A continuous lattice is a complete lattice $L$ in which every element $y$ is equal to $\bigvee${$x \in L \mid x \ll y$} where $x \ll y$ ("x approximates y" or "x is way below y") if for any directed set $D \subseteq L$, $y \leq \bigvee D$ implies that there is a $d \in D$ such that $x \leq d$.

It is known that the set of projections in an arbitrary W*-algebra is a complete orthomodular lattice. I would like to know for which kind of W*-algebras this lattice is also continuous.

  • $\begingroup$ Is the article by Kaplansky "Any orthocomplemented complete modular lattice is a continuous geometry." useful? $\endgroup$
    – Abel Stolz
    Jun 7, 2013 at 7:38
  • $\begingroup$ Theorem 6 of this article states that any orthocomplemented modular lattice of type II is a continuous geometry. A continuous geometry is a lattice L that is complemented, modular, meet continuous, and join continuous. I don't know how it relates to continuous lattices but I know that continuous lattices can be characterized equationally : encyclopediaofmath.org/index.php/Continuous_lattice $\endgroup$
    – Rennela
    Jun 9, 2013 at 18:34
  • $\begingroup$ According to Gierz' Compendium on continuous lattices, continuous lattices form an overlapping subclass of continuous geometry. $\endgroup$
    – Rennela
    Jun 10, 2013 at 9:28

2 Answers 2


I'm going to say the von Neumann algebra has to be either atomic abelian or finite dimensional for this to happen. If $M$ is nonatomic then I don't think any projection besides 0 is "way below" any other projection, because if $p$ is nonzero then we can write $p = \bigvee p_\alpha$ where the $p_\alpha$ are directed and all strictly less than $p$, so if $p \leq q$ we have $q = \bigvee (p_\alpha + q-p)$ and this shows that $p$ is not way below $q$. So you need minimal projections to even have a chance. But $B(H)$ fails for infinite dimensional $H$: take $H = l^2$, let $p_1$ be the rank one projection whose range is spanned by the vector $e_1$, and for $n \geq 2$ let $p_n$ be the rank one projection whose range is spanned by the vector $e_1 + \frac{1}{n}e_n$. Then $e_1$ is not contained in the span of $p_2 \vee \cdots \vee p_n$ because if it were then $e_j$ would be too for $2 \leq j \leq n$, but the span of $p_2 \vee \cdots \vee p_n$ only has dimension $n-1$. However, the range of $\bigvee p_n$ contains vectors arbitrarily close to $e_1$, hence it contains $e_1$, hence it contains $e_n$ for $n \geq 2$, hence it is everything. Thus $I$ is the join of the directed sequence $p_2$, $p_2 \vee p_3$, $p_2 \vee p_3 \vee p_4$, $\ldots$, but $p_1$ is not less than any of these projections. This shows that $p_1$ is not way below $I$, and by symmetry no nonzero projection is way below $I$.

  • $\begingroup$ The lattice of projection of a finite von Neumann algebra is isomorphic to a lattice of closed subspaces $\mathbb{C}^n$. I found out that the lattice of the closed balls of a separable Hilbert space, ordered by reversed inclusion, is a continuous lattice (Edalat, Heckmann, "A computation model for metric spaces"). But if I follow your argument, for the usual order on von Neumann algebras, I can only hope to get a continuous lattice with projections in a finite-dimensional von Neumann algebra, right? $\endgroup$
    – Rennela
    Jun 9, 2013 at 19:21
  • 1
    $\begingroup$ No, I think it either has to be finite dimensional or atomic abelian, like $l^\infty$. $\endgroup$
    – Nik Weaver
    Jun 10, 2013 at 3:00
  • 1
    $\begingroup$ Is it possible to "combine" these two cases, like a C*-sum of finite dimensional algebras? $\endgroup$ Jun 17, 2013 at 19:15
  • $\begingroup$ Oh yeah, good point. A direct sum of the two types would work. $\endgroup$
    – Nik Weaver
    Jun 17, 2013 at 20:45

The already given answer and comments have essentially solved the problem; however I
will show the way the question can be studied from the point of view of
"classical" lattice theory (say, the times of von Neumann and
Birkhoff, then Halperin, Kaplansky, F. and M. Maeda)

For complete lattices, it is well known that algebraic implies
Scott-continuous implies meet-continuous.

{Remember the
definitions: a complete lattice is a partial order where every subset
has a supremum (equivalently, every subset has a infimum). It is
meet-continuous when increasing joins distribute over meet
(dually for join continuity); it is Scott continuous when every element
$a$ is the join of elements $b$ "way below" $a$ (meaning: every
increasing join that is at least $a$ has a member that is at least $b$);
it is algebraic when meet-continuous and each $a$ is the join of
join-inaccessible elements $b$ (elements that are "way below"

The term "continuous" tout-court might seem suitable for
Scott-continuity, since the latter coincides with embeddability (with
preservation of all meets and increasing sups) in a power of the real
interval $[0,1]$ (the "archetipe" of the continuum), in the same ways
as algebraicity uses the two-element chain $\{0,1\}$ (the "archetipe"
of discreteness) instead of $[0,1]$. However, this also means that such
a concept of continuity is a generalization of (and not a opposition to)

For classical objects of lattice theory (like complemented modular
lattices or orthomodular lattices) which are related to (discretely or
continuously valued) dimension functions, the term "continuity" is
traditionally associated with meet-continuity and its dual (von
Neumann's [meet and/or join] "continuous geometries").

{For the concept of semiorthogonality (used below) and related
ones, see S. Maeda papers, especially the last paper (1961, freely
available inside projecteuclid) about dimension lattices.}

The point (to be proved below): for such classical (complete, relatively
semi-orthocomplemented) lattices, Scott-continuity coincides with
algebraicity and even "compactly atomistic" (hence usually even more:
direct product of "finite discrete factors": the discrete subcase of
continuous geometries).

The lemma: in a complete (relatively) semiorthocomplemented lattice,
"$b$ way below $a$" implies "$b$ join inaccessible" (hence compact
in the meet-continuous case).

Proof: suppose $b$ inceasing join of the $b_i$. Fix a
semi-orthocomplement $c$ of $b$ in $a$; then $b_i\oplus c$ is a
increasing family, with join $a$ (it contains $c$ and the join $b$ of
the $b_i$; conversely, each $b_i\oplus c$ is contained in $b\oplus c=a$). By definition of "$b$ way below $a$", $\exists i:b_i\oplus c\geq b$. Adding $c$, $b_i\oplus c\geq b\oplus c$. Now, write
$b=b_i\oplus c_i$; so: $b_i\oplus c\geq b_i\oplus c_i\oplus c$; hence:
$c_i=0$, $b_i=b$.

From the lemma it follows that a Scott-continuous relatively
semiorthocomplemented lattice is algebraic (it is meet-continuous, hence
"$b$ way below $a$" implies "$b$ compact" and so each element is
join of compact elements); then it is also atomistic (i.e. sectionally
semicomplemented and atomic: use relatively complemented [whci follows
from semi-ortho-complemented] and weakly atomic [which follows from
algebraic]: every interval $[a,b]$ contains a covering $\{a',b'\}$ and
the realtive orthocompement $c'$ of $a'$ in $b'$ is an atom in $[a.b]$:
if not, $c'$ properly splits in $x\oplus y$ and so between $a'$ and $b'$
one has $a'\oplus x$).

Once one has a compactly atomistic lattice, a weak (semimodularity)
condition (like the covering property, which is satisfied by
meet-continuous geometries and projection ortholattices of
AW$^*$-algebras and their Jordan analogues) is well known to imply a
decomposition into a direct product of subdirectly irriducible factors
of the same kind (see "matroid lattices" in F. Maeda - S. Maeda book,
theory of symmetric lattices).

In the semimodular and orthomodular case (i.e. dimension ortholattices),
the irreducible factors are exactly the finite-dimensional
orthocomplemented projective geometries; conclusion: Scott-continuous
semimodular otholattices are exactly the direct products of
(irreducible) orthocomplemented finite-dimensional projective
geometries. [The W$^*$ or AW$^*$ algebras that give such ortholattices
are exactly the direct products, in the category of such algebras, of
matrix $*$-algebras over the real, complex or quaternional $*$-field.]

{The direct product as (algebras or) rings (with involution) of
matrix $*$-algebras with scalar entries is a $*$-regular ring hence it
is not even a C$^*$-algebra (unless there are only finitely many
nontrivial factors); the direct product in the suitable category is
smaller (one takes only the bounded elements of the $*$-regular ring,
see Berberian's book "Baer$^*$-rings").}

In the complemented modular case (i.e. meet-continuous geometries), the
irreducible factors are exactly the (possibly infinite dimensional)
projective geometries (associated to vector spaces over a sfield, except
possibly nonarguesian planes, and lines and points). Conclusion: the
Scott-continuous complemented modular lattices are exactly the (possibly
reducible) projective geometries; subcase: the Scott-continuous
continuous geometries are exactly the direct products of
finite-dimensional projective geometries.

Note: this generalizes the characteriazion of Scott-continuous boolean
algebras in the "compendium of continuous lattices".


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.