>Assume  that $A$ is  a  Banach  algebra with two closed two sided ideals  $I$  and  $J$ such that $I$  and  $J$ are  commutative  and $A=I+J$. Does this implies that $A$ is  commutative? For the  $C^{*}$  algebra, the  answer is  ["Yes"](http://math.stackexchange.com/questions/998702/a-question-on-non-commutative-ring-or-algebra).

**The motivation:** A (noncommutative) compact n dimensional topological  manifold could  be  defined as  follows:

A  (non  commutative)  $C^{*}$  algebra  $A$  such that there  are ideals  $I_{k}$, $k=1,2,\ldots,n$ such that $A=I_{1}+I_{2}+\ldots I_{k}$ and  each $I_{j}$ is  isomorphic to $C_{0}(\mathbb{R}^{n})$. 


But  the  above  answer in MSE shows that, in the context of  $C^{*}$  algebras, this  definition does  not  give  any non  commutative  example,.


>So  we  search  for  a non commutative  example  in  Banach  algebras.

**Note:** According to the comment on my MSE question: To what extent  Banach or  $C^{*}$  algebras whose underline Lie  algebras are  metabelian  are studied and  classified?