Let $(M,\tau)$ be a finite von Neumann factor (in my case $M=R^\omega$, but I don't think this additional hypothesis might be useful for this particular problem) and fix a projection $p$. Let $\tau\_p$ (resp. $\tau\_{1-p}$) denote the unique normalized trace on $pMp$ (resp. $(1-p)M(1-p)$.


**Convention.** Given two projections $q\_1,q\_2$ in some finite factor, I can do several operations between them; for instance, $q\_1\vee q\_2$, but also $q\_1\wedge q\_2^{\perp}$. I will describe these operations using *formulas*. So when I will say that a certain property of two projections is true for all formula $f$, will mean that this property is true for any projection obtained by $q\_1,q\_2$ doing operations using $\vee, \wedge$ and $\perp$. I will use the formulas as functions; namely, $f(q\_1,q\_2)$ is the projection obtained by $q\_1,q\_2$ when the formula $f$ is applied. For instance, $f(q\_1,q\_2)=q\_1\vee q\_2$.


Suppose that we have the following data:


* projections $p\_1,p\_2,q\_1,q\_2\in pMp$
* projections $p\_1',p\_2',q\_1',q\_2'\in (1-p)M(1-p)$


such that for all formula $f$ one has


$$
\tau\_p(f(p\_1,p\_2))\sim\tau\_p(f(q\_1,q\_2)) \qquad\text{and}\qquad\tau\_{1-p}(f(p\_1',p\_2'))\sim\tau\_{1-p}(f(q\_1',q\_2'))
$$



> 
> **Question:** Is it true that for all formula $f$, one has
>  $$
> \tau(f(p\_1+p\_1',p\_2+p\_2')\sim\tau(f(q\_1+q\_1',q\_2+q\_2')) ?
> $$
> 
> 
> 


It should be reasonable. Basically I have two pairs of component-wise orthogonal projections $(p\_1,p\_2)$ and $(p\_1',p\_2')$ and then I have another two pairse of component-wise orthogonal projections that have approximately the same geometry as the first two pairs. I want to conclude that the sums still have approximately the same geometry...
I'd been trying a bit but I am in trouble, basically because there is no way to distribute the sum of projections with $f$. One needs maybe some clever idea (or a counterexample), but I'm stuck now.


Thanks in advance for any help,


Valerio



Geometric approximation of projections in a finite von Neumann factor
The point here is that you can distribute the sum over any function $f$:


This is almost obvious for the elementary operations:
Let $p\_1,p\_2\leq p$ and $q\_1,q\_2\leq 1-p$, then we compute that
$(p\_1+q\_1)\wedge (p\_2+q\_2)=(p\_1\wedge p\_2)+(p\_1\wedge q\_2)+(q\_1\wedge p\_2)+(q\_1\wedge q\_2)=(p\_1\wedge p\_2)+(q\_1\wedge q\_2)$. Moreover, $(p\_1\wedge p\_2)\leq p$ and $(q\_1\wedge q\_2)\leq 1-p$. A similar thing is true for the operation $\vee$, and $(p\_1+q\_1)^\perp=(p-p\_1) + (1-p - p\_2)=p\_1^\perp + p\_2^\perp$.


By induction, we find that, for every function $f$, we have $f(p\_1+q\_1,p\_2+q\_2)=f(p\_1,p\_2)+f(q\_1,q\_2)$, and $f(p\_1,p\_2)\leq p$, $f(q\_1,q\_2)\leq 1-p$.


