Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f(c)\to 0 ~~as~~ c\to 0$.


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.

share|cite|improve this question

For your question about projections, I think it is true even in an arbitrary C$^*$-algebra. Such results are often called "perturbations". You can find a good account on Section III.3 in Davidson's "C$^*$-algebras by example".

I cannot make sense of your second question: $u=I$ and $f(c)=c$ will always work.

share|cite|improve this answer
Thanks, I would check that, and I fixed the typoes in second question. – ougao Mar 26 '14 at 14:57
I am afraid the perturbation you mentioned is a different issue from the question I asked, my question asks whether we can find some abelian algebra close to some given almost commuting system, while perturbation says if we already know this, then blabla... – ougao Mar 26 '14 at 15:07
up vote 0 down vote accepted

we can take $q_2=p_1\vee p_2-p_1$, and so on, then you check this works.


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 $||p_i-q_i||_2$.

share|cite|improve this answer
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – naslundx Mar 26 '14 at 17:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.