A $C^*$ algebraic analogy of the concept of complemented subspace in the particular case of $\ell^\infty$

Question

Let $A$ be a $C^*$ algebra. A $C^*$ subalgebra $C\subset A$ is said to be $C^*$ algebraic complemented of $A$ if there exist a $C^*$ subalgebra $D\subset A$ with $A=C\oplus D$ and the obvios mapping $A\mapsto C\oplus D$ preserves all algebraic structure.

In this question we consider the particular case $A=\ell^\infty$. Inspired by the result of Joram Lindenstrauss On complemented subspace of m, Israel Journal of Mathematics, volume 5, pages153{156, 1967. which says that the only infinit dimensional complemented subspace of $\ell^\infty$ are those Banach subspace which are isomorphism to $\ell^\infty$ we ask the following $C^*$ algebraic question:

Is it true to say that an infinite dimensional $C^*$ subalgebra of $\ell^\infty$ is $C^*$ algebraic complemented if and only if it is $C^*$ isomorphic to $\ell^\infty$?

Note: This would suggest characterization of all $C^*$ algebras whose only infinite dimensional $C^*$ algebraic complemented subalgebras are isomorphic to itself.

Answer 1

Suppose $l^\infty \cong C \oplus D$ (a C${}^*$-direct sum). Then both $C$ and $D$ must have units, which will be projections in $l^\infty$, and hence must be of the form $1_X$ and $1_{\mathbb{N}\setminus X}$ for some $X \subseteq \mathbb{N}$. Then every element of $C$ must be supported on $X$ and every element of $D$ on $\mathbb{N}\setminus X$, so $C = l^\infty(X) \subseteq l^\infty$. If $C$ is infinite-dimensional it clearly must be C${}^*$-isomorphic to $l^\infty$.

The converse is false: let $C \subset l^\infty$ consist of all those sequences satisfying $a_{2n} = a_{2n+1}$ for all $n$. This is clearly C${}^*$-isomorphic to $l^\infty$ but it does not have the above form and therefore is not C${}^*$-complemented.