# Direct Integral Decomposition

Let $\{H_x\}_{x\in X}$ be a collection of Hilbert spaces such that $(X,\Sigma,\mu)$ is a measure space.  Now define

$\displaystyle H=\left\{s\in\bigoplus_{x\in X} H_x:\int_X |s(x)|^2\,d\mu<\infty\right\}$

where $s(x)=s_x$ (the $x$th component of $s$).  Then

$\displaystyle\langle s,t\rangle=\int_X\langle s_x,t_x\rangle_x\,d\mu$

defines a pre inner product on $H$.  Now let $H_0=span\{s-t:s=t\,a.e.\}$.  Then $H/H_0$ is an inner product space, and its completion is a Hilbert space called the direct integral of $\{H_x\}_{x\in X}$.  We denote the direct integral by

$\displaystyle H_X^\oplus=\int_X^\oplus H_x\,d\mu.$

Now suppose $\{T_x\}_{x\in X}$ is a collection of linear operators where $T_x\in L(H_x)$ such that $\{\|T_x\|\}$ is uniformly bounded.  Then there is an operator $T\in L(H_X^\oplus)$ where $T(s)_x=T_x(s_x)$ and where we can define

$\displaystyle \|T\|={\mbox{ess}\sup}_{x\in X}\{\|T_x\|\}.$

This gives a representation $\rho:L^\infty(X,\mu)\to L(H_X^\oplus)$ defined by

$\left(\rho(f)(s)\right)_x=f(x)s_x,$

which is essentially just a component-wise left action of $f$ on $s$$\rho(L^\infty(X,\mu))$ is called the algebra of diagonalizable operators of $H_X^\oplus$, which we will denote $D(H_X^\oplus)$.

Theorem 1.  Let $\rho:A\to L(H)$ be a representation of von Neumann algebra $A$ on a separable Hilbert space $H$ such that $B$ is a von Neumann subalgebra of $A'$.  Then there exists a measure space $(X,\Sigma,\mu)$, a collection of Hilbert spaces $\{H_x\}_{x\in X}$, and a  unitary map $U:H\to H_X^\oplus$ such that $U\left(\rho(B)(s)\right)=D(H_X^\oplus)(s')$ for all $s\in H$ and corresponding $s'\in H_X^\oplus$ and

$\displaystyle UTU^*=\int_X^\oplus T_x\,d\mu\in L(H_X^\oplus)$

for all $T\in\rho(B')$.

Thus if we let $B=A'$ from above, then $L(H_x)$ is a factor and we also write

$\displaystyle\rho(A')=\int_X^\oplus\rho(A')_x\,d\mu=\int_X^\oplus\rho(A_x)'\,d\mu.$

This is called the central decomposition of $A$.  This also gives a representation of $A$ on $H_X^\oplus$ defined by $a\mapsto U\rho(a)U^*$.

[1]  Blackadar, Bruce.  Operator Algebras.  Encyclopedia of Mathematical Sciences.  Springer-Verlag.  2006.

# von Neumann Algebras

Let $M$ be a magma and $S\subseteq M$.  The commutant of $S$ is defined as:

$S'=\{x\in M: sx=xs\}$

for all $s\in S$.  We have that $S'=S^{(2n+1)}$ and $S''=S^{(2n)}$ for $1\leq n\in\mathbb{N}$ and that $M=M'$ iff $M$ is abelian.  We can also refer to the commutant of the whole structure $M$ as the center of $M$ (i.e. $M'=Z(M)$).

Definition 1.  A von Neumann algebra is a C*-algebra $A$ such that $A=A''$.

A projection in a *-algebra is an element $p$ such that $p^2=p=p^*$.  A partial isometry is an element $u$ such that $u^*u$ is a projection. Recall an element $x$ is positive, denoted $x\geq 0$, if $x^*=x$ and $\sigma_A(x)\subseteq [0,\infty)$, and that $x\leq y$ if $y-x\geq 0$.  We will also say that two elements $x,y$ are orthogonal, denoted $x\perp y$, if $xy=0$.

Definition 2.  Two projections $p,q$ are Murray-von Neumann equivalent, denoted $p\sim q$, if there is a partial isometry $u$ such that $p=u^*u$ and $q=uu^*$.  We say $p$ is subordinate to $q$, denoted $p\preceq q$, if there is a projection $q'$ such that $p\sim q'$ and $q'\leq q$.

Proposition 3.  Let $A$ be a *-algebra and $\{(p_i,q_i)\}$ be a sequence of pairs of projections such that $p_i\perp p_j$ and $q_i\perp q_j$ for $i\neq j$ and $p_i\sim q_i$ for all $i$.  Then $\sum p_i\sim\sum q_i.$  Also, if $p_i\preceq q_i$ for all $i,$ then $\sum p_i\preceq\sum q_i.$

Proof.  If $p_i\sim q_i$, then $p_i=u_i^*u_i$ and $q_i=u_iu_i^*.$  Hence $\sum p_i=\sum u_i^*u_i$ and $\sum q_i=u_i^*u_i$.  Now define $u=\sum u_i$ and $u^*=\sum u_i^*.$  Then we have

$\displaystyle u^*u=\left(\sum u_i^*\right)\left(\sum u_i\right)=\sum u_i^*u_j=\sum u_i^*u_i$

where the last equality follows from orthogonality.  Hence $\sum p_i\sim\sum q_i.$

Now suppose $p_i\preceq q_i.$  Then there are $q_i'$ such that $p_i\sim q_i'$ and $q_i'\leq q_i.$  Then since $\sum p_i\sim\sum q_i'$ by the previous claim, it remains to show that $\sum q_i'\leq\sum q_i.$  But $\sum q_i-\sum q_i'=\sum (q_i-q_i')\geq 0$ since $q_i'\leq q_i$ for all $i.$  So we have the result.

Proposition 4 (Schroder-Bernstein).  Let $p$ and $q$ be projections in a *-algebra such that $p\preceq q$ and $q\preceq p.$  Then $p\sim q$.

It turns out that if a unital von Neumann algebra $A$ is a factor ($Z(M)=1$), then $\preceq$ is a total order on the projections.

Definition 5.  A projection $p\in A$ is

1. abelian if $pAp$ is commutative;
2. finite if $p\sim p'$ where $p'\leq p$ implies that $p=p'$;
3. infinite if it is not finite;
4. properly infinite if $p\sim p_1$ and $p\sim p_2$ where $p_1,p_2\leq p$ and $p_1\perp p_2.$

Lemma 6.  If $p$ is an infinite projection in a von Neumann algebra $A$, then there is a projection $z\in Z(A)$ such that $pz$ is nonzero and properly infinite.

This allows for a somewhat complicated decomposition (see [1] for details) of $A$

$A=Az_1\oplus Az_{2_1}\oplus Az_{2_\infty}\oplus Az_3$

where $z_1$ is a discrete central projection, $z_{2_1}$ is the largest finite continuous central projection, $z_{2_\infty}$ is the largest properly infinite semifinite continuous projection, and $z_3$ is a purely infinite projection.  The algebra $A$ is said to be of pure type $\alpha$ if $z_\beta=0$ for all $\beta\neq\alpha.$

[1]  Blackadar, Bruce.  Operator Algebras.  Encyclopedia of Mathematical Sciences.  Springer-Verlag.  2006.