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Let $H$ be a Hilbert space. Any unitary operator $u\in B(H)$ induces an spatial isomorphism $\phi_u(x)=uxu^*$ on $B(H)$.

Question: Let $\phi:B(H)\to B(H)$ be a surjective *-ismorphism. Is $\phi$ implemented by a unitary operator $u\in B(H)$ i,e. $\phi=\phi_u$?

Remark. As for this problem, let us review two facts:

i) It is well-known that any surjective *-isomorphism $\phi$ on $B(H)$ (and in general on any von Neumann algebra) is $w^*$-continuous.

ii) Let $\psi$ be the restriction of $\phi$ on compact operators $K(H)$. Assume $\psi$ forms an isomorphism from $K(H)$ onto $K(H)$ then $\psi$ is implemented by a unitary operator and then i) implies that $\phi$ is also implemented by a unitary operator.

This remark says that, my problem may be re-written in this form too: Does there exist any surjective $*$-isomorphism $\phi$ on $B(H)$ such that $K(H)\neq\phi(K(H))$.

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Fix an orthonormal basis $\xi_j$, and consider the associated matrix units $\{e_{kj}\}$, i.e. $e_{kj}\xi=\langle \xi,\xi_j\rangle\,\xi_k$. The matrix units are characterized by $$ e_{kj}e_{st}=\delta_{js}\,e_{kt}.$$ It is easy to check that $\{\phi(e_{kj})\}$ is also a system of matrix units. If we let $\eta_j$ denote a unit vector in the range of the projector $\phi(e_{jj})$, then $\{\eta_j\}$ is an orthonormal basis (because $\sum\phi(e_{jj})=1$).

Let $u\in B(H)$ be the unitary that sends $\xi_j\longmapsto \eta_j$. Then, for any $\xi\in H$, $$ \phi(e_{kj})\xi=\langle\xi,\eta_j\rangle\,\eta_k =\langle\xi,u\xi_j\rangle\,u\eta_k=u\,\langle u^*\xi,\xi_j\rangle\,\xi_k =ue_{kj}u^*\xi. $$ So $\phi(e_{kj})=ue_{kj}u^*$. Now one can extend by linearity and continuity to all of $K(H)$ (and all of $B(H)$ by using wot-continuity.

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  • $\begingroup$ Dear Martin, if $\phi(e_{jj})$ was not a rank one projection(or even be infinite rank), then why does the sequence $\{\eta_j\}$ form a basis? $\endgroup$ – Ali Bagheri Jan 29 '16 at 10:15
  • $\begingroup$ That's not possible. If $q\leq\phi(e_{jj})$, then $\phi^{-1}(q)\leq e_{jj}$. So $\phi(e_{jj})$ is a minimal projection. $\endgroup$ – Martin Argerami Jan 29 '16 at 13:34

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