B(H) as a direct sum of a closed two sided ideal and a subalgebra

Question

Let $B(H)$ is the C*-algebra of all bounded linear operators on Hilbert space $H$. Are there a closed two-sided ideal $I$ and a subalgebra $A$ of $B(H)$ such that $B(H)=I \oplus A$ (direct sum I and A)?

Answer 1

$K(H)$, the compact operators on $H$, is the only proper closed ideal in $B(H)$ when $H$ is a separable infinite dimensional Hilbert space, and $K(H)$ is not complemented in $B(H)$ (because if it were the diagonal compact operators would be complemented in the diagonal bounded operators, which is to say that $c_0$ would be complemented in $\ell_\infty$).

Answer 2

Let me complement Bill's answer. It is basically the same idea.

Gramsch and Luft described the lattice of closed ideals of $B(H)$ where is $H$ a non-separable Hilbert space.

B. Gramsch, Eine Idealstruktur Banachscher Operatoralgebren, J. Reine Angew. Math. 225 (1967), 97–115.

E. Luft, The two-sided closed ideals of the algebra of bounded linear operators of a Hilbert space, Czechoslovak Math J. 18 (1968), 595–605.

They proved that for a non-separable Hilbert space $H$ with density character ${\rm dens}\; H$ (this is the minimal cardinality of a dense subset) all the closed ideals of $B(H)$ are of the form $\{0\}$, $K(H)$ (the compact operators) and

$$K_\lambda(H) = \{T\in B(H)\colon \mbox{dens }T[H]< \lambda\}$$

where $\lambda\leqslant\kappa^+$. Certainly, $$B(H) = K_{({\rm dens}\; H)^+}(H).$$ Hence, we are interested in the case only where $\lambda\leqslant {\rm dens}\; H$.

Fix an orthonormal basis for $H$ and identify operators on $H$ with matrices with respect to this basis. Consider the Banach space $\ell_\infty^\lambda({\rm dens}\; H)$ of all bounded complex-valued functions on the cardinal ${\rm dens}\; H$ with at most $<\lambda$ non-zero entries, endowed with the sup-norm.

If you intersect the ideal $K_\lambda(H)$ with the diagonal masa (that is, $\ell_\infty({\rm dens}\; H)$) then you'll get precisely $\ell_\infty^\lambda({\rm dens}\; H)$. The diagonal masa is complemented because $\ell_\infty({\rm dens}\; H)$ is an injective Banach space. Consequently, if $K_\lambda(H)$ was complemented in $B(H)$ then $\ell_\infty^\lambda({\rm dens}\; H)$ would be complemented in $\ell_\infty({\rm dens}\; H)$. This is however a contradiction because $\ell_\infty^\lambda({\rm dens}\; H)$ is not injective (see this paper for a discussion of this space and its injectivity-like properties).

Proof of non-injectivity of $\ell_\infty^\lambda(\kappa)$: Assume $\ell_\infty^\lambda(\kappa)$ is injective. Manifestly, it contains $c_0(\kappa)$. Then by Rosenthal's theorem

H.P. Rosenthal, On relatively disjoint families of measures, with some applications to Banach space theory, Studia Math., 37 (1970), 13–36.

it would contain a copy of $\ell_\infty(\kappa)$ and by Pełczyński's decomposition method, it would be isomorphic to $\ell_\infty(\kappa)$; a contradiction.

EDIT 05.05.15: This is actually an ancient observation by Pełczyński and Sudakov:

A. Pełczyński and V. N. Sudakov, Remark on non-complemented subspaces of the space $m(S)$, Colloq. Math. 19 (1962), 85-88.

EDIT: As Bill pointed out we have to be careful why this is indeed impossible. Under GCH this is evident but life would be too easy with GCH. Actually, to see this is a result in ZFC, one has to tweak the argument in Paragraph d) on page 12 of the hyperlinked paper by replacing $\mathbb{N}$ with $\lambda$ and $\aleph_1$ with $\lambda$. Then the whole proof carries over. (Frankly, I learnt it from one of the authors of this paper some time ago and presumed that it is well-known. My apologies for that.)