# How is $\Bbb K$ well defined, operator algebra

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.

• I found the answer on page 53 just by using the symbol index in the book. – Randall Mar 6 at 16:10
• E.g., "We denote by $\mathbb{K}$ the C*-algebra of compact operators on a separable, infinite-dimensionalHilbert space. " – Randall Mar 6 at 16:10
• Ok, thanks a lot, now I reformulated my problem. – CL. Mar 6 at 16:29

(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.

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