Question on matrix decomposition in a C*-algebra.

Question

I am reading a book entitled an introduction to the classification of amenable C*-algebras, and the notion of matrix decomposition has come up several times(without any definition or proof).

Such as on page 146, it reads,

Define $b=f(a)+(\epsilon/4)(v+v^*)+(\epsilon/4)(p-q)$. It has the following matrix decomposition corresponding to the decomposition $1=(1-p)\oplus q\oplus (p-q)$: $$b=\left(\begin{array}{}f(a)&\epsilon/4&0\\\epsilon/4&0&0\\0&0&\epsilon/4\end{array}\right)$$ It is clear that $b\in A_{sa}$ and $b$ is invertible.

$(1-p)$ and $q$ are equivalent here. ($u^*u=1-p$ and $uu^*=q$)

Also $p$ is equivalent to $1$.

It seems to me that it is somehow embedding $A$ into $M_n(A)$, or at least constructing a homomorphism from some subalgebra of $M_n(A)$ into $A$. But I can not figure out how it is doing it.

score 1 · Accepted Answer · answered Sep 23 '20 at 00:32

1

You are not mentioning that $q(1-p)=0$, which is essential for this to work.

In general, given pairwise orthogonal projections $p_1,\ldots,p_n\in A$ with $\sum_jp_j=1$, the monomorphism you are looking for is $\pi:A\to M_n(A)$ given by $$ \pi(a)=\big[p_kap_j\big]_{k,j}. $$ This is obviously linear, and it is easy to check that it preserves adjoints. To see that $\pi$ is multiplicative, $$ \pi(a)\pi(b)=\big[\sum_h(p_kap_h)(p_hbp_j)\big]_{k,j}=\big[p_ka\big(\sum_hp_h\big)bp_j)\big]_{k,j}=\big[p_kabp_j)\big]_{k,j}=\pi(ab). $$ Finally, if $\pi(a)=0$, then $p_kap_j=0$ for all $k,j$, so $$ a=\sum_{k,j}p_kap_j=0. $$

answered Sep 23 '20 at 00:32

Martin Argerami

205,756

But where does $[\pi(a)]_{1,2}=\epsilon$ come from then? It should be $v\epsilon/4$ or $v^*\epsilon/4$. – Sui Sep 23 '20 at 01:11
Yes, a 1 in the 1,2 entry is $v^$ (note that the 1,2 entry has $(1-p)$ on the left and $q$ on the right, so it has to have a partial isometry involved). Another way to see it is that you need $E_{1,2}q=1-p$, so it has to be $E_{1,2}=v^$. – Martin Argerami Sep 23 '20 at 04:11
So you mean the book is wrong? – Sui Sep 23 '20 at 04:13
No. $E_{1,2}$ is $v^$, since $E_{12}^E_{12}=E_{22}=q$, and $E_{12}E_{12}^*=E_{11}=1-p$. – Martin Argerami Sep 23 '20 at 04:46
But the book says it's $\epsilon/4$... – Sui Sep 23 '20 at 04:47
Or do you mean $\epsilon/4$ means $v^*\epsilon/4$ here> – Sui Sep 23 '20 at 04:49

Question on matrix decomposition in a C*-algebra.

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