Is $K_0(Q(B))=0$? $Q(B)=M(B\otimes\mathcal K)/B\otimes\mathcal K(\mathscr H)$,$M$ denotes the multiplier algebra.

Question

$$M(A)=\{x\in A'':xA,Ax\subseteq A\}$$ $\mathscr H$ is an infinite dimensional separable Hilbert space and $\mathcal K$ is the compact operator algebra.

$$ Q(B)=M(B\otimes\mathcal K)/B\otimes\mathcal K $$

I think I only need to show $K_0(M(B\otimes \mathcal K))=0$, since $Q(B)$ is a unital quotient algebra of $M(B\otimes \mathcal K$).

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First suppose $p$ is a projection in $M(B\otimes \mathcal K)$. Let $f_1,f_2,...$ be orthogonal projections such that each of them is equivalent to $1_M=\text{diag}(1,1,...)$.

Then there is a projection $q\leq f_1$ which is equivalent with $p$. It sufficient to show $q\oplus 1_M\sim 1_M\oplus 0$ in $M_2(M(B\otimes\mathcal K))$.

Write $$u=\left(\begin{array}{}\left(\begin{array}{}q&&&\\ &0\\&&0\\ &&&...\end{array}\right)\left(\begin{array}{}1-q&&&\\q&1-q\\&q&1-q\\ &&&...\end{array}\right)\\\left(\begin{array}{}0&&&\\ &0\\&&0\\ &&&...\end{array}\right)\left(\begin{array}{}0&&&\\ &0\\&&0\\ &&&...\end{array}\right)\end{array}\right) $$

whose real meaning is

$$u=\left(\begin{array}{}q&(f_1-q)+f_{21}q+f_{21}(f_1-q)f_{12}+f_{31}qf_{12}+f_{31}(f_1-q)f_{13}+...\\0&0\end{array}\right)$$

where $f_{ij}^*f_{ij}=f_i$ and $f_{ij}f_{ij}^*=f_j$.

Then by calculation, $u^*u=q\oplus 1_M$ and $uu^*=1_M\oplus 0$. This shows both $u\in M_2(M(B\otimes\mathcal K))$ (since it is bounded) and $q\oplus 1_M\sim 1_M\oplus 0$. Therefore $[q]=0$.

Generally for projection $p\in M_n(M(B\otimes\mathcal K))$, since $1_M\oplus 0\sim 1_{M}\oplus 1_M$, there is a projection $q\leq 1_M$ which is equivalent with $p$. Therefore $[p]=0$.

Thus $K_0(M(B\otimes\mathcal K))=0$ and so $K_0(M(B\otimes\mathcal K)/B\otimes\mathcal K)$ should be.

Am I right?

I am asking this question because the book I am reading is discussing properties of $K_0(Q(B))$ which seems $0$ to me.

Answer 1

The fact that a C*-algebra has trivial $K_0$ does not mean that every quotient of it has the same property! You are correct in saying that $K_0(M(K\otimes B))$ is zero, so you can plug this informtation in the six-term exact sequence for $$0 \to K\otimes B\to M(K\otimes B) \to M(K\otimes B)/K\otimes B\to0 $$ and conclude that $$K_0(Q(B))=K_1(K\otimes B) = K_1(B).$$