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".

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

A (non commutative) $C^{*}$ or Banach 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?

nota good way to try to define a NC manifold. Recall that the Gelfand-Naimark correspondence only works for commutative C*-algebras. Also please read about general Banach algebras to see that they behave very very very very very very very very differently from the C*-case and to call a Banach algebra a noncommutative space is IMHO extremely tendentious $\endgroup$Thenyou can start to ask about gluing, as in your original question. $\endgroup$4more comments