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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 $k:Calk(H) \to B(H)$?

edited Jan 9 at 12:51

Davide Giraudo
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asked Jan 9 at 12:31

truebaran
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Nik Weaver · Answer 1 · 2013-01-09 13:22:26Z UTC

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(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.

answered Jan 9 at 13:22

Nik Weaver
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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 at 13:34

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@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 at 13:51

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@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 at 14:25

Calkin Algebra and the embedding

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