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:



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



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


