Let X be a Banach space. $B(X)$ is the algebra of all bounded linear operators on X. $\phi: B(X)\rightarrow B(X)$ is a isomoprphisn, $\varphi: B(X)\rightarrow B(X)$ is a isomorphism or negative anti-isomorphism, $\tau$ is a linear map from $B(X)$ to its center. If $\phi(A)+\tau(A)=\varphi(A)$ for all $A\in B(X)$ is commutator, I would like to know if we can get $\tau(A)=0$ for all commutator $A\in B(X)$. If $\varphi$ is a isomorphism, since isomorphism preserveing spectrum and spectrum set is a compact set, we get the conclusion. If $\varphi$ is negative anti-isomophismt, how to prove?
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$\begingroup$ I notice no-one has answered this. I think it might help if you explain some of your terms: e.g. what do you mean by "$A\in B(X)$ is a commutator", and what is a "negative anti-isomorphism"? $\endgroup$– OllieCommented Apr 16, 2013 at 14:04
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$\begingroup$ I must admit I don't quite understand the question but perhaps this might be helpful to you. For $X$ being the James space or $X=C[0,\omega_1]$ there is a unique trace $\tau$ on $B(X)$. Hence $\tau$ is centre-valued and kills all the commutators. Obviously, the kernel of $\tau$ has codimension one in $B(X)$. Also, every Banach-algebra isomorphism $B(X)\to B(X)$ is implemented by an isomorphism $V$ on the Banach-space level. Thus, $\psi(T) = V^{-1}TV$. $\endgroup$– Tomasz KaniaCommented Apr 18, 2013 at 16:15
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$\begingroup$ "$A\in B(X)$ is a commutator" means that there exist $B,C\in B(X)$ such that $A=BC-CB$. "$\phi$ is a isomorphism" means that there exist $T\in B(X)$ such that $\phi(A)=T^{-1}AT$ for all $A\in B(X)$."negative anti-isomorphism" mean sthere exist $T\in B(X)$ such that $\phi(A)=-T^{-1}A^{*}T$ for all $A\in B(X)$. thank you Tomek! you give an example under condition that $\phi$ is isomorphism. $\endgroup$– LingChengCommented Apr 23, 2013 at 2:54
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