most recent 30 from http://mathoverflow.net 2013-05-24T21:13:41Z http://mathoverflow.net/feeds/question/44081 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44081/projection-in-hereditary-c-subalgebra Qingyun 2010-10-29T07:48:06Z 2010-11-01T21:10:48Z

<p>This is actually something in a paper but the author claimed it without proof. Let x be a positive elment of norm 1 in a $C^*-$algebra A, and Her(x) is the hereditary subalegbra generated by x. Given $\epsilon>0$,let $f_\epsilon$ be thecontinuous function on R defined as follow:</p> <p>$f_\epsilon \equiv 0 \quad on \quad [-\infty,\epsilon/2]$</p> <p>$f_\epsilon \quad is \quad linear \quad on\quad [\epsilon/2,\epsilon]$</p> <p>$f_\epsilon \equiv 1 \quad on\quad [\epsilon, +\infty]$</p> <p>So $f_\epsilon$ increase to the identiy function on [0,1] when $\epsilon$ decrease to 0, and $Her(x)=\overline{\cup_{\epsilon>0} f_\epsilon(x)Af_\epsilon(x)}$. Let p be a projection in Her(x). then how do we know that there must exist a $\epsilon$ such that $p\in \overline{f_\epsilon(x)Af_\epsilon(x)}$? Or more generally, Let A be the inductive limit of {$A_n$}, and p is a projection in A,does it follow that p is actually in some $A_n$?</p>

http://mathoverflow.net/questions/44081/projection-in-hereditary-c-subalgebra/44122#44122 Narutaka OZAWA 2010-10-29T12:46:03Z 2010-10-29T12:46:03Z

<p>No. Let $A$ be the C$^*$-algebra of compact operators on $\ell_2$ and $x$ is the diagonal operator $(1,1/2,1/3,\ldots)$. Then, $f_\epsilon(x)Af_\epsilon(x)$ is a matrix algebra in the left upper corner. The rank one projection corresponding to any vector of infinite support does not belong to $\bigcup f_\epsilon(x)Af_\epsilon(x)$. However, it is a standard fact that if $a$ is a positive element such that $\| p - a \| &lt; 1/2$, then the spectrum of $a$ has a gap around $1/2$ and $q=\chi_{[1/2,3/2]}(a)$ is a projection in $C^*(a)$ such that $\| p - q \| &lt; 1$, which implies that $p$ and $q$ are unitarily equivalent.</p>