1
$\begingroup$

Let $A$ be a Baer*-ring with the unit $1$. An important point that holds in every Baer*-ring is this: for every element $y\in A$ there is an smallest projection $e_y\in A$, called the left projection of $y$, satisfying in $e_yy=y$.

We say $x\in A$ is a shift if $x^*x=1$ and there exists a sequence of pairwise orthogonal projections $\{p_n\}$ such that:

  1. $\sup_{n\geq1} \sum_1^n p_k=1$.

  2. $p_n=e_{xp_{n-1}}$.

Q. Let $x$ be a shift. For a finite projection $q$ in $A$, I guess $q\wedge e_{x^*q}$ should be strictly less than $q$, however I have no clear proof!

$\endgroup$
2
  • $\begingroup$ Wouldn't $q=0$ be a counter-example to the question? Or does "finite projection" somehow imply that $A$ is infinite-dimensional, and you want the kernel to be finite dimensional? $\endgroup$ Commented Nov 2, 2017 at 21:15
  • $\begingroup$ @Pace Nielsen Thanks. Actually $q$ is a non-zero finite projection. $\endgroup$
    – ABB
    Commented Nov 3, 2017 at 19:30

0

You must log in to answer this question.