Commutator ideal in nonunital C*-algebra

Question

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.

Answer 1

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 $x = \sum_{k=1}^\infty a_k[b_k,c_k]d_k$.