# Calkin Algebra and the embedding

Let $H$ be a separable, infinite dimensional Hilbert Space and $Calk(H):=B(H)/K(H)$ denotes the Calkin algebra. There is obvious surjection $\pi: B(H) \to Calk(H)$ but I'm interested in somehow opposite question: is it possible to construct embedding $j_1:Calk(H) \to B(H)$? The same question for embedding $j_2:B(H) \to Calk(H)$ and finally is there epimorphism $k:Calk(H) \to B(H)$?

-

(1) No. The Calkin algebra contains an uncountable family of mutually orthogonal nonzero projections.

(2) Yes. Embed $B(H)$ into $B(H \otimes H)$ by the map $A \mapsto A \otimes I$, then pass to $Q(H\otimes H) \cong Q(H)$.

(3) No. The Calkin algebra is simple.

-
If you don't care about separability, then you can of course embed $Calk(H)$ into $B(H')$ for some other (non-separable) Hilbert space $H'$, because $Calk(H)$ is a C*-algebra. –  Ulrich Pennig Jan 9 '13 at 13:34
@Nik: Could you give some reference about your answer in (1)? I am interested in to read more in this subject. –  Vahid Shirbisheh Jan 9 '13 at 13:51
@Vahid: there is an uncountable family of subsets $A$ of ${\bf N}$, any two of which have finite intersection. For each such $A$ let $P_A$ be the orthogonal projection of $l^2$ onto the sequences supported on $A$. When you pass to the Calkin algebra these become mutually orthogonal projections. –  Nik Weaver Jan 9 '13 at 14:25
(1) One cannot embed $Q(H)$ into $B(H)$ as a Banach space either. Indeed, $Q(H)$ contains $\ell_\infty/c_0$ (as a diagonal masa). J. Bourgain gave a very clever proof of the fact that $\ell_\infty/c_0$ has no strictly convex renorming:
J. Bourgain, $\ell_\infty/c_0$ has no equivalent strictly convex norm, Proc. Amer. Math. Soc. 78 (1980), 225-226.
On the other hand, $B(H)$ is a dual space of a separable Banach space (the space of nuclear operators on $H$), so it does have a strictly convex renorming. The property of having a strictly convex renorming passes to subspaces.
There's something wrong here: The calkin algebra is a C* algebra and can therefore be embedded into a $B(\tilde{H})$ for some big enough hilbert space $\tilde{H}$. The point of the question is that one cannot take $\tilde{H}=H$. –  Johannes Hahn Dec 15 '13 at 12:36
Johannes, $H$ is a separable Hilbert space, as defined in truebaran's post. –  Tomek Kania Dec 15 '13 at 12:40
Oh yes, if $K$ is a non-separable Hilbert space, then $B(K)$ contains $\ell_\infty(\omega_1)$ which has no strictly convex renorming by a result of Mahlon Day. –  Tomek Kania Dec 15 '13 at 12:42