.label Let $(M, \tau)$ be a $\vnII_{1}$ factor and $s, t > 0$, then $(M^{s})^{t} \iso M^{st}$.
Proof, [Lemma 4.2.3, AP]. Let $m \in \natp$ and $p \in \text{Proj}(M_{m}(M))$ with $(\text{Tr}_{m} \otimes \tau)(p) = s$, so that $M^{s} = pM_{m}(M)p$, and $\tau_{s} = s^{-1}(\text{Tr}_{m} \otimes \tau)|_{M^s}$ denotes the normalised tracial state on $M^{s}$.
Let $n \in \natp$ and $q \in \text{Proj}(M_{n}(M^{s}))$ with $(\text{Tr}_{n} \otimes \tau_{s})(q) = t$, so that $(M^{s})^{t} = qM_{n}(M^{s})q$. Identify $M_{n}(M^{s}) \subset M_{n}(M_{m}(M))$, so that $q \in M_{n}(M_{m}(M))$, then
Since $q \in M_{n}(M^{s})$, $p$ is the identity on entries of $q$, so
and $(M^{s})^{t} = qM_{n}(M_{m}(M))q = qM_{mn}(M)q$. Finally,
Therefore $(M^{s})^{t} = M^{st}$.$\square$
Definition 45.2.7 (Fundamental Group).label Let $M$ be a $\vnII_{1}$ factor, then the fundamental group of $M$ is the family
Proof. By Lemma 45.2.6, $\fF(M)$ is a group.$\square$
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