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?



Are these ideals in rings of operators on Hilbert space unique?
No proper ideal satisfies your condition. Take the Hilbert space to be $ \ell\_2\oplus \ell\_2$ and define $f(x,y)=(o,x)$; $g(x,y)= (x,0)$.


