I'm interested in knowing the requirements for a $*$-representation, $\pi_{\omega}$, of a C*-algebra, $\mathbb{C}(\mathcal{G})$, (or equivalently the requirements for the unitary representation, $U_{\mathcal{G}}$, corresponding to $\pi_{\omega}$) to have a unique cyclic and separating vector $\xi$.

P.S. I'm a physicist and I'm new to this field so I would appreciate a more detailed answer. Thank you so much in advance!!


There is an obvious and unavoidable lack of uniqueness for a cyclic and separating vector, in that any nonzero scalar multiple will also be cyclic and separating.

If $\mathcal A$ is a C$^*$-algebra, $\omega$ a representation, and $\xi$ a cyclic and separating vector for $\omega$, let $U\in\omega(A)'$ be a unitary. Then $$ \overline{\omega(\mathcal A)U\xi}=U\overline{\pi(\mathcal A)\xi}=UH_\omega=H_\omega, $$ so $U\xi$ is cyclic. And if $\omega(A)U\xi=0$, then $$ 0=\omega(A)U\xi=U\omega(A)\xi, $$ so $\omega(A)\xi=0$ and thus $\omega(A)=0$. Then $U\xi$ is separating. It follows that $U\xi$ is cyclic and separating. With the hypothesis that $\xi$ is unique, we get $U\xi=\lambda_U\,\xi$ for some $\lambda_U\in\mathbb T$. Then $$ U\omega(A)\xi=\omega(A)U\xi=\lambda_U\,\omega(A)\xi. $$ From $\xi$ separating (for the commutant, which we get since $\xi$ is cyclic), we get that $U=\lambda_U\,I$. As $\omega(\mathcal A)'$ is the span of its unitaries, $\omega(\mathcal A)'=\mathbb C\,I$. So $\omega$ is an irreducible representation.

Now, for an irreducible representation, every vector is cyclic. What about separating? A separating vector for an algebra, is cyclic for the commutant. That is, $$ H_\omega=\overline{\omega(\mathcal A)'\xi}=\overline{\mathbb C\xi}=\mathbb C\,\xi. $$ So $H_\omega$ is one-dimensional.

In summary, the only way to have unique (up to scalar multiples) cyclic and separating vector, occurs when $\omega$ is a character (i.e., a one-dimensional representation).


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