# $C^*$ algebra of compact operators as a direct limit of matrix algebras?

This was written in Page 92, of Higson's Analytic $$K$$-Homology book.

Let $$H$$ be a hilbert space. The $$C^*$$ algebra $$K(H)$$ of compact operators is the direct limit of a sequence $$M_2(\Bbb C) \subseteq M_4(\Bbb C) \subseteq \cdots$$

How is this so? I suppose that $$H$$ is an infinite dimensional $$\Bbb C$$ vector space.

I know any compact operator is the limits (in operator norm) of finite rank operator. But this still doesn't really explain the direct limit...

Also why are we only considering matrices of size $$2^n$$?

• What does direct limit mean here? – SmileyCraft Mar 5 at 10:52
• It is the colimit of the sequence where the maps between each inclusion is by the diagonal inclusion, $$a \mapsto \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}$$ – CL. Mar 5 at 10:56

The direct limit of the sequence $$M_2(\mathbb{C}) \subset M_4(\mathbb{C}) \subset \cdots$$ with diagonal maps $$a \mapsto \left(\begin{array}{cc} a & 0 \\ 0 & a\end{array} \right)$$ will give you the $$2^{\infty}$$ UHF algebra (CAR algebra), not the compact operators. Perhaps you mean $$a \mapsto \left(\begin{array}{cc} a & 0 \\ 0 & 0\end{array} \right)$$? In that case everything in the limit is the limit of finite rank operators so must be contained in $$\mathcal{K}$$.
On the other hand, observe that you can get any finite rank operator in the limit since, for any $$n$$, you can find $$k$$ such that $$M_n(\mathbb{C}) \subset M_{2^k}(\mathbb{C})$$, so in fact the limit must be $$\mathcal{K}$$.
• Thanks a lot, I am still a little confused. Firstly, this requires $H$ to be infinite dimensional hilbert space. And it seems that your argument works for any such $H$. Are there any further requirmenets - i.e. $H$ being separable? So does this imply any $C^*$ algebra of compact operators on a hiblert space (inf. dim. ) is isomoprhic to $M_\infty(\Bbb C)$?? – CL. Mar 6 at 15:33
• As for your first question, no, the $C^*$-algebra of compact operators will not be isomorphic to $M_{\infty}(\mathbb{C})$, because $M_{\infty}(\mathbb{C})$ is not closed and hence is not itself a $C^*$-algebra. However, the closure of $M_{\infty}(\mathbb{C})$ is indeed $\mathcal{K}$. (Assuming, again, that $H$ is separable.) – Prita Santesh Apr 9 at 11:45