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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

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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 $f(p_1,p_2)\leq p$, $f(q_1,q_2)\leq 1-p$.

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  • $\begingroup$ It's what I thought, actually. However I was little scared by the "general $f$". Also, I have actually $k$ projections, and not only $2$, but I think that nothing changes.. right? $\endgroup$
    – Valerio Capraro
    Commented May 8, 2012 at 12:25
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    $\begingroup$ It certainly does not change anything. The property $f(p_1+q_1,\ldots,p_n+q_n)=f(p_1,\ldots,p_n)+f(q_1,\ldots,q_n)$ is valid, at least if all $p_i\leq p$, all $q_i\leq 1-p$, and in the right hand side, $p_i^\perp$ means $p-p_i$ and similarly for the $q_i$ $\endgroup$
    – Steven Deprez
    Commented May 8, 2012 at 23:38

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