In page 42, lectures on operator $K$-theory

The writer defines a map from

$$\Phi: K_0(A) \rightarrow [\mathcal S,A\otimes \mathcal K]$$

Notation: $\mathcal S:=C_0(\Bbb R)$ and $\mathcal K$ is a graded $C^*$ algebra of compact operators on graded Hilbert space $H=H^0\oplus H^1$.

He claims to define the map as follows:

given a general element $[p]-[q] \in K_0(A)$. We may define, a graded $*$-homomorphisms, $$\phi_{p,q}: \mathcal S \rightarrow M_2(A)$$ by $$\phi_{p,q}(f) = \begin{pmatrix} pf(0) & 0 \\ 0& qf(0) \end{pmatrix} $$

Then he says,

by $\otimes \mathcal K$ this allows us to define for matrix algebras. Inducing the map $\Phi$.

I believe there are some typos.

  1. The first part we define for $p,q \in A$.

  2. But how does tensoring with $\otimes \mathcal K$ allow us to generalize the map? In particular, is $\mathcal K = M_\infty(\Bbb C)$?


If you read it more carefully, they write

Let $p,q$ be projections in $A$, whose formal difference defines a $K$-theory class $[p]-[q]\in K_0(A)$. Define a graded $*$-homomorphism $\phi_{p,q}:\mathcal S\to M_2(A)$ by $$\phi_{p,q}(f)=\begin{pmatrix}pf(0)&0\\0&qf(0) \end{pmatrix}.$$

So for your first question, it is not a typo, they are really considering $p,q\in A$ at first.

Secondly, we have $\mathcal K=M_\infty(\mathbb C)=\lim_{n\to\infty}M_n(\mathbb C)$, with maps $M_n(\mathbb C)\to M_{n+1}(\mathbb C)$ being inclusion in the top left corner. The idea is that you fix an orthonormal basis $\{e_n\}_{n=1}^\infty$ of the hilbert space $H$, and embed $M_n(\mathbb C)$ into $\mathcal K$ by sending the matrix unit $E_{j,k}\in M_n(\mathbb C)$ to to rank-one operator $e_k\otimes e_j$ in $\mathcal K$. It's easy to show that this yields an injective homomorphism from the algebraic direct limit $M_\infty(\mathbb C)$ to $\mathcal K$, and then show the range is dense in $\mathcal K$ to complete the proof.

Then tensoring with the $C^*$-algebra $A$ shows that $A\otimes\mathcal K$ is the direct limit of the algebras $M_n(A)$ with top left corner inclusion (the details are essentially the same as the case $A=\mathbb C$).

If now $\mathcal K$ is graded by a grading on the Hilbert space $H=H_0\oplus H_1$, then any operator $T\in\mathcal K$ can be decomposed as $$T=\begin{pmatrix}T_{00} & T_{01}\\ T_{10} & T_{11}\end{pmatrix}$$ with $T_{ij}\in\mathcal K(H_j,H_i)$ for $i,j\in\{0,1\}$.

Let $p$ be a projection in $M_n(A)$, and let $q$ be a projection in $M_m(A)$, so that $[p]-[q]$ is an arbitrary element of $K_0(A)$. By what's stated above, $M_n(A)$ embeds in $A\otimes \mathcal K(H_0)$ and $M_m(A)$ embeds in $A\otimes \mathcal K(H_1)$, and we have $$\begin{pmatrix}p & 0 \\ 0 & q\end{pmatrix}\in A\otimes\mathcal K,$$ and furthermore this lies in the even part of $\mathcal K$, under the grading defined above. Thus the map $\phi_{p,q}:\mathcal S\to A\otimes\mathcal K$ given by $$\phi_{p,q}(f)=\begin{pmatrix} pf(0) & 0\\ 0 & qf(0)\end{pmatrix}$$ is a well-defined graded $*$-homomorphism.

Also, showing the map $K_0(A)\ni[p]-[q]\mapsto \phi_{p,q}\in[\mathcal S,A\otimes\mathcal K]$ is well-defined is a good exercise (if you haven't realized it by now, Higson and Roe enjoy leaving many details to the reader), and if you have any questions about proving this, please let me know.


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