Let $M$ be a von Neumann algebra, and let $\mathcal{P}$ be the set of nontrivial (not equal to $0$ or $e$) projections of $M$. Define $p,q \in \mathcal{P}$ to be equivalent if there exist projections $p_1, \ldots, p_n \in \mathcal{P}$ with $p_1=p$, $p_n = q$, $p_i \perp p_{i+1}$ and $p_i + p_{i+1} < e$ for $1 \leq i < n$.

If $p \in \mathcal{P}$ is maximal, then clearly no other projection is equivalent to it. The question is: are every two nonmaximal projections $p$ and $q$ equivalent?

This is true in $B(H)$. If $\dim(H) \leq 2$, then all projections in $\mathcal{P}$ are maximal, so we may assume $\dim(H) \geq 3$. Clearly $p$ and $q$ are equivalent to rank 1 projections, so we may assume that $p$ and $q$ are rank 1. Then the orthogonal complements of the range of $p$ and $q$ are codimension 1 and hence have nontrivial intersection: take $p_2$ the projection onto (a subspace of) this intersection and $p_3 = q$.

It is also true in $L^\infty$; the problem becomes a set-theoretic one on the $\sigma$-algebra. If $p \vee q \not= e$, then $p$ and $q$ are equivalent to $e-p \vee q$, so assume $p \vee q = e$; this implies that $p$ and $q$ are incomparable. Now if $p \perp q$, then by the nonmaximality of $p$, $p$ is equivalent to a subprojection $p_2$ of $q$, and $q= p_4$ is equivalent to a subprojection $p_3$ of $p$. If $p$ and $q$ are not orthogonal, then $p \sim q - p \wedge q \not= 0$ and $q \sim p - p \wedge q \not= 0$. This proof uses the fact that if $p$ and $q$ are not orthogonal, then $p \wedge q \not= 0$, which is unfortunately no longer true if $M$ is not abelian.

Is it true for arbitrary von Neumann algebras? If not, what conditions (e.g., type I) on $M$ are needed?

P.S. I am not an expert on von Neumann algebras.

  • 1
    $\begingroup$ Is there any motivation for defining this equivalence relation? $\endgroup$
    – Rasmus
    Mar 10, 2015 at 9:03
  • 2
    $\begingroup$ I'm looking at triangles with the corners projections $p,q,r$ that are orthogonal and satisfy $p+q+r=e$. Given two such triangles, I need to connect them with a sequence of triangles that have one corner point in common. This can be done if each nonmaximal projection is equivalent to each other one in my sense. This is needed to classify Hilbert's metric isometries on cones of von Neumann algebras. $\endgroup$ Mar 10, 2015 at 10:13

2 Answers 2


This is true. Since $p, q \in \mathcal P(M)$ are non-trivial and not maximal, then under your definition they are equivalent to any non-trivial subprojection. Also, we have $M \not= \mathbb M_2(\mathbb C)$, and so there exists a non-trivial projection $z \in \mathcal P(M)$ commuting with $p$ and $q$, such that $zp \not= 0$ (see, e.g., Theorem 1.41 in Volume 1 of Takesaki's books).

Then we can consider separate cases:

If $q = p^\perp$, then $q \sim p_0 \sim p$, where $p_0$ is any projection $0 \not= p_0 \lneq p^\perp$.

If $z = q$ then $q \sim zp \sim p$.

If $z^\perp = q$ and $z \not= p$, then $q \sim zp \sim p$.

If $zq \not= 0$ and $z \not= q$, then $q \sim zq \sim z^\perp \sim zp \sim p$.

If $zq = 0$, $z^\perp \not= q$, then $q = z^\perp q \sim zp \sim p$.

  • $\begingroup$ Your $z$ need not be non-trivial; in Takesaki Theorem V.1.41, $M$ is not the entire algebra but the algebra generated by the projections. An example where $z$ is the identity can be constructed by taking projections $e$ and $f$ (Takesaki's notation) that satisfy equations $(*)$ on page 306. Then, if one traces the proof of V.1.41, $z= e + e^\perp = 1$. An example of projections that satisfy $(*)$ can be constructed inside $M_4(\mathbb{C})$ by taking projections onto 2-dimensional subspaces $E$ and $F$ such that $E \cap F = E \cap F^\perp = E^\perp \cap F = E^\perp \cap F^\perp = \{0\}$. $\endgroup$ Mar 11, 2015 at 11:31
  • 1
    $\begingroup$ The $z$ will not be the same as the $z$ in Takesaki's book, sorry for the confusion. Theorem V.1.41 in Takesaki shows that either $W^*(p,q)$ is not a factor, in which case $W^*(p, q)' \cap M \not= \mathbb C$. Or else $W^*(p, q)$ is $\mathbb C$ or $\mathbb M_2(\mathbb C)$, and then we again have $W^*(p, q)' \cap M \not= \mathbb C$ since $M \not\cong \mathbb C$ and $M \not\cong \mathbb M_2(\mathbb C)$ (this is an easy exercise to show). If $z$ is any non-trivial projection in $W^*(p, q)' \cap M$ then either $zp \not= 0$, or else $z^\perp p \not= 0$. $\endgroup$ Mar 11, 2015 at 18:32
  • $\begingroup$ I understand your approach now, except that I don't see the solution to your easy exercise... $\endgroup$ Mar 13, 2015 at 12:12
  • 1
    $\begingroup$ Suppose $\mathbb M_n(\mathbb C) \subset M$, and let $e_{i j} \in M$ denote the matrix with $1$ in the $ij$th position and $0$ elsewhere. Set $N = e_{1 1} M e_{1 1}$. Then show that the map $\theta: M \to \mathbb M_{n}(N)$ given by $\theta(x)_{ij} = e_{1 i} x e_{j 1}$ is a $*$-isomorphism with inverse $\theta^{-1}( ( a_{i j} )_{ij} ) = \sum_{i j = 1}^n e_{i 1} a_{i j} e_{1 j}$. Considering diagonal matrices then gives an isomorphism $\mathbb M_n(\mathbb C)' \cap M \cong N$. $\endgroup$ Mar 15, 2015 at 14:13

This is a fully worked out answer based mostly on the answer and comments by Jesse Peterson.

If two non-maximal projections commute, then they are equivalent; this follows from the same $L^\infty$-proof as in the question (it is a set-theoretic thing, drawing a Venn diagram makes it clear).

Claim: it suffices to find a non-trivial projection $z \in M$ that commutes with $p$ and $q$ (or equivalently, the von Neumann algebra $W^*(p,q)' \cap M$ is not $\mathbb{C}$).

Proof(claim): In this case $z^\perp$ also commutes with $p$ and $q$, so by the above we are done if either $z$ or $z^\perp$ is non-maximal. Suppose they are both maximal; then they are both minimal as well, and therefore $z \geq pz \in \{0, z\}$ and $z^\perp \geq p z^\perp \in \{0, z^\perp\}$, and so $p = p z + p z^\perp \in \{0,z,z^\perp, e\}$, contradicting the non-maximality of $p \in \mathcal{P}$.

Consider $W^*(p,q)$. If it is not a factor, then $\mathbb{C} \not= W^*(p,q)' \cap W^*(p,q) \subset W^*(p,q)' \cap M$. If it is a factor, then Theorem V.1.41(ii) in Takesaki I shows that either $W^*(p,q)$ is of type $I_2$, or $W^*(p,q)$ is abelian. In the second case $p$ and $q$ commute and we are done, and in the first case $ \mathbb{M}_2(\mathbb{C}) \cong W^*(p,q) \subset M$.

Let $e_{ij} \in \mathbb{M}_2(\mathbb{C}) \subset M$ be the matrix with $1$ in the $ij$-th position and $0$ elsewhere. Let $N := e_{11} M e_{11}$. Consider the map $\phi \colon M \to \mathbb{M}_2(N)$ given by $\phi(x)_{ij} := e_{1i} x e_{j1}$. Then, using $\sum_{k=1}^2 e_{kk} = e$, $$ (\phi(x) \phi(y))_{ij} = \sum_{k=1}^2 \phi(x)_{ik} \phi(y)_{kj} = \sum_{k=1}^2 e_{1i} x e_{k1} e_{1k} y e_{j1} = \sum_{k=1}^2 e_{1i} x e_{kk} y e_{j1} = e_{1i} x e_{j1} = \phi(xy)_{ij}, $$ $$ (\phi(x)^*)_{ij} = (\phi(x)_{ji})^* = (e_{1j}x e_{i1})^* = e_{1i} x^* e_{j1} = \phi(x^*)_{ij}. $$ So $\phi$ is a *-homomorphism. Moreover, the map $\theta \colon \mathbb{M}_2(N) \to M$ given by $\theta(a_{ij}) := \sum_{ij=1}^2 e_{i1} a_{ij} e_{1j}$ satisfies $$ \theta(\phi(x)) = \sum_{ij=1}^2 e_{i1}x e_{j1} e_{1j} = \sum_{ij=1}^2 e_{ii} x e_{jj} = exe = x, $$ $$ \phi(\theta(a_{ij}))_{kl} = e_{1k} \sum_{ij=1}^2 e_{i1} a_{ij} e_{1j} e_{l1} = e_{11} a_{kl} e_{11} = a_{kl}. $$ Hence $\phi$ is invertible and so it is a *-isomorphism. Under this *-isomorphism, $\mathbb{M}_2(\mathbb{C})$ is identified with $\mathbb{M}_2(e_{11} \mathbb{C} e_{11}) \subset \mathbb{M}_2(N)$, since $\phi(e_{kl})_{ij} = e_{1i} e_{kl} e_{j1} = \delta_{ik} \delta_{lj} e_{11} = e_{11} (e_{kl})_{ij} e_{11}$. It is now clear that $W^*(p,q)' \cap M \cong \mathbb{M}_2(\mathbb{C})' \cap \mathbb{M}_2(N)$ is the set of matrices with a single element from $N$ on the main diagonal and $0$ elsewhere, and so it is isomorphic to $N$. $N$ cannot be trivial, otherwise $M \cong \mathbb{M}_2(\mathbb{C})$, contradicting the non-maximality of $p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.