Suppose $\{p_i\}_{i=1}^{m}$ are projections in the d by d matrix algebra $A$ over the complex numbers and satisfy the following condition:
$||Id-\sum_{i=1}^m{p_i}||_2<c$, $||p_ip_j||_2<c, \forall i\neq j$
My first question is
Can we find $q_i, i=1,\cdots, m$ projections in $A$ and a function $f=f(c)$ such that
$$Id=\sum_{i=1}^m{q_i},$$ $$ ||q_i-p_i||_2<f(c), $$ $$q_iq_j=q_jq_i, \forall i\neq j$$ $$f(c)\to 0 ~~as~~ c\to 0$$?
A more general question is:
If we are given positive operators $p_i,i=1,.., m$ in A, such that they almost commute with each other in the sense that $||p_ip_j-p_jp_i||_2<c$, can we find a unitary u(May depend on c)in A, operators $q_j,j=1,.., m$ which commutes with each other and a function $f(c)$ such that $$||p_j-q_j||_2<f(c)$$ for all j. And $f(c)\to 0 ~~as~~ c\to 0$.
Remarks:
1, Clearly, this is true for $m=2$.
2, if it holds in general in question 1, then we could find a unitary $u\in A$ Such that $up_ju^*$ is close to some diagonal matrix for all j.
we can take $q_2=p_1\vee p_2-p_1$, and so on, then you check this works.
Add:
Set $q_1=p_1, q_i=p_1\vee \cdots\vee p_{i}-p_1\vee \cdots\vee p_{i-1}, 1<i<m$, $p_m=1-p_1\vee \cdots\vee p_{m-1}$.
Then use the fact $p\vee q-p\sim q-p\wedge q$ so they have the same trace to get an upper bound for $||p_i-q_i||_2$.