I have some questions about Joachim Cuntz’s proof of the $ K_{1} $-injectivity of purely infinite simple unital $ C^{*} $-algebras, which is found in this paper.
For this post, let us adopt the following definitions and terminology for a $ C^{*} $-algebra $ A $.
- If $ A $ is unital, then $ \mathcal{U}(A) $ denotes the set of unitaries of $ A $. Observe that $ \mathcal{U}(A) $ is a group.
- If $ A $ is unital, then $ {\mathcal{U}_{0}}(A) $ denotes the set of unitaries of $ A $ that are connected to $ 1_{A} $ by a path in $ \mathcal{U}(A) $. Observe that $ {\mathcal{U}_{0}}(A) $ is a subgroup of $ \mathcal{U}(A) $.
- If $ A $ is unital, then $ \sim $ is an equivalence relation on $ \mathcal{U}(A) $, where $ u \sim v \stackrel{\text{df}}{\iff} u v^{-1} \in {\mathcal{U}_{0}}(A) $.
- $ \sim_{\text{M-vN}} $ is the Murray-von Neumann equivalence relation between projections of $ A $.
- A projection $ p \in A $ is called infinite if and only if there exists another projection $ p' \in A $ such that $ p \sim_{\text{M-vN}} p' \lneq p $.
- $ A $ is called purely infinite if and only if $ \overline{x A x} $ contains an infinite projection for every positive element $ x \in A $.
Let me first state two important lemmas found in the paper.
Lemma 1. Let $ A $ be a simple $ C^{*} $-algebra. Then the set $ S $ of infinite projections of $ A $ satisfies: $$ (\star) \quad \text{For all $ p,q \in S $, there exists a $ p' \in S $ such that $ p \sim_{\text{M-vN}} p' \lneq q $ and $ q - p' \in S $}. $$ If $ A $ is furthermore purely infinite, then every non-zero projection of $ A $ is infinite.
Lemma 2. Let $ A $ be a purely infinite (not-necessarily-simple) unital $ C^{*} $-algebra. Then for every $ u \in \mathcal{U}(A) $, there exist a projection $ p \in A $ with $ 0_{A} \lneq p \lneq 1_{A} $ and a $ u' \in \mathcal{U}(p A p) $ such that $$ u \sim u' + (1_{A} - p). $$ Note: The corner $ p A p $ is a unital $ C^{*} $-subalgebra of $ A $ with $ p $ as its identity.
I was able to verify the proofs of these lemmas (after much difficulty), but I was unable to do so for the next one.
Lemma 3. Let $ A $ be a purely infinite simple unital $ C^{*} $-algebra. Let $ u \in \mathcal{U}(A) $ and $ e $ a minimal projection of $ \mathscr{K} $, the compact operators on a separable infinite-dimensional Hilbert space. Then $ u \in {\mathcal{U}_{0}}(A) $ if and only if $$ e \otimes u + (\mathbf{1} - e \otimes 1_{A}) \sim \mathbf{1} $$ in $ (\mathscr{K} \otimes A)^{+} $, where $ (\mathscr{K} \otimes A)^{+} $ is the unitization of $ \mathscr{K} \otimes A $ and $ \mathbf{1} $ denotes its identity.
Reproduction of Cuntz’s proof
Of course, we only have to show the ‘if’ part. By Lemma $ 2 $, we may assume that $$ u = p u p + (1_{A} - p) \qquad (\clubsuit) $$ for a proper projection $ p \in A $. By $ (\star) $, there is, then, a projection $ q \in A $ such that
- $ 0_{A} \lneq q \lneq 1_{A} - p $ and
- $ q \sim_{\text{M-vN}} 1_{A} - p $.
Again by $ (\star) $, we can find a sequence $ (r_{i})_{i \in \Bbb{N}} $ of mutually orthogonal projections of $ A $ such that $$ \forall i \in \Bbb{N}: \quad r_{i} \leq 1_{A} - p - q \quad \text{and} \quad r_{i} \sim_{\text{M-vN}} 1_{A}. $$ Set
- $ r_{0} \stackrel{\text{df}}{=} p + q $ (note that $ r_{0} \sim_{\text{M-vN}} 1_{A} $) and
- $ f_{k} \stackrel{\text{df}}{=} r_{0} + \cdots + r_{k} $ for all $ k \in \Bbb{N} $.
Then $ f_{k} A f_{k} \cong {\text{M}_{k}}(\Bbb{C}) \otimes A $, and the $ C^{*} $-algebra generated by $ \displaystyle \bigcup_{k \in \Bbb{N}} f_{k} A f_{k} $ and $ 1_{A} $ is isomorphic to $ (\mathscr{K} \otimes A)^{+} $. This isomorphism carries $ e \otimes u + (\mathbf{1} - e \otimes 1_{A}) $ to $ u $. $ \quad \blacksquare $
Questions. Why may we assume $ (\clubsuit) $ by Lemma $ 2 $, and more importantly, how do we establish that $ f_{k} A f_{k} $ is isomorphic to $ {\text{M}_{k}}(\Bbb{C}) \otimes A $?
Thank you for your help!
I will write $f_k=r_1+\cdots+r_k$, because otherwise we get $M_{k+1}(\mathbb C)$.
The projections $r_j$ are each equivalent to $1_A$, so there exist partial isometries $e_{1,j},\ldots,e_{1,k}$ with $$ e_{1,j}^*e_{1,j}^\vphantom{*}=r_j,\ \ \ \ e_{1,j}^\vphantom{*}e_{1,j}^*=r_1. $$ Now define $$ e_{k,j}^\vphantom{*}=e_{1,k}^*e_{1,j}^\vphantom{*}. $$ This is a system of matrix units in $A$, i.e. $$\tag{1} e_{k,j}^\vphantom{*}e_{s,t}^\vphantom{*}=\delta_{j,s}^\vphantom{*}\,e_{k,t}^\vphantom{*}. $$ Now define $\varphi:f_kAf_k\to M_k(\mathbb C)\otimes r_1Ar_1$ by $$ \varphi(x)=\sum_{k,j=1}^k E_{k,j}\otimes e_{1,k}\,x\,e_{j,1}. $$ This maps is obviously linear, it is multiplicative and preserves adjoints by using the properties $(1)$. It is one-to-one, because if $e_{1,k}\,x\,e_{j,1}=0$ for all $k,j$, we get $r_kxr_j=0$, and by adding over all $k,j$ we get $0=f_kxf_k=x$. It is onto, because for $y_{k,j}^\vphantom{*}\in r_1Ar_1$, $$ \sum_{k,j=0}^kE_{k,j}\otimes y_{k,j}^\vphantom{*}=\varphi(\sum_{k,j}e_{k,1}^\vphantom{*}y_{k,j}^\vphantom{*}e_{1,j}^\vphantom{*}). $$ So it remains to show that $A\simeq r_1Ar_1$. This follows from the fact that $r_1\sim 1_A$. Indeed, there exists a partial isometry $w$ with $w^*w=1_A$, and $ww^*=r_1$. Now define $\psi:r_1Ar_1\to A$ by $$ \psi(x)=w^*xw. $$ A straighforward check shows that $\psi$ is a $*$-homomorphism. If $\psi(x)=0$, then $w^*xw=0$; then $0=ww^*xww^*=r_1xr_1=x$. And given any $y\in A$, we have $wyw^*\in r_1Ar_1$ (because $w=r_1w$), and so $\psi(wyw^*)=y$. So $\psi$ is bijective.