Complementary $C^*$-subalgebra

Question

Does there exist a closed subspace of $\ell^{\infty}$ complementary to $c_0$?

If $A$ is a $C^*$-algebra, $B$ is a $C^*$-subalgebra of $A$, under which condition can one ensure that there exists a $C^*$-subalgebra $C$ of $A$ which is complementary to $B$, i.e. $A=B\oplus C$.

Answer 1

The question has already been answered, but there is a perspective I'd like to add. We have $\ell^\infty(\mathbb N) \cong C(\beta \mathbb N)$, where $\beta \mathbb N$ is the Stone-Cech compactification of $\mathbb N$, i.e: a universal compact space in wihch $\mathbb N$ is an open dense subset. $c_0$ is isomorphic to the functions $\varphi \in C(\beta \mathbb N)$ that vanish over $\beta \mathbb N \setminus \mathbb N$. If there were a complement, it will be given by continuous functions on $\beta \mathbb N $ that are null over $\mathbb N$, but that is impossible by density.

Equivalently, if $y \in \ell_\infty$ satisfies that $x y = 0$ for every $x \in c_0$, then $y = 0$. ($c_0$ is an essential ideal of $\ell_\infty$)

Answer 2

The complement of $c_0$ in $\ell^\infty$ is $\{0\}$. Indeed, let $x \in \ell^\infty$ and assume $x \bot c_0$. Let $\bar x^{(n)}$ denote the element $(\bar x_0, \bar x_1,\cdots,\bar x_n,0,0,\cdots)$. Then ${\bar x}^{(n)} \in c_0$ and hence $x\bar x^{(n)} = 0$. It follows that $\lvert x_i \rvert^2 = x_i \bar x_i = 0$ for $i \leq n$. Hence $x_i = 0$ for $i \leq n$. Since $n$ was arbitrary, $x = 0$.

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One answer two your question is given in [Theorem 2.9, Injective Hilbert $C^*$-moudles, Lin]:

Theorem Assume $E$ is a full Hilbert module over a $C^*$-algebra $A$, such that $\mathscr L(E) = B(E)$, i.e. the set of bounded module maps on $E$ is equal to the set of adjointable maps. Then $E$ is orthogonally complementary, i.e. whenever $E$ is a Hilbert submodule of $F$, there exists a submodule $E'$ of $F$ such that $$ E \oplus E' = F. $$

Answer 3

No, $c_0$ is not complemented in $\ell^\infty$ (in the sense of Banach spaces). The idea is to split $\mathbb N$ into an uncountable family of subsets $\{E_i\}_{i\in I}$ such that $E_i\cap E_j$ is finite for each $i\neq j$. Then use this to show that if $T:\ell^\infty\to\ell^\infty$ is a bounded linear map with $c_0\subset\ker(T)$, then for some $E_i$ we have $\ell^\infty(E_i)\subset\ker(T)$ (not entirely trivial). Thus if there exists a projection $P:\ell^\infty\to c_0$, then by considering $1-P$ the above shows that $\ell^\infty(E_i)\in\operatorname{Im}(P)$, a contradiction. The full details are in section 2.4 of Albiac and Kalton's excellent book Topics in Banach Space Theory. If you like, I can expand on the details.

I claim that $B$ is complemented in $A$ (in the sense of $C^*$-algebras) if and only if $B$ is an ideal in $A$, and there is a $*$-homomorphism $\varphi:A\to B$ such that $\varphi(b)=b$ for all $b\in B$. The forward direction is clear, for if $A=B\oplus C$, just consider the projection onto the first factor. Conversely, if $B\subset A$ is an ideal and such a $\varphi:A\to B$ exists, then let $\pi:A\to A/B$ be the quotient $*$-homomorphism and define a $*$-homomorphism $\theta:A\to B\oplus(A/B)$ by $\theta(a)=(\varphi(a),\pi(a))$. Then $\theta$ is a $*$-isomorphism, and the result follows.