Do projections in an $AW^\ast$-algebra form an orthomodular lattice?

Question

I’m currently studying orthomodular lattices arising out of operator algebras. One of the most standard examples is the projection lattice of a von Neumann algebra - if $M$ acts on a Hilbert space $H$, then $P(M)$ is a complete sub-ortholattice of $P(H)$, the lattice of closed subspaces of $H$, which is well-known to be orthomodular. So, $P(M)$ is always a complete orthomodular lattice.

I’ve recently found out that the projections in an $AW^\ast$-algebra form a complete ortholattice as well. See, for example, Corollary 3 in Section 2 of this paper by Kaplansky (which is one of the papers where he originally defined $AW^\ast$-algebras). However, I’m unable to find a reference or a proof on whether these lattices are orthomodular. To summarize:

Let $M$ be an $AW^\ast$-algebra. Is the ortholattice of projections of $M$ an orthomodular lattice?

The result is clearly true when $M$ is commutative, in which case $P(M)$ is a Boolean algebra. More generally, $P(M)$ is modular when $M$ is finite. See Theorem 6.3 in the paper by Kaplansky cited above.

As an additional question, I’ve seen it implied somewhere before (I don’t remember where) that the projections in an arbitrary unital $C^\ast$-algebra form an ortholattice as well. Is this true? In addition, when $P(A)$ is an ortholattice for some unital $C^\ast$-algebra $A$, is it necessarily orthomodular?

It seems that there are unital C*-algebra whose projections do not form a lattice (see mathoverflow.net/questions/86370/… and [mathoverflow.net/questions/172073/…). — Jordi
– Jordi, Commented Oct 2, 2024 at 12:43
But the projections do always form an orthocomplemented poset (this is for instance stated in Proposition 3.3 here math.nmsu.edu/people/personal-pages/files/…, I cannot find a better reference now). — Jordi
– Jordi, Commented Oct 2, 2024 at 13:17
The readers of this post may find this article interesting & somewhat related, so I'm just dropping the link here doi.org/10.1016/j.jmaa.2014.02.041 — Onur Oktay
– Onur Oktay, Commented Oct 2, 2024 at 18:36

Jordi · Accepted Answer · 2024-10-03 13:30:36Z

I think we have the following:

The projections in a unital C$^*$-algebra always form an orthomodular poset. Indeed, observe that $p\leq q$ means that $pq=p.$ Taking $^*$ we see that $p$ and $q$ commute and hence also $p$ and $1-q$ commute. Thus $p\vee (q\wedge (1-p))=p\vee(q-p)=p+q-p+p(q-p)=q$ (see Proposition 3 in Chapter 1, section 1 here).
If $A$ is a weakly Rickart C$^*$-algebra (defined bellow) then its projections form a lattice (see Proposition 7 in Chapter 1, section 5 here).
If $A$ is a Rickart C$^*$-algebra (defined bellow) then its projections form an orthomodular lattice by 1 and 2 (a Rickart C$^*$-algebra is weakly Rickart).

Let $A$ be a C$^*$-algebra and $a\in A$. The right annihilator of $a$ is defined as $R(a):=\{x\in A:ax=0\}.$ Rickart C$^*$-algebra is a C$^*$-algebra such that for every $a\in A$ there exists a projection $p\in A$ such that $R(a)=pA.$ Every Rickart C$^*$-algebra is unital (see definition 2 and Proposition 2 in Chapter 1, section 3 here).

A weakly Rickart C$^*$-algebra is a C$^*$-algebra such that every $a\in A$ has an annihilating right projection, that is, a projection $p$ such that $ap=a$ and $ab=0$ implies $pb=0.$ A Rickart C$^*$-algebras are unital weakly Rickart C$^*$-algebras (see Proposition 2 in Chapter 1, section 5 here).

AW$^*$-algebras are are precisely those Rickart C$^*$-algebras whose projections form a complete lattice (see Proposition 1 in Chapter 12, section 4 here).

Remark: Actually the paper I linked in the comments above already claims that the projections form an orthomodular poset. In (1) I have checked the (ortho)modular condition because I was thinking that we only could assume that we had an orthomodular poset. I will leave the computations in case someone finds them useful. Another reference for that is this, however one needs to take into account that if $A$ is a C$^*$-algebra then $A_{sa}$ is a JB-algebra and $\leq$ is the same in the C$^*$-algebra sense and in the Jordan sense.

Thanks! I was wondering if I had missed some quick proof of orthomodularity. Should have realized it just follows because the condition of orthomodularity only deals with commuting projections… And thanks for the references both in your answer and in your comments! — David Gao
– David Gao, Commented Oct 2, 2024 at 18:57
Btw, in your paragraph defining weakly Rickart $C^\ast$-algebras, did you mean $ap = a$ instead of $ap = p$? — David Gao
– David Gao, Commented Oct 2, 2024 at 18:59

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