commutator of $C^*$ algebra

Question

If $A$ is a non-commutative $C^*$ algebra,can we define $[A,A]$ as the ideal generated by the set $\{xy-yx:x,y\in A\}$such that $A/[A,A]$ is commutative?

Answer 1

Indeed, $A/[A,A]$ is always commutative. Let $\pi:A\to A/[A,A]$ be the quotient map. If $x,y\in A/[A,A]$, fix $a,b\in A$ such that $\pi(a)=x$, $\pi(b)=y$. Then $[x,y]=\pi([a,b])=0$, since $[a,b]\in [A,A]$.

But it may well happen that $[A,A]=A$, and it may not. For example, if $A=K(H)$ or $B(H)$ for a Hilbert space $H$, then $[A,A]=A$. But it may not be the whole algebra, as in the case that $A=C([0,1])\oplus M_n(\mathbb C)$ we have $[A,A]=0\oplus M_n(\mathbb C)$.