"adding" a projection to von Neumann algebra This is a question about what happens when you "add" a new projection $p$ to a von Neumann algebra $\mathcal{R}$ to generate a larger v.N. algebra $(\mathcal{R} \cup \{p\})''$.
Suppose that $\mathcal{R}$ is a non-type-I factor acting on Hilbert space $H$. When $p \not \in \mathcal{R}$ is any (bounded) projection operator on $H$, call $p$ "$\mathcal{R}$-finite" if it is the join of finitely many minimal projections in $(\mathcal{R} \cup \{p\})''$.
If $p$ and $q$ are $\mathcal{R}$-finite, must their join $p \vee q$ also be $\mathcal{R}$-finite? ($p \vee q$ meaning the projection onto the closed span of $p$ and $q$.)
$p \vee q$ will be trivially $\mathcal{R}$-finite if the ranges of $p, q$ are finite-dimensional subspaces of $H$, so assume this is not the case. (For example $p$ could be a minimal projection in some type I algebra $\mathcal{S} \supseteq \mathcal{R}$ that is not the whole algebra $B(H)$ of all bounded operators on $H$, and $q$ could be a minimal projection in another such algebra.)
 A: [Update: the following makes use of an invalid premise but I'm leaving it up because it may be possible to patch it up.]
I believe I can construct a counterexample assuming the existence of a separating and cyclical vector $v$ for $\mathcal{R}$, and assuming the truth of Claim 1 below.
Call a projection $a \not \in \mathcal{R}$ a "separating projection" for $\mathcal{R}$ if every nonzero vector in $a$'s range in separating for $\mathcal{R}$.
Claim 1: if $a$ is a separating projection for $\mathcal{R}$ then it is $\mathcal{R}$-minimal (i.e. it is minimal in $(\mathcal{R} \cup \{a\})''$).
(I do not have a proof of this yet but I strongly suspect it's true; in fact I suspect it generalizes to a characterization of the $\mathcal{R}$-minimal projections as follows. For a projection $p$, let $p \uparrow \mathcal{R}$ denote the least projection in $\mathcal{R}$ whose range includes $p$'s, and similarly for a vector $v$ let $v \uparrow \mathcal{R}$ denote the least projection in $\mathcal{R}$ with $v$ in its range. Then I suspect $p$ is $\mathcal{R}$-minimal iff $v \uparrow \mathcal{R} = p \uparrow \mathcal{R}$ for every nonzero $v$ in $p$'s range.)
Fix a vector $v$ that is separating and cyclical for $\mathcal{R}$. Let $\{ p_n : n \geq 0 \} \subseteq \mathcal{R}$ and $\{ q_n : n \geq 0 \} \subseteq \mathcal{R}'$ be two resolutions of the identity, that is, maximal sets of pairwise-orthogonal projections.
Proposition: for our separating/cyclical $v$, if $p \in \mathcal{R}$ and $q \in \mathcal{R}'$ are nonzero, then $pqv (=qpv)$ is nonzero. [this is the invalid premise -- see Can a vector separating for both $\mathcal{R}$ and $\mathcal{R}'$ be annihilated by successively applying projections from each? . hopefully it can be patched up.]
For $n \geq 0$, let $v_n$ be the vector $q_{2n} v$; and let $w_n$ be the vector $p_n q_{2n+1} v + (1-p_n) q_{2n} v$.
Let $a$ be the projection onto the closed span of all the $v_n$'s, and let $b$ be the projection onto the closed span of all the $w_n$'s.
Claim: $a$ and $b$ are both $\mathcal{R}$-finite (indeed $\mathcal{R}$-minimal) but $a \vee b$ is not.
To show $a$ is $\mathcal{R}$-finite we will show it is a separating projection for $\mathcal{R}$ and invoke Claim 1 above. Let $z = c_0 v_0 + c_1 v_1 + \ldots$ be an arbitrary nonzero member of $a$'s range. Let $n$ be arbitrary such that $c_n \neq 0$. Let $T \in \mathcal{R}$ be an arbitrary nonzero operator. Consider $q_{2n} T z$. Since $q_{2n} \in \mathcal{R}'$,
$q_{2n} T z = T q_{2n} z$,
and since $z = c_0 q_0 v + c_1 q_2 v + c_2 q_4 v + \cdots$, with all the $q_{2i}$'s mutually orthogonal, $T q_{2n}z = T c_n q_{2n} v$. By the Proposition above, this is nonzero. And since $q_{2n} T z$ is nonzero, $T z$ is nonzero, so $z$ is indeed separating for $\mathcal{R}$.
The proof that $b$ is a separating projection for $\mathcal{R}$ is basically identical. Let $z$ now be an arbitrary vector in $b$'s range, namely $z = c_0 w_0 + c_1 w_1 + \ldots$, with some particular $c_n$ nonzero. Again fix nonzero $T \in \mathcal{R}$. The ranges of $p_n$ and $(1-p_n)$ cannot both be in the kernel of $T$. Suppose $p_n$ is not, so $T p_n \neq 0$. Then as before we have:
$q_{2n+1} T z = T q_{2n+1} z = T q_{2n+1} (p_n q_{2n+1} v) = T p_n q_{2n+1} v$,
and our Claim 1 above entails $q_{2n+1} T z \neq 0$, so $T z \neq 0$. The case where $T (1-p_n) \neq 0$ is symmetrical.
To show that $a \vee b$ is not $\mathcal{R}$-finite, we will prove that $\{ p_n \wedge (a \vee b) : n \geq 0 \}$ is an infinite set of orthogonal subprojections of $a \vee b$ that exist in $(\mathcal{R} \cup \{a \vee b\} )''$. (It should however be justified that this entails $\mathcal{R}$-finiteness as we defined it.)
Let $x_n$ denote the vector $v_n - w_n$, which is clearly nonzero and in the range of $a \vee b$. Expanding using our definitions above, we get
$x_n = q_{2n} v - (p_n q_{2n+1} v + (1-p_n) q_{2n} v)$
$ = (p_n q_{2n} v + (1-p_n) q_{2n}v) - (p_n q_{2n+1} v + (1-p_n) q_{2n} v)$
$ = p_n q_{2n} v - p_n q_{2n+1} v $
$= p_n (q_{2n} - q_{2n+1}) v.$
Thus $x_n$ is also in the range of $p_n$, so $p_n \wedge (a \vee b) \neq 0$.
