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Let $H$ be an infinite dimensional (separable if necessary) complex Hilbert space, and denote by $K(H)$ the ideal (in $B(H)$) of compact operators on $H$. Let $G_c=\{I+K\in B(H): I+K \text{ is invertible and } K\in K(H)\}$, and $U_c=\{I+K\in B(H): I+K \text{ is unitary and } K\in K(H)\}$.

$G_c$ and $U_c$ act by left multiplication on $K(H)$. Are these actions transitive?

Additionally, any references which delve into actions of these groups would be appreciated.

EDIT: Clearly, these actions preserve rank, so the answer to the general question is no. But, what I am actually interested in is if they are transitive on each rank-class. In particular, are they transitive on the infinite rank compact operators.

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  • $\begingroup$ Maybe I've misunderstood, but isn't there a rank obstruction? Namely, if $F$ is a finite rank operator (say rank $n$) then then rank of $(I+K)F$ is at most $n$, for any $K$. Thus e.g. one cannot move a rank one operator to a rank two. $\endgroup$
    – Mike Jury
    Commented Jan 21, 2014 at 20:15
  • $\begingroup$ Yes, I've edited to reflect that issue. I am really concerned about the infinite rank compact operators. $\endgroup$
    – Iian Smythe
    Commented Jan 21, 2014 at 20:15

1 Answer 1

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No. Let $e_n$ be a basis of the Hilbert space, and let $\alpha_n$ be any sequence converging to zero. Then you have a compact operator $K_0$ defined by $K_0e_n=\alpha_ne_n$. Now let $K_1$ be any other compact operator. Since compact operators are continuous weak to strong, $K_1e_n$ converges to zero. Consequently $(I+K_1)K_0e_n=\alpha_n(e_n+K_1e_n)$ converges to zero at the same rate as $\alpha_n$.

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