Could somebody please give me an example, assuming one exists, of an element $a$ of a C*-algebra $A$ such that $f(a^*a) = 0$ but $f(aa^*) \neq 0$ for some positive linear functional $f:A \to \mathbb{C}$? I tried to prove there was no example, but that didn't seem to work out so now I'm guessing it's false. Thanks in advance.
Element $a$ of a C*-algebra such that $f(a^*a) = 0$ but $f(aa^*) \neq 0$ for some positive linear functional $f$
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examples-counterexamples
c-star-algebras
1 Answers
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Let $e_0=(1,0,0,0,0,\ldots)\in\ell^2$, let $f:B(\ell^2)\to\mathbb C$ be defined by $f(T)=\langle Te_0,e_0\rangle$, and let $S\in B(\ell^2)$ be the "backward shift" defined by $S(a_0,a_1,a_2,a_3,\ldots)=(a_1,a_2,a_3,a_4,\ldots)$. Then $f(SS^*)=1$ but $f(S^*S)=0$.
Similar but simpler: $A=M_2(\mathbb C)$, $f(T)=\left\langle T\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}1\\0\end{bmatrix} \right\rangle$, $a=\begin{bmatrix}0&1\\0&0\end{bmatrix}$.
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0Mike: $B(H)$ is the dual of the space of [trace class operators](http://en.wikipedia.org/wiki/Trace_class) on $H$, and the corresponding weak* topology on $B(H)$ is defined as you indicated. (So we're considering $B(H)=X^*$, not $B(H)=X$. The predual is unique, so the weak* topology is unique. Positive normal functionals can also be defined in terms of the order structure in $B(H)$.) – 2011-12-05