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)?

  • $\begingroup$ The tag is inappropriate; I changed it. $\endgroup$ – Yemon Choi Mar 12 '13 at 0:33

$K(H)$, 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 $\ell_\infty$).

  • $\begingroup$ Do you know anything about general case of H? Would you please let me know. $\endgroup$ – Ali Mar 11 '13 at 23:17
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    $\begingroup$ For general $H$ the ideals are $K(H)$ and, for each cardinal '$\aleph' \le$' the dimension $\aleph$ of $H$, the set of operators whose ranges have dimension less than $\aleph$. If such an ideal were complemented in $B(H)$ you argue that you would have the functions in '$\ell_\infty(\aleph')$' whose support has cardinality less than $\aleph$ is complemented in '$\ell_\infty(\aleph')$', which it is not. But I don't recall an easy argument that it is not complemented. In the separable case, it is basically Phillips' lemma that $c_0$ is not complemented in $\ell_\infty$ and the result is in $\endgroup$ – Bill Johnson Mar 12 '13 at 0:44
  • $\begingroup$ ...text books (e.g., Albiac-Kalton). $\endgroup$ – Bill Johnson Mar 12 '13 at 0:46

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.)

  • $\begingroup$ @Tomek: The argument you suggest at the end only shows that $\ell_\infty^\lambda({\rm dens}\; H)$ is not $1$-injective. $\endgroup$ – Bill Johnson Mar 12 '13 at 1:03
  • $\begingroup$ Thank you, Bill. I hope my answer is now fairly complete. $\endgroup$ – Tomek Kania Mar 12 '13 at 1:18
  • $\begingroup$ @Tomek: Yes, I saw that argument, too, but why is $\ell_\infty^\lambda(\kappa)$ not isomorphic to $\ell_\infty(\kappa)$? The latter space has density character $2^\kappa$ and the former density character sup $\{ 2^\alpha: \alpha < \lambda\}$, so non isomorphism is easy under the generalized continuum hypothesis. Absent GCH, I did not at first see a simple reason. Now I do, but I'll wait a bit before posting in case you want to think about it some more. $\endgroup$ – Bill Johnson Mar 12 '13 at 16:15
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    $\begingroup$ Nice answer, Tomek. For the sake of completeness$^\ast$, I mention that, with the definitions given in your answer, $K_\lambda(H)$ is not closed in $B(H)$ when $\lambda< dens(H)$ is of cofinality $\omega$; similarly, $\ell_\infty^\lambda (dens(H))$ is not closed in $\ell_\infty(dens(H))$ when $\lambda< dens(H)$ is of cofinality $\omega$ (these spaces are otherwise closed in their respective overspaces). The definition of $\ell_\infty^\lambda (dens(H))$ should presumably be those $f\in \ell_\infty(dens(H))$ with $\vert\{ \alpha\in dens(H)\mid\vert f(\alpha)\vert>\epsilon\}\vert <\lambda$. $\endgroup$ – Philip Brooker Mar 13 '13 at 12:05
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    $\begingroup$ For my "simple" argument that $\ell_1(2^\kappa)$ does not embed into (the closure of) $\ell_\infty^\lambda(\kappa)$ I implicitly used $\lambda < \gamma$ implies $2^\lambda < 2^\gamma$. (Correcting typo in earlier comment.) $\endgroup$ – Bill Johnson Mar 14 '13 at 15:24

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