Is it true to say that every nontrivial idempotent in the Cuntz algebra $\mathcal{O}(n)$ is a commutator element?(Or a linear combination of commutator elements?)
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1$\begingroup$ Do you have any results pointing in this direction? $\endgroup$– MTysonCommented May 14, 2018 at 1:03
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$\begingroup$ @MTyson No I have no any result. But an obvious observation is that there are some particular projections in the Cuntz algebra which are commutator elements. Because $1-xx^*=x^* x-xx^*$.So for $n=2$ both projections are commutator elememts. $\endgroup$– Ali TaghaviCommented May 14, 2018 at 6:02
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1 Answer
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There are a few things you could do to simplify this.
- Kaplansky's theorem (Reference needed for: every idempotent in a C*-algebra is similar to a hermitian one) lets you show that every idempotent in a $C^*$-algebra is similar to a projection. Thus to prove your claim, you just need to show that every projection in $\mathcal{O}_n$ is a linear combination of commutator elements, because similarity preserves the property of being a commutator or linear combination of commutators.
- In $\mathcal{O}_n = C^*(s_1,\ldots,s_n)$ (here $s_1,\ldots,s_n$ are the usual generating isometries), the projections $s_i s_i^*$ are linear combinations of commutators. For any $i=1,\ldots,n$ you can take the isometry $s_i$ and then the commutator $[s_i^*,s_i]$ is $Q_i=\sum_{k \neq i} s_k s_k^*$. The set of all these projections $\{Q_i\}_{i =1}^n$ forms a basis for the vector space spanned by $\{s_1 s_1^*, \ldots, s_n s_n^*\}$. (This is just like showing the matrix with zeros on diagonal and $1$ everywhere else is invertible.)
- Every projection in $\mathcal{O}_n$ is in the same $K_0$-class as a projection of the form $s_1 s_1^* + \ldots +s_j s_j^*$ for some $j \leq n$. So all you need to do is show that belonging to the same $K_0$-class preserves being a linear combination of commutators. I think that the similarity characterization of deciding when two projections are in the same $K_0$-class works.
Sorry I realize that I haven't provided a definite answer, it just seems like all this would be quite cluttered as a comment.
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$\begingroup$ Thank you. It is not a comment but is a very helpful answer. Just a question is not possible that two projections $e,f$ are not Murray Von Neuman equivalent but $e\oplus 0\simeq f\oplus 0$? $\endgroup$ Commented May 17, 2018 at 4:59
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1$\begingroup$ It is possible, that's one of the details about the semigroup $V(A)$ of projections that can't be overlooked. However, you can check (the details are a little long in notation, but mechanical) that if two projections $p,q $ in a $C^*$-algebra are stably similar, in the sense that $P = p \oplus 0_{k-1}$ and $Q = q \oplus 0_{k-1}$ are conjugates $P = UQU^*$ via some unitary $U$ in $M_k(A)$, then $p$ is a linear combination of commutator elements if and only if $q$ is such an element. It's a little lengthy to check but I'm fairly certain it's just a matrix multiplication problem. $\endgroup$ Commented May 17, 2018 at 17:46
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$\begingroup$ please see my comment to the following answer in MSE. I think that there are some contradictory situation. what is my mistake math.stackexchange.com/questions/2785117/… $\endgroup$ Commented May 17, 2018 at 20:18
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$\begingroup$ Oops! The statement that if $p \oplus 0_{k-1}$ and $q \oplus 0_{k-1}$ are M-vN equivalent in $M_k(A)$, then $p$ and $q$ are M-vN equivalent in $A$ is True. I forgot that if $[p]_0=[q]_0$, then there exists $k$ such that $p \oplus 1_{k-1}$ and $q \oplus 1_{k-1}$ are M-vN equivalent in $M_k(A)$. This implies that $p \oplus 1_{k-1} \oplus 0_k$ and $q \oplus 1_{k-1} \oplus 0_k$ are unitarily equivalent in $M_{2k}(A)$. (All this can be found in Sec 2.2 and 3.1 of Rordam, Larsen, and Laustsen.) I believe the algebra still works to show that $q$ a lin comb of commutators implies $p$ the same. $\endgroup$ Commented May 23, 2018 at 23:55