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The letter $\Bbb K$ in Bruce Blackadar, on operator algebra denotes the algebra of compact operators on a separable infinite dimensional hilbert space, $H$.


In my other post, it is shown that $M_\infty(\Bbb C)$ is isomoprhic to $\Bbb K$ set-wise.

My questions are as follows:

(i) We can give $M_\infty(\Bbb C)$ a $*$-algebra structure. If each embedding is an isometric isomorphism, in the colimit diagram, we can also give it the norm. Does this also make $M_\infty(\Bbb C)$ a $C^*$ algebra?

(ii) How is $\Bbb K$ independent of choice of $H$? Where is separablility used?


Partial replies or references are appreciated.

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  • $\begingroup$ I found the answer on page 53 just by using the symbol index in the book. $\endgroup$ – Randall Mar 6 at 16:10
  • $\begingroup$ E.g., "We denote by $\mathbb{K}$ the C*-algebra of compact operators on a separable, infinite-dimensionalHilbert space. " $\endgroup$ – Randall Mar 6 at 16:10
  • $\begingroup$ Ok, thanks a lot, now I reformulated my problem. $\endgroup$ – CL. Mar 6 at 16:29
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(i) It does make $M_\infty(\mathbb{C})$ a C$^*$-algebra, but it might be worth mentioning that $M_\infty(\mathbb{C})$ is to be interpreted as the direct limit described in that post and not as the collection of all infinite matrices $\{(a_{ij})_{i, j \in \mathbb{N}}: a_{ij} \in \mathbb{C}\}$. The latter has pathological unbounded examples like $a_{ij} = \delta_{ij}j$. Also, the isomorphism described in your post is an isomorphism of C$^*$-algebras (not just sets).

(ii) We're using the separability assumption because there's only one separable infinite-dimensional Hilbert space up to unitary equivalence, and any unitary $U: H \rightarrow H'$ induces a $*$-isomoprhism (Ad $U$) between $K(H)$ and $K(H')$. So $K$ is independent of $H$ up to isomorphism.

In general, there's one Hilbert space for each cardinal (representing the cardinality of a basis for $H$). The separable infinite-dimensional Hilbert space is the one with a countably infinite basis, so as long as we restrict to the separable case, $K$ is well-defined.

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  • $\begingroup$ Thanks a lot Josh. I hope you can comment on my newest post. It is quite related to this problem. $\endgroup$ – CL. Mar 7 at 8:07

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