# Smallness/ Rigidity of $\kappa(\mathcal{H})$ without using minimal projections?

Let $\mathcal{H}$ be a Hilbert space and $\kappa(\mathcal{H})$ the $C^*$-algebra of compact operators on $\mathcal{H}$. By smallness/ rigidity of $\kappa(\mathcal{H})$ I am referring to the following collection facts about $\kappa(\mathcal{H})$.

Let $\mathcal{A}$ be a $C^*$-subalgebra of $\kappa(\mathcal{H})$. If $\mathcal{A}$ is irreducible then $\mathcal{A}=\kappa(\mathcal{H})$.

The only closed ideals in $\kappa(\mathcal{H})$ are $\{0\}$ and $\kappa(\mathcal{H})$.

Let $\mathcal{B}$ be a $C^*$-algebra in $\mathcal{L}(\mathcal{H})$. If $\mathcal{B}\cap\kappa(\mathcal{H})$ is non-trivial, then $\mathcal{B}\supset\kappa(\mathcal{H})$.

$\kappa(\mathcal{H})$ is so small that it cannot contain proper irreducible $C^*$-algebras, or proper ideals; and it only touches those $C^*$-algebras that fully contain itself.

Also we have the following facts about representations of $\kappa(\mathcal{H})$:

Let $\mathcal{A}$ be a $C^*$-algebra in $\kappa(\mathcal{H})$. Each nondegenerate representation of $\mathcal{A}$ is the sum of orthogonal irreducible representations that are equivalent to the identity representation.

And in particular,

The only irreducible representation of $\kappa(\mathcal{H})$ is the identity.

These are describing the very rigid structure of $\kappa(\mathcal{H})$.

In all the books I read, the proof of these series of facts is based on the notion of minimal projections, in particular the following

Let $\mathcal{A}$ be a $C^*$-subalgebra of $\kappa(\mathcal{H})$. A projection $E$ in $\mathcal{A}$ is minimal if and only if $E\mathcal{A}E=\mathbb{C}E$.

This line of argument seems standard now and indeed it is very clever and clear. But it is also a little tricky at least to me. So I wonder whether we can prove these facts independent of minimal projections (or at least try to hide them to the backstage). If this can be done, then I really hope to see the proof (a sketch or some references would be enough). If this cannot be done, then I'd like to know the reason.

Thanks very much!

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I just saw your question and I think I can introduce a reference for one of the above statements. The fact that $K(H)$ is simple is proved in Corollary 5.7.6 of my lecture notes on $C^*$-algebras available at arXiv:1211.3404. The proof rests on the facts that $K(H)$ is the closure of the ideal of finite rank operators and this latter ideal is contained in every non-zero ideal of $B(H)$.