Question Regarding the Standard Picture of $K_{0}(A)$ for a C$^{*}$-Algebra $A$

Question

I am working through Rordam's book and I am stuck on exercise 4.4. Suppose we have a (not necessarily unital) C$^{*}$-algebra $A$. We are asked to prove that every element of $K_{0}(A)$ can be written as $$ [p]_{0}-\left[\begin{pmatrix}1_{n} & 0_{n}\\ 0_{n} & 0_{n}\end{pmatrix}\right]_{0}, $$ for some projection $p\in M_{2n}(\widetilde{A})$ satisfying $$ p-\begin{pmatrix}1_{n} & 0_{n}\\ 0_{n} & 0_{n}\end{pmatrix}\in M_{2n}(A). $$

Following one of the proofs in the book (a quarter-way down page 64), I took an arbitrary $g\in K_{0}(A)$ and expressed it as $[e]_{0}-[f]_{0}$ for some projections $e,f\in M_{n}(\widetilde{A})$. Then, I put $$ p=\begin{pmatrix}1_{n}-f & 0\\ 0 & e\end{pmatrix}\qquad\text{and}\qquad q=\begin{pmatrix}1_{n} & 0_{n}\\ 0_{n} & 0_{n}\end{pmatrix}. $$ As in the proof in the book, it follows that $g=[p]_{0}-[q]_{0}$ and that $$ [s(p)]_{0}=[q]_{0}, $$ where $s\colon\widetilde{A}\to\widetilde{A}$ is the map given by $s(a+\lambda\cdot1_{\widetilde{A}})=\lambda\cdot 1_{\widetilde{A}}$. So, we have $g=[p]_{0}-[s(p)]_{0}$, but to guarantee that $p-\begin{pmatrix}1_{n} & 0_{n}\\ 0_{n} & 0_{n}\end{pmatrix}\in M_{2n}(A)$, I need $s(p)=q$, which obviously does not hold in general.

I've been thinking for hours how to modify this construction to produce the result, but I haven't been able to come up with anything else. Any help is really appreciated.

Thank you.

Answer 1

Take some element $x \in K_0(A)$ with $$ x = [p]_0 - [q]_0, $$ where lets say $p,q \in M_n(\widetilde A)$. By $\pi : \widetilde A \to \mathbb C$ I denote the map which sends $a+\lambda 1_{\widetilde A}$ to $\lambda$.

By definition of $K_0$ we know that $\pi(p)$ and $\pi(q)$ define the same element in $V(M_n(\mathbb C)) = \mathbb N$, i.e. these projections have the same rank $k \leq n$. So both of them are unitarily equivalent to $1_k$ That means, there exist unitaries $u_1,u_2 \in M_n(\mathbb C)$ with $$ u_1 \pi(p)u_1^* = 1_k, \quad u_2 \pi(q)u_2 = 1_k. $$ Replacing $p$ by $u_1pu_1^*$ and $q$ by $u_2qu_2^*$ we see that we may assume that $p-q \in M_n(A)$ whenever $[p]_0-[q]_0 \in K_0(A)$.

Now, given $[p]_0-[q]_0 \in K_0(A)$ with $p-q \in M_n(A)$, we do the following (as in Wegge-Olsens book):

Since $q \in M_n(\widetilde A)$ is a projection, we know $q \leq 1_n$. Let $p' \in M_{2n}(\widetilde A)$ be the projection $0_n \oplus p$. Then $$ [p]_0 = [p']_0 \quad \text{and} \quad \ p' \bot 1_n-q. $$ With $p'' := (1_n-q) + p' \in M_{2n}(\widetilde A)$, we get

\begin{align*} [p'']_0 - [1_n]_0 & = [1_n-q]_0+[p']_0-[1_n]_0 \\ & = [1_n-q]_0+[q]_0-[q]_0+[p']_0-[1_n]_0 \\ & = [1_n]_0 - [1_n]_0 + [p']_0-[q]_0 \\ & = [p]_0-[q]_0. \end{align*} Now, we can again replace $p''$ be $upu^*$ for some suitable unitary $u \in M_{2n}(\mathbb C)$ such that $\pi(p'') = 1_n$.

Answer 2

I would like to give another answer $\underline{\text{which is essentially the same with the one above}}$, but phrased in a different manner and more fit to the approach of Rordam's book.

So let $x\in K_0(A)$ and write $x=[q]_0-[s(q)]_0$ for some projection $q\in P_n(\tilde{A})$. Mimicking the proof of the standard picture of $K_0$, we set $p'=q\oplus(1_n-s(q))$, which is a projection in $P_{2n}(\tilde{A})$ and note that $s(p')=s(q)\oplus(1_n-s(q))\sim_01_n\sim_01_n\oplus0_n$. It is very easy to verify that $x=[p']_0-[s(p')]_0$. Observe that $s(p')$ and $1_n\oplus0_n$ are projections of the C*-algebra $M_{2n}(\mathbb{C}\cdot1_{\tilde{A}})$ and they are Murray-von Neumann equivalent.

Now for any $k\geq1$ we have an obvious *-isomorphism $M_k(\mathbb{C})\cong M_k(\mathbb{C}1_{\tilde{A}})$. It is true that when we have two projections $r_1,r_2\in M_k(\mathbb{C})$ of the same rank (i.e. of the same trace) then they are similar, i.e. we have a unitary $u\in M_k(\mathbb{C})$ so that $ur_1u^*=r_2$. Passing this result through our *-isomorphism, since $s(p')$ and $1_n\oplus0_n$ are Murray-von Neumann equivalent (thus the corresponding projections through the isomorphism have the same trace, i.e. they are of the same rank), there exists a unitary $u\in M_{2n}(\mathbb{C}\cdot1_{\tilde{A}})$ so that $u\cdot s(p')\cdot u^*=1_n\oplus0_n$.

Note that $u$ is a scalar element, i.e. $u=s(u)$. We set $p=u\cdot p'\cdot u^*$. Now $p$ is a projection in $M_{2n}(\tilde{A})$, we have $s(p)=s(up'u^*)=s(u)s(p')s(u^*)=us(p')u^*=1_n\oplus0_n$ and it is obvious that $p\sim_0 p'$ and $s(p)\sim_0 s(p')$, so $[p']_0-[s(p')]_0=[p]_0-[s(p)]_0$, as we wanted.