Projections and types of von Neumann algebras

Originally written by Claude Opus 5.5 (Anthropic), September 2026, with a separate historical AI spot-check. GPT-6.1 Sol (OpenAI), at the Ultra setting, read and replayed the entire lesson and all eight solutions, compared the full programme prerequisite proofs, and completed the boundary and construction details, October 2026. Public domain (CC0).

In a von Neumann algebra, projections play the part of closed subspaces. Two projections are equivalent when a partial isometry of the algebra maps the range of one isometrically onto the range of the other. So equivalence says that two projections have the same size, measured inside the algebra. This lesson develops the comparison theory that this notion leads to, and uses it to sort von Neumann algebras into types.

The main results are these. Any two projections become comparable after they are cut by a suitable central projection (the comparison theorem), and in a factor one of any two projections is subequivalent to the other. Every von Neumann algebra splits in exactly one way into central summands of types I, II$_1$, II$_\infty$ and III. Type I algebras are described completely: each is a direct sum of algebras $A\bar\otimes B(\ell^2(\alpha))$ with $A$ commutative, and a matching spatial form describes the commutant as well. Properly infinite projections can be halved and absorb countable sums. Finite projections satisfy the modular law, and equivalent finite projections are unitarily equivalent. We also determine the von Neumann algebra generated by two projections. The last two sections compare the cyclic projections of an algebra with those of its commutant, and apply the comparison to vector states and to spatial isomorphisms.

This comparison theory underlies traces and dimension functions, the multiplicity theory of representations, and the finer classification of von Neumann algebras.

The exact full proofs in Hilbert spaces, continuous functional calculus, the spectral theorem, the double commutant theorem, operator preduals and topologies, and the universal enveloping algebra supply the required inputs. Section 2 identifies their precise proof locations, including the Banach-space and set-theoretic prerequisites. The references identify freely readable comparison sources.

Equivalence compares projections through partial isometries. Central supports and finite versus infinite behavior then organize the algebra into its type components. The full decomposition is proved here; a freely readable treatment is [Blackadar].

1. Conventions

$H$ is a Hilbert space with inner product $\langle\cdot,\cdot\rangle$, linear in the first variable. For $S\subseteq B(H)$, $S'$ is the commutant. For $S\subseteq B(H)$ and $X\subseteq H$, $[SX]$ is the closed linear span of $\{x\xi:x\in S,\ \xi\in X\}$. We often identify a closed subspace with the projection onto it.

A von Neumann algebra on $H$ is a $*$-subalgebra $M\subseteq B(H)$ with $M=M''$. A commutant of a set that is closed under adjoints is a $*$-algebra that contains $1$ and is closed in the weak operator topology. So $M$ is weakly closed, and hence strongly closed. The closed unit ball of $B(H)$ is weakly compact by the coefficient argument immediately below, and therefore so is the weakly closed unit ball of $M$. The centre is $Z=M\cap M'$, and a nonzero $M$ is a factor if $Z=\mathbb C1$. We write $\mathcal P(M)$ for the set of projections of $M$, ordered by $e\le f\iff ef=e$, and $e^\perp=1-e$. An isomorphism of von Neumann algebras is a $*$-isomorphism. It carries projections, partial isometries, central projections and the order of projections to the same objects, so every notion below that is defined through them is invariant. The notation $\{M,H\}\cong\{N,K\}$ means a spatial isomorphism: a unitary $W:H\to K$ with $WMW^*=N$.

Why the operator ball is compact. Put each coefficient in its compact disk $\{z:|z|\le\|\xi\|\|\eta\|\}$. The set of coefficient families that are linear in $\xi$, conjugate-linear in $\eta$, and satisfy the displayed bound is closed in the product: all linearity identities and scalar bounds are closed conditions on finitely many coordinates. Every such family is a bounded sesquilinear form, hence is uniquely $\langle T\xi,\eta\rangle$ for a contraction $T$, by the full Hilbert-space form representation, Theorem 3.1. Conversely every contraction gives such a family. The coordinate topology is exactly the weak operator topology. The full Tychonoff proof therefore proves compactness, including $H=0$. This supplies the norm-attainment input in Section 17.

Vector functionals are $\omega_\xi(x)=\langle x\xi,\xi\rangle$ and $\omega_{\eta,\xi}(x)=\langle x\eta,\xi\rangle$.

Throughout, $M$ is a von Neumann algebra on $H$ with centre $Z$, unless another algebra is named. Representations are nondegenerate unless explicitly described otherwise.

2. Full proof prerequisites

The nine inputs below have full earlier programme proofs. The stated corner, support, continuity and cardinality conclusions are included in those proofs; no normality or countability hypothesis is suppressed.

Fact 2.1 (Reduced and induced algebras). For $e\in\mathcal P(M)$, the restriction of $eMe$ to $eH$ (the reduced algebra) and the restriction of $M'e$ to $eH$ (the induced algebra) are von Neumann algebras on $eH$, and each is the commutant of the other there. Applied to $M'$, whose commutant is $M$: for $e'\in\mathcal P(M')$, the algebras $Me'$ and $e'M'e'$ on $e'H$ are mutual commutants. We usually omit the restriction from the notation. This is proved in The double commutant theorem.

Fact 2.2 (Polar decomposition). Every $x\in M$ is $x=u|x|$ with $|x|=(x^*x)^{1/2}\in M$ and $u\in M$ a partial isometry; $u^*u$ is the projection onto $[x^*H]=(\ker x)^\perp$ and $uu^*$ the projection onto $[xH]$. This is proved in The double commutant theorem.

Fact 2.3 (Cyclic and separating sets). $X\subseteq H$ is cyclic for $M$ if $[MX]=H$, and separating for $M$ if $x\in M$ and $xX=\{0\}$ force $x=0$. A set is cyclic for $M$ exactly when it is separating for $M'$. This is proved in The double commutant theorem.

Fact 2.4 ($\sigma$-finiteness). $M$ is $\sigma$-finite if every family of mutually orthogonal nonzero projections in $M$ is countable. This holds if and only if $H$ has a countable subset that is separating for $M$, and if and only if $M$ has a faithful normal positive functional. A projection $e$ is $\sigma$-finite when $eMe$ is. $\sigma$-finite algebras are also called countably decomposable. This is proved in The double commutant theorem.

Fact 2.5 (Normal functionals). $M_*$ is the space of $\sigma$-weakly continuous linear functionals on $M$, called normal. Every positive $\varphi\in M_*$ equals $\sum_n\omega_{\xi_n}$ for some $\xi_n\in H$ with $\sum_n\|\xi_n\|^2<\infty$. Its support $s(\varphi)$ is the least projection $p\in M$ with $\varphi(1-p)=0$. The full sum formula is proved in The double commutant theorem, and the least-support and faithfulness proof is given in the tools before Section 1 of The universal enveloping von Neumann algebra of a $C^*$-algebra, and $W^*$-algebras.

Fact 2.6 (Automatic normality and normal images). An isomorphism of one von Neumann algebra onto another is $\sigma$-weakly bicontinuous, by the full Corollary 11.4 of the universal-enveloping lesson. The image under a nondegenerate normal representation is a von Neumann algebra, by its full Proposition 12.1. These are distinct results, with the normal extension and central-corner inverse proved before them.

Fact 2.7 (The universal representation, extensions and density). For a nonzero $C^*$-algebra $A$, take the Hilbert direct sum of its state GNS representations, denoted $\{\pi_u,H_u\}$, and put $\tilde A=\pi_u(A)''$. The full construction and its isometric predual pairing are proved before Section 2 of the universal-enveloping lesson; Section 3 identifies the universal algebra. Every positive functional $\varphi$ has a unique positive normal extension $\tilde\varphi$, with the same norm. For $\varphi\ne0$, the state $\varphi/\|\varphi\|$ supplies a cyclic summand and the vector $\sqrt{\|\varphi\|}\xi_{\varphi/\|\varphi\|}$; for $\varphi=0$, take the zero vector. We write $s(\varphi)=s(\tilde\varphi)$. For $A=0$ all these spaces and functionals are zero. The full double commutant density theorem proves strong and sigma-weak density for nondegenerate algebras; the norm-bounded Kaplansky theorem, together with the full bounded-topology comparison, Lemma 8.5, proves uniqueness of the normal extension.

Fact 2.8 (Functional calculus and spectral approximation). The full continuous calculus, Theorem 4.1 applies to normal elements. For $h=h^*\in M$ and bounded Borel $k:\mathbb R\to\mathbb C$, $k(h)\in M$, and $k\mapsto k(h)$ is a unital $*$-homomorphism with $\|k(h)\|\le\sup|k|$; the full Borel calculus, Theorem 3.1 and spectral projections, Proposition 5.1 prove this. In particular spectral projections belong to $M$. To see the norm approximation explicitly, partition $[-\|h\|,\|h\|]$ into finitely many Borel sets of diameter at most $\varepsilon$, choose a real value in each nonempty set, and approximate $t$ by the resulting real simple function $s$. Then $\|h-s(h)\|\le\varepsilon$. Each $s(h)$ is a real linear combination of spectral projections. Write $x=(x+x^*)/2+i(x-x^*)/(2i)$ to conclude that $M$ is the norm-closed complex linear span of its projections.

Fact 2.9 (Set theory). We work with the axiom of choice and Zorn's lemma, as in the full Hahn–Banach lesson, Section 1. Its full Theorem 8.1 proves Cantor–Schröder–Bernstein; Theorem 8.4 proves $\kappa\cdot\aleph_0=\kappa$, $\kappa+\kappa=\kappa$, and the countable-union estimate for infinite $\kappa$. The proof also gives a bijection from an infinite set to itself minus one point, by shifting a countably infinite subset. Hilbert bases and their cardinalities have the full Hilbert-space proof in Section 4, and the tensor product used in Section 8 has the full construction in Section 8.

3. The projection lattice and equivalence of projections

Proposition 3.1 (The projection lattice). Let $\{e_i\}_{i\in I}$ be any family in $\mathcal P(M)$. The projection $\bigwedge_ie_i$ onto $\bigcap_ie_iH$ and the projection $\bigvee_ie_i$ onto $[\bigcup_ie_iH]$ lie in $M$. They are the greatest lower bound and the least upper bound of the family, both in $\mathcal P(M)$ and in $\mathcal P(B(H))$. So $\mathcal P(M)$ is a complete lattice whose meets and joins are those of $B(H)$.

Proof. Put $L=\bigcap_ie_iH$. Take $x'\in M'$. It commutes with every $e_i$, so it maps each $e_iH$ into itself, and therefore maps $L$ into $L$. So does $x'^*$, since $M'$ is closed under adjoints. A closed subspace that is invariant under an operator and under its adjoint reduces the operator. Hence the projection onto $L$ commutes with every $x'\in M'$ and lies in $M''=M$. A projection $g$ lies below every $e_i$ exactly when $gH\subseteq L$, so this projection is the greatest lower bound in $\mathcal P(B(H))$, and a fortiori in $\mathcal P(M)$. Apply this to the family $1-e_i$. The projection onto $\bigcap_i(e_iH)^\perp=(\bigcup_ie_iH)^\perp$ lies in $M$, and so does its complement $\bigvee_ie_i$. Taking complements reverses the order, so $\bigvee_ie_i$ is the least upper bound. $\square$

$\mathcal P(M)$ is the projection lattice of $M$. For mutually orthogonal $e_i$, $\bigvee_ie_i=\sum_ie_i$, the strong sum. De Morgan's laws $(\bigvee_ie_i)^\perp=\bigwedge_ie_i^\perp$ hold.

Definition 3.2 (Equivalence of projections). Let $e,f\in\mathcal P(M)$.

$e$ and $f$ are equivalent, $e\sim f$, if some $u\in M$ satisfies $u^*u=e$ and $uu^*=f$. Then $u$ is a partial isometry with initial projection $e$ and final projection $f$, and we say that $u$ implements $e\sim f$.
$e\precsim f$, also written $f\succsim e$, if $e\sim f_1$ for some $f_1\in\mathcal P(M)$ with $f_1\le f$.
$e\prec f$ if $e\precsim f$ and $e$ is not equivalent to $f$.

These relations are also applied to closed subspaces whose projections lie in $M$.

Lemma 3.3.

If $u\in M$ and $u^*u=e$ is a projection, then $u=ue$, $uu^*$ is a projection, and $u$ maps $eH$ isometrically onto $uu^*H$.
$\sim$ is an equivalence relation, and $\precsim$ is reflexive and transitive.
(Additivity) Let $\{e_i\}_{i\in I}$ and $\{f_i\}_{i\in I}$ be families of mutually orthogonal projections. If $e_i\sim f_i$ for every $i$, then $\sum_ie_i\sim\sum_if_i$. The same holds with $\precsim$ in place of $\sim$.
(Central cuts) If $e\sim f$ (or $e\precsim f$) and $z$ is a central projection, then $ze\sim zf$ (or $ze\precsim zf$).

Proof. (1) $(u-ue)^*(u-ue)=e-e-e+e=0$, so $u=ue$. Then $(uu^*)^2=ueu^*=uu^*$, and $\|u\xi\|^2=\langle u^*u\xi,\xi\rangle=\|e\xi\|^2$. So $u$ is isometric on $eH$ and zero on $e^\perp H$. Its range $u(eH)$ is closed, and $uu^*$ is the projection onto it.

(2) $e$ implements $e\sim e$. If $u$ implements $e\sim f$, then $u^*$ implements $f\sim e$. If $u$ implements $e\sim f$ and $v$ implements $f\sim g$, then $fu=u$ because $u=uu^*u$. Hence $(vu)^*(vu)=u^*fu=e$ and $(vu)(vu)^*=vfv^*=g$. For $\precsim$, let $e\sim f_1\le f$, and let $v$ implement $f\sim g_1\le g$. The operator $vf_1$ has initial projection $f_1v^*vf_1=f_1$ and final projection $vf_1v^*\le vfv^*=g_1\le g$. So $e\sim f_1\sim vf_1v^*\le g$.

(4) If $u$ implements $e\sim f$, then $zu$ implements $ze\sim zf$, because $z$ is central. For $\precsim$, cut the subprojection by $z$ as well. $\square$

Definition 3.4 (Central support). For $e\in\mathcal P(M)$, the central support $c(e)$ is the least central projection that majorizes $e$. It exists: $Z=(M\cup M')'$ is a von Neumann algebra whose projections are the central projections, so by Proposition 3.1, applied to $Z$, the meet of all central projections that majorize $e$ is again central. It majorizes $e$, and it is the least such projection.

Proposition 3.5 (Central supports).

$c(e)$ is the projection onto $[MeH]$. For a central projection $z$, $ze=0$ if and only if $zc(e)=0$. Moreover $c(ze)=zc(e)$.
If $e\sim f$ then $c(e)=c(f)$. If $e\precsim f$ then $c(e)\le c(f)$.
(Induction) The map $x'\mapsto x'e|_{eH}$ is a $*$-homomorphism of $M'$ onto the induced algebra $M'e$, with kernel $M'(1-c(e))$. Likewise, for $e'\in\mathcal P(M')$, the map $x\mapsto xe'|_{e'H}$ is a $*$-homomorphism of $M$ onto $Me'$ with kernel $M(1-c(e'))$; here $c(e')$ is the central support of $e'$ in $M'$, the projection onto $[M'e'H]$.
(Centre of a reduced algebra) The centre of $eMe$ (on $eH$) is $Ze=\{ae:a\in Z\}$, and $a\mapsto ae|_{eH}$ is an isomorphism of $Zc(e)$ onto it. In particular every projection in the centre of $eMe$ is $ze$ for a central projection $z$.

Proof. (1) The subspace $[MeH]$ is invariant under $M$. It is also invariant under $M'$, since $x'ye=yex'$ for $x'\in M'$ and $y\in M$. Both algebras are closed under adjoints, so the projection $q$ onto $[MeH]$ commutes with $M$ and with $M'$. Hence $q\in M''\cap M'=Z$, and $q\ge e$ because $1\in M$. If $z$ is central and $z\ge e$, then $zye=yze=ye$ for $y\in M$, so $[MeH]\subseteq zH$ and $q\le z$. So $q=c(e)$. If $z$ is central and $ze=0$, then $zye=yze=0$, so $z$ vanishes on $[MeH]$ and $zc(e)=0$. The converse follows from $e=c(e)e$. Finally, a central projection $w$ satisfies $w(ze)=0$ iff $(wz)c(e)=0$, iff $w\le1-zc(e)$. The largest such $w$ is $1-c(ze)$ by what was just shown, so $c(ze)=zc(e)$.

(2) Let $u$ implement $e\sim f$, and let $z$ be central. From $u=ue$, $e=u^*u$, $u=fu$ and $f=uu^*$ we get $ze=0\iff zu=0\iff zf=0$. By (1), $zc(e)=0\iff zc(f)=0$. Taking $z=1-c(e)$ and $z=1-c(f)$ gives $c(f)\le c(e)\le c(f)$. The second claim follows, since $f_1\le f$ implies $c(f_1)\le c(f)$.

(3) The map is a $*$-homomorphism because $e$ commutes with $M'$, and its image is $M'e$ by definition. Further, $x'e=0$ iff $x'$ vanishes on $eH$, iff $x'$ vanishes on $[MeH]$ (as $x'ye=yx'e$), iff $x'c(e)=0$. The second statement is the first one with the roles of $M$ and $M'$ exchanged.

(4) By Fact 2.1, $eMe$ and $M'e$ are each other's commutants on $eH$, so they have the same centre. By (3), the induction restricts to an isomorphism of $M'c(e)$ onto $M'e$, and an isomorphism maps centre onto centre. The centre of $M'c(e)$ is $Zc(e)$: if $y'c(e)$ ($y'\in M'$) commutes with $M'c(e)$, it also commutes with $M'(1-c(e))$ (both products vanish), hence with all of $M'$, so it lies in $M'\cap M''=Z$. Therefore the centre of $M'e$ is $\{ac(e)e:a\in Z\}=Ze$. On $Zc(e)$ the map $a\mapsto ae|_{eH}$ is the restriction of the induction, which is injective there. An isomorphism maps projections to projections. $\square$

4. Supports, cyclic projections and the parallelogram law

Every operator in $M$ has a left and a right support in $M$, and the two supports are equivalent. This is the basic way to produce equivalent projections.

Definition 4.1 (Left and right supports). For $x\in M$, the left support $\mathrm l(x)$ is the least projection $p\in M$ with $px=x$, and the right support $\mathrm r(x)$ is the least projection $q\in M$ with $xq=x$.

Lemma 4.2. $\mathrm l(x)$ is the projection onto $[xH]$, and $\mathrm r(x)=\mathrm l(x^*)$ is the projection onto $[x^*H]=(\ker x)^\perp$. Both lie in $M$.

Proof. For a projection $p$, $px=x$ iff $xH\subseteq pH$, iff $p$ majorizes the projection onto $[xH]$. That projection lies in $M$: it is $uu^*$ for the partial isometry $u$ of the polar decomposition (Fact 2.2). Next, $xq=x$ iff $qx^*=x^*$, iff $q\ge\mathrm l(x^*)$. Finally $[x^*H]^\perp=\ker x$. $\square$

Proposition 4.3. $\mathrm l(x)\sim\mathrm r(x)$ for every $x\in M$.

Proof. In the polar decomposition $x=u|x|$ of Fact 2.2, $u^*u=\mathrm r(x)$ and $uu^*=\mathrm l(x)$. $\square$

Proposition 4.4 (The parallelogram law). For $e,f\in\mathcal P(M)$, $(e\vee f)-e\sim f-(e\wedge f)$.

Proof. Put $x=e^\perp f\in M$. A vector $\xi$ lies in $\ker x$ iff $f\xi\in eH$. Writing $\xi=f^\perp\xi+f\xi$, this gives $\ker x=f^\perp H\oplus(eH\cap fH)$, with projection $f^\perp+e\wedge f$. Hence $\mathrm r(x)=f-e\wedge f$. In the same way, $\xi\in\ker x^*=\ker(fe^\perp)$ iff $e^\perp\xi\in f^\perp H$. Writing $\xi=e\xi+e^\perp\xi$, we get $\ker x^*=eH\oplus(e^\perp H\cap f^\perp H)$, with projection $e+e^\perp\wedge f^\perp=e+(e\vee f)^\perp$. Hence $\mathrm l(x)=(e\vee f)-e$. Now apply Proposition 4.3. $\square$

Vector functionals also have supports in $M$: they are the cyclic projections.

Definition 4.5 (Cyclic projections). For $\xi\in H$, $p_\xi$ is the projection onto $[M'\xi]$ and $p'_\xi$ the projection onto $[M\xi]$. These are the cyclic projections of $M$ and of $M'$ defined by $\xi$.

Lemma 4.6. $p_\xi\in M$, and $p_\xi$ is the least projection $p\in M$ with $p\xi=\xi$. Equivalently, $p_\xi$ is the support of the restriction of $\omega_\xi$ to $M$. The same holds for $p'_\xi$ with $M'$ in place of $M$.

Proof. $[M'\xi]$ is invariant under $M'$, which is closed under adjoints, so $p_\xi$ commutes with $M'$ and lies in $M$. As $1\in M'$, $p_\xi\xi=\xi$. If $p\in\mathcal P(M)$ and $p\xi=\xi$, then $px'\xi=x'p\xi=x'\xi$ for $x'\in M'$, so $p\ge p_\xi$. Finally $\omega_\xi(1-p)=\|(1-p)\xi\|^2$, which vanishes iff $p\xi=\xi$. $\square$

Exercise 4.7. (medium) Let $\mathfrak m$ be a left ideal of $M$, not necessarily closed, and $\mathcal P=\mathcal P(M)$.

(a) $\mathfrak m=\{xh:\ x\in M,\ h\in\mathfrak m\cap M_+\}$.
(b) The linear span of $\{xe:\ x\in M,\ e\in\mathcal P\cap\mathfrak m\}$ is norm dense in $\mathfrak m$. So a norm-closed left ideal is determined by its projections.

Solution. (a) If $y\in\mathfrak m$ has polar decomposition $y=u|y|$ (Fact 2.2), then $|y|=u^*y\in\mathfrak m\cap M_+$ and $y=u|y|$. Conversely $xh\in\mathfrak m$ for $h\in\mathfrak m$.

(b) Each $xe$ with $e\in\mathcal P\cap\mathfrak m$ lies in $\mathfrak m$. Let $y=u|y|\in\mathfrak m$, put $h=|y|\in\mathfrak m\cap M_+$, and let $\varepsilon>0$. The spectral projection $p=\chi_{(\varepsilon,\infty)}(h)$ equals $k(h)h$ with $k(t)=t^{-1}\chi_{(\varepsilon,\infty)}(t)$, a bounded Borel function (Fact 2.8), so $p\in\mathcal P\cap\mathfrak m$, and $\|h-hp\|=\|h\chi_{[0,\varepsilon]}(h)\|\le\varepsilon$. Then $yp=u(hp)$, so $\|y-yp\|\le\|h-hp\|\le\varepsilon$, and $yp$ has the form $xe$. If $\mathfrak m$ is norm closed, it is the closed linear span of these products, which depends only on $\mathfrak m\cap\mathcal P$. $\square$

5. The comparison theorem

Two ingredients lead to the comparison theorem: a Schröder–Bernstein theorem for projections, and a test for when two projections have equivalent nonzero subprojections.

Proposition 5.1 (Schröder–Bernstein for projections). If $e\precsim f$ and $f\precsim e$, then $e\sim f$.

The proof makes one "Hilbert hotel" shift inside $e$.

Proof. Step 1. Let $e_2\le e_1\le e$ be projections with $e\sim e_2$. We show $e\sim e_1$. Let $w$ implement $e\sim e_2$. Then $w=we$ and $ew=w$, so the powers $w^n$ have initial projection $e$, and their final projections $w^nw^{*n}$ decrease, since $\begin{gathered} w^{n+1}w^{*(n+1)}\\ =w^n(ww^*)w^{*n}\\ \le w^new^{*n}\\ =w^nw^{*n}. \end{gathered}$ Put $g=e-e_1$ and $g_n=w^ngw^{*n}$ for $n\ge0$. As $g\le e-e_2=e-ww^*$, we get $g_n\le w^nw^{*n}-w^{n+1}w^{*(n+1)}$. The right-hand sides are differences of a decreasing sequence, so the $g_n$ are mutually orthogonal. Put $G=\sum_ng_n\le e$ and $t=wG+(e-G)$. Since $wg_nw^*=g_{n+1}$, the operator $wG$ maps $GH$ isometrically onto $\sum_{n\ge1}g_nH\subseteq GH$, while $e-G$ is the identity on $(e-G)H$, which is orthogonal to $GH$. The cross terms in $t^*t$ and $tt^*$ vanish, and $\begin{gathered} t^*t\\ =G+(e-G)\\ =e,\\ tt^*\\ =\sum_{n\ge1}g_n+(e-G)\\ =e-g_0\\ =e_1 . \end{gathered}$ So $e\sim e_1$.

Step 2. Let $u$ implement $e\sim f_1\le f$ and $v$ implement $f\sim e_1\le e$. Put $e_2=vf_1v^*$. Then $e_2\le vfv^*=e_1$, and $vf_1$ implements $f_1\sim e_2$, so $e\sim e_2$. Step 1 gives $e\sim e_1$, and $e_1\sim f$. $\square$

Definition 5.2. Projections $e,f\in\mathcal P(M)$ are centrally orthogonal if $c(e)c(f)=0$.

Lemma 5.3 (Central orthogonality). For $e,f\in\mathcal P(M)$ the following are equivalent.

$c(e)c(f)\ne0$.
$eMf\ne\{0\}$.
There are nonzero projections $e_1\le e$ and $f_1\le f$ in $M$ with $e_1\sim f_1$.

Proof. (3)⇒(2). If $u$ implements $e_1\sim f_1$, then $u=f_1ue_1\in fMe$, so $0\ne u^*\in eMf$.

(2)⇒(1). If $c(e)c(f)=0$, then $eyf=ec(e)\,y\,c(f)f=ey\,c(e)c(f)\,f=0$ for all $y\in M$.

(1)⇒(2). Suppose $eMf=\{0\}$. Then $fMe=(eMf)^*=\{0\}$, so $f$ vanishes on $[MeH]=c(e)H$ (Proposition 3.5(1)). Thus $fc(e)=0$, so $f\le1-c(e)$ and $c(f)\le1-c(e)$, which contradicts (1).

(2)⇒(3). Take $x=eyf\ne0$. Then $ex=x$ and $xf=x$, so $\mathrm l(x)\le e$ and $\mathrm r(x)\le f$. These supports are nonzero because $x\ne0$, and they are equivalent by Proposition 4.3. $\square$

Corollary 5.4. If $e\ne0$ and $c(e)\le c(f)$, there are nonzero $e_1\le e$ and $f_1\le f$ with $e_1\sim f_1$. Indeed $c(e)c(f)=c(e)\ne0$.

Theorem 5.5 (The comparison theorem). For $e,f\in\mathcal P(M)$ there is a central projection $z$ with $ze\precsim zf\qquad\text{and}\qquad(1-z)f\precsim(1-z)e .$ If $M$ is a factor, exactly one of $e\prec f$, $e\sim f$, $f\prec e$ holds.

Proof. Call a family $\{(e_i,f_i)\}_{i\in I}$ admissible if the $e_i$ are mutually orthogonal nonzero subprojections of $e$, the $f_i$ are mutually orthogonal subprojections of $f$, and $e_i\sim f_i$ for each $i$. The union of a chain of admissible families is admissible, so Zorn's lemma gives a maximal one. Put $e_0=\sum_ie_i$ and $f_0=\sum_if_i$. Then $e_0\sim f_0$ by additivity (Lemma 3.3(3)). If $e-e_0$ and $f-f_0$ were not centrally orthogonal, Lemma 5.3 would give nonzero $e'\le e-e_0$ and $f'\le f-f_0$ with $e'\sim f'$. Adding the pair $(e',f')$ would contradict maximality. So $c(e-e_0)\,c(f-f_0)=0$.

Put $z=1-c(e-e_0)$. Then $z(e-e_0)=0$, and $\begin{gathered} (1-z)(f-f_0)\\ =c(e-e_0)c(f-f_0)(f-f_0)\\ =0. \end{gathered}$ Cutting by a central projection preserves equivalence (Lemma 3.3(4)), so $\begin{gathered} ze\\ =ze_0\sim zf_0\\ \le zf,\\ (1-z)f\\ =(1-z)f_0\sim(1-z)e_0\\ \le(1-z)e . \end{gathered}$

Let $M$ be a factor. Then $z\in\{0,1\}$, so $e\precsim f$ or $f\precsim e$. If both hold, $e\sim f$ by Proposition 5.1. If only $e\precsim f$ holds, then $e$ is not equivalent to $f$ (else $f\precsim e$), so $e\prec f$. Symmetrically, if only $f\precsim e$ holds, then $f\prec e$. The three cases exclude each other: $e\prec f$ rules out $e\sim f$ by definition, and $e\prec f$ together with $f\prec e$ would give $e\sim f$ by Proposition 5.1. $\square$

Exercise 5.6. (hard) Let $\alpha$ be an automorphism of $M$. Suppose that some $e\in\mathcal P(M)$ with $c(e)=1$ and some $u\in M$ satisfy $\alpha(x)=uxu^*$ for all $x\in eMe$. Show that $\alpha$ is inner.

Solution. Normalizing $u$. Put $v=ue$. Then $vv^*=ueu^*=\alpha(e)$ is a projection, so $v$ is a partial isometry (Lemma 3.3(1) applied to $v^*$), and $q=v^*v=eu^*ue$ is a projection below $e$. For $y\in eMe$, $\alpha(y)=vyv^*$. Since $vq=v$, the element $e-q\in eMe$ satisfies $\alpha(e-q)=vv^*-vqv^*=0$; as $\alpha$ is injective, $q=e$. So $v^*v=e$, $vv^*=\alpha(e)$, and $\alpha(y)v=vyv^*v=vy$ for $y\in eMe$.

Building the unitary. Since $c(e)=1$, a maximal family of mutually orthogonal projections each subequivalent to $e$ has sum $1$ (otherwise Corollary 5.4, applied to the remainder and $e$, would enlarge it). So there are partial isometries $v_i$ with $v_i^*v_i\le e$, mutually orthogonal final projections $v_iv_i^*$, and $\sum_iv_iv_i^*=1$. Put $t_i=\alpha(v_i)vv_i^*$. Using $\alpha(y)=vyv^*$ for $y=v_i^*v_i\in eMe$ and $v_i=v_ie$, one finds that $t_i$ has initial projection $v_iv_i^*$ and final projection $\alpha(v_iv_i^*)$. The initial projections are mutually orthogonal with sum $1$. So are the final ones: $\alpha$ preserves orthogonality and the order of projections, hence least upper bounds (Proposition 3.1), so $\sum_i\alpha(v_iv_i^*)=\alpha(1)=1$. By the proof of Lemma 3.3(3), $w=\sum_it_i$ converges strongly to a unitary in $M$.

It implements $\alpha$. As $v_i^*v_j=0$ for $i\ne j$, $\begin{gathered} wv_j\\ =t_jv_j\\ =\alpha(v_j)vv_j^*v_j\\ =\alpha(v_j)\alpha(v_j^*v_j)v\\ =\alpha(v_j)v. \end{gathered}$ For $x\in M$, $xv_j=\sum_iv_iv_i^*xv_j$ strongly, and $v_i^*xv_j\in eMe$, so $\begin{gathered} wxv_j\\ =\sum_i\alpha(v_i)v(v_i^*xv_j)\\ =\sum_i\alpha(v_i)\alpha(v_i^*xv_j)v\\ =\sum_i\alpha(v_iv_i^*)\,\alpha(xv_j)v\\ =\alpha(xv_j)v\\ =\alpha(x)wv_j . \end{gathered}$ The ranges of the $v_j$ span a dense subspace of $H$, so $wx=\alpha(x)w$, that is, $\alpha(x)=wxw^*$. No normality of $\alpha$ was used. $\square$

6. Finite, infinite and abelian projections

Finiteness is the analogue of finite dimension: a finite projection is not equivalent to a proper subprojection of itself.

Definition 6.1. A projection $e\in\mathcal P(M)$ is

finite if $e\sim f\le e$ implies $f=e$, and infinite otherwise;
purely infinite if $0$ is its only finite subprojection in $M$;
properly infinite if each nonzero central cut $ze$ ($z$ a central projection) is infinite;
abelian if $eMe$ is commutative.

The same four words are applied to $M$ itself when the projection $1$ has the property. By these definitions, $0$ is finite, abelian, properly infinite and purely infinite.

Lemma 6.2.

A subprojection of a finite (or abelian) projection is finite (or abelian). A projection equivalent to a finite (or abelian) projection is finite (or abelian). A projection that is subequivalent to a finite projection is finite.
Abelian projections are finite. Minimal projections, meaning nonzero projections whose only subprojections are $0$ and themselves, are abelian.
$e$ is finite (or abelian) in $M$ if and only if $eMe$ is a finite (or commutative) algebra. A nonzero central cut $ze$ of a properly infinite $e$ is properly infinite, and a central cut of $e$ that is finite is $0$. A purely infinite projection is properly infinite.
If $e$ is abelian, then $eMe=Ze$, and every subprojection $f$ of $e$ satisfies $f=c(f)e$.

Proof. (1) Let $e$ be finite, $f\le e$, and $f\sim f_1\le f$. Additivity gives $e=(e-f)+f\sim(e-f)+f_1\le e$, so $(e-f)+f_1=e$ and $f_1=f$. Next let $v$ implement $e\sim g$, and let $g\sim g_1\le g$. The operator $v^*g_1$ has initial projection $g_1vv^*g_1=g_1$ and final projection $v^*g_1v\le v^*gv=e$. So $e\sim g\sim g_1\sim v^*g_1v\le e$. As $e$ is finite, $v^*g_1v=e$, and then $g_1=vv^*g_1vv^*=vev^*=g$. So $g$ is finite. If $e\sim f_1\le f$ with $f$ finite, then $f_1$ is finite, and so is $e$. For abelian projections: if $f\le e$, the elements of $fMf$ lie in $eMe$, so they commute; and if $u$ implements $e\sim g$, then $y\mapsto uyu^*$ is an isomorphism of $eMe$ onto $gMg$, with inverse $y\mapsto u^*yu$.

(2) Let $e$ be abelian and let $u$ implement $e\sim f\le e$. Then $u=fue\in eMe$, so $u^*u=uu^*$, that is, $e=f$. Let $e$ be minimal. The algebra $eMe$ on $eH$ is a von Neumann algebra (Fact 2.1) whose only projections are $0$ and $e$. The spectral projections of a self-adjoint element of $eMe$ lie in $eMe$ (Fact 2.8), so each self-adjoint element is a real multiple of $e$. Hence $eMe=\mathbb Ce$.

(3) A partial isometry $u$ with $u^*u=e$ and $uu^*\le e$ satisfies $u=uu^*u\in eMe$. So finiteness of $e$ means the same in $M$ and in $eMe$, and so does commutativity. If $e$ is properly infinite and $z,z'$ are central with $z'(ze)\ne0$, then $z'ze=(z'z)e$ is infinite. A finite nonzero central cut $ze$ would contradict the definition. If $e$ is purely infinite and $ze\ne0$, then $ze$ is a nonzero subprojection of $e$, hence infinite.

(4) $eMe$ is a commutative von Neumann algebra on $eH$, so it is its own centre, which is $Ze$ by Proposition 3.5(4). A subprojection $f$ of $e$ lies in $eMe$, so $f=ze$ for a central projection $z$, again by Proposition 3.5(4). Then $c(f)=zc(e)$ by Proposition 3.5(1), and $c(f)e=zc(e)e=ze=f$. $\square$

Lemma 6.3 (Centrally orthogonal sums). The sum of a family of mutually centrally orthogonal abelian (or finite) projections is abelian (or finite).

Proof. Let $e=\sum_ie_i$ with $c(e_i)c(e_j)=0$ for $i\ne j$, and write $c_i=c(e_i)$. Then $c_ie=e_i$, because $c_ie_j=c_ic_je_j=0$ for $j\ne i$. For $x\in M$ and $i\ne j$, $e_ixe_j=e_ic_i\,x\,c_je_j=e_ixe_jc_ic_j=0$. Hence $exe=\sum_ie_ixe_i$, a strong sum. If every $e_i$ is abelian, then for $x,y\in M$, $(exe)(eye)=\sum_i(e_ixe_i)(e_iye_i)$, and the terms commute inside $e_iMe_i$; so $eMe$ is commutative. If every $e_i$ is finite and $e\sim f\le e$, then $c_ie\sim c_if\le c_ie$ by Lemma 3.3(4), that is, $e_i\sim c_if\le e_i$. So $c_if=e_i$. As $f\le e\le\sum_ic_i$, we get $f=\sum_ic_if=\sum_ie_i=e$. $\square$

For a central decomposition $1=\sum_i z_i$, the map $x\mapsto(xz_i)_i$ identifies $M$ with the algebra of bounded families $x_i\in Mz_i$, with norm $\sup_i\|x_i\|$. Indeed, the finite sums of such a family converge strongly: for each $\xi$, their orthogonal components have total squared norm at most $(\sup_i\|x_i\|)^2\|\xi\|^2$. Their bounded strong limit belongs to $M$, has the prescribed central components, and has that supremum norm. Thus algebra direct sums $\bigoplus_i Mz_i$ below mean this bounded product, acting on the Hilbert direct sum $\bigoplus_i z_iH$.

7. The type decomposition

Definition 7.1 (Types). $M$ is

of type I if below each nonzero central projection there is a nonzero abelian projection;
of type II if $0$ is its only abelian projection, while below each nonzero central projection there is a nonzero finite projection;
of type III if $M$ is purely infinite: $0$ is its only finite projection;
of type II$_1$ if it is of type II and finite, and of type II$_\infty$ if it is of type II and $0$ is its only finite central projection;
semifinite if it has no nonzero central summand of type III.

For a central projection $z$, the projections of the central summand $Mz$ (on $zH$) are the projections of $M$ below $z$, its central projections are the central projections of $M$ below $z$, and equivalence among them is the same as in $M$. So each type passes to central summands. Each type is also invariant under isomorphism.

Theorem 7.2 (The type decomposition). There are unique mutually orthogonal central projections $z_{\rm I}$, $z_{{\rm II}_1}$, $z_{{\rm II}_\infty}$, $z_{\rm III}$ with sum $1$ such that $Mz_{\rm I}$, $Mz_{{\rm II}_1}$, $Mz_{{\rm II}_\infty}$ and $Mz_{\rm III}$ are of types I, II$_1$, II$_\infty$ and III. Every projection $e\in M$ can be written in exactly one way as $e=e_1+e_2$ with $e_1$ and $e_2$ centrally orthogonal, $e_1$ finite and $e_2$ properly infinite.

Proof. Existence. Zorn's lemma gives a maximal family $\{a_k\}$ of mutually centrally orthogonal nonzero abelian projections. Their sum $a$ is abelian (Lemma 6.3); put $z_{\rm I}=c(a)$. If $z\le z_{\rm I}$ is central and nonzero, then $za\ne0$ (Proposition 3.5(1)), and $za$ is abelian. So $Mz_{\rm I}$ is of type I. The summand $M(1-z_{\rm I})$ has no nonzero abelian projection $q$: such a $q$ would have $c(q)\le1-z_{\rm I}$, so it would be centrally orthogonal to every $a_k$, against maximality.

Next take a maximal family $\{b_k\}$ of mutually centrally orthogonal nonzero finite projections below $1-z_{\rm I}$. Their sum $b$ is finite (Lemma 6.3); put $z_{\rm II}=c(b)\le1-z_{\rm I}$. The summand $Mz_{\rm II}$ has no nonzero abelian projection, and each nonzero central $z\le z_{\rm II}$ majorizes the nonzero finite projection $zb$. So $Mz_{\rm II}$ is of type II. Put $z_{\rm III}=1-z_{\rm I}-z_{\rm II}$. A nonzero finite projection below $z_{\rm III}$ would be centrally orthogonal to all $b_k$, against maximality. So $Mz_{\rm III}$ is of type III.

Finally take a maximal family $\{c_k\}$ of mutually orthogonal nonzero finite central projections below $z_{\rm II}$. Orthogonal central projections are centrally orthogonal, so $z_{{\rm II}_1}=\sum_kc_k$ is finite (Lemma 6.3), and $Mz_{{\rm II}_1}$ is of type II$_1$. By maximality, $Mz_{{\rm II}_\infty}$, with $z_{{\rm II}_\infty}=z_{\rm II}-z_{{\rm II}_1}$, has no nonzero finite central projection, so it is of type II$_\infty$.

Uniqueness. A nonzero algebra of type I has a nonzero abelian projection, and one of type II or III has none. A nonzero algebra of type I or II has a nonzero finite projection, and one of type III has none. Let $z'_{\rm I},z'_{{\rm II}_1},z'_{{\rm II}_\infty},z'_{\rm III}$ be a second family as in the theorem. The algebra $Mz'_{\rm I}(1-z_{\rm I})$ is a central summand both of $Mz'_{\rm I}$ and of $M(z_{\rm II}+z_{\rm III})$, so it is $0$. Hence $z'_{\rm I}\le z_{\rm I}$, and by symmetry $z'_{\rm I}=z_{\rm I}$. Next, $Mz'_{\rm III}z_{\rm II}$ is a summand of an algebra of type III and of one of type II, so it is $0$; and $z'_{\rm III}z_{\rm I}=z'_{\rm III}z'_{\rm I}=0$. So $z'_{\rm III}\le z_{\rm III}$, and by symmetry they are equal. Then $z'_{{\rm II}_1}+z'_{{\rm II}_\infty}=z_{{\rm II}_1}+z_{{\rm II}_\infty}$. The central projection $z'_{{\rm II}_1}z_{{\rm II}_\infty}$ lies below the finite projection $z'_{{\rm II}_1}$, so it is finite, and it lies in a summand of type II$_\infty$; so it is $0$. Hence $z'_{{\rm II}_1}\le z_{{\rm II}_1}$. By symmetry they are equal, and then $z'_{{\rm II}_\infty}=z_{{\rm II}_\infty}$.

Projections. Let $e\in\mathcal P(M)$. Take a maximal family $\{c_k\}$ of mutually orthogonal central projections below $c(e)$ such that each $c_ke$ is nonzero and finite. Put $c=\sum_kc_k$, $e_1=ce$ and $e_2=(1-c)e$. The projection $e_1=\sum_kc_ke$ is finite by Lemma 6.3. It is centrally orthogonal to $e_2$, since $c(e_1)\le c$ and $c(e_2)\le1-c$. If $z$ is central and $ze_2$ is finite and nonzero, then $z'=z(1-c)c(e)$ is a central projection below $c(e)$, orthogonal to $c$, and $z'e=ze_2$ is finite and nonzero, against maximality. So $e_2$ is properly infinite. For uniqueness, let $e=e_1'+e_2'$ be a second such decomposition and put $a=c(e_1')$. Then $e_1'=ae$, since $ae=e_1'+ae_2'$ and $ae_2'=ac(e_2')e_2'=0$; and $e_2'=(1-a)e$. The projection $a(1-c)e$ equals $(1-c)e_1'\le e_1'$, so it is finite; it also equals $ae_2$, a central cut of the properly infinite $e_2$; so it is $0$ (Lemma 6.2(3)). Likewise $c(1-a)e=(1-a)e_1$ is finite and is a central cut of $e_2'$, so it is $0$. Hence $ae=ace=ce$, that is, $e_1'=e_1$. $\square$

Corollary 7.3. A nonzero factor is of exactly one of the types I, II$_1$, II$_\infty$, III.

Proof. Its only central projections are $0$ and $1$, so exactly one of the four projections of Theorem 7.2 equals $1$. By the facts listed at the start of the uniqueness step in that proof, a nonzero algebra cannot be of two of these types. $\square$

Lemma 7.4 (Good projections).

In a type I algebra, every nonzero projection majorizes a nonzero abelian projection, and some abelian projection has central support $1$.
In a semifinite algebra, below each nonzero central projection there is a nonzero finite projection, and some finite projection has central support $1$.

Proof. (1) Let $q\ne0$. The central projection $c(q)$ majorizes a nonzero abelian projection $e_1$, and $c(e_1)\le c(q)$. Corollary 5.4 gives nonzero $e_2\le e_1$ and $q_1\le q$ with $e_2\sim q_1$, and $q_1$ is abelian by Lemma 6.2(1). For the second claim, take a maximal family $\{e_k\}$ of mutually centrally orthogonal nonzero abelian projections. Their sum $e$ is abelian (Lemma 6.3), and $c(e)=\bigvee_kc(e_k)$, because a central projection majorizes $e$ iff it majorizes every $e_k$. If $c(e)\ne1$, the central projection $1-c(e)$ majorizes a nonzero abelian projection, which is centrally orthogonal to every $e_k$, against maximality. (2) If $M$ is semifinite, then $z_{\rm III}=0$ in Theorem 7.2. A nonzero central $z$ has $zz_{\rm I}\ne0$ or $zz_{\rm II}\ne0$. In the first case it majorizes a nonzero abelian projection, which is finite (Lemma 6.2(2)); in the second case it majorizes a nonzero finite projection by the definition of type II. The rest is the argument of (1), with Lemma 6.3 for finite projections. $\square$

Example 7.5 ($B(H)$). Assume $H\ne0$; the zero algebra has only its zero projection and is covered separately by the conventions. Here $M'=\mathbb C1$. Two projections are equivalent iff their ranges have the same Hilbert dimension, since a partial isometry is a unitary between its initial and final spaces. A projection is finite iff it has finite rank: an infinite-dimensional $eH$ is unitarily equivalent to a proper subspace of itself (shift a countable part of an orthonormal basis), while in finite dimension $e\sim f\le e$ forces $f=e$. The abelian projections are those of rank at most $1$, and $B(H)$ is of type I$_{\dim H}$ in the sense of Definition 10.1.

Example 7.6 (Atomic algebras). For $H\ne0$, $B(H)$ is a type I factor: its commutant is $\mathbb C1$, and a rank-one projection is minimal, hence abelian (Lemma 6.2(2)). More generally, call $M$ atomic if every nonzero projection majorizes a minimal projection. An atomic algebra is of type I, because minimal projections are abelian. Equivalently, $1$ is a sum of minimal projections. Indeed, if $M$ is atomic, a maximal orthogonal family of minimal projections has sum $1$. Conversely, let $1=\sum_ip_i$ with minimal $p_i$, and let $e\ne0$. Then $ep_i\ne0$ for some $i$, and Lemma 5.3 gives nonzero $e_1\le e$ and $f_1\le p_i$ with $e_1\sim f_1$. Then $f_1=p_i$. If $u$ implements $p_i\sim e_1$ and $g\le e_1$, then $u^*gu\le p_i$ is $0$ or $p_i$, so $g=u(u^*gu)u^*$ is $0$ or $e_1$: the projection $e_1$ is minimal.

Example 7.7 (Commutative algebras). In a commutative $A$, $e\sim f$ iff $e=f$ (since $u^*u=uu^*$), every projection is abelian and finite, $c(e)=e$, and $A$ is of type I$_1$ in the sense of Definition 10.1.

8. Matrix units and tensor splitting

If the unit of $M$ is a sum of mutually equivalent projections, then $M$ is a matrix algebra over one of its corners. This section makes that precise. We first describe operators on $K\otimes\ell^2(I)$ by matrices.

Let $N$ be a von Neumann algebra on $K$ and $I$ a nonempty set. On $K\otimes\ell^2(I)$ put $V_i\zeta=\zeta\otimes\delta_i$. The entries of $T\in B(K\otimes\ell^2(I))$ are $T_{ij}=V_i^*TV_j\in B(K)$. An operator is determined by its entries, $(T^*)_{ij}=(T_{ji})^*$, and $(ST)_{ij}=\sum_kS_{ik}T_{kj}$, the sum converging strongly. Put $\begin{gathered} \mathbb M_I(N)\\ =\{T\in B(K\otimes\ell^2(I)):\\ \ T_{ij}\in N\ \text{for all }i,j\},\\ N\otimes1\\ =\{x\otimes1:\ x\in N\}. \end{gathered} \tag{8.1}$ The operator $x\otimes1$ has entries $\delta_{ij}x$, and $E_{ij}$ are the matrix units of $B(\ell^2(I))$. We write $\mathbb M_n(N)$ when $I=\{1,\ldots,n\}$, and $\mathbb M_\alpha(N)$ when only the cardinal $\alpha=|I|$ matters; $\ell^2(\alpha)$ is $\ell^2(I)$ for a set of that cardinality.

Lemma 8.2 (Matrix algebras).

$(N\otimes1)'=\mathbb M_I(N')$.
$\mathbb M_I(N)'=N'\otimes1$.
$\mathbb M_I(N)$ and $N\otimes1$ are von Neumann algebras, and $N\otimes1$ together with $1\otimes B(\ell^2(I))$ generates $\mathbb M_I(N)$ as a von Neumann algebra. We also write $N\bar\otimes B(\ell^2(I))$ for it.
The centre of $\mathbb M_I(N)$ is $(N\cap N')\otimes1$.

Proof. (1) $(T(x\otimes1))_{ij}=T_{ij}x$ and $((x\otimes1)T)_{ij}=xT_{ij}$. So $T$ commutes with $N\otimes1$ iff all its entries lie in $N'$.

(2) The operators $1\otimes E_{kl}$ lie in $\mathbb M_I(N)$, since their entries are $0$ or $1$. Suppose $S$ commutes with all of them. Comparing entries, $(S(1\otimes E_{kl}))_{ij}=\delta_{lj}S_{ik}$ and $((1\otimes E_{kl})S)_{ij}=\delta_{ik}S_{lj}$. Taking $j=l$ gives $S_{ik}=\delta_{ik}S_{ll}$ for all $i,k,l$. So $S=s\otimes1$ for a single $s\in B(K)$. If $S$ also commutes with every $x\otimes1$, $x\in N$, then $s\in N'$. Conversely, $s\otimes1$ with $s\in N'$ commutes with every $T\in\mathbb M_I(N)$, entry by entry.

(3) By (2), and then (1) for $N'$: $\mathbb M_I(N)''=(N'\otimes1)'=\mathbb M_I(N'')=\mathbb M_I(N)$. By (1), and then (2) for $N'$: $(N\otimes1)''=\mathbb M_I(N')'=N''\otimes1=N\otimes1$. The von Neumann algebra $R$ generated by $N\otimes1$ and $1\otimes B(\ell^2(I))$ lies in $\mathbb M_I(N)$. By the computation in (2), the commutant of $1\otimes B(\ell^2(I))$ is $B(K)\otimes1$, so $R'=\mathbb M_I(N')\cap(B(K)\otimes1)=N'\otimes1$, and $R=R''=\mathbb M_I(N)$.

(4) The centre is $\mathbb M_I(N)\cap(N'\otimes1)=\{s\otimes1:\ s\in N\cap N'\}$. $\square$

Definition 8.3 (Matrix units). A matrix unit in $M$ indexed by $I$ is a family $\{w_{ij}\}_{i,j\in I}\subseteq M$ with $w_{ij}^*=w_{ji}$, $w_{ij}w_{kl}=\delta_{jk}w_{il}$, and $\sum_iw_{ii}=1$.

Proposition 8.4 (Splitting off a matrix algebra). Let $\{e_i\}_{i\in I}$ be mutually orthogonal, mutually equivalent projections with $\sum_ie_i=1$. Fix $i_0\in I$, put $e=e_{i_0}$, and choose $u_i$ implementing $e\sim e_i$, with $u_{i_0}=e$. Then $w_{ij}=u_iu_j^*$ is a matrix unit, and $W:eH\otimes\ell^2(I)\to H$, $W(\zeta\otimes\delta_i)=u_i\zeta$, is a unitary with $\begin{gathered} W^*MW\\ =\mathbb M_I(eMe)\\ =eMe\bar\otimes B(\ell^2(I)),\\ W^*M'W\\ =M'e\otimes1 . \end{gathered}$ So $\{M,H\}\cong\{eMe,eH\}\otimes\{B(\ell^2(I)),\ell^2(I)\}$.

Proof. We have $u_j^*u_k=u_j^*e_je_ku_k=0$ for $j\ne k$, $u_j^*u_j=e$, and $u_ie=u_i$. Hence $w_{ij}^*=w_{ji}$, $w_{ij}w_{kl}=u_iu_j^*u_ku_l^*=\delta_{jk}u_ieu_l^*=\delta_{jk}w_{il}$, and $w_{ii}=e_i$. For $\zeta,\zeta'\in eH$, $\langle u_i\zeta,u_k\zeta'\rangle=\langle u_k^*u_i\zeta,\zeta'\rangle=\delta_{ik}\langle\zeta,\zeta'\rangle$, so $W$ is isometric. Its range contains every $u_i(eH)=e_iH$, so $W$ is onto. Since $W^*\eta=\sum_iu_i^*\eta\otimes\delta_i$, the entries of $W^*xW$ are $u_i^*xu_j|_{eH}$. For $x\in M$ they lie in $eMe$, so $W^*MW\subseteq\mathbb M_I(eMe)$. For $x'\in M'$ they are $x'u_i^*u_j|_{eH}=\delta_{ij}x'e|_{eH}$, so $W^*M'W=M'e\otimes1$. By Fact 2.1, $M'e=(eMe)'$ on $eH$, and by Lemma 8.2(2), $W^*M'W=(eMe)'\otimes1=\mathbb M_I(eMe)'$. Hence $\begin{gathered} W^*MW\\ =(W^*M'W)'\\ =\mathbb M_I(eMe)''\\ =\mathbb M_I(eMe), \end{gathered}$ where the last step is Lemma 8.2(3). $\square$

Lemma 8.5 (Equivalent projections have isomorphic corners). If $u$ implements $e\sim f$, then $x\mapsto uxu^*$ is an isomorphism of $eMe$ onto $fMf$, implemented by the unitary $u|_{eH}:eH\to fH$. So $\{eMe,eH\}\cong\{fMf,fH\}$.

Proof. $u(eMe)u^*\subseteq fMf$ and $u^*(fMf)u\subseteq eMe$, and the two maps are inverse to each other. For $\zeta\in eH$ and $x\in eMe$, $(uxu^*)(u\zeta)=ux\zeta$. $\square$

9. Comparing and counting abelian projections

An abelian projection is subequivalent to every projection whose central support is at least as large. So two abelian projections with the same central support are equivalent, and the number of such projections needed to fill a central projection is well defined. Matrix algebras over commutative algebras supply the examples and the counting tool.

Proposition 9.1 (Matrices over a commutative algebra). Let $A$ be a commutative von Neumann algebra acting on $K$.

For every nonzero Hilbert space $L$, $A\bar\otimes B(L)$ is of type I. For a rank-one projection $q$ on $L$, $1\otimes q$ is abelian with central support $1$.
For finite $n$, the matrix algebra $\mathbb M_n(A)$ of (8.1) is finite and of type I. The linear map $\Phi(x)=\frac1n\sum_ix_{ii}$, with values in $A$ (identified with the diagonal copy $A\otimes1$), satisfies $\begin{gathered} \Phi(a)\\ =a,\\ \Phi(axb)\\ =a\Phi(x)b\ \ (a,b\in A),\\ \Phi(x^*x)\\ =\Phi(xx^*)\\ \ge0, \end{gathered}$ and $\Phi(x^*x)=0$ only if $x=0$.

Proof. (1) Choose an orthonormal basis of $L$ that contains a unit vector of $qL$; then $L=\ell^2(I)$ and $q=E_{i_0i_0}$. For $T\in\mathbb M_I(A)$, $(1\otimes q)T(1\otimes q)=T_{i_0i_0}\otimes q$, so $(1\otimes q)\mathbb M_I(A)(1\otimes q)=A\otimes q$ is commutative. The centre is $A\otimes1$ by Lemma 8.2(4). If $(a\otimes1)(1\otimes q)=a\otimes q=0$, then $a=0$; so no nonzero central projection is orthogonal to $1\otimes q$, and $c(1\otimes q)=1$. Every nonzero central projection $z$ therefore majorizes the nonzero abelian projection $z(1\otimes q)$.

(2) $\mathbb M_n(A)$ is of type I by (1). The first two properties hold because $a\in A$ acts as $\mathrm{diag}(a,\ldots,a)$ and $A$ is commutative, so $(axb)_{ii}=ax_{ii}b$. Next, $n\Phi(x^*x)=\sum_{i,k}x_{ki}^*x_{ki}$ and $n\Phi(xx^*)=\sum_{i,k}x_{ik}x_{ik}^*$. These agree, since every entry is a normal operator ($A$ is commutative), and they are positive. If $\Phi(x^*x)=0$, each $x_{ki}^*x_{ki}$ vanishes, so $x=0$. Finally, if $v^*v=1$ and $vv^*=f$, then $\Phi(1-f)=\Phi(v^*v)-\Phi(vv^*)=0$ and $1-f=(1-f)^*(1-f)$, so $f=1$: the algebra is finite. $\square$

Definition 9.2 (Algebra-valued traces). Let $A$ be a $*$-subalgebra of $M$. An $A$-valued trace on $M$ is a linear map $\Phi:M\to A$ with $\Phi(a)=a$, $\Phi(axb)=a\Phi(x)b$ for $a,b\in A$, and $\Phi(x^*x)=\Phi(xx^*)\ge0$ for $x\in M$. It is faithful if $\Phi(x^*x)=0$ only for $x=0$. When $A$ is the centre of $M$, $\Phi$ is a centre-valued trace.

An $A$-valued trace is positive, since every positive element of $M$ has the form $x^*x$; hence it is monotone. It is constant on equivalence classes: if $u$ implements $e\sim f$, then $\Phi(e)=\Phi(u^*u)=\Phi(uu^*)=\Phi(f)$. Centre-valued traces on finite algebras are the subject of the lesson Traces on von Neumann algebras.

Lemma 9.3 (Comparing abelian projections). Let $e$ be abelian and $f\in\mathcal P(M)$ with $c(e)\le c(f)$ (equivalently, $e\le c(f)$). Then $e\precsim f$. If $f$ is also abelian and $c(e)=c(f)$, then $e\sim f$.

The direction of subequivalence matters: Remark 9.4 gives the rank-one matrix test.

Proof. First, if $f\precsim e$ and $c(e)\le c(f)$, then $f\sim e$. Indeed $f\sim e_1\le e$, where $e_1=c(e_1)e$ by Lemma 6.2(4) and $c(e_1)=c(f)\ge c(e)$ by Proposition 3.5(2); so $e_1\ge c(e)e=e$, and $e_1=e$. Now take $z$ from the comparison theorem (Theorem 5.5) with $ze\precsim zf$ and $(1-z)f\precsim(1-z)e$. The projection $(1-z)e$ is abelian (Lemma 6.2(1)), and by Proposition 3.5(1), $\begin{gathered} c((1-z)e)\\ =(1-z)c(e)\\ \le(1-z)c(f)\\ =c((1-z)f). \end{gathered}$ By the first step, $(1-z)f\sim(1-z)e$. Adding (Lemma 3.3(3)), $e=ze+(1-z)e\precsim zf+(1-z)f=f$. If $f$ is abelian with $c(f)=c(e)$, exchange the roles and use Proposition 5.1. $\square$

Remark 9.4. The relation in Lemma 9.3 cannot be reversed. In $M=M_2(\mathbb C)$, the projection $e=E_{11}$ is abelian and $c(e)=1=c(1)$, but $1\precsim E_{11}$ fails, since $1$ is not equivalent to any projection of rank one.

Lemma 9.5 (Counting abelian projections). Let $c\ne0$ be a central projection. Let $\{e_i\}_{i\in I}$ be mutually orthogonal abelian projections with $c(e_i)=c$ and $\sum_ie_i=c$. Let $\{f_j\}_{j\in J}$ be mutually orthogonal abelian projections with $c(f_j)=c$ and $\sum_jf_j\le c$. Then $|J|\le|I|$. Consequently, if also $\sum_jf_j=c$, then $|I|=|J|$.

Proof. Replacing $M$ by $Mc$, we may assume $c=1$. By Lemma 9.3, all the $e_i$ and $f_j$ are equivalent to one another.

$I$ finite, $|I|=n$. By Proposition 8.4, $W^*MW=\mathbb M_n(A)$ with $A=e_{i_0}Me_{i_0}$ commutative, and $W^*e_iW=1\otimes E_{ii}$. Carry the trace $\Phi$ of Proposition 9.1(2) over to $M$. Then $\Phi(e_i)=\frac1n1$, and $\Phi(f_j)=\Phi(e_{i_0})=\frac1n1$ because $f_j\sim e_{i_0}$. For a finite set $F\subseteq J$, $\sum_{j\in F}f_j\le1$, and $\Phi$ is monotone, so $\frac{|F|}n1\le1$ and $|F|\le n$. Hence $|J|\le n$.

$I$ infinite. Fix $i_0$, a unit vector $\zeta\in e_{i_0}H$, and partial isometries $v_i$ implementing $e_{i_0}\sim e_i$. Put $\varphi_i=\omega_{v_i\zeta}$ and $z_0=c(p_\zeta)$, which is nonzero because $p_\zeta\zeta=\zeta\ne0$. Since $p_\zeta\le e_{i_0}$ (Lemma 4.6) and $e_{i_0}$ is abelian, $p_\zeta=z_0e_{i_0}$ (Lemma 6.2(4)). Suppose $\varphi_i(f_j)=\|f_jv_i\zeta\|^2=0$. Then the operator $f_jv_i\in M$ vanishes on $[M'\zeta]=p_\zeta H$, so $f_jv_iz_0e_{i_0}=0$. As $v_i=v_ie_{i_0}$, this gives $z_0f_jv_i=0$, and so $z_0f_je_i=z_0f_jv_iv_i^*=0$. Now fix $j$. Since $c(f_j)=1$, we have $z_0f_j\ne0$ (Proposition 3.5(1)), while $\sum_iz_0f_je_i=z_0f_j$. So $z_0f_je_i\ne0$, and hence $\varphi_i(f_j)>0$, for some $i$. Thus $J=\bigcup_iJ_i$ with $J_i=\{j:\varphi_i(f_j)>0\}$. Each $J_i$ is countable, because $\sum_{j\in F}\varphi_i(f_j)=\varphi_i(\sum_{j\in F}f_j)\le1$ for every finite $F\subseteq J$. Therefore $|J|\le|I|\cdot\aleph_0=|I|$ (Fact 2.9).

The last statement follows by symmetry and the Cantor–Schröder–Bernstein theorem for cardinals. $\square$

10. Homogeneous algebras and the structure of type I algebras

Definition 10.1 (Homogeneous algebras). Let $\alpha$ be a nonzero cardinal. A central projection $z$ is $\alpha$-homogeneous if $z=\sum_{i\in I}e_i$ for mutually orthogonal abelian projections $e_i$ with $c(e_i)=z$ and $|I|=\alpha$. A type I algebra is of type I$_\alpha$ if $1$ is $\alpha$-homogeneous.

Lemma 10.2.

A central projection below an $\alpha$-homogeneous one is $\alpha$-homogeneous. The sum of mutually orthogonal $\alpha$-homogeneous central projections is $\alpha$-homogeneous.
A nonzero central projection is $\alpha$-homogeneous for at most one $\alpha$.
If $M$ is of type I, then below each nonzero central projection there is a nonzero central projection that is $\alpha$-homogeneous for some $\alpha$.
If $M\cong A\bar\otimes B(L)$ with $A$ abelian and $\dim L=\alpha$, then $1$ is $\alpha$-homogeneous in $M$.

Proof. (1) Let $z=\sum_ie_i$ as in the definition and let $z_1\le z$ be central. Then $z_1e_i$ is abelian, $c(z_1e_i)=z_1c(e_i)=z_1$, and $\sum_iz_1e_i=z_1$. For a sum $\sum_kz_k$, index all the families by one set $I$ with $|I|=\alpha$, and put $e_i=\sum_ke_i^{(k)}$. The summands have mutually orthogonal central supports $z_k$, so $e_i$ is abelian (Lemma 6.3), $c(e_i)=\sum_kz_k$, and $\sum_ie_i=\sum_kz_k$.

(2) This is Lemma 9.5 in $Mz$.

(3) Let $c\ne0$ be central. It majorizes a nonzero abelian projection $e$; put $c_1=c(e)\le c$. Take a maximal family $\{e_i\}_{i\in I}$ of mutually orthogonal abelian projections with $c(e_i)=c_1$ (it may start with $e$), and put $r=c_1-\sum_ie_i$. Suppose $c(r)=c_1$. Take a maximal family of mutually centrally orthogonal nonzero abelian projections below $r$, and let $p$ be its sum, which is abelian (Lemma 6.3). If $d=c(r)-c(p)$ were nonzero, then $dr\ne0$ (Proposition 3.5(1)), and by Lemma 7.4(1), $dr$ would majorize a nonzero abelian projection centrally orthogonal to that family, against maximality. So $c(p)=c(r)=c_1$. Then $p$ is an abelian projection with central support $c_1$, orthogonal to every $e_i$, against the maximality of $\{e_i\}$. Hence $z=c_1-c(r)\ne0$. As $zr=0$, $z=zc_1=\sum_ize_i$, where each $ze_i$ is abelian with $c(ze_i)=z$. So $z\le c$ is $|I|$-homogeneous.

(4) By Proposition 9.1(1), the projections $1\otimes E_{ii}$ for an orthonormal basis of $L$ are abelian, have central support $1$, are mutually orthogonal and add up to $1$. An isomorphism preserves all of this, including the sum (it preserves the order of projections, hence least upper bounds). $\square$

Theorem 10.3 (Structure of type I algebras). Let $M$ be of type I. For each nonzero cardinal $\alpha$ there is a largest $\alpha$-homogeneous central projection $z_\alpha$. The $z_\alpha$ are mutually orthogonal, $\sum_\alpha z_\alpha=1$, and $z_\alpha=0$ unless $\alpha\le\dim H$. For each $\alpha$ with $z_\alpha\ne0$, $\{Mz_\alpha,z_\alpha H\}\cong\{A_\alpha\bar\otimes B(\ell^2(\alpha)),\,K_\alpha\otimes\ell^2(\alpha)\},$ where $A_\alpha$ is commutative and isomorphic to $Zz_\alpha$. Hence $M\cong\bigoplus_\alpha A_\alpha\bar\otimes B(\ell^2(\alpha))$. The family $\{z_\alpha\}$ is unique: if $\{c_\alpha\}$ are mutually orthogonal central projections with $\sum_\alpha c_\alpha=1$ and $Mc_\alpha\cong B_\alpha\bar\otimes B(\ell^2(\alpha))$ with $B_\alpha$ abelian, then $c_\alpha=z_\alpha$ for every $\alpha$. Finally, $M$ is finite if and only if $z_\alpha=0$ for every infinite $\alpha$.

The sum and uniqueness are proved explicitly below.

Proof. Set of indices. Every nonzero $\alpha$-homogeneous projection is a sum of $\alpha$ nonzero orthogonal projections. Choosing a unit vector in each range gives an orthonormal set, so $\alpha\le\dim H$ by the Hilbert basis theorem. Consequently all nonzero terms in the sum are indexed by the set of nonzero cardinals at most $\dim H$; for larger cardinals the largest homogeneous projection is zero. All sums below are over this set.

Largest homogeneous projections. Take a maximal family of mutually orthogonal nonzero $\alpha$-homogeneous central projections, and let $z_\alpha$ be its sum. It is $\alpha$-homogeneous by Lemma 10.2(1). If $z$ is $\alpha$-homogeneous and $z(1-z_\alpha)\ne0$, then $z(1-z_\alpha)$ is $\alpha$-homogeneous and orthogonal to the family, against maximality. So $z\le z_\alpha$.

Orthogonality and sum. For $\alpha\ne\beta$, $z_\alpha z_\beta$ is both $\alpha$- and $\beta$-homogeneous (Lemma 10.2(1)), so it is $0$ (Lemma 10.2(2)). If $1-\sum_\alpha z_\alpha\ne0$, Lemma 10.2(3) gives a nonzero $\gamma$-homogeneous $z\le1-\sum_\alpha z_\alpha$. But $z\le z_\gamma$, so $z=zz_\gamma=0$, a contradiction. If $z_\alpha\ne0$, it is a sum of $\alpha$ mutually orthogonal nonzero projections, so $\alpha\le\dim H$.

Structure. Write $z_\alpha=\sum_{i\in I}e_i$ as in the definition. The $e_i$ are mutually equivalent by Lemma 9.3, so Proposition 8.4 in $Mz_\alpha$ gives $\{Mz_\alpha,z_\alpha H\}\cong\{A_\alpha\bar\otimes B(\ell^2(I)),K_\alpha\otimes\ell^2(I)\}$ with $A_\alpha=e_{i_0}Me_{i_0}$ on $K_\alpha=e_{i_0}H$. This algebra is commutative, and by Lemma 6.2(4) and Proposition 3.5(4) it equals $Ze_{i_0}\cong Zc(e_{i_0})=Zz_\alpha$.

Uniqueness. By Lemma 10.2(4), carried through the isomorphism $Mc_\alpha\cong B_\alpha\bar\otimes B(\ell^2(\alpha))$, the projection $c_\alpha$ is $\alpha$-homogeneous, so $c_\alpha\le z_\alpha$. Then $z_\alpha=\sum_\beta z_\alpha c_\beta=z_\alpha c_\alpha\le c_\alpha$, because $z_\alpha c_\beta\le z_\alpha z_\beta=0$ for $\beta\ne\alpha$.

Finiteness. Let $\alpha$ be infinite with $z_\alpha\ne0$. Choose $i_0\in I$ and a bijection $\sigma:I\to I\setminus\{i_0\}$. Additivity gives $z_\alpha=\sum_ie_i\sim\sum_ie_{\sigma(i)}=z_\alpha-e_{i_0}<z_\alpha$, so $z_\alpha$ is infinite, and $M$ is not finite. Conversely, if $z_\alpha=0$ for every infinite $\alpha$, each $Mz_n\cong\mathbb M_n(A_n)$ is finite (Proposition 9.1(2)), and $1=\sum_nz_n$ is a sum of centrally orthogonal finite projections, which is finite (Lemma 6.3). $\square$

Corollary 10.4 (Type I factors). Every type I factor is isomorphic to $B(L)$ for some Hilbert space $L$, and $B(L_1)\cong B(L_2)$ if and only if $\dim L_1=\dim L_2$.

Proof. The central projections are $0$ and $1$, so $1=z_\alpha$ for exactly one $\alpha$. In a factor, a nonzero abelian $e$ satisfies $eMe=Ze=\mathbb Ce$, so $e$ is minimal. With Lemma 6.2(2), the abelian projections are exactly the minimal ones (and $0$). So $A_\alpha=e_{i_0}Me_{i_0}=\mathbb C$ and $M\cong B(\ell^2(\alpha))$. If $B(L_1)\cong B(L_2)$, the isomorphism carries the rank-one projections of an orthonormal basis of $L_1$, which are mutually orthogonal abelian projections with central support $1$ and sum $1$, to a family of the same kind in $B(L_2)$ of cardinality $\dim L_1$. Lemma 9.5 compares it with the family given by a basis of $L_2$, so $\dim L_1=\dim L_2$. The converse is clear. $\square$

Remark 10.5. By Corollary 10.4, a type I$_n$ factor with $n$ finite is a copy of the matrix algebra $M_n(\mathbb C)$.

Example 10.6 (A type decomposition). Let $M=\mathbb C\oplus M_2(\mathbb C)\oplus B(\ell^2(\mathbb N))\oplus\mathbb M_3(L^\infty[0,1])$ on $\mathbb C\oplus\mathbb C^2\oplus\ell^2(\mathbb N)\oplus(L^2[0,1]\otimes\mathbb C^3)$. It is of type I, with $z_1=1\oplus0\oplus0\oplus0$, $z_2=0\oplus1\oplus0\oplus0$, $z_3=0\oplus0\oplus0\oplus1$ and $z_{\aleph_0}=0\oplus0\oplus1\oplus0$ in Theorem 10.3. It is not finite, and $z_{\aleph_0}$ is the properly infinite part of $1$. For $e=0\oplus E_{11}\oplus p\oplus0$ with $p$ of infinite rank, the decomposition of Theorem 7.2 is $e_1=0\oplus E_{11}\oplus0\oplus0$ (finite) and $e_2=0\oplus0\oplus p\oplus0$ (properly infinite). If $p$ has finite rank, $e$ is finite.

Exercise 10.7. (medium) Let $M$ be of type I. Show that an automorphism $\alpha$ of $M$ that fixes every element of the centre is inner.

Solution. Let $e$ be abelian with $c(e)=1$ (Lemma 7.4(1)). The projection $\alpha(e)$ is abelian, since $\alpha$ maps $eMe$ onto $\alpha(e)M\alpha(e)$. It has central support $\alpha(c(e))=1$, because $\alpha$ maps the central projections above $e$ onto the central projections above $\alpha(e)$, preserving order, and fixes the centre. By Lemma 9.3, some $v$ implements $e\sim\alpha(e)$. For $y\in eMe=Ze$ (Lemma 6.2(4)), write $y=ae$ with $a\in Z$; then $\alpha(y)=a\alpha(e)=avv^*=v(ae)v^*=vyv^*$. So $\alpha$ agrees with $x\mapsto vxv^*$ on $eMe$, and Exercise 5.6 shows that $\alpha$ is inner. $\square$

11. Type I algebras and their commutants

A von Neumann algebra with a commutative commutant is of type I, and the commutant of a type I algebra is again of type I. With both algebras in view, a type I algebra has a spatial normal form.

Lemma 11.1 (Cyclic commutative algebras are maximal abelian). Let $A$ be a commutative von Neumann algebra on $K$, and $\xi\in K$ a cyclic vector for it. Then $A$ is maximal abelian: $A'=A$.

Proof. We know $A\subseteq A'$, and $A'$ is spanned by its positive contractions. So let $S\in A'$ with $0\le S\le1$; we show $S\in A$. Choose $a_n\in A$ with $a_n\xi\to S\xi$. For $b\in A$, $\begin{gathered} \langle a_n^*\xi,b\xi\rangle\\ =\langle\xi,a_nb\xi\rangle\\ =\langle\xi,ba_n\xi\rangle\to\langle\xi,bS\xi\rangle\\ =\langle\xi,Sb\xi\rangle\\ =\langle S\xi,b\xi\rangle , \end{gathered}$ and $\|a_n^*\xi\|=\|a_n\xi\|$ because $a_n$ is normal. So $a_n^*\xi\to S\xi$ weakly, with $\|a_n^*\xi\|\to\|S\xi\|$, hence in norm. The self-adjoint elements $h_n=\frac12(a_n+a_n^*)\in A$ satisfy $h_n\xi\to S\xi$. Let $g(t)=\min(\max(t,0),1)$. Then $|g(s)-g(t)|\le|s-t|$ and $g(S)=S$. The operators $h_n$ and $S$ commute, since $S\in A'$, so they lie in a commutative C$^*$-algebra $C(\Omega)$, where $\begin{gathered} (g(h_n)-g(S))^*(g(h_n)-g(S))\\ \le(h_n-S)^*(h_n-S) \end{gathered}$ holds pointwise. Therefore $\|(g(h_n)-S)\xi\|\le\|(h_n-S)\xi\|\to0$. The operators $b_n=g(h_n)$ lie in $A$, $\|b_n\|\le1$, and $b_n\xi\to S\xi$. For $c\in A$, $b_nc\xi=cb_n\xi\to cS\xi=Sc\xi$. So $b_n\to S$ strongly on the dense set $A\xi$, and, being bounded, strongly on $K$. As $A$ is strongly closed, $S\in A$. $\square$

Proposition 11.2 (Type I through the commutant). $M$ is of type I exactly when some faithful normal representation $\pi$ of $M$ has a commutative commutant $\pi(M)'$.

Proof. "⇐". By Fact 2.6, $\pi(M)$ is a von Neumann algebra isomorphic to $M$, and types are invariant under isomorphism. So we may assume that $M$ acts on $H$ with $M'$ commutative. Then $M'\subseteq M''=M$, so $M'=Z$. For $\xi\in H$, $p_\xi$ is the projection onto $[Z\xi]$. The induced algebra $M'p_\xi=Zp_\xi$ on $p_\xi H$ is commutative and has the cyclic vector $\xi$, so it is maximal abelian by Lemma 11.1. By Fact 2.1 its commutant on $p_\xi H$ is $p_\xi Mp_\xi$. Hence $p_\xi Mp_\xi=Zp_\xi$ is commutative: $p_\xi$ is abelian. If $p\in\mathcal P(M)$ is nonzero, pick a nonzero $\xi\in pH$; then $0\ne p_\xi\le p$ (Lemma 4.6). So every nonzero projection majorizes a nonzero abelian one, and $M$ is of type I.

"⇒". By Corollary 11.3 below, whose proof uses only "⇐", the commutant $M'$ is of type I. So $M'$ has an abelian projection $e'$ with central support $1$ in $M'$ (Lemma 7.4(1)). By Proposition 3.5(3), $x\mapsto xe'|_{e'H}$ is a $*$-homomorphism of $M$ onto the induced algebra $Me'$, and it is injective, because its kernel is $M(1-c(e'))=\{0\}$. It is normal, being a restriction. By Fact 2.1, the commutant of $Me'$ on $e'H$ is $e'M'e'$, which is commutative. $\square$

Corollary 11.3. If $M$ is of type I, so is $M'$.

Proof. Let $e\in M$ be abelian with $c(e)=1$ (Lemma 7.4(1)). By Proposition 3.5(3), the induction $x'\mapsto x'e|_{eH}$ is an isomorphism of $M'$ onto $M'e$, a von Neumann algebra on $eH$ whose commutant $eMe$ is commutative (Fact 2.1). By "⇐" of Proposition 11.2, applied to the identity representation of $M'e$, the algebra $M'e$ is of type I, and so is $M'\cong M'e$. $\square$

Theorem 11.4 (The spatial form of a type I algebra). Let $M$ be of type I on $H$. By Corollary 11.3, $M'$ is of type I as well, and its centre is $Z$, so Theorem 10.3 applies to both algebras. Let $z_\alpha$ be the largest $\alpha$-homogeneous central projection of $M$ and $z'_\beta$ the largest $\beta$-homogeneous central projection of $M'$. Put $z_{\alpha,\beta}=z_\alpha z'_\beta$. These central projections are mutually orthogonal with sum $1$. For each $(\alpha,\beta)$ with $z_{\alpha,\beta}\ne0$ there are a Hilbert space $K_{\alpha,\beta}$, a maximal abelian algebra $A_{\alpha,\beta}$ on it, and a unitary $W:K_{\alpha,\beta}\otimes\ell^2(\alpha)\otimes\ell^2(\beta)\to z_{\alpha,\beta}H$ such that $W^*(Mz_{\alpha,\beta})W=\mathbb M_\alpha(A_{\alpha,\beta})\otimes1_{\ell^2(\beta)}$, and $W^*(M'z_{\alpha,\beta})W$ is the algebra of operators whose matrix entries with respect to $\ell^2(\beta)$ lie in $A_{\alpha,\beta}\otimes1_{\ell^2(\alpha)}$. That is, $\begin{gathered} \{M,H\}\\ \cong\bigoplus_{\alpha,\beta}\{A_{\alpha,\beta},K_{\alpha,\beta}\}\\ \otimes\{B(\ell^2(\alpha)),\ell^2(\alpha)\}\\ \otimes\{\mathbb C,\ell^2(\beta)\}, \end{gathered}$ $\begin{gathered} \{M',H\}\\ \cong\bigoplus_{\alpha,\beta}\{A_{\alpha,\beta},K_{\alpha,\beta}\}\\ \otimes\{\mathbb C,\ell^2(\alpha)\}\\ \otimes\{B(\ell^2(\beta)),\ell^2(\beta)\}, \end{gathered}$ with the same unitary. The decomposition is unique: the projections $z_{\alpha,\beta}$ are determined by $M$, and each $\{A_{\alpha,\beta},K_{\alpha,\beta}\}$ is determined up to spatial isomorphism.

Proof. The $z_{\alpha,\beta}$ are central and mutually orthogonal, and $\sum_{\alpha,\beta}z_{\alpha,\beta}=(\sum_\alpha z_\alpha)(\sum_\beta z'_\beta)=1$. Fix $(\alpha,\beta)$ with $z=z_{\alpha,\beta}\ne0$. Cutting the homogeneous decompositions by $z$ (Lemma 10.2(1)), $z=\sum_{i\in I}e_i$ with abelian $e_i\in M$, $c(e_i)=z$, $|I|=\alpha$, and $z=\sum_{j\in J}e'_j$ with $e'_j\in M'$ abelian in $M'$, of central support $z$, $|J|=\beta$. Fix $i_0$ and $j_0$, put $e=e_{i_0}$ and $e'=e'_{j_0}$, and choose $u_i\in M$ implementing $e\sim e_i$ and $v_j\in M'$ implementing $e'\sim e'_j$ (Lemma 9.3 in $M$ and in $M'$). Let $K=ee'H$; the projections $e$ and $e'$ commute.

The algebra $A$. Let $A=\{ye'|_K:\ y\in eMe\}$. This is the induced algebra of the reduced algebra $eMe$ (on $eH$) by the projection $e'|_{eH}$, which lies in its commutant $M'e$. By Fact 2.1, used twice, $A$ is a von Neumann algebra acting on $K$, with commutant $\{y'e|_K:\ y'\in e'M'e'\}$. Both are images of commutative algebras, $eMe$ and $e'M'e'$, under $*$-homomorphisms, so both are commutative. Hence $A\subseteq A'\subseteq A''=A$: the algebra $A$ is maximal abelian.

The unitary. Put $W(\zeta\otimes\delta_i\otimes\delta_j)=u_iv_j\zeta$ for $\zeta\in K$. The $u$'s commute with the $v$'s, $u_k^*u_i=\delta_{ik}e$, and $v_l^*v_j=\delta_{jl}e'$. So $\langle u_iv_j\zeta,u_kv_l\zeta'\rangle=\langle u_k^*u_iv_l^*v_j\zeta,\zeta'\rangle=\delta_{ik}\delta_{jl}\langle\zeta,\zeta'\rangle ,$ and $W$ is isometric. The partial isometry $u_iv_j$ has initial projection $ee'$ and final projection $e_ie'_j$, so $u_iv_jK=e_ie'_jH$. As $\sum_{i,j}e_ie'_j=z$, $W$ is a unitary onto $zH$.

The two algebras. Let $x\in M$ and $\zeta\in K$. Then $xu_kv_l\zeta=\sum_{i,j}e_ie'_jxu_kv_l\zeta$. Here $e'_jxu_kv_l\zeta=xu_ke'_jv_l\zeta=\delta_{jl}xu_kv_l\zeta$, and $e_ixu_kv_l\zeta=u_iv_l(u_i^*xu_k)\zeta$ with $u_i^*xu_k\in eMe$. Hence $\begin{gathered} W^*xW(\zeta\otimes\delta_k\otimes\delta_l)\\ =\sum_i(u_i^*xu_k)e'\zeta\otimes\delta_i\otimes\delta_l , \end{gathered}$ so $W^*(Mz)W$ lies in $R_1=\mathbb M_I(A)\otimes1_{\ell^2(J)}$. In the same way, for $y'\in M'$, $W^*y'W(\zeta\otimes\delta_k\otimes\delta_l)=\sum_j(v_j^*y'v_l)e\zeta\otimes\delta_k\otimes\delta_j$, so $W^*(M'z)W$ lies in the algebra $R_2$ of operators whose $\ell^2(J)$-entries lie in $A'\otimes1_{\ell^2(I)}=A\otimes1_{\ell^2(I)}$. By Lemma 8.2(2), $\mathbb M_I(A)'=A'\otimes1_{\ell^2(I)}$, and then Lemma 8.2(1) for the amplification by $\ell^2(J)$ gives $R_1'=R_2$. Since $Mz$ and $M'z$ are each other's commutants on $zH$, $\begin{gathered} W^*(Mz)W\\ =\bigl(W^*(M'z)W\bigr)'\\ \supseteq R_2'\\ =R_1''\\ =R_1\\ \supseteq W^*(Mz)W . \end{gathered}$ So $W^*(Mz)W=R_1$ and $W^*(M'z)W=R_1'=R_2$.

Uniqueness. The $z_\alpha$ are determined by $M$ and the $z'_\beta$ by $M'$ (Theorem 10.3 for each), so the $z_{\alpha,\beta}$ are determined by $M$. Let $\{c_{\alpha,\beta}\}$ be the central projections of the summands in a second decomposition of this form. Then $Mc_\alpha$, with $c_\alpha=\sum_\beta c_{\alpha,\beta}$, is isomorphic to an abelian algebra tensor $B(\ell^2(\alpha))$, so $c_\alpha=z_\alpha$ by the uniqueness in Theorem 10.3. In the same way $\sum_\alpha c_{\alpha,\beta}=z'_\beta$ (Theorem 10.3 in $M'$), and therefore $c_{\alpha,\beta}=z_\alpha z'_\beta=z_{\alpha,\beta}$. Next, let $e_1,e_2\in M$ be abelian with central support $z$, and $e'_1,e'_2\in M'$ abelian in $M'$ with central support $z$. Choose $u\in M$ implementing $e_1\sim e_2$ and $v\in M'$ implementing $e'_1\sim e'_2$ (Lemma 9.3). The unitary $uv:e_1e'_1H\to e_2e'_2H$ carries $ye'_1$ to $(uyu^*)e'_2$ for $y\in e_1Me_1$. So the algebra $A$ does not depend on the choices, up to spatial isomorphism. Finally, in a model $\mathbb M_\alpha(B)\otimes1$ on $L\otimes\ell^2(\alpha)\otimes\ell^2(\beta)$ with $\{B,L\}$ maximal abelian and the commutant as described, the projection $1\otimes E_{11}\otimes1$ is abelian with central support $1$, and so is $1\otimes1\otimes E_{11}$ in the commutant (Proposition 9.1(1) for each). The algebra built from this pair is $\{B,L\}$ itself. So any such model has $\{B,L\}$ spatially isomorphic to $\{A_{\alpha,\beta},K_{\alpha,\beta}\}$. $\square$

Corollary 11.5 (Commutative von Neumann algebras and multiplicity). A commutative von Neumann algebra $\{A,H\}$ is spatially isomorphic to $\bigoplus_\beta\{A_\beta\otimes1,K_\beta\otimes\ell^2(\beta)\}$ with each $\{A_\beta,K_\beta\}$ maximal abelian. The central projections of the summands are unique, and each $\{A_\beta,K_\beta\}$ is unique up to spatial isomorphism.

Proof. $A$ is of type I, and $1$ is abelian with central support $1$, so $1$ is $1$-homogeneous: $z_1=1$ and $z_{1,\beta}=z'_\beta$. Apply Theorem 11.4 with $\alpha=1$, where $B(\ell^2(1))=\mathbb C$. The cardinal $\beta$ is the multiplicity of the summand: the homogeneity of $A'$ there. $\square$

Example 11.6 (Multiplicity $n$). For $A=L^\infty[0,1]\otimes1$ on $L^2[0,1]\otimes\mathbb C^n$, Lemma 8.2(1) gives $A'=\mathbb M_n(L^\infty[0,1])$, because $L^\infty[0,1]$ is maximal abelian on $L^2[0,1]$ (Lemma 11.1, with the cyclic vector $1$). This algebra is of type I$_n$ (Proposition 9.1(1) and Lemma 10.2(4)). So in Corollary 11.5 only $\beta=n$ occurs: the multiplicity is $n$.

12. The algebra generated by two projections

Let $e$ and $f$ be projections on $H$, and $M=\{e,f\}''$ the von Neumann algebra they generate. This section describes $M$ completely. It is the direct sum of a commutative algebra of dimension at most $4$ and an algebra of type I$_2$. On the second summand the pair has a normal form given by two commuting operators, which play the parts of a cosine and a sine.

Lemma 12.1 (The diagonal part). The four projections $\begin{gathered} p_{11}\\ =e\wedge f,\\ p_{10}\\ =e\wedge f^\perp,\\ p_{01}\\ =e^\perp\wedge f,\\ p_{00}\\ =e^\perp\wedge f^\perp \end{gathered}$ lie in $M$, are mutually orthogonal, and are central in $M$. Let $z=1-(p_{11}+p_{10}+p_{01}+p_{00})$, $H_0=zH$, $e_0=ez$ and $f_0=fz$.

$e=p_{11}+p_{10}+e_0$ and $f=p_{11}+p_{01}+f_0$. In particular $e$ commutes with $e\wedge f+e^\perp\wedge f$, and $f$ commutes with $e\wedge f+e\wedge f^\perp$.
(Generic position) On $H_0$, with complements taken in $H_0$: $\begin{gathered} e_0\wedge f_0\\ =e_0\wedge f_0^\perp\\ =e_0^\perp\wedge f_0\\ =e_0^\perp\wedge f_0^\perp\\ =0. \end{gathered}$ Hence all four joins $e_0\vee f_0$, $e_0^\perp\vee f_0$, $e_0\vee f_0^\perp$, $e_0^\perp\vee f_0^\perp$ equal $z$.
$M(1-z)$, on $(1-z)H$, is the linear span of the nonzero $p_{ab}$; it is commutative and has dimension at most $4$. And $Mz$, on $H_0$, equals $\{e_0,f_0\}''$.

Proof. The $p_{ab}$ lie in $M$ by Proposition 3.1. Each lies below or is orthogonal to $e$, and likewise for $f$, so it commutes with $e$ and $f$, hence with $M=\{e,f\}''$; so it is central. Two different $p_{ab}$ differ in at least one index and are then orthogonal.

(1) $ep_{11}=p_{11}$, $ep_{10}=p_{10}$ and $ep_{01}=ep_{00}=0$, so $\begin{gathered} e\\ =e(p_{11}+p_{10}+p_{01}+p_{00})+ez\\ =p_{11}+p_{10}+e_0. \end{gathered}$ The formula for $f$ is the same. So $e$ commutes with $p_{11}+p_{01}$, and $f$ with $p_{11}+p_{10}$.

(2) A vector of $e_0H_0\cap f_0H_0$ lies in $eH\cap fH=p_{11}H$, which is orthogonal to $H_0$; so it is $0$. The other three cases are the same, with $p_{10}$, $p_{01}$ and $p_{00}$. The joins follow by De Morgan's laws in $H_0$.

(3) The span $S$ of the nonzero $p_{ab}$ is a commutative $*$-algebra on $(1-z)H$ spanned by orthogonal projections with sum $1-z$; its commutant consists of the operators that commute with each $p_{ab}$, and $S$ equals its own bicommutant. $S$ contains $e(1-z)=p_{11}+p_{10}$ and $f(1-z)=p_{11}+p_{01}$, and each $p_{ab}$ is a product of $e(1-z)$ or its complement with $f(1-z)$ or its complement; so $S=\{e(1-z),f(1-z)\}''$. Now $Mz\supseteq\{e_0,f_0\}''$, because $Mz$ acts on $H_0$ as a von Neumann algebra (Fact 2.1) and contains $e_0$ and $f_0$. Conversely, if $y\in B(H_0)$ commutes with $e_0$ and $f_0$, then $y\oplus0$ on $H=H_0\oplus(1-z)H$ commutes with $e$ and $f$, hence with $M$, so $y$ commutes with $Mz$. Thus $Mz\subseteq\{e_0,f_0\}''$. The same argument gives $M(1-z)=\{e(1-z),f(1-z)\}''=S$. $\square$

Proposition 12.2 (A pair in generic position). Suppose $e\wedge f=e\wedge f^\perp=e^\perp\wedge f=e^\perp\wedge f^\perp=0$ (so $z=1$ above) and $H\ne0$. Let $M=\{e,f\}''$.

$e\sim f\sim e^\perp\sim f^\perp$ in $M$.
The operator $a=e^\perp fe$ is injective on $eH$ and has dense range in $e^\perp H$. In its polar decomposition $a=u|a|$, $u\in M$, $u^*u=e$ and $uu^*=e^\perp$.
The map $W:eH\otimes\mathbb C^2\to H$, $W(\zeta_1\otimes\delta_1+\zeta_2\otimes\delta_2)=\zeta_1+u\zeta_2$, is unitary, and $\begin{gathered} W^*eW\\ =\begin{pmatrix}1&0\\0&0\end{pmatrix},\\ W^*fW\\ =\begin{pmatrix}C^2&CS\\CS&S^2\end{pmatrix}, \end{gathered} \tag{12.3}$ where $C=(efe|_{eH})^{1/2}$ and $S=(1-C^2)^{1/2}$ are commuting, injective, positive contractions on $eH$ with $C^2+S^2=1$.
$W^*MW=\mathbb M_2(A)$ with $A=\{C^2\}''$ on $eH$, a commutative von Neumann algebra. So $M$ is of type I$_2$.
$W^*|e-f|W=S\otimes1$ and $W^*|e-f^\perp|W=C\otimes1$.

Proof. (1) Apply the parallelogram law (Proposition 4.4) to the pair $(f^\perp,e)$: $(f^\perp\vee e)-f^\perp\sim e-e\wedge f^\perp$. Here $f^\perp\vee e=(f\wedge e^\perp)^\perp=1$ and $e\wedge f^\perp=0$, so $f\sim e$. The pair $(e,f)$ gives $e^\perp=(e\vee f)-e\sim f-e\wedge f=f$. The pair $(e^\perp,f^\perp)$ gives $e=(e^\perp\vee f^\perp)-e^\perp\sim f^\perp-e^\perp\wedge f^\perp=f^\perp$.

(2) Let $\zeta\in eH$ with $e^\perp f\zeta=0$. Then $f\zeta\in eH$, so $\zeta-f\zeta\in eH\cap f^\perp H=\{0\}$, and $\zeta=f\zeta\in eH\cap fH=\{0\}$. The same argument, with $e$ and $e^\perp$ exchanged, shows that $a^*=efe^\perp$ is injective on $e^\perp H$. As $a$ vanishes on $e^\perp H$ and maps into $e^\perp H$, we get $\mathrm r(a)=e$ and $\mathrm l(a)=e^\perp$, and Fact 2.2 gives $u$.

(3) $u$ maps $eH$ onto $e^\perp H$, so $W$ is unitary. The entries of $W^*XW$ are $eXe$, $eXu$, $u^*Xe$ and $u^*Xu$, restricted to $eH$. For $X=e$ this gives the first matrix. For $X=f$, the $(2,1)$ entry is $u^*fe=u^*e^\perp fe=u^*u|a|=|a|$, using $u^*=u^*e^\perp$; the $(1,2)$ entry is its adjoint $|a|$; and the diagonal entries are $F_{11}=efe$ and $F_{22}=u^*fu$. Comparing entries in $f^2=f$ gives $\begin{gathered} F_{11}^2+|a|^2\\ =F_{11},\\ F_{11}|a|+|a|F_{22}\\ =|a|,\\ |a|^2+F_{22}^2\\ =F_{22}. \end{gathered}$ The first identity shows that $|a|^2$ commutes with $F_{11}$, so its square root $|a|$ does too; the third gives the same for $F_{22}$. The second then reads $|a|(F_{11}+F_{22}-1)=0$. Since $|a|$ is injective on $eH$ ($\ker|a|=\ker a$), $F_{11}+F_{22}=1$. So $F_{11}=C^2$, $F_{22}=1-C^2=S^2$, and $|a|=(F_{11}(1-F_{11}))^{1/2}=CS$. $C$ and $S$ are injective because $CS=|a|$ is.

(4) The entries of $W^*eW$ and $W^*fW$ lie in $A$, since $C$, $S$ and $CS$ are continuous functions of $C^2$. By Lemma 8.2(3), $\mathbb M_2(A)$ is a von Neumann algebra; it contains $W^*eW$ and $W^*fW$, and so it contains $W^*MW=\{W^*eW,W^*fW\}''$. Conversely, $W^*MW$ contains $W^*uW=1\otimes E_{21}$, hence all the $1\otimes E_{ij}$. Its corner $(1\otimes E_{11})W^*MW(1\otimes E_{11})$ acts on $eH\otimes\delta_1$ as a von Neumann algebra (Fact 2.1), and it contains $C^2\otimes E_{11}$, hence $A\otimes E_{11}$. So $a\otimes E_{ij}=(1\otimes E_{i1})(a\otimes E_{11})(1\otimes E_{1j})$ lies in $W^*MW$ for $a\in A$, and $W^*MW=\mathbb M_2(A)$. By Proposition 9.1(1), $1\otimes E_{11}$ and $1\otimes E_{22}$ are orthogonal abelian projections with central support $1$ and sum $1$, so $M$ is of type I$_2$.

(5) $W^*(e-f)W=\begin{pmatrix}S^2&-CS\\-CS&-S^2\end{pmatrix}$, whose square is $S^2\otimes1$, because $C$ and $S$ commute and $C^2+S^2=1$. Likewise $\begin{gathered} W^*(e-f^\perp)W\\ =W^*(e+f-1)W\\ =\begin{pmatrix}C^2&CS\\CS&-C^2\end{pmatrix}, \end{gathered}$ whose square is $C^2\otimes1$. Take positive square roots. $\square$

Theorem 12.4 (The algebra of two projections). Let $e$ and $f$ be projections on $H$, and $M=\{e,f\}''$.

$M$ is of type I.
There is exactly one central projection $z$ of $M$ such that $Mz$ is of type I$_2$ and $M(1-z)$ is commutative. Moreover $\dim M(1-z)\le4$.

Proof. Take $z$ from Lemma 12.1. By its part (3), $M(1-z)$ is commutative of dimension at most $4$. If $z\ne0$, the pair $e_0,f_0$ is in generic position on $H_0$ and $Mz=\{e_0,f_0\}''$, so $Mz$ is of type I$_2$ by Proposition 12.2. (If $z=0$, then $Mz=\{0\}$, which is of every type.) So $M$ is the direct sum of a commutative algebra and an algebra of type I$_2$, hence of type I. For uniqueness, let $z'$ be a second such projection. The algebra $Mz'(1-z)$ is a central summand of the type I$_2$ algebra $Mz'$ and of the commutative algebra $M(1-z)$. A nonzero algebra of type I$_2$ is not commutative: cutting $1=e_1+e_2$ (abelian, central support $1$, and $e_1\sim e_2$ by Lemma 9.3) by a nonzero central projection gives two orthogonal, nonzero, equivalent projections, while in a commutative algebra equivalent projections are equal. So $z'(1-z)=0$, $z'\le z$, and by symmetry $z'=z$. $\square$

Definition 12.5 (Sine and cosine). For projections $e$ and $f$, we call $\sin(e,f)=|e-f|$ the sine of the pair and $\cos(e,f)=|e-f^\perp|$ its cosine.

Remark 12.6. The identity $(e-f)^2+(e+f-1)^2=1$, which holds for any two projections, gives $\sin(e,f)^2+\cos(e,f)^2=1$. Moreover $(e-f)^2$ commutes with $e$ and with $f$ (both products equal $e-efe$ for $e$, and similarly for $f$), so $\sin(e,f)$ and $\cos(e,f)$ are central in $\{e,f\}''$. On the generic part they are $S\otimes1$ and $C\otimes1$, by Proposition 12.2(5). On $(1-z)H$ they are $p_{10}+p_{01}$ and $p_{11}+p_{00}$. If $H=\mathbb C^2$ and $eH$, $fH$ are lines at an angle $\theta$ with $0<\theta<\pi/2$, the pair is in generic position, $C=\cos\theta$ and $S=\sin\theta$. So $C$ and $S$ generalize $\cos\theta$ and $\sin\theta$ for a pair of lines.

Example 12.7 (Two lines in $\mathbb C^2$). Let $eH=\mathbb C\delta_1$ and $fH=\mathbb C(\cos\theta\,\delta_1+\sin\theta\,\delta_2)$ with $0<\theta<\pi/2$. The four corner projections vanish, so $z=1$ and $M=M_2(\mathbb C)$, of type I$_2$. Here $efe=\cos^2\theta\,e$, so in (12.3) $C=\cos\theta$ and $S=\sin\theta$ (scalars on the line $eH$), $|e-f|=\sin\theta\cdot1$ and $|e-f^\perp|=\cos\theta\cdot1$. For $\theta=0$ or $\theta=\pi/2$ the projections commute, $z=0$, and $M$ is commutative.

13. Orthogonal families of equivalent projections, and halving

This section shows when projections can be halved. In an algebra without a type I part every projection is the sum of two equivalent halves, and a properly infinite projection is the sum of two, or of countably many, copies of itself. The tool is a maximal family of mutually orthogonal, mutually equivalent projections.

Proposition 13.1 (Orthogonal families of equivalent projections). Let $\{e_i\}_{i\in I}$, with $I\ne\emptyset$, be mutually orthogonal, mutually equivalent, nonzero projections. There are a nonzero central projection $z$ and mutually orthogonal, mutually equivalent projections $\{f_j\}_{j\in J}$, with $J\supseteq I$, such that $f_i=ze_i$ for $i\in I$ and $f_0:=z-\sum_{j\in J}f_j\prec f_j\qquad\text{for every }j\in J .$ If $J$ is infinite, the $f_j$ can be replaced by mutually orthogonal, mutually equivalent projections with $f_i\sim ze_i$ for $i\in I$ and $f_0=0$. The $e_i$ must be nonzero: for a family of zero projections the conclusion fails (Example 16.1).

Nonzero projections are essential for the strict comparison conclusion, as Example 16.1 verifies.

Proof. Fix $i_1\in I$. By Zorn's lemma the family extends to a maximal family $\{e_j\}_{j\in J}$ of mutually orthogonal projections equivalent to $e_{i_1}$. Put $e_0=1-\sum_je_j$. The comparison theorem (Theorem 5.5), applied to $e_0$ and $e_{i_1}$, gives a central $z$ with $ze_0\precsim ze_{i_1}$ and $(1-z)e_{i_1}\precsim(1-z)e_0$. Central cuts preserve equivalence (Lemma 3.3(4)), so the same holds with any $e_j$ in place of $e_{i_1}$. If $z$ were $0$, then $e_{i_1}\precsim e_0$: some nonzero $g\le e_0$ would satisfy $g\sim e_{i_1}$, and adding $g$ would contradict maximality. So $z\ne0$. Put $f_j=ze_j$ for $j\in J$. Then $f_0=ze_0=z-\sum_jf_j$, and $f_0\precsim f_j$. If also $f_j\precsim f_0$, then together with $(1-z)e_j\precsim(1-z)e_0$, additivity gives $e_j\precsim e_0$, which contradicts maximality as before. So $f_0\prec f_j$.

Let $J$ be infinite. Fix $j_0\in J$ and a bijection $\sigma:J\to J\setminus\{j_0\}$. Then $\sum_jf_j\sim\sum_jf_{\sigma(j)}=\sum_{j\ne j_0}f_j$, and $f_0\precsim f_{j_0}$. Additivity gives $\begin{gathered} z\\ =f_0+\sum_{j\in J}f_j\ \precsim\ f_{j_0}+\sum_{j\ne j_0}f_j\\ =\sum_{j\in J}f_j\\ \le z , \end{gathered} \tag{13.2}$ so $z\sim\sum_jf_j$ by Proposition 5.1. If $w$ implements $\sum_jf_j\sim z$, the projections $wf_jw^*$ are mutually orthogonal, each is equivalent to $f_j$ (through $wf_j$), and they add up to $z$. $\square$

Proposition 13.3 (Algebras without a type I part). The following are equivalent.

$M$ has no nonzero central summand of type I, that is, $z_{\rm I}=0$ in Theorem 7.2.
$M$ has no nonzero abelian projection.
Every projection of $M$ is the sum of two orthogonal, equivalent projections.

Proof. (1)⇔(2). In the proof of Theorem 7.2, $z_{\rm I}=c(a)$ for the sum $a$ of a maximal family of mutually centrally orthogonal nonzero abelian projections, and $z_{\rm I}$ is unique. If $M$ has a nonzero abelian projection, that family is not empty, so $z_{\rm I}\ne0$. Conversely, if $z_{\rm I}\ne0$, the type I summand $Mz_{\rm I}$ has a nonzero abelian projection.

(3)⇒(2). Let $e$ be abelian, and $e=e_1+e_2$ with $e_1\perp e_2$ and $e_1\sim e_2$. By Lemma 6.2(4), $e_k=c(e_k)e$, and $c(e_1)=c(e_2)$ by Proposition 3.5(2). So $e_1=e_2$, and orthogonality forces $e_1=e_2=0$, hence $e=0$.

(2)⇒(3). First, every nonzero projection $e$ majorizes two orthogonal, equivalent, nonzero projections. The algebra $eMe$ is not commutative, and it is the norm-closed span of its projections (Fact 2.8), so some projection $p\in eMe$ fails to commute with some element of $eMe$. Then $pM(e-p)\ne\{0\}$: otherwise, for $x\in eMe$, $px(e-p)=0$ and $(e-p)xp=(px^*(e-p))^*=0$, so $px=pxp=xp$. Lemma 5.3 gives nonzero $e_1\le p$ and $e_2\le e-p$ with $e_1\sim e_2$. Next, let $e$ be any projection, and take a maximal family of pairs $(a_k,b_k)$ of nonzero projections below $e$ such that all the $a_k$ and $b_k$ are mutually orthogonal and $a_k\sim b_k$. If $r=e-\sum_k(a_k+b_k)$ were nonzero, the first step inside $r$ would enlarge the family. So $e=\sum_ka_k+\sum_kb_k$, and $\sum_ka_k\sim\sum_kb_k$ by additivity. $\square$

Proposition 13.4 (Properly infinite algebras halve). If $M$ is properly infinite, there is $e\in\mathcal P(M)$ with $e\sim1-e\sim1$.

Proof. Step 1. Every nonzero central $z$ majorizes a nonzero central $z_1$ that contains a projection $a$ with $a\sim z_1-a\sim z_1$. Indeed, $z$ is infinite, because $M$ is properly infinite, so some $u$ has $u^*u=z$ and $uu^*<z$. The powers $u^n$ have initial projection $z$ and decreasing final projections, so the projections $\begin{gathered} p_n\\ =u^{n-1}(z-uu^*)u^{*(n-1)}\\ =u^{n-1}u^{*(n-1)}-u^nu^{*n}, \end{gathered}$ $n\ge1$ (with $u^0=z$, the unit of $Mz$), are mutually orthogonal; they are nonzero and equivalent to $p_1$ through $u^{n-1}p_1$. Proposition 13.1 in $Mz$ gives a nonzero central $z_1\le z$ and an infinite family $\{q_j\}_{j\in J}$ of mutually orthogonal, mutually equivalent projections with $\sum_jq_j=z_1$. Split $J=J_1\sqcup J_2$ with $|J_1|=|J_2|=|J|$ (Fact 2.9), and put $a=\sum_{j\in J_1}q_j$. Additivity gives $a\sim\sum_{j\in J_2}q_j=z_1-a$ and $a\sim\sum_{j\in J}q_j=z_1$.

Step 2. Take a maximal family $\{e_k\}$ of nonzero projections with mutually orthogonal central supports such that $e_k\sim c(e_k)-e_k\sim c(e_k)$ for each $k$. If $c=\sum_kc(e_k)$ were not $1$, Step 1 with $z=1-c$ would give a projection $a$ with $c(a)=c(z_1)=z_1\le1-c$ and $a\sim z_1-a\sim z_1$, enlarging the family. So $\sum_kc(e_k)=1$. Put $e=\sum_ke_k$. By additivity, $1=\sum_kc(e_k)\sim\sum_ke_k=e$ and $e\sim\sum_k(c(e_k)-e_k)=1-e$. $\square$

Corollary 13.5 (Division by $\aleph_0$). If $e$ is properly infinite, then $e=\sum_{n\ge1}e_n$ with mutually orthogonal $e_n\sim e$.

Proof. The algebra $eMe$ is properly infinite: its central projections are the $ze$ (Proposition 3.5(4)), and each nonzero $ze$ is infinite. Proposition 13.4 in $eMe$ gives $e=a_1+b_1$ with $a_1\sim b_1\sim e$. The projection $b_1$ is properly infinite as well: its central cuts $zb_1\sim ze$ are infinite when nonzero. So $b_1=a_2+b_2$ with $a_2\sim b_2\sim b_1$, and so on. The $a_n$ are mutually orthogonal and equivalent to $e$, and $q=\sum_na_n$ satisfies $e\sim a_1\le q\le e$, so $q\sim e$ (Proposition 5.1). If $w$ implements $q\sim e$, the projections $e_n=wa_nw^*$ are mutually orthogonal (conjugation by $w$ is an isomorphism of $qMq$ onto $eMe$), $e_n\sim a_n\sim e$, and $\sum_ne_n=wqw^*=e$. $\square$

14. Finite projections: the modular law and unitary equivalence

Finite projections behave like finite-dimensional subspaces: they satisfy the modular law, and equivalent finite projections are unitarily equivalent.

Theorem 14.1 (Finite projections form a modular lattice). If $e$ and $f$ are finite, then $e\vee f$ and $e\wedge f$ are finite, so the finite projections form a sublattice of $\mathcal P(M)$. This sublattice is modular. More generally, the modular law $(e\vee f)\wedge g=e\vee(f\wedge g)\qquad(e\le g)$ holds for all $e,f,g\in\mathcal P(M)$ with $e\le g$ as soon as $g$ is finite.

Proof. The projection $e\wedge f\le e$ is finite (Lemma 6.2(1)). By the parallelogram law (Proposition 4.4), $(e\vee f)-e\sim f-e\wedge f\le f$, so $g'=(e\vee f)-e$ is finite, and $e\vee f=e+g'$ with $e\perp g'$. It remains to show: if $a$ and $b$ are orthogonal finite projections, then $q=a+b$ is finite.

Write $q=q_1+q_2$ as in Theorem 7.2, with $q_1$ finite, $q_2$ properly infinite and the two centrally orthogonal, and put $c=c(q_2)$, so that $q_2=cq$. Suppose $c\ne0$. The projections $ca$ and $cb$ are finite and orthogonal, and $cq=ca+cb$ is properly infinite and nonzero. We work in $N=cqMcq$, whose unit $cq$ is properly infinite, and write $a,b,q$ for $ca,cb,cq$. By Proposition 13.4 in $N$, $q=r+s$ with $r\sim s\sim q$. The comparison theorem (Theorem 5.5) in $N$ gives a central projection $w$ of $N$ with $w(a\wedge r)\precsim w(b\wedge s)$ and $(q-w)(b\wedge s)\precsim(q-w)(a\wedge r)$.

On $w$: $wr=w(a\wedge r)+w(r-a\wedge r)$, and $r-a\wedge r\sim(a\vee r)-a$ by Proposition 4.4. The projections $b\wedge s$ and $(a\vee r)-a$ are orthogonal, because $b\wedge s$ is orthogonal to $a$ and to $r$, hence to $a\vee r$; and both lie below $b=q-a$. Hence $wr\ \precsim\ w(b\wedge s)+w((a\vee r)-a)\ \le\ wb ,$ so $wr$ is finite. As $wr\sim wq=w$, the central projection $w$ of $N$ is finite, and since the unit of $N$ is properly infinite, $w=0$.

Hence $b\wedge s\precsim a\wedge r$. The same computation with the roles exchanged gives $\begin{gathered} s\\ =b\wedge s+(s-b\wedge s)\ \\ \sim\ b\wedge s+((b\vee s)-b)\ \\ \precsim\ a\wedge r+((b\vee s)-b)\ \\ \le\ a , \end{gathered}$ since $a\wedge r$ is orthogonal to $b$ and to $s$, and both summands lie below $a=q-b$. So $s\precsim a$ is finite, while $s\sim q$ is properly infinite and nonzero. This contradiction shows $c=0$, so $q$ is finite.

The modular law. Let $e\le g$ with $g$ finite, and put $h=(e\vee f)\wedge g$ and $k=e\vee(f\wedge g)$. Then $k\le h$, since $e$ and $f\wedge g$ lie below both $e\vee f$ and $g$. Next, $h\vee f=e\vee f=k\vee f$, because $e\le h\le e\vee f$ and $e\le k\le e\vee f$. And $h\wedge f=f\wedge g=k\wedge f$, because $h\wedge f=(e\vee f)\wedge g\wedge f=g\wedge f$ and $f\wedge g\le k\wedge f\le h\wedge f$. By Proposition 4.4, $\begin{gathered} h-f\wedge g\\ =h-h\wedge f\ \sim\ (h\vee f)-f\\ =(e\vee f)-f , \end{gathered}$ and in the same way $k-f\wedge g\sim(e\vee f)-f$. Adding $f\wedge g$ gives $h\sim k$. As $k\le h\le g$ and $g$ is finite, $h$ is finite, and so $h=k$. $\square$

Proposition 14.2 (Unitary equivalence of finite projections). If $e\sim f$ and $e$ is finite, then $1-e\sim1-f$, and $ueu^*=f$ for some unitary $u\in M$.

Proof. $f$ is finite (Lemma 6.2(1)), and so is $g=e\vee f$ (Theorem 14.1). In the finite algebra $N=gMg$ apply the comparison theorem (Theorem 5.5) to $g-e$ and $g-f$: a central projection $w$ of $N$ has $w(g-e)\precsim w(g-f)$ and $(g-w)(g-f)\precsim(g-w)(g-e)$. Choose $h\le w(g-f)$ with $w(g-e)\sim h$. Then $w=we+w(g-e)\sim wf+h\le w$, because $we\sim wf$ and $wf\perp h$. As $w$ is finite, $wf+h=w$, so $h=w(g-f)$ and $w(g-e)\sim w(g-f)$. In the same way $(g-w)(g-e)\sim(g-w)(g-f)$. Adding, $g-e\sim g-f$; adding $1-g$, $1-e\sim1-f$. Let $v$ implement $e\sim f$ and $v'$ implement $1-e\sim1-f$. Then $v=fve$ and $v'=(1-f)v'(1-e)$, so the cross terms vanish, and $u=v+v'$ satisfies $u^*u=e+(1-e)=1$, $uu^*=f+(1-f)=1$ and $ueu^*=vv^*=f$. $\square$

Exercise 14.3. (easy) Show that $M$ is finite if and only if, for all $e,f\in\mathcal P(M)$, $e\precsim f$ implies $1-f\precsim1-e$.

Solution. Let $M$ be finite and $e\sim e_1\le f$. All projections are finite (Lemma 6.2(1)), so Proposition 14.2 gives $1-e\sim1-e_1\ge1-f$; hence $1-f\precsim1-e$. Conversely, if $1\sim f$, then $1\precsim f$, so $1-f\precsim1-1=0$ and $f=1$. $\square$

Exercise 14.4. (hard) Projections $e,f$ are unitarily equivalent if $ueu^*=f$ for a unitary $u\in M$; then $e\sim f$. Show that if $e\sim f$, there are orthogonal decompositions $e=\sum_ie_i$ and $f=\sum_if_i$ such that $e_i$ and $f_i$ are unitarily equivalent for each $i$.

Solution. Unitary equivalence of $e_i$ and $f_i$ holds exactly when $e_i\sim f_i$ and $1-e_i\sim1-f_i$: one direction is clear, and the other is the last step of the proof of Proposition 14.2. Let $v$ implement $e\sim f$. Write $e=e_1+e_2$ as in Theorem 7.2, with $e_1$ finite and $e_2$ properly infinite, centrally orthogonal, and put $c=c(e_1)$; then $e_1=ce$ and $e_2=(1-c)e$, and $f=cf+(1-c)f$ with $cf\sim ce$ and $(1-c)f\sim(1-c)e$.

The finite pair $ce$, $cf$ is unitarily equivalent by Proposition 14.2.
For the properly infinite pair, put $f_2=(1-c)f$. The algebra $e_2Me_2$ is properly infinite, as in the proof of Corollary 13.5, so Proposition 13.4 in $e_2Me_2$ gives $e_2=a_1+a_2$ with $a_1\sim a_2\sim e_2$. The partial isometry $w=(1-c)v$ implements $e_2\sim f_2$; put $b_k=wa_kw^*$, so $f_2=b_1+b_2$ and $a_k\sim b_k$. If $t$ implements $a_2\sim e_2$, then $(1-e_2)+t$ is a partial isometry with initial projection $(1-e_2)+a_2=1-a_1$ and final projection $1$ (the cross terms vanish). So $1-a_1\sim1$, and in the same way $1-b_1=(1-f_2)+b_2\sim1$. Hence $1-a_1\sim1-b_1$, and $a_1$, $b_1$ are unitarily equivalent. By symmetry, so are $a_2$ and $b_2$.

So $e=ce+a_1+a_2$ and $f=cf+b_1+b_2$ is a decomposition of the required kind, with at most three nonzero pieces. $\square$

15. Countable sums and properly infinite algebras

A properly infinite projection absorbs countably many projections that are subequivalent to it. Under a countability hypothesis this gives the normal form $M\cong N\bar\otimes B(\ell^2(\mathbb N))$ for properly infinite algebras.

Definition 15.1 (Locally $\sigma$-finite projections). A projection $f$ is locally $\sigma$-finite if every central projection $z$ with $zf\ne0$ majorizes a central projection $z'$ such that $z'f$ is nonzero and $\sigma$-finite. $M$ is locally $\sigma$-finite if $1$ is. Every $\sigma$-finite projection is locally $\sigma$-finite.

By Zorn's lemma, $M$ is locally $\sigma$-finite exactly when $1$ is a sum of mutually orthogonal central projections $z_k$ with every $Mz_k$ $\sigma$-finite. Indeed, if $M$ is locally $\sigma$-finite, a maximal family of mutually orthogonal nonzero central projections $z_k$ with $Mz_k$ $\sigma$-finite has sum $1$, since otherwise $1-\sum_kz_k$ would majorize a further member. Conversely, if $1=\sum_kz_k$ with every $Mz_k$ $\sigma$-finite, and $z$ is a nonzero central projection, then $zz_k\ne0$ for some $k$, and $Mzz_k$ is $\sigma$-finite, since orthogonal families in $Mzz_k$ are orthogonal families in $Mz_k$. Such algebras are also called locally countably decomposable.

Proposition 15.2 (Properly infinite projections absorb countable sums). Let $e$ be a properly infinite projection.

If $f_1,f_2,\ldots$ are mutually orthogonal projections with $f_n\precsim e$, then $\sum_nf_n\precsim e$.
If $f$ is $\sigma$-finite and $c(f)\le c(e)$, then $f\precsim e$.
If $f$ is locally $\sigma$-finite and $c(f)\le c(e)$, then $f\precsim e$.
If $M$ is locally $\sigma$-finite and $c(e)=1$, then $e\sim1$. In particular, in a $\sigma$-finite factor any two infinite projections are equivalent.

The countability conditions cannot be dropped (Example 16.4).

Proof. (1) By Corollary 13.5, $e=\sum_ne_n$ with mutually orthogonal $e_n\sim e$. Then $f_n\precsim e\sim e_n$, and additivity gives $\sum_nf_n\precsim\sum_ne_n=e$.

(2) Take a maximal family $\{f_k\}$ of mutually orthogonal nonzero subprojections of $f$ with $f_k\precsim e$. It is countable, because $f$ is $\sigma$-finite. If $r=f-\sum_kf_k$ were nonzero, then $c(r)\le c(f)\le c(e)$, and Corollary 5.4 would give a nonzero $r_1\le r$ with $r_1\precsim e$, against maximality. So $f=\sum_kf_k$, and (1) applies (a finite family is padded with zeros).

(3) Take a maximal family $\{z_k\}$ of mutually orthogonal central projections below $c(f)$ such that each $z_kf$ is nonzero and $\sigma$-finite. If $z=c(f)-\sum_kz_k$ were nonzero, then $zf\ne0$ (Proposition 3.5(1)), and the hypothesis would give a further member of the family. So $f=\sum_kz_kf$. Each $z_ke$ is nonzero, because $0\ne z_k\le c(e)$, and it is properly infinite (Lemma 6.2(3)); moreover $c(z_kf)=z_kc(f)\le z_kc(e)=c(z_ke)$. By (2), $z_kf\precsim z_ke$, and additivity over the orthogonal central pieces gives $f\precsim\sum_kz_ke\le e$.

(4) The projection $1$ is locally $\sigma$-finite and $c(1)=1=c(e)$, so $1\precsim e$ by (3), and $e\sim1$ by Proposition 5.1. In a factor, infinite projections are properly infinite, nonzero projections have central support $1$, and in a $\sigma$-finite factor every projection is $\sigma$-finite; so two infinite projections are subequivalent to each other, hence equivalent. $\square$

Proposition 15.3 (Properly infinite semifinite algebras). Let $M$ be properly infinite and semifinite, and let $f$ be a finite projection with $c(f)=1$ (Lemma 7.4(2)).

There are mutually orthogonal central projections $z_\alpha$, indexed by the infinite cardinals $\alpha\le|M|$, some possibly $0$, with $\sum_\alpha z_\alpha=1$ and $\begin{gathered} \{Mz_\alpha,z_\alpha H\}\\ \cong\{N_\alpha\bar\otimes B(\ell^2(\alpha)),\,fz_\alpha H\otimes\ell^2(\alpha)\}, \end{gathered}$ where $N_\alpha=fz_\alpha Mfz_\alpha$ is finite. The $N_\alpha$ are not unique (Example 16.5).
If $M$ is locally $\sigma$-finite, then $M\cong fMf\bar\otimes B(\ell^2(\mathbb N))$.
More generally, if $M$ is properly infinite and locally $\sigma$-finite (not necessarily semifinite) and $p$ is any projection with $c(p)=1$, then $1=\sum_{n\in\mathbb N}p_n$ with mutually orthogonal $p_n\sim p$, and $M\cong pMp\bar\otimes B(\ell^2(\mathbb N))$.

Reference: The decomposition of $1$ in part (3) is [Blackadar, III.1.3.6].

The further trace theory studies uniqueness of the family $\{z_\alpha\}$ in (1); see Traces on von Neumann algebras. That extension is not proved or used here. The present proof establishes existence; when the countability hypothesis in (2) holds, it constructs a single amplification by $\ell^2(\mathbb N)$.

Proof. (1) Take a maximal family $\{c_k\}$ of mutually orthogonal nonzero central projections such that each $c_k=\sum_{j\in J_k}f_{k,j}$ for mutually orthogonal projections $f_{k,j}\sim fc_k$. Suppose $c=1-\sum_kc_k\ne0$; then $fc\ne0$ since $c(f)=1$. Apply Proposition 13.1 in $Mc$ to the one-element family $\{fc\}$. It gives a nonzero central $c''\le c$ and mutually orthogonal, mutually equivalent projections $\{g_j\}_{j\in J}$, one of which is $fc''$, with $g_0=c''-\sum_jg_j\prec g_j$. If $J$ were finite, $c''=g_0+\sum_jg_j$ would be a finite sum of finite projections (each $g_j\sim fc''$ is finite, and $g_0\precsim g_j$), hence finite by Theorem 14.1. But $c''\ne0$ is infinite, because $M$ is properly infinite. So $J$ is infinite, and Proposition 13.1 lets us take $g_0=0$ with every $g_j\sim fc''$. Then $c''$ can be added to the family, a contradiction. Hence $\sum_kc_k=1$, and by the same finiteness argument every $J_k$ is infinite. For an infinite cardinal $\alpha$ put $z_\alpha=\sum\{c_k:\ |J_k|=\alpha\}$. Fix a set $S_\alpha$ with $|S_\alpha|=\alpha$ and bijections $\beta_k:S_\alpha\to J_k$ for the $k$ with $|J_k|=\alpha$, and put $f_{\alpha,s}=\sum_kf_{k,\beta_k(s)}$. These are mutually orthogonal, $f_{\alpha,s}\sim\sum_kfc_k=fz_\alpha$ (additivity over orthogonal central pieces), and $\sum_sf_{\alpha,s}=z_\alpha$. By Proposition 8.4 in $Mz_\alpha$ and Lemma 8.5, $\begin{gathered} \{Mz_\alpha,z_\alpha H\}\\ \cong\{fz_\alpha Mfz_\alpha\bar\otimes B(\ell^2(S_\alpha)),fz_\alpha H\otimes\ell^2(S_\alpha)\}. \end{gathered}$ The algebra $fz_\alpha Mfz_\alpha$ is finite, because $fz_\alpha$ is (Lemma 6.2(1) and (3)). If $z_\alpha\ne0$, then $fz_\alpha\ne0$ because $c(f)=1$, so the $f_{\alpha,s}$ are nonzero and mutually orthogonal; they are $\alpha$ distinct elements of $M$, and $\alpha\le|M|$.

(3) Take a maximal family $\{c_k\}$ of mutually orthogonal nonzero central projections such that each $c_k=\sum_{n\in\mathbb N}p_{k,n}$ for mutually orthogonal projections $p_{k,n}\sim c_kp$. Suppose $c=1-\sum_kc_k\ne0$, and choose a nonzero central $c'\le c$ with $Mc'$ $\sigma$-finite. In $N=Mc'$, whose unit $c'$ is properly infinite, the projection $q=c'p$ is nonzero. By Corollary 13.5, $c'=\sum_ng_n$ with $g_n\sim c'\ge q$, so each $g_n$ contains a copy of $q$; thus $N$ contains an infinite sequence of mutually orthogonal projections equivalent to $q$. Extend it to a maximal family $\{q_i\}_{i\in I}$ of mutually orthogonal projections equivalent to $q$. The set $I$ is infinite, and countable because $N$ is $\sigma$-finite; put $r=\sum_iq_i$. The comparison theorem (Theorem 5.5) in $N$ gives a central $w\le c'$ with $w(c'-r)\precsim wq$ and $(c'-w)q\precsim(c'-w)(c'-r)$. If $w=0$, then $q\precsim c'-r$, and a copy of $q$ orthogonal to all the $q_i$ would contradict maximality. So $w\ne0$. Fix $i_0\in I$ and a bijection $\sigma:I\to I\setminus\{i_0\}$. Since $w(c'-r)\precsim wq\sim wq_{i_0}$ and $wr=\sum_iwq_i\sim\sum_iwq_{\sigma(i)}$, additivity gives $\begin{gathered} w\\ =w(c'-r)+wr\ \precsim\ wq_{i_0}+\sum_{i\ne i_0}wq_i\\ =wr\\ \le w , \end{gathered}$ so $w\sim wr$ (Proposition 5.1). If $t$ implements $wr\sim w$, the projections $t(wq_i)t^*$ are mutually orthogonal, equivalent to $wq=wp$, and add up to $w$. After relabeling $I$ as $\mathbb N$, $w\le c$ can be added to $\{c_k\}$, a contradiction. So $\sum_kc_k=1$. The projections $p_n=\sum_kp_{k,n}$ are mutually orthogonal, $p_n\sim\sum_kc_kp=p$, and $\sum_np_n=\sum_kc_k=1$. Proposition 8.4 and Lemma 8.5 give $M\cong pMp\bar\otimes B(\ell^2(\mathbb N))$.

(2) This is (3) with $p=f$. $\square$

Exercise 15.4. (hard) Let $M$ on $H$ and $N$ on $K$ be von Neumann algebras, with $H$ and $K$ separable, and let $\pi:M\to N$ be an isomorphism. Show that $x\otimes1\mapsto\pi(x)\otimes1$ is a spatial isomorphism of $\{M\otimes1,H\otimes\ell^2(\mathbb N)\}$ onto $\{N\otimes1,K\otimes\ell^2(\mathbb N)\}$. Show that separability can be replaced by $\sigma$-finiteness of $M'$ and $N'$, and that it cannot simply be dropped.

Solution. On $L=(H\otimes\ell^2)\oplus(K\otimes\ell^2)$ put $\sigma(x)=(x\otimes1)\oplus(\pi(x)\otimes1)$. By Fact 2.6, $\pi$ is $\sigma$-weakly continuous, so $\sigma$ is a faithful normal representation, and $P=\sigma(M)$ is a von Neumann algebra. The projections $p=1\oplus0$ and $q=0\oplus1$ lie in $P'$.

Central supports. The centre of $P'$ is that of $P$, namely $\sigma(Z)$. If $c_0\in Z$ is a projection with $\sigma(c_0)p=0$, then $c_0\otimes1=0$ and $c_0=0$. So the central projection $1-c(p)$, which is orthogonal to $p$, is $0$, and $c(p)=1$. As $\pi$ is injective, $c(q)=1$ in the same way.
Proper infiniteness. By Fact 2.1 for $P'$ and $p$, the reduced algebra $pP'p$ on $H\otimes\ell^2$ is the commutant of $Pp=M\otimes1$, which is $\mathbb M_{\mathbb N}(M')$ by Lemma 8.2(1). Its central projections are the $c_0\otimes1$ with $c_0\in Z$, and each nonzero one is infinite: $c_0\otimes s$, with $s$ the unilateral shift, implements $c_0\otimes1\sim c_0\otimes(1-E_{00})$. So $p$ is properly infinite (Lemma 6.2(3)), and so is $q$.
$\sigma$-finiteness. If $M'$ is $\sigma$-finite, then by Facts 2.3 and 2.4 there is a countable set $\{\kappa_m\}$ that is separating for $M'$, hence cyclic for $M$. The countable set $\{\kappa_m\otimes\delta_n\}$ is cyclic for $M\otimes1$, hence separating for its commutant $pP'p$, so $p$ is $\sigma$-finite (Fact 2.4). When $H$ is separable, a countable dense subset of $H$ is cyclic for $M$. The same holds for $q$.

Proposition 15.2(2) gives $q\precsim p$ and $p\precsim q$, and Proposition 5.1 gives $p\sim q$ in $P'$. If $v$ implements $p\sim q$, then $V=v|_{pL}:H\otimes\ell^2\to K\otimes\ell^2$ is unitary, and $V(x\otimes1)=(\pi(x)\otimes1)V$ because $v$ commutes with $\sigma(x)$.

Countability cannot be dropped. Let $M=\mathbb C1$ on $H=\mathbb C$, $N=\mathbb C1$ on $K$ with $\dim K=\aleph_1$, and $\pi(\lambda1)=\lambda1$. A unitary $H\otimes\ell^2\to K\otimes\ell^2$ would require $\aleph_0=\aleph_1$. Here $N'=B(K)$ is not $\sigma$-finite. $\square$

Exercise 15.5. (hard) Let $\mathcal J$ be a two-sided ideal of $M$, not necessarily closed, and $\mathcal P=\mathcal P(M)$.

(a) If $e,f\in\mathcal P$, $e\precsim f$ and $f\in\mathcal J$, then $e\in\mathcal J$.
(b) $\mathcal P\cap\mathcal J$ is a sublattice of $\mathcal P$.
(c) If $\mathcal P_0$ is a nonempty sublattice of $\mathcal P$ such that $e\precsim f\in\mathcal P_0$ implies $e\in\mathcal P_0$, then $\mathcal J_0=\{x\in M:\mathrm l(x)\in\mathcal P_0\}$ is a two-sided ideal and $\mathcal J_0\cap\mathcal P=\mathcal P_0$.
(d) For every two-sided ideal $\mathcal J$, the ideal $\mathcal J_0=\{x:\mathrm l(x)\in\mathcal P\cap\mathcal J\}$ satisfies $\mathcal J_0\subseteq\mathcal J\subseteq\overline{\mathcal J_0}$ (norm closure).
(e) The norm-closed two-sided ideals of a factor are totally ordered by inclusion.
(f) A factor that is finite, or $\sigma$-finite and of type III, has no two-sided ideals other than $\{0\}$ and itself; in particular it is simple.
(g) Conversely, a factor that is infinite and semifinite, or of type III and not $\sigma$-finite, has a proper nonzero norm-closed two-sided ideal. So a factor is simple exactly when it is finite, or $\sigma$-finite of type III.

Solution. (a) Let $u$ implement $e\sim f_1\le f$. Then $f_1=f_1f\in\mathcal J$ and $e=u^*u=u^*f_1u\in\mathcal J$.

(b) If $e,f\in\mathcal P\cap\mathcal J$, then $e\wedge f=(e\wedge f)e\in\mathcal J$. By Proposition 4.4, $(e\vee f)-e\sim f-e\wedge f\le f$, so $(e\vee f)-e\in\mathcal J$ by (a), and $e\vee f=e+((e\vee f)-e)\in\mathcal J$.

(c) Choose $f\in\mathcal P_0$, using the nonempty hypothesis. Then $0\precsim f$ gives $0\in\mathcal P_0$. An empty lattice would give an empty $\mathcal J_0$, which is not an ideal. For $x,y\in M$ and $\lambda\ne0$: $\mathrm l(x+y)\le\mathrm l(x)\vee\mathrm l(y)$; $\mathrm l(\lambda x)=\mathrm l(x)$; $\mathrm l(xy)\le\mathrm l(x)$; and $\mathrm l(yx)=\mathrm l(y\,\mathrm l(x))\sim\mathrm r(y\,\mathrm l(x))\le\mathrm l(x)$ by Proposition 4.3, since $[yxH]=[y\,\mathrm l(x)H]$. As $\mathcal P_0$ is a lattice and closed under subequivalence (in particular under $\le$), $\mathcal J_0$ is closed under sums, scalar multiples and products on either side. For $e\in\mathcal P$, $\mathrm l(e)=e$, so $\mathcal J_0\cap\mathcal P=\mathcal P_0$.

(d) By (a) and (b), $\mathcal P_0=\mathcal P\cap\mathcal J$ satisfies the hypothesis of (c), so $\mathcal J_0$ is an ideal. If $x\in\mathcal J_0$, then $x=\mathrm l(x)x\in\mathcal J$. Now let $x\in\mathcal J$ and $\varepsilon>0$. The spectral projection $p=\chi_{(\varepsilon,\infty)}(xx^*)$ equals $k(xx^*)\,xx^*$ with $k(t)=t^{-1}\chi_{(\varepsilon,\infty)}(t)$ bounded (Fact 2.8), so $p\in\mathcal J\cap\mathcal P$. The element $y=px$ has $\mathrm l(y)\le p$, so $\mathrm l(y)=\mathrm l(y)p\in\mathcal J$ and $y\in\mathcal J_0$. Finally $\|x-y\|^2=\|(1-p)xx^*(1-p)\|\le\varepsilon$. So $\mathcal J\subseteq\overline{\mathcal J_0}$.

(e) By (d), a closed ideal $\mathcal J$ equals $\overline{\mathcal J_0}$, and $\mathcal J_0$ depends only on $\mathcal P\cap\mathcal J$. So $\mathcal P\cap\mathcal J\subseteq\mathcal P\cap\mathcal K$ implies $\mathcal J\subseteq\mathcal K$ for closed ideals $\mathcal J,\mathcal K$. Suppose $\mathcal P\cap\mathcal J\not\subseteq\mathcal K$, and pick $e\in\mathcal P\cap\mathcal J$ outside $\mathcal K$. For every $f\in\mathcal P\cap\mathcal K$, $e\precsim f$ is impossible by (a), so $f\prec e$ by Theorem 5.5 ($M$ is a factor), and $f\in\mathcal J$ by (a). Hence $\mathcal P\cap\mathcal K\subseteq\mathcal P\cap\mathcal J$, and $\mathcal K\subseteq\mathcal J$.

(f) Let $\mathcal J\ne\{0\}$ and $0\ne x\in\mathcal J$. As in (d), for small $\varepsilon$ the projection $p=\chi_{(\varepsilon,\infty)}(xx^*)$ is a nonzero element of $\mathcal J$. If $M$ is $\sigma$-finite of type III, $p$ is infinite, so $p\sim1$ by Proposition 15.2(4), and $1\in\mathcal J$ by (a). If $M$ is finite, apply Proposition 13.1 to the family $\{p\}$. As $M$ is a factor, $z=1$, and we get mutually orthogonal projections $\{f_j\}_{j\in J}$ equivalent to $p$ with $f_0=1-\sum_jf_j\prec f_j$. The set $J$ is finite: otherwise $\sum_jf_j\sim\sum_{j\ne j_0}f_j<\sum_jf_j$ would be an infinite projection in a finite algebra. So $1=f_0+\sum_jf_j$ is a finite sum of projections subequivalent to $p$, all in $\mathcal J$ by (a), and $1\in\mathcal J$.

(g) Let $\mathcal P_0$ be the set of finite projections (if $M$ is infinite and semifinite) or of $\sigma$-finite projections (if $M$ is of type III and not $\sigma$-finite). In both cases $\mathcal P_0$ is a sublattice closed under subequivalence. For finite projections this is Theorem 14.1 with Lemma 6.2(1). For $\sigma$-finite ones: a projection subequivalent to a $\sigma$-finite one is $\sigma$-finite; and $e\vee f=e+((e\vee f)-e)$ with $(e\vee f)-e\precsim f$ (Proposition 4.4), where a sum of two orthogonal $\sigma$-finite projections $e,h$ is $\sigma$-finite, because faithful normal positive functionals $\varphi_1$ on $eMe$ and $\varphi_2$ on $hMh$ (Fact 2.4) give the faithful normal functional $x\mapsto\varphi_1(exe)+\varphi_2(hxh)$ on $(e+h)M(e+h)$. For faithfulness, if $x\ge0$ has value zero, faithfulness on the two corners gives $exe=hxh=0$. Hence $x^{1/2}e=x^{1/2}h=0$, so $x^{1/2}(e+h)=0$ and $x=0$. Normality follows from the normality of the two corner functionals and fixed compression. By (c), $\mathcal J_0=\{x:\mathrm l(x)\in\mathcal P_0\}$ is an ideal. It is nonzero: a semifinite algebra has nonzero finite projections, and every cyclic projection $p_\xi$ with $\xi\ne0$ is $\sigma$-finite, because $\omega_\xi$ is faithful on $p_\xi Mp_\xi$ (Fact 2.4). Its norm closure is a closed ideal, and it is proper: if $\|1-x\|<1$ for some $x\in\mathcal J_0$, then $x$ is invertible, $\mathrm l(x)=1$, and $1$ would be finite (respectively $\sigma$-finite). $\square$

16. Counterexamples

The examples below show that several hypotheses in Sections 13–15 cannot be dropped, and that the corners in Proposition 15.3(1) are not unique.

Example 16.1 (A family of zero projections). If $I\ne\emptyset$ and all $e_i=0$, then $f_i\sim ze_i=0$, so every $f_j=0$, and $f_0=z\prec0$ is impossible ($z\precsim0$ forces $z=0\sim0$). This is why Proposition 13.1 asks for nonzero $e_i$.

Example 16.2 (Modularity needs finiteness). Let $K=\ell^2(\mathbb N)$ and $T\delta_n=\delta_n/(n+1)$. Then $T$ is injective with dense range, and $w=\sum_n\delta_n/(n+1)$ is not in the range of $T$ (its preimage would be $\sum_n\delta_n\notin K$). On $H=K\oplus K$, let $e$ be the projection onto $K\oplus0$, $f$ the projection onto the graph $\{(\xi,T\xi)\}$, and $g$ the projection onto $(K\oplus0)+\mathbb C(0,w)$; these subspaces are closed, and $e\le g$. If $(\xi,T\xi)=(x,\lambda w)$, then $T\xi=\lambda w$ forces $\lambda=0$ and $\xi=0$; so $f\wedge g=0$ and $e\vee(f\wedge g)=e$. But $(K\oplus0)+\text{graph}=K\oplus T(K)$ is dense in $H$, so $e\vee f=1$ and $(e\vee f)\wedge g=g\ne e$. So the modular law fails in $B(H)$, with $g$ infinite. Exercise 16.6 extends this to every algebra that is not finite.

Example 16.3 (The shift). On $\ell^2(\mathbb N)$ the unilateral shift $s$ satisfies $s^*s=1$ and $ss^*=1-p_0$, where $p_0$ is the projection onto $\mathbb C\delta_0$. So $1\sim1-p_0$, and $1$ is infinite. The projections $1$ and $1-p_0$ are equivalent but not unitarily equivalent, since their complements $0$ and $p_0$ are not equivalent. So Proposition 14.2 needs finiteness.

Example 16.4 (Countability in Propositions 15.2 and 15.3). Let $H$ have Hilbert dimension $\aleph_1$, $M=B(H)$, and $e$ the projection onto a separable infinite-dimensional subspace. Then $e$ is properly infinite ($M$ is a factor and $e$ is infinite), and $c(e)=1=c(1)$. But $1\not\precsim e$, since an isometry of $H$ into $eH$ would give $\dim H\le\aleph_0$. So the conclusions of Proposition 15.2(2)–(3) fail for $f=1$, which is neither $\sigma$-finite nor locally $\sigma$-finite. Also $M$ is properly infinite and semifinite, and a rank-one $f$ is finite with $c(f)=1$, but $M\not\cong fMf\bar\otimes B(\ell^2(\mathbb N))=B(\ell^2(\mathbb N))$ by Corollary 10.4. So Proposition 15.3(2) needs its countability hypothesis. In Proposition 15.3(1), $z_{\aleph_1}=1$.

Example 16.5 (Non-uniqueness in Proposition 15.3). $B(\ell^2(\mathbb N))\cong\mathbb C\bar\otimes B(\ell^2(\mathbb N))\cong M_2(\mathbb C)\bar\otimes B(\ell^2(\mathbb N))$, because $\mathbb C^2\otimes\ell^2(\mathbb N)\cong\ell^2(\mathbb N)$. Both $\mathbb C$ and $M_2(\mathbb C)$ are finite; they are the corners $fMf$ for an $f$ of rank one and of rank two.

Exercise 16.6. (hard) Show that the projection lattice $\mathcal P(M)$ is modular if and only if $M$ is finite.

Solution. If $M$ is finite, every projection is finite, and Theorem 14.1 applies. Suppose $M$ is not finite. By Theorem 7.2 the properly infinite part $z$ of the projection $1$ is a nonzero central projection, and $Mz$ is properly infinite. By Corollary 13.5, $z=\sum_{n\in\mathbb N}q_n$ with mutually orthogonal $q_n\sim z$, and Proposition 8.4 gives a unitary $W$ with $W^*(Mz)W=\mathbb M_{\mathbb N}(N)$, $N=q_1Mq_1$ on $L=q_1H$. So $Mz$ contains $W(1\otimes B(\ell^2(\mathbb N)))W^*$. For projections $a,b$ on $\ell^2(\mathbb N)$, the range of $1\otimes a$ is $L\otimes a\ell^2$, and $(L\otimes A)\cap(L\otimes B)=L\otimes(A\cap B)$ for closed subspaces $A,B$ (decompose along an orthonormal basis of $L$). So $a\mapsto W(1\otimes a)W^*$ preserves meets, and by De Morgan's laws joins, which in $\mathcal P(M)$ are those of $B(H)$ (Proposition 3.1). Identify $\ell^2(\mathbb N)$ with $K\oplus K$ as in Example 16.2, and take the projections $e\le g$ and $f$ found there. Their images violate the modular law in $\mathcal P(M)$. $\square$

17. Comparison with the commutant: the transfer theorem

This section compares the cyclic projections of $M$ (Definition 4.5) with those of $M'$. The main result, the transfer theorem, says that $p_\xi\succsim p_\eta$ in $M$ exactly when $p'_\xi\succsim p'_\eta$ in $M'$. Its key step rotates a vector functional into a positive one. This is the vector case of the polar decomposition of normal functionals. Two elementary lemmas prepare it.

Lemma 17.1 (A positivity test). Let $A$ be a unital C$^*$-algebra and $\omega$ a bounded linear functional on $A$. If $\omega(h)=\|\omega\|$ for some $h\in A$ with $0\le h\le1$, then $\omega$ is positive.

Proof. Put $R=\|\omega\|$; we may assume $R>0$. For $-1\le s\le1$, the self-adjoint element $h+s(1-h)$ has its spectrum in $[-1,1]$: each point $\lambda+s(1-\lambda)$ with $\lambda\in[0,1]$ is a convex combination of $1$ and $s$. So $|R+s\,\omega(1-h)|=|\omega(h+s(1-h))|\le R$. Writing $a=\omega(1-h)$, this says $2sR\operatorname{Re}a+s^2|a|^2\le0$ for all $s\in[-1,1]$. Small $s$ of both signs force $\operatorname{Re}a=0$, and then $a=0$. So $\omega(1)=R$.

Now let $x=x^*$, $\|x\|\le1$, and $\omega(x)=\alpha+i\beta$. For real $t$, $\|1+itx\|^2=\|1+t^2x^2\|\le1+t^2$, so $(R-t\beta)^2+t^2\alpha^2\le R^2(1+t^2)$. This reads $-2tR\beta+t^2(\alpha^2+\beta^2)\le R^2t^2$. Divide by $|t|$ and let $t\to0$ from each side: $\beta=0$. So $\omega$ is real on self-adjoint elements. If $0\le x\le1$, then $\|1-x\|\le1$, so $|R-\omega(x)|\le R$ with $\omega(x)$ real, and hence $\omega(x)\ge0$. $\square$

Lemma 17.2 (Dropping a projection). Let $\varphi$ be a weakly continuous linear functional on $M$, let $e\in\mathcal P(M)$, and put $\varphi_e(x)=\varphi(xe)$. If $\|\varphi_e\|=\|\varphi\|$, then $\varphi_e=\varphi$.

Proof. We may assume $\|\varphi\|=1$. Put $f=1-e$, and suppose that $\varphi(yf)\ne0$ for some $y\in M$. Rescaling $y$ gives $b$ in the unit ball with $\varphi(bf)=\delta>0$. The functional $\varphi_e$ is weakly continuous and the unit ball of $M$ is weakly compact, so $|\varphi_e|$ attains its norm there. After a rotation, $\varphi(ae)=1$ for some $a$ in the unit ball. For $t>0$, $(ae+tbf)(ae+tbf)^*=aea^*+t^2bfb^*$, because $ef=0$. So $\|ae+tbf\|^2\le1+t^2$. But $\varphi(ae+tbf)=1+t\delta$. Hence $(1+t\delta)^2\le1+t^2$, that is, $2\delta\le t(1-\delta^2)$ for every $t>0$. This is impossible for $\delta>0$. So $\varphi(yf)=0$ for all $y$, which says $\varphi=\varphi_e$. $\square$

Lemma 17.3 (Rotating a vector functional). Let $\xi,\eta\in H$ and $\varphi(x)=\langle x\eta,\xi\rangle$ for $x\in M$. There is a partial isometry $u\in M$ such that $\begin{gathered} \varphi(xuu^*)\\ =\varphi(x)\\ (x\in M)\\ \text{and}\\ x\mapsto\varphi(xu^*)\ \text{is positive on }M. \end{gathered} \tag{17.4}$ If $\eta\in[M\xi]$, then $uu^*\eta=\eta$, and the vector $\eta_2=u^*\eta$ lies in $[M'\xi]\cap[M\xi]$.

Proof. If $\varphi=0$, take $u=0$. Then $\eta\perp x^*\xi$ for all $x\in M$, so an $\eta\in[M\xi]$ must be $0$, and the last statement holds trivially. Otherwise let $R=\|\varphi\|>0$. The functional $\varphi$ is weakly continuous and the unit ball of $M$ is weakly compact, so $\varphi(a)=R$ for some $a$ in the unit ball. Let $a^*=u|a^*|$ be the polar decomposition (Fact 2.2). Then $a=|a^*|u^*$ with $u\in M$ and $0\le|a^*|\le1$. The functional $\omega(x)=\varphi(xu^*)$ has norm at most $R$, and $\omega(|a^*|)=\varphi(a)=R$. By Lemma 17.1, $\omega$ is positive. Next put $e=uu^*$. Then $ae=|a^*|u^*uu^*=|a^*|u^*=a$, so $\varphi_e(a)=R$ and $\|\varphi_e\|=\|\varphi\|$. Lemma 17.2 gives $\varphi(xuu^*)=\varphi(x)$. This proves (17.4).

Now let $\eta\in[M\xi]$. The first property in (17.4) says $\langle uu^*\eta-\eta,x^*\xi\rangle=0$ for all $x\in M$. The vector $uu^*\eta-\eta$ lies in $[M\xi]$, which is invariant under $M$, and it is orthogonal to $[M\xi]$. So it is zero. Put $\eta_2=u^*\eta\in[M\xi]$. The functional $\psi(x)=\langle x\eta_2,\xi\rangle=\varphi(xu^*)$ is positive, hence hermitian: $\psi(x^*)=\overline{\psi(x)}$. As $p_\xi\xi=\xi$, we have $\psi(x)=\langle p_\xi x\eta_2,\xi\rangle=\psi(p_\xi x)$, and therefore $\psi(x)=\overline{\psi(x^*)}=\overline{\psi(p_\xi x^*)}=\psi(xp_\xi).$ So $\langle x(\eta_2-p_\xi\eta_2),\xi\rangle=0$ for every $x\in M$. The vector $\eta_2-p_\xi\eta_2$ lies in $[M\xi]$ and is orthogonal to it, so it vanishes. Thus $\eta_2=p_\xi\eta_2\in[M'\xi]$. $\square$

Theorem 17.5 (The transfer theorem). For $\xi,\eta\in H$, $p_\xi\succsim p_\eta\ \text{ in } M\quad\Longleftrightarrow\quad p'_\xi\succsim p'_\eta\ \text{ in } M'.$

Proof. Step 1 (moving a vector inside its orbit). If $v\in M'$ is a partial isometry with $v^*v\eta=\eta$, then $p_{v\eta}=p_\eta$. Indeed $[M'v\eta]\subseteq[M'\eta]$ because $v\in M'$, and $\eta=v^*(v\eta)$ gives the reverse inclusion.

Step 2 (the core). If $\eta\in[M\xi]$, then $p_\eta\precsim p_\xi$. Take $u$ and $\eta_2=u^*\eta$ from Lemma 17.3. The projection $uu^*\in M$ fixes $\eta$ and commutes with $M'$, so it fixes $[M'\eta]$ pointwise, and $p_\eta\le uu^*$. The operator $w=u^*p_\eta$ therefore has initial projection $p_\eta uu^*p_\eta=p_\eta$. Its final projection is the projection onto $u^*[M'\eta]$. Since $u^*$ is isometric on $uu^*H\supseteq[M'\eta]$, this subspace is closed and equals $[u^*M'\eta]=[M'u^*\eta]=[M'\eta_2]$. So $p_\eta\sim p_{\eta_2}$. Finally $\eta_2\in[M'\xi]$ gives $p_{\eta_2}\le p_\xi$ (Lemma 4.6).

Step 3. Suppose $p'_\xi\succsim p'_\eta$: some $v\in M'$ has $v^*v=p'_\eta$ and $vv^*\le p'_\xi$. Put $\eta_1=v\eta$. Then $\eta_1\in vv^*H\subseteq[M\xi]$, and $v^*v\eta=p'_\eta\eta=\eta$, so $p_{\eta_1}=p_\eta$ by Step 1. Step 2 gives $p_\eta=p_{\eta_1}\precsim p_\xi$. This proves "⇐". The implication "⇒" is the same statement for the von Neumann algebra $M'$, whose commutant is $M$, with the roles of $p$ and $p'$ exchanged. $\square$

Example 17.6 ($B(H)$ and the scalars). For $M=B(H)$ and $\xi\ne0$, $p_\xi$ is the projection onto $\mathbb C\xi$ and $p'_\xi=1$; Theorem 17.5 reads: $p_\xi\succsim p_\eta$ iff $\xi\ne0$ or $\eta=0$, iff $p'_\xi\succsim p'_\eta$. The algebra $M=\mathbb C1$ on $H\ne0$ is this example with $M$ and $M'$ exchanged: for $\xi\ne0$, $p_\xi=1$ and $p'_\xi$ is the projection onto $\mathbb C\xi$. Theorem 17.5 is symmetric under this exchange.

18. Vector states and spatial isomorphisms

The transfer theorem answers three natural questions: when one cyclic representation of a C$^*$-algebra is contained in another; which normal functionals are vector functionals; and, for algebras with cyclic and separating vectors, whether an isomorphism is implemented by a unitary.

Proposition 18.1 (Subrepresentations and cyclic projections). Let $A$ be a C$^*$-algebra, $\{\pi,H\}$ a nondegenerate representation of $A$, $\xi,\eta\in H$, $\varphi=\omega_\xi\circ\pi$, $\psi=\omega_\eta\circ\pi$, and $N=\pi(A)''$. The following are equivalent.

The cyclic representation $\pi_\varphi$ of $\varphi$ embeds in $\pi_\psi$: some isometry $H_\varphi\to H_\psi$ intertwines $\pi_\varphi$ with $\pi_\psi$. (Its range is then an invariant subspace, so this says that $\pi_\varphi$ is carried by a unitary onto a subrepresentation of $\pi_\psi$.)
$p'_\xi\precsim p'_\eta$ in $N'=\pi(A)'$.
$p_\xi\precsim p_\eta$ in $N$.

In the universal representation (Fact 2.7), $p_\xi=s(\varphi)$ and $p_\eta=s(\psi)$. So $\pi_\varphi$ embeds in $\pi_\psi$ exactly when $s(\varphi)\precsim s(\psi)$ in $\tilde A$.

Proof. Since $\pi$ is nondegenerate, $\pi(A)$ is strongly dense in $N$ by the double commutant theorem (Fact 2.7). So every vector $a\xi$ with $a\in N$ is a limit of vectors $\pi(b)\xi$ with $b\in A$, and $[N\xi]=[\pi(A)\xi]$. Thus $p'_\xi$ is the projection onto $[\pi(A)\xi]$. The map $\pi_\varphi(a)\xi_\varphi\mapsto\pi(a)\xi$ is isometric, since both sides have squared norm $\varphi(a^*a)$. It carries a dense subspace of $H_\varphi$ onto a dense subspace of $p'_\xi H$, so it extends to a unitary $U_\xi:H_\varphi\to p'_\xi H$ that intertwines $\pi_\varphi$ with the restriction of $\pi$ to $p'_\xi H$. The same holds for $\eta$ and $\psi$.

(2)⇒(1). Let $v\in\pi(A)'$ have $v^*v=p'_\xi$ and $vv^*\le p'_\eta$. The range $L=vv^*H$ is invariant under $\pi(A)$, and $v$ restricts to a unitary $p'_\xi H\to L$ that intertwines the restrictions of $\pi$. Composing with $U_\xi$ and $U_\eta^{-1}$ shows that $\pi_\varphi$ is equivalent to the restriction of $\pi_\psi$ to the invariant subspace $U_\eta^*L$.

(1)⇒(2). Let $T:H_\varphi\to L_0$ be a unitary onto a closed $\pi_\psi$-invariant subspace, intertwining $\pi_\varphi$ with the restriction of $\pi_\psi$. Define $v=U_\eta TU_\xi^*$ on $p'_\xi H$ and $v=0$ on its orthogonal complement. On $p'_\xi H$, $v$ intertwines. On $(p'_\xi H)^\perp$, both $v\pi(a)$ and $\pi(a)v$ vanish, because $\pi(a)$ leaves that space invariant. So $v\in\pi(A)'$, $v^*v=p'_\xi$ and $vv^*\le p'_\eta$.

(2)⇔(3) is Theorem 17.5 in $N$, whose commutant is $\pi(A)'$.

In the state-indexed universal representation, normalize each nonzero functional and rescale its cyclic vector as in Fact 2.7; use the zero vector for a zero functional. This gives $\xi,\eta\in H_u$ with $\varphi=\omega_\xi\circ\pi_u$ and $\psi=\omega_\eta\circ\pi_u$. The restriction of $\omega_\xi$ to $\tilde A$ is $\sigma$-weakly continuous and extends $\varphi$, so it is $\tilde\varphi$, and its support is $p_\xi$ (Lemma 4.6). The same holds for $\psi$. $\square$

Proposition 18.2 (Which normal functionals are vector functionals). Let $\xi_0\in H$ and let $\varphi$ be a positive normal functional on $M$. Then $\varphi=\omega_\xi|_M$ for some $\xi\in[M\xi_0]$ if and only if $s(\varphi)\precsim p_{\xi_0}$. In particular, if $M$ has a separating vector $\xi_0$, then $p_{\xi_0}=1$, and every positive normal functional on $M$ is a vector functional $\omega_\xi$ with $\xi\in[M\xi_0]$.

Proof. If $\varphi=\omega_\xi$ with $\xi\in[M\xi_0]$, then $s(\varphi)=p_\xi$ (Lemma 4.6), and $p_\xi\precsim p_{\xi_0}$ by Step 2 of the proof of Theorem 17.5.

Conversely, suppose $s(\varphi)\precsim p_{\xi_0}$. By Fact 2.5, $\varphi=\sum_n\omega_{\xi_n}$ with $\sum_n\|\xi_n\|^2<\infty$. On $K=H\otimes\ell^2(\mathbb N)$, with orthonormal basis $(\delta_n)$ of $\ell^2(\mathbb N)$, let $N=M\otimes1$. By Lemma 8.2, $N$ is a von Neumann algebra, its commutant $N'$ consists of the operators whose matrix entries lie in $M'$, and $x\mapsto x\otimes1$ is an isomorphism of $M$ onto $N$. Put $\zeta=\sum_n\xi_n\otimes\delta_n$ and $\zeta_0=\xi_0\otimes\delta_1$. Then $\omega_\zeta(x\otimes1)=\varphi(x)$. A projection $p\otimes1$ of $N$ fixes $\zeta$ iff $\varphi(1-p)=0$, so the cyclic projection of $N$ at $\zeta$ is $p^N_\zeta=s(\varphi)\otimes1$. Next, $[N'\zeta_0]=[M'\xi_0]\otimes\ell^2(\mathbb N)$: the inclusion "⊇" holds because $N'$ contains the operators $x'\otimes1$ and the matrix units $1\otimes E_{n1}$, and "⊆" holds because every $T\in N'$ maps $\zeta_0$ to $\sum_nT_{n1}\xi_0\otimes\delta_n$ with $T_{n1}\in M'$. So $p^N_{\zeta_0}=p_{\xi_0}\otimes1$. The isomorphism $x\mapsto x\otimes1$ turns $s(\varphi)\precsim p_{\xi_0}$ into $p^N_\zeta\precsim p^N_{\zeta_0}$. By Theorem 17.5 in $N$, $p'^N_\zeta\precsim p'^N_{\zeta_0}$ in $N'$: some $v\in N'$ has $v^*v=p'^N_\zeta$ and $vv^*\le p'^N_{\zeta_0}$. The range of $p'^N_{\zeta_0}$ is $[N\zeta_0]=[M\xi_0]\otimes\delta_1$. Hence $v\zeta=\xi\otimes\delta_1$ with $\xi\in[M\xi_0]$, and for $x\in M$ $\begin{gathered} \omega_\xi(x)\\ =\langle(x\otimes1)v\zeta,v\zeta\rangle\\ =\langle(x\otimes1)v^*v\zeta,\zeta\rangle\\ =\omega_\zeta(x\otimes1)\\ =\varphi(x). \end{gathered}$ If $\xi_0$ is separating for $M$, it is cyclic for $M'$ (Fact 2.3), so $p_{\xi_0}=1$ and the condition holds for every $\varphi$. $\square$

Example 18.3 (A normal state that is not a vector state). Let $M=M_2(\mathbb C)$ on $\mathbb C^2$ and $\varphi=\frac12\mathrm{tr}$. Its support is $1$. Each vector functional $\omega_\xi$ has support $p_\xi$, the rank-one projection onto $\mathbb C\xi$ (here $M'=\mathbb C1$). So $\varphi$ is not a vector functional, in line with Proposition 18.2: $s(\varphi)=1\not\precsim p_{\xi_0}$ for every $\xi_0$. Accordingly $M$ has no separating vector. On $\mathbb C^2\otimes\mathbb C^2$, with $M$ acting as $M\otimes1$, the vector $(\delta_1\otimes\delta_1+\delta_2\otimes\delta_2)/\sqrt2$ is separating and gives $\varphi$.

Lemma 18.4 (Interpolating two vectors). Let $\xi_1,\xi_2\in H$ with $p'_{\xi_1}\sim p'_{\xi_2}$ in $M'$; by Theorem 17.5 and Proposition 5.1 this is the same as $p_{\xi_1}\sim p_{\xi_2}$ in $M$. Then some $\xi_0\in H$ has $p'_{\xi_0}=p'_{\xi_1}$ and $p_{\xi_0}=p_{\xi_2}$.

Proof. Let $u\in M'$ implement $p'_{\xi_2}\sim p'_{\xi_1}$, and put $\xi_0=u\xi_2$. Since $u$ commutes with $M$ and maps $[M\xi_2]$ isometrically onto $p'_{\xi_1}H$, we get $[M\xi_0]=u[M\xi_2]=p'_{\xi_1}H$. Since $u^*u\xi_2=\xi_2$, Step 1 of the proof of Theorem 17.5 gives $p_{\xi_0}=p_{\xi_2}$. $\square$

Proposition 18.5 (A cyclic and separating vector). If some vector is cyclic for $M$ and some vector is separating for $M$, then a single vector is both cyclic and separating.

Proof. Let $\xi_1$ be cyclic and $\xi_2$ separating for $M$. Then $p'_{\xi_1}=1$, and $p_{\xi_2}=1$ by Fact 2.3. As $p_{\xi_1}\le1=p_{\xi_2}$, Theorem 17.5 gives $1=p'_{\xi_1}\precsim p'_{\xi_2}$, and Proposition 5.1 gives $p'_{\xi_2}\sim1=p'_{\xi_1}$. Lemma 18.4 yields $\xi_0$ with $p'_{\xi_0}=1$ and $p_{\xi_0}=1$. So $\xi_0$ is cyclic for $M$ and for $M'$, hence also separating for $M$ (Fact 2.3). $\square$

Proposition 18.6 (Isomorphisms are spatial). Let $\{M_1,H_1\}$ and $\{M_2,H_2\}$ be von Neumann algebras with cyclic and separating vectors $\xi_1$ and $\xi_2$. Every isomorphism $\pi:M_1\to M_2$ is spatial.

Proof. By Fact 2.6, $\pi$ is $\sigma$-weakly continuous, so $\psi=\omega_{\xi_2}\circ\pi$ is a positive normal functional on $M_1$. It is faithful: $\psi(x^*x)=\|\pi(x)\xi_2\|^2=0$ forces $\pi(x)=0$ and $x=0$. By Proposition 18.2, applied to $M_1$ and its separating vector $\xi_1$, $\psi=\omega_\zeta$ for some $\zeta\in H_1$. Faithfulness makes $\zeta$ separating for $M_1$, so $p_\zeta=1=p_{\xi_1}$ (Fact 2.3), and Theorem 17.5 with Proposition 5.1 gives $p'_\zeta\sim p'_{\xi_1}=1$. Lemma 18.4 gives $\zeta_0=u\zeta$, with $u\in M_1'$ isometric on $[M_1\zeta]$, such that $p'_{\zeta_0}=1$. Then $\omega_{\zeta_0}=\omega_\zeta=\psi$ on $M_1$, because $\langle xu\zeta,u\zeta\rangle=\langle u^*u\,x\zeta,\zeta\rangle=\langle x\zeta,\zeta\rangle$.

Define $U$ on $M_2\xi_2=\pi(M_1)\xi_2$ by $U\pi(x)\xi_2=x\zeta_0$. It is well defined and isometric, since $\|x\zeta_0\|^2=\psi(x^*x)=\|\pi(x)\xi_2\|^2$. Its domain is dense in $H_2$ and its range is dense in $H_1$, so $U$ extends to a unitary $H_2\to H_1$. For $x,y\in M_1$, $U\pi(y)\pi(x)\xi_2=yx\zeta_0=yU\pi(x)\xi_2$. So $U\pi(y)=yU$, that is, $\pi(y)=U^*yU$. $\square$

Where this leads

The following topics extend this lesson. Their further assertions and constructions are not proved here and are not inputs to any proof or solution above.

Traces. Traces and dimension functions are built on the comparison theory of this lesson. The lesson Traces on von Neumann algebras constructs the centre-valued trace of a finite algebra, of which the map $\Phi$ of Proposition 9.1(2) is the simplest case. It also proves that the family $\{z_\alpha\}$ of Proposition 15.3(1) is unique.
Types II and III. This lesson gives examples of type I only. Factors of types II and III exist, but their construction needs other tools, such as group von Neumann algebras, infinite tensor products or crossed products.
Multiplicity. The spatial form of type I algebras (Theorem 11.4) and the normal form of properly infinite algebras (Proposition 15.3) are the starting points of the multiplicity theory of normal representations.

References

[Kostecki] R. P. Kostecki, W$^*$-algebras and noncommutative integration, arXiv:1307.4818, 2013.
[Blackadar] B. Blackadar, Operator Algebras: Theory of C*-Algebras and von Neumann Algebras, revised author edition, 8 February 2017.

Freely accessible reading: B. Blackadar, Operator Algebras, III.1.1–III.1.4 gives a route through projection comparison and type decomposition; all cardinal and boundary cases stated here retain their complete proofs. The lesson includes its own complete proofs at the stated hypotheses; references to human sources do not imply permission to adapt their expression.

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