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.



Commutator ideal in nonunital C*-algebra
The answer is NO. Rordam and Robert [MR3072284](https://mathscinet.ams.org/mathscinet-getitem?mr=3072284) 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 $x = \sum\_{k=1}^\infty a\_k[b\_k,c\_k]d\_k$.


