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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 $H$.

Question: When do such two-sided ideals $I$ satisfy the following:
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 $f \circ g \in I(K)$?

Taking $I(H)$ to be the trace class operators gives one example. Is this the unique one?

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?

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up vote 1 down vote accepted

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

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Ah, I see I missed something crucial, many thanks for the enlightening counterexample! So $g \circ f$ is trace class, but $f \circ g$ is not. Probably I'm a bit slow on the uptake, but: you can conclude from this one example that any ideal as in the question must consist of all operators because $f \circ g$ is not even compact, correct? –  Chris Heunen Mar 4 '11 at 22:18
    
Right, Chris. On a separable infinite dimensional Hilbert space, the identity operator factors through every non compact operator. For the example, you don't need this general fact, since it is clear that the ideal generated by $f(g)$ is the set of all bounded linear operators on the space. –  Bill Johnson Mar 5 '11 at 4:57
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