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I am having trouble with Exercise 4.9 (ii) from Rordam's book. Let $A$ be a unital C$^{*}$-algebra. I am trying to prove the following.

Suppose $p\in \mathcal{P}_{n}(A)$ and $q\in\mathcal{P}_{m}(A)$ satisfy $[p]_{0}=[q]_{0}$. Then, $p\oplus e\sim_{0} q\oplus e$ for any properly infinite full projection $e\in\mathcal{P}_{\infty}(A)$.

This is the last question in a sequence of related exercises. In a previous exercise, the relation $\precsim$ on $\mathcal{P}_{\infty}(A)$ was introducted. We say $p\precsim q$ iff $\exists q_{0}\in \mathcal{P}_{\infty}(A)$ such that $p\sim_{0}q_{0}\leq q$ iff $q\sim_{0}p\oplus p_{0}$ for some $p_{0}\in\mathcal{P}_{\infty}(A)$. Some facts from previous exercises that I think might be useful are:

  1. a non-zero projection $r$ is properly infinite iff $r\oplus r\precsim r$.

  2. $1_{A}$ is properly infinite if $A$ contains a properly infinite full projection.

  3. for every projection $r\in \mathcal{P}_{\infty}(A)$ and every properly infinite full projection $s\in\mathcal{P}_{\infty}(S)$, we have $r\precsim s$.

Since $[p]_{0}=[q]_{0}$, we know that $p\oplus 1_{s}\sim_{0} q\oplus 1_{s}$ for some $s\in\mathbb{N}$. Hence, $p\oplus e\oplus 1_{s}\sim_{0} q\oplus e\oplus 1_{s}$, but this is not quite what we want. I have tried playing around with the $\precsim$ relation, obtaining results like $p\precsim e$, $q\precsim e$, $p\oplus e\precsim e$, and $q\oplus e\precsim e$ using fact 3, but I have been unsuccessful following this trail and would really appreciate some help.

Thank you!

ervx
  • 12,208

1 Answers1

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As pointed out here, the algebraic order on $V(A)$ is induced by $\lesssim$. Now, let $e$ be a properly infinite full projection in some $M_n(A)$.

Since $[p]_0 = [q]_0$ in $K_0(A)$, it follows that $[p]+x = [q]+x$ for some $x \in V(A)$. As pointed out by you, $x \leq [e]$ in $V(A)$. That means, there is $x' \in V(A)$ with $x +x' = [e]$. Then $$ [p]+[e] = [p]+x+x' = [q]+x+x' = [q]+[e]. $$

Notation: $[p]$ denotes the class of a projection in $V(A)$ and $[p]_0$ the class in $K_0(A)$.