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Let A be a $C^*$-algebra, unital or not.

  1. I want to show that each element in $K_0(A)$ is of the form

$$[p]_0 - \bigg[ \begin{pmatrix} 1_n & 0_n \\ 0_n & 0_n \\ \end{pmatrix} \bigg]_0$$ for some projection $p \in M_{2n}(\tilde A)$ satisfying the following which I will call (A):

$$p - \begin{pmatrix} 1_n & 0_n \\ 0_n & 0_n \\ \end{pmatrix} \in M_{2n}(A)$$

  1. And I want to show that an element $p$ in $M_{2n}(\tilde A)$ satisfies (A) if and only if $s(p)=diag(1_n , 0_n)$.

Idea:

  1. By definition $K_0(A)= \lbrace [p]_0 - [s(p)]_o : p \in \mathcal{P}_\infty (\tilde A) \rbrace$ where as far as I can tell $s(a +\alpha 1)= \alpha 1$ for all $a \in A$ and all $\alpha \in \mathbb{C}$. My book says that "the image of $s_n$ is the subset $M_n(\mathbb{C}$ of $M_n(\tilde A)$ consisting of all matrices with scalar entries, and $x-s_n (x)$ belongs to $M_n(A)$ for all x in $M_n(\tilde A)$" so my question is what exactly this means. Does this mean that:

$$s(p)= \begin{pmatrix} 1_n & 0_n \\ 0_n & 0_n \\ \end{pmatrix}$$

As this seems too easy I don't think it is true. Or is there another way to show this?

  1. I think that the $\Leftarrow$ should follow from the passage in my book, but I am not quite sure..
Miep
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  • Which book is that, please cite it correctly so that we can help you. What is $s(x)$? The support of $x$? The support of a projection is the same projection. I do not have a sense of why the statement 1 should be true for general $A$. – Adrián González Pérez May 14 '20 at 11:44
  • I am using the book "introduction to K-theory" by Rørdam, and s(x) is defined as the scalar mapping from $\tilde A \to \tilde A$ and then there is an induced map $s_n : M_n(\tilde A) \to M_n(\tilde A) $ – Miep May 14 '20 at 12:21

1 Answers1

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  1. An element $x\in M_n(\tilde A)$ is a matrix of the form $x=[x_{ij}]$, where $x_{ij}\in\tilde A$ for each $i,j$, and $s_n(x)$ is the matrix $[s(x_{ij})]$. Thus we have $$x-s_n(x)=[x_{ij}-s(x_{ij})]\in M_n(A),$$ since $a-s(a)\in A$ for all $a\in\tilde A$.
    For a projection $p\in M_{n}(\tilde A)$, it isn't necessarily true that $s(p)\in M_n(\mathbb C)$ is diagonal. However, it is a projection, hence is diagonalizable. So there is some unitary $u\in M_{n}(\mathbb C)$ such that $$us(p)u^*=\begin{pmatrix}1_k&0\\0&0_{n-k}\end{pmatrix}$$ for some $k\leq n$. By adding an appropriately sized identity matrix and zero matrix, we obtain the form you're looking for.
  2. Indeed, this does follow from the passage. If $s(p)=1_n\oplus0_n$, then $$p-\begin{pmatrix}1_n&0_n\\0_n&1_n\end{pmatrix}=p-s(p)\in M_{2n}(A).$$
Aweygan
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