# Commutator ideal in nonunital C*-algebra

Let $$A$$ be a C*-algebra that has no one-dimensional irreducible representations, that is, there is no (closed) two-sided ideal $$I\subseteq A$$ such that $$A/I\cong\mathbb{C}$$.

Let $$J$$ denote the (not necessarily closed) two-sided ideal generated by additive commutators in $$A$$: $$J:=\{ \sum_{k=1}^n a_k[b_k,c_k]d_k : a_k,b_k,c_k,c_k\in A\}.$$

Question: Is $$A=J$$?

The answer is `Yes', if $$A$$ is unital and in some other cases, but does it hold in general? Note that $$J$$ is a dense, two-sided ideal (thus contains the Pedersen ideal of $$A$$), and that $$A/J$$ is a commutative algebra.

• Is this even known about $K(H)$? Aug 28, 2022 at 22:23
• Yes, this follows from Corollary 3 in : Every compact hermitian operator is a sum of two self-commutators of compact operators.  Fan, Fong. Which operators are the self-commutators of compact operators? Proc. Amer. Math. Soc. 80 (1980), no. 1, 58–60 Aug 28, 2022 at 23:04
• What is an example where commutator is not closed? Aug 30, 2022 at 6:05

The answer is NO. Rordam and Robert MR3072284 have found a sequence $$(A_n)_n$$ of simple unital infinite dimensional C*-algebras such that $$\prod A_n$$ has a nonzero character. (Thanks are due to Yasuhiko Sato for informing me of this.) Thus the following is true: For every $$m$$ and $$C>1$$, there is $$n=n(m,C)$$ such that $$1=\sum_{k=1}^m a_k[b_k,c_k]d_k$$ in $$A_n$$ implies $$\sum_{k=1}^m\|a_k\| \|b_k\| \|c_k\| \|d_k\| > C$$. Now consider the $$c_0$$-sum $$A:=\bigoplus_m A_{n(m,m^2)}$$. Then $$(m^{-1})_m \in A$$ cannot be expressed as $$\sum_{k=1}^l a_k[b_k,c_k]d_k$$ in $$A$$, because it would imply $$\sum_{k=1}^l \|a_k(m)\| \|b_k(m)\| \|c_k(m)\| \|d_k(m)\| \geq m$$ for every $$m\geq l$$.
On the the hand, it is easy to show that there are $$m$$ and $$C>1$$ that satisfies the following: For every von Neumann algebra without nonzero abelian direct summand, one has $$1=\sum_{k=1}^m a_k[b_k,c_k]d_k$$ for some $$a_k,b_k,c_k,d_k$$ with $$\sum_{k=1}^m\|a_k\| \|b_k\| \|c_k\| \|d_k\| < C$$. By the Hahn--Banach separation theorem, this implies the following: For every $$A$$ without nonzero characters and every $$x\in A$$, there are infinite sequences $$a_k,b_k,c_k,d_k$$ such that $$\sum_{k=1}^\infty\|a_k\| \|b_k\| \|c_k\| \|d_k\| \le C\|x\|$$ and $$x = \sum_{k=1}^\infty a_k[b_k,c_k]d_k$$.