### Introduction to von Neumann algebras, Lecture 5 (comparison of projections and classification into types of von Neumann algebras)

In the previous lecture we discussed the group von Neumann algebras, and we saw that they can never be isomorphic to $B(H)$. There is something fundamentally different about these algebras, and this was manifested by the existence of a trace. von Neumann algebras with traces are special, and the existence or non-existence of a trace can be used to classify von Neumann algebras, into rather broad “types”. In this lecture we will study the theory of Murray and von Neumann on the comparison of projections and the use of this theory to classify von Neumann algebras into “types”. We will also see how traces (or generalized traces) fit in. (For preparing these notes, I used Takesaki (Vol I) and Kadison-Ringrose (Vol. II).)

Most of the time we will stick to the assumption that all Hilbert spaces appearing are separable. This will only be needed at one or two spots (can you spot them?).

In addition to “Exercises”, I will start suggesting “Projects”. These projects might require investing a significant amount of time (a student is not expected to choose more than one project).

#### 1. Murray- von Neumann equivalence

Definition 1: Let $A$ be a von Neumann algebra. Two projections $p,q$ in $A$ are said to be equivalent (or Murray-von neumann equivalent) if there exists a partial isometry $v \in A$ such that $v v^* = p$ and $v^*v = q$. If $p$ and $q$ are equivalent, then we write $p \sim q$.

Note: a crucial part of the definition is that $v \in A$. One can think of the subspaces corresponding to $p$ and $q$ as subspace the “look the same in the eyes of $A$.”

Exercise A: Murray-von Neumann equivalence is an equivalence relation.

Exercise B: Describe when two projections are equivalent in (i) $B(H)$, and (ii) $L^\infty(\mu)$.

Recall that in Lecture 3 (Definition 2), we defined the range projection $R(T)$ of an operator $T \in B(H)$ to be equal to the orthogonal projection onto $\overline{Im T}$; equivalently $R(T)$ is the smallest projection $P \in B(H)$ such that $PT = T$. One sometimes denotes $s_l(T) = R(T)$, and calls it the left support of $T$. Similarly, the right support of $T$ is the projection onto $ker(T)^\perp$, it is equal to $s_r(T) := R(T^*)$, and is the smallest projection $P$ such that $TP = T$. In Exercise B of Lecture 3, you proved that if $T$ is an element of a von Neumann algebra, then $R(T) \in A$ (and therefore also $R(T^*) \in A$).

Proposition 2: If $A \subseteq B(H)$ is a von Neumann algebra and $T \in R(T)$, then $R(T) \sim R(T^*)$

Proof: Use the polar decomposition of $T$ to find the partial isometry that provides the equivalence.

Definition 3: Let $A$ be a von Neumann algebra, and let $p,q$ be projections in $A$. We write $p \preceq q$, and say that $p$ is (Murray-von Neumann) sub-equivalent to  $q$, if there exists a partial isometry $v \in A$ such that $v v^* = p$ and $v^*v \leq q$ (in other words, if $p$ is equivalent to a sub-projection of $q$). If $p$ is sub-equivalent but not equivalent to $q$, then we write $p \prec q$.

Murray-von Neumann sub-equivalence is a partial ordering on the set of projections in a von Neumann algebra. The reflexivity and transitivity of this relation is straightforward. We will soon prove that it is anti-symmetric.

Lemma 3: Let $\{p_i\}$ and $\{q_i\}$ be two families of orthogonal projections. If $p_i \sim q_i$ (respectively, $p_i \preceq q_i$) for all $i$, then $\sum_i p_i \sim \sum_i q_i$ (respectively, $\sum_i p_i \preceq \sum_i q_i$) for all $i$

Proof: If $u_i u_i^* = p_i$ and $u_i^* u_i = q_i$ [respectively, $u_i^* u_i \leq q_i$] then $u = \sum_i u_i$ converges strongly to a partial isometry and $u u^* = \sum_i u_i u_i^* = \sum_i p_i$, while $u^* u = \sum_i u_i^* u_i = \sum_i q_i$ [respectively, $\sum_i u^*_i u_i \leq \sum_i q_i$] (note that as $p_i \perp p_j$ and $q_i \perp q_j$ for $i \neq j$, so $u_i u_j^* = u_i u_i^* u_i u_j^* u_j u_j^* = u_i p_i p_j u_j^*= 0$, and likewise $u_i^*u_j = 0$).

The next proposition is an analogue of the Cantor-Bernstein-Schroder theorem, and shows that sub-equivalence is indeed a partial ordering on $P(A)$.

Proposition 4: If $p \preceq q$ and $q \preceq p$, then $p \sim q$

Proof: Suppose that

$p = u^* u$  and  $u u^* = q_1 \leq q$

and

$q = v^* v$  and  $v v^* = p_1 \leq p$.

We define two decreasing sequences of projections $\{p_n\}$ and $\{q_n\}$ by induction. Write $p_0 = p$ and $q_0 = q$, and define

$p_{n+1} = v q_n v^*$   and   $q_{n+1} = u p_n u^*$   for all   $n \geq 1$.

We have that $p_1 \leq p_0 = p$, $q_1 \leq q_0$ by assumption, and $p_{n+1} = v q_n v^* \leq v q_{n-1} v^* = p_n$, by induction (likewise for $q_n$).

Now, since $\sum_{n=0}^N (p_n - p_{n+1}) = p_0 - p_{N+1}$ for all $N$, we have

$p = \sum_{n=0}^\infty (p_n - p_{n+1}) + p_\infty$,

where $p_\infty = \lim_n p_n$. Likewise,

$q = \sum_{n=0}^\infty (q_n - q_{n+1}) + q_\infty$.

Now, for every $n$, we have by definition that $u (p_n - p_{n+1}) u^* = q_{n+1} - q_{n+2}$. But then $U:= u(p_n - p_{n+1})$ is a partial isometry setting up an equivalence between $U^*U = (p_n - p_{n+1}) u^*u (p_n - p_{n+1}) = (p_n - p_{n+1})$ and $U U^* = u(p_n - p_{n+1}) u^* = q_{n+1} - q_{n+2}$. Thus, $p_n - p_{n+1} \sim q_{n+1} - q_{n+2}$.

Likewise, $q_n - q_{n+1} \sim p_{n+1} - p_{n+2}$ and $p_\infty \sim q_\infty$. Now we cleverly put this together, obtaining

$p = \sum_{n=0}^\infty (p_{2n} - p_{2n+1}) + \sum_{n=0}^\infty (p_{2n+1} - p_{2n+2}) + p_\infty \sim$

$\sim \sum_{n=0}^\infty (q_{2n+1} - q_{2n+2}) + \sum_{n=0}^\infty (q_{2n} - q_{2n+1}) + q_\infty = q$.

Proposition 5: For two projections $p$ and $q$ in a von Neumann algebra $A$, TFAE:

1. $C_p \perp C_q$ (i.e., $C_p C_q = 0$, where $C_p$ and $C_q$ are the central covers of $p$ and $q$).
2. $pA q = 0$
3. For all nonzero $p_1 \leq p$ and $q_1 \leq q$, $p_1$ and $q_1$ are not equivalent.

Proof: If 1 holds, then $p a q = C_p p a q C_q = p a q C_p C_q = 0$ for all $a \in A$, so 2 holds. On the other hand, if $pAq = 0$, then we set $J = \{a \in A : pAa = 0\}$. Then $J$ is weakly closed ideal, so by Theorem 6 in Lecture 3 $J = cA$ for some $c \in Z(A)$. But $q \in J$, so $q \leq c$, and therefore $C_q \leq c$. But then $p c = 0$, and it follows that $p C_q = 0$. Therefore $p \leq 1 - C_q$ so $C_p \leq 1 - C_q$, and it follows that $C_p C_q = 0$. So 2 implies 1.

Next, if $p_1 = u u^*$ and $q_1 = u^* u$ where $p_1 \leq p$ and $q_1 \leq q$, then $p u q = pp_1 u q_1 q = p_1 u q_1 = u u^* u u^* u=u \neq 0$, thus 2 implies 3.

Finally, suppose that 2 fails. If $a \neq 0$ is in $pAq$, then $a = paq = pa = aq$, so $R(a) \leq p$ and $R(a^*) \leq q$. By Proposition 2, $R(a) \sim R(a^*)$, so 3 fails as well.

Definition 6: Two projections in a von Neumann algebra satisfying the conditions of the previous proposition are said to be centrally orthogonal.

The relation of Murray-von Neumann sub-equivalence is a partial ordering, but it is not full: if $p$ and $q$ are projections in a von Neumann algebra $A$, it may happen that neither $p \preceq q$ nor $q \preceq p$ hold. The following comparison theorem shows how one may always bring projections to a position where they are comparable.

Theorem 7 (the comparison theorem): If $p$ and $q$ are projections in a von Neumann algebra $A$, then there exists a projection $z \in Z(A)$ such that $pz \preceq qz$ and $q(1-z) \preceq p(1-z)$

Proof: Let $(p_0,q_0)$ be a maximal pair of projections that satisfy $p_0 \leq p$, $q_0 \leq q$ and $p_0 \sim q_0$. Well, if $p_0 = p$ or $q_0 = q$ then we are done. Otherwise, consider the projections $p - p_0$ and $q - q_0$; these do not have any subprojections $p_1 \leq p - p_0$ and $q_1 \leq q - q_0$ such that $p_1 \sim q_1$, for otherwise the pair $(p_0, q_0)$ would not be maximal. By the previous proposition, $C_{p-p_0}$ and $z:=C_{q-q_0}$ are orthogonal. We find that $(p-p_0) z = 0$, so

$pz = p_0 z \sim q_0 z \leq qz$,

where we used the fact $P \sim Q \Rightarrow PC \sim QC$ for all central $C$. Likewise, $(q - q_0)z = q-q_0$, so

$q(1-z) = q_0(1-z) \sim p_0 (1-z) \leq p(1-z)$.

Corollary: In a factor, every two projections are comparable.

#### 2. Types of projections and types of vN algebras

Definition 8: Let $p$ be a projection in a von Neumann algebra $A$. $p$ is said to be:

1. abelian if $pAp$ is abelian.
2. finite if $p \sim q \leq p$ implies that $p = q$.
3. infinite if it is not finite.
4. properly infinite if for every central projection $z$, $ze$ is either infinite or zero.
5. purely infinite if for every projection $q \leq p$, $q$ is either infinite or zero.

If the identity $1$ in a von Neumann algebra $A$ is finite/infinite/properly infinite/purely infinite, then $A$ is said to be finite/infinite/properly infinite/purely infinite.

Examples:

1. In an abelian von Neumann algebra, all projections are abelian.
2. Perhaps surprisingly, $1 \in \ell^\infty(\mathbb{Z})$ is finite. In fact, every abelian projection is finite (why?).
3. It is easy to see which projections in $B(H)$ are finite, which are abelian, which are infinite.
4. $I_H \in B(H)$ is properly infinite, but not purely infinite. Can you find an example of a projection in a von Neumann algebra that is infinite but not properly infinite? (I bet you can).
5. Can you find an example of a projection in a von Neumann algebra that is purely infinite? (I bet you can’t).

Corollary (to Proposition 5): A nonzero projection $p$ in a factor is abelian if and only if it is minimal.

Proof: Let $p$ be a nonzero abelian projection. Then it must be minimal, because if $0 \neq p_1 < p$, then $p_1$ and $p-p_1$ are not centrally orthogonal, so by Proposition 5 they dominate a pair of equivalent projections, and this would show that $pAp$ is not abelian.

Conversely, if $p$ is minimal, then $pAp = \mathbb{C}p$, so $p$ is abelian.

Exercise C: Let $\{p_i\}$ be a family of centrally orthogonal projections (i.e., $C_{p_i} C_{p_j} = 0$ for $i \neq j$). If every $p_i$ is abelian (finite), then $\sum_i p_i$ is abelian (finite).

Definition 9: A von Neumann algebra $A \subseteq B(H)$ is said to be (of):

1. Type I   if for every nonzero central projection $z$, there exists a nonzero abelian $p \leq z$ in $A$.
2. Type II   if $A$ has no nonzero abelian projections, but for every nonzero central projection $z$, there exists a nonzero finite $p \leq z$ in $A$.
3. Type III   if $A$ has no nonzero finite projections (i.e., if $A$ is purely infinite).

For the sake of addressing an issue that better not be addressed, let us say that the algebra $\{0\}$ acting on the Hilbert space $\{0\}$ is a von Neumann algebra of any type.

Theorem 10: Let $A$ be a von Neumann algebra. Then there exists a unique decomposition of $A$ into a direct sum

$A = A_I \oplus A_{II} \oplus A_{III}$

of a type $I$, a type $II$ and type $III$ von Neumann algebra.

Proof: If there are no abelian projections in $A$, let $z_I = 0$. Otherwise, let $\{p_n\}$ be a maximal family of centrally orthogonal abelian projections. Then, by Exercise C $p:=\sum p_n$ is abelian. Let $z_I$ be the central cover of $p$. Then $A_I := A z_I$ is a von Neumann algebra, and we claim that it is of type I. Indeed, if $z \leq z_I$, that is, if $z$ is a nonzero central projection in $A_I$, then $zp$ is a nonzero abelian projection dominated by $z$ (if it was zero, then $z_I - z would contradict that fact that $z_I$ is the central cover of $p$).

By design, $(1-z_I)A$ is a von Neumann algebra with no abelian projections. Let $\{q_i\}$ be a maximal family of centrally orthogonal finite projections in $(1-z_I)A$, and let $q = \sum q_i$, which is finite thanks to Exercise C. Now let $z_{II}$ be the central cover of $q$ in $(1-z_I)A$. Then $A_{II} = z_{II}A = z_{II}(1-z_I)A$ is a von Neumann algebra, and as in the previous paragraph, one shows that it is of type II.

Finally, letting $z_{III} = 1 - z_I - z_{II}$, we find that $z_{III}$ is central, and $A_{III} = z_{III}A$ is a type III von Neumann algebra.

We leave it to the reader to check that the decomposition $A = A_I \oplus A_{II} \oplus A_{III}$ is unique.

Thus, a von Neumann algebra in general does not have to be of a particular type. But for factors, things are nicer.

Corollary: A factor is either of type I, type II, or type III.

There is a theory, going back to von Neumann, that describes how every von Neumann algebra can be decomposed uniquely into a direct integral of factors. We shall not go into that direction. Since a considerable amount of interesting work on classification theory is concentrated on factors, and there are many interesting examples, we shall mostly speak about factors.

#### 3. Type I algebras and factors

Example 11: As our first example, we note that, trivially, commutative von Neumann algebras are always type I. A slightly deeper fact is this: if $A \subseteq B(H)$ ($H$ separable) is an abelian type I algebra, then $A'$ is also of type I. To see this, let $z \in Z(A') = A$ be a central projection. We need to show that $z$ dominates an abelian projection in $A'$. For this end, let $h \in zH$ be a nonzero vector, and let $M = [Ah]$. If $p = p_M$, then $p \in A'$ is nonzero and $p \leq z$. Then $A_p = pAp$ is a cyclic and commutative von Neumann algebra on $M$, and we may also assume that it is singly generated. Therefore, $A_p$ is unitarily equivalent to $L^\infty(\mu)$, so $A'_p = A_p$ is abelian. Therefore, $p$ is abelian, and this shows that $A'$ is type I.

Example 12: As an example at the opposite extreme, let us consider $B(H)$, where $H$ is a Hilbert space. We have already seen that this is a factor, and it is a type I factor as a special case of the previous example, since $B(H) = (\mathbb{C}I)'$. Alternatively, to see that it is type I, one needs to show that it contains a nonzero abelian projection. But clearly, if $p$ is a minimal projection (which must have the form $v v^* = p_{\mathbb{C}v}$), then $pAp = \mathbb{C} p$ is abelian.

In fact, the previous example contains all type I factors (up to isomorphism, not up to unitary equivalence), but we will have to wait a little bit for this. Before the following lemma, the reader might want to review Proposition 9 in Lecture 3.

Lemma: Let $A$ be a von Neumann algebra, and let $p \in A$ such that $C_p = 1$. Then $\varphi : b \mapsto bp = pb\big|$ is a *-isomorphism from $A'$ onto $A'p$

Proof: We know that $\varphi$ is a WOT continuous and surjective *-homomorphism, and so the kernel of this isomorphism is $zA'$ for a central projection $z \in Z(A)$. Therefore, $(1-z)p = p \neq 0$, so $1-z$ is a central element dominating $p$. Since $C_p = 1$ we must have $z = 0$, and $\varphi$ is injective.

Theorem 13: Let $A$ be a von Neumann algebra. The following conditions are equivalent:

1. $A$ is type I.
2. $A'$ is type I.
3. There exists a faithful and WOT continuous representation $\pi : A \to B(H)$ such that $\pi(A)'$ is abelian.

Remark: Before the proof, let us remark that in 3 above, $\pi(A)$ will be a von Neumann algebra, because the unit ball of $\pi(A)$ will be WOT compact.

Remark: One last remark before the proof: it also true that $A$ is type II (respectively, type III) if and only if its commutant $A'$ is type II (respectively, type III). You may take this as a challenging:

Exercise D: Take care of the remark above (for reference to start with, see Section 9.1 in Kadison-Ringrose, Vol II).

Proof of Theorem 13: Suppose that $A$ is type I. As in the proof of Theorem 10, let $p$ be a maximal sum of abelian projections. We then have $C_p = z_I = 1$. The lemma now implies that $A'_p \cong A'$. But $A'_p = (A_p)'$ is the commutant of an abelian von Neumann algebra, so by Example 11, $A'_p$ is type I, and therefore $A'$ is type I.

If $A'$ is type I, then the previous paragraph (with the roles of $A$ and $A'$ reversed) shows that there exists some $p \in A'$ so that the map $a \mapsto ap \in A_p$ is WOT continuous *-isomorphism onto the commutant of an abelian von Neumann algebra.

Finally, if $\pi(A)'$ is abelian, then $\pi(A) = \pi(A)''$ is type I, so $A$ is type I.

Corollary : If $A$ is a type I factor, then there is a Hilbert space $K$ such that $A$ is isomorphic to $B(K)$

Proof: If $A$ is a type I factor, then $A'$ is clearly a factor too, and it is type I by the theorem. Let $p \in A'$ be an abelian projection. By the corollary to Proposition 5, $p$ is minimal, so $A'_p = \mathbb{C}p$ as a von Neumann subalgebra of $B(pH)$. Thus $A_p = (A'_p)' = B(pH)$. Since $C_p = 1$ ($A$ being a factor), the lemma shows that $A \cong A_p = B(pH)$.

Definition 14: A type I factor is said to be of type $I_n$ if it is isomorphic to $B(H)$ where $\dim H = n$. One write  $I_\infty$ if $n = \aleph_0$.

The classification problem of type I factors up to isomorphism is therefore settled: there is exactly one type I factor of type $I_n$ for every cardinal $1, 2, \ldots, \aleph_0 = \infty$, and there aren’t any other examples up to isomorphism (except non-popular examples $B(H)$ living on non-separable Hilbert spaces $H$).

We also see that the equivalence classes of projections in a type I factor, as a partially ordered space, is isomorphic either to $\{1, \ldots, n\}$ for some $n$ or to $\mathbb{N}$.

If one wants to classify type I factors up to unitary equivalence, there is another issue that comes in, which are not technically prepared to handle at the moment. Roughly, type I factors look like $B(H) \otimes I_K \subseteq B(H \otimes K)$, where $H$ and $K$ are Hilbert spaces. The meaning of the tensor notation will be made precise in the upcoming lectures.

Finally, we mention that one can describe all type I algebras on separable Hilbert spaces. Roughly, these are just direct sums of “matrix algebras with coefficients in commutative von Neumann algebras”. We leave it to the interested student to work or dig this out.

Project 1: Determine the structure theory of type I von Neumann algebras. You might be able to go a significant part of the way on your own. Once stuck, help can be found in the following references: Conway (A Course in Operator Theory), Kadison-Ringrose (Fundamentals of the Theory of Operator Algebras, Vol. II), or Takesaki (Theory of Operator Algebras, Vol. I).

#### 4. Existence of type II factors

Definition 15: A type II factor is said to be a $II_1$ factor if it is finite (that is, if $1$ is a finite projection); otherwise it is said to be a $II_\infty$ factor.

In this section, we will show that the group von Neumann factors are type $II_1$. Let us recall some notation. If $G$ is a countable group, we let $\{\delta_g : g \in G\}$ be the standard orthonormal basis of $\ell^2(G)$, and let the left and right regular representations be given by

$\lambda(g) \delta_h = \delta_{gh}$

and

$\rho(g) \delta_h = \delta_{hg^{-1}}$.

The (left and right) group von Neumann algebras are defined to be $L(G) = \lambda(G)''$ and $R(G) = \rho(G)''$. In the previous lecture, we saw that $R(G) = L(G)'$ and vice versa, and that $L(G)$ is a factor if and only if $G$ was an ICC group (the conjugacy class of every element, except the identity, is infinite).

Recall that on $L(G)$ we defined the trace

$\tau(a) = \langle a \delta_e , \delta_e \rangle$.

“The trace” is a WOT continuous, faithful tracial state, a notion we recall in the following definition:

Definition 16: Let $A$ be a C*-algebra. A trace $\tau$ on $A$ is a positive ($\tau(a) \geq 0$ for $a \in A$) and tracial ($\tau (ab) = \tau (ba)$ for all $a,b \in A$) linear functional. A trace is called a tracial state if $\tau(1) = 1$. A trace $\tau$ is said to be faithful if $\tau(a) = 0$ for $a \in A_+$ implies $a = 0$.

Sometimes we will just say trace instead of “tracial state.”

Theorem 17: Let $G$ be a countable ICC group. Then $L(G)$ is a type $II_1$ factor.

Proof: We already know that $L(G)$ is a factor. Now, $L(G)$ is finite, which just means that $1$ is a finite projection. Indeed, if $u u^* = 1$ and $u^*u \leq 1$, then $uu^* - u^* u \in L(G)_+$ and $\tau(uu^* - u^* u) = 0$, so $u^*u = 1$. This argument shows that $L(G)$ is finite.

Being finite, $L(G)$ cannot be type $I_\infty$, $II_\infty$, or $III$. The only remaining possibilities are $I_n$ for $n \in \mathbb{N}$, or $II_1$. Since $L(G)$ is infinite dimensional, only the $II_1$ case remains.

To argue a little more “constructively”, we have to show simply that $L(G)$ has no abelian projections (since we already know that it is a finite factor). But if it had an abelian projection, the arguments used in the type I case would show that $L(G)$ is isomorphic to $B(H)$, which we have seen cannot happen.

Note that the above proof actually shows that any infinite dimensional factor with a tracial state is a type $II_1$ factor. Let us record this.

Corollary: If $A$ is an infinite dimensional factor, and if $A$ has a faithful tracial state, then $A$ is a type $II_1$ factor.

Nice, we see that there exist $II_1$ factors. Are there many of them? Yes. We will be able to show that there are some, not just there is one. Dusa McDuff proved that there are uncountably many non isomorphic ones, in fact uncountably that arise as group von Neumann algebras.

Project 2: Read and present McDuff’s paper “A countable infinity of $II_1$ factors” (there is also a second paper “Uncountably many $II_1$ factors”, if you are ambitious).

What about $II_\infty$ factors? It turns out that a von Neumann algebra $B$ is a (separably acting) type $II_\infty$ factor, if and only if there exists a type $II_1$ factor such that

$B = A \overline{\otimes} B(H)$

for $H$ an infinite dimensional separable Hilbert space. We plan to discuss tensor products in the next lecture.

#### 5. A little more on $II_1$ factors

In the previous section we saw that ICC groups give rise to type $II_1$ factors. The fact that these factors are type $II_1$ followed from the existence of a faithful (and WOT continuous) tracial state. It can in fact be shown that the existence of such a tracial state characterizes type $II_1$ factors.

Theorem 18: An infinite dimensional factor is of type $II_1$ if and only if it has a faithful tracial state (“trace”, for short). In this case, the trace is unique, and is in fact WOT continuous.

Before sketching the idea of the proof of the Theorem, we collect some more definitions and propositions.

Definition 19: A von Neumann algebra is said to be diffuse if it contains no minimal projections.

Thus a factor is diffuse if and only if it is type II or type III.

Proposition 20 (the halving lemma): Let $A$ be a diffuse factor. For every $p \in P(A)$, there exist $r \sim q$ in $P(A)$, such that $r + q = p$

Proof: Since $p$ is not minimal, there is some $p' < p$. By Proposition 5, there are mutually equivalent nonzero projections $r_1 \leq p'$ and $q_1 \leq p-p'$, and these satisfy $r_1 + q_1 \leq p$.

Now we consider a maximal family $\{(r_i,q_i)\}$ of pairs such that $r_i \sim q_i$ and such that all $\{p_i, q_i\}_i$ are mutually orthogonal. Set $r = \sum r_i$ and $q = \sum q_i$. Then $r \sim q$, and by maximality (and the first part of the proof) $r + q = p$.

Exercise E: Use the halving lemma to show that if $A$ is a factor with a tracial state $\tau$, then $\tau$ is the unique tracial state, and it is faithful. Conversely, prove that if $A$ is a von Neumann algebra with tracial state $\tau$, and if $\tau$ is the unique tracial state, then $A$ must be a factor (hence a $II_1$ factor).

Exercise F: Show that if $p$ is an infinite projection, then the halving lemma can be improved: there exist $r \sim q \sim p$ in $P(A)$, such that $r + q = p$ (note the difference: $r$ and $q$ are also equivalent to $p$).

Idea of the proof of  Theorem 18: Since all the examples come equipped with such a trace, we will not prove this theorem (at least for now). But let us go over the idea of the proof. The corollary to Theorem 17 says that the existence of a trace implies type $II_1$.

Conversely, let $A$ be a type $II_1$ factor. The factor $A$ is finite, and the equivalence classes of projections form a totally ordered set. Since there are no minimal projections, we might think of it as being something like $[0,1]$ – which is indeed what it turns out to be. Using the halving lemma, we construct inductively a sequence of orthogonal projections $\{p_n\}$ such that $\sum_{k=1}^\infty p_k = 1$ and $p_{n} \sim 1 - \sum_{k=1}^n p_k$ (equivalently, $p_n \sim \sum_{k=n+1}^\infty p_k$ ). [Indeed, we start by finding $p_1 \sim p_1'$ such that $p_1 + p_1' =1$, then we throw $p_1'$ away and find $p_2 \sim p_2'$ such that $p_2 + p_2' = 1- p_1$, so $p_2 \sim p_2' = 1- p_1 - p_2$ and so forth. ]

One then proceeds to show that every the sequence $\{p_n\}$ can be used to give a “binary expansion” for every projection, i.e., every $p$ is equivalent to sum partial sum $\sum_{k} p_{n_k}$ (this requires work). One then defines $\tau(p_n) = 2^{-n}$, and uses the binary expansion to define

$\tau(p) = \sum_k \tau(p_{n_k}) = \sum_k 2^{-n_k}$.

The function $\tau$, currently defined on $P(A)$, is call the dimension function. If this can be extended to a WOT continuous state, there is only one way in which it could, since $A$ is generated by its projections. One then works and works to show that this indeed extends to a WOT continuous, faithful tracial state.

Uniqueness you have already shown in Exercise E (by slightly less sophisticated technology) basically follows by the same ideas: the value of a (normalized) trace $\tau'$ on $p_n$ must be $\tau'(p_n) = \frac{1}{2^n}$ (because $\tau'(p_1) = \frac{1}{2}$ and induction), and this determines that value of $\tau'$ on any projection , hence on $A$.

We finish this section by showing that the equivalence classes of projections in a $II_1$ factor is isomorphic (as a partially ordered set) to $[0,1]$.

Theorem 21: Let $A$ be a $II_1$ factor, and let $\tau$ be the tracial state on $A$. Then $\tau(P(A)) = [0,1]$, and for any pair of projections, $p,q \in P(A)$, $p \sim q$ (resp. $p \prec q$) if and only if $\tau(p) = \tau(q)$ )resp. $\tau(p) < \tau(q)$).

Proof: The first assertion really follows from the proof of the above theorem. Next, if $p \sim q$ then clearly $\tau(p) = \tau(uu^*) = \tau(u^*u) = \tau(q)$. If $p \prec q$, then $p \sim q_1 < q$, so $\tau(p) = \tau(q_1) < \tau(q)$ because of positiveness and faithfulness. This basically finishes the proof.

The (non-normalized) trace on a matrix algebra, when evaluated on a projection, gives the dimension of the range of the projection. The trace on type II factor therefore serves as a kind of generalized “dimension function”. von Neumann was fascinated by the fact the dimension of projections in a type II factor can vary continuously.

#### 6. Semifinite tracial weights on type $I_\infty$ and $II_\infty$ factors

Definition 22: Let $A \subseteq B(H)$ be a von Neumann algebra. A tracial weight on $A$ is a map $\rho : A_+ \to [0,\infty]$ such that

1. $\rho(ca+b) = c\rho(a) + \rho(b)$ for all $a,b \in A_+$ and $c \in [0,\infty)$.
2. $\rho(a^*a) = \rho(aa^*)$ for all $a \in A$.

Some immediate consequences: $\rho(uhu^*) = \rho(h)$, for all $h \in A_+$ and $u \in U(A)$, and $p \sim / \preceq q$ implies $\rho(p) = / \leq \rho(q)$.

Definition 22 (continued): A tracial weight $\rho$ is said to be normal if $\lim\rho(a_\alpha) = \rho(\lim a_\alpha)$ (equivalently, $\sup\rho(a_\alpha) = \rho(\sup a_\alpha)$) for every increasing net $a_\alpha \in A_+$. It is said to be faithful if, as usual, $\rho(a) \neq 0$ if $0 \neq a \in A_+$. It is semi-finite if every nonzero $a \in A_+$, there is some $c>0$ and $0 \neq p \in P(A)$ such that $cp \leq a$ and  $\rho(p) < \infty$.

Sometimes, we will abbreviate semi-finite normal trace instead of the longer “semi-finite normal tracial weight”.

Example: Let $A$ be a von Neumann algebra with a tracial state $\tau$. Then $\tau$ is a tracial weight. (A semi-finite tracial weight $\tau$ for which $\tau(1)<\infty$ is said to be a finite weight.)

Example: Let $A = L^\infty(\mathbb{R})$ (with Lebesgue measure), and define $\rho :A_+ \to [0, \infty]$ by

$\rho (f) = \int_\mathbb{R} f(x) dx$.

Then $\rho$ is indeed a tracial  weight (obvious). It is semi-finite because the Lebesgue measure is regular, but it is not finite. It is normal because of the monotone convergence theorem.

Example: Let $A = B(H)$, and let $\{e_n\}$ be an orthonormal basis for $H$. Define $Tr: B(H)_+ \to [0,\infty]$ by

$Tr(T) = \sum_{n} \langle T e_n , e_n \rangle$.

When $\dim H < \infty$, this is just the usual trace. When $\dim H = \aleph_0$, this is just the sum over the diagonal elements in the matrix representation of $T$ in the basis $\{e_n\}$. This is, too, a normal, faithful and semi-finite tracial weight, which is not finite (you can prove this with your bare hands; it will also follow from the proof of Proposition 24 below).

Proposition 23: Let $\rho$ be a nonzero normal semi-finite trace on a factor $A$. Then

1. $\rho$ is faithful.
2. For every $p \in P(M)$, $p$ is infinite if and only if $\rho(p) = \infty$

Remark: Before the proof, note that this proposition also shows that a normal trace on a $II_1$ factor is faithful.

Proof: For (1), we will show that if $\rho$ is normal, semi-finite, and not faithful, then it is zero. It suffices to show that $\rho(1) = 0$, for then positivity implies that $0 \leq \rho(a) \leq \|a\|\rho(1) = 0$ for every $a \in A_+$, giving $\rho \equiv 0$.

If $\rho$ is not faithful, then there is some $0 \neq a \in A_+$ such that $\rho(a) = 0$. Then there is also some nonzero $p \in P(M)$ such that $\rho(p) = 0$. Now let $\{p_n\}$ be a maximal family of projections equivalent to $p$. Then $p \npreceq 1 - \sum p_n$, because $\{p_n\}$ is maximal. Since $A$ is a factor, we have by the corollary to the comparability theorem (Theorem 7) that $1 - \sum p_n \prec p$. Thus

$\rho(p_n) = \rho(p) = 0$,

and

$\rho(1 - \sum p_n) \leq \rho(p) = 0$.

But now additivity and normality of $\rho$ implies that

$\rho(1) = \rho(1-\sum p_n) + \rho(\sum p_n) = 0 + \sum 0 = 0$.

That concludes the proof of (1).

For (2), first note that an infinite projection $p$ can be written as $p = q+r$, where $q$ and $r$ are orthogonal projections equivalent to $p$ (the case of type I is immediate, and the case of types II and III is taken care of by Exercise F). But then

$\rho(p) = \rho(q) + \rho(r) = \rho(p) + \rho(p)$ .

Since part (1) rules out $\rho(p) = 0$, we must have $\rho(p) = \infty$.

Finally, let $p$ be a finite projection. By semi-finiteness, there is a nonzero  $q \in P(M)$ such that $q \leq p$ and $\rho(q) < \infty$. Let $\{q_n\}$ be a maximal family of subprojections of $p$, such that $q_n \sim q$ for all $n$. Since $p$ is finite, this family has finitely many elements, say $\{q_1, \ldots, q_N\}$. As above,  $p - \sum q_n \prec q$ by maximality, so $\rho(p - \sum q_n) \leq \rho(q)$. Therefore we find

$\rho(p) = \rho(1 - \sum q_n) + \rho(\sum q_n) \leq \rho(q) + N \rho(q) < \infty$.

Proposition 24 (existence of tracial weights): Every type $I_\infty$ factor and every type $II_\infty$ factor have a faithful, normal, semi-finite trace. This trace is unique up to a scalar factor.

Proof: Since every factor type I factor has the form $B(H)$, the example given above (the usual trace $Tr$) shows that it carries such a tracial weight when $\dim H = \infty$ (and if $\dim H < \infty$, then the usual trace is a finite tracial state satisfying all conditions). Thus, we need only consider the case of a type $II_\infty$ factor $A$ (the reader will notice though, that the proof could work for type $I_\infty$ just as well). Moreover, the previous proposition shows that faithfulness is immediate, so we only have to prove that there exists a nonzero normal semi-finite weight.

Since $A$ is type II, it has a nonzero finite projection $p$.

Claim: Let $p$ be a nonzero finite projection in $II_\infty$ factor. Then there exists a family $\{p_n\}$ of orthogonal and projections, such that $p_n \sim p$ for all $n$, and such that $\sum_n p_n = 1$

Assuming the claim for the moment, we prove the existence of a semi-finite normal tracial weight as follows. The von Neumann algebra $pAp$ is finite, so it has a WOT continuous trace $\tau$ defined on it. Let $v_n \in A$ be partial isometries so that $v_n v_n^* = p$, and $v_n^* v_n = p_n$. We define

$\rho : A_+ \to [0,\infty]$

by

$\rho(a) = \sum_n \tau(v_n a v_n^*)$.

First of all, this is well defined, because $v_n a v_n^* \in pAp$, and the summands are all non-negative. So we have a map $\rho : A_+ \to [0,\infty]$, and by properties of infinite summation of non-negative numbers, we have the first item of Definition 22.

Let us drop the habit of skipping details that we have picked up, and show that $\rho$ is normal, semi-finite, tracial weight.

We begin by showing $\sup\rho(a_\alpha) = \rho(\sup a_\alpha)$ for every increasing net $a_\alpha \in A_+$.  Write $a:= \sup a_\alpha$. By positivity,

$\rho(a) = \rho(\sup a_\alpha) \geq \rho(a_\alpha)$

for all $\alpha$, so $\rho(a) \geq \sup \rho(a_\alpha)$.

For the reverse, suppose that $M \in \mathbb{R}_+$, and that $\rho(a) > M$. Our goal is to show that $\rho(a_\alpha) > M$ for all “sufficiently large” $\alpha$.

On the one hand, there exists a finite set of indices $F$ such that

$\sum_{n\in F} \tau_{n}(a) > M$.

(In case that $A$ acts on a separable Hilbert space, the family $\{p_n\}$ is an infinite sequence, and we could say that there exists an integer $N$ such that $\sum_{n=1}^N \tau_n(a) > M$.)

On the other hand, we have

$\tau_n(a_\alpha) : = \tau(v_n a_\alpha v_n^*) \to \tau_n(a)$

for all $n$, because $\tau$ is WOT continuous. So there is some $\alpha_0$, such that

$\sum_{n\in F} \tau_n(a_\alpha) > M$

for all $\alpha \geq \alpha_0$. For such $\alpha$, we find

$\rho(a_\alpha) = \sum_n \tau_n(a_\alpha) \geq \sum_{n\in F} \tau_n(a_\alpha) > M$.

This shows that $\sup_\alpha \rho(a_\alpha) \geq \rho(a)$, and normality is established.

Next, let us show that $\rho$ is tracial. Since we have already dealt with normality, the following formal calculations are legal:

$\rho(aa^*) = \sum_n \tau(v_n a a^* v_n^*)$

$(*) = \sum_n \sum_m \tau(v_n a p_m a^* v_n^*) = \sum_{m,n} \tau(v_n a v_m^* v_m a^* v_n^*)$

$(**) = \sum_{m,n} \tau (v_m a^* v_n^* v_n a v_m^*)$

$(*) = \sum_m \tau(v_m a^* a v_m^*) = \rho(a^* a)$.

Equations (*) follow from $\sum p_n = 1$ (SOT) and (**) follows from $\tau$ being a trace on $pAp \ni v_n a v_m^*$.

It remains to show that $\rho$ is semi-finite. For every finite subset $F$ of indices $n$, let $P_F := \sum_{n \in F} p_n$. Now, for every $F$, $P_F$ is a finite projection. If $0 \neq a \in A_+$, then the net $a_F := P_F a P_F$ converges SOT to $a$. Therefore, there is some $F$ such that $a_F \neq 0$. Now $a_F \in P_F A P_F$ – a finite von Neumann algebra. Therefore, there is a projection $0 \neq q \in P_F A P_F \in A$, such that $cq \leq a_F$ and $\rho(q) = \sum_{n \in F} \tau(v_n q v_n^*) < \infty$. This shows that $\rho$ is semi-finite.

Finally, the uniqueness of $\rho$ follows from uniqueness of the trace on a finite type II factor. It seems like the good time to revert to the habit of skipping details 🙂

Definition 25: A von Neumann algebra is said to be semi-finite if it is type I or type II.

Thus, a factor is semi-finite if and only if it is not type III. We have seen that semi-finite algebras have normal, faithful, semi-finite traces. The converse will be established below. In the meanwhile, I did not forget that we owe ourselves the following:

Proof of claim: Let $\{p'_n\}$ be a maximal family of orthogonal projections such that $p'_n \sim p$. Then $p \npreceq 1 - \sum p'_n$ by maximality, so $1 - \sum p'_n \prec p$. If $1 = \sum p'_n$, then we put $p_n = p_n'$, and we are done.

Assume that $1 \neq \sum p'_n$. Let $v$ be a partial isometry such that $v^*v = 1 - \sum p'_n$ and $vv^* = q < p$. Note that $\sum p'_n + v^* v = 1$, so this implies that the family $\{p_n\}$ is infinite (see Exercise G below). For the proof, we will assume that this family is an infinite sequence $\{p_n\}_{n=1}^\infty$ (by the end of the proof, it should be clear what to do if the cardinality of the family is greater than $\aleph_0$; if $A$ acts on a separable Hilbert space, then of course the cardinality cannot be strictly greater than $\aleph_0$).

Now, being equivalent to $p = q + (p-q)$, every $p'_n$ breaks up as $p'_n = r_n + q_n$, where $q_n \sim q$ and $r_n \sim p - q$. Now we define a new family of orthogonal projections, $\{p_n\}$ by

$p_1 = r_1 + 1 - \sum p_n'$

and

$p_{n+1} = r'_{n+1} + q_n$.

Then $p_n \sim p$ and $\sum p_n = \sum p_n' + 1 - \sum p_n' = 1$.

Exercise G: Prove that if $P \leq Q$ and $Q$ is finite, then $P$ is finite. Prove that if $P \sim Q$ and $Q$ is finite, then $P$ is finite. Prove that if $P,Q$ are finite and orthogonal projections, then $P+Q$ is finite.

#### 7. III factors

After Murray and von Neumann initiated the program of classification into types, they determined all type I algebras and gave examples of type II factors, but at first it was not known whether there exist type III algebras. Then von Neumann provided an example, and later Powers found uncountably many examples, and the classification problem for type III von Neumann algebras is still today a very active field of research. We will see examples of type III factors later on in this course. For now, we record the following result that is one of the technical keys for showing that a factor is type III.

Proposition 26: A factor $A$ is of type III if and only if there does not exist a semi-finite normal trace on $A$

Proof: We already know, by the previous proposition, that if $A$ is not type III, then there exists a normal semi-finite trace on it. On the other hand, if $A$ is type III, then all projections in $A$ are infinite. Proposition 23(2) now tells us that if there was a semi-finite normal trace $\rho$ on $A$, then necessarily $\rho(p) = \infty$ for all $p \in P(A)$, but such a weight cannot be semi-finite. This completes the proof.