most recent 30 from http://mathoverflow.net 2013-05-18T14:35:51Z http://mathoverflow.net/feeds/question/57388 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57388/are-these-ideals-in-rings-of-operators-on-hilbert-space-unique Chris Heunen 2011-03-04T19:13:19Z 2011-03-04T19:54:57Z

<p>Suppose that, for every Hilbert space $H$, we have a subset $I(H) \subseteq B(H)$ of bounded linear operators on $H$, and that together all $I(H)$ form a two-sided ideal, in the sense that whenever $h \in I(H)$, also $f \circ h \circ g \in I(K)$ for any bounded linear maps $f \colon H \to K$ and $g \colon K \to H$. To prevent degeneration, additionally assume $I(\mathbb{C})=B(\mathbb{C})$ and $I(H) \neq B(H)$ for some $H$.</p> <blockquote> <p><strong>Question:</strong> When do such two-sided ideals $I$ satisfy the following: <br> if $f \colon H \to K$ and $g \colon K \to H$ are bounded linear maps, and $g \circ f \in I(H)$, then also $f \circ g \in I(K)$?</p> </blockquote> <p>Taking $I(H)$ to be the trace class operators gives one example. Is this the unique one?</p> <p>I know that $I(H)$ at least has to contain the finite rank operators, and has to be contained in the compact operators. Finite rank operators also form a two-sided ideal, but do they satisfy the requirement, i.e. if $g \circ f$ is of finite rank, is $f \circ g$, too?</p>

http://mathoverflow.net/questions/57388/are-these-ideals-in-rings-of-operators-on-hilbert-space-unique/57393#57393 Bill Johnson 2011-03-04T19:54:57Z 2011-03-04T19:54:57Z

<p>No proper ideal satisfies your condition. Take the Hilbert space to be $ \ell_2\oplus \ell_2$ and define $f(x,y)=(o,x)$; $g(x,y)= (x,0)$.</p>